Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the smallest number such that when you divide with $4,6,8,10,12$ the remainder is $2,4,6,8,10$
Find the smallest number such that when you divide with $4,6,8,10,12$ the remainder is $2,4,6,8,10$.
This is the solution:
Let $n \in \mathbb N$ such that $n = 4q_1 + 2$, for some $q_1$, $n = 6q_2 + 4$, for some $q_2$,... | $$n+2=4q_1+2+2=4(q_1+1)\\n+2=6q_2+4+2=6(q_2+1)\\n+2=8q_3+6+2=8(q_3+1)\\\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the radius of convergence for $\frac{1}{(3 - x)(1 + x)^{2}}$ at $x_{0} = 0$ and $x_{0} = 1$ I am having trouble with the problem
Find the radius of convergence for $f(x) = \frac{1}{(3 - x)(1 +
x)^{2}}$ at $x_{0} = 0$ and $x_{1} = 1$.
Usually, when I'm trying to find radius of convergence, I'm given some sor... | Here are my thoughts. This might not be the original purpose of the exercise, but this is my guess.
According to your partial fraction decomposition, and using the well-know power series $(1-x)^{-1} = \sum_0^\infty x^n$ for $x \in (-1,1)$, at $x=0$, we have
\begin{align*}
f(x)&= \frac 1{16}\sum_0^\infty (-1)^n x^n - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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What are the possible values of these letters? Out of all the questions I answered in a math reviewer, this one killed me (and 7 more).
Let $J, K, L, M, N$ be five distinct positive integers such that
$$
\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1.
$$
Then, what is $J + K +... | Induction could lead you to the answer.
The equation is :
$$
\frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_{n}} + \frac 1 { x_1 x_2 \dots x_{n}} = 1
$$
Case $ n = 0 $: the empty set solves the equation as an empty product is 1
Case $ n = 1 $: the obvious solution is $ x_1 = 2 $.
Case $ n = 2 $: a bit more dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
} |
Verification of Laurent series for the function $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$ Suppose that $$f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
\begin{align}
f(z)&=-2\ \frac{d}{dz}\left(\frac{1}{z+2}\right)... | After the last $=$ sign, what you got was:$$-2\sum_{n=1}^\infty(-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$\sum_{n=1}^\infty(-1)^{n-1}\frac{2n(z-2)^{n-1}}{4^{n+1}}=\sum_{n=0}^\infty(-1)^n\frac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theore... | You're on the right track. By Fermat's little theorem, $2^n\pmod{5}$ and $2^n\pmod{13}$ depend only on $n\pmod{4}$ and $n\pmod{12}$, respectively. You've shown that $2^n\equiv 3 \pmod{5}$ if $n\equiv 3\pmod{5}$, but that's the opposite of what you want to prove; you want to show that $2^n\equiv 3\pmod{5}$ only if $n\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Limit of a sequence of integrals involving continued fractions The following question was asked in a calculus exam in UNI, a Peruvian university. It is meant to be for freshman calculus students.
Find $\lim_{n \to \infty} A_n $ if
$$ A_1 = \int\limits_0^1 \frac{dx}{1 + \sqrt{x} }, \; \; \; A_2 =
\int\limits_0^1 \fra... | $\sqrt{x}$
is a McGuffin.
More generally,
let
$f_1 = \frac{1}{1 + g(x) }
$
where
$g'(x) > 0,
g(0) = 0
$,
$f_n(x)
=\frac{1}{1+f_{n-1}(x)}
$,
and
$A_n = \int_0^1 f_n(x) dx
$.
Then
$f_n(x)
\to \dfrac{\sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{
\begin{array}{c}
|x-3|<2x \\
|2x+5|>3
\end{array}
\right.
$$
The steps I'm taking are:
Finding the absolute values sings, so for
$x-3 \geq 0$ we have $x \geq 3$ therefore
$$|x-3| =
\left\{
\begin{a... | There is actually only the first inequality to be solved:
$$ |2x+5| \stackrel{\color{blue}{0}\leq|x-3| \color{blue}{< 2x}}{>} |x-3|+5 > 3$$
So, you only need to solve $|x-3| < 2x$ while $x>0$:
$$
-2x < x-3 < 2x \Leftrightarrow
\begin{cases}
3x > 3 \Leftrightarrow \color{blue}{x> 1} \\
x > -3 \mbox{ does not extend t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Why does $\int_1^\sqrt2 \frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ equal to $0$? In this question, the OP poses the following definite integral, which just happens to vanish:
$$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$
As noticed by one commenter to the question, the... | Put in other words we desire to show that:
$$I=\int_1^\sqrt2 \ln\left(\frac{2-2x^2+x^4}{2-2x+x^2}\right)\frac{dx}{x}=\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2$$
Let's take the LHS integral and split it in two parts.
$$I=\underbrace{\int_1^\sqrt2 \frac{\ln(2-2x^2+x^4)}{x}dx}_{\mathcal K}-\underbrace{\int_1^\sqrt2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 1,
"answer_id": 0
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What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1... | hint
Take logarithm and use
$$
\sum_{k=2}^{n+1}\ln\left(1-\frac{1}{k^2}\right)
= \sum_{k=2}^{n+1}\left(\ln\left(\frac{k-1}{k}\right)+\ln\left(\frac{k+1}{k}\right)\right)
=\ln\left(\frac{1}{n+1}\right)+\ln\left(\frac{n+2}{2}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Sequence in which adding 2 produces a square Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 \cdot 7 - 2 = 47, x_5=47 \cdot 7 - 7=322, x_6=322 \cdot 47 - 7 =15127, x_7=15127 \cdot 322 -47=4870847,x_8=4870847 \cdot 15127-322=73681302247.$
My spreadsheet cannot calculate... | We have
\begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\\
\end{align*}
then
\begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\\
\end{align*}
then by telescoping
\begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\\
x_{n+1}x_{n-2}&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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If $2n+1$ and $3n+1$ are perfect squares, then prove that $8|n$.
If for some number $n\in \mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ wi... | It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ I need to find
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5} \frac{2... | As an alternative by $y=x+5 \to 0$
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}=\lim_{y\rightarrow 0} \frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=\lim_{y\rightarrow 0} \frac{2y^2-20y}{2y^2-17y}=\lim_{y\rightarrow 0} \frac{2y-20}{2y-17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^... | Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = \frac{(k+1)(2k+3)(7k+8)}{6}$$.
We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = \frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
$$ = \frac{(k+1)(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the general solution in terms of Bessel functions: $t^2x'' + x' + x = 0, \quad t < 0, \text{ Hint: } s = 2\sqrt{t}$ I was asked the following question:
Find the general solution in terms of Bessel functions:
$$t^2x'' + x' + x = 0, \quad t < 0, \text{ Hint: } s = 2\sqrt{t}$$
My approuch
I think that what I have to ... | We see that $t=0$ is not a regular singularity of the equation
$$x''+\dfrac{1}{t^2}x'+\dfrac{1}{t^2}x=0$$
then this is not a Bessel differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$ I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that
$$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\fra... | *
*The equation $$\;x(16x^4 - 20x^2 + 5) = 0\; \tag 1$$ can have $5$ real solutions (incl. multiplicity).
