Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin... | $F=\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
$=\dfrac1{abc}\left|
\begin{array}{cc} ab^2c^2 & abc & a(b+c) \\
c^2a^2b & bca & b(c+a) \\
a^2b^2c & abc & c(a+b) \\
\end{array}
\right|$
$=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\
c^2a^2b &1& b(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Expanding an expression in a certain field If $\mathbb F_2$ is a field of characteristic $2$, then we have $x+x=y+y=z+z=0$ for all $x,y,z \in \mathbb F_2$. When I expand $(x+y)(y+z)(z+x)$, I get
\begin{align}
(x+y)(y+z)(z+x) &= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+2xyz \\
&= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+(x+x)yz \\
&= xz^2+... | In $\mathbb{F}_2$, we also have $x^2=x$ and so on. So your expression will result in $0$. Another way to look at it is that at least two of $x,y$ or $z$ will take the same value, in which case at least one of $x+y, y+z, x+z$ will be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An integral arising from Kepler's problem $\frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$ I'm dealing with this integral in my spare time, since days and days, and it's really interesting. I'll provide to write what I tried until now, and I would really appreciate some hel... | I am assuming $0 < \epsilon < 1$ in the following. The cases
$\epsilon = 0$ and $\epsilon = 1$ have to be handled separately,
see below.
There is a small error in your calculation, the root of
$2z + \epsilon(z^2+1) = 0$ are
$$
z_1 = \frac{-1-\sqrt{1 - \epsilon^2}}{\epsilon} \, , \quad z_2 = \frac{-1+\sqrt{1 - \epsilon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.
| Hint:
$$\int \frac { x^{ 2 } }{ 4x^{ 4 }+25 } dx=\int { \frac { x^{ 2 } }{ 4x^{ 4 }-20{ x }^{ 2 }+25-20{ x }^{ 2 } } dx } =\int { \frac { { x }^{ 2 } }{ { \left( 2{ x }^{ 2 }-5 \right) }^{ 2 }-20{ x }^{ 2 } } dx } =\int { \frac { x^{ 2 }dx }{ \left( 2x^{ 2 }-2\sqrt { 5 } x-5 \right) \left( 2x^{ 2 }+2\sqrt { 5 } x-5 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| Notice, $$\frac{s-1}{s+1}\frac{s}{s^2+1}=\frac{A}{s+1}+\frac{Bs+C}{s^2+1}$$
$$s^2-s=(A+B)s^2+(B+C)s+(A+C)$$
comparing the corresponding coefficients on both the sides, one should get $$A+B=1\tag 1$$
$$B+C=-1\tag 2$$
$$A+C=0\tag 3$$
on solving (1), (2) & (3), one can easily get $\color{red}{A=1}, \color{blue}{B=0}, \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find the antiderivative of this function $\frac{1}{1+x^4}$? How to find the antiderivative of the function $f(x)=\frac{1}{1+x^4}$?
My math professor suggested to use the method of partial fractions, but it doesn't seem to work, because the denominator cannot be factored at all.
Attempting to integrate with other... | It's easy to see factorization in the first line.
\begin{align*}
\int \frac{1}{1+x^{4}}\mathrm{d}x &=\int \left ( \frac{\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}+\sqrt{2}x+1}+\frac{-\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}-\sqrt{2}x+1} \right )\mathrm{d}x\\
&=\int \left [ \frac{\dfrac{\sqrt{2}}{4}\left ( x+\dfrac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\s... | For large $x$ you have $\sqrt x<x$ and so
$$
\sqrt{x+\sqrt{x+\sqrt{x}}}<\sqrt{x+\sqrt{2x}}<\sqrt{3x}.
$$
Since $\sqrt{x+\sqrt{x+\sqrt{x}}}/x<\sqrt{3x}/x\to0$ when $x\to\infty$, your limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^... | The polynomial $x^2+3x+5$ is irreducible over $\mathbb{Q}$. Hence, if $x^2+3x+5$ and $ax^2+bx+c$, with $a,b,c\in\mathbb{Q}$, have a common root, they must be proportional. That is, $$ax^2+bx+c=a\left(x^2+3x+5\right)\,.$$
The problem would be more challenging if $x^2+3x+5$ is replaced by $x^2+3x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Solve $z^4+2z^3+3z^2+2z+1 =0$ Solve $z^4+2z^3+3z^2+2z+1 =0$ with $z$: a complex variable.
Attempt at solving the problem:
We divide the polynom by $z^2$ and we get:
$z^2+2z+3+\dfrac{2}{z}+ \dfrac{1}{z^2}=0 $ $ $ We set $w=z+ \dfrac{1}{z}$
We now have $w^2+2w+5=0$
$\bigtriangleup = -16$
Let's find $\omega$ s... |
"I don't know how to find z"
Sure you do!
....
If you are correct in what you have done so far and you have
$z + \frac 1z = w$
And $w_1 = -1-2i$ and $w_2 = -1 + 2i$ then
you need to solve $z +\frac 1z = (-1-2i)$ or $z^2 +(1+2i)z + 1=0$
ANd $z + \frac 1z = (-1+2i)$ or $z^2 + (1-2i)z + 1 = 0$.
Both of which can be solv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Matrix induction proof Given the following
$\lambda_{1}=\frac{1-\sqrt{5}}{2}$ and $\lambda_{2}=\frac{1+\sqrt{5}}{2}$
How do I prove this using induction:
$\begin{align*}
A^k=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^... | 1) A straightforward proof which is more natural than recursion, in my opinion (for a recursion proof see 2).)
Use diagonalization identity $A=P\Lambda P^{-1}$ from which $A^k=P\Lambda^kP^{-1} \ \ (1)$
where $\Lambda$ is the diagonal matrix diag$(\lambda_1,\lambda_2)$.
Here is an extension of my first explanation:
Inde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$? The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.
I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:
$a = \cos A \\
b = \sin A \cos B \\
c = \sin A ... | From what you've done, $ab+bc+ca = \dfrac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} \geq \dfrac{0 - 1}{2} = \dfrac{-1}{2}$, and this is the minimum value you sought. The minimum occurs when $a+b+c = 0, a^2+b^2+c^2 = 1$. To solve for $a,b,c$ you only need to find one solution of the system of $2$ equations above, then you are done.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The nature of roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0,$ If the expression $ax^2+2bx+c$, where $a$ is a non-zero real number, has the same sign as that of $a$ for every real value of $x$, then roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0$ are:
(A) real and equal
(B) real and unequal
(C) non-real h... | Since $ax^2+2bx+c$ has always the same sign as $a$ for any real $x$, it has no real roots, so $4b^2 - 4ac < 0$.
