Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Deriving Taylor series without applying Taylor's theorem. First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-... | Alan Turing, at a young age, derived the series expansion of $\arctan$ without using (and, purportedly without knowing) calculus whatsoever.
Using the identity
$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x},$$
he obtained
$$\tan(2 \arctan x) = \frac{2x}{1-x^2},$$
and
$$2 \arctan x = \arctan\left( \frac{2x}{1-x^2}\right).$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Even though this might be too much for simple task you want (you can assume the polynomial of fifth degree and calculate coefficients) I think it is worth mentioning that there is a general formula for finding polynomial of sum of $p$-th powers $1^p+2^p+\dots+n^p$, and it is called Faulhaber's formula.
In short
$$1^p+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$? If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$?
I am getting the bounds of the fractions $\frac{a}{b},\frac{b}{c},\frac{... | This follows up on my former comment and proves $413=139+138+136$ is the minimal sum.
$92 c \gt 90 a \implies 2c > 90(a-c) \ge 90 \cdot 2 = 180$ therefore $c \gt 90 \iff c \ge 91$.
$91 b \gt 92 c \implies 91(b-c) \gt c \ge 91$ therefore $b-c \gt \frac{91}{91} = 1 \iff b-c \ge 2 \iff b \ge c+ 2$.
Since $b \ge c+2$ and $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving ${\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)}$ This was given to me by my Math Teacher almost a year ago and I've not been able to make much progress on it. I am hoping to see it resolved by our community members. $$\large{\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log... | Hint
If $x^2-4x-2<0$ or $x^2-4x-3<0$ then the corresponding $x$ cannot be a solution. But forget about that for a second and explain the meaning of the equation below:
$$\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3).$$
If for an $x$ the equation holds then there must exist some $A$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$. Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$.
My attempt:I put the value of $x^2$ in the second equation.I get:
$(2y-1)^2+y^4=2 \Rightarrow [(2y-1)^2-1^2]+(y^4-1^4)=0 \Rightarrow 4y(y-1)+(y-1)(y+1)(y^2+1)=0 \Rightarrow (y-1)(y^3+y^2+5y+... | There are no other real solutions.
From $2y-1=x^2\ge 0,$ $y$ has to satisfy $y\ge \frac 12$.
However, if $y\gt 0$, then $y^3+y^2+5y+1\gt 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rewrite rational function $f(x)$ as a series if the quadratic expression in the denominator has no roots A function of the type
$$f(x)=\frac{ex+f}{ax^2+bx+c}$$
with $b^2-4 a c \geq 0$ can be written as a series using partial fraction decomposition and geometric series.
But if one has the same function with $b^2-4 a c ... | More than likely, too simplistic !
You wrote the function $$F(x)=\frac{ex+f}{ax^2+bx+c}$$ Since you look for an expansion around $x=0$, let us rewrite it as $$F(x)=\frac{f+ex}{c+bx+ax^2}$$ and now use the long division to get $$F(x)=\frac{f}{c}+\frac{ c e-b f}{c^2}x+\frac{ b^2f-a c f-b c
e}{c^3}x^2+O\left(x^3\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doolittle transformation is non-unique for singular matrices Decomposing the singular matrix $$A = \begin{bmatrix}
1 & 2 \\
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}=LU$$
by Doolittle decomposition seems to be unique for this case. But how to proove tha... | The row of zeroes in your $U $ allows you to play with the second column of $L $. You have
$$
\begin{bmatrix}
1 & 2 \\
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & x\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}
$$ for any choice of $x $.
If you need $x=1$, there is no other choice and the decompositi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the integral: $\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}$ What is
$$\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}?$$
$a,b \in \mathbb{R}$ appropriate fixed numbers.
| $$\mathcal{I}\left(\text{a},\text{b}\right)=\int_0^{\frac{\pi}{4}}\frac{\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=$$
$$\frac{\text{a}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to ... | Hint. We have that for $x\not=1$, and $N\geq 2$,
$$\frac{d^2}{dx^2}\left(\frac{1-x^{N+1}}{1-x}\right)=\frac{d^2}{dx^2}\left(\sum_{n=0}^N x^n\right)=\sum_{n=2}^N n(n-1)x^{n-2}.$$
P.S. for the downvoters. I considered a finite sum because it is not so straightforward to say that we can interchange the differentiation and... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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An differential equation$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$ How to solve this ODE $$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$$
I tried to find its integral factor, but failed.
Many thanks for your help.
| $(y^2+x^2+3)(y^2)'=4x(2y^2-x^2)$
$(y^2+x^2+3)(y^2+x^2+3)'=4x(2y^2-x^2)+2x(y^2+x^2+3)$
$(y^2+x^2+3)(y^2+x^2+3)'=10x(y^2+x^2+3)-4x(3x^2+6)$
With $\enspace z:=y^2+x^2+3\enspace$ we get $\enspace zz'=10xz-12x^3-24x$ .
I don't know how to solve this, but Wolfram solves it in a closed form with
http://www.wolframalpha.com... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \fr... | Clearly $\sin(A)\sin(C)\geq 0$ since $A$ and $C$ are between $0^{\circ}$ and $180^{\circ}$. Let $A$ approach $0^{\circ}$. Then $\sin(A)$ approaches $0$ as well, while $\sin(C)$ is bounded above by $1$. This shows that
$$\sin(A)\sin(C)\rightarrow 0$$
if $A\rightarrow 0^{\circ}$ so that $\sin(A)\sin(C)$ can be as small a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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For $f(x)=4x^3+3x^2-x-1$, The Range of values $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ can take is- My Attempt :-
$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that
$$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$ But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$
So my ... | Intuitively, as long as $f'(x)$ attains its minimum at one single point, there won't exist two distinct points where the secant matches that slope.
To formalize, assume $f'(x)$ has a global minimum $m$, then it follows from MVT that:
$$\frac{f(b)-f(a)}{b-a} \ge m \quad \text{for} \;\;\forall a \ne b$$
If $a,b$ points e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as... | Here's a start that shows that
any other solutions
would have to have
distinct $a, b, $ and $c$.
In
$a^2+b^2+c^2 =1$
and
$a+b+c= \pm 1$,
if $a=b$,
these become
$2a^2+c^2 = 1,
2a+c = \pm 1$.
Then
$c = -2a\pm 1$,
so
$1
= 2a^2+(-2a\pm 1)^2
=2a^2+4a^2\pm 4a+1
=6a^2\pm 4a+1
$
so
$0 = 6a^2\pm 4a
=2a(3a\pm 2)
$.
Therefore
$a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find a general form of a sequence and its sum I have a problem to find a general form of the sequence
\begin{align}
- \frac{{n\left( {n - 1} \right)}}{{2\left( {2n - 1} \right)}},\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{2 \cdot 4 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)}... | Here is a more compact representation as sum formula, most of it was already stated in the comment section.
Since
\begin{align*}
a_n(k)=\frac{(-1)^kn(n-1)\cdots (n-2k+1)}{2\cdot4\cdots (2k)\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\qquad\qquad 1\leq k\leq n
\end{align*}
We obtain
\begin{align*}
a_n(k)&=(-1)^k\frac{n!}{(n-2... | {
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"url": "https://math.stackexchange.com/questions/2018992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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line integral hard function to differentiate $$\int_\gamma \frac{(x^2+y^2-2)\,dx+(4y-x^2-y^2-2) \, dy}{x^2+y^2-2x-2y+2}$$
where $\gamma = 2\sin\left(\frac{\pi x}{2}\right)$ from $(2,0)$ to $(0,0)$.