*From $\;\cos 5\theta = 16\cos^5 \theta - 20\cos^3 \theta + 5\;$ we deduce that we will look for solutions of $(1)$ in the form of $\cos \theta\;$ where $\;\cos 5\theta = 0.$
*$\cos 5\theta = 0$ holds for $10$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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How to get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$? How would I get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$?
It make sense to me logically. When one multiplies it out, it would condense to $x^{p-1}-1$. But it's just not clicking. What is the arithmetic between these steps?
| First of all, we might as well set
$n = p - 1, \tag 1$
and then investigate the equation
$x^n - 1 \overset{?}{=} (x - 1)(x^{n - 1} + x^{n - 2} + \ldots + x + 1) = (x - 1)\displaystyle \sum_0^{n - 1} x^i. \tag 2$
One may in fact use a simple induction to validate (2); we check
$n = 1: \; x^1 - 1 = x^1 - 1 = (x - 1) \dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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find all $k$ such that $k^5+3$ is divisible by $k^2+1$ That's it. I have a solution with substitutions, but it's tedious and not too generic. Is there a solution using some cool theorem for polynomials divisibility?
| Solution with substitutions.
If $k^5+3$ is divisible by $k^2+1$, it could be expressed as $(k^2+1)(k^3+n)$, which if expanded, leads us to $k^3+nk^2+n-3$.
Now let's make another substitution by introducing another variable $a: k=-(n+a)$, which will get us $-an^2+(1-2a^2)n-(a^3+3)$. It is solvable for $n$ only when $D\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\sqrt{2}\sec x+\tan x=1$
Solve $\sqrt{2}\sec x+\tan x=1$
I understand it can be very easily solved by expanding in terms of $\sin x$ and $\cos x$, gives $x=2n\pi-\frac{\pi}{4}$. But, what if I do the following:
$$
\sqrt{2}\sec x+\tan x=1\\
\text{Differentiating}\implies\sqrt{2}\sec x\tan x+\sec^2 x=0\implies\s... | The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Congruence proof. Show $a$ is odd, $b$ is even, and $ a \equiv 1$ mod $4 $ under certain conditions. If $ p \equiv$ $1$ mod $4$, $ p = a^2 + b^2$, and $a + bi \equiv 1$ mod $2+2i$, then $a$ is odd and $b$ is even. Moreover, if $4|b$, then $ a \equiv 1$ mod $4$, and if $ 4 \nmid b$, then $ a \equiv -1$ mod $4$.
Proof:
$... | First, $a+bi \equiv 1\pmod{2+2i}$ means that $a+bi = 1 + k(2+2i) = 1+2k(1+i)$ for some integer $k$, so that $a+bi \equiv 1\pmod{2}$. As for your second question, $4 = -2(i-1)(i+1) = (1-i)(2+2i)$, so $4\mid b$ implies that $2+2i\mid b$ and thus $bi\equiv 0\pmod{2+2i}$. For your final question, I think the correct statem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the exp... | Write $a:=\arctan x$ etc. so $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan a+\tan b + (1-\tan a\tan b)\tan c}{1-\tan a\tan b - (\tan a+\tan b)\tan c}.$$If $xy\ne 1$, we can cancel $1-\tan a\tan b$ to get $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan (a+b)+\tan c}{1-\tan (a+b)\tan c}.$$If $\tan (a+b)\tan c\ne 1$ i.e. $(x+y)z\ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the residue theorem to evaluate an integral $$I = \int_0^{\infty} \frac {dx}{x^6+1}$$
My thinking is that I can use the property of even functions to integrate across the whole domain from negative infinity to positive infinity.
Can the residue theorem be used to evaluate this integral?
Can you use a pie-shaped... | First approach:
$$\begin{eqnarray*}\int_{\mathbb{R}^+}\frac{dx}{x^6+1}&=&\frac{1}{2}\int_{\mathbb{R}}\frac{dx}{x^6+1}=\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\operatorname*{Res}_{z=\xi}\frac{1}{z^6+1}\\&=&\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\frac{1}{6\xi^5}=-\frac{\pi i}{6}\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A functional equation (another) I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $x\in [1/2,1]$
$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right).$$
I am actually not sure about existence of a solution.
| $$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right) \tag 1$$
As already pointed out this implies $f(1)=\frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The distributive law $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
| Do it in steps.
You accept that $M(x+y) = Mx + My$
So replace $M$ with $(x+y)$ and you get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y$
Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y=$
$(x+y)A + (x+y)B=$
$xA + yA + xB +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$
Finding sum of series $$\displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$
Try: Let $$S = \displaystyle \sum^{\infty}_{k=1}\frac{k^2}{(2k-1)(2k)(2k+1)(2k+2)}$$
So, $$S =\sum^{\infty}_{k=1}\frac{k^2\cdot (2k-2)!}{(2k+2)!}=\frac1{3!}\... | Applying partial fractions along with the digamma function, we get that:
\begin{align*}
\sum_{k=1}^{\infty} \frac{k}{2\left ( 2k-1 \right )\left ( 2k+1 \right )\left ( 2k+2 \right )} &= \sum_{k=1}^{\infty} \left ( \frac{1}{24\left ( 2k-1 \right )}+ \frac{1}{8\left ( 2k+1 \right )} -\frac{1}{12\left ( k+1 \right )} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Method to factor an expression As the title says, i want to factorize an expression, but i don't have any clue how to proceed.
Here is the expression :
$2x² -7x +3$
And here is the factorized form :
$(x-3)(2x - 1)$
My question is, which method or rule to use to go from first to second ?
please note that I am a beginner... | In general ,if $x_1$ and $x_2$ are roots of $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$
So your case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate the area of trapezoid. In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $\angle BAD=\theta,$
$BD=h$,
$\angle ABD=90^\circ$
$\angle CBD=90^\circ-\theta$
$CD=2$because trapezoid is isosceles
Apply cosin... | I've got $3\sqrt 3$ using a slightly different method. $\angle BAD=\theta$, $\angle ABD =90^\circ$ means $\angle CBD=\angle BDA=90^\circ-\theta$. Since the trapezoid is isosceles, $\angle CDA=\theta$, and you can get $\angle CDB=2\theta -90^\circ$. Since $BC=AB=CD$ you get $\angle CDB=\angle CDB$ or $$2\theta-90^\circ=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate the integral $\int_{-1}^1 \frac{1}{(2-x)\sqrt{1-x^2}}\,dx$ I started with the substitution $x=\sin(\theta)$ with $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$
And I got the integral $$\int_{-\pi/2}^{\pi/2} \frac{1}{2-\sin(\theta)}\,d\theta$$
which I am again not able to evaluate.
