Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$ has all roots real Given the equation:
$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$
For what values of $m$ are all the roots real?
I've rewritten the equation as: $x^4-6x^3+11x^2-6x-m=0$
I'm quite sure this is done with Vieta's but didn't really figure out ... | Start from $x^4-6 x^3+11 x^2-6 x-m=0$ and substitute $x=z+\dfrac{3}{2}$
we get
$\left(z+\frac{3}{2}\right)^4-6 \left(z+\frac{3}{2}\right)^3+11 \left(z+\frac{3}{2}\right)^2-6 \left(z+\frac{3}{2}\right)-m=0$
Expand and reorder
$z^4-\frac{5 }{2}z^2+\frac{9}{16}-m=0$
substitute $z^2=w$
$w^2-\frac{5 }{2}w^2+\frac{9}{16}-m=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_0^1 \mathrm e^{-x^2} \,\mathrm dx$ using power series? I'm trying to evaluate
$$\int_0^1 \mathrm e^{-x^2} \, \mathrm dx$$
using power series.
I know I can substitute $x^2$ for $x$ in the power series for $\mathrm e^x$:
$$1-x^2+ \frac{x^4}{2}-\frac{x^6}{6}+ \cdots$$
and when I calculate the ant... | $$\left[ x-\frac{x^3}{3}+ \frac{x^5}{5*2}-\frac{x^7}{7*6}+ \dots \right]_0^1 = \left( 1-\frac{1}{3} + \frac{1}{5 \cdot 2} - \frac{1}{7 \cdot 6} + \dots \right) - 0 = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$$
So $\int_{0}^{1}e^{-x^2}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$, that is, the answer is whatever the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
| $$ \sqrt{3+\sqrt{13+4\sqrt3}}= \sqrt{3+\sqrt{1+2\times2\sqrt{3}+(2\sqrt{3})^2}}$$
$$=\sqrt{3+\sqrt{(1+2\sqrt{3})^2}}$$
$$=\sqrt{3+1+2\sqrt{3}}$$
$$=\sqrt{(1+\sqrt{3})^2}$$
$$=1+\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a cle... | Factor from $\dfrac{1}{\sqrt{x^2+1}}$. You will have:
$$\frac{1}{\sqrt{x^2+1}} \bigg(1 - \frac{x^2}{x^2+1}\bigg) = \frac{1}{\sqrt{x^2+1}} \frac{1}{x^2+1} = \frac{1}{(x^2+1)^\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find A and B in this limit Can you find $a$ and $b$? In how many ways can I find them?
$$\lim_{x\to0} \frac{a+\cos(bx)}{x^2}=-8$$
| For
$\lim_{x\to 0}\frac{\ a+\cos bx}{x^2}=-8
$,
note that,
for small $x$,
$\cos x
\approx 1-\frac{x^2}{2}
$.
Therefore
$\dfrac{\ a+\cos bx}{x^2}
\approx
\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
$.
If
$a+1 \ne 0$,
then
$\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
\to \infty$
as $x \to 0$.
Therefore,
to have the limit exist,
we must ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Better way to reduce $17^{136}\bmod 21$? What I have done:
Note $17\equiv -4$ mod 21, and $(-4)^2 \equiv 5$ mod 21. So $17^{136} \equiv (-4)^{136} \equiv 5^{68}$ mod 21. Also note $5^2 \equiv 4$ mod 21 and $4^3 \equiv 1$ mod 21, so $5^{68} \equiv 4^{34} \equiv (4^3)^{11}\cdot4 \equiv 4$ mod 21. I feel this is rather co... | You could also reduce modulo each of the factors of $21$ and then use the Chinese Remainder Theorem to recover the result modulo $21$.
$$
17^{136} \cong 2^{136} \cong (2^2)^{68} \cong 1^{68} \cong 1 \pmod{3}
$$
and
$$
17^{136} \cong 3^{136} \cong 3^{6 \cdot 22 + 4} \cong (3^6)^{22} \cdot 3^4 \cong 1^{22} \cdot 3^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate a limit using l'Hospital rule Evaluate $$\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}$$
I tried to apply l'Hospital rule in order to get the limit to be equal to
$$\lim_{x\to0}\frac{e^x-1}{2(e^x-\frac{x^2}{2}-x-1)^{-\frac{1}{3}}(e^x-x-1)}$$
but the new denominator has an indeterminate fo... | You can use $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)$
$$\qquad{\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-x-1}{3((1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-\frac{x^2}{2}-x-1)^{\frac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A question about asymptotic notations with sums. I need to prove that $$ \sum_{k=0}^{n-2017} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}} = \Theta(3^{n})$$
I'm pretty sure it's straightforward to prove that it's $\Omega(3^{n})$ but I'm not sure how to prove the $O(3^{n})$ part. Maybe with using square root and de... | *
*For the upper bound,, note that
$$\begin{align}
\sum_{k=0}^{n-2017} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}}
&\leq \sum_{k=0}^{n} \binom{n}{k} \binom{2k}{k} \frac{ \sqrt{k}}{2^{k}}
\leq \sum_{k=0}^{n} \binom{n}{k} \frac{2^{2k}}{\sqrt{k}} \frac{ \sqrt{k}}{2^{k}}
\\&= \sum_{k=0}^{n} \binom{n}{k} 2^k
= (1+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing the local maximum or minimum, while the function changes sign infinitely often Please I need a hand in solving this problem:
These 3 functions' values at $0$ are all $0$ and for $x\ne0$, $$f(x)=x^4\sin\frac{1}{x}, \, g(x)=x^4\left(2+\sin\frac{1}{x}\right), \, h(x)=x^4\left(-2+\sin\frac{1}{x}\right)$$
b- Show th... | HINT: we have for 1) $$f'(x)=4x^3\sin\left(\frac{1}{x}\right)+x^4\cos\left(\frac{1}{x}\right)\cdot \left(-\frac{1}{x^2}\right)$$
and for 2)$$g'(x)=4x^3\left(2+\sin\left(\frac{1}{x}\right)\right)+x^4\cdot\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$
and 3)$$h'(x)=4x^3\left(-2+\sin\left(\frac{1}{x}\right)\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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System of differential equations $dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$ This is the first time I have seen system of differential equations in this form: $$dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
Can you please help me solve it because I don't even know where to start?
| $$\frac{dx}{1}=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
This system looks like to be involved in solving a PDE with the method of characteristics. The PDE should be :
$$\frac{\partial z(x,y)}{\partial x}+(y+z(x,y))\frac{\partial z(x,y)}{\partial y}=x+y+z(x,y)$$
$\underline{\text{If this supposition is true}}$,
unfortunately t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, How many solutions are there if n is odd? $X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, $n \in N^*$
How many solutions are there if n is odd?
From the powers of $\begin{pmatrix}3&6\\ 2&4\end{pmatrix}$ I got that $X=\begin{pmatrix}\frac{3}{\sqrt[n]{7^{n-1}}}&\frac{6}{\sqrt[n... | $\det(X^n)=(\det X)^n=\det \begin{pmatrix}3&6\\ 2&4\end{pmatrix} = 0$, so $\det X=0$, and $X$ is singular.
$X$ cannot be the zero matrix, so it has two real eigenvalues: $0$ and $a\neq 0$. $X$ is diagonalizable: there is some non-singular $P$ such that $X=P\begin{pmatrix}0&0\\ 0&a\end{pmatrix}P^{-1}$, yielding $X^n=P\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proof verification: $x^3 + px - q = 0$ has three real roots iff $4p^3 < -27q^2$ I'm trying to prove the problem in the title. I'm not sure if my proof is clear, or even correct. It seems rather long, so I am not sure. Any advice would be great, especially on parts I might need to expand.
Let $f(x) = x^3 + px - q = 0$... | I think your statement is wrong.
Try, $p=-3$ and $q=-2$.
