Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The largest integer dividing these $(n+19)(n+18)(n+17)(n+16)$
The largest integer that divides $(n+19)(n+18)(n+17)(n+16) ~ \forall ~ n \in \mathbb{N}? $ is?
It struck me that we should write this as:
$\dbinom{n+19}{4}\times 4!$
which is always divisible by $4!$. Now I am confused as to how do I take it from here? ... | Let $f(n) = (n + 16)(n + 17)(n + 18)(n + 19)$.
Then we have that
$$
\gcd(f(1), f(4), f(5)) = \gcd(17 \cdot 18 \cdot 19 \cdot 20, 20 \cdot 21 \cdot 22 \cdot 23, 21 \cdot 22 \cdot 23 \cdot 24) = \gcd(2^3 \cdot 3^2 \cdot 5 \cdot 17 \cdot 19, 2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 23, 2^4 \cdot 3^2
\cdot 7 \cdot 11 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the system of linear equations for all values of a When solving the following matrix for $a$:
$$ \left[
\begin{array}{ccc|c}
x&2y&az&3\\
(2a-4)x&(2-a)y&(-2a+4)z&0\\
4x&4y&2az&8
\end{array}
\right]
$$
When do I put restrictions on which values $a$ can take? Say I divide the second row by $(2-a)$, does this ... | I didn't check the detail of calculation but, if it is correct, the solution you have found is valid for $a\neq 2$ (i.e. the system has exactly a solution $\forall a\neq2$).
Now you need also to verify the solution/s for the case $a=2$ in the original system, that is
$$\left[
\begin{array}{ccc|c}
1&2&2&3\\
4&4&4&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose... | Note that for $x\ge0$, $1+x\le x^2+x+1\le(x+1)^2$.
Hence, we have
$$\frac13<\frac12=\int_0^1 \frac{1}{(1+x)^2}\,dx\le \int_0^1 \frac{1}{x^2+x+1}\,dx\le \int_0^1 \frac{1}{1+x}\,dx=\log(2)<\frac\pi4$$
which provide tighter bounds then those that were requested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4... | The entries are all between 0 and 1 and in each row the entries add up to 1. They are called stochastic matrices. They have 1 as an eigen value with $(1,1,\ldots,1)^T$ as corresponding eigenvector. This property is inherited by all the powers of that matrix.
| {
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"url": "https://math.stackexchange.com/questions/2699410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 1
} |
Given the function $f(x)=\frac {1}{\sqrt[3] {1-x^3}}$ find $\underbrace {f(f(\cdots f(19)\cdots))}_{95}$ Given the function $$f(x)=\frac {1}{\sqrt[3] {1-x^3}}$$ find $$\underbrace {f(f(\cdots f(19)\cdots))}_{95}$$
My try:
Define $$f^n(x)=\underbrace {f(f(\cdots f(x)\cdots))}_{n}$$
We see that $$f(x)=\frac {1}{\sqrt[3]... | As you have already said there are only 3 interactions, then it goes back to the original function:
\begin{align}
f_0 = \underbrace{f(x)}_1&=\frac{1}{\sqrt[3] {1-x^3}} = \left(\frac{1}{1-x^3}\right)^{\frac{1}{3}}\\
f_1 = \underbrace{f(f(x))}_2&= \left(\frac{1}{1-\left(\frac{1}{1-x^3}\right)}\right)^{\frac{1}{3}} = \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Algebraic Proof on equivalence $$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$
Demonstrate the identities above considering a, b & c real numbers and distinct from each other.
I'm stucked on this problem. Take a look on what I did:
$x=(a-b)\therefore -x=(b-a)$; $y=(a-c);-y=(c-a)$;... | Assuming $b\neq c$ the function
$$ f(z) = \frac{z^3}{(z-b)(z-c)}+\frac{b^3}{(b-c)(b-z)}+\frac{c^3}{(c-z)(c-b)} $$
is holomorphic over $\mathbb{C}\setminus\{b,c\}$. In the worst scenario $b$ and/or $c$ are simple poles, but $\operatorname*{Res}_{z=c}f(z)=\operatorname*{Res}_{z=b}f(z)=0$ imply that $f$ is entire (once it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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On $3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$ and friends Let $\phi$ be the golden ratio. We know it has a beautiful infinite nested radical,
$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\phi$$
However, it is also the case that,
$$3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$$
$$5+\sqrt{31+\sqrt{... | If we plug in the given formulas we get the famous formula. We note:
$$\sqrt{b_n+x}=x \Rightarrow x=\frac{1+\sqrt{1+4b_n}}{2}=\frac{1+F_n\sqrt{5}}{2};$$
Hence:
$$\frac{L_n-1}{2}+\frac{1+F_n\sqrt{5}}{2}=\frac{L_n+F_n\sqrt{5}}{2}=\phi^n.$$
This is a famous formula that relates the two sequences. See: https://en.m.wikiped... | {
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"url": "https://math.stackexchange.com/questions/2703402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$? The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?
I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place
My solution:
\begin{align}
\left(\frac{1+i}{1-i}.\frac{1+i}{1+i... |
The result $k=4$ is correct. The given multiple choice answer $k=2$ is wrong, since
\begin{align*}
\left(\frac{1+i}{1-i}\right)^2=i^2=-1\ne 1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
If $\gcd(m,15)=\gcd(n,15)=1,$ show that $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. If $\gcd(m,15)=\gcd(n,15)=1$ show $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$.
This is what I have so far but I'm stuck on essentially the last step.
Proof: Assume $\gcd(m,15)=\gcd(n,15)=1$. By the Euler's Phi Function: If $\gcd(a,m)=1$ th... | Since none of the existing answers is built-up on OP's work, I posted another one.
OP has proved $15\mid m^8-n^8 = (m^4-n^4)(m^4+n^4)$.
Since $15$ is a composite number, to apply Euclid's Lemma, we have to do so in two steps (on prime number $3$ and $5$).
Note that (without Fermat's Little Theorem)
*
*$a^2 \equiv 0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\... | $$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$
$$(1+x^2)y\,dy =-(1+y^2)x\,dx $$
If $x^2,y^2\ne-1$, above equation can be written as (after multiplying both side by $2$):
$$\frac{2y}{1+y^2}\,dy =-\frac{2x}{1+x^2}\,dx $$
$$\int \frac{2y}{1+y^2}\,dy =-\int \frac{2x}{1+x^2}\,dx $$
$$\ln (1+y^2)=-\ln (1+x^2)+\ln c=\ln \frac {c}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Use Ramanujan’s method to denest $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{7\sqrt[3]{20}-19}$ A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$x^3-1=1\implies x-1=\frac 1{1+x+x^2}=\frac 3{1+3x+3x^2+x^3}$$Multiply both sides by $9$ so$$9(x-1)=\left(\frac 3{1+x}\right)^3\implies\sqrt[3... | $x^3 +a x^2 +bx +c $ is called a Ramanujan cubic polynomial if
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
Its three roots then satisfy
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
Note that (2) contains nested cubic radicals of the form $\sqrt[3]{\sqrt[3] A -B}$ and it can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Help with a trigonometry problem. I have encountered some problems to solve the left side to the right.