Even integration by parts is not w... | Here is a slightly different approach.
Let
$$I = \int_{-1}^1 \frac{dx}{(2 - x) \sqrt{1 - x^2}} \qquad (1)$$
Enforcing a substitution of $x \mapsto - x$ gives
$$I = \int_{-1}^1 \frac{dx}{(2 + x) \sqrt{1 - x^2}} \qquad (2)$$
Adding (1) to (2) one obtains
$$I = 2 \int_{-1}^1 \frac{dx}{(4 - x^2) \sqrt{1 - x^2}} = 4 \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\int_0^\infty \ln(1+\frac{z}{\cosh(x)})dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$ and a closed form for $-1I observed graphically that $$f(z)=\int_0^\infty \ln\left(1+\frac{z}{\cosh(x)}\right)dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$$
Can anyone explain why this holds? I tried diffe... | Here is a more direct way.
The first thing to note is the function $f(z)$ is continuous for all $z > - 1$, including at the point $z = 1$.
From
$$f(z) = \int_0^\infty \ln (1 + z \operatorname{sech} x) \, dx,$$
on differentiating with respect to $z$ we have
\begin{align}
f'(z) &= \int_0^\infty \frac{\operatorname{sech} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Extract the Square Root of $ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $
Extract the Square Root of $$ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $$
I found a method to do... | $$a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)+\frac{41a}{16}-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+1$$
For convenience, let $x=\sqrt a$. Then we get
$$x^4-\dfrac 32x^3 +\frac{41}{16}x^2-\dfrac 32x+1$$
Use of the square root algorithm would produce this.
\begin{array}{r}
&& x^2 & -\dfrac 34x & +1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{... | You could have continue your way.
Starting from
$$\sum_{n=0}^\infty a_nx^{n+2}+\sum_{n=0}^\infty a_nx^{n+1}+\sum_{n=0}^\infty a_nx^n=1$$ you already found $a_0=1$ and $a_1=-1$.
Now, consider a given degree $m >1$. You then have the recurrence relation
$$a_{m-2}+a_{m-1}+a_m=0$$ and the roots of the characteristic equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove Nice Determinant Equations I often come across these kinds of problems in A-level exam papers:
Prove that
$$
\begin{vmatrix}
(a+b)^2 & 1 & 1 \\
a^2 & (1+b)^2 & a^2 \\
b^2 & b^2 & (1+a)^2
\end{vmatrix}
= 2ab(1+a+b)^3
$$
or
Prove that
$$
\begin{vmatrix}
1 & 1 ... | By subtracting the first column from the second and the third we get that :
$$
\begin{split}
\begin{vmatrix}
1 & 1 & 1 \\
x & y & z \\
yz & xz & xy
\end{vmatrix}
&=\begin{vmatrix}
1 & 0 & 0 \\
x & y-x & z-x \\
yz & z(x-y) & y(x-z)
\end{vmatrix}\\
\\
&=(x-y)(z-x)\begin{vmatrix}
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$??? I saw in a paper by @Jack D'aurizio the following integral
$$I=\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$$
Below is my attempt.
$$I=4\int_0^{\pi/2}\log^2(\cos x)\mathrm{d}x$$
Then we define
$$F(a)=\int_0^{\pi/2}\log^2(a... | \begin{align}J=\int_0^{\frac{\pi}{2}} \ln^2\left(\cos x\right)\,dx\end{align}
Observe that,
\begin{align}I&=4J\\
J&=\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)\,dx\\
\int_0^{\frac{\pi}{2}} \ln\left(\sin x\right)\,dx&=\int_0^{\frac{\pi}{2}} \ln\left(\cos x\right)\,dx
\end{align}
(change of variable $y=\dfrac{\pi}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}... | The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.
$\sqrt{x^2+3}- \sqrt{x^2+2} = x(\sqrt{1+3/x^2}- \sqrt{1+2/x^2})=$
$= x(1+\frac{1}{2}\cdot(3/x^2)+O(1/x^4)- 1-\frac{1}{2}\cdot(2/x^2)+O(1/x^4))= \frac{1}{2x}+O(1/x^3)$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my questio... | Let $\dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=\dfrac{am+b}{m(m+1)}$
$\implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$
Set $2b=2\iff b=1,ar=-r\iff a=-1$
$$\implies\sum_{r=1}^n\dfrac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(f(r)-f(r+1)\right)=f(1)-f(n+1)$$
Set $n\to\infty$ and show that $n\to\infty f(n)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
Ho... | Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$ or $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and a... | Note that we have
$$\psi'(z)=\sum_{n=0}^\infty \frac{1}{(n+z)^2}$$
where $\psi'(z)$ is the derivative of the digamma function. Hence, we can write
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^2}=\frac19 \psi'(1/3)$$
and
$$\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac19 \psi'(2/3)$$
Interestingly, since we have
$$\sum_{n=0}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$AC^T = \det(A)I$ Let A = $\begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} &
\dots & a_{nn} \end{bmatrix}$
and the matrix of cofactors of $A$ is
$$C=\begin{bmatrix} C_{11} & C_{12} & \dots & C_{1n} \\ C_{21} & C_{2... | The expression
$$
\color{red}{a_{11}}C_{21} + \color{red}{a_{12}}C_{22} + \dots + \color{red}{a_{1n}}C_{2n}
$$
is the Laplace expansion along the second row of
$$
\begin{vmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\ \color{red}{a_{11}} & \color{red}{a_{12}} & \color{red}\dots & \color{red}{a_{1n}} \\ a_{31} & a_{32} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate: $\lim\limits_{x\to0^-} \frac1{\ln(1-x)}+\frac1x$ without LHR/Expansions
How to calculate $$\lim\limits_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)$$ without using L'Hopital, expansions nor integration?
I found the answer:
Using the Mean value theorem on:
$f(x)=e^x-\frac{x^2}2-x-1$
We get:
$0\le\frac{e... | Note that
$$\begin{eqnarray}
\lim_{x\to0^-} \left(\frac1{\ln(1-x)}+\frac1x \right)&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x\ln(1+x)} \\&= &\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}\cdot \lim_{x\to0^+}\frac{x}{\ln(1+x)}\\&=&\lim_{x\to0^+}\frac{x-\ln(1+x)}{x^2}.
\end{eqnarray}$$Now, let $$g(x) =\begin{cases}\frac{x-\ln(1+x)}{x},\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let $A= \begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}$. Find the entry in the first row and second column of $A^{2014}$ I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=\begin{pmatrix} 1&1 \\ -4&-1 \end{pmatrix}$
$D=\begin{pmatrix} 0&0 \\ 0&6 \end{pmatrix}$
$P^{-1}=\frac{1}{3}\begin{pmat... | Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=\left(\begin{bmatrix}1\\-1\end{bmatrix}\begin{bmatrix}8&2\end{bmatrix}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$? I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p \mid (a-b)$ or $p \mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ an... | If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=\frac{p+1}{2}$ for every prime p besides $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Minimising the Sum of distance of two line formed by joining some point on a line to two fixed points Consider the xy coordinate system.