Now try writing
\begin{align}
(b-c)^2 + 4a(2b+c+a) &= (b-c)^2 + 4(2ab+ac+a^2) \\
&= (b-c)^2 + 4(2ab+b^2+a^2) + 4(ac - b^2) .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $x$ which is correct I have one exercise which is $$(x+2013)(x+2014)(x+2015)(x+2016)+1=0$$
I tag $A=x+2013$ or other for many ways but still can not find the first $x$ value. please help.
| If you let $y=x+2014$, then the equation becomes
$$(y-1)y(y+1)(y+2)+1=0 \Leftrightarrow (y^2+y-1)^2=0$$
So
$$y = \frac{-1\pm\sqrt{5}}{2} \Rightarrow x = \ldots$$
\begin{align}
(y-1)y(y+1)(y+2)+1 & = y^4+2 y^3-y^2-2 y+1 \\
{} & = \left( y^4+y^3-y^2 \right) + y^3-2y+1 \\
{} & = y^2 \left( y^2+y-1 \right) + \left(y^3+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Image of a family of circles under $w = 1/z$ Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.
However, I am having difficu... | Write $z=x+iy$, so $x^2+y^2=z\bar{z}$, and $x=\frac{z+\bar{z}}{2}$. Thus the circles can be described by
$$
z\bar{z}=a\frac{z+\bar{z}}{2}
$$
Upon doing $z=1/w$, you get
$$
\frac{1}{w\bar{w}}=\frac{a}{2}\frac{\bar{w}+w}{w\bar{w}}
$$
that becomes
$$
a\frac{\bar{w}+w}{2}=1
$$
Writing $w=X+iY$, you get
$$
aX=1
$$
More gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prove that the sequence of combinations contains an odd number of odd numbers
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+... | Suppose $n=2k+1$
Note that $\binom{2k+1}{1}=\binom{2k+1}{2k}$, $\binom{2k+1}{2}=\binom{2k+1}{2k-1}$,...,$\binom{2k+1}{k}=\binom{2k+1}{k+1}$.
Thus
$$\binom{2k+1}{1}+\binom{2k+1}{2}+\cdots+\binom{2k+1}{k}+\binom{2k+1}{k+1}+\cdots+\binom{2k+1}{2k-1}+\binom{2k+1}{2k}=2^{2k+1}-2$$
From the above considerations,
$$2\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2})... | $3(x - 1)(x + {2 \over 3}) = (x -1)(3x + 2) = (3x - 3)(x + {2 \over 3}) =...$ etc. are all valid factoring. The leading coefficient is just a constant.
And if $(x - 1)(x + {2 \over 3}) = 0$ then $3(x - 1)(x + {2 \over 3}) = 0 = (x - 1)(x + {2 \over 3}) $.
If you are concerned about going from roots to factoring think... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequa... | $f(x) = 1/x$ is strictly convex, therefore
$$
\frac{1}{2n} < \frac 12 \left( \frac{1}{n+k} + \frac{1}{3n-k} \right)
$$
for $k = 1, ..., n-1$, or
$$
\frac{1}{n+k} + \frac{1}{3n-k} > \frac {1}{2n} + \frac {1}{2n}
$$
Combining terms pairwise from both ends of the sum shows that
$$
\frac{1}{n+1} + \frac{1}{n+3}+\dots+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\... | I think one could do this in a conceptually simpler way:
Since
$$
\frac{y}{1+y}<\log(1+y)<y
$$
your integrand is bounded as
$$
\frac{1}{x^3}<\frac{1}{nx^2\ln(1+x/n)}<\frac{1}{nx^2(x/n)/(1+x/n)}=\frac{1}{x^3}(1+x/n).
$$
By monotonicity, your integral satisfies
$$
\frac{1}{2}=\int_1^{+\infty}\frac{1}{x^3}\,\mathrm dx<\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Cauchy like inequality $(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$ Problem: Prove that for real $x, y, \alpha, \beta$,
$(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$.
I am looking for an elegant (non-bashy... | By C-S $(5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)=$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)^2+\frac{14\beta^2}{5}\right)\left(\left(\sqrt5x+\frac{y}{\sqrt5}\right)^2+\frac{14y^2}{5}\right)\geq$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)\left(\sqrt5x+\frac{y}{\sqrt5}\right)+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from... | One has $$\int \frac{u+\frac{3}{2}}{u^2+\frac{59}{4}} du = \int \frac{u}{u^2+\frac{59}{4}}du + \int \frac{\frac{3}{2}}{u^2+\frac{59}{4}}du.$$ The first term can be computed be setting $v = u^2+\frac{59}{4}$ and the second thanks to the $\arctan$ function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability: Finding the Number of Pears Given Two Scenarios
You have a bag containing 20 apples, 10 oranges, and an unknown number
of pears. If the probability that you select 2 apples and 2 oranges is
equal to the probability that you select 1 apple, 1 orange, and 2
pears, then what is the number of pears orig... | It is good that you were alert. I believe everything you are doing is correct. If we continue, then
$$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$
I think it is ok to have a decimal number here.
This gives
\begin{align*}
\frac{n!}{2!(n-2)!} &= \frac{171}{4}\\
\implies \frac{n!}{(n-2)!} &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the the integral of $\sqrt{x^a + b}$? How do you evaluate $\displaystyle\int\sqrt{x^a + b}\,\,\text{dx}$, where $a \neq 0$ and $a \neq 1$?
For example, how do you evaluate $\displaystyle\int\sqrt{x^2 + 1}\,\text{dx}$? If we let $u=x^2+1$, then $du=2x\,\text{dx}$. We cannot do this because there is no $2x$ in th... | Kim Peek's "funny" hypergeometric solution is really the series solution near $x=0$. We have, for $|x^a/b| < 1$,
$$ \sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} =
\sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$
so integrating term-by-term
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Distance between two circles on a cube I found this problem in a book on undergraduate maths in the Soviet Union (http://www.ftpi.umn.edu/shifman/ComradeEinstein.pdf):
A circle is inscribed in a face of a cube of side a. Another circle is circumscribed about a neighboring face of the cube. Find the least distance betw... | I placed the inscribed circle on the top face (+y direction) and the circumscribed circle on the front face (+z direction). Their locus of points is
$$ \begin{align}
\vec{r}_1 & = \begin{bmatrix} r_1 \cos \theta_1 & \frac{a}{2} & r_1 \sin \theta_1 \end{bmatrix} \\
\vec{r}_2 & = \begin{bmatrix} r_2 \cos \theta_2 & r_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac... | Using just Macaulay2, you can do the following
Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
ReesAlgebra, TangentCone
i1 : R=QQ[a,b,c]
o1 = R
o1 : PolynomialRing
i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to g... | We have $\sqrt[6]{3} \approx 1.2009$ and $\sqrt[6]{4} \approx 1.2599$. Let $f(x)=x^3+x^2-x-2$. Then $f(1.2) \approx -0.032$ and $f(1.25) \approx 0.2656$. So, $a$ must be between $1.2$ and $1.25$.
EDIT: As I said in the comments below, I see no way of showing that there is only one root using only precalculus. But,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many integer-sided right triangles are there whose sides are combinations?