$$ \cos^ 2x \sin x = \frac{\sin 3x + \sin x}{4}$$
I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$
Have t... | Consider
\begin{align}
\cos^2x\sin x&=(1-\sin^2x)\sin x\\
&=\sin x-\sin^3 x\\
&=\sin x-\frac14(3\sin x-\sin 3x) &(*)\\
&=\frac{\sin3x-\sin x}{4}
\end{align}
For $(*)$,
\begin{align}
\sin 3x&=\sin(x+2x)\\
&=\sin x\cos 2x+\sin 2x\cos x \\
&= \sin x (1-2\sin^2x)+2\sin x\cos^2x\\
&= \sin x-2\sin^3x+2\sin x(1-\sin^2x)\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $ $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $
Answer:
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \... | Note that
$$\frac{x^2+1}{x+1}-ax-b=\frac{x^2+1-ax^2-bx-ax-b}{x+1}=\frac{x^2(1-a)-x(a+b)-b+1}{x+1}$$
and in order to have limi zero we need
*
*$(1-a)=0 \implies a=1$
*$(a+b)=0\implies b=-1$
indeed
$$\frac{x^2(1-1)-x(1-1)-(-1)+1}{x+1}=\frac{2}{x+1}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Use Triangle Inequality to Solve Inequality $\left|x+\frac{1}{2}\right| > \frac{\sqrt{5}}{2}$ So this is a really simple question, but I can't seem to work it out. I started with the equation
$$
x^2 + x + 1 > 2
$$
and, by completing the square and taking the square root, was able to simplify it to
$$
\left|x+\frac{1}{2... | You shouldn't use the triangle inequality here. If $|x+\frac{1}{2}|>\frac{\sqrt{5}}{2}$, then you are really considering two inequalities:
$$
x+\frac{1}{2}>\frac{\sqrt{5}}{2} \text{ when } x+1/2\geq 0
$$
$$
-\left(x+\frac{1}{2}\right)>\frac{\sqrt{5}}{2} \text{ when } x+1/2< 0
$$
These two inequalities then read $x>\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713586",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the value of the constant c such that $\sum_{n=2}^\infty(1+c)^{-n} = 2$
Find the value of the constant c such that $$\sum_{n=2}^\infty(1+c)^{-n} = 2 $$
For this question,I'm not sure if I'm doing it right. If I am doing it right, I'm not sure how to get further. Here is what I have so far. Can anyone please help... | HINT: you must solve the equation $$\frac{1}{c(1+c)}=2$$ for $c$
this comes from $$\sum_{i=2}^n(1+c)^{-i}=\frac{(c+1)^{-n-1} \left((c+1)^n-c-1\right)}{c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively.
| The given rational function can be written as
$$(1+x^6+x^{12}+\ldots)(1+x^6+x^{12}+\ldots)(1+x^3+x^6+\ldots)(1+x^2+x^3+\ldots)\ .$$
The coefficient $c_{24}$ we are after therefore is the number of nonnegative integer solutions to
$$6k_1+6k_2+3k_3+k_4=24\ ,$$
or $6k_1+6k_2+3k_3\leq24$, which is the same as
$$2(k_1+k_2)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show $Z_{ab}\simeq Z_a\oplus Z_b$ by elementary row and column operations Let $a,b\in\mathbb N$ be coprime. By manipulating with the matrix $\operatorname{diag}(a,b)$, prove that the cyclic group $Z_{ab}$ is isomorphic to the direct sum $Z_a\oplus Z_b$.
I guess I should obtain the matrix $\operatorname{diag}(1,ab)$ fro... | You can certainly multiply rows and columns by $a$ and $b$, if all you do is add that multiple to another row or column. Less cryptically put:
$$
\det\begin{pmatrix}
1 & a \\ 0 & 1
\end{pmatrix} = 1
$$
so it is safe to multiply by it. First however, let $s,t\in\Bbb Z$ be so that $sa+tb=1$, which you can do because they... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Arrangement of definite integral in increasing order
If $\displaystyle I=\int^{\frac{\pi}{2}}_{0}\cos(\cos x)dx$ and
If $\displaystyle J=\int^{\frac{\pi}{2}}_{0}\sin(\cos x)dx$ and
If $\displaystyle K=\int^{\frac{\pi}{2}}_{0}\cos xdx$. Then Arrangement of $I, J,K$ in increasing order is
Try: Using $\sin x<x$ fo... | Let $\displaystyle x=\frac{\pi}{2}-u$. Then
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(\cos(x))dx=\int_{\frac{\pi}{4}}^0\cos(\sin u)(-1)du=\int_0^\frac{\pi}{4}\cos(\sin(x))dx$$
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(x)dx=\int_{\frac{\pi}{4}}^0\sin u(-1)du=\int_0^\frac{\pi}{4}\sin(x)dx$$
For $\displaystyle x\in\left(0,\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integration Problem: $\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw$ I know that:$$\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw=\frac{1}{{(x^2+2)}^{3/2}}$$
I've attempted to solve this with integration by parts $\int{fdg}=fg-\int{dfg}$
I let
$$\begin{array} ff=\sqrt{w} & dg=... | Hint
you have nearly with $w=u^2 \implies dw=2udu$ this:
$$I=\int \sqrt w e^{-w}dw=2\int u^2e^{-u^2}du$$
$$I=2\int u\times ue^{-u^2}du=\int u\times (-e^{-u^2})'du$$
Integrate by part once...
$$\int_0^{\infty} u\times (-e^{-u^2})'du=(u\times (-e^{-u^2}))|_0^{\infty}-\int_0^{\infty} (-e^{-u^2})du$$
And use gaussian integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$
I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\... | Based on DeepSea's hint, I've managed to solve by myself.
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integral part of sum of huge powers
Question: What is the integral part of the following expression?$$(a+\sqrt{b})^{2n}+(a-\sqrt{b})^{2n}$$
The question has specific values of $a=2,b=5$ and $2n=2016$.
I was able to simplify (or complexify) it to:$$\sum_{i=0}^n\binom{2n}{2i}a^{2i}b^{n-i}.$$
I think I need to use $$\bi... | Define $c_0 = 2$, $c_1 = 4$, and $c_{n + 2} = 4c_{n + 1} + c_n$ for $n \geqslant 0$, then$$
c_n = (2 + \sqrt{5})^n + (2 - \sqrt{5})^n. \quad \forall n \in \mathbb{N}
$$
Note that$$
\begin{pmatrix}c_{n + 2}\\c_{n + 1}\end{pmatrix} = \begin{pmatrix}4 & 1\\ 1 & 0\end{pmatrix} \begin{pmatrix}c_{n + 1}\\c_n\end{pmatrix}.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Is it possible to use partitions of an odd integer to generate primes in a given interval? We start with the partition of $N=5$.
$$5$$
$$4+1$$
$$3+2$$
$$3+1+1$$
$$2+2+1$$
$$2+1+1+1$$
$$1+1+1+1+1$$
Then we form the sum of squares (no limit on the number of elements) to get:
$$4^2+1^2=17$$
$$3^2+2^2=13$$
$$3^2+1^2+1^2=11... | One counterexample is when $p=31$ and $q=853.$ No partition of $31$ or $33$ yields the prime $q=853.$
For $p\geq 7,$ the only values greater than $(p-2)^2$ you can get from partitions of $p$ are:
$$(p-2)^2+2,(p-2)^2+4,(p-1)^2+1$$
This means that $31$ cannot be gotten from a partition of $p=7$, for example.
(This is ac... | {
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"url": "https://math.stackexchange.com/questions/2718875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to demonstrate this inequality? I have this statement:
If $x > 0 , y > 0$, prove that $\frac{1}{x}+\frac{1}{y} > \frac{2}{x+y}$
I will get the hypothesis? (That's what my teacher calls him, I do not know if it's correct)
$\frac{1}{x}+\frac{1}{y} - \frac{2}{x+y}> 0$
$\frac{x^2 + y^2 + xy + xy - 2xy}{(x+y)xy} > 0$
... | Use a better hypothesis, and get a better inequality.
\begin{align}
\frac{(x-y)^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 - 2xy + y^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 + 2xy + y^2}{(x+y)xy} &\ge \frac{4xy}{(x+y)xy} \\
\frac{(x+y)^2}{(x+y)xy} &\ge \frac{4}{x+y} \\
\frac{x+y}{xy} &\ge \frac{4}{x+y} \\
\frac{1}{x} + \frac{1}{y} &\ge \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\fra... | Move $a(x-y)$ to the LHS to make $F(x,y)=0$. Then:
$$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{-\frac{x}{\sqrt{1-x^2}}-a}{-\frac{y}{\sqrt{1-y^2}}+a}=\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}\cdot \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}},$$
because:
$$\begin{align}&\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$... | $$F=\sin^2A+\sin^2B+\sin^2C=1-\cos^2A+1-(\cos^2B-\sin^2C)$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
and $\cos(B+C)=\cos(\pi-A)=?$
$$2-F=\cos^2A+\cos(B+C)\cos(B-C)$$
$$=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\{\cos A-\cos(B-C)\}$$
$$=-\cos A\{\cos(B+C)+\cos(B-C)\}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Proving any integer cube is equal to a difference of two squares I know this problem has been asked before by someone. However, my problem is a bit different. I found the following problem in Elementary number theory by Burton:
Prove that the cube of any integer can be written as the difference of two squares. Notice ... | Well, what happens when you try?