I think it should be a shortcut to this problem that I cannot see , if that is not the case I will keep trying to simplify... | Complete the squares in the denominator $(x-1)^2+(y-1)^2$ and change the variables $x-1\mapsto x$ and $y-1\mapsto y$. You get the vector field
\begin{align}
&\left[\frac{x^2+2x+y^2+2y}{x^2+y^2},\frac{4y-x^2-2x-y^2-2y}{x^2+y^2}\right]=\\
&=\left[1+\frac{2x}{x^2+y^2},-1+\frac{2y}{x^2+y^2}\right]-2\left[\frac{-y}{x^2+y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | Any symmetric polynomial in $X$ and $Y$ can be expressed as a polynomial in $S=X+Y$ and $P=XY$. If $X=\cos^2x$ and $Y=\sin^2x$, then $S=\cos^2x+\sin^2x=1$, so a symmetric polynomial expression in $\cos^2x$ and $\sin^2x$ can be written as a polynomial in $P=\cos^2x\sin^2x$.
If the symmetric polynomial is also homogeneou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 8
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | \begin{align}
\underbrace{1^3+2^3+\ldots+n^3}_{\left[\frac{n(n+1)}{2} \right]^2}+(n+1)^3
=&
\left[\frac{n(n+1)}{2} \right]^2+(n+1)^3
\\
=&
\frac{n^2\color{red}{(n+1)^2}}{4}
+
(n+1)\color{red}{(n+1)^2}
\\
=&
\left\lgroup
\frac{n^2}{4}
+
(n+1)
\right\rgroup
\color{red}{(n+1)^2}
\\
=& \left\lgroup \frac{n^2 +4(n+1)}{4} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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heterogeneous recurrence with f(n) as constant How to solve this $s_{n+1}=4s_{n-1}-3s_n+5$ where f(n)=5 conditions $s_0=-3$ $s_1=3$
I calculated the general solution $s_n=c_1*(-4)^n+c_2*1^n$ of this recurrence. The roots are $q_1=-4$ and $q_2=1$ but I have problem with particular solution with method of prediction .
I ... | $s_{n+1}=4s_{n-1}-3s_n+5$
$q^2-4+3q=0$
$\Delta=3^2-4*1*(-4)=25$
$\sqrt\Delta=5$
$q_1=\frac{-3-5}2=-4$
$q_2=\frac{-3+5}2=1$
homo general $s_n=c_1*1^n+c_2*(-4)^n$
$k=1$ where $k$ is multipiclity of root
hetero particular $s_n=Q(n)*q^n*n^k$
$s_n=A*1^n*n^1=An$; $1^n=1$ because $1^0=1$, $1^1=1$ and so on
$A(n+1)=4(A(n-1))... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with $x,y,z\geq 0$, $x+y+z=1.$ With $x,y,z\geq 0$, $x+y+z=1$.Prove that $$\sqrt{x+y^2}+\sqrt{y+z^2}+\sqrt{z+x^2}\geq 2 \tag{i}$$
The hint is using a lemma:
If $a,b,c,d\geq 0 $satisfying $a+b=c+d$ and$|a-b|\leqslant|c-d|$ then we have $\sqrt{a}+\sqrt{b}\geq \sqrt{c}+\sqrt{d}$
How to prove this lemma? And is ... | There is the following Vo Quoc Ba Can's solution.
We need to prove that
$$\sum\limits_{cyc}\left(\sqrt{x+y^2}-y\right)\geq1$$ or
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}\geq1.$$
Now, by AM-GM
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}=\sum\limits_{cyc}\frac{x(x+y)}{(x+y)\sqrt{x+y^2}+y(x+y)}\geq\sum\limits_{cyc}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdo... | You could remark some of the easy steps you did, making the proof more understandable yet (even though they are very simple). For example, you could write $3=3(2-1)$ (same with the other 2) to show that the formula holds for the smaller values of $n$.
Moreover, and more important, you should explain how you used Lapla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)}) } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \fra... | In your solution, when substituting the already known expression for $I_4$ (in the third line from the bottom), you forgot to multiply it by $5$. That's the only error there. Put it back in there, and you'll have a correct answer. Your answer would still look different from the output of that online integrator, but the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of angles of a triangle In a triangle $ABC$, if $\sin A+\sin B+\sin C\leq1$,then prove that $$\min(A+B,B+C,C+A)<\pi/6$$
where $A,B,C$ are angles of the triangle in radians.
if we assume $A>B>C$,then $\sum \sin A\leq 3 \sin A$,and $ A\geq \frac{A+B+C}{3}=\pi/3$.also $\sum \sin A\geq 3\sin C$ and $ C\leq \... | Since you assumed $A\geq B\geq C$, it must be that $\dfrac{A}{2}+C\leq\dfrac{\pi}{2}.$ Hence, $\sin\tfrac{A}{2}<\sin(\tfrac{A}{2}+C) = \cos(\tfrac{B-C}{2})$.
Finally, $$1\geq \sin A+\sin B+\sin C = \sin A+2\sin\tfrac{B+C}{2}\cos\tfrac{B-C}{2} = 2\cos\tfrac{A}{2}\big(\sin\tfrac{A}{2}+\cos\tfrac{B-C}{2}\big)>4\cos\tfrac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$ I don't know how to deal with this integral
$$I=\displaystyle\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$$
I reached the step
$$I
=\displaystyle\ \int_0^{\pi/2}\frac{1}{a + b\tan^2x}dx$$
Now what should I do? Please help.
| Now substitute
$\displaystyle\ u=\tan x \implies x= tan^{-1} u \implies dx=\frac{1}{1+u^2} du$
$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du$
Partial Fractions
$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du= \frac{1}{a-b} \left(\int_{0}^{\infty} \frac{du}{1+u^{2}} - b \int_{0}^{\infty} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
4th root question/guidance Find all 4th roots of $-8 + 8i\sqrt 3$
$a=-8$
$b=8\sqrt 3$
$r= \sqrt{a^2+b^2}= \sqrt {(-8^2)+(8\sqrt{3})^2)}=\sqrt{64+192}=\sqrt {256} =16$
$\frac ar= cos\theta=\frac{-1}{2}$ $\space $ $\frac br= sin\theta$=$\frac {\sqrt3}{2}$
This gives me a different $\theta$ one being 120 degrees... | Polar form:
$z_1= 2(\cos 30 + i \sin 30)$
$z_2= 2(\cos 120 + i \sin 120)$
$z_3= 2(\cos 210 + i \sin 210)$
$z_4= 2(\cos 300 + i \sin 300)$
Rectangular Form:
$z_1= 2(\frac {\sqrt{3}}{2} + i \frac 12) = \sqrt {3} + i$
$z_2= 2(-\frac 12 + i\frac {\sqrt{3}}{2} ) =-1 +i \sqrt {3}$
$z_3= 2(-\frac {\sqrt{3}}{2} - i \frac 12)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
| Both geometric series are convergent, so
$$S=\sum_{n=3}^{+\infty}(\frac{2}{3})^n-\sum_{n=3}^{+\infty}\frac{1}{3^n}$$
$$\frac{2^3}{3^3}\frac{1}{1-\frac{2}{3}}-\frac{1}{3^3}\frac{1}{1-\frac{1}{3}}$$
$$=\frac{8}{9}-\frac{1}{18}=\frac{5}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given a series defined by recursion. Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$ I have this problem from an old exam that I can't solve.