Let $l$ be the line $y=-x$.
Consider two points $A=(0,2)$ and $B=(2,3)$
The Question asks to find a point $C$ on the line $l$ such that $|AC|+|BC|$ is minimum.
The only method I got ... | Let's see. We have $A = (0, 2)$, $B = (2, 3)$, and the point $C$ (as a function of $t$) $C(t) = ( t, -t)$. The Euclidean distance between two points is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, so the sum of distance from $A$ to $C$, and the distance from $C$ to $B$, as a function of $t$ is
$$\bbox{
L(t) = \sqrt{t^2 + (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many pairs $(x,y)$ such that $x+2=y$ have an $x$ or $y$ divisible by 5 and not 3? Let the set $S_{n}$ = {$(x,y):x,y \in \mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $\mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of p... | Integers to be avoided are multiples of $15$ so you have the following sequence for $g(\infty$): $$\{\color{red}{(23,25),(25,27)},\color{blue}{(33,35),(35,37)},\color{green}{(53,55),(55,57)},\cdots\}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find minimum value of $a^2+b^2$ Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coeffi... | $$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2-\frac{b^2}{4}x^2+(\frac{b}{2}x+1)^2=$$
$$=(x^2+\frac{a}{2}x)^2+x^2(2-\frac{a^2+b^2}{4})+(\frac{b}{2}x+1)^2$$
If $2-\frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. Th... | For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is... | Let $a=x+1$, $b=y+1$ and $c=z+1$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$3((x+1)^2+(y+1)^2+(z+1)^2+2(x-y+1))-(x+y+z+3)^2\geq0$$ or
$$x^2+y^2+z^2-xy-xz-yz+3x-3y+3\geq0$$ or
$$z^2-(x+y)z+x^2+y^2-xy+3x-3y+3=0,$$ for which it's enough to prove that
$$(x+y)^2-4(x^2+y^2-xy+3x-3y+3)\leq0$$ or
$$(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Distinct prime number solution of an equation I am trying to solve the following question:
$$\textrm{Find all distinct prime numbers } p, q \textrm{ and } r \textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
... | If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 \equiv 1 \pmod{5}$$
or $$r^2 \equiv -2 \pmod{5}$$
which has no solution. So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$. If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$. Now solv... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | $$a_n=\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$
$$=\frac{n^3(\sqrt{n^4+1}-n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$
$$=\frac{n^3}{(\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n)(\sqrt{n^4+1}+n^2)}$$
$$=\frac{1}{(\sqrt{1+\sqrt{1+\frac{1}{n^4}}}+\sqrt{2})(\sqrt{1+\frac{1}{n^4}}+1)}$$
Now, apply $\lim_n\... | {
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"source": "stackexchange",
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"answer_id": 1
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Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$. Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$.
So far I have that $n=m(m-3)(m+3)$, which are $3$ consecutive integers so at least one must be a multiple of 2. I am not sure how to get any further? Maybe use 2 cases?
... | This can be proved by induction. If $m=1$, then your number is $0$, which is a multiple of $6$. And if $m^3-3m^2+2m$ is a multiple of $6$; then, since\begin{align}(m+1)^3-(m+1)^2+2(m+1)-(m^3-3m^2+2m)&=3m^2-3m\\&=3m(m-1),\end{align}which is clearly a multiple of $6$, $(m+1)^3-(m+1)^2+2(m+1)$ is a multiple of $6$ too.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k... | Using the fact that $$\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$$
multiply both sides by $\ \displaystyle-\frac{2}{n4^n}{2n \choose n}$ then take the sum, we get,
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n}}{n^24^n}{2n \choose n}=-2\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(x^2)^n}{n4^n}{2n \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Infinite series problem
The sum of $$\frac{2}{4-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\cdots \cdots $$
Try: write it as $$S = \sum^{\infty}_{r=0}\frac{2^{2^{r}}}{2^{2^{r+1}}-1}=\sum^{\infty}_{r=0}\frac{2^{2^r}-1+1}{(2^{2^r}-1)(2^{2^r}+1)}$$
d not know how to solve from here, could some help me
to solve it, Thanks
| You may continue as:
$$\frac{2^n}{(2^n-1)(2^n+1)} =\frac{(2^n+1)-1}{(2^n-1)(2^n+1)} =\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}$$
Where $n = 2^r$. Now write the sum as:
$$\left(\frac 11 - \frac 13\right) + \left(\frac 13 - \frac 1{15}\right) + ... + \left(\frac 1{2^n-1} - \frac 1{(2^n-1)(2^n+1)}\right)$$
After cancellin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Compute $\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$ Compute $$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}\implies I = 3\int_0^{\infty}\frac {u}{u^6+1}du=\frac 32\int_0^\infty \frac {1}{u^3+1}du=\frac 32 \int_0^\infty \frac {1}{(x+1)(x^2+x+1)}dx\\\implies I=\frac 32 \int_0^{\infty} \frac 1{... | Yet another approach.
\begin{align}
\int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &\overset{\large x \mapsto x^{3/2}}=\frac{3}{2} \int_0^\infty \frac{dx}{1 + x^3}\\
&= \frac{3}{2} \int_0^1 \frac{dx}{1 + x^3} + \frac{3}{2} \int_1^\infty \frac{dx}{1 + x^3}.
\end{align}
Enforcing a substitution of $x \mapsto 1/x$ in the... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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On the evalution of an infinite sum
I wish to show that
$$\sum_{n = 0}^\infty (-1)^n \left [\frac{2n + 1/2}{(2n + 1/2)^2 + 1} + \frac{2n + 3/2}{(2n + 3/2)^2 + 1} \right] = \frac{\pi}{\sqrt{2}} \frac{\cosh \left (\frac{\pi}{2} \right )}{\cosh (\pi)}.$$
The reason I wish to find such a sum is as follows.
The question... | $$I=\int_0^\infty \frac{\sqrt{x} \cos (\ln x)}{x^2 + 1} \, dx$$
Here is an easy way to compute this integral.
Let's start with a classic integral:
$$\int_0^\infty \frac{y^{a-1}}{1+y}dy=\frac{\pi}{\sin(\pi a)};0<a<1$$
(First calculated by Euler)
The change of variable $y=x^2$ gives
$$I(a)=\int_0^\infty \frac{x^a}{1+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088089",
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"source": "stackexchange",
"question_score": "21",
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Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
| Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+\lambda(a^2+b^2+c^2-1) \tag 1$$ Computing the partial derivatives
$$\frac{\partial F}{\partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a \lambda \tag 2$$
$$\frac{\partial F}{\partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b \lambda\tag 3$$
$$\frac{\par... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089074",
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"source": "stackexchange",
"question_score": "2",
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Where to start with: $\lim_{n\to\infty} (\sqrt[3]{n^{48}+n} - \sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$ I have limit:
$\lim_{n\to\infty} (\sqrt[3]{n^{48}+n} - \sqrt[3]{n^{48}+n^2}) ((n^3 +3)^{12} - (n^4+4n)^9)$
I have to find that it is equal to -6 but I do not know how.