$(n+1)^3 = n^3 + 3n^2 + 3n + 1 =$
$(\frac {n(n+1)}2)^2 - (\frac {n(n-1)}2)^2 + 3n^2 + 3n + 1$
Meanwhile $(\frac{(n+1)(n+2)}2)^2 - (\frac {(n+1)n}2)^2 =$
$(\frac{(n+1)n}2 + \frac {2(n+1)}{2})^2 - (\frac {n(n-1)}2 + \frac {2n}{2})^2=$
$(\frac {(n+1)n}2)^2 + (n+1)^2n +(n+1)^2 - (\frac {n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Divisibility property for sequence $a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$ Let $(a_n)$ be the sequence uniquely defined by $a_1=0,a_2=1$ and
$$
a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}
$$
Can anybody show (or provide a counterexample) that $p|a_{p-2}$ and
$p|a_{p-1}$ for any prime $p\geq 5$ ? I have checked this fact fo... | The sequence starts
$$0, 1, -5, 25, -105, 105, 5355, \dots$$
We can observe that the statement is true not only for primes, but for odd numbers in general. Even though recurrences might be better for solving this kind of problem, here we can go for closed form formula (luckily there is one!). I have used approach inspi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
} |
Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$ Let $a,b,c $ be positive real numbers such that $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$ then find $\frac{b+c}{a}$
So given $\sqrt{a}+9\sqrt{b}+44\sqrt{c}=\sqrt{2018(a+b+c)}$
So squaring both sides $(\sqrt{a}+... | By Cauchy Inequality,
$\sqrt{a}+9\sqrt{b}+44\sqrt{c}\le\sqrt{1^2+9^2+44^2}\sqrt{(\sqrt{a})^2+(\sqrt{b})^2+(\sqrt{c})^2}=\sqrt{2018(a+b+c)}$
with equality holds if and only if $\displaystyle \frac{\sqrt{a}}{1}=\frac{\sqrt{b}}{9}=\frac{\sqrt{c}}{44}$, i.e. $$a:b:c=1:81:1936$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Multiplier: Distance to the Origin Find the points on the curve $x^2+xy+y^2=2$ that are closest to the origin.
Is there a way to use Lagrange multipliers to answer this question?
| Using Lagrange multipliers, consider the function
$$f=x^2+y^2+\lambda \left(x^2+x y+y^2-2\right)$$ Compute the partial derivatives and set them equal to $0$ to get the equations
$$\frac{\partial f}{\partial x}=2x+\lambda (2 x+y)=0$$
$$\frac{\partial f}{\partial y}=2y+\lambda ( x+2y)=0$$
$$\frac{\partial f}{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Degree of splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$ I'm trying to find the degree splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$. After substituting $z= x^3$, and using the qudratic formula, and thensubstituting $x$ back in, I get the roots
$$\frac{\sqrt[3]{-t\pm\sqrt{t^2-4 t}}}{\sqrt[3]{2}},$$
b... | As @jyre says, the polynomial $X^6 + tX^3 + t$ is irreducible, so you can make a first field extension of degree $6$. $\Bbb F_3(t) \to k = \Bbb F_3(t)[x]/(x^6+tx^3+t)$.
Over this field $k$, $x$ is a root of $X^6 + tX^3 + t$ so the polynomial factors a bit. It turns out $x$ is a multiple root :
$X^6 + tX^3 + t = X^6 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$ I've been working on this interesting problem for a while already, and here it is:
In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.
I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart... | Are you sure? $p = \dfrac{7+15+8\sqrt{2}}{2} = 11+4\sqrt{2}\implies S^2 = p(p-15)(p-7)(p-8\sqrt{2}) = (11+4\sqrt{2})(11-4\sqrt{2})(-4+4\sqrt{2})(4+4\sqrt{2}) = (121-32)(32-16) = 89\cdot 16.$
So $S = 4\sqrt{89}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this se... | You have already shown that your $N$ works, I think it's all the notation that is giving you a headache. I would write like this (these are your words, but rewritten):
Given $\varepsilon$, take any $N > \frac{3}{2\epsilon}$. Now, for any $n >N$, we have $$ \frac{3}{2n} < \varepsilon$$
and so,
$$\left|\frac{2+\frac{3}n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the extrema of $\sin(x)\cos(y)$ with the Hessian I've got the following function: $f(x,y) = \sin(x)\cos(y)$.
I performed the Hessian matrix correctly, with the second derivatives:
\begin{bmatrix}
-\sin(x)\cos(y) & -\cos(x)\sin(y)\\
-\cos(x)\sin(y) & -\sin(x)\cos(y)
\end{bmatrix}
But the trouble comes when I want t... | Find the extrema of $f(x,y) = \sin(x)\cos(y)$.
$$\begin{cases} f_x=\cos x\cos y=0 \\ f_y=-\sin x\sin y=0\end{cases} \Rightarrow 1)\begin{cases}\cos x=0 \\ \sin y=0\end{cases} \ \ \text{or} \ \ 2)\begin{cases}\cos y=0 \\ \sin x=0\end{cases}.$$
The Hessian is:
$$H=\begin{bmatrix}
-\sin x \cos y & -\cos x \sin y \\
-\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Factoring and expanding $1-x^8$ It's been a while since I've studied factoring, and I need it for a question. How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to $(1-x)(1+x+...+x^7)$? I remember studying this a few years go but unfortunately I don't remember.
$$1-x^8=(1-x^4)(1+x^4)=(1-x)(1+x+\dots + x^7)$$
| You may remember $$(a+b)(a-b) = a^2 - b^2.$$ To see this, just multiply out. You get $a^2 + ba - ab - b^2$, and the middle terms cancel, leaving you with $a^2-b^2$.
Then if you take $a=1, b=x^4$, you get $$(1+x^4)(1-x^4) = 1-x^8.$$ That's your first identity.
You may also remember $$(1+x+x^2+\ldots+x^n)(x-1) = x^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Kummer solution to second order ODE I need to solve a second order linear ODE with non-constant coefficients of the form
$$
\frac{d^2Z}{dt^2}+(a+be^{-ct})\frac{dZ}{dt}+dZ=0
$$
where $a,b,c,f$ are positive real constants.
Mathematica provides me with this solution
$$
i^{(a - \sqrt{a^2 - 4 d})/c}\, b^{(a - \sqrt{a^2 - 4 ... | For
$$y'' + (a + b e^{-ct}) \, y' + d \, y = 0$$
let $c x = b \, e^{-c t}$ to obtain
\begin{align}
x &= \frac{b}{c} \, e^{-c t} \\
\frac{dx}{dt} &= - c x \\
\frac{dy}{dt} &= - c x \, \frac{dy}{dx} = - c x y' \\
\frac{d^{2} y}{d x^{2}} &= c^{2} \, x \, \frac{d}{dx} \left( x \, \frac{dy}{dx} \right) = c^{2} \, (x^2 y'' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Generating function of $\log(\log(n))$ What do we know about the following function?
$$f(x)=\sum_{n=2}^\infty \log{(\log{(n)})}x^n$$
Do we know any closed form for it?