Let $\{a_i\}_{i \geq 0}$ be the series define by recursion as:
$a_0 = 2$
$a_1 = 3$
$a_{n+1} = \frac{a_n^2 +5}{a_{n-1}}$ , $\forall n \in \Bbb N... | Define a series $a_{n+1} = \frac{a_n^2 +k}{a_{n-1}}$ with some $a_0, a_1$ for $n\ge2$. Assume there are $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$.
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n + T a_{n-1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n a_{n-1} + T a_{n-1}^2 = a_n^2+k \tag{1}
$$
now
$$a_{n+1} = \frac{a_n^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Elementary Algebra Problem (in 8th grade) The exercise is to prove that $$ \forall x \in [0,3] $$ : $$ f(x)=\sqrt{18 + 3x -x^{2}} + \sqrt{9-x^{2}} + \sqrt{9-6x+x^{2}} + \sqrt{9x-3x^{2}} \le 12 $$
I notice that when $$ x=0 \implies f(x) = 3\sqrt{2} + 3 + 3 \le 12 $$
and when $$ x=3 \implies f(x) = 3\sqrt{2} \le 12 $$, ... | $$f(x)=\sqrt{(6-x)(3+x)}+\sqrt{(3-x)(3+x)}+\sqrt{(3-x)^2}+\sqrt{3x(3-x)}$$
$\forall x\in [0,3]$, the factors $6-x$, $3+x$, $3-x$ and $3x$ in the radicands are non-negative.
By $GM\le AM$,
$$f(x)\le \frac{(6-x)+(3+x)}{2}+\frac{(3-x)+(3+x)}{2}+3-x+\frac{3x+(3-x)}{2}=12$$
Note that $12$ is not the global maximum, but a r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating the convergence radius of a power series I've tried to calculate the convergence radius of the following power series:
$$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$
The Cauchy–Hadamard theorem doesn't help in this situation (I think).
So what I did is I tried to apply the d'Alembert ratio test to it and... | $$\begin{align}
\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})} &=\frac{6\cdot24^n+6\cdot18^n+5\cdot20^n+5\cdot15^n}{4\cdot24^n+4\cdot20^n+3\cdot18^n+3\cdot15^n}\\
&=\frac{6+6(3/4)^n+5(5/6)^n+5(5/8)^n}{4+4(5/6)^n+3(3/4)^n+3(5/8)^n}\to\frac32
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove using mean value theorem. $|\arctan(\frac{a}{4})^{4}-\arctan(\frac{b}{4})^{4}|\leq \pi^{3}\cdot|a-b|$
I started with $f(x)=\arctan(\frac{x}{4})^{4}$.
The function is continuous on $[a,b]$ and differentiable in (a,b)
So there exists a $c\in (a,b)$ such that $f'(c)=\frac{\arctan(\frac{b}{4})^{4}-\arctan(\frac{a}{... | Let $f(x) = x^4$, and suppose $|x| \le M$, then $|f'(x)| \le 4 M^3$ for $|x| \le M$.
Using the mean value theorem
we have $|f(x)-f(y)| \le 4 M^3 |x-y|$.
Let $M = {\pi \over 2}$ and note that $|\arctan x| \le M$ and
$|\arctan' x| \le 1$, hence
$|(\arctan x)^4 - (\arctan y)^4| \le {1 \over 2} \pi^3 |\arctan x -\arctan y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\... | Our hint is: $a\cos \theta +b\sin \theta =c$.
Given: $(\sqrt{3}-1)\cos \theta +(\sqrt{3}+1)\sin \theta =2$.
Let $(\sqrt{3}-1) = r\cos \alpha$ and $(\sqrt{3}+1) =r\sin \alpha$.
Then $r\cos \alpha \cos \theta + r\sin \alpha \sin \theta =2 \Rightarrow r\cos(\theta-\alpha) =2 \Rightarrow \cos(\theta-\alpha) =\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
On solvable octic trinomials like $x^8-5x-5=0$ Solvable quintic trinomials
$$x^5+ax+b=0$$
have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as,
$$x^6+3x+3=0$$
which factors over $\sqrt{-3}$. No $7$th-deg are known, but surprisingly there are octic ones, such as the simple,
$$x... | A result of Harris [1] is that every monic palindromic polynomial of degree-8 can be factored into two monic palindromic polynomials of degree-4.
$$
\begin{align}
f(x) & = x^8 + ax^7 + bx^6 + cx^5 + dx^4 + cx^3 + bx^2 + ax + 1 \\
& = (x^4 + px^3 + qx^2 + px + 1)(x^4 + rx^3 + sx^2 + rx + 1) \\
& = x^8+x^7 (p+r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
\lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks... | From the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^6)
\\\tanh x&=x-\frac{x^3}{3}+\frac{2 x^5}{15}+O(x^6)
\end{align}
$$ ones gets
$$
\begin{align}
\left(\sin x\right)^2&=x^2-\frac{x^4}{3}+O(x^6)
\\\left(\tanh x\right)^2&=x^2-\frac{2 x^4}{3}+O(x^6)
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Integration with Half space Gaussian I have a problem to solve and I have something that i don't know how to do.
The half-space Gaussian integral is given :
$$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$
I have to calculate
$$\int_{0}^\infty \exp \left(-ax^2 - \frac{b}{x^2} \right)dx$$
a and b are ... |
Note that in THIS ANSWER, I presented a solution to a more general version of the integral of interest herein.
Let $I(a,b)$ be the integral given by
for $a>0$ and $b>0$.
$$I(a,b)=\int_0^\infty e^{-\left(ax^2+\frac{b}{x^2}\right)}\,dx \tag 1$$
Enforcing the substitution $x\to \sqrt[4]{b/a}x$ into $(1)$ reveals
$$\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this ... | √(31+√(31+√(31....))) = s
s = √(31+s)
s² = s+31
s = (5√5 + 1)/2 (by the quadratic formula)
√(1+√(1+√(1....))) = k
k = √(1+k)
k² = k+1
k = (√5 + 1)/2 (by the quadratic formula)
s/k = (5√5 + 1)/(√5 + 1)
= (5√5 + 1)(√5 - 1)/4 (multiplying the numerator and denominator both by conjugate √5 - 1)
= (24-4√5)/4
= 6-√5
Since 6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x... | While I believe that the "right" answers are those already given, I would like to add yet another one, based on polar coordinates. Introduce
$$
\begin{cases}
x=r\cos \phi\\
y=r\sin \phi
\end{cases}
$$
The given function $f(x, y)=\frac{y}{x^2+y^2}$ is harmonic if and only if
$$
\left(\partial_r^2 +r^{-1}\partial_r +r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Showing $\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
How is the proof. If I separate the denominator with triangle inequality,
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le \frac{\lvert x^2-2x+3\rvert}{\lvert x^2-2x+3 \rvert-\lvert 2x\rvert}\... | What you want is
$$\left|\frac{x^2-2x+3}{x^2-4x+3}\right|\le1\iff-1\le\frac{x^2-2x+3}{x^2-4x+3}\le1$$
Beginning with the left inequality:
$$-1\le\frac{x^2-2x+3}{x^2-4x+3}\iff\frac{x^2-2x+3}{x^2-4x+3}+1\ge0\iff\frac{2x^2-6x+6}{x^2-4x+3}\ge0\iff$$
$$\frac{x^2-3x+3}{(x-1)(x-3)}\ge 0\iff (x-1)(x-3)>0\;\text{ (why?)}\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $a(n)=n^2+1$ then $\gcd(a_n,2^{d(a_n)})=1\text{ or }2$? Let $n\in\mathbf{N}$. I write $a_n=n^2+1$ and let $d(a_n)$ count the number of divisors of $a_n$. Set $$\Phi_n=\gcd\left(a_n,2^{d\left(a_n\right)}\right)$$ I would like to show and I believe it to be true that
$$\Phi_n =
\begin{cases}
1, & \text{if $n$ is eve... | Note that $2^{d(a_n)}$ can only be divisible by $1$ and powers of $2$.