How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?
Attempt:
This seems like a hard question, since I can't even think of... | Solving $(1)$ for $z$, we have,
$$z = \frac{1\pm\sqrt{1\pm4w}}{2}\tag3$$
where,
$$w^2 = (x^2-x)^2+(y^2-y)^2\tag4$$
It can be shown that $(4)$ has infinitely many integer solutions. (Update: Also proven by Sierpinski in 1961. See link given by MXYMXY, Pythagorean Triples and Triangular Numbers by Ballew and Weger, 1979.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1655884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Prove that if $ 2^n $ divides $ 3^m-1 $ then $ 2^{n-2} $ divides $ m $ I got a difficult problem. It's kind of difficult to prove. Can you do it?
Let $ m,n\geq 3 $ be two positive integers. Prove that if $ 2^n $ divides $ 3^m -1$ then $ 2^{n-2} $ divides $ m $
Thanks :-)
| Because $n \geq 3$ we get $8 \mid 3^m-1$ and so $m$ must be even .
Let $m=2^l \cdot k$ with $k$ odd . Now use the difference of squares repeatedly to get :
$$3^m-1=(3^k-1)(3^k+1)(3^{2k}+1)\cdot \ldots \cdot (3^{2^{k-1} \cdot l}+1)$$
Each term of the form $3^s+1$ with $s$ even has the power of $2$ in their prime factori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but... | \begin{align}x^2-4x+3&=x^2-3x-x+3\\
&=x(x-3)-(x-3)\\
&=(x-1)(x-3)
\end{align}
Note: You could also see that the sum of coefficients is zero, hence one root is $x=1$. Now divide the quadratic by $x-1$ to get the other factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$
Evaluate :$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)$$
$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)=\lim_{x\to \infty} \left[(\sqrt{4x^2+x}-2x)\frac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x}\right]=\lim_{x\to \infty}\frac{{4x^2+x}-4x^2}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}... | Hint :
$\displaystyle\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$ , dividing numerator and denominator by $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | Let $T_{m}$ be $a^m+b^m+c^m$.
Let $k=-ab-bc-ca$, and $l=abc$.
Note that this implies $a,b,c$ are solutions to $x^3=kx+l$.
Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)
It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$.
From her... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 4
} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions)... | One can rewrite:
$$\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^2 \sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{1-\cos x}{x^2 \cos x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\sin^2 x}{x^2}\frac{1}{(1+\cos x)\cos x}$$
$$=2^{\sin x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
How to evaluate this limit? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to 0+}\left(\frac{\left[\ln\left(\frac{5+x^2}{5+4x}\right)\right]^6\ln\left(\frac{5+x^2}{1+4x}\right)}{\sqrt{5x^{10}+x^{11}}-... | Let's try the elementary way. We have
\begin{align}
L &= \lim _{x \to 0^{+}}\left(\dfrac{\left[\log\left(\dfrac{5 + x^{2}}{5 + 4x}\right)\right]^{6} \log\left(\dfrac{5 + x^{2}}{1 + 4x}\right)}{\sqrt{5x^{10} + x^{11}} - \sqrt{5}x^5}\right)\notag\\
&= \lim _{x \to 0^{+}}\left(\dfrac{\left[\log\left(1 + \dfrac{x^{2} - 4x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\te... | From here and here we learn that (mistake 1)
\begin{align}
\int \frac{x}{\sqrt{1+ x^4}}dx&=\frac12\arcsin x^2\\
&=\frac12\ln(x^2+\sqrt{1+x^4})
\end{align}
Hence (for part 2 I basically take your solution multiplied by $-1$: mistake 2)
\begin{align}
I&=\frac12\ln x^2+\sqrt{1+x^4})-\frac14 \ln \frac{ \sqrt{1+ x^4}-1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Cauchy Residue Theorem Integral I have been given the integral $$\int_0^ {2\pi} \frac{sin^2\theta} {2 - cos\theta} d\theta $$ I have use the substitutions $z=e^{i\theta}$ |$d\theta = \frac{1}{iz}dz$ and a lot of algebra to transform the integral into this $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{z^2-4z+1}dz$$ I... | There was an error in the original post. We have
$$\int_0^{2\pi}\frac{\sin^2(\theta)}{2-\cos(\theta)}d\theta=-\frac i2\oint_{|z|=1}\frac{(z^2-1)^2}{z^2(z^2-4z+1)}\,dz$$
There are two poles inside $|z|=1$. The first is a second order pole at $z=0$ and the second is a first order pole at $z=r_2$.
To find the reside of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| Here is a very late answer since I just saw the problem:
By brute force, we may check that $n=12$ is the smallest possible integer such that $2^8+2^{11}+2^n$ is a perfect square. We also claim that this is the only integer.
To see why the above is true, let $2^8+2^{11}+2^n=2^8(1+2^3+2^k)=2^8(9+2^k), k \ge 4$. Now, we o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
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How to prove the following binomial identity How to prove that
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
| Suppose we seek to verify that
$$\sum_{q=0}^n {2q\choose q} 4^{-q} =
(2n+1) {2n\choose n} 4^{-n}$$
using a method other than induction.
Introduce the Iverson bracket
$$[[0\le q\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^q}{w^{n+1}} \frac{1}{1-w}
\; dw$$
This yields for the sum (we extend the sum to infin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | $$\lim _{t\to 0}\left(\left[\left(\frac{1}{t}+1\right)\left(\frac{1}{t}+2\right)\left(\frac{1}{t}+3\right)\left(\frac{1}{t}+4\right)\left(\frac{1}{t}+5\right)\right]^{\frac{1}{5}}-\frac{1}{t}\right) = \lim _{t\to 0}\left(\frac{\sqrt[5]{1+15t+85t^2+225t^3+274t^4+120t^5}-1}{t}\right) $$
Now we use the Taylor's developmen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Find all possible values of $c^2$ in a system of equations. Numbers $x,y,z,c\in \Bbb R$ satisfy the following system of equations:
$$x(y+z)=20$$
$$y(z+x)=13$$
$$z(x+y)=c^2$$
Find all possible values of $c^2$.
To try to solve this, I expanded the equations:
$$xy+xz=20$$
$$yz+xy=13$$
$$xz+yz=c^2$$
Then I subtracted the f... | Let $c^2 = s$. Eliminating $x$ and $y$, you get an equation in $s$ and $z$:
$$ s^2 + 2 z^2 s - 66 z^2 - 49 $$
Thus
$$ z^2 = \dfrac{s^2 - 49}{66 - 2 s}$$
Since $z^2 \ge 0$, we need either $s \le -7$ or $7 \le s < 33$. This corresponds to $\sqrt{7} \le c < \sqrt{33}$. We then have
$$ \eqalign{y &= \dfrac{z (33 - c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Summing the terms of a series I have a really confusing question from an investigation. It states-
Find the value of:
$$\sqrt{1^3+2^3+3^3+\ldots+100^3}$$
How would I go about answering this??
| And if you don't know the formula
and don't need it exactly,
$\sum_{k=1}^{100} k^3
\approx \int_0^{100} x^3 dx
=\frac{100^4}{4}
$
so the result is
$\sqrt{\frac{100^4}{4}}
=\frac{100^2}{2}
=5000
$.