I have tried to work with it through different means, but, as it is difficult both to calculate sums over $\log{(\log{(n)})}$ and evaluate integrals in... | We can write
$$
\eqalign{
& F(x) = \sum\limits_{2\, \le \,n} {\ln \left( {\ln \left( n \right)} \right)x^{\,n} } = x^{\,2} \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 2} \right)} \right)x^{\,n} } = \cr
& = x^{\,2} \ln \left( {\ln \left( 2 \right)} \right) + x^{\,3} \sum\limits_{0\, \le \,n} {\ln \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is the sum of outer products equal to the matrix product of a matrix and its transpose , so $A^TA = \sum_{i=1}^n a_i a_i^T$? Why is the sum of outer products equal to the matrix product of a matrix and its transpose? So $A^TA = \sum_{i=1}^n a_i a_i^T$, where $A = [ a_0, a_1 , ... , a_n ] $, $a_i \in \mathbb{R}^k$. ... | I think you got the order wrong. $ A A^T = \sum_{i=1}^n a_i a_i^T $. What you should do is to is to look at a sequence of examples:
$$ \left( \begin{array}{c c c}
1 \\
2 \\
3
\end{array} \right)
\left( \begin{array}{c c c}
1 \\
2 \\
3
\end{array} \right) ^T =
$$
$$ \left( \begin{array}{c c c}
1 & -1\\
2 & -2\\
3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all positive integers $x$ such that $[\frac{x}{5}]-[\frac{x}{7}]=1$, where, for any real number $t$, $[t]$ denotes the greatest integer $\le t$ I have tried the following steps:
$[\dfrac{x}{5}]-[\dfrac{x}{7}]=1$
@Berci suggested this:
$[\dfrac{x}{5}]\le\dfrac{x}{5}$
$-[\dfrac{x}{7}]<-\dfrac{x}{7}+1$
$\implies [\df... | To obtain a meaningful bound, you need the inequalities in the other way: $\left[\dfrac{x}{5}\right]\ge \dfrac{x}{5}-1$ and $\left[\dfrac{x}{7}\right]\le \dfrac{x}{7}$. Hence, $$\left[\dfrac{x}{5}\right]-\left[\dfrac{x}{7}\right]\ge \frac{x}{5}-1-\frac{x}7=\frac{2x}{35}-1$$ Now, for any $x\ge 36$, you have that $$x\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\... | Note
$$ \sin(2\theta)=2\sin\theta\cos\theta,\cos(2\theta)=2\cos^2\theta-1.$$
Let $\theta=2\arcsin(\frac35)$. It is easy to get
$$ \sin\theta=\frac{24}{25},\cos\theta=\frac{7}{25} $$
and hence
$$\sin(2\theta)=2\cdot\frac{24}{25}\cdot\frac{7}{25}=\frac{336}{625},\cos(2\theta)=2(\frac7{25})^2-1=\frac{527}{625} $$
and
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to find the area of triangle? $A =(3, 1)$ is reflected over line $y = 2x$, become $A'$.
$O = (0,0)$
Find area of triangle OAA'
I find that $A'$ is $(-1, 3)$
I find it by mutiply $A$ to matrix of $y = 2x$
There is rule Area = $$\sqrt{s(s-a)(s-b)(s-c)}$$
$S = \frac{a+b+c}{2}$
$a$ = distance between $A$ and $O=\sqrt {... | The long way:
The formula for the area of a triangle with sides $a$, $b$ and $c$ is given by
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where
$$s=\frac{a+b+c}{2}$$
With $a=\sqrt{10}$, $b=\sqrt{10}$ and $c=\sqrt{20}=2\sqrt{5}$,
$$s=\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}=\frac{2\sqrt{10}+2\sqrt{5}}{2}=\sqrt{10}+\sqrt{5}$$
The area ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Weird infinite sum Evaluate
$\sum_{n=0}^{\infty}{\frac{(-1)^n}{3n+1}=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+...}
$
This looks a lot like the series expansion for $\ln(1+x)$ when $x=1$, but I cannot find the relationship.
| We will use imprecise argument to guess the value of the sum. We know that
$$\sum_{n=0}^\infty x^{3n} = \frac{1}{1-x^3}.$$
By integrating it both sides we have $$\sum_{n=0}^\infty \frac{x^{3n+1}}{3n+1} = \int_0^x \frac{1}{(1-u)(1+u+u^2)}du = \frac{1}{6}\left(\log\left(\frac{x^2+x+1}{(1-x)^2}\right)+ 2\sqrt{3}\arctan\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Legendre symbol of $(-2/p)$. Question. Show $(-2/p)$ equals $1$ when $p\equiv 1,3\bmod 8$ and $-1$ when $p\equiv 5,7\bmod 8$.
So using the multiplicativity of the symbol; we have $$\Big(\frac{-1}{p}\Big)\Big(\frac{2}{p}\Big),$$ and I know the rules for $(2/p)$ and $(-1/p)$. But then what? Usually I use Chinese remainde... | The proof that
*
*$\Big(\frac{-1}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1 \pmod 4 \\ -1 & \text{if } p \equiv 3 \pmod 4 \end{cases}$ and
*$\Big(\frac{2}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv \pm1 \pmod 8 \\ -1 & \text{if } p \equiv \pm3 \pmod 8 \end{cases}$
are classic. The first bullet point... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why did not I find all the solutions to this equation? $\sin{3x}+\cos{2x}-\sin{x}=1$ $x$ needs to be in $[0,\pi]$
What I tried
$\sin{3x}+\cos{2x}-\sin{x}=1$
$\sin{3x}-\sin{x}=1-\cos{2x}$
$2\sin{x}\cos{2x}=1-(\cos^2{x}-\sin^2{x})$
$2\sin{x}\cos{2x}=\sin^2{x}+\cos^2{x}-\cos^2{x}+\sin^2{x}$
$2\sin{x}\cos{2x}=2\sin^2{x}$
$... | Your idea is correct except when you simplify $\sin x$ in
$$2\sin{x}\cos{2x}=2\sin^2{x}$$
A good way is:
$$2\sin{x}\cos{2x}-2\sin^2{x}=0$$
$$(\sin x)(2\cos{2x}-2\sin{x})=0$$
what give you
$$\sin x=0$$
or
$$2\cos{2x}-2\sin{x}=0$$
then, solve both equations and get the complete solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues for $4\times 4$ matrix Show that $0,2,4$ are the eigenvalues for the matrix $A$:
$$A=\pmatrix{
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2 \\
}$$
and conclude that $0,2,4$ are the only eigenvalues for $A$.
I know that you can find the eigenvalues by finding ... | You can easily guess some eigenvectors:
$$
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -1 & 2
\end{pmatrix}
\begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} =
\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}
\\
\begin{pmatrix}
2 & -1 & -1 & 0 \\
-1 & 3 & -1 & -1 \\
-1 & -1 & 3 & -1 \\
0 & -1 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 4
} |
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
| You wrote $$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
which is not true because
$$\sqrt[3]{16}=16^\frac{1}{3}$$
That there is the third root of $16$, not $2^{4\cdot 3}$. Also you can not write $$\sqrt[5]{something\cdot \sqrt[3]{something}}=\sqrt[5\cdot 3]{something\cdot something}$$
The way to do it i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Legendre symbol of $(-3/p)$. Question. Show $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1\bmod 3,\\
-1 & p\equiv 2\bmod 3. \end{cases}$$
Attempt. So, using the established results of $$\Big(\frac{3}{p} \Big) = \begin{cases}+1 & p\equiv\pm1\bmod 12,\\
-1 & p\equiv\pm5\bmod 12,\end{cases}$$ $$\Big(\frac{-1}{p}\Big... | You have that $p\equiv 1,7\pmod{12}$ if and only if $p\equiv 1\pmod{3}$ and $p$ odd.
Similarly for $p\equiv 5,11\pmod{12}$ and $p$ odd with $p\equiv 2\pmod{3}.$\
There is an easier approach if you know quadratic reciprocity:
$$\begin{align}\left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probabili... | Various answers have been already provided, but let me show a different approach to get it.
At the first toss you have $p_1=0$ and you end up for sure with 1 one-ball bin.
At the second, either you hit the one-ball (prob. $p_2=1/n=(1-p_1)1/n$) or you end up with 2 one-ball bins (prob. $q_2=1-p_2=(n-1)/n$).
At this poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\frac{1}{n + 1}{2n\choose n}$ is a positive integer for $n \ge 0$. I attempted to use Pascal's triangle identity to help out, but I do not know how to deal with $\frac{1}{n+1}$.
| Approach 1:
Note that
$$
\frac{2n+1}{n+1}\overbrace{\ \ \binom{2n}{n}\ \ }^{\large\frac{(2n)!}{n!\,n!}}=\overbrace{\binom{2n+1}{n+1}}^{\large\frac{(2n+1)!}{(n+1)!\,n!}}
$$
Then, because $\frac1{n+1}=2-\frac{2n+1}{n+1}$, we have
$$
\begin{align}
\frac1{n+1}\binom{2n}{n}
&=2\binom{2n}{n}-\frac{2n+1}{n+1}\binom{2n}{n}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Probability of drawing from a urn with and without replacement An urn contains $4$ balls: $1$ red, $1$ black, $1$ white, $1$ yellow.