If $n$ is even then $n^2+1$ is odd and in that case $\gcd=1$.
If $n$ is odd, then $n^2+1 \equiv 2 \pmod{4}$. Thus $n$ is not divisible by $4$, hence the $\gcd=2$.
Added explanation:
Using the division algorithm, we can write any integer $n=4k+r$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$
Show that
$\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$,
when $n\to+\infty$
I'm learning Taylor's Formula. The given solution is:
$\sqrt[n+1]{n+1}-\sqrt[n]{n}=e^{\frac{\ln(n+1)}{n+1}}-e^{\frac{\ln{n}}{n}}$
and use Taylor's Formula:
$e^{\frac{\... | You just need to handle the subtraction of series for $\sqrt[n+1]{n+1}$ and $\sqrt[n] {n} $ term by term in a proper manner upto 3 terms. The first term $1$ cancels out in both series. The second terms upon subtraction lead to
\begin{align}
A &= \frac{\log(n+1)}{n+1}-\frac{\log n} {n} \notag\\
&= \frac{\log(n+1)}{n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The sum of the fourth powers of the first $n$ positive integers I am studying mathematical induction and most of the times I have to prove something. Like, for example:
$1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}$
This time I found a question that ask me to find a formula for
$1 + 16 + 81 + .... + n^4$
How can I do... | As $S_0=0$ and $S_n-S_{n-1}=n^4$, $S_n$ must be a polynomial of the fifth degree with no independent term, let
$$S_n=an^5+bn^4+cn^3+dn^2+en.$$
Then
$$S_n-S_{n-1}=\\
a(n^5-n^5+5n^4-10n^3+10n^2-5n+1)+
\\b(n^4-n^4+4n^3-6n^2+4n-1)+\\
c(n^3-n^3+3n^2-3n+1)+\\
d(n^2-n^2+2n-1)+\\
e(n-n+1)=\\
a(5n^4-10n^3+10n^2-5n+1)+
\\b(4n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$... | EDIT: This answer is wrong, because I mixed up my left and right-hand sides right at the end. I think it is salvageable, but it'll be quite a bit of work.
I'll do it without induction.
Rearrange: we want $\left(\frac{(n+1)(n+2)}{3}\right)^{n-1} \frac{n+2}{3} > (n!)^2$
We'll show that this actually holds if we remove th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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A natural number is a perfect square as well as a perfect cube. Show that it is $0$ or $1$ $ ($mod $7$ $)$. A natural number is a perfect square as well as a perfect cube. Show that it is $0$ or $1$ $($mod $7$$)$.
I tried the following.
There are integers numbers $x,y$ such that $n=x^{2}=y^{3}.$ By using Euclidean div... | You know that all the squares are equals to $0,1,2$ or $4$ mod $7$ because:
$$0^2=0\pmod 7$$
$$1^2=1\pmod 7$$
$$2^2=4\pmod 7$$
$$3^2=2\pmod 7$$
$$4^2=2\pmod 7$$
$$5^2=4\pmod 7$$
$$6^2=1\pmod 7.$$
And all the cubes are equals to $0,1$ or $6$ mod $7$ because:
$$0^3=0\pmod 7$$
$$1^3=1\pmod 7$$
$$2^3=1\pmod 7$$
$$3^3=6\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determinant of a non-square block matrix $M_{n\times k}$ is defined as a matrix whose all elements are '-1'.
The following block matrix is as such:
$A =\begin{bmatrix}m\cdot I_{n-1} & M_{n-1\times m}\\M_{m\times n-1} & n\cdot I_{m}\end{bmatrix}$
prove the following: $det A = n^{m-1}\cdot m^{n-1}$
| Let more generally let $C(\gamma) = \gamma 1_{n-1}1_{m}^T $ denote the ${n-1}\times m$ matrix with each element equal to $\gamma$ (here $1_k$ denotes the k-dimensional column vector of all ones) and let $A(\gamma) = \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix}$. Your problem is to compute th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\int_{-\infty}^x\frac{1}{\sqrt{2\pi}} e^{-(\frac{x^2}{2}+2x+2)}\,dx$ I want to solve this integral from $-\infty$ to $x$ .
$$f_X(x)=\frac{1}{\sqrt{2\pi}} e^{-(\frac{x^2}{2}+2x+2)}, -\infty<x<\infty$$
I have searched as much as I could and I found a solution in wikipedia
$$\int_{-\infty}^{\infty} x e^{-a(x-... | Complete the square in the exponential function, then let $y=\frac{x+2}{\sqrt{2}}$
\begin{align}
\frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{z} \mathrm{e}^{-(\frac{1}{2}x^{2}+2x+2)} dx &=
\frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{z} \mathrm{e}^{-\frac{1}{2}(x+2)^{2}} dx \\
&= \frac{1}{\sqrt{\pi}} \int\limits_{-\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ i... | Assuming distinct teachers, distinct schools, and having identified the $5$ patterns, a foolproof mechanical way is to sum up the product of two multinomial coefficients for each case, one for the pattern, the other for the frequencies of singletons, doubles, triples, etc, viz.
$\binom{8}{1,1,1,5}\binom{4
}{3,1} + \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Given $\lim_{n\to\infty} a_n = a$ what is the limit $\lim \limits_{n \to \infty}\frac{a_n}{3^1}+\frac{a_{n-1}}{3^2}+\ldots+\frac{a_1}{3^n}$?
Given $\lim \limits_{n \to \infty}a_n = a$ then I need to find the limit $\lim \limits_{n \to \infty} \frac{a_n}{3} + \frac{a_{n-1}}{3^2} + \frac{a_{n-2}}{3^3} + \dotso + \frac{a... | Fix $\epsilon > 0$. Since $\displaystyle\lim_{n \to \infty}a_n = a$, there exists an $N \in \mathbb{N}$ such that $|a_n-a| < \epsilon$ for all $n \ge N$.
