Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am... | Like Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$,
as $2$ is a primitive root $\pmod{37}$ as
$2^5\equiv-5\pmod{37}\implies2^{10}\equiv(-5)^2\equiv-12\implies2^{12}\equiv-12\cdot2^2\equiv26\not\equiv1$
and $2^{18}\equiv(-10)^3\equiv-1$
using Discrete Logarithm , $3$ind$_2x\equiv0\pmod{36}\iff$ind$_2x\equiv0\pmod{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Find limit of $\frac{2x+7}{\sqrt{x^2+2x-1}}$ (check my steps please..)
Compute the limit $$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}$$
Here are my steps:
$$\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}=\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}\cdot\frac{-1/x}{-1/x}=\frac{\displaystyle \lim_{x\to-\infty}... | To avoid confution the best way when we deal with limit $\to -\infty $ is to change variable and set $x=-y$ with $y\to +\infty$ then
$$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}=\lim_{y\to+\infty}{\frac{-2y+7}{\sqrt{y^2-2y-1}}}=\lim_{y\to+\infty}{\frac{y}{y}\frac{-2+7/y}{\sqrt{1-2/y-1/y^2}}}=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How prove this inequality $H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$
Prove that: There exists a constant $C>0$ such that
$$H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$$
holds for arbitrary positive integer $m$ and any $m$ positive integers $a_1,a_2,\cdots,a_m$,
where $H(n)=\su... | In view of the rearrangement inequality, it suffices to check the inequality when $(a_i)$ is decreasing, i.e., $a_1 \geq a_2 \geq \cdots \geq a_m$. Also, it is no harm to introduce $a_{m+1} = 0$. Then
$$ \sum_{i=1}^{m} H(a_i)
= \sum_{i=1}^{m} \sum_{k=i}^{m} \left( H(a_k) - H(a_{k+1}) \right)
= \sum_{k=1}^{m} k \left( H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
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,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\... | The DE is
$$\frac12d(x^2)+\frac12d(y^2)+\frac12d(x^2y^2)=0$$
Then, the solution is
$$\boxed{\frac12x^2+\frac12y^2+\frac12x^2y^2=c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Use Ramanujan’s method to denest $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{7\sqrt[3]{20}-19}$ A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$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... | HINT:
A method that proves the first equality but is not able to find a denesting.
One checks that both $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$ are roots of the polynomial $x^9 + 3 x^6 + 3 x^3 - 6859$, a polynomial with a unique real root.
ADDED: Note that th... | {
"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": 0
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The convergence of a recursive sequence
The sequence $(a_n)$ is defined by $a_1 = 1$ and
$a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. A sequence $(b_n)$ is defined by
$b_n = \dfrac{a_n}{2^n}$. It is easy to see $(b_n)$ is monotone but is
it convergent?
Suppose $(b_n)$ be bounded. Then $(b_n)$ is convergent by Monotone C... | $$
b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}
$$
where $b_1 = 1/2$. If you try to bound the relation, you get
$$
\dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}} \le b_n + x_n
$$$$
\dfrac{1}{2^{2n+2}} \le b_nx_n + x_n^2
$$
so $x_n = 2^{-n}$ is enough. This proves that
$$b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the value of the constant c such that $\sum_{n=2}^\infty(1+c)^{-n} = 2$
Find the value of the constant c such that $$\sum_{n=2}^\infty(1+c)^{-n} = 2 $$
For this question,I'm not sure if I'm doing it right. If I am doing it right, I'm not sure how to get further. Here is what I have so far. Can anyone please help... | When the series converges,
$$\sum_{n=2}^\infty(1+c)^{-n}=\frac{(1+c)^{-2}}{1-(1+c)^{-1}}=\frac1{(1+c)\,c}=2.$$
Solve $$c^2+c-\frac12=0.$$
After solving, check the convergence condition, $|1+c|>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8.... | Here is a suggested alternate approach.
Consider the recursive sequence $a_k\in\mathbb{R}$
\begin{eqnarray}
a_0&=&\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+...}}}}}\\
a_{k+1}&=&\frac{a_k^2+1}{2(k+1)}-1 \tag{1}
\end{eqnarray}
This gives the increasing unbounded sequence
\begin{eqnarray}
a_1&=&\sqrt{3+4\sqrt{5+6\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively.
| For convenience, in the index notation, I only sum over nonnegative integers.
\begin{align}
\sum_{6r_1+3r_2+r_3=24} (1+r_1) &= \sum_{2r_1+r_2+k=8} (r_1+1) \text{, we let } r_3 = 3k \\
&= \sum_{2r_1+2w=8} (2w+1)(r_1+1) \text{, we let } r_2+k = 2w \\
&= \sum_{r_1+w=4} (2w+1)(r_1+1)\\
&=\sum_{w=0}^4(2w+1)(5-w) \\
&= -2 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove the relation is an equivalence relation. Problem
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \vert(a^2 + b) $. Prove that $R$ is an equivalence relation.