Two balls are drawn one at a time, compute the probability that:
*
*first is red OR second is white (with replacement)
*first is red OR second is white (without replacement)
(1) $P(... | Using the law of total probability, we have $P(2W)=P(2W|1W)P(1W) + P(2W|\overline{1W}) P(\overline{1W}) = 0 + \frac{1}{3}\frac{3}{4}$, where $1W$ denotes the event "first taken is white" and $\overline{1W}$ its negation. Plugging that into the equation, you get $\frac{5}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Surface Integrals - Parametric Representation
Find the parametric representation for the parts of the plane $$2x+3y+z=4$$ where $$1\leq x+y+z\leq 7$$ and $$2\leq x-y\leq4$$.
My attempt:
I thought to let $u=x+y+z$ and $v=x-y$ such that $1\leq u \leq 7$ and $2\leq v\leq4$. But I'm unable to find a suitable parametric r... | All the points that lie on the relevant parts of the plane satisfy the following equations,
\begin{align*}
2x + 3y + z &= 4 \\
x+y+z &= u \\
x-y &= v
\end{align*}
where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.
If we can find expressions for $x$, $y$ and $z$ in terms of $u$ and $v$, we have found a parametrisation. Doi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
The correct answer is $\dfrac13$ while... | It is immediate (by using arithmetic progression) to get that there are equally $668$ numbers of residues $0,1,2$ modulo $3$. We have
$$(3n)^3+(3n)^3\equiv 0\pmod3\\(3n)^3+(3n+1)^3\equiv 1\pmod3\\(3n)^3+(3n+2)^3\equiv 2\pmod3\\(3n+1)^3+(3n+1)^3\equiv 2\pmod3\\(3n+1)^3+(3n+2)^3\equiv 0\pmod3$$
It follows$$\binom{668}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rotate the standard basis and determine the bases that arise from it
Given is $\mathbb{R}^2$ with standard basis $B^2_0$
A "new" basis $B=\left\{\vec{b_1}; \vec{b_2}\right\}$ arises from
$B^2_0$ by rotation of $30°$ (clockwise). Determine the change of
basis $T^{B}_{B^2_0}$ and the basis vectors $\vec{b_1}$ and
... | To calculate the rotation matrix, with reference to the standard basis, it suffices to consider what are the transformed vectors for $\vec e_1$ and $\vec e_2$ after the rotation, that is
$$\vec{e_1} = \begin{pmatrix}
1\\
0
\end{pmatrix}\to \vec{b_1} = \begin{pmatrix}
\frac{\sqrt{3}}{2}\\
\frac{1}{2}
\end{pmatrix}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $f$ be a real polynomial function find relation between the coefficients such that its roots are in an progression
Let $f(x)=ax^3+bx^2+cx+d$, be a polynomial function, find relation between $a,b,c,d$ such that it's roots are in an arithmetic/geometric progression. (separate relations)
So for the arithmetic progre... | Using your notations
$$ax^3+bx^2+c x+d=a(x-\frac \alpha q)(x-\alpha)(x-\alpha q)$$
Expand the rhs to get after simplifications
$$a x^3 -\frac{a \alpha \left(q^2+q+1\right)}{q}x^2+\frac{a \alpha ^2 \left(q^2+q+1\right)}{q}x-a \alpha ^3$$ Compare the coefficients to get
$$b=-\frac{a \alpha \left(q^2+q+1\right)}{q}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2... | Both answers are correct, that is,$$A.\begin{pmatrix}1\\-1\end{pmatrix}=5\begin{pmatrix}1\\-1\end{pmatrix}\iff A.\begin{pmatrix}-1\\1\end{pmatrix}=5\begin{pmatrix}-1\\1\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
| Note that by product to sum formula
$$\cos x \cos 3x = \frac12 (\cos 2x+\cos 4x)\\\implies \cos x \cos 2x \cos 3x = \frac12 (\cos^2 2x+\cos 2x \cos 4x)=\frac12(2\cos^3 2x+\cos^2 2x - \cos 2x )$$
then
$$\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=
\frac{1 - \cos^3 2x-\frac12\cos^2 2x + \frac12\cos 2x }{x^2}=\\\frac{\frac12(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prov... | By Taylor's expansion
$$\cos(x) \le 1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48} \le e^{-x^2/2}$$
indeed
$$1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48}\iff \frac{x^4}{12}-\frac{x^6}{48}\ge 0 \iff x^4(4-x^2)\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
inclusion-exclusion: class distribution so that professor teaches the same two courses both semesters
How many ways are there to assign each of five professors in a math department to two courses in the fall semester (i.e., $10$ different math courses in all) and then assign each professor two courses in the spring se... | The number of ways five professors can each be assigned two courses in the fall semester is
$$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
To see this, list the professors in some order, say by seniority. Assign the first professor on the list two of the ten courses, which leaves eight courses avai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Limit of a Function involving tangent function and limits at infinity Determine $$\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$.
Attempt
Let $$y=\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
Put $\frac{1}{x}=p$.
$$\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
We have
$$\lim_{x \... | Note that
$$\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\frac{\pi x+\frac{\pi}2-\frac{\pi}2}{2x+1}}\right)^\frac{1}{x}=\left(\tan{\left(\frac{\pi}2-\frac{\frac{\pi}2}{2x+1}\right)}\right)^\frac{1}{x}=\left(\tan{\frac{\pi}{4x+2}}\right)^{-\frac{1}{x}}=e^{-\frac{\log{\left(\tan{\frac{\pi}{4x+2}}\right)}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
What order of operation should I use calculating $2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4$? I am struggling if I should use PEMDAS or BODMAS in this equation. What is the right method to get around these type of equations?
$$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 = \ ?$$
| You silly humans with your mnemonics that just wind up confusing you even more! Mwahahaha! Mwahahahahahahahahaha!
Now, let's see: PEMDAS is Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. And BODMAS is Brackets, Orders, Division, Multiplication, Addition, Subtraction.
Parentheses are pretty muc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the ... | Rearrange
$B=\sqrt{ab}$
to get
$B^2/a=b.$
Substitute in to get
$2A=a+B^2/a.$
Rearrange and solve:
$a^2-2Aa+B^2=0$
$(a-A)^2-A^2+B^2=0$
$a=A\pm\sqrt{A^2-B^2}.$
So we need $A^2-B^2=C^2$ for some $C$.
$(10x+y)^2-(10y+x)^2=C^2$
$((10x+y)-(10y+x))((10x+y)+(10y+x))=C^2$
$(9x-9y)(11x+11y)=C^2$
$3^211(x-y)(x+y)=C^2$
So we need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(... | $$
\begin{align}
\color{#C00}{\sin(x)+\sin(y)}+\color{#090}{\sin(x+y)}
&=\color{#C00}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)}+\color{#090}{2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x+y}2\right)}\\
&=2\sin\left(\frac{x+y}2\right)\left[\cos\left(\frac{x-y}2\right)+\cos\left(\frac{x+y}2\right)\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Chole... | Notice $A$ can be rewritten as a sum of 5 matrices.
$$A =
\left[\begin{smallmatrix}
2 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 3
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
2 & 2 & 0 & 0 & 0\\
2 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 2
} |
Simplifying $\operatorname{tanh}(\operatorname{arsinh}(x))$ So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$
In general,
$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$
therefore
\begin{align}\tanh(\operatorname{arsinh}(x)) & =\... | Your route is fine, but you have just made a mistake from the beginning, observe that
$$
\operatorname{arsinh} x =\ln \left ( x + \sqrt{x^2 \color{red}{+} 1} \right )\ne \ln(x+\sqrt{x^2-1}).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_... | I came up with the solution my professor was looking for and figured I would share it here for both my own clarification and for anyone else in the future who wants to find the solution via a Laurent expansion.