We have $$x^3+px-q=x^3-3x+2=(x-1)^2(x+2),$$
which says that the equation $$x^3+px-q=0$$
has three real roots, but $4p^3<-27q^2$ gives $-108<-108$, which is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n -... | Let $\epsilon>0$
$$\left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right|\leq\frac{4n+8-4n-1}{16n+4}=\frac{7}{16n+4} \leq \frac{7}{16n}$$
We have that $\frac{7}{16n} \to 0$
Thus exists $n_0 \in \mathbb{N}$ such that $\frac{7}{16n}< \epsilon, \forall n \geq n_0$
So $\frac{1}{n}<\frac{16\epsilon}{7} \Rightarrow n> \frac{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$? How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}... | To address your question of how to handle loops when integrating by parts, let $I=\int e^x\sin x \ dx$. Both functions are transcendental. We'll try using $u_1=e^x$ and $dv_1=\sin x \ dx$. These give $du_1=e^x \ dx$ and $v_1=-\cos x$. Thus, $$I=-e^x\cos x+\int e^x\cos x \ dx.$$
Now we have another integral. We'll try b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With ... | Hint:
You have done a good substitution so here you are
$$\int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta=\dfrac{1}{9}\int\dfrac{\cos^2\theta}{\sin^4\theta}\cos\theta\,d\theta=\dfrac{1}{9}\int\dfrac{1-\sin^2\theta}{\sin^4\theta}\cos\theta\,d\theta$$
now let $\sin\theta=u$ and c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$ I was looking for the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$
My work:
Dividing the $\frac{8t^3 +13}{(t+2)(4t^2+1)}$, I get $2 + \frac{-16t^2 -2t + 9}{4t^3 + 8t^2 + t + 2}$ or $2 + \frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$
Using the partial fraction dec... | You know that
$$
\int \frac{1}{t^2+1}dt=\arctan t + C.
$$
Thus, let $u=2t$, then
\begin{align}
\int \frac{6}{4t^2+1}dt &= \int \frac{6}{u^2+1}\frac{1}{2}du\\
&=\int \frac{3}{u^2+1}du\\
&=3\arctan u + C\\
&=3\arctan 2t + C.
\end{align}
Thus $\int \frac{6}{4t^2+1}dt = 6\arctan (2t) + c$, you found, is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Binomial Expansion - Simple application of the formula I found this one question in an old book:
What is the coefficient of $x^{n+1}$ in the expansion of $(x+2)^n \cdot x^3$?
Answer (according to my book): $(n^2-n) \cdot 2^{n-3}$
Here is my work: Since $\ T_{k+1} = \binom{n}{k}\cdot a^k\cdot b^{n-k}$, we can obtain a... | The binomial theorem yields
$$
(x+2)^n = \sum_{k=0}^n \binom nk x^k 2^{n-k},
$$
and so the coefficient of $x^{n+1}$ in $(x+2)^nx^3$ is
$$
\binom n{n-2}2^{n-(n-2)}) = \binom n2 2^2 = 2n(n-1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequalities involving geometry but I can't post a picture yet How do I show that
$$ \frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right) < \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots \quad ?$$
| After moving the odd terms from the LHS to the RHS, we obtain the following equivalent inequality,
$$\frac 12 \left(\frac 1{4^2}+ \frac 1{6^2}+ \frac 1{8^2}+\dots\right) < \left(1-\frac 12\right)\left( \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots\right).$$
Then note that for all positive integer $n$, each term $\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Some formula related with factor of (a+b+c+d) I am looking for some math formula
For example
\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}
First one related with factor a+b and the second one related with factor a+b+c
then
How about some formula relat... |
$$\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}$$
The two relations are not quite "alike" since the second one is symmetric in $\,a,b,c\,$ (i.e. stays invariant if you permute the variables), while the first one is not (both sides change sign). Maybe a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1 $ is divisible by $ \ \large 2^{n+2} $ .
Answer:
For $ n=1 \ $ we have
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1.
For $ n=2 $ we have
$ \large 3^{2^2}-1=81-1=80 \ \... | $$\begin{array}{} &3^{2^m} -1 &=x 2^{m+2} & \text{assumed and checked by } m=1 \text{with $x$ odd}\\
&3^{2^{m+1}} -1 &= 3^{2 \cdot 2^m} -1 & \text{general step in induction} \\
& &= (3^{2^m})^2 -1\\
& &= (3^{2^m} -1)(3^{2^m} +1) \\
& &=x 2^{m+2}(3^{2^m} +1) & \text{ with $x$ odd }\\
& &=x 2^{m+2}(3 \cdot 3^{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How to Solve Non-Homogeneous Recurrence Relations : $r_n = 2\left(r_{n-1} - \binom{n-1}{2}\right) + \binom{n-1}{2}$? $$r_n = 2\left(r_{n-1} - \binom{n-1}{2}\right) + \binom{n-1}{2}$$
which is equal to $$r_n - 2r_{n-1} = -\frac{n^2-3n+2}{2}$$
This given recurrence relation is derived from the question "How many regions ... | Using generating functions
$$f(x)=\sum\limits_{n=0}^{\infty}r_nx^n=r_0+r_1x+r_2x^2+\sum\limits_{n=3}^{\infty}r_nx^n=r_0+r_1x+r_2x^2+\sum\limits_{n=3}^{\infty}\left(2r_{n-1}-\binom{n-1}{2}\right)x^n=\\
r_0+r_1x+r_2x^2+2\left(\sum\limits_{n=3}^{\infty}r_{n-1}x^n\right)-\sum\limits_{n=3}^{\infty}\binom{n-1}{2}x^n=\\
r_0+r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Multi variable function and its corresponding range? How do we find the range of
$z=f(x,y)=1/\ln(4−x^2−y^2)$
1) Replace $t=\ln(4−x^2−y^2)$ and $t\in (−\infty,\ln4)$ and there is a DNE at ln('')=0
2) Find range of $1/t$ ; How is this done?
Answers: $(−\infty,0)∪(1/\ln4,\infty)$
| We know the domain of $f(x,y)=\dfrac{1}{\ln(4-x^2-y^2)}$ is
$$\color{blue}{D_f=\{(x,y)\in\mathbb{R}^2:x^2+y^2<4~,~x^2+y^2\neq3\}}$$
and in this domain $0\leq x^2+y^2<4$ so
$$0< 4-x^2-y^2\leq4~,~(x^2+y^2\neq3)$$
the function $\ln x$ is increasing then
$$-\infty< \ln(4-x^2-y^2)\leq\ln4~,~(x^2+y^2\neq3)$$
with reciprocati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$... | Multiply (1) by $y$ to get:
$$xy + 1 = 4y \implies y = \frac{1}{4-x}$$
Multiply (2) by $z$ to get:
$$yz + 1 = z \implies z = \frac{1}{1-y}$$
Multiply (3) by $x$ to get:
$$zx + 1 = \frac{7z}{3} \implies x = \frac{3}{7-3z}$$
Plug the equation for $z$ into the equation for $x$. This will give you $x$ in terms of $y$. Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using Lagrange multiplier to find the shortest distance from the origin to a given Set An exercise of an old exam wants me to find the point with the shortest distance to the origin which is in $M=\{(x,y): x^2y=2, x>0\}$.
So I think the function I have to minimize is $f(x,y)=\sqrt{(x^2+y^2)}$ with the condition $g(x,y)... | If I were you, I will use a simpler way. Since $y=\frac{2}{x^2}$, the distance is
$d(x)^2=x^2+\frac{4}{x^4}$. Applying the AM-GM inequality, we have
$$
d(x)^2=\frac{x^2}{2}+\frac{x^2}{2}+\frac{4}{x^4}\geq 3 \sqrt[3]{\frac{x^2}{2}\cdot \frac{x^2}{2}\cdot \frac{4}{x^4}}\geq 3.
$$
The identity holds when $x^6=8$ and he... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find mistake in solving $\sin 2x=\sin x + \cos x$ I am solving $\sin 2x = \sin x +\cos x $ for $0\le x \le 360$
$$\sin 2x = \sin x +\cos x$$
$$2 \sin x \cos x =\sin x + \cos x$$
$$(\cos x + \sin x) ^{2} - (\sin x)^{2} - (\cos x)^{2} =(\cos x + \sin x)^{2} - 1=\sin x + \cos x$$
Let $\cos x + \sin x = y$
$$y^{2} - 1= y$$... | Consider these three equations:
\begin{align}
\sin 2x = \sin x + \cos x \tag{1} \\
(\sin x + \cos x)^2 - 1 = \sin x + \cos x \tag{2} \\
(\sin 2x)^2 -1 = \sin 2x \tag{3}
\end{align}
You were correct in determining that (1) and (2) are equivalent. Also, (1) does imply (3).