If you add in the
usual correction of
$\frac12 f(n)$,
the result is
$\sqrt{\frac{100^4}{4}+\frac12 100^3}
=\frac{100^2}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Proof that $\sum_{i=0}^n 2^i = 2^{n+1} - 1$ $\sum_{i=0}^n 2^i = 2^{n+1} - 1$
I can't seem to find the proof of this. I think it has something to do with combinations and Pascal's triangle. Could someone show me the proof? Thanks
| Mathematical induction will also help you.
*
*(Base step) When $n=0$, $\sum_{i=0}^0 2^i = 2^0 = 1= 2^{0+1}-1$.
*(Induction step) Suppose that there exists $n$ such that $\sum_{i=0}^n 2^i = 2^{n+1}-1$. Then $\sum_{i=0}^{n+1}2^i=\sum_{i=0}^n 2^i + 2^{n+1}= (2^{n+1}-1)+2^{n+1}=2^{n+2}-1.$
Therefore given identity ho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Limit of sum of the series What would be the sum of following ?
$$\lim_{n\to\infty} \left[\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + \cdots + \frac{1}{(n+n)^{2}}\right]$$
I tried to turn it into integral :
$\displaystyle\int \frac{1}{(1+\frac{r}{n})^{2}}\frac{1}{n^{2}} $ but
I can't figure out ... | $$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{n}{(n+1)^2}$$
$$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Integrating $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$ I came across a question today...
Find $$\displaystyle\int\dfrac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$$
How to do this? I tried to take $x^4+1=u^2$ but no result. Then I tried to take $x^2+1=\frac{1}{u}$, but even that didn't work. Then I manipulated it to $\int \dfr... | Hint Divide the numerator and denominator by $x^2$, to get:
$$\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x})(\sqrt {(x+\frac{1}{x})^2-2} )}$$
Then put $x+\frac{1}{x}=t$
$$\int \frac{dt}{t(\sqrt{t^2-2})}$$
Which is easily taken care of by putting $t=\sqrt2 \sec\theta$,
$$\int \frac{(\sqrt2 \sec\theta\tan \theta)d\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-... | You need to first find the square root of $1 - 4 \sqrt{3}i$ using the same method: let $(c+di)^2 = 1 - 4 \sqrt{3} i$ and then compare real and imaginary parts to find $c$ and $d$ explicitly. This will give you two different answers. Then assume $(a+bi)^2 = c + di$, and compare real and imaginary parts to find $a$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How would one solve the following equation? This equation is giving me a hard time.
$$e^x(x^2+2x+1)=2$$
Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calcu... | The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is $\color{red}{x=0.249}$. Now we have
$$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Weird Inequality that seems to be true Is it true that:
$$\left (3x+\frac{4}{x+1}+\frac{16}{y^2+3}\right )\left (3y+\frac{4}{y+1}+\frac{16}{x^2+3}\right )\geq 81,\ \forall x,y\geq 0$$
I have proved that $3x+\frac{4}{x+1}+\frac{16}{x^2+3}= 9 +\frac{(x-1)^2 (3x^2+1)}{(x+1)(x^2+3)}, \ \forall x\geq 0$, but I did not succe... | This inequality is false, e.g. $x=2.5$, $y=0.5$. You will obtain a value of $79.9901<81$.
| {
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"url": "https://math.stackexchange.com/questions/1682088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number. Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number.
| $$\sin^8 x+\cos^8x \ge \frac 18;$$
$$ \left (\frac{1-\cos2x}{2} \right )^4+ \left (\frac{1+\cos2x}{2} \right )^4\ge \frac 18;$$
$$(1-\cos2x)^4+(1+\cos2x)^4 \ge2$$
$$1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x+$$
$$+1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x\ge 2$$
$$12\cos^22x+2\cos^42x \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\... | Continuing from what you got:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
and since $\sec \theta \cos \theta = 1, \cot \theta \tan \theta = 1$, expand the brackets:
$$\sec\theta \cot\theta = \frac{\cot \theta + \sec \theta}{\cos \theta + \tan \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximum minimum values in trigonometry
Find minimum value of $2\sin^2a+3\cos^2a$
Solving it we get $2+ \cos^2a$
Answer: $3$ (taking $\cos a$ as $-1$)
Why are we using the minimum cosine value as $-1$ instead of using the cosine as $0$?
This can make the minimum value as $2$.
| $$2 \sin^2a+3 \cos^2 a= 3\cos^2a+2-2\cos^2 a=\cos^2a+2$$
$$0\le \cos^2a \le 1 \Rightarrow 2\le \cos^2a+2 \le 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt... | The coefficient of $x^7$ is
$$\binom{n}{7}\frac{2^{n-7}}{3^7}$$
And the coefficient of $x^8$ is
$$\binom{n}{8}\frac{2^{n-8}}{3^8}$$
Comparing them we get:
$$\binom{n}{8}=\binom{n}{7}\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, other... | Let me try do derive that antiderivative. You computed:
$$f(x)=\underbrace{-2\arcsin\sqrt{\frac{2-x}{2}}}_{f_1(x)}\underbrace{-\sqrt{2x-x^2}}_{f_2(x)}.$$
The easiest term is clearly $f_2$:
$$f_2'(x)=-\frac{1}{2\sqrt{2x-x^2}}\frac{d}{dx}(2x-x^2)=\frac{x-1}{\sqrt{2x-x^2}}.$$
Now the messier term. Recall that $\frac{d}{dx... | {
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"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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What's the formula for this series for $\pi$? These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}
= \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1
$$
$$\small
\pi = 3 + \cf... | The third one should be obtained from $4.1.40$ in A&S p.68 using $z:=ix$ (from Euler I think not sure) :
$$-2\,i\,\log\frac{1+ix}{1-ix} = \cfrac{4x} {1+\cfrac{(1x)^2} {3+\cfrac{(2x)^2} {5+\cfrac{(3x)^2} {7+\ddots}}}}
$$
Except that the expansion of the function at $x=1$ is simply your expansion for $(1)$.
Some neat var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
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Proof of an identity that relates hyperbolic trigonometric function to an expression with euclidean trigonometric functions.