For our function $f(z) = \frac{1}{z^3 \sin{(z)}}$, we can solve for the residue of $f$ at $z = 0$ by doing th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \fr... | By starting with $g(x) = \ln(1 + x)$ it can be determined that $g^{(n)}(x) = (-1)^{n-1} \, (n-1)! (1+x)^{-(n+1)}$ for $n \geq 1$ along with $g(0) = \ln(1 + 0) = 0$ and $g^{(n)}(0) = (-1)^{n-1} \, (n-1)!$. From this it is developed
$$\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n}.$$
Now, as Mark Viola ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \... | The "identity"
$$\arcsin\left(2x\over1+x^2\right)=2\arctan x$$
does not hold for all $x$. You can see this by comparing the ranges of the two sides. The arcsine function has $[-\pi/2,\pi/2]$ as its range, but twice the arctangent function has $(-\pi,\pi)$ as its range. In particular, if $x\gt1$, then $2\arctan x\gt\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Sum of $1+4\epsilon +9\epsilon^2 +16\epsilon^3+...+2018^2 \epsilon^{2017}$ I have this problem from a college exam: Let $\epsilon$ be a non-real root of unity of order 2018, find the sum$$S=1+4\epsilon +9\epsilon^2 +16\epsilon^3+...+2018^2 \epsilon^{2017}$$Here is my try. First I considered $$S_1=\sum_{k=0}^{2018} x^k=... | Using the operator $\delta = x \, \frac{d}{dx}$ leads to
\begin{align}
\sum_{k=0}^{n} x^{k} &= \frac{1 - x^{n+1}}{1-x} \\
\sum_{k=0}^{n} k \, x^{k} &= \delta \left(\frac{1 - x^{n+1}}{1-x} \right) = \frac{x - (n+1) x^{n+1} + n x^{n+2}}{(1-x)^2} \\
\sum_{k=0}^{n} k^{2} \, x^{k} &= \delta^{2} \left(\frac{1 - x^{n+1}}{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$. I need to find the Laurent series of $f(z)=\frac{z-1}{z(z^3-1)}$ for $0<|z|<1$ and $|z|>1$.
I believe I am nearly there but I have a confusion.
Notice $f(z)=\frac{z^2-1}{1-z^3}+\frac{1}{z}$ so if $0<|z|<1$ we have
$$f(z)=\frac{1}{z}+(z^2-1)\sum_{n... | If $|z|<1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=(1-z)(1+z^3+z^6+z^9+\cdots)\\&=1-z+z^3-z^4+z^6-z^7+\cdots\end{align}and therefore$$f(z)=z^{-1}-1+z^2-z^3+z^5-z^6+\cdots$$If $|z|>1$, then\begin{align}\frac{z-1}{z^3-1}&=\frac{1-z}{1-z^3}\\&=-(1-z)(z^{-3}+z^{-6}+z^{-9}+\cdots)\\&=z^{-2}-z^{-3}+z^{-5}-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the values of $(2\sin x-1)(\cos x+1)=0$ How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?
The solution shows that one of these is true:
$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$
$\cos x = -1$ and thus $x = 180^\circ$
Question: Inserting the $\arcsin$ of $1... | If $ab = 0$, then either $a=0$ or $b=0$.
Given $(2\sin x - 1)(\cos x + 1) = 0$, we can separate these and solve separately.
So... $\ \ \ \ \ 2\sin x -1 = 0$
$\Rightarrow \sin x = \frac12 \Rightarrow x = 30^\circ, 150^\circ$
We get those two values because $\sin x$ is symmetrical across the $y$-axis.
Also... $\ \ \ \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|... | Yet another way:
\begin{align*}
\frac{a_{k+1}}{a_{k}} & =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{(1+k)^{2}}\\
& =\frac{1}{3}\left(1+\frac{1}{k}\right)^{-k^{2}}\left(1+\frac{1}{1+k}\right)^{k^{2}+k}\left(1+\frac{1}{1+k}\right)^{k+1}.
\end{align*}
Now, if
$$
b_{k}\equiv\left(1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve the indefinite integral $\int \frac{1}{x^{1/2}+x^{1/3}} dx$
Solve $\int \frac{1}{x^{1/2}+x^{1/3}} dx$
My Attempt
$$
\int \frac{1}{x^{1/2}+x^{1/3}} dx=\int\frac{dx}{x^{1/2}(1+x^{-1/6})}
$$
Put $t=x^{1/2}\implies dt=\frac{dx}{2.x^{1/2}}\implies\frac{dx}{x^{1/2}}=2dt$
$$
\int \frac{1}{x^{1/2}+x^{1/3}} dx=2\int\fra... | $$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=6\left(\frac{(x^{\frac{1}{6}}+1)^3}{3}-\frac{3(x^{\frac{1}{6}}+1)^2}{2}+3(x^{\frac{1}{6}}+1)-\ln \left|x^{\frac{1}{6}}+1\right|\right)+k, \,k\in \mathbb Z$$
Proof:
If I take this substitution $t=x^{\frac{1}{6}}$ we have:
$$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=\int \frac{6t^3}{t+... | {
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"source": "stackexchange",
"question_score": "2",
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Can I Systematically Determine the Cardinality of a Finite Set? Is there any way you can, in a systematically manner, determine the cardinality of a finite set in which the elements are given by a specific expression?
For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set defined by:
$$B=\biggl\{\frac{a-b... | It's really dependent on how you built the set. The thing that can make things rather tricky is when things in the set are equal only one thing is counted. So usually inequalities are easy but the exact number you'd have to eleminate all equivalent elements.
For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \... | An elementary evaluation:
$$\int_0^1 \frac{x-x^2}{\sin \pi x} dx \overset{ibp}
=\int_0^1 \frac{2x-1}\pi\ln \tan \frac{\pi x}2 dx
\overset{t=\tan^2\frac{\pi x}2}=
\frac2{\pi^3}\int_0^\infty \frac{\ln t\tan^{-1}\sqrt t}{\sqrt t(1+t)}dt
$$
Let $J(a)=\int_0^\infty \frac{\ln t\tan^{-1}a\sqrt t}{\sqrt t(1+t)}dt$
$$J’(a)= \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 1
} |
Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
| (i) $5<2.2+1.7+1.4<5^{1/2}+5^{1/3}+5^{1/4}$.
(ii) $8>2.9+2+1.7>8^{1/2}+8^{1/3}+8^{1/4}$.
(iii) $n\ge3\sqrt n>n^{1/2}+n^{1/3}+n^{1/4}$ because $n \ge 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem on series . I found this problem on series in my school textbook.
If one of the two series below is 2,what is the value of the other series?
The series are:
$$(1+x+x^{2}+x^{3}+\cdots+x^{n})$$
and the other one is
$$(x+2x^{2}+3x^{3}+\cdots+nx^{n})$$
The four options are:
A) 1
B) 2
C) 3
D) 4
I did it i... | In case the sum is infinite, that is, you're considering $$1+x+x^2+\dots\tag{1}$$
Assume that $(1)$ is equal to $2$, so that $$x+x^2+x^3+\dots=1\tag{2}$$ Then look at: $$\begin{align}x+x^2+x^3+\dots\\x^2+x^3+\dots\\x^3+\dots\\\\x+2x^2+3x^3+\dots\tag{3}\end{align}$$
The desired sum $(3)$ is equal to $$x(1+x+x^2+\dots)+x... | {
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"url": "https://math.stackexchange.com/questions/2798266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left... | Divide numerator and denominator by $a$ (where $a>0$). The denominator becomes
$$\frac{1}{a}\sqrt{a^2-b^2}=\sqrt{\frac{1}{a^2}}\times\sqrt{a^2-b^2}=\sqrt{1-\frac{b^2}{a^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the Eigenvalues and its Eigen vectors. Consider the $ \ n \times n \ $ matrix $$ \begin{pmatrix} a & -1 & & & & \\ -1 & a & -1 & & & \\ & -1 & a & -1 & & & \\ & & -1 & a & -1 & \\ & & ..... & .... & .... \\ & & && a & -1 \\ & & & &-1 & a \end{pmatrix} $$
Find the Eigenvalues and its Eigen vectors.
Also d... | You're not quite right about the eigenvalues for the $3\times 3$ matrix since you're missing a $-1$ in the $(2,3)$ position.