However, what you failed to notice is that (3) d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=4x^2.$ I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welc... | The allowed values for $x$ are in $[-1,1]$, because of $\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\tan\arcsin x$.
(Note: your $x\in[-\pi/4,\pi/4]$ is wrong and the probable cause for the low grade.)
The equation remains the same if we change $x$ into $-x$, so we can lim... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the probability of getting the sum $26$ when $7$ chips are taken out? Suppose you have a bag in which there are $10$ chips numbered $0$ to $9$. You take out a chip at random, note its number and then put it back. This process is done $7$ times and after that the numbers are added.
What is the probability that... | The number of ways of drawing 7 chits with sum equal to 26 is the coefficient of $x^{26}$ in the expansion
\begin{align*}
(1+x+x^2+\cdots +x^9)^7 &= \left(\frac{1-x^{10}}{1-x}\right)^7 \\
&=(1-7x^{10} + 21x^{20} - \cdots)\left(1+\binom{7}{1}x + \binom{8}{2}x^2 + \cdots\right)
\end{align*}
Coefficient of $x^{26}$ is
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality on tangent and secant function
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$
| let $\tan\alpha =a, \tan\beta =b, \tan\gamma =c, \tan\delta =d\implies a,b,c,d>0$
Also $\alpha+\beta+\gamma+\delta=180^{\circ}\implies \alpha+\beta=180^{\circ}-(\gamma+\delta)$
$\implies \tan (\alpha+\beta)=-\tan (\gamma+\delta)$
$\implies \frac{a+b}{1-ab}=-\frac{c+d}{1-cd}$
$\implies a+b+c+d=abc+abd+acd+bcd$ ___(i)
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve differential equation $y''-2y'+3y=\sin x$ using invers operator method.
$$y''-2y'+3y=\sin x$$
$s^2-2s+3=0$
$s=1 \pm i\sqrt {2}$
uisng operator $(D^2-2D+3)y_1=\sin x$
\begin{aligned}
y_1 &= \frac{1}{D^2-2D+3}\sin x \\
&= \Im\left(\frac{1}{D^2-2D+3} e^{ix}\right) \\
&= \Im \left(\frac{1}{i^2-2i+3}e^{ix}\right) ... | The starred part consists in basic algebraic manipulations of complex numbers, with $i^2 = -1$ and $e^{ix} = \cos x + i \sin x$. Indeed,
\begin{aligned}
\frac{1}{i^2 -2i + 3} & = \frac{1}{2 -2i}\\
& = \frac{1}{2} \frac{1}{1 - i} \frac{1+i}{1+i} \\
& = \frac{1}{2} \frac{1 + i}{2} \, .
\end{aligned}
Now, multiplying by $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\int_{0}^\pi \frac{2\cos 2\theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$ Given that $$\int_{|z|=1|}\frac{z^2}{2z+1} dz = \frac{i\pi}{4}$$,
show $$\int_{0}^\pi \frac{2\cos 2 \theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$$.
I saw the bounds of the latter integral and thought that I should try a... | With $\theta\to-\theta$
$$
I=\int_0^{-\pi} \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,(-)d\theta=\int_{-\pi}^0 \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta
$$
\begin{align}
2I
&=\int_{-\pi}^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\, d\theta\\
&=\int_{|z|=1} \frac{{\bf Re\,}(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answe... | As $\alpha+\beta=\dfrac32, \alpha\beta=\dfrac62$
let $y=\dfrac\alpha\beta\iff y+1=\dfrac3{2\beta}\iff\beta=\dfrac3{2(y+1)}$
But as $\beta$ is a root of $$2x^2-3x+6=0$$
$$2\left(\dfrac3{2(y+1)}\right)^2-3\left(\dfrac3{2(y+1)}\right)+6=0$$
As $y+1\ne0,$ multiply both sides by $\dfrac{2(y+1)^2}3$ to find $$0=3-3(y+1)+4(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| Using the general form of quadritic equation, $x^2 -(a+b) x + ab$,
we get the values of $a+b$ and $ab$.
Now, the expression $a^3+b^3$ can be reduced to $(a+b)^3 -3ab(a+b)$.
Substitute the value of $a+b$ and $ab$ in the above equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find minimum and maximum value of x, if x+y+z=4 and $x^2 + y^2 + z^2 = 6$? I just know that putting y=z, we will get 2 values of x. One will be the minimum and one will be the maximum. What is the logic behind it?
| $z=4-x-y$
$x^2+y^2+(4-x-y)^2-6=0$
Differentiate wrt $x$ and $y$
$2x-2(4-x-y)=0$
$2y-2(4-x-y)=0$
Gives $x=y=4/3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Show that $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$ If $\Delta$ is the area of a triangle with side lengths a, b, c, then show that: $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$. Also show that equality occurs in the above inequality if and only if a = b = c.
I am not able to prove the inequality.
| One has $\Delta = \frac{1}{4}\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ (See this link).
Moreover, one has $(a+b-c)(b+c-a) \leq (\frac{a+b-c + b+c-a}{2})^2 = b^2$
So $[(a+b+c)(b+c-a)(c+a-b)(a+b-c)]^2 \leq a^2b^2c^2$.
Thus, $\Delta \leq \frac{1}{4}\sqrt{abc(a+b+c)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all roots of $2x^3+16$ in $\mathbb C$
Find all roots of $p(x) = 2 x^3 + 16$ in $\mathbb C$.
I found my answers to be x = 2, -1+i$\sqrt{3}$, -1+i$\sqrt{3}$.
But when I put the expression into Symbolab, it gives me -2, 1+i$\sqrt{3}$, 1+i$\sqrt{3}$ as roots of p(x) in C.
Can someone explain where I went wrong?
This... | Another way you can calculate this is to use De Moivre's theorem, which states that if $z = r(\cos \theta + i\sin \theta)$, the $n$th roots of $z$ are $$r^{1/n}\left(\cos\frac{\theta+2\pi k}{n} + i \sin \frac{\theta+2\pi k}{n}\right)$$
From $(1)$, which is $x^3=-8$, we find that $z = 8(\cos \pi + i \sin \pi)$, so $r=8$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Conditional probability of multivariate gaussian I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.
Let $\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda ) $ , where
$$\begin{align}
\mu &= \begin{pmatrix}
1 \\
1
... | The covariance between $X_1 + \lambda (3 X_1 + X_2)$ and $3 X_1 + X_2$ is $10 + 35 \lambda$, therefore if we take $\lambda = -2/7$, we get
$$\operatorname{P}(X_1 \geq 2 \mid 3 X_1 + X_2 = 3) =
\operatorname{P} \left(
X_1 -\frac 2 7 (3 X_1 + X_2 - 3) \geq 2 \mid 3 X_1 + X_2 = 3 \right) = \\
\operatorname{P} \left( X_1 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Solving $n\times n$ determinant using triangular shape I have just started learning to solve nth order determinants by getting it into the triangular shape ( in this way the determinant is equal to the multiple of main or additional diagonal + the determination of the sign ). I have solved a couple of easy ones, but go... | First, you can subtract the last line from the others which gives you :$$\begin{vmatrix}
2 & 0 &\ldots&0&-3 \\
0 & 3 &\ddots&\vdots& \vdots \\
\vdots &\Large{0}&\ddots&0& \vdots \\
0&\ldots&0&3&-3\\
3&\ldots&\ldots&3&6
\end{vmatrix}$$
Then subtract $\frac{3}{2} L_{1}$ from $L_{n}$:
$$\begin{vmatrix}
2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\... | Hint: We can express
$$\frac{1}{2^n-1}=\sum_{k=1}^{\infty} \frac{1}{2^{kn}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Can someone tell me what should i do next? Should I use the inequality between Arithmetic and Geometric mean? Problem 1: Let $a,b,c> 0$
$ab+3bc+2ca\leqslant18$
Prove that:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$.