What I did was to get rid of cube root... | Expanding the 2nd term
$$((n^3+3)^{12}-(n^4+4n)^{9})\\
=((n^{36}+36n^{33}+594n^{30}+\cdots)-(n^{36}+36n^{33}+576n^{30}+\cdots))\\
=18n^{30}+\cdots$$
where the dots represent lower order terms. Expanding the 1st term
$$(\sqrt[3]{n^{48}+n}-\sqrt[3]{n^{48}+n^{2}})\\
=n^{16}(\sqrt[3]{1+n^{-47}}-\sqrt[3]{1+n^{-46}})\\
=n^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rewriting $2x^2-12x+11$ by completing the square I want to rewrite $2x^2-12x+11$ in square form. According to my book, the answer is $2(x-3)^2-7$. However, I get $2(x-3)^2+2$ for some reason.
My Steps:
$2x^2-12x+11$
$2x^2-12x=-11$
$2(x^2-6x)=-11$
$2(x^2-6x+9)=-11+9$
$2(x - 3)^2=-2$
$2(x-3)^2+2$
What am I doing wrong?
| An alternative way to see how this all works is to note that completing the square doesn't change the value of the expression. What you are essentially doing is adding a cleverly-written $0$ to it so that it becomes easier to work with.
Observe, $$\begin{align}2x^2-12x+11 & = 2x^2 - 12x \qquad\qquad\qquad\qquad\qquad\q... | {
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"source": "stackexchange",
"question_score": "2",
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$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
My working:
$$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\int_{0}^{1}\frac{\ln x}{x^2-1}dx=-\sum_{r\ge 1}\int_0^{1}x^{2r}\ln x\,dx =\sum_{r\ge 1}\frac {1}{(2r-1)^2}=
\frac{\pi^2}{8}$$
Is there any other appr... | \begin{align}J&=\int_{0}^{1}\frac{\ln x}{x^2-1}dx\\
&=\frac{1}{4}\int_{0}^{1}\frac{2x\ln(x^2)}{1-x^2}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
&=-\frac{3}{4}\... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Is there any other way to establish this trig identity? $\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $ I needed to verify this trig identity: $$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $$
what I did is I worked on both sides individually:
LHS: $$\frac{\frac1{\cos(x)} + 1}{\frac {\sin(x)}{... | Working both sides of
$$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)}$$
First multiply both sides by $\sin(x)$ giving
$$\frac{\tan(x) + \sin(x)}{\tan(x)} = \frac{\sin^2(x)}{1 - \cos(x)}$$
Since $\sin^2(x) = 1-\cos^2(x)$ we immediately have
$$1 + \cos(x) = 1 + \cos(x)$$
| {
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"source": "stackexchange",
"question_score": "4",
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mes... | Hint:
It is a telescoping sum since, for any $k>0$,
$$\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \time... | Nice observation!
It is indeed true that $ \sum_{k=0}^{n - 1} (2 k + 1) (n - k) = \sum_{k=1}^{n} k^2$.
This can be proved by using the formulas for $\sum 1, \sum k, \sum k^2$.
| {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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If $y=\frac {a+bz}{c+dz}$, $z=\frac{a+bx}{c+dx}$, $x=\frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$ I need to solve this problem and I don’t know how.
If $x, y, z$ are unequal and $y = \frac {a + bz}{c + dz}, z = \frac {a + bx}{c + dx}, x = \frac {a + by}{c + dy},$ then $ad + bc + b^2 + c^2 = 0$.
I see that I nee... | Substitute formula for x into formula for z we have
$$z = \frac{ac + ady + ab + b^2y}{c^2 + cdy + ad + bdy}$$
Substitute this into formula for y we have
$$(c^3 + acd)y + (c^2d + bcd + ad^2 + b^2d)y^2 = (ac^2 + a^2d + abc + ab^2) + (bad + b^3)y$$
$$(c^3 + acd - abd - b^3)y + d(c^2 + bc + ad + b^2)y^2 = a(ad + bc + b^2 +... | {
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Find the power series expansion of $f(x)= \frac{e^x - 1}{x}$
Prove that the following function is analytical in 0 and find its power series centered in 0
$$f(x)= \frac{e^x - 1}{x}, f(0)=1$$
I'm trying to write $f$ as some kind of combination of function with known Taylor series expansion in an open interval around ze... | You know that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot$, so
$$
e^x- 1= \sum_{n=1}^\infty \frac{x^n}{n!}= x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot.
$$
thus
$$
\frac{e^{x}- 1}{x}= \sum_{n= 1}^\infty \frac{x^{n-1}}{n!}=\su... | {
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"url": "https://math.stackexchange.com/questions/3101571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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This sum $\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}g(n)$ and the golden ratio We got this strange sum?:
$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\left[8\cdot\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\frac{(n-\alpha)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(\alpha+1)^{1/5... | This solution is only the prooving of the statement, I did not look for $f(a), f(b)$.
Let's split the original sum into two pieces:
$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n}}\frac{(n-\alpha)^{1/5}}{(n+1)^2}\tag1$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Why is $\sum_{r=1}^n r^{\overline{a+1}}(r+b)=\big(\sum_{r=1}^n r^{\overline{a+1}}\big)\big(An+B\big)$?
Why is
$$\sum_{r=1}^n r^{\overline{a+1}}(r+b)=\bigg(\sum_{r=1}^n r^{\overline{a+1}}\bigg)\bigg(An+B\bigg)=\frac{n^{\overline{a+2}}}{a+2}\cdot (An+B)$$
where $A, B$ are rational numbers?
?
It is clear that LHS... | $$\begin{align}
\sum_{r=1}^n r^{\overline{a+1}}(r+b)
&=\sum_{r=1}^n r^{\overline{a+1}}(r+a+1)+ r^{\overline{a+1}}(b-a-1)\\
&=\sum_{r=1}^n r^{\overline{a+2}}+(b-a-1)\sum_{r=1}^n r^{\overline{a+1}}\\
&=\frac {n^{\overline{a+3}}}{a+3}+(b-a-1)\cdot \frac {n^{\overline{a+2}}}{a+2}\\
&=(n+a+2)\cdot \frac {a+2}{a+3}\cdot \fra... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of $x+y$ if x and y are co-primes and $PR=\dfrac{x}{y}$ in the diagram?
$ST$ is the perpendicular bisector of $PR$ and $SP$ is the angle
bisector of $\angle QPR$. If $QS=9cm$ and $SR=7cm$ then
$PR=\dfrac{x}{y}$ where x, y are co-primes. $x+y$=?