Let $S_n := \dfrac{a_n}{3}+\dfrac{a_{n-1}}{3^2}+\cdots+\dfrac{a_2}{3^{n-1}}+\dfrac{a_1}{3^n}$. Then, $S_{n+1} = \dfrac{1}{3}S_n+\dfrac{1}{3}a_{n+1}$ for all $n \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $\sqrt{1000} < x < 1000$? I have been given that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac {996}{995} \times \frac{998}{997} \times \frac {1000}{999}$$
How can I prove that $\sqrt{1000} < x < 1000$?
| \begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{1000}{999} \times \frac {1000}{999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{1000}{999} \time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solve the inequality and show solution sets on the real line $$-\frac{x+5}{2} \le \frac {12+3x}{4}$$
I always have issues with problems like this. I chose to ignore the negative sign at the beginning and got the answer. Is that a good method? When solving this inequality what is the best method for no mistakes?
My answ... | $$-\frac{x+5}{2} \le \frac {12+3x}{4}$$
Multiply $4$ both sides:
$$-2(x+5) \le 12+3x$$
$$-2x-10 \le 12 +3x$$
$$-22 \le 5x$$
$$x \geq \frac{-22}{5}$$
Your mistakes:
$-2(12+3x)=-24-6x$ rather than $-24+6x$. If this is just a typo, the next line is fine.
After this mistake, surprisingly in the next line, you corrected y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A question about neighboring fractions. I have purchased I.M. Gelfand's Algebra for my soon-to-be high school student son, but I am embarrassed to admit that I am unable to answer seemingly simple questions myself.
For example, this one:
Problem 42. Fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are called neighbor fract... | We have to prove that for positive integers $a,b,c,d$, we either have
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
or
$$\frac{c}{d}<\frac{a+c}{b+d}<\frac{a}{b}$$
First of all, $\frac{a}{b}=\frac{a+c}{b+d}$ is equivalent to $ab+ad=ab+bc$, hence $ad=bc$, which contradicts the assumption $ad-bc=\pm 1$. We can
disprove $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Integrating $\int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx $ In the following integral:
$$I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx $$
I thought of making partial fractions , then solve it .
But I am not able to make partial fractions.
| $\displaystyle \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}(1+x^{-4}+x^{-5})^2}dx = \int\frac{5x^{-6}+4x^{-5}}{(1+x^{-4}+x^{-5})^2}dx$
put denominator is $=t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\cos{\frac{\pi}{n}}\notin\mathbb Q$ for any postive integer $n\ge 4$ Let $n\ge 4$ postive integer,show that
$$\cos{\dfrac{\pi}{n}}\notin\mathbb Q$$
Now I have solve for a case:
Assmue that
$$\cos{\dfrac{\pi}{n}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$ use Chebyshev polynomials?
$$T_{n}(\cos{x})=\cos{(nx)}$$
so we ha... | An elementary proof. From deMoivre's Theorem, if $k\in \Bbb N$ then $\cos kx=\sum_{(0\le 2j\le k)}\binom {k}{2j}(-1)^j(\cos x)^{k-2j}(1-\cos^2 x)^j.$ So if $\cos (\pi/n)\in \Bbb Q$ and $m|n$ then $\cos (\pi/m)\in \Bbb Q.$ So it suffices to show $\cos (\pi/m)\not \in \Bbb Q$ when (i)$\,m=4,$ or (ii) $\,m=6,$ or (iii) $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Interesting log series Calculate the sum of series:
$$\sum_{k=1}^{\infty} k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right)$$
| We write
$$ k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right) = (k+1) \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-k \log\left(\frac{k}{\sqrt{k^2+1}}\right) +\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right).$$
The sum of the first two terms is telescoping and we obtain t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simple inequality ( or false) Given $a$, $b$ and $c > 0$ such that $abc=1$, prove that
$$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8abc$$
I don't know if its true or not but i need help in both cases.
Thanks
| If $a$ and $b$ are "small" and very very close together (but $a$ slightly larger), then $c$ is "large" and $c/(a-b)$ can be made arbitrarily large.
$a/(b-c)$ is negative but a it can be a "small" negative (as $c$ is "large" compared to $a$ and $b$. $b/(c-a)$ is positive (but small). So that sum can be arbitrarily l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trigonometry and Quadratics If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
My Attempt:
Using the sum and product formulae we have,
$q=\tan A\tan B, $ $-p=\tan A+\tan B$
And,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \Right... | You can proceed easily after finding $\tan (A+B)$.
So we have $$p =\tan (A+B)(q-1) $$ So substituting into the LHS we get, $$\sin^2 (A+B)+ (q-1)\tan (A+B)\sin (A+B)\cos (A+B) +q \cos^2 (A+B) = \sin ^2 (A+B) +(q-1)\sin ^2(A+B) +q\cos ^2 (A+B) =q\sin ^2 (A+B) +q\cos ^2 (A+B) =q $$ Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90... | Ignore $n = 0, 1$ since they're kind of irrelevant. Then $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So,
$$4a_n + 1 = 4(n^2 - n) + 1 = (2n-1)^2$$
So, then $\sqrt{4a_n + 1} = 2n - 1$, which is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The le... | We can prove this by using properties of complex numbers:
$$ \begin{align}a\cos(x) + b\sin(x) &= a \cos(x) + b \cos\left(x-\frac{\pi}{2}\right)\\
&=a \operatorname{Re}\left\{e^{ix}\right\} + b \operatorname{Re}\left\{e^{i(x-\pi/2)}\right\}\\
&= \operatorname{Re}\left\{ae^{ix} + be^{i(x-\pi/2)}\right\}\\
&= \opera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Find all positive integer solutions to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ Find all positive integers to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
Multiply both sides with $(abc)^2$ to get $(bc)^2 + (ac)^2 = (ab)^2$.
I then tried some pythagorean triples and nothing worked so I assu... | Rewrite the equation as
$$
\frac{c^2}{a^2}+\frac{c^2}{b^2}=1
$$
and set $c/a=x/z$, $c/b=y/z$, where $x$ and $y$ are coprime.
Then $(x,y,z)$ is a primitive Pythagorean triple, so, without loss of generality,
$$
x=u^2-v^2,\quad y=2uv,\quad z=u^2+v^2
$$
where $\gcd(u,v)=1$, one among $u$ and $v$ being even.
Thus
$$
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \... | You tried to solve for both $x$ in terms of $y$ and $y$ in terms of $x$.
Try expressing everything as either a function of $x$ or of $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...
| Note that $\sqrt{2} = \frac{2}{\sqrt{2}}$. Hence,
$$f(x + 1) + f(x - 1) = \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}f(x)$$
$$\begin{align}f(x + 1) &= \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(f(x) - \sqrt{2}f(x - 1)\right)\\
&=\frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(-f(x - 2)\right)\\
&= \frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Prove that the sequence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$ is not increasing Let $x_1\ge1$ and $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$. Prove that the sequence $\{x_n\}$ is not increasing, that is, $x_{n+2}\le x_{n+1}$.
My attempt was as follows: we have to prove that $1\le \frac{x_{n+1}}{x... | The reasoning seems correct until the end when you write
$$x_n+\frac{1}{x_n}\geq 2$$ Now here you should prove that $$x_n+\frac{1}{x_n}\geq 2$$
Which is trivial by AM-GM.
Here's a similar approach note that by AM-GM
$$x_{n+1}=\frac{1}{2}(x_n+\frac{1}{x_n})\geq 1$$
hence we have that $x_n\geq 1$.