I started on this way:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$
$\frac{3bc+2ac+ab}{... | By Holder:
$$18\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)^2\geq(ab+3bc+2ac)\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)\left(\frac{2}{b}+\frac{1}{c}+\frac{3}{a}\right)\geq$$
$$\geq\left(\sqrt[3]{ab\cdot\frac{3}{a}\cdot\frac{2}{b}}+\sqrt[3]{3bc\cdot\frac{2}{b}\cdot\frac{1}{c}}+\sqrt[3]{2ac\cdot\frac{1}{c}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the volume between $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral So the volume of $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral shall look similar to $$\int^2_0\int^{y=?}_{y=?}\int^{4-x^2-y^2}_{4-2x} dz dy dx$$ but how do I find the limits on $y$?
| The point is that you are integrating over a domain in which $4-2x<4-x^2-y^2$. In that domain,
$$
-2x < -x^2-y^2\\
2x>x^2+y^2\\
y^2 < 2x -x^2 \\
-\sqrt{2x-x^2} < y < +\sqrt{2x-x^2}
$$
So the integral is
$$
\int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (4- x^2 -y^2 -(4-2x) )dy\,dx = \int_{x=0}^2 \int_{y=-\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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The number of terms in the Multinomial Expansion $(x+\frac{1}{x}+x^2+\frac{1}{x^2})^n$ I am aware that there is a formula to calculate the number of terms in a multinomial expression $(x_1+x_2+x_3+...x_r)^n$, i.e. $^{n+r-1}C_{r-1}$. However, this is in the case when the terms $x_1, x_2, x_3 ... x_r$ are different varia... |
We obtain
\begin{align*}
\color{blue}{\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)}&=\frac{1}{x^{2n}}(1+x+x^2+x^3)^n\\
&=\frac{1}{x^{2n}}\left(1+x+x^3\left(1+x\right)\right)^n\\
&=\frac{1}{x^{2n}}(1+x)^n(1+x^3)^n\\
&\color{blue}{=\frac{1}{x^{2n}}\sum_{j=0}^n\binom{n}{j}x^j\underbrace{\sum_{k=0}^n\binom{n}{k}x^{3k}}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculus of variation: Lagrange's equation A particle of unit mass moves in the direction of $x$-axis such that it has the Lagrangian $L= \frac{1}{12}\dot x^4 + \frac{1}{2}x \dot x^2-x^2.$ Let $Q=\dot x^2 \ddot x$ represent a force (not arising from a potential) acting on the particle in the $x$-direction. If $x(0)=1$ ... | well, continue
\begin{align}
\frac{d}{dt}(x \dot x ) &=\frac{1}{2} \dot x^2-2x \\
\implies \dot{x}^2+x \ddot{x} &= \frac{1}{2}\dot{x}^2-2x \\
\implies 0 &= x \ddot{x}+\frac{1}{2}\dot{x}^2+2x \\
&\vdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove a 3 parameter integral identity I have stumbled upon the following identity:
$$\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt$$
It appears to work for all $b \in \mathbb{R}$, $c \geq 0$ and $|z|<1$.
The identity is related to ... | We want to show
\begin{eqnarray*}
\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt.
\end{eqnarray*}
Make the substitution $t=\frac{1}{u}$ into the third integral ($dt=-\frac{du}{u^2}$)
\begin{eqnarray*}
\int_0^1 \frac{t^c(1+t)^{b-c-2}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:
For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$
I attempted to rewrite and simplify the left hand side ... | $$\scriptsize\begin{align}
&\;\;\;\frac 1{2m-1}+\frac {2m-2}{(2m-1)(2m-3)}+\frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)}+\cdots+\frac {(2m-2)!!}{(2m-1)!!}\\\\
&=\frac 1{2m-1}\left(1+\frac {2m-2}{2m-3}\left(1+\frac {2m-4}{2m-5}\left(1+\;\;\cdots\;\;\ \left(1+\frac 65\left(1+\frac 43\left(1+\frac 21\right)\right)\right)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Closed form of the elliptic integral $\int_0^{2\pi} \sqrt{1+\cos^2(x)}\,dx $ I want to prove the closed form shown in Wikipedia for the arc length of one period of the sine function.
Source of wikipedia
$$\int_0^{2\pi} \sqrt{1+\cos^2(x)} \ dx= \frac{3\sqrt{2}\,\pi^{\frac32}}{\Gamma\left(\frac{1}{4}\right)^2}+\frac{\Gam... | $$\int_0^{2\pi} \sqrt{1+\cos^ 2 x} dx = 4 \int_0^1 \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} dx = 4\int_0^1 \frac{1+x^2}{\sqrt{1-x^4}} dx $$
Now use the beta function.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$
I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm m... | We know that, $\sin^2\theta+\cos^2\theta = 1$ and
$a^2-b^2=(a-b)(a+b$
then
\begin{split} \frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} &= &\frac{\color{red}{\sin^2\theta+\cos^2\theta} +\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta} \qquad\quad\\\\&=& \frac{\color{red}{\sin^2\theta+(-i\cos\theta)(i\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Find two arithmetic progressions of three square numbers I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:
\begin{align}
\ & a^2 +r = b^2 \\
& b^2 +r = c^2 \\
& a^2 +c^2 = 2\,b^2 \\
\end{align}
and
\begin{align}
\ & d^2 +r = e^2 \\
& e... | There are infinitely many solutions to the system,
\begin{align}
\ & a^2 +r_1 = b^2 \\
& b^2 +r_1 = c^2 \\
& a^2 +c^2 = 2b^2 \\
\hline
\ & d^2 +r_2 = e^2 \\
& e^2 +r_2 = f^2 \\
& d^2 +f^2 = 2e^2 \\
\end{align}
with $\color{blue}{r_1=r_2}$. Eliminating $r_1$ between the first two equations (and similarly for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Differential equations for chemical reaction $\mathrm{A + 2B \to 3C}$
In a chemical reaction $\mathrm{A + 2B \to 3C}$, the concentrations $a(t)$, $b(t)$ and $c(t)$ of the three substances A, B and C measure up to the differential equations
$$
\begin{align}
\frac{da}{dt} &= -rab^2\tag{1}\\
\frac{db}{dt} &= -2rab^2\ta... | Notice that, by your equations,
$\dfrac{d(2a - b)}{dt} = 2\dfrac{da}{dt} - \dfrac{db}{dt} = -2rab^2 -(-2rab^2) = 0; \tag 1$
hence $2a(t) - b(t)$ is constant. Now
$2a(0) - b(0) = 2(1) - 2 = 0; \tag 2$
the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain ... | By your work
$$f(x)=x^4-2(a^2+b^2)x^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-(a^2+b^2)^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-4a^2b^2.$$
Now, we see that
$$f(x)\geq-4a^2b^2$$ and the equality occurs for $x^2=a^2+b^2$, which is possible.
Thus, $$\min_{\mathbb R}f=-4a^2b^2.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to integrate this kind of functions? I was wondering how to evaluate $$\int\frac{sin^4 x}{cos^7 x}dx$$
I tried the usual method of writing the expression in terms of powers of $tan(x)$ and $sec(x)$, but nothing useful came out of it.
My attempt
$$\int\frac{sin^4 x}{cos^7 x}dx$$$$=\int({tan^4x}) ({sec^3x})dx$$$$=\in... | \begin{align*}
\int{\frac{sin^{4}x}{cos^{7}x}}\,dx &= \int{tan^{4}x \cdot sec^{3}x}\,dx\\
&= \int{(sec^{2}x - 1)^2 \cdot sec^3{x}}\,dx\\
&= \int{sec^{7}x}\,dx -2\int{sec^{5}x}\,dx + \int{sec^{3}x}\,dx
\end{align*}
Now,
\begin{align*}
I_{2n+1} &= \int{sec^{2n+1}x}\,dx = \int{sec^{2n-1}x\cdot \sec^{2}x}\,dx\\
&= sec^{2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.