For the higher dimensional cases let $A_n$ be the $n\times n$ matrix with $a-\lambda$ along the diagonal and $-1$s above and below the diagonal. Then $A_n$ has the form
\begin{pmatrix}
a-\lambd... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help finding the limit of the following function as x tends to 0 $$
\lim_{x\to0}\left(\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}\,\right)
$$
From Demidovich 5000 problems in mathematical analysis
| You can try with $x=t^2$ (the domain is $x>0$, we assume also $t>0$), so your function becomes
$$
\sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} - \sqrt{\frac{1}{t^2} - \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2}}}} =
\frac{\sqrt{1+\sqrt{t^2+t^3}}-\sqrt{1-\sqrt{t^2+t^3}}}{t}
$$
and your limit is reduc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \frac{x^n}{x^2-x+1} dx$ Greetings I desire to find a closed form for $$I=\int_0^1 \frac{x^n}{x^2-x+1} dx$$ My try was to use $x-\frac{1}{2}=t$ to get $$\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{(t+\frac{1}{2})^n}{t^2+\frac{3}{4}}dt$$ then $\frac{\sqrt 3}{2}\tan u=t$ $$I=\frac{2}{\sqrt 3}\int_{-\frac{\p... | Sketch: Write $I_n = \int_0^1 \frac{x^n}{x^2-x+1}\,dx$, then consider $f(t) = \sum_{n=0}^\infty I_n t^n$:
\begin{align}\sum_{n=0}^\infty I_n t^n&=\sum_{n=0}^\infty \int_0^1 \frac{(xt)^n}{x^2-x+1}\,dx
\\&=\int_0^1\frac{1}{1-xt}\frac{1}{x^2-x+1}\,dx\\
&=-\frac{\sqrt{3} \pi (t-2) + 9t\log(1-t)}{9t^2-9t+9}\end{align}
by pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possib... | I show here that the only correct answer is, as it maintain @fleablood and @farruhota in their answers.
$$\frac{x^3+y^3}{3}=\frac{(x+y)^3}{3}-xy(x+y)\Rightarrow xy+\frac{(x+y)^3}{3}-xy(x+y)=2007$$ It follows $x+y\equiv 0\pmod3$.
The cubic $$xy+\frac{x^3+y^3}{3}=2007$$ is symmetric respect to the diagonal $y=x$ then i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\tan\frac{\alpha}{2}=\pm\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$ based on given condition
If $$\cos\alpha=\cos\beta\cos\phi=\cos\gamma\cos\theta$$
and $$\sin\alpha=2\sin\frac{\phi}{2}\sin\frac{\theta}{2}$$
then prove that $$\tan\frac{\alpha}{2}=\pm\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$$
My approach:... | $$\sin^2\alpha=2\sin^2\dfrac\phi2\cdot2\sin^2\dfrac\theta2$$
$$=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$
$$\cos\beta\cos\gamma(1-\cos^2\alpha)=(\cos\beta-\cos\alpha)(\cos\gamma-\cos\alpha)$$
$$-\cos^2\alpha\cos\beta\cos\gamma=-\cos\alpha\cos\beta-\cos\alpha\cos\gamma+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a coefficient of $x^{12}$ in $(x + x^{2} +x^{3} +x^{4})^{5}$ Find a coefficient of $x^{12}$ in $$(x + x^{2} +x^{3} +x^{4})^{5}$$
Using formula for the sum of a geometric series I have turned it into $$x^5(1-x^4)(1-x)^{-4} $$
So now I see that in $(1-x^4)(1-x)^{-4}$ we need to find coefficient of $x^7$ as we actual... | you can have a look at this post too : Coefficient of $x^{25}$
$$(x+x^2+x^3+x^4)^5=\sum\limits_{i+j+k+l=5} \binom{5}{i,j,k,l}x^{i+2j+3k+4l}$$
The coefficient $x^{12}$ is obtained when $i+2j+3k+4l=12$.
$\begin{array}{cccc|c}
i & j & k & l & \dfrac {5!}{i!j!k!l!}\\\hline
0 & 3 & 2 & 0 & 10\\
0 & 4 & 0 & 1 & 5\\
1 & 1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\... | Using complex numbers we can derive $\sin(\frac{\pi}{5})$ and $\cos(\frac{\pi}{5})$
$$ z^5=-1 \implies z^4-z^3+z^2-z+1=0 \implies z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0 $$
By substitution $t=z+\frac{1}{z}$ we get:
$$ t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2} $$
And because $\cos(\frac{\pi}{5})$ is positive, we may consi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
differentiation under the integral sign $\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx$ Using differentiation with respect to the parameter,show that for $|r|<1$
$$\mathbf{F}(r)=\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx =0$$
my attempt is
$$\mathbf{F}'(r)=\int_{0}^{\pi}{-2\cos x + 2r \over1-2r\cos x +r^2}dx$$... | A slightly different approach.
It is clear that $F'(0)=0$. We assume $r\neq 0$ from now on. Notice that for $|r|<1$ one has:
\begin{align}
\sum_{n=0}^\infty r^n \cos(nx)& = \frac{1-r\cos(x)}{1-2r\cos(x)+r^2}\\\sum_{n=0}^\infty r^n \sin(nx)&= \frac{r\sin(x)}{1-2r\cos(x)+r^2}
\end{align}
One can use that to write $F'(r)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
My attempts:
Attempt 1:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\sin^2 x+\cos^2 x-2\sin x \cos x}{\sin^2 x+\cos^2 x+2\sin x \cos x}}$
$\implies y=... | The statement that you are trying to prove is not true for every $x$. If it was, we would always have $\frac{\mathrm dy}{\mathrm dx}\leqslant0$ and therefore $y$ would be decreasing. But it is increasing in $\left[\frac\pi4,\frac\pi2\right]$, for instance.
The statement holds in $\left(-\frac\pi4,\frac\pi4\right]$. You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it ... | Perhaps another method to notice is that, by Euler's Generalization of Fermat's Little Theorem, we have that:
$$1^n-3^n-6^n+8^n \equiv 1^{n \mod \phi(10)} - 3^{n \mod \varphi(10)} - 6^{n \mod \varphi(10)} + 8^{n \mod \varphi(10)} \mod 10$$
Since $\varphi(10) = 4$, you really only need to check the cases of $n = 0, 1, 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find out if polynomial has inverse in quotient ring Find out if polynomial $x^3-x^2+x-1$ has inverse element in quotient ring ${\displaystyle \mathbb {Z} }_{11}/(x^4+3x^3-3x^2-4x-1)$, if yes find this inverse.
I know that $x^3-x^2+x-1$ has inverse element when $gcd(x^4+3x^3-3x^2-4x-1;x^3-x^2+x-1)$~$1$. I also know that... | The extended Euclidean algorithm over $\mathbb Q$ gives:
$$
232 = (49 x^3 + 168 x^2 - 75 x - 195)(x^3-x^2+x-1) +(-49 x^2 + 28 x - 37)(x^4+3x^3-3x^2-4x-1)
$$
Mod $11$, this becomes
$$
1 \equiv (5 x^3 + 3 x^2 - 9 x - 8)(x^3-x^2+x-1) \bmod (x^4+3x^3-3x^2-4x-1)
$$
and so the inverse is $5 x^3 + 3 x^2 - 9 x - 8 \equiv 5x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Integral $\int_0^{\pi} \frac{\cos(2018x)}{5-4\cos{x}}dx$ I wish to evaluate $$I(2018)=\int_{0}^{\pi}\frac{\cos(2018x)}{5-4\cos x} dx$$ Considering $$X=I(k)+iJ(k)=\int_{-\pi}^{\pi}\frac{\cos{kx}}{5-4\cos x} dx +i\int_{-\pi}^{\pi}\frac{\sin{kx}}{5-4\cos x} dx=\int_{-\pi}^{\pi}\frac{e^{ikx}}{5-4\cos x} dx$$ let us substi... | I am going to show the general case:
$$
I_n=\int_{0}^{\pi} \frac{\cos nx}{5-4 \cos x} d x= \frac{\pi}{3 \cdot 2^{n}}
$$
using contour integration on the unit circle $|z|=1$.