Given a line $r$ and a (superior) semicircle perpendicular to $r$, and an arc $[AB]$ in the semicircle, I need to prove that
$$
\sinh(m(AB)) = \frac{\cos(\alpha)+\cos(\beta)}{\sin(\alpha)\sin(\b... | Using the definition of the hyperbolic sine and hyperbolic cosine functions, we have
$$
\sinh m(AB)
= \frac{e^{m(AB)} - e^{-m(AB)}}{2}
= \frac{(AA'\cdot BB')^2 - (BA'\cdot AB')^2}{2(AA'\cdot BB')(BA'\cdot AB')} \\
\cosh m(AB)
= \frac{e^{m(AB)} + e^{-m(AB)}}{2}
= \frac{(AA'\cdot BB')^2 + (BA'\cdot AB')^2}{2(AA'\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Does $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$ converge? $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$
and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using... | On the one hand,
$$a_n \ge \mbox{smallest summand} \times \mbox{number of summands}= \frac{1}{\sqrt{n^2+2n-1}}\times n .$$
To deal with the denominator, observe that
$$n^2+2n-1 \le n^2+2n+1=(n+1)^2.$$
On the other hand,
$$a_n \le \mbox{largest summand}\times \mbox{number of summands} = \frac{1}{\sqrt{n^2+n}}\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$? How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$,$n\in N$ I know $\lim_{n\to \infty } (1+ \frac{7}{7n+5} )^{ n+1}=e$ and $\lim_{n\to \infty } \sqrt[n+1]{n+1} =1$.
| By AM-GM
$$\frac{1+2 + 3 + \cdots + n}{n} \ge \sqrt[n]{1 \times 2 \times 3 \times \cdots \times n}$$
$$\implies \frac{n+1}2 \ge \sqrt[n]{n!} \implies 6n+6 \ge 12\sqrt[n]{n!}$$
But $7n+5 \ge 6n+6$ for $n \ge 1$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solving for $k$ when $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ Consider $$|z|=|z-3i|$$
We know that if $z=a+bi\Rightarrow b=\frac{3}{2}$
$z_1$ and $z_2$ will represent two possible values of $z$ such that $|z|=3$. We are given $\arg(z_1)=\frac{\pi}{6}$
The value of $k$ must be found assuming $\arg\left(\frac{z_1^kz_2... | You already know that $\arg(z_1)=\frac{\pi}{6}$, and moreover $\arg(z_2)=\frac{5\pi}{6}$ and $\arg \left(\frac{1}{2i}\right)=\frac{-\pi}{2}$. Multiplying complex numbers results in adding their arguments (modulo $2\pi$) so you get the equation
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=k\frac{\pi}{6}+\frac{5\pi}{6}-\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
closed form for $I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$ $$I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$$
for $n=1$ I tried to use $\arctan x=u$
and by notice that $$\frac{1+\tan u}{1-\tan u}=\cot\left ( \frac{\pi}{4}-u \right )$$... | For $n=2$ we have, integrating by parts, $$I\left(2\right)=\int_{0}^{\pi/4}x^{2}\cot\left(x\right)dx=\frac{\pi^{2}}{16}\log\left(\frac{1}{\sqrt{2}}\right)-2\int_{0}^{\pi/4}x\log\left(\sin\left(x\right)\right)dx
$$ and now we can use the Fourier series of $\log\left(\sin\left(x\right)\right)$ $$\log\left(\sin\left(x\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}... | Expand the expression to get
$$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)=1+\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x\cos x}$$
Then using the identity $\sin x\cos x = \frac{1}{2}\sin 2x$, rewrite as
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) &=
& 1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
My book tells me to use tchebycheff's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \... | As @ChenJiang stated, its a case of cauchy's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$
$$(a+b+c+d)^2\le 4(a^2+b^2+c^2+d^2)$$
$$(8-e)^2\le 4(16-e^2)$$
$$5e^2-16e\le 0$$
$$e(5e-16)\le 0$$
$$\implies 0\le e\le \frac{16}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Use Cauchy Product to Find Series Representation of $[\ln(1 + x)]^2$ Problem:
Let $f(x) = [\ln(1 + x)]^2$. Use the series for the logarithm to compute that
\begin{align*}
f(x) = [\ln(1 + x)]^2 = \sum_{n = 2}^{\infty}(-1)^n\Bigg(\sum_{k = 1}^{n - 1} \frac{1}{(n - k)k}\Bigg) x^n.
\end{align*}
Use this to evaluate th... | Note that for $-1<x< 1$ we have
$$\begin{align}
\log^2(1+x)&=\sum_{k=1}^\infty\sum_{m=1}^\infty\frac{(-1)^{k+m}x^{k+m}}{k\,m}\tag 1\\\\
&=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 2\\\\
&=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 3\\\\
\end{align}$$
as was to be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Linear system 2 unknowns There are $x$ white and $y$ black pearls and their ratio is $z$. If I add six black and six white pearles, the ratio doubles.
I did the following:
$ \frac{x+6}{y+6} = \frac{2x}{y}$ and then I get
$xy -6(2x-y)=0$
I can find solutions by guessing. Is there any other way?
ADDED:
Now I have to so... | $\frac{x}{y} = z$ and $\frac{x+6}{y+6} = 2z$
$\implies x = zy$ and $\frac{zy+6}{y+6} = 2z$
$\implies 6 = z(y+12) \implies z = \frac{6}{y+12}$
$\implies x = zy = \frac{6y}{y+12}$
Since $x$ and $y$ are the number of pearls, they must be integers. That is
$(y+12) | (6y)$.
There are only 3 possible $y$ values by giving $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $p>3$ and $p+2$ are twin primes then $6\mid p+1$ I have to prove that if $p$ and $p+2$ are twin primes, $p>3$, then $6\ |\ (p+1)$.
I figure that any prime number greater than 3 is odd, and therefore $p+1$ is definitely even, therefore $2\ |\ (p+1)$. And if I can somehow prove $3\ |\ (p+1)$, then I would be done. Bu... | The easy way.
Note that one of $p,p+1,p+2$ must be divisible by $3$, since they are three consecutive numbers, and since $p$ and $p+2$ are prime, that must be $p+1$. We can do the same to show that $p+1$ is divisible by $2$.
Looking modulo $6$.
We can look $\mod 6$. We see that
\begin{align}
6k+0\equiv 0\mod 6&\Righta... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | HINT
Perhaps the other methods are easier but to continue where you left off,
Realize that $$\sin 2x=7\cos 2x+5$$
Use $$\sin^2 2x+\cos^2 2x=1$$
To make your last equation into a quadratic for $\cos 2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 2
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If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then show that $a+b$ is a square.
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.
This can be simplifie... | To show this, we note that $c(a+b)=ab$. Now let $g$ be the gcd of $a$ and $b$, which need not necessarily be $1$. Denote $a=a'g$ and $b=b'g$ so that we get $c(a'+b') = a'b'g$.