Source: Bangladesh Math Olympiad 2015 junior categ... | Here's a solution with less algebra:
Note first that $R$ is the mirror image of $P$ across $ST$. Consider now the point $Q'$ obtained by reflecting $Q$ across $ST$. We obtain an isosceles trapezoid $PQQ'R$, which is a cyclic quadrilateral. We know that $|PS| = |RS| = 7$ and $|QS| = |Q'S| = 9$, and that $\angle QPS = \... | {
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"source": "stackexchange",
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Find upper and lower bounds for the finite sum Find upper and lower bound for the following finite sum:
$\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$
My attempt is:
Using the integral test:
we know that $\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$ = $\sum_{i=1}^n 1/i^3$ = $... | You do not know that $\sum_{i=1}^n 1/i^3= \int_1^n1/x^3\, dx$ as the equality is not exact. In fact $$\int\limits _1^{n+1}\frac{1}{x^3}\, dx \le \sum_{i=1}^n \frac{1}{i^3}\le 1+ \int\limits_1^n \frac{1}{x^3}\, dx $$
and using your earlier analysis, this becomes $$\frac12 - \frac{1}{2(n+1)^2} \le \sum\limits _{i=1}^n ... | {
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Given that A, B and C do not lie on the same line...find the area of triangle ABC.
\begin{array} { c } { \text { Given that } A , B \text { and } C \text { do not lie on the same line.If } \vec { O A } + \vec { O B } + \vec { O C } = 0 , | \vec { O A } | = \sqrt { 6 } \text { , } } \\ { | \vec { O B } | = 2 \text { an... | Wlog $B=(2,0)$, $A=(x,y)$ with $x^2+y^2=6$, $C=(u,v)$ with $u^2+v^2=14$. Moreover, $x+u+2=0$ and $y+v=0$. This allows us to eliminate $u,v$:
$$ \begin{align}x^2+y^2&=6\\(x+2)^2+y^2&=14\\\implies\qquad 4x+4=14-6\end{align}$$
and thereby
$$ x=1,\;u=-3,\;y=\pm\sqrt 5,v=\mp\sqrt 5.$$
From $y+v=0$, we conclude that $BC$ int... | {
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"source": "stackexchange",
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Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| You can always pass to polar coordinates to compute 2d limits:
Set $x=r \cos\theta , y=r\sin \theta$, and now we need to compute
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)}$$
Since this is a limit for the variable $r$, you can use L'Hôpital taking derivatives w.r.t. $r$. Note that if we... | {
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} |
Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$ Let $a,b,c$ be reals with $a+b+c=1$. Show that :
$$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$
I have tried to something like this:
$$18a^4-24a^3+6a^2-12a+12\geq 0$$
$$18b^4-24b^3+6b^2-12b+12\geq 0$$
$$18c^4-24c^3+6c^2-12c+12\g... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$
Thus, since our inequality is fourth degree, we see that it's equivalent to $f(w^3)\geq0,$
where $f$ is a linear function.
But the linear function gets a minimal value for an extreme value of $w^3$,
which happens for equality case of two vari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^... | By your work $$a+b+c\leq13,$$ but since
$$3(a+b+c)^2=2(a^3+b^3+c^3)=2(a^3+b^3+x^3-3abc+3abc)=$$
$$=2(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+6abc,$$ we obtain that $a+b+c$ is divisible by $3$.
Thus, we see that $$a+b+c\leq12.$$ But $$(a,b,c)=(3,4,5)$$ is valid, which says that $12$ is a maximal value.
Actually, your inequality we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$... | You can draw a line from your point to the plane and then find the length of its projection onto the unit normal vector to the plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Solving the equation $2^x+2^{-x} = 10$ I'm trying to figure out how to solve this equation in terms of $x$.
Mistake I made so far is:
$$\log(2^x) + \log(2^{-x}) = \log(10)$$
$$x \cdot \log(2) - x\cdot \log(2) = 1$$
$$0 = 1$$
| Let $2^x=y$
The equation becomes: $y^2-10y+1=0$. This quadratic equation has two solutions:
$$y = \frac{10 \pm \sqrt{100-4}}{2} = 5 \pm 2\sqrt{6}.$$
So you get $x = \log_2(5+2\sqrt{6})$ and $x = \log_2(5-2\sqrt{6})$ as solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find some digits of $17!$
$17!$ is equal to $$35568x428096y00$$
Both $x$ and $y$, are digits. Find $x$ and $y$.
So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$
If there`s a product of $(2\times 5)^3$
Then this nu... | The alternating sum of digits must be divisible by $11$, i.e., $11\mid 18-x$. It follows that $x=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$. I am new to multivariable limits and I am having a hard time arguing this limit:
Give a $\delta$-$\varepsilon$ proof that $f(x,y)=\sqrt{x+y-2}$ is continuous at $(2,1)$.
I can work out the overarching context of the proof as foll... | Let $\varepsilon > 0$ and set $\delta = \min\left\{\frac{\varepsilon^2}{6}, \frac12\right\}$.
Let $\|(x,y) - (2,1)\| < \delta$. This implies $|x-2| < \delta$ and $|y-1| < \delta$. Also
$$x+y-2 = 1 + (x-2) + (y-1) \ge 1-|x-2|-|y-1| > 1-2\delta \ge 0$$
so $\sqrt{x+y-2}$ is well-defined.
We have
$$|x+y-2| \le |x-2| + |y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please... | For $a=b=c=\frac{1}{3}$ we get the value $\frac{7}{27}.$
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(a+b+c)(ab+ac+bc)-2abc\leq\frac{7(a+b+c)^3}{27}$$ or
$$\sum_{cyc}(7a^3-6a^2b-6a^2c+5abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(a^2b+a^2c-2abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Triangle Geometry question find minumum value of $n-m$ In the triangle shown, for $\angle A$ to be the largest angle of the triangle, it must be that $m<x<n$. What is the least possible value of $n-m$, expressed as a common fraction?
I found that $x<4.5$ and $x>1.6$ so I thought the answer was $4.5-1.6=\frac{29}{10}$ ... | By triangle inequality we know that
1) $3x + (x+4) > x+9$
2) $(x+4)+(x+9) > 3x$
3) $(x+9) + 3x > x+4$.
1) gives us $3x > 5$ or $x > \frac 53$.
2) gives us $x < 13$
And 3) gives us $3x > -5$ which is a no brainer as $x > 0$.
The fact that A is the largest angle gives us that $x+9$ is the largest side. Obviously $x + 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c... | Consider three cases.
*
*$a=\max\{a,b,c\}$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)\geq0;$$
*$b=\max\{a,b,c\}$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
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If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite
If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite.