First, I tried to simplify the proof statement but I got an even mo... | AM-GM helps!
$$\sum_{cyc}\frac{ab}{c^3}=\frac{1}{4}\sum_{cyc}\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)\geq\frac{1}{4}\sum_{cyc}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}\right)=\sum_{cyc}\frac{1}{c}.$$
Done!
Without $cyc$ we can write the solution so:
$$\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proving equivalence of norms in $\mathbb{R}^2$
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) ... | Here's a version that's more explicitly geometric, but whose underlying mathematics resemble Roberto's matrix diagonalization.
Rewrite the norm in rotated coordinates $(x', y')$, where $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$. We'll choose $\theta$ at our convenience—specifically,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How should one proceed in this trigonometric simplification involving non integer angles? The problem is as follows:
Find the value of this function
$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$
when $\omega=33^{\circ}{20}'$ and $\phi=... | Hint: You did a miscalculation:
$$A=\left (\cos\frac{\omega}{2}+\cos\frac{\phi}{2} \right )^{2}+\left (\sin\frac{\omega}{2}-\sin\frac{\phi}{2} \right )^{2}$$
$$=\cos ^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin ^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sq... | Use change of variable
$$\theta=\arctan\sqrt{\frac{1+x}{1-x}}\in[0,\frac{\pi}{2}).$$
Then we have
$$x=\frac{\tan^2\theta-1}{\tan^2\theta+1}=\sin^2\theta-\cos^2\theta=-\cos2\theta.$$
Therefore
\begin{align}
\int\arctan\sqrt{\frac{1+x}{1-x}}dx&=-\int\theta\,d\cos 2\theta=-\theta\cos 2\theta+\int\cos 2\theta d\theta\\
&=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Strange symmetry regarding sum $\sum_{n=0}^\infty\frac{n^ne^{-bn}}{\Gamma(n+1)}$ and integral $\int_{0}^\infty\frac{x^xe^{-bx}}{\Gamma(x+1)}dx$ One can show by computation the following for $b>1$
$$\sum_{n=0}^\infty\frac{n^ne^{-b n}}{\Gamma(n+1)}=\frac{1}{1+W_{\color{blue}{0}}(-e^{-b})},\tag{1}$$
(here one assumes that... |
Question 2. Is it possible to alter (1) and (2) to obtain a function
for which sum equals integral?
A simpler form for $z\in[0,\mathrm{e}^{-1})$:
\begin{align}
\sum_{n=0}^\infty
\frac{(z\,n)^n}{\Gamma(n+1)}
&=
\frac1{1+\operatorname{W}_{0}(-z)}
\tag{1}\label{1}
,\\
\int_0^\infty
\frac{(z\,x)^x}{\Gamma(x+1)}\,dx
&=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Integrate $f(x)=\sqrt{x^2+2x+3}.$ Completing the square and letting $t=x+1$, I obtain $$\int\sqrt{(x+1)^2+2} \ dx=\int\sqrt{t^2+2}\ dt.$$
Letting $u=t+\sqrt{t^2+2},$ I get
\begin{array}{lcl}
u-t & = & \sqrt{t^2+2} \\
u^2-2ut+t^2 & = & t^2+2 \\
t & = & \frac{u^2-2}{2u} \\
dt &=& \frac{u^2+2}{2u^2}du
\end{array}
Thus ... | Here is how I would work it.
$t = \sqrt 2 \tan \theta\\
dt = \sqrt 2 \sec^2 \theta$
$\int 2\sec^3 \theta\\
\sec\theta\tan\theta + \ln [\sec\theta+\tan\theta]+C\\
\frac {1}{2} t\sqrt {t^2 + 2} + \ln \frac 12 (t+\sqrt{t^2 + 2}+ C\\
\frac {1}{2} (x+1)\sqrt {x^2 + 2x + 3} + \ln [x+1 + \sqrt {x^2 + 2x + 1}]+ C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$
Solve the equation $\dfrac{x^3+y^3}{x^3+z^3} = \dfrac{1006}{1001}$ for $x,y,z \in \mathbb{Z}$.
We must have $x^3+y^3 = 1006d$ and $x^3+z^3 = 1001d$ where $d$ is an integer. This means that $x^3+y^3 \equiv 0 \pmod{1006}$ and $x^3+z^3 \equiv 0 \pmod{100... | Well, at least we have some solutions, like
$$\frac{669^3 + 337^3}{669^3 + 332^3} = \frac{1006}{1001}$$
This should be studied with elliptic curves.
May assume $x$, $y$, $z$, rationals, and then may even assume $x=1$. We get
$$\frac{y^3+1}{z^3+1} = \frac{1006}{1001}\\
z^3 +1 = \frac{1001}{1006}(y^3 + 1)$$
This is an e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$ Question:
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$
My attempts:
*
*Here's the expanded form:$$x^4-6x^3+12x^2-12x+4=0$$
*I've plugged this into several online "math problem solving" websites, all claim that "solution could not be determined algebraically, he... | $$(x^2+2)^2+8x^2=6x(x^2+2)$$
$$(x^2+2)^2-6x(x^2+2)+8x^2=0$$
Let $x^2+2=U, x=V$. Then
$$U^2-6UV+8V^2=0$$
Then
$$\left(\frac UV\right)^2-6\left(\frac UV\right)+8=0$$
Then
$\frac UV=2$ or $\frac UV=4$
$\frac {x^2+2}{x}=2$ or $\frac {x^2+2}{x}=4$
$x^2-2x+2=0$ or $x^2-4x+2=0$
$$x=2\pm \sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given equations for the side-lines of a parallelogram, why are these the equations for the diagonal-lines? My book, for a parallelogram $ABCD$ with sides as
$$\begin{align}
AB&\;\equiv\; a\phantom{^\prime}x+b\phantom{^\prime}y +c\phantom{^\prime}=0 \\
BC&\;\equiv\; a^\prime x +b^\prime y +c^\prime=0... | As Michael Rozenberg has written, they should be
$$\small BD\equiv (ax+by+c)(a'x+b'y+c)-(a'x+b'y+c')(ax+by+c')=0\tag1$$$$\small AC\equiv (ax+by+c)(a'x+b'y+c')-(a'x+b'y+c)(ax+by+c')=0\tag2$$
why the equations of diagonals (taking AC for instance) is for the entire line AC and not just points A and C?
$(1)$ can be writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Odd number proof
Prove that for every odd number $n$, it holds that $n^2+3$ is not not divisible
by $8$.
My idea:
Let $n=2k+1$ for $k \in \mathbb{N}$, which implies
$$n^2+3=(2k+1)^2+3=4k^2+4k+1+3=4k^2+4k+4$$
How can I conclude that I cannot divide $4k^2+4k+4$ by $8$?
| Just for fun, a slightly different approach:
If $n^2+3$ is divisible by $8$, then so is $(n-4)^2+3=n^2+3-8n+16$, hence, by induction, $8$ divides $n^2+3$ for at least one $n$ between $-2$ and $2$. But $0^2+3=3$, $(\pm1)^2+3=4$, and $(\pm2)^2+3=7$ are not divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
... | Note that $1$ and $\sqrt{2}$ is a basis for $\mathbb{Q}(\sqrt{2}).$
Hence from
$$3=a^2+b^2+2\sqrt{2}ab$$
We have $a^2+b^2=3$ and $2ab=0$.
From there, you should be able to see a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i could... | $S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +... $
Take 1/4 common from the terms whose denominator is even.
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} ( 1 + \frac{1}{4} + \frac{1}{9} + ... ) $
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} S$
$ 1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine constant c so that g(x,y) is continuous at every point $$g(x,y)=\begin{cases} \frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2} & \text{if} & (x,y) \neq (0,0) \\
c & \text{if} & (x,y) = (0,0) \end{cases}$$
Should I set the first function equal to c and then solve using polar coordinates?
| Hint. By using polar coordinates one gets, for $(x,y)\ne (0,0)$,
$$
g(x,y)=\frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2}=\frac{r^3\cos^3 \theta+r^3\cos\theta \sin^2\theta +2r^2}{r^2},\quad r\ne0,
$$ that is
$$
g(x,y)=r\cos^3 \theta+r\cos\theta \sin^2\theta +2,\quad r\ne0,
$$ then this tends to $2$ as $r \to 0^+$.