Now $$\frac{x_n}{x_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the coefficient of $x^m$ in $(x+x^2+x^3+.....+x^i)^n$ EDIT:
My doubt may be silly but if the expansion of $(1-x)^{-n}$ is infinite, how come wolfram displayed 11 terms?
Wolfram result
Can anybody help to solve the whole equation?
I'm trying to solve this question.
I have an equation $(x+x^2+x^3+x^4+x^5+x^6)^2$... | This GitHub repository solved my problem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ordered triples with given condition We are given that there are $6250k$ ordered triples $(m,n,p)$ where $m,n,p$ are integers such that $$1≤m≤100,\;1≤n≤50,\;1≤p≤25\quad \& \quad 3\,|\,2^m+2^n+2^p$$ Find $k$.
With the given range of values of $m, n,p$, I guess the possible numbers of ordered triples without... | Given that $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ and $ 3\,|\,2^m+2^n+2^p$ where $m,n,p$ are integers.
Now if $2^m+2^n+2^p$ is divisible by $3$, then each of $2^m$, $2^n$, $2^p$ is divided by $3$.
Remainder of ${2^m}$ when it is divisible by $3$ is $(-1)^m$.
So remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\ta... | Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .
$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is the... | By AM-GM $$\left(1+\cos{x}\right)\sin\frac{x}{2}=2\cos^2\frac{x}{2}\sin\frac{x}{2}=2\left(\sin\frac{x}{2}-\sin^3\frac{x}{2}\right)=$$
$$=-2\left(-\sin\frac{x}{2}+\sin^3\frac{x}{2}+\frac{1}{3\sqrt3}+\frac{1}{3\sqrt3}\right)+\frac{4}{3\sqrt3}\leq$$
$$\leq-2\left(-\sin\frac{x}{2}+3\sqrt[3]{\sin^3\frac{x}{2}\cdot\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Complex Roots with improper fraction I'm having trouble with the following:
$(-16i)^{5/4}$
My calculations for the Principal root is:
$32(\cos (3\pi/2) * 5/4) + i \sin (3\pi/2)* 5/4))$
$=32(Cis (15\pi/8))$
This answer does not agree with the online calculators. It gives a positive real value and the online calculators... | Let $z=-16i$ and $n=\dfrac54$. For solving you have to compute $r=|z|$ and argument $\theta$ where $\tan\theta=\dfrac{y}{x}$.
then
$r=|z|=|-16i|=16$ and argument $\theta=\dfrac{3\pi}{2}$. Then write
$$z_k=r^n(\cos n\theta+i\sin n\theta)$$
But argument adds with $2k\pi$ so we have
$$z_k=r^n\Big(\cos n(\theta+2k\pi)+i\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A Ramanujan infinite series $$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$
I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.
B... | Modifying @Claude Leibovici's answer a little.
$\frac{1}{\left(1-z\right)^{a}}=\,_{1}F_{0}\left(a;;z\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}z^{n}}$
and
$\frac{1}{\sqrt{1-4\cdot x}}={\displaystyle \sum_{n=0}^{\infty}\left(\begin{array}{c}
2n\\
n
\end{array}\right)x^{n}}$
Let $z=4x,x=\frac{z}{4}$
$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this que... | Consider the polynomial
$$p(t) = (1-x t)(1-y t)(1-z t)$$
Let's consider the series expansion of $\log\left[p(t)\right]$:
$$\log\left[p(t)\right] =-\sum_{k=1}^{\infty}\frac{S_k}{k} t^k$$
where
$$S_k = x^k + y^k + z^k$$
Since we're given the $S_k$ for $k$ up to $3$ we can write down the series expansion of $\log\left[p(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Proving that $\cos(\arcsin(x))=\sqrt{1-x^2}$ I am asked to prove that $\cos(\arcsin(x)) = \sqrt{1-x^2}$
I have used the trig identity to show that $\cos^2(x) = 1 - x^2$
Therefore why isn't the answer denoted with the plus-or-minus sign?
as in $\pm \sqrt{1-x^2}$.
Thank you!
| Let $\arcsin x = \theta$. Then, by definition of the arcsine function, $\sin\theta = x$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and
$\cos(\arcsin x) = \cos\theta$. Using the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$, we obtain
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\cos^2\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Find all complex numbers $z$ satisfying the equation $z^{4} = -1+\sqrt{3}i$ Since any complex number can be of polar form. We set that $z = r\cos \theta +ir\sin \theta$. Now by de Moivre's Theorem, we easily see that $$z^{4} = r^{4}\cos4 \theta + ir^{4}\sin4 \theta$$
Since $$z^{4} = -1+\sqrt{3}i$$
We equip accordingly ... | Observe that $-1+\sqrt{3}i=2e^{\frac{2\pi i}{3}}$. Then the roots of the equation are
$$
z=\sqrt[4]{2}e^{\frac{\pi i}{6}+\frac{k\pi}{2}},
$$
where $k=0,1,2,3$. Thus $r=\sqrt[4]{2}$. Substitute $k$, then we get $z=\sqrt[4]{2}e^{\frac{\pi i}{6}}$, $z=\sqrt[4]{2}e^{\frac{2\pi i}{3}}$, $z=\sqrt[4]{2}e^{\frac{7\pi i}{6}}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of the area of a Koch Snowflake this is my first question for this site and I made this account specifically for help with the following topic.
I am doing a research presentation on the Koch Snowflake, specifically, the area.
So far, I have been attempting to generalize a formula for finding the area... | The following forms a GP with $a=\frac{1}{9}$ and $r=\frac{4}{9}$.
$$\begin{align}
&\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}\\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\lim_{n\to\infty} \sum_{r=2}^{n} \frac{ 4^{r-2}}{9^{r-1}} \\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $y'$ given $y\,\sin\,x^3=x\,\sin\,y^3$? The problem is
$$y\,\sin\,x^3=x\,\sin\,y^3$$
Find the $y'$
The answer is
Can some explain how to do this, please help.
| I have found that it helps students to understand implicit differentiation if first they think of both $x$ and $y$ as functions of some third variable such as $t$ and take the derivative of both sides with respect to $t$, being careful to use the product rule, chain rule, etc when needed. Then as a final step, multiply... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $ \sin\theta + \cos\theta = \frac 1 2$, what does $\tan\theta + \cot\theta$ equal? A SAT II question asks:
If $ \sin\theta + \cos\theta = \dfrac 1 2$, what does $\tan\theta + \cot\theta$ equal?
Which identity would I need to solve this?
| Hint
$$\sin\theta+\cos\theta=\frac{1}{2} \implies \left( \sin\theta+\cos\theta \right)^2 = \frac{1}{4} \iff \color{blue}{\cos\theta\sin\theta} = \cdots$$
and
$$\tan\theta+\cot\theta = \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} = \frac{\cos^2\theta+\sin^2\theta}{\cos\theta\sin\theta}= \frac{1}{\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total nu... | Remember that you are given $t=12$.
Also,
$$\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$ is not correct, as the order in which you draw the balls doesn't matter. There are 6 total ways to draw three distinctly colored balls, so your equation should be:
$$6*\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Combinatorics: throwing a dice three times to get an even number. Suppose you throw a six face dice three times, how many times will be the sum of the faces even?
I approached it this way:
You either get all three times even face, or twice odd and once even.