Because $a' + b'$ is relatively prime to both $a'$ and $b'$, it follows that it divides $g$. But g also divides $c(a'+b')$. Further, note that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively and $CY$ is perpendicular o... | Equation of $Tt$ is $bx\cos \theta + ay\sin \theta - ab = 0$. Hence $C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$ and $C{P^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta $. Now,
$$P{Y^2} = C{P^2} - C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }} - \left( {{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I calculate this limit $\lim_{(x,y)\to(0,0)} \frac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$? How can I calculate this limit $$\lim_{(x,y)\to(0,0)} \dfrac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$$ at the origin?
I tried to use the substitution $x ^2 + y^2=t$ but how can I evaluate the value of $xy$? I... | Outline: Note that $(x-y)^2\ge 0$, so $|xy|\le \frac{1}{2}(x^2+y^2)$. One can also get this from polar coordinates, for $|xy|=r^2|\cos\theta\sin\theta|=\frac{1}{2}r^2|\sin(2\theta)|\le \frac{r^2}{2}$.
Now you can comfortably let $t=x^2+y^2$. You will need to look at the behaviour of $1-\cos t$ near $0$. This can be ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expansion and factorisation I have a little problems with a few questions here and I need help.. Thanks ...
*
*Factorise completely
$$9x^4 - 4x^2 - 9x^2y^2 + 4y^2 $$
My workings ..
$$ (3x^2+2x)(3x^2-2x) - y^2 (9x^2-4) = (3x^2 + 2x)(3x^2 -2x) - y^2 (3x+2)(3x-2) $$
*Factorise $3x^2 + 11x - 20$ and , hence Factori... | *
*$$9x^4 - 9x^2y^2 - 4x^2 + 4y^2$$
Group in paris such that; $9x^2(x^2-y^2) - 4(x^2 - y^2) = (x^2 - y^2)(9x^2 - 4)$
Then using the difference of 2 squares we get: $$(x-y)(x+y)(3x-2)(3x+2)$$
2.$3x^2 + 11x - 20$, factorise to $(x-5)(3x-4)$. As @mathlove pointed out $11(a-b) - 20 + 3(?)^2$, where $ ? = a-b$. Which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\s... | multiply and divide $ \frac{1}{\sin{x}+\cos{x}}$ by $ \frac{1}{\sin{x}\cos{x}}$ then in the numerator substitute $1$ by $sin^2{x} + cos^2{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Unknown Inequality $$ \left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right) \leq 8(y+x)(x+z)(y+z)$$
I can prove this inequality, but i need know if this inequaliy is known...
| We need to prove that
$$\prod\limits_{cyc}(a\sqrt{3(a^2+b^2+c^2)}+bc)\leq8\prod\limits_{cyc}(a^2+b^2)$$
which is true even for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, it's obvious that the last inequality is equivalent yo $f(w^3)\leq0$, whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to determine if the following series converge or not? $\Sigma_{n=1}^{\infty} a_n $ where:
*
*$ a_n = \frac{1}{\ln(n)^{\ln(n)}}$
*$a_n = \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)$
in the first case, I really have no idea
in the second case, is it correct to say that for $ \frac{1}{n }-\ln\left( 1+\frac{1}{n}\r... | For
2.
$\begin{array}\\
a_n
&= \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)\\
&= \frac{1}{n }-\int_1^{1+1/n} \frac{dx}{x}\\
&= \frac{1}{n }-\int_0^{1/n} \frac{dx}{1+x}\\
&= \int_0^{1/n} (1-\frac{1}{1+x})dx\\
&= \int_0^{1/n} (\frac{x}{1+x})dx\\
&< \int_0^{1/n} x\,dx\\
&= \frac{x^2}{2}|_0^{1/n}\\
&= \frac{1}{2n^2}\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24... | AM/GM. $\frac{x}{2}+\frac{1}{x}<x$ iff $x^2>2$. So it is enough to show that if $x^2>2$ then $(\frac{x}{2}+\frac{1}{x})^2>2$ or $\frac{x^2}{4}+\frac{1}{x^2}>1$. But by AM/GM $(\frac{x^2}{4}+\frac{1}{x^2})/2>\sqrt{\frac{1}{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \f... | $$\sum_{i=1}^\infty \frac{2^n + 3^n}{6^n}=\sum_{i=1}^\infty (\frac{2}{6})^n + \sum_{i=1}^\infty (\frac{3}{6})^n=\frac{1}{2}+1=\frac{3}{2}$$
sums from forumla of geometric series $1/3^n$ and $1/2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Probability Problem - Finding a pdf Below is a problem I did. However, I did not come up with the answer in book.
I am thinking that I might have the wrong limits for the integral. I am hoping
somebody can point out what I did wrong.
Bob
Problem
Let $Y = \sin X$, where $X$ is uniformly distributed over $(0, 2 \pi)$. Fi... | your answer is correct indeed just a little mistake. in the first step:
$$P(Y \le y) = P( \sin x \le y ) \Rightarrow P(x_1 \le x \le x_2 )$$ as shown in figure below (sorry figure is for $y=\sin(x+\theta)$ but still is useful .just draw the figure for $\sin(x)$ in your imagination. also figure is from "Probability, Ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the stationary point of $f(x,y)=\sin x \sin y \sin (x+y)$ How to find the stationary point(s) of $f(x,y)=\sin x \sin y \sin (x+y)$
With $x,y\in(0,\pi)$
So far I have found $$\nabla f =(\color{red}{\sin x\cos (x+y)+\cos x\sin (x+y)\sin y},\color{blue}{\sin y\cos (x+y)+\cos y\sin (x+y)\sin x)}$$
So we need
$... | $\left(\sin x \cos(x+y) + \cos x\sin(x+y)\right) \sin y = 0$ when $\sin x \cos(x+y) + \cos x\sin(x+y) = 0$ or when $\sin y = 0$. I trust that you know when $\sin y = 0$. What about that first one?
\begin{align}
\sin x \cos(x+y) + \cos x\sin(x+y) &= 0 \\
\sin x \cos(x+y) &= -\cos x\sin(x+y)\\
-\tan x &= \tan(x+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do I find the vector $T\begin{pmatrix} 5 & 0 \\ -10 & -13 \end{pmatrix}$? I defined a function $T: M^R_{2x2} \rightarrow R_4[x]$ and I defined:
$T\begin{pmatrix} 2 & 3 \\ 1 & 0 \end{pmatrix} = x^2$
$T\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = 3x - 4$
$T\begin{pmatrix} 0 & 2 \\ 4 & 5 \end{pmatrix} = 2x^2 - 7$
Ho... | Hint: You'll need to determine whether there exist $a,b,c$ for which
$$
\pmatrix{5&0\\-10&-13} =
a\pmatrix{2&3\\1&0} +
b\pmatrix{1&0\\0&2} +
c \pmatrix{0&2\\4&5}
$$
and, if such $a,b,c$ exist, find them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\fra... | Assuming your calculations so far are correct (I didn't check), you are almost there. Divide both sides by $y$, you will get
$$\frac {3x^2} {y^2} + \frac {2.5x} y = 3,$$
a quadratic equation for $\frac xy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k}$
Question: The curve in the figure is the parabola $y=kx^2$ where $k>0$.