Can you take it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a g... | Hint. Note that by letting $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, we have that as $(x,y)\to(0,0)$ then $\rho\to 0$ and
$$0\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right|\leq
\frac{(|\rho\cos\theta|^5+|\rho\sin\theta|^5)|\ln(\rho^2)|}{\rho^4}
\leq \frac{\rho^5(1+1)|\ln(\rho^2)|}{\rho^4}={4\rho|\ln(\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Inequality proof as a part of calculus lesson As part of a calculus lesson I was required to prove that:
(1) if $\ |x-3| < \frac{1}{2},\ $ then $\ \bigg|\displaystyle{\frac{\sin(x^2 -8x+15)}{4x-7}}\bigg| < \frac{1}{2}$
So, by using $|\sin(t)| \le |t|,$ I can prove that:
$$\bigg|\frac{\sin(x^2 -8x+15)}{4x-7}\bigg| \le \... | Because, $x^{2}-8x+15=(x-3)(x-5)\ $, you can start saying that:
First,
$$ \vert{\sin(x^{2}-8x+15)}\vert\leq {x^{2}-8x+15} $$
and then you start like this
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert=\frac{\vert{\sin(x^{2}-8x+15)}\vert}{\vert{4x-7}\vert}\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
$$\overline{abc}=100a+1... | We show a more general fact:
There is no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=d^2$ where $d$ is a non-negative integer.
We first note that $d\leq b\leq 9$. Since $b^2-4ac=d^2$, we have that
$$P(x):=ax^2+bx+c=a\left(x-\frac{-b+d}{2a}\right)\left(x-\frac{-b-d}{2a}\right)$$
Assume that $P(10)=\overline{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the limit of the complex function. $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)}.$$
I have simplified this limit to this extent :
$$e^{ \lim_{x\to a} \left(\left(1- \frac{x}{a}\right){\left(\tan \frac{\pi x}{2a}\right)}\right)}$$
I don't know how to simplify the limit after that... | \begin{align*}
\log\left(2-\frac{x}{a}\right)^{\tan(\pi x/2a)}&=\left(\tan\frac{\pi x}{2a}\right)\left(\log\left(2-\frac{x}{a}\right)\right)\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\log\left(1+\left(1-\dfrac{x}{a}\right)\right)}{\cos\left(\dfrac{\pi}{2}\left(\dfrac{x}{a}-1\right)+\dfrac{\pi}{2}\right)}\\
&=\sin\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Frobenius method series solution Q: $x^2y^{''}-(x^2+2)y=0$ [1]
Solving using frobenius
$y=\sum_{n=0}^{\infty}a_nx^{x+r}$ [2]
$ y'=\sum_{n=0}^{\infty}(n+r)a_nx^{x+r-1}$ [3]
$ y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}$ [4]
inserting [2,4] into [1]
$x^2 \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}-(x^2+2)\sum_{n=... | Your approach is fine. There is just a small mistake which causes the problem. In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
Let's consider again the series equation
\begin{align*}
\sum_{k=0}^\infty[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k=0\ta... | {
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Three couples sit at random in a line of six seats, probability that no couple sits together? If three married couples (so 6 people) sit in a row of six seats at random, what is the probability that no couples sit together?
Another way to think about it (couples are AB, CD, and EF)
| There are $6!$ possible seating arrangements. From these, we must exclude those in which one or more couples sit in adjacent seats.
There are three ways to select a couple who sit in adjacent seats. That gives us five objects to arrange, the couple and the other four people. The objects can be arranged in $5!$ ways.... | {
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l'Hôpital vs Other Methods Consider the first example using repeated l'Hôpital:
$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} =... | After doing derivative one more time you get $12x^2 +2 $ which is not $0$ when $x$ goes to $0$.
| {
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Prove 11 does not divide $3^{3k-1}+5*3^k$ for any odd k. First I did an induction proof that it does work for even k. Then I started the proof as so.
Suppose there exists a k of the form 2n+1, s.t 11 divides $3^{3k-1}+5*3^k$.
After some algebra I can arrive at this point $5*2^{6n-1}+3(2^{6n-1}+5*3^{2n})$
Since I proved... | $3^{3k-1}+5\times 3^k=3^{k-1}(3^{2k}+15)=3^{k-1}(x^2+15)$ with $x=3^k$
Also $x^2+15\equiv x^2+4\pmod{11}$
$\begin{array}{l}
k=0: & 3^0\equiv 1\pmod{11} & x^2+4\equiv 5\pmod{11}\\
k=1: & 3^1\equiv 3\pmod{11} & x^2+4\equiv 13\equiv 2\pmod{11}\\
k=2: & 3^2\equiv 9\pmod{11} & x^2+4\equiv 85\equiv 8\pmod{11}\\
k=3: & 3^3\eq... | {
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"question_score": "2",
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How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$
I try to use this well known: if $a>b>0,c>0$,then we have
$$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
$$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
so we have
$$\sum_{k=2}^{n}\dfrac{1}{3^... | Since $3^k > 6$ for $k \geq 2$, $$\frac{1}{3^k-1}<\frac{1}{3^k-\frac{3^k}{6}}=\frac{6}{5}\cdot\frac{1}{3^k}$$ so $$\sum_{k=2}^{n}\dfrac{1}{3^k-1}<\frac{6}{5}\sum_{k=2}^{n}\dfrac{1}{3^k}<\frac{6}{5}\sum_{k=2}^\infty\dfrac{1}{3^k}=\dfrac{1}{5}$$ as required.
| {
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show that $\sum_{k=1}^{n}(1-a_{k})<\frac{2}{3}$ Let $a_{1}=\dfrac{1}{2}$, and such $a_{n+1}=a_{n}-a_{n}\ln{a_{n}}$,show that
$$\sum_{k=1}^{n}(1-a_{k})<\dfrac{2}{3}$$
My attemp: let $1-a_{n}=b_{n}$,then we have
$$b_{n+1}=b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}<\cdots<(b_{1})^{2^{n}}=\dfrac{1}{2^{2^n}}$$
where use $\ln{(1+... | We can prove that $0< a_{n}< 1$ inductively by making use of the graph$:\quad y= x\left ( 1- \ln x \right ).$ We let
$$b_{n}:= 1- a_{n}, \left \{ b_{n} \right \}_{n= 1}^{\infty}\Leftrightarrow b_{1}= \frac{1}{2}, b_{n+ 1}= b_{n}+ \left ( 1- b_{n} \right )\ln\left ( 1- b_{n} \right )$$
Well$,\quad a_{n+ 1}- a_{n}= -a_{n... | {
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"source": "stackexchange",
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Prove that: $1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$ Prove that:
$$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$
I know only this method:
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4... | My method :
$$\left\{ 1-\frac {1}{2}-\frac {1}{4}-...-\frac {1}{128} \right\}+\left\{ \frac {1}{3}-\frac {1}{6}- \frac{1}{12}-...- \frac{1}{192}\right\}+\left\{\frac {1}{5}-\frac{1}{10}-\frac{1}{20}-...- \frac{1}{160}\right\}+...+\left\{ \frac{1}{99}-\frac{1}{198}\right\}+\left\{ \frac{1}{101}+\frac{1}{103}+\frac{1}{1... | {
"language": "en",
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Help solving variable separable ODE: $y' = \frac{1}{2} a y^2 + b y - 1$ with $y(0)=0$ I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve:
$$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$
The solution is given as
$$y(x) =... | You complete the square $\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}$ and use this to inspire the change of coordinates $u=ay+b$ leading to
$$
\int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2}
$$
and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you p... | {
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Divergence of reciprocal of primes, Euler On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \righ... | Hint :
$$\sum_p \frac {1}{p^s} < \sum_{n=1}^{\infty} \frac 1{n^s} = \zeta(s) $$
$\zeta(s)$ is Riemann Zeta function, and $\zeta(s)$ converges for each $s \in \Bbb R, s>1$
| {
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Trying to solve differential equation $y'=\frac{3x-y+4}{x+y}$ ÊDIT: Found the second mistake (failed by calculating $u_{1,2}$)!