My attempt so far:
$$a^2+b^2+c^2+d^2$$
$$=a^2+d^2+2ad+b^2+c^2+2bc-2ad-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+bc+c^2)-2bc$$
$$=(a+d)^2+(b+c)... | Firstly, if at least one of $a,b,c,d$ is greater than one than
$$a^2+b^2+c^2+d^2\gt a+b+c+d$$
$$\therefore a-b-c+d\gt 1$$
If they are all equal to $1$, then $1^2+1^2+1^2+1^2=4$ which is composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor I have to calculate some limits and try to solve them in use of taylor.
$$
\lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right)
$$
In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace i... | Proceeding with your substitution, since $\log \left( 1+y\right) =y-\frac{1}{
2}y^{2}+O\left( y^{3}\right) $, we have:
\begin{eqnarray*}
\frac{1}{y}-\frac{1}{y^{2}}\log \left( 1+y\right) &=&\frac{1}{y}-\frac{1}{
y^{2}}\left( y-\frac{1}{2}y^{2}+O\left( y^{3}\right) \right)
&=&\frac{1}{2}+O\left( y\right) \overset{y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum value of $PA+PB$ is If $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$
$PA+PB=\sqrt{(2\... | Note that
$$
|PA|+|PB|=3\left(\sqrt{(\lambda-4)^2+1}+\sqrt{(\lambda-2)^2+1}\right).
$$ The expression in the parenthesis is equal to the sum of distance between $X=(\lambda,0)$ and $Y=(4,1)$ and distance between $X=(\lambda,0)$ and $Z=(2,-1)$. Then, by the triangle inequality, $|XY|+|XZ|
$ is minimized when $X$ is on t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve the following problem by using the Binomial formula If $n=10$ and $p=0.60$ find $P(X\geq 3)$.
The formula we've been given: $P(X=x) = \binom{n}{x} \cdot p^x \cdot q^{n-x}$, where $n$ is the number of trials, $p$ is the probability of success and $q = 1-p$.
I know that I could do $P(X\geq 3) = \binom{10}3 \cdot... | You have exchanged $p$ and $1-p$. As you said $p=0.6$. Therefore
$P(X\leq 2)={10 \choose 2} \cdot 0.6^2 \cdot 0.4^8 + {10 \choose 1} \cdot 0.6^1 \cdot 0.4^9 + {10 \choose 2} \cdot 0.6^0 \cdot 0.4^{10}=0.01229$
Thus $P(X\geq 3)=1-P(X\leq 2)=1-0.01229=0.98771$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometry high school math competition question
Let $ABC$ be an equilateral triangle with side length $2$.
Let the circle with diameter AB be $\gamma$.
Consider the two tangents from $C$ to $\gamma$, and let the tangency point closer to $A$ be $D$.
Find the area of triangle $CAD$.
I was able to figure out that $... | Let $O$ be a center of the circle $\Gamma$.
Thus, by the Pythagoras's theorem
$$CD^2=CO^2-DO^2=\left(\sqrt3\right)^2-1^2=2$$ and $$CD=\sqrt2.$$
Id est, $$S_{\Delta ACD}=\frac{1}{2}CD\cdot AC\sin\measuredangle ACD=$$
$$=\frac{1}{2}\cdot\sqrt2\cdot2\sin\left(\arctan\frac{1}{\sqrt2}-30^{\circ}\right)=$$
$$=\sqrt2\left(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 +... | One example is 729+492=1221
a=2,b=9
7 a b
+4 b a
1 a 2 1
Hint: b+a is not equal to 1. So b+a=11 hence the solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3145603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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If $85_b = 58_c$, what is the smallest possible value of $b$? I have searched and watched videos online and can't find a method to solve this problem:
If $85_b=58_c$ for some positive integer bases $b$ and $c$, what is the smallest possible value for $b$?
| $85_b = 58_c$ means $8b + 5 = 5 c + 8$. Thus, $b = (3+5c)/8$. This is an integer whenever $c = 8n + 1$ for some integer $n$. In this case, $b = 5n+1$. Now, since $8$ appears as a digit, we must have $b > 8$. The smallest integer of the form $5n+1$ that is larger than $8$ is $11$.
So the smallest solution is $b = 11$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Easy exponential equation but… $$\left(\sqrt{5+2\sqrt6}\right)^x-\left(\sqrt{5-2\sqrt6}\right)^x=96$$
I was trying to use that $$(5+2\sqrt6)(5-2\sqrt6)=1$$ but I end up with $$t^2-98t+1=0$$ which has some ugly solutions and I think I should have some nice ones there. Thx
| $$\left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x = 96$$
Take natural log of both sides
$$\ln \left[ \left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x \right]= \ln96$$
Use the $\ln(a+b) = \ln(a) + \ln(1+b/a)$ property
$$\ln \left(\sqrt{5+2\sqrt{6}}\right)^x + \ln \left(1-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of t... | Let $a=b=c=\frac{2}{3}$. Thus, $P=8.$
We'll prove that it's a minimal value of $P$.
Indeed, we need to prove that
$$\sum_{cyc}(4a^3+5abc)\geq(a+b+c)^3$$ or
$$3\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is true by Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Product of pairwise sum is perfect square For which $n$ can we divide $1,2,\ldots,2n$ into $n$ pairs so that the product of the sum of the $n$ pairs is a perfect square?
If $n$ is even, this is possible: match the first number with the last, second number with the second-to-last, etc., so every pair has sum $2n+1$ and ... | You're almost done just take for $n$ odd $(1+5)(2+4)(3+6)(7+2n)(8+2n-1)\cdots=6^2\cdot 3^2\cdot (2n+7)^{n-3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\int_0^{n^2} \left\lfloor\sqrt t\right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$ I want to prove
$$\int_{0}^{n^{2}} \left\lfloor \sqrt{t} \right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$$
Is it correct to say that $\left\lfloor \sqrt{t} \right\rfloor=\sqrt{(k-1)}$ and $(k-1)^{2} < t < k^{2}$, where
$$\int_{0}^{n^{2}} \... | You have
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} \int_{k^2}^{(k+1)^2} \lfloor \sqrt{t} \rfloor \mathrm{dt}$$
But if $k^2 < t < (k+1)^2$, then $\lfloor \sqrt{t} \rfloor = k$. So
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} k((k+1)^2-k^2) = \sum_{k=0}^{n-1} k(2k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all positive powers. Determine all positive powers of the $5 \times 5$ matrix...
$$C=\begin{bmatrix}
0&0&0&0&0 \\
1&0&0&0&0 \\
0&1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&1&0
\end{bmatrix}$$
I know that taking powers of $C$ just shifts each $1$ down a row, replacing each $1$ with a $0$. The right most $1$ is shifted out ... | $C$ is the adjacency matrix of a directed graph $5 \to 4 \to 3 \to 2 \to 1$. $(C^k)_{ij}$ is then the number of paths of length $k$ starting at $i$ and ending at $j$. This is $1$ if $i-j = k$, otherwise $0$. Of course if $k \ge 5$ all entries are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the functio... | With
$y = Ax^2 + Bx + C, \tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, \tag 2$
viz,
$y' = 2Ax + B, \tag 3$
$y'' = 2A, \tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; \tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, \tag 5$
from which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to justify the convergence of series $\left(\frac{1}{2}\right)^{2(2k-1)}$? I see this formula somewhere in a book, though the book doesn't provide the justification.
$$
\sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2(2k-1)}
= \frac{4}{15}
$$
Any clue would be appreciated.
| Here is a way you may understand:
For any $x\in \Bbb R$ s.t. $|x|<1$ we have $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$
Proof is here.