Several normal lines to this parabola are also shown. Consider the points in the first quadrant from which the normal lines are dr... | You did almost all the work. From
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$
the $x$ you are looking for is the one with the '$-$':
$$x={-\frac{1}{2kx_0}- { (2kx_0 + \frac{1}{2kx_0})} \over 2k}=-\frac{1}{2k^2x_0}-x_0$$
(the other one is simply $x=x_0$).
Now
$$y=k\left(-\frac{1}{2k^2x_0}-x_0\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.
| This seems like an exercise in Lagrange Multipliers. You need to minimize the distance function $f(x,y,z) = (x-1)^2 + (y-4)^2 + z^2$ (which is the square distance from $(x,y,z)$ to $(1,4,0)$) subject to the constraint that $g(x,y,z) = x^2 + y^2 - z^2 = 0$. Any point which minimizes such $f$ subject to the constraint wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$ Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$If $k,w,f$ are relatively prime positive integers,then find the value of $k+w+f.$
The given equation is $\log_8(kx^2+wx+f)=2$ i.e. $kx^2+wx+f=64$
Since $x=\frac{1}{3}... | Eliminating $f$ gives $44k=3w$. Take $w=44,k=3$. Then we get $f=49$. Note that the general solution is $k=3h,w=44h,f=64-15h$, but the requirement that the numbers are relatively prime positive integers forces $h=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
| You may use common denominator technique:
$$
\frac 1{x+3} + \frac 4{x^2+2x-3} = \frac {1}{x+3} + \frac {4}{(x+3)(x-1)} = \frac {1 \times (x-1) + 4 \times 1}{(x+3)(x-1)} = \frac{x-1+4}{(x+3)(x-1)}
$$
Take a denominator which divisible to both denominators as the common denominator and divide that by each denominator se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | An even simpler way is to substitute $x \rightarrow y - \frac{\pi}{4}$
You equation is now $$\sqrt{2} \sin(y) = \frac{1}{3}$$
The solutions are
$$\begin{aligned} x & = \arcsin \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2\pi n \\ x & = \arccos \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2 \pi n
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
} |
Finding ALL solutions of the modular arithmetic equation $25x \equiv 10 \pmod{40}$ I am unsure how to solve the following problem. I was able to find similar questions, but had trouble understanding them since they did not show full solutions.
The question:
Find ALL solutions (between $1$ & $40$) to the equation $25x \... | Let's use the definition of congruence. $a \equiv b \pmod{n} \iff a = b + kn$ for some integer $k$. Hence, $25x \equiv 10 \pmod{40}$ means $$25x = 10 + 40k$$ for some integer $k$. Dividing each side of the equation $25x = 10 + 40k$ by $5$ yields $$5x = 2 + 8k$$
for some integer $k$. Thus,
$$5x \equiv 2 \pmod{8}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | The reciprocal of the term of interest is
$$\begin{align}
\frac{(2n-1)!!}{n!}&=\left(\frac{(2n-1)}{n}\right)\left(\frac{(2(n-1)-1)}{(n-1)}\right)\left(\frac{(2(n-2)-1)}{(n-2)}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\
&=\left(2-\frac{1}{n}\right)\left(2-\frac{1}{n-1}\right)\left(2-\frac{1}{n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow ... | The slope of the normal line is going to be: $-\frac{1}{2a}$.
set $g(x)=-\frac{1}{2a}x+(a^2+\frac{1}{2})$
You want to solve $g(x)=f(x)$.
$-\frac{1}{2a}x+(a^2+\frac{1}{2})=x^2$
$x^2+\frac{1}{2a}x+\frac{1}{16a^2}=(a^2+\frac{1}{2})+\frac{1}{16a^2}$
$x=\pm\sqrt{(a^2+\frac{1}{2})+\frac{1}{16a^2}}-\frac{1}{4a}$
Personally, I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arct... | Here is a straightforward (though long) derivation of the piece wise function description of $\arctan(x)+\arctan(y)$.
We will show that:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$ If $a>1$ then $\frac{1}{x}+\ln x=a$ has two distinct roots($x_1$ and $x_2$, Assume $x_1<x_2$). Show that $$x_1+x_2+1<3\exp(a-1)$$
First I tried to estimate the place of the roots separately. I have got that $x_1\leq \frac{1}{a}$ and $\exp(a-1)<x_... | I can show the inequality when $a$ is close enough to $1$ (namely, $a\leq 1+ln(5/4)\approx 1.22$) or when $a$ is big enough (namely, $a \geq 1+\ln(5) \approx 2.6$). In the sequel $f(x)$ denotes $x+\ln(\frac{1}{x})$.
When $a$ is close to $1$. Let us put $w=\sqrt{e^{a-1}-1}$. The inequality then becomes $x_1+x_2\leq 2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Taking inverse Fourier transform of $\frac{\sin^2(\pi s)}{(\pi s)^2}$ How do I show that
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$
I know that $\sin^2(\pi s)=\frac{1-\co... | Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\
&=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\... | For every $x\in \mathbb R$ which $|x|\lt 1$, we have: $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ From here: $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ Now, multiplying $x^2$ in both side we get that: $$\sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}$$ And so: $$\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n+1}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limiting question:$\displaystyle \lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}$ How do I find the value of $$\lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x - \sin\ x}$$
in easy way.
| As Henry W; commented, Taylor series make thigs quite simple.
$$A=a \tan(x) \implies \log(A)=\tan(x)\log(a)=\Big(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)\Big)\log(a)$$ $$A=e^{\log(a)}\implies A=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)+2 \log
(a)\right)+O\left(x^4\right)$$ So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solve: $\int\frac{\sin 2x}{\sqrt{3-(\cos x)^4}}$ (and a question about $t=\tan \frac{x}{2}$) I tried substituting $$t=\tan \frac{x}{2}$$ but the nominator is $\sin {2x}$, so is there a way to get from $$\sin x=\frac{2x}{1+x^2}$$ to an expression with $\sin 2x$?
| $$\int\frac{\sin(2x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Use $\sin(2x)=2\sin(x)\cos(x)$:
$$2\int\frac{\sin(x)\cos(x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:
$$-2\int\frac{u}{\sqrt{3-u^4}}\space\text{d}u=$$
Substitute $s=u^2$ and $\text{d}s=2u\space\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$... | Let $P$ be the expression we want to maximise.
Using the following notation: $S_1=x+y+z$, $S_2=xy+xz+yz$ and $S_3=xyz$.
From the hypothesis we get that, $S_1^2=4S2$.