I might have made a mistake, however, I am not able to detect it. Here we go:
$$y'=\frac{3x-y+4}{x+y}, \quad y(1)=1$$
1.) set $x = X+a$ and $y=Y+b$, so $$\frac{dY}{dX} = \frac{3X-Y+(3a-b+4)}... | Your mistake started from 6 (I'm writing $c=C^2$ here)
$$ u^2 + 2u = 3 - \frac{1}{cX^2} $$
$$ (u+1)^2 = 4 - \frac{1}{cX^2} $$
$$ u= -1 \pm \sqrt{4-\frac{1}{cX^2}} $$
or
$$ \frac{y-1}{x+1} =-1 \pm \sqrt{4-\frac{1}{c(x+1)^2}} $$
Using the condition $x = 1, y = 1$ we get
$$ -1 \pm \sqrt{4-\frac{1}{4c}} = 0 $$
Only the pl... | {
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Determine galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ I've had some hard time determining Galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ because I didn't know exactly how to compute the order of the elements. See here for the computation of the o... | Wouldn't it be simpler to remark that by definition, $\omega$ is a primitive cubic root of $1$, so that your field $K$ is just the splitting field of the polynomial $X^3 - 3$ ? As such, $K/\mathbf Q$ is normal, with Galois group $G$ isomorphic to the permutation group of the roots, so $G\cong S_3 \cong D_6$, generated ... | {
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"source": "stackexchange",
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How to show that $\frac{\pi}{3}\le \iint_D \left(x^2+(y-2)^2\right)^{-1/2}\,dx\,dy\le \pi$ where $D$ is the unit disc.
How to show that $$\frac{\pi}{3}\le \iint_D \frac{dx\ dy}{\sqrt{x^2+(y-2)^2}}\le \pi$$ where $D$ is the unit disc centered at the origin?
I was trying to integrate it using polar coordinates but got ... | We have $D=\{(x,y)\ |\ x^2+y^2\leq 1\}$. Clearly $x^2+(y-2)^2 \geq 1$. So we get:
\begin{align}
\iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \iint_D dx dy = \pi
\end{align}
For the other inequality we note that the function $f(x,y)=x^2 + (y-2)^2$ can only have its maximum on the boundary of $D$ (why?). Let $\phi(... | {
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"source": "stackexchange",
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Completing squares with three variables. I want to complete the squares for this polynomial
$2x^2+2y^2-z^2+2xy+3xz-4yz$
Is there any kind of easy and non-confusing way to solve it?
I’ve done this up until now:
$$2x^2+2y^2-z^2+2xy+3xz-4yz$$
$$2x^2+2xy+3xz-4yz+2y^2-z^2$$
$$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$
$$2(x^2+... | We can do it simply using $(a \pm b )^2 = a^2 \pm 2ab + b^2$
$$2x^2 + 2y^2 -z^2 + 2xy +3xz - 4yz $$
$$ x^2 + y^2 + 2xy + x^2 +2(x)(\frac{3}{2}z)+(\frac{3}{2}z)^2 -(\frac{3}{2}z)^2+y^2 -2(y)(2z)+(2z)^2 -(2z)^2-z^2$$
$$(x+y)^2 +(x+\frac{3}{2}z)^2 +(y-2z)^2 - (\frac{\sqrt{29}}{2}z)^2 $$
| {
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Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
| $$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$
$$ =\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4} $$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}\right) .\frac{2}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{n+4-n}{n+4}\right).\frac{1}{2}$$
$$=\l... | {
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Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \ri... | Note that $A^4=0$. Thus all eigenvalues of $A$ must be $0$ thus its Jordan normal form has one of the following forms
$$
A_1=\left( \begin {array}{ccc} 0&0&0\\ 0&0&0 \\ 0&0&0\end {array} \right) \text{ or }
A_2=\left( \begin {array}{ccc} 0&1&0\\ 0&0&0 \\ 0&0&0\end {array} \right)
\text{ or }
A_3=\left( \begin {array}... | {
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$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.
So the pdf of $(X,Y)$ is $f(x,y)=\f... | $X$ and $Y$ are standard Cauchy hence $\frac{X+Y}{2}$ is standard Cauchy (what easily can be seen using characteristic functions)
So the pdf of $\frac{3}{2}Z$ is also $$f(t) = \frac{1}{\pi}\frac{1}{1+t^2}$$
Hence the distribution of $Z$ is: $$f(z) = \frac{d}{dz}P(Z \le z) = \frac{d}{dz}F\left(\frac{3}{2}Z \le \frac{3}{... | {
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Evaluate the limit $\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$ Calculate the following limit :
$$\lim_{x\rightarrow 0}\frac{1-\sqrt{1+x^2}\cos x}{\tan^4(x)}$$
This is what I have tried:
Using Maclaurin series for $ (1+x)^a $:
$$(1+x^2)^{1/2}=1+\frac{1}{2!}x^2\quad \text{(We'll stop at order 2)}$$
Usi... | Without using L'Hospital & Taylor's Expansion,
$$=\dfrac{1-(1+x^2)(1-\sin^2x)}{1+\cos x\sqrt{1+x^2}}\cdot\dfrac{\cos^4x}{\sin^4x}$$
$$=\dfrac{x^2\sin^2x+(\sin x-x)(\sin x+x)}{x^4}\cdot\dfrac{\cos^4x}{\left(\dfrac{\sin x}x\right)^4(1+\cos x\sqrt{1+x^2})}$$
Now $\dfrac{(\sin x-x)(\sin x+x)}{x^4}=\left(\dfrac{\sin x}x+1\r... | {
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Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working
tell me if I did correct or not thanks
My working:
EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$:
$$2... | It's hard to read your handwriting but it looks like you have the right idea but were sloppy in your execution and made so distributive error.
Assume if $n=k$ then statement is true and
$2^{3k} - 3^k = 5P$ for some integer $P$. (Always a good idea to specify what a variable is whenever you introduce it.)
$2^{3k+1} -3... | {
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Proving that all the real roots of Hermite polynomials are in $(-\sqrt{4n+1}, \sqrt{4n+1})$ The Hermite polynomials are given by:
$H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$
There is the proof that all the roots are real: https://math.stackexchange.com/a/104875/504137.
And I know the fact that they all are bounde... | It doesn't look like there are many questions on this topic on Math.SE, so just for fun let's use some tricks from matrix analysis to slightly improve the bound $\sqrt{2n-2}$ in Jack's answer.
Let $A = [a_{ij}]$ be a real $n \times n$ matrix. Define $|A| = [|a_{ij}|]$. We will write $A \geq 0$ if all $a_{ij} \geq 0$. ... | {
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How to figure out if there is an actual horizontal tangent without a graph There is this practice problem that asks to determine the points at which the graph of $y^4=y^2-x^2$ has a horizontal tangent.
So I did implicit differentiation to find that
$$\displaystyle\frac{dy}{dx} = \frac{-x}{2y^3-y}$$
To find the horizon... | If
$y^4=y^2-x^2
$,
then
diffing implicitly,
$4y^3y'
=2yy'-2x
$
or
$2y^3y'
=yy'-x
$.
If $y' = 0$,
then
$x = 0$.
Putting this in,
$y^4 = y^2$
so the possible values are
$y = 0, \pm 1$.
At $x=y=0$,
suppose $y' = c$.
For small $x$ and $y$,
$y \approx cx$
so
$c^4x^4 \approx c^2x^2-x^2$
or,
dividing by $x^2$,
$c^4 x^2 \appro... | {
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Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$ The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I n... |
And I'm not sure if this is right. How does one check whether their summation is correct?