This is what I have so far.
Claim:
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \mid(a^2... | Your approach is too technical.
Note that $$2| (a^2 + b)$$ if and only if both $a$ and $b$ are odd or both are even.
Reflexivity: $a$ and $a^2$ are either both even or both odd.
Symmetry: If both $a$ and $b$ are odd then both $b$ and $a$ are also odd. Similarly for the even case.
Transitivity: If $a$ and $b$ are r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\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\... | hint: replace the denominator on the third line of your proof $\sin^2(2x)$ by $(2x)^2$ and apply L'hopitale rule three times to the expression: $\dfrac{4x^2-\sin^2(2x)}{16x^4}$. I did it and it works. Try it. Note that "replace" here means you write: $\sin^2 (2x) = 4x^2\cdot \left(\dfrac{\sin(2x)}{2x}\right)^2$, and th... | {
"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": 0
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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
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\fra... | After my comment you will get $$a\sqrt{1-x^2}+x=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and
$$a\sqrt{1-y^2}-y=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and you will get the desired result!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
What is the sum of series $1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$ I am trying to solve this series question which is
$$1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$$
Till now I've only been able to write the general form of this series which is
$$\sum_{n=1}^\infty\frac1{(2n-... | Hint Reindexing your expression for the series for convenience gives
$$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k .$$
Now, the factors $\frac{1}{2 k + 1}$ suggest considering the power series
$$\operatorname{artanh} x \sim \sum_{k = 0}^\infty \frac{1}{2 k + 1} x^{2 k + 1} .$$
Rewrite the given se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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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,
$$a=2R \sin A$$
$$... | By the law of cosines,
\begin{align}
a^2+b^2+c^2
&=
2ab\cos\gamma+
2bc\cos\alpha+
2ca\cos\beta
\tag{1}\label{1}
\end{align}
Using expressions for the area $S$ of triangle
\begin{align}
S&=\tfrac12ab\sin\gamma=
\tfrac12bc\sin\alpha=
\tfrac12ca\sin\beta
,\\
S&=2R^2\sin\alpha\sin\beta\sin\gamma
,
\end{align}
we have
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$ If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$.
Find the limit of the sequence.
Can somebody he... | $X_n=Aa^n+Bb^n$ for some constants $A$ and $B$, where $a$ and $b$ are the roots of the equation $x^2-\alpha x+(1-\alpha)=0$. The sequence is convergent if both $|a|\le 1$ and $|b|\le 1$.
When $\alpha^2-4(1-\alpha)<0$, i.e. when $0<\alpha<2\sqrt{2}-2$, the quadratic equation has unreal roots, $|a|=|b|$ and $|a||b|=|ab|=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Limit of equation regarding ratio of gamma function I get the result from Wolfram that
$$\lim_{a\to \infty}a-\frac{\Gamma(a+1/2)^2}{\Gamma(a)^2}=\frac{1}{4}.$$
I am trying to prove it. It seems the Stirling's formula cannot be used here.
Can anyone help me on this?
Thank you very much!
I tried to use Stirling's approxi... | $\Gamma$ is log-convex by the Bohr-Mollerup theorem/characterization, hence it is enough to prove
$$ \lim_{n\to +\infty} n-\left(\frac{n\sqrt{\pi}}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}.\tag{1} $$
On the other hand
$$ \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),\qquad \left(\frac{1}{4^n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Equation in 3d-Space Consider we have 3 fixed points A,B and C also,We have this Equation :
$MA^2+MB^2+MC^2=30$.
What does this Equation represent in 3d-Space,( where M is such a Non-fixed point in that space)?
Suppose $A(a1,b1,b1),B(a2,b2,c2),C(a3,b3,c3),M(x,y,z)$
any help will be appreciated.
| Given $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3),M(x,y,z)$ and $r>0$.