As there are only 3 faces that are even, you have $3^3$ possi... | P(all even)$=\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{1}{8}$
P(one even and two odd)$=\dbinom{3}{2}\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{3}{8}$
Required probability $=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... | It might just be easier to use that $a^2-b^2=(a-b)(a+b)$.
So if $a-b=0$ then $(a-b)^2=(a-b)(a+b)$, and if $a-b\neq 0$ then $(a-b)^2=(a-b)(a+b)$ if and only if $a-b=a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
| $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\Leftrightarrow \left|\frac{n^3+3n}{2n^4-n+1}\right|<\epsilon.$$ Now, $4n^3\ge n^3+3n$ and $2n^4-n+1\ge n^4$ for all $n$ positive integer. So $$\left|\frac{n^3+3n}{2n^4-n+1}\right|\le \frac{4n^3}{n^4}=\frac{4}{n},$$ and choosing $n_0=\lfloor 4/\epsilon \rfloor+1$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculating the $100th$ term of a series I came across a series
$1,5,13,25,41,61.......$
I have to calculate the $100^{th}$ term of this series. How do I do it?
Is there any formula to calculate the $n^{th}$ term of a series?
| The difference of successive terms forms an Arithmetic progression: $4,8,\cdots $.
We can write the terms of our sequence as: $$0\times 4+1,1\times 4+1, 3\times 4+1, 6\times 4+1, 10\times 4+1, \cdots $$ $$=\frac {0\times 1}{2}\times 4+1, \frac {1\times 2}{2}\times 4+1, \frac {2\times 3}{2}\times 4+1, \frac {3\times 4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find coefficient of generating function f(x). Find coefficient of generating function.
$ f(x) = \frac{2x}{1-x^{2}} +x$
MY WAY OF SOLVING SIMILAR PROBLEM:
1) $ g(x) = \frac{2x}{1-x^{2}}$
2) partial fraction $g(x) = \frac{A}{1-x} + \frac{B}{1+x} $
3) $ g(x) = \sum\limits_{n=0}^\infty Ax^{n} + \sum\limits_{n=0}^\infty... | There is no reason to add up the right hand terms, since the term $x$ is simple and convenient. The other term can be expanded using the geometric series expansion
\begin{align*}
\frac{1}{1-y}=\sum_{n=0}^\infty y^n\qquad\qquad |y|<1
\end{align*}
with $y=x^2$.
We obtain
\begin{align*}
f(x)&=\frac{2x}{1-x^2}+x\\
&=2x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find two numbers whose $AM+...$ Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.
My Attempt;
Given,
$\frac {AM}{GM}=\frac {5}{3} = k (let) $
$\AM=5k,
GM=3k$.
Also,
$AM+GM=25$
$5k+3k=25$
$8k=25$
$k=\frac {25}{8}$.
Am I going right? Or, is there any other simple alternative.?
| we have $$AM=\frac{5}{3}GM$$ from here we get with the first equation:
$$\frac{5}{3}GM+GM=25$$ thus we have $$\frac{8}{3}GM=25$$ and $$GM=\frac{75}{8}$$
from here we get $$ab=\left(\frac{75}{8}\right)^2$$ and $$a+b=\frac{375}{12}$$ you can solve one equation e.g. for $a$ and plug these equation in the other one
ok with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} +\tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
In a triangle $ABC$ with one angle exceeding $\frac {2}{3} \pi$, prove that
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} + \tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
I tried expanding that half angle, applying AM-GM on various sets, using Si... | Let $\gamma\geq\frac{2\pi}{3}$ and $\tan\frac{\gamma}{4}=x$.
Hence, $\frac{1}{\sqrt3}\leq x<\frac{1}{\sqrt2}$ and since $\tan$ is a convex function on $\left[0,\frac{\pi}{2}\right)$, by Jensen we obtain:
$$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}\geq2\tan\frac{\alpha+\beta}{4}=2\tan\left(\frac{\pi}{4}-\frac{\gamma}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}... | The Taylor Theorem tells us that the estimation error after $n$ terms is given by $f'(c)\frac{(x-a)^{n+1}}{(n+1)!}$, for som $c \in (x-a)$. Now you should be able to find an upper bound on the derivative in that interval, which should give you an upper bound on the error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ w... | Try working mod $3$ and mod $8$. Assuming $x, y>3$, we have $x,y = \pm 1$ mod $3$. Since $x, y$ are odd we have $x^2, y^2=1$ mod $8$, so
$$x^2, y^2 = 1 \text{ mod } 24.$$
Substituting in the equation gives $$x = 24k+11 $$ for some integer $k$.
Rearranging the original equation we get
$$x(5x-7)=(y-1)(y+1), \tag{1}$$
the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?
|
Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!
A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$
Is to first move all the constants to the RHS, so $(1)$ becomes... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$ Prove that:$$\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$$.
My Attempt,
$$L.H.S= \frac {\sec (16A)-1}{\sec (8A)-1}$$
$$=\frac {\frac {1}{\cos (16A)} -1}{\frac {1}{\cos (8A)} -1}$$
$$=\frac {(1-\cos (16A)).(\cos (8A)... | $\frac{\sec 16A -1}{\sec 8A -1}$
= $\frac{\frac{1}{\cos 16A}-1}{\frac{1}{\cos 8A}-1}$
= $\frac{\frac{1 - \cos 16A}{\cos 16A}}{\frac{1 - \cos 8A}{\cos 8A}}$
= $\frac{2 \sin^2 8A}{\cos 16A} × \frac{\cos 8A}{2 \sin^2 4A}$
= $\frac{2 \sin 8A \cos 8A}{\cos 16A} × \frac{\sin 8A}{2 \sin^2 4A}$
= $\frac{\sin 16A}{\cos 16A} × \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.
| Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) ^{ \frac { 1 }{ x } }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Quadratic equation, find $1/x_1^3+1/x_2^3$ In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.
It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?
| If $x_{1}$ and $x_{2}$ are the roots then
$x_{1} + x_{2} = -\frac{b}{a}$ and $x_{1} \cdot x_{2}=\frac{c}{a}$, now $\frac{1}{x_{1}}+\frac{1}{x_{2}} = -\frac{b}{c}$ and $$\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}} = \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)^3- 3\cdot\frac{1}{x_{1}.x_{2}}\left(\frac{1}{x_{1}}+\frac{1}{x_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite diffe... | Treat your numbers as a sequence with $g(0)$ being the first term, corresponding to $f(-3)$,
$$f(x-3):=g(x) : \color{green}{-9},3,3,-3,-9,-9,3$$
$$\Delta g=g(x+1)-g(x) : \color{green}{12},0,-6,-6,0,12$$
$$\Delta^2 g : \color{green}{-12},-6,0,6,12$$
$$\Delta^3 g : \color{green}{6},6,6$$
Now assume all else is $0$. To ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Fourier Series/ fourier transform What is the Fourier series of the following piece-wise function?