Replace "2015" with "10", do it by hand, and see if your results match with your general formula.
| {
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Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad... | Hint:
Let $u=\sqrt{x-y},v=\sqrt{x+y}$. The system now reads
$$u+v=a,\\\sqrt{\frac{u^4+v^4}2}-uv=a^2$$
Raising the first equation to the fourth power,
$$a^4=u^4+v^4+4uv(u^2+v^2)+6u^2v^2.$$
Then using $u^4+v^4=2(a^2+uv)^2$ and $u^2+v^2=a^2-2uv$, you get an equation in $uv$, which simplifies:
$$a^4=2(a^2+uv)^2+4uv(a^2-2uv... | {
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Smallest possible value of expression involving greatest integer function If $a, b, c \gt 0$ then what is the smallest possible value of
$$\left[\frac{a+b}{c}\right]+ \left[\frac{b+c}{a}\right] + \left[\frac{c+a}{b}\right]$$
where $[.]$ denotes greatest integer function.
I tried using the AM GM inequality at first b... | Wolog $a \le b \le c$
$[\frac {b+c}a] \ge [\frac {a+a}a] =2$
And $[\frac {a+c}b] \ge [\frac {a + b}b] \ge [\frac bb] = 1$.
So you can not get less than $3$
If $\frac {a+b}{c} < 1$ then $c > a+b$ and $\frac {c+a}b > \frac {2a + b}b= \frac {2a}b + 1$. So If $\frac {c+a}b < 2$ then $\frac {2a}b < 1$ so $b > 2a$. So $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.
2.) Then $\cos(x) = ... | If $\tan(x)=3$, then $\tan^2(x)=9$. This means that $\frac{\sin^2x}{1-\sin^2x}=9$. So, $\sin^2(x)=\frac9{10}$; in other words (at least if we're on the first quadrant), $\sin(x)=\frac3{\sqrt{10}}$.
Your error lies in item 4: $\sin^2(x)+\left(\frac1{3\sin(x)}\right)^2=\frac{10}9\sin^2(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$ I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$.
So I pose the following problem:
Given $a \in \mathbb{R}-\{0\}$, fin... | A note about Case $1$ with $\,a:= 2^{-\frac{2}{3}}\,$ and $\,\displaystyle x\to 2^{\frac{1}{3}}\,$ .
Left side $\,=0\,$ for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\, ax^2-1=0\,$ and $\,\displaystyle ax^2-a^2x+1=\frac{3}{2}\ne 0\,$ .
Right side $\,=0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>1\,$:
$\,\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?
How do I show $f_n(x)=n^2 x^n(1-x)$ pointwise converges to $0$ on $[0,1]$?
I first started with the case $x=1/2$ to try out. We see that $n^2(1/2)^n(1/2)=n^2/2^{n+1}$ which goes to $0$ as $n$ goes to infinity since $n^2 < 2^{n+1}$.
Now I need... | For $0 < x < 1$ we have $y = 1/x - 1 > 0$ and $ x = 1/ (1 + (1/x - 1))= 1/(1+y).$
Thus,
$$n^2 x^n = \frac{n^2}{(1 + y)^n}.$$
Using the binomial expansion $(1+y)^n = 1 + ny + \frac{1}{2}n(n-1)y^2 + \frac{1}{6}n(n-1)(n-2)y^3 + \ldots,$we have
$$0 \leqslant n^2 x^n = \frac{n^2}{(1 + y)^n} < \frac{n^2}{n(n-1)(n-2)}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. O... | We have that $$a^4+b^4=(a+b)(a^3-a^2b+ab^2-b^3)+2b^4$$ so if $a+b$ is a factor of $a^4+b^4$ it is also a factor of $2b^4$ and (by symmetry) $2a^4$
If the highest common factor of $a$ and $b$ is $y$ so that $a=py$ and $b=qy$ we find that $(p+q)y$ is a factor of $2q^4y^4$. Now $p+q$ can have no factor in common with $q$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute the sum fast. How can I compute the following sum in the fastest way possible?
$y = 1 + x + ... + {x}^{{n}^{3}}=\sum_{i = 0}^{n}{x}^{{i}^{3}}$
I wrote that $n^3 - (n-1)^3 = 3n^2-3n+1$, but so far it does not help a lot.
| This is a modified version of my answer here, which was the same question only with square exponents rather than cubed ones.
You have already calculated $x^{(n+1)^3} = x^{n^3}\cdot x^{3(n+1)^2-3(n+1)+1}$ (I shifted the indices by one to conform with my other answer). We also have $x^{3(n+1)^2-3(n+1)+1}=x^{3n^2-3n+1}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$
Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$
I am getting:
$$\sin(x) = x - \frac{x^3}{3!} + R_4(x)$$
where $R_4(x) = \frac{\cos(c)x^5}{5!}$ for some $c$ between $0$ and $x$
I want to prove $R_4(x)\geq 0$ to arrive at the... | The aim is to show that:
$$ x - \frac{x^3}{3!} \leq \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + ...\leq x $$
For $sin x\leq x$ it suffices to use MVT:
$$\cos c =\frac{\sin x- \sin 0}{x-0}=\frac{\sin x}{x}\implies -1\leq \frac{\sin x}{x}\leq \implies \frac{\sin x}{x}\leq 1 \implies sin x\leq x$$
For ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to find the shortest distance from a line to circle while their equations are given Consider a line $L$ of equation $ 3x + 4y - 25 = 0 $ and a real circle $C$ of real center of equation $ x^2 + y^2 -6x +8y =0 $
I need to find the shortest distance from the line $L$ to the circle $C$.
How do I find that?
I am new t... | Hint:
Any point on the circle can be set as $$P(3+5\cos t,5\sin t-4)$$
The distance of this point from $L$ will be $$\dfrac{|3(3+5\cos t)+4(5\sin t-4)-25|}{\sqrt{3^2+4^2}}$$
$3(3+5\cos t)+4(5\sin t-4)-25=5(3\cos t+4\sin t)-32$
Now $-\sqrt{3^2+4^2}\le3\cos t+4\sin t\le\sqrt{3^2+4^2}$
$\iff-5\cdot5-32\le5(3\cos t+4\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Ratio of an inscribed circle's tangent to original square In the diagram, the circle is inscribed within square $PQRS,$ $\overline{UT}$ is tangent to the circle, and $RU$ is $\frac{1}{4}$ of $RS.$ What is $\frac{RT}{RQ}$?
| Suppose that the circle touches $RS$, $TU$ and $RQ$ at $X$, $Y$ and $Z$ respectively.
If $RQ=4$, $RU=1$. Let $RT=x$.
Then $TU=TY+YU=TZ+XU=2-x+1$.
So $1^2+x^2=(3-x)^2$ and thus $1+x^2=9-6x+x^2$.
$\displaystyle x=\frac{4}{3}$.
$\displaystyle \frac{RT}{RQ}=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find out all solutions for the system Given the system
$$ \left[
\begin{array}{ccc|c}
x_1&x_2&x_3&k\\
x_1&x_2&kx_3&1\\
x_1&kx_2&x_3&1\\
kx_1&x_2&x_3&1\\
\end{array}
\right] $$
I tried to solve this...It looks simple but I found a problem at the end...
$$ \left[
\begin{array}{ccc|c}
1&1&1&k\\
1&1&k&1\\
1&k... | The system has solutions only if
$\det(A|B)=0$, otherwise $\text{rank }(A|B)=4>\text{rank }(A)$
that is $k^4-6 k^2+8 k-3=0\to (k-1)^3 (k+3)=0$
for $k=-3$ and $k=1$
for $k=-3$ we get the system
$\begin {cases}
x+y+z=-3\\
x+y-3 z=1\\
x-3 y+z=1\\
-3x+y+z=1\\
\end{cases}
$
which has solution $(-1,-1,-1)$
for $k=1$ we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions a... | Hint: Your expression equals
$$x^2[(1+1/x^2)^{1/2} - (1+1/x^3)^{1/3}].$$
Now use the fact that $(1+h)^p = 1 + ph +o(h)$ as $h\to 0.$ (This fact is equivalent to the statement that the derivative of $(1+x)^p$ at $x=1$ is $p.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
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.