Now in your case, let $\displaystyle S=\sum_{k=1}^\infty \frac{1}{4^{2k-1}}=\frac{1}{4}+\frac{1}{4^3}+\frac{1}{4^5}+\dots=\frac{1}{4}(1+\frac{1}{4^2}+\frac{1}{4^4}+\dots)=\frac{1}{4}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x -... | $ t = \sqrt{3-x}\geq 0$, then we get $$\sqrt{5-t^2}-t=t^4$$
so $$5-t^2 = t^2+2t^5+t^8$$
so we have $$t^8+2t^5+2t^2-5=0$$
Since $f(t)=t^8+2t^5+2t^2$ is strictly increasing for positive $t$, given equation has at most one real (positive) solution and that is $t=1$, or $x= 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
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minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is
what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please
| Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)\in P$ for $t\in\Bbb R$, then we get $... | {
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Prove that there exists an integer $ x$ such that $ \frac{x^2-2}{p}$ is the square of an integer
Let $ p\equiv -1\pmod 8$ be a prime number. Prove that there
exists an integer $ x$ such that $ \frac{x^2-2}{p}$ is the square of
an integer
This problem is from :
AOPS,there are some In the discussion, there was one ... | Let $p \equiv 7 \pmod 8$ be prime. Let $u,v > 0$ give the smallest positive solution to $u^2 - p v^2 = 1.$ Indeed $u \geq 3.$ Then $u^2 - 1 = p v^2,$ so $(u+1)(u-1) = p v^2.$ As $\gcd(u+1, u-1)$ is either $1$ or $2,$ we have four possibilities with some $\gcd(g,h) = 1:$
(I)
$$ u+1 = p g^2 \; , \; \; u-1 = h^2 $$
Subtra... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Series $\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$ Alright, this is another problem that I have been stuck on. The goal is to determine whether it is convergent or divergent.
$$\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$$
So to start off, Integral Test seems rough as the denominator is not factorable for p... | For $n \ge 5$, $$\frac{n^2 - 5n}{n^3 + n + 1} > \frac{(n^2 + 2n + 1) - (7n + 7) + 6}{n^3 + 3n^2 + 3n + 1} > \frac{(n+1)^2 - 7(n+1)}{(n+1)^3} = \frac{1}{n+1} - \frac{7}{(n+1)^2}.$$ Since $$\sum_{n=5}^\infty \frac{1}{n+1}$$ does not converge but $$\sum_{n=5}^\infty \frac{7}{(n+1)^2}$$ is convergent, the original sum doe... | {
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"url": "https://math.stackexchange.com/questions/3167462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R... | Big wall of algebra incoming:
\begin{align*}
\frac{a\cdot2^{x}+b\cdot2^{-x}}{a+b}\le\cosh\left(x\ln2\right)&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+2^{-x}}{2}\\
&\implies\frac{a\cdot2^{x}+\frac{b}{2^{x}}}{a+b}\le\frac{2^{x}+\frac{1}{2^{x}}}{2}\\
&\implies\frac{a\cdot2^{2x}+b}{2^{x}\left(a+b\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Partial fraction decomposition of even function I need to do partial fraction decomposition of this function (to solve its integral):
$\frac{t^2}{t^4+4}$
Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$
And then cont... | Let $t=2u$
$$\dfrac{2u^2}{u^4+1}=\dfrac{u^2-1+u^2+1}{u^4+1}=\dfrac{1-1/u^2}{(u+1/u)^2-2}+\dfrac{1+/u^2}{(u-1/u)^2+2}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Bounded recursive sequence - Proof by induction
Given the sequence $(x_n)$ defined by
\begin{cases}
x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right),
\end{cases}
prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.
I verified the base case for $n=1$ and $n=2$.
Assumed that the boundaries hold f... | Let $a \gt 0$.
Exercise 1: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 = 2 \Leftarrow\Rightarrow a = \sqrt 2$.
Exercise 2: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 \ge 2$.
Hint: Apply the quadratic formula to a quartic polynomial.
Using the exercises, if $n \gt 1$ then $x_n \gt \sqrt 2 \gt 1$, and we get 'one half' (an... | {
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"url": "https://math.stackexchange.com/questions/3171307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $f(p) = (\frac{3p-3}{p^2-1}) + (\frac{1}{p-1}) \times \sum_{q=3}^{l(p), with \ q\ prime} (\frac{q+1}{q-1})$ is always <= 1? Let function $l(p)$ be defined as the largest prime number less than $p$.
For example: $l(7)=5$, $l(11)=7$, $l(17)=13$
Let the function $f(p)$ be defined as follows:
$$f(p) = \le... | The essence of this problem is finding a good upper bound for the sum over the primes. Let $\pi(n)$ denote the prime-counting function, i.e. $\pi(n)$ is the number of primes less than or equal to $n$, and let $H_n:=\sum_{k=1}^n\tfrac1k$ denote the $n$-th harmonic sum. Then
\begin{eqnarray*}
\sum_{\substack{q=3\\q\ \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Remainder of polynomial product, CRT solution via Bezout Given:
$$f(x) \pmod{x^2 + 4} = 2x + 1$$
$$f(x) \pmod{x^2 + 6} = 6x - 1$$
Define r(x) as:
$$f(x) \pmod{(x^2 + 4)(x^2+6)} = r(x)$$
What is $r(4)$?
The 3 equations can be restated as quotient · divisor + remainder:
$$f(x) = a(x)(x^2 + 4) + 2x + 1 $$
$$f(x) = b(x)(x... | Hint
Observe that $\gcd(x^2+4, x^2+6)=1$ and
$$\frac{1}{2}(x^2+6)-\frac{1}{2}(x^2+4)=1.$$
Now apply Chinese remainder theorem to the system
\begin{align*}
f(x) & \equiv 2x+1 \pmod{x^2+4}\\
f(x) & \equiv 6x-1 \pmod{x^2+6}
\end{align*}
To get something like:
$$f(x) \equiv \underbrace{(2x+1)(\ldots) + (6x-1)(\ldots)}_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$... | The issue :
when you write out $\sin x = \cos x$ implies $\sin^2 x = \cos^2 x$, there is no problem with the implication. However, an implication leads to a weaker statement, which may admit "extraneous" solutions. For example, the above implication is got by multiplying by squaring both sides, but then that ends up i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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.