Noting $I_n= \frac{1}{2} \int_{0}^{2\pi} \frac{\cos nx}{5-4 \cos x} dx=\frac{1}{2} Re(J_n)$, where
$$J_n=\int_{0}^{2\pi} \frac{e^{nxi}}{5-4 \cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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How do you read $\pm$? Why does $|x|=3$ imply $x=\pm 3$? Q1) How is this symbol read as $±$?
"Plus AND Minus" or Plus OR Minus"?
All this time I've been reading it as Plus-Minus.
I asked my math teacher and he said it's the former one. Then my question is if it's "Plus AND Minus" then when we take square roots on both... | The symbol $\pm$ should be read plus or minus.
Background
By convention, if $x \geq 0$, then $\sqrt{x}$ is the principal square root of $x$, meaning the nonnegative square root of $x$. Hence, $\sqrt{9} = 3$. If we want to denote the negative square root, we write $-\sqrt{x}$. Thus, the negative square root of $9$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 day... | The alternative problem is this: one has 3 shirts of red color, 2 of green and 1 of white of which he chooses arbitrarily. What is the probability that he puts on all of the 3 colors within 3 days?$$P=\dfrac{N}{N_{\text{tot}}}\\N=\binom{3}{1}\binom{2}{1}\binom{1}{1}=6\\N_{\text{tot}}=\dfrac{6!}{1!2!3!}=60\\P=0.1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Limit of sequence as $n$ goes to infinity with $n$ in every term I am studying for an exam in introductory analysis and came across this exercise in a practice exam:
Determine the limit of $(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$.
I know that every single fr... | We have,
$(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2})\le(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})\le(\frac{1}{n^2-n}+\frac{2}{n^2-n}+\frac{3}{n^2-n}+ \cdots + \frac{n}{n^2-n})\ \forall \ n \in N$
$\Rightarrow a_n= \frac{(n)(n+1)}{2(n^2)} \le (\frac{1}{n^2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$
If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$
I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+... | Let $y=x+a$, $z=x+a+b$, $w=x+a+b+c$ and $u=x+a+b+c+d$.
Thus, $a$, $b$, $c$ and $d$ are non-negatives and we need to prove that:
$$(x+y+z+u+w)^2\geq5(xw+wz+zy+yu+ux)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(x(x+a+b+c)+(x+a+b+c)(x+a+b)+(x+a+b)(x+a)+(x+a)(x+a+b+c+d)+(x+a+b+c+d)x)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(5x^2+(8a+6b+4c+2d)x+(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding $C_n=4C_{n-1}+6n-1$ 2.$$\begin{cases}
C_n=4C_{n-1}+6n-1\\
C_0=2\\
\end{cases} $$
$\begin{cases}
C_n=D_n+dn+e\\
D_n=\lambda D_{n-1}\end{cases}$
$D_n+dn+e=4C_{n-1}+6n-1$
But $C_{n-1}=D_{n-1}+d(n-1)+e$
So $D_n+dn+e=4(D_{n-1}+d(n-1)+e)+6n-1$
$D_n+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$
But for $\lambda=4$ we get $4D_{n-1}+... | We start by solving the homogenous part: letting $x^n=C_n$ gives the characteristic equation $$x^n=4x^{n-1}\Rightarrow x=4,$$ which gives $$C_n=A(4^n).$$ Now, for the the inhomogenous part, let us guess that the solution is of the form $$p(n)=d_0+d_1n,$$ substituting into the recurrence gives $$d_0+d_1n=4(d_0+d_1(n-1))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Consider this system of congruence equations. \begin{cases}
4x \equiv 14 \pmod m \\
3x \equiv 2 \pmod 5
\end{cases}
I want to prove that for $m \in 4\mathbb{Z}$ there are no solutions(1). Moreover, I want to determine all m for which I have solutions(2).
First of all, the second equation is equivalent to $ x \equi... | If $m=4n$, then
\begin{align}
4x \equiv 14 \pmod m
&\implies 4x \equiv 14 \pmod{4n} \\
&\implies 4n \mid 14-4x \\
&\implies 2 \mid 7-2x \\
&\implies 2 \mid 7
\end{align}
which is false.
$3x \equiv 2 \pmod 5 \implies x \equiv 4 \pmod 5 \implies x = 4 + 5u$ for some integer, $u$.
\begin{align}
4x \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim\limits_{n→∞}\frac{3^n(\overbrace{\sin(\sin(\cdots\sin(x)))}^{n\text{ times}}+(\sqrt2\cos x+2)^n+2^n\cos x}{3^n+\sin x(\sqrt2\cos x+2)^n}$
Let$$f(x)= \lim_{n \to \infty}\dfrac{3^n(\overbrace{\sin(\sin(...\sin(x)))}^{\text{n times}}+(\sqrt 2 \cos x+2)^n+2^n\cos x}{3^n+\sin x(\sqrt 2\cos x+2)^n}.$$
If $l = \... | Keep in mind that you are first taking the limit with respect to $n$ and only then taking the limit with respect to $x$. So in particular, you care about the behavior of $\dfrac{(\sqrt 2 \cos x+2)^n}{3^n}$ for $x<\pi/4$ for the limit from below and for $x>\pi/4$ for the limit from above.
When $x<\pi/4$, the terms with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the differential equation $x^5\frac{d^2y}{dx^2}+3x^3\frac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$ Solve the differential equation $$x^5\frac{d^2y}{dx^2}+3x^3\frac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$$
In the solution of the above the question the author is trying to convert it into an exact differential equation by multiplying $x^m... | Hint:
$x^5\dfrac{d^2y}{dx^2}+3x^3\dfrac{dy}{dx}+(3-6x)x^2y=x^4+2x-5$
$x^3\dfrac{d^2y}{dx^2}+3x\dfrac{dy}{dx}-(6x-3)y=x^2+\dfrac{2}{x}-\dfrac{5}{x^2}$
Let $y=x^nu$ ,
Then $\dfrac{dy}{dx}=x^n\dfrac{du}{dx}+nx^{n-1}u$
$\dfrac{d^2y}{dx^2}=x^n\dfrac{d^2u}{dx^2}+nx^{n-1}\dfrac{du}{dx}+nx^{n-1}\dfrac{du}{dx}+n(n-1)x^{n-2}u=x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m... | Empirically:
Consider the function
$$p(n):=\sqrt{n(n+1)(n+2)(n+3)+1}.$$
For $n=0,1,2,3,\cdots$ we have $p(n)=1,5,11,19,29,\cdots$ a sequence with constant second order differences ($2$), and we can postulate the polynomial
$$n^2+3n+1$$
(because $p(0)=1$, the coefficient of $n^2$ must be $1$, and $p(n)-n^2-1=0,3,6,9,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 7
} |
Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
... | Here a rather geometric way:
$$y = x^2+x+1 = x(x+1) + 1$$
So, $y = x^2+x+1$ is the parabola $y=x(x+1)$ shiftet by $1$ upwards.
$y=x(x+1)$ has its vertex at $x_V = -\frac{1}{2} \Rightarrow y_V = -\frac{1}{4}$
So, the vertex of $y= x(x+1) + 1$ is also at $x_V = -\frac{1}{2}$ with a minimum value of $y_{min}= -\frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 21,
"answer_id": 4
} |
Facing difficulty in working $\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}$ I would like to evaluate this integral,$$\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}\tag1$$
This is the approach I will take:
We can begin with a sub: $y=\sqrt{x}$, ... | $\newcommand{\Im}{\operatorname{Im}}$I don't think your last integral is correct. You have made a mistake somewhere which I could not find.
Assume wolog $a\geq 0$ (because the case $a<0$ follows from the case $a>0$ by noticing that the arctangent is an odd function). Let the original integral be $F(a)$. Set $u=\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization... | Write $J=\pmatrix{1&1\\1&1}$. The equation $X^2+X=J$ implies $XJ=JX$. The matrices satisfying $XJ=JX$ are those of the form $X=aI+bJ$. Then $X^2=a^2I+(2ab+2b^2)J$, so we get
$$(a^2+a)I+(b+2ab+2b^2)J=J.$$
Therefore $a^2+a=0$, entailing $a\in\{0,-1\}$ and $b+2ab+2b^2=1$,
which gives a quadratic equation for each of the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
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