$$
f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
$$
| Given f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
Its nth Fourier polynomial is $S_n(x)=\sum_{v=-n}^{n}\alpha_ve^{ivx}$, where $\alpha_v=\int_{-\pi}^{\pi}f(x)e^{ivx}dx=\int_{-1/2}^{1/2} \cos(3\pi x)e^{ivx}dx$. Notice that $\dfrac{e^{3\pi ix}+e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate a limit involving definite integral Evaluate the following limit:
$$\lim_{n \to \infty} \left[n - n^2 \int_{0}^{\pi/4}(\cos x - \sin x)^n dx\right]$$
I've tried to rewrite the expression as follows:
$$\lim_{n \to \infty} \left[n - n^2 \sqrt{2}^n \int_{0}^{\pi/4}\sin^n \left( \frac{\pi}{4} - x \right) dx\right]... | Writing
$$ (\cos x - \sin x)^n = \color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}} \cdot \color{red}{ (\cos x + \sin x)(\cos x - \sin x)^{n-1}} $$
and applying integrating by parts, we have
$$ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx
= \frac{1}{n} - \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding a geometric interpretation I recently solved a question of complex numbers which was this:
$A\left( \frac{2}{\sqrt{3}} e^\frac{i\pi}{2}\right)$, $B\left( \frac{2}{\sqrt{3}} e^\frac{-i\pi}{6}\right)$, $C\left( \frac{2}{\sqrt{3}} e^\frac{-5i\pi}{6}\right)$ are the vertices of an equilateral triangle. If $P$ be a ... | The sides of your triangle are $2$. E.g. the distance of $A$ and $B$ is the length of $$\frac{2}{\sqrt{3}}(e^{-\frac{i\pi}{6}}-e^{\frac{i\pi}{2}})=\frac{2}{\sqrt{3}}(\frac{\sqrt{3}}{2}-\frac{3}{2}i)=1-\sqrt{3}i.$$
We can calculate $PA^2+PB^2+PC^2$ using elementary methods, if $P$ is a point on the incircle of an equila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How many distinct groups of 5 possible from 10 seniors and 6 freshman? A school is forming a group of 5. There are 10 freshman and 6 seniors, and the group must have at least 2 freshman and at least 1 senior. How many distinct groups are possible?
My approach (that I think is definitely flawed) -
This would be a combi... | As you say the order doesn't matter. Then you have to think of all allowed groups, that is 2 freshmen 3 seniors, 3 freshmen 2 seniors, and 4 freshmen 1 senior.
2 freshmen 3 seniors:
$\dfrac{10 \cdot 9}{2!} \cdot \dfrac{6 \cdot 5 \cdot 4}{3!} = \dfrac{10!}{2!(10-2)!}\cdot \dfrac{6!}{3!(6-3)!} = 900 $ possible combinat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\en... | you start with a $2×2$ matrix and see the eigenvalues they are $4,6$.
Then see for $3×3$ matrix the eigenvalues are $9,6,-6$
for $4×4$ they are $14,6,-6,6.$.
Hence whenever the order is even the eigenvalues 6 exceeds the eigenvalue $-6$ by 1 in multiplicity.
and hence for even $n$ the snswer is
$det= (5(n-1)-1).6^{n/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\f... | Continuing from where you left, expressing terms of the numerator as:
$$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$
$$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$
$$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$
$$x+y+z=xyz $$
Now add everything and the result follows. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$. Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$.
Using Euclidean algorithm:
$$2^{2004}-1=4(2^{2002}-1)+3$$
$$2^{2002}-1=x\cdot 3+y$$
The solution manual says that $2^{2002}-1$has the remainder $0$ when divided by $3$, that is $y=0$ so... | We have,
$2^{2002} - 1$
$= (2^4)^{500}.2^2 - 1$
Now $2^4 \equiv 1 (\mod 3)$
From above,
$= (1)^{500} .2^2 - 1$
$= 1.2^2 - 1$
$= 4 - 1 = 3$
Divisible by 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt,
$a,b,c$ are real and positive numbers, then
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$... | By AM-GM $$6(a^2+b^2+c^2)=\sum_{cyc}(4a^2+b^2+c^2)\geq6\sum_{cyc}\sqrt[6]{a^8b^2c^2}=6(a+b+c)$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope ... | by AM-GM we get $$\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2}{6}\geq \sqrt[6]{a^{12}b^{12}c^{12}}=a^2b^2c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the serie... | First, the partial fraction of the summand can be written
$$\begin{align}
\frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\
&=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right)
\end{align}$$
Second, we note that
$$\begin{align}
\sum_{n\,\,\text{odd}}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| If $(9+x)^\frac 13 + (9-x)^\frac 13 = 3$,
then there is a $y$ such that $(9+x)^\frac 13 = (\frac 32+y)$ and $(9-x)^\frac 13 = (\frac 32-y)$.
Taking cubes of both equations you get $9 \pm x = (\frac {27} 8+\frac 92y^2) \pm (\frac {27} 4 +y^2)y$,
and so $9 = \frac {27} 8+\frac 92y^2$ and $x = (\frac {27} 4 +y^2)y$.
This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there i... | Another approach to break down the last integral might be to consider the integral of $\displaystyle \frac{\log^3 (1-iz)}{z^2}$ along a positively oriented semi-circular contour $\gamma_R = [-R,R]\cup Re^{i[0,\pi]}$ in the upper half-plane. (We choose the branch of logarithm $\log (1-iz)$ in the lower half-plane along ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
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Proving non-existence of rational points in a simple equation Working on Chapter 6.20 of Hammack Book of Proof
Show that $x^{2} + y^{2} - 3 = 0$ has no rational points.
First prove: IF $3\not\vert m$ THEN $m^{2}\equiv1(mod 3)$
If $3\not\vert m$ then either of two cases are true:
CASE $m\equiv 1 (mod3)$
$m=3k+1$ where... | Just do the hint.
If $n=2k $ then $n^2=4k^2\equiv 0\mod 4$
If $n=2k+1$ then $n^2=4k^2+4k+1\equiv 1 \mod 4$
So $3c^2 \equiv 0|3\mod 4$
And $a^2+b^2=0,1,2\mod 4$
So if $a^2 +b^2 =3c^2$ then all $a,b,c$ are even. But if we let $a=\gcd (a,b,c)a';b=\gcd (a,b,c)b';c=\gcd (a,b,c)c'$. $a',b',c'$ can't all be even (unless the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lower bound for $\frac{(x+y+z)^3}{xy+yz+zx}$ Let $x,y,z\geq 0$ and $x^2+y^2+z^2\geq 3$. What is the minimum value of $$D(x,y,z)=\frac{(x+y+z)^3}{xy+yz+zx}?$$
When $x=y=z=1$, $D(x,y,z)=9$. We have $(x+y+z)^2\leq 3(x^2+y^2+z^2)$ and $xy+yz+zx\leq x^2+y^2+z^2$, but these do not help directly with bounding $D(x,y,z)$.
| Let $x^2+y^2+z^2=k(xy+xz+yz)$.
Hence, by AM-GM we obtain:
$$\frac{(x+y+z)^3}{xy+xz+yz}\geq\frac{(x+y+z)^3}{xy+xz+yz}\sqrt{\frac{3}{x^2+y^2+z^2}}=$$
$$=\sqrt{\frac{3(x+y+z)^6}{(xy+xz+yz)^2(x^2+y^2+z^2)}}=\sqrt{\frac{3(k+2)^3}{k}}\geq\sqrt{\frac{3(3\sqrt[3]k)^3}{k}}=9.$$
The equality occurs for $x=y=z=1$, which says tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.