So the expression we want to maximise is:
$P=\dfrac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}=\dfrac{S_1^3-3S_1S_2+3S3}{2S_1S_2}$
Then, simplify it using the h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given three coordinates (a,b,c), (d,e,f), and (l,m,n), what is the center of the circle in the 3D plane (h,k,i) that contains these three points. I have tried the following:
$$(a-h)^2+(b-k)^2+(c-i)^2=r^2$$
$$(d-h)^2+(e-k)^2+(f-i)^2=r^2$$
$$(l-h)^2+(m-k)^2+(n-i)^2=r^2$$
Subtracted equation 2 from 1, equation 3 from equa... | For points $(a,b,c)$ and $(d,e,f)$, their perpendicular bisector can be found by:
$$\begin{align*}
(x-a)^2 + (y-b)^2+(z-c)^2 &= (x-d)^2 + (y-e)^2 + (z - f)^2\\
a^2-2ax+b^2-2by+c^2 -2cz &= d^2 - 2dx + e^2 - 2ey + f^2 - 2fz\\
2(a-d)x +2(b-e)y + 2(c-f)z &= a^2+b^2+c^2-d^2-e^2-f^2
\end{align*}$$
Do the same and find the pe... | {
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"url": "https://math.stackexchange.com/questions/1741496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve $\cos 2x - 3\sin x - 1 = 0$ using addition formula
Solve $\cos 2x - 3\sin x - 1 = 0, \quad 0^{\circ} \le x \le 360^{\circ}$
\begin{align} \cos 2x - 3\sin x - 1 = 0
&\iff 1 - 2\sin^2 x - 3\sin x - 1 = 0 \\
&\iff- 2\sin^2 x - 3\sin x = 0 \\
&\iff2\sin^2 x + 3\sin x =0\\
&\iff\sin x(2\sin x + 3) =0 \\
&\iff\s... | $\sin x =0 \Leftrightarrow x =\pi n (0^{\circ}, 180^{\circ}, 360^{\circ})$
$2\sin x=-3 \Rightarrow \sin x = -\frac 32 - $impossible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Rolling a die until two rolls sum to seven Here's the question:
You have a standard six-sided die and you roll it repeatedly, writing
down the numbers that come up, and you win when two of your rolled
numbers add up to $7$. (You will almost surely win.) Necessarily, one of the
winning summands is the number roll... | While not exactly stated, it seems that you win only if the sum of consecutive numbers sums to 7.
Obviously you can't win on the first roll. For every roll thereafter, you have a $1/6$ chance of winning on the next roll, (and a $5$ in $6$ chance of the game continuing).
Now, $P(N=2) = \frac{1}{6}$, $P(N=3) = \frac{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Roots of a Quartic (Vieta's Formulas)
Question: The quartic polynomial $x^4 −8x^3 + 19x^2 +kx+ 2$ has four distinct real roots denoted
$a, b, c,d$ in order from smallest to largest. If $a + d = b + c$ then
(a) Show that $a + d = b + c = 4$.
(b) Show that $abcd = 2$ and $ad + bc = 3$.
(c) Find $ad$ and $bc.$
(d) F... | Hint If you know that $ad = 2$ and $a + d = 4$, then $$(x - a)(x - d) = x^2 - (a + d) x + ad = x^2 - 4 x + 2 ,$$ so finding $a, d$ is just finding the roots of that quadratic. Of course, finding $b, c$ is analogous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Express $1/(x-1)$ in the form $ax^2+bx+c$
Let $x$ be a root of $f=t^3-t^2+t+2 \in \mathbb{Q}[t]$ and $K=\mathbb{Q}(x)$. Express $\frac{1}{x-1}$ in the form $ax^2+bx+c$, where $a,b,c\in \mathbb{Q}$.
I have proved that $f$ is the minimal polynomial of $x$ over $\mathbb{Q}$ but I am stuck showing the above claim. I tri... | Using the Extended Euclidean Algorithm as implemented in this answer, we get
$$
\begin{array}{r}
&&x^2&1&-(x+2)/3\\\hline
1&0&1&-1&(1-x)/3\\
0&1&-x^2&x^2+1&(x^3-x^2+x+2)/3\\
x^3-x^2+x+2&x-1&x+2&-3&0\\
\end{array}
$$
which means that
$$
\left(\vphantom{x^2}x-1\right)\left(x^2+1\right)+\left(x^3-x^2+x+2\right)\cdot\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
| Hint. You may use the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\cos x&=1-\frac{x^2}2+O(x^4)\\
\sin x&=x-\frac{x^3}6+O(x^4)
\end{align}
$$ giving
$$
\begin{align}
\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}&=\frac{(1+a-b) x+\frac16 (-3 a+b) x^3+O(x^5)}{(f(x))^3}
\\\\&=\frac{(1+a-b) x+\frac169 (-3 a+b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^... | On your way to the last line, you're tacitly assuming that, for example, $\frac1x\sqrt{x^2+x} = \sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}$. But that is only true when $x$ is positive!
When $x$ is negative, $\frac1x\sqrt{\cdots\vphantom{x}}$ will be negative, and thus it can never be written as a sum of square roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tridiagonal matrix inner product inequality I want to show that there is a $c>0$ such that
$$
\left<Lx,x\right>\ge c\|x\|^2,
$$
for alle $x\in \ell(\mathbb{Z})$ where
$$
L=
\begin{pmatrix}
\ddots & \ddots & & & \\
\ddots & 17 & -4 & 0 & \\
\ddots & -4 & 17 & -4 & \ddots \\
& 0 & -4 & 17 & \ddots \\
... | You can also finish your proof by noting that $k \le 2 \, \|x\|^2$ (by applying Hölder's inequality). Hence,
$$\langle L \, x , x \rangle \ge -4 \, k + 17 \, \|x\|^2 \ge 9 \, \|x\|^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What's the value of $x$ in the following equation?
So this is how I approached this question, the above equations could be simplified to :
$$a = \frac{4(b+c)}{b+c+4}\tag{1!}$$
$$b = \frac{10(a+c)}{a+c+10}\tag{2}$$
$$c=\frac{56(a+b)}{a+b+56}\tag{3}$$
From above, we can deduce that $4 > a$ since $\frac{(b+c)}{b+c+4} < 1... | Your deductions are wrong and that is what is misleading you. Integers can be both positive and negative.
If you solve equations (1), (2) and (3) simultaneously you can find a, b and c.
I did this to find $$a=3$$
$$b=5$$
$$c=7$$
You can then plug this into the forth equation given in the problem to solve for x.
$$x = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Picking two random points on a disk I try to solve the following:
Pick two arbitrary points $M$ and $N$ independently on a disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2 \leq 1\}$ that is unformily inside. Let $P$ be the distance between those points $P=d(M,N)$. What is the probabilty of $P$ being smaller than the radius of the ... | Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is
$$P(R\le r) = \begin{cases}
\frac{\pi r^2}{\pi\cdot 1^2} = r^2, & 0\le r\le 1\\
1, &r>1\\
0, &\text{otherwise}
\end{cases}$$
The probability density function of $R$ is
$$f_R(r) = \frac{d}{dr}P(R\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.