What subset of $\mathbb{R}^3$ is specified by the equation
\begin{equation} |MA|^2+|MB|^2+|MC|^2=r^2 \tag{1}\end{equation}
\begin{eqnarray}
|MA|^2&=&(x-x_1)^2+(y-y_1)^2+(z-z_1)^2\\
|MB|^2&=&(x-x_2)^2+(y-y_2)^2+(z-z_2)^2\\
|MC|^2&=&(x-x_3)^2+(y-y_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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... | Define $$a_1=0,\; a_2=1,\;\text{and}\quad a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}.\tag{0}$$
Define $$\;R(x) := \sqrt{1-4x} = 1-2\sum_{n=1}^\infty C_{n-1}x^n = \sum_{n=0}^\infty {2n \choose n}\frac{x^n}{1-2n}\;$$ which is the generating function of the OEIS integer sequence A002420.
Define $$\;B(x) := \frac{ (1+4x+8x^2)R(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 0
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Solve $(1+x)y’=y$ by power series Solve $(1+x)y’=y$ by power series. $$$$Start with $y$ and $y’$:
$y=a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n$
$y’=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}$
Then $(1+x)y’=a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}$
Then $(1+x)y’-y=[a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}]-[a_0... | Are you required to "use series"? Here's a very quick solution. Rewrite it as
$$ \frac{y'}{y} = \frac{1}{1+x}. $$
Note that
$$ \frac{y'}{y} = \frac{d\phantom{x}}{dx}\ln(y) $$
and
$$ \frac{1}{1+x} = \frac{d\phantom{x}}{dx}\ln(1+x). $$
Integrating gives
$$ \ln(y) = \ln(1+x) + C. $$
Therefore $y=c(1+x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion I want to solve this problem using Taylor expansions.
I tried
\begin{align*}
1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;,
\end{align*}
where $R_{2n}(x)$ is the remainder function.
For $n=2... | I think I found an appropriate method.
The formula for the remainder term of an $n$-degree Taylor polynomial around $a$ is given by
\begin{align*}
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \; ,
\end{align*}
for some constant $c$ in the open interval between $a$ and $x$, so $c\in(a,x)$ or $c\in(x,a)$.
The remaind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Conceptual problem in Theory Of Equations While doing some self-study from the book: Higher Algebra by Hall & Knight, I encountered some articles that I could not understand properly. Those articles were Art. 562 & Art. 563.
Art 562: The result of the preceding article(given below) enables us very easily to find the su... | For all $x$ such that $|x|>\max (|a|,|b|,|c|)$ and for each $d\in \{a,b,c\}$ we have $$\frac {1}{x-d}=(1/x)\cdot\frac {1}{1-d/x}=(1/x)\sum_{n=0}^{\infty}(d/x)^n=$$ $$=(1/x)+(d/x^2)+(d^2/x^3)+...$$ If we sum this over each of the 3 series with $d=a, d=b$ and $d=c,$ the resulting co-efficient of $x^{-5}$ is $a^4+b^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}≥ \frac {(a+b+c)^2}{2}$
EDİTED:
I tried to solve the last question.I can not get the state of equality and I can not find my mistake. Can you show me my mistake(s)?
4.
Let, $a,b,c$ be the lengths of sides of a triangle. Prove the inequality:
$$(a+b) \sqrt {ab... | Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}(2x+y+z)\sqrt{(x+y)(x+z)}\geq2(x+y+z)^2$$ and since by C-S $$\sqrt{(x+y)(x+z)}\geq x+\sqrt{yz},$$ it's enough to prove that
$$\sum_{cyc}(2x+y+z)(x+\sqrt{yz})\geq2(x+y+z)^2$$ or
$$\sum_{cyc}\left(\sqrt{x^3y}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Conditions on coefficients of positive semidefinite matrix with certain symmetries I have a real matrix, with certain symmetries, defined as
$
A = \left( {\begin{array}{*{20}{c}}
1-x&a&b&c\\
a&x&d&b\\
b&d&x&a\\
c&b&a&1-x
\end{array}} \right),
$
with $x,a,b,c,d \in \mathbb{R},{\rm{~ }}0 \le x \le 1$.
I want to obtain co... | You have both $1-(c+d)-\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ and $1-(c+d)+\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ non negative iff:
$$1-(c+d)\geq 0$$
$$(1-(c+d))^2 \geq 4(a-b)^2 +(1-2x+d-c)^2$$
the second condition can be simplified as:
$$((1-x)-c)(x-d) \geq (a-b)^2$$
In particular $(1-x)-c$ and $x-d$ have the same sign.
As the sum $(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I determine this integral? $\int_{0}^{+\infty}\sin^2(1/x)\frac{dx}{(4+x^2)^2}$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}$$
$${\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1-\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1\over 2(4a^2+x^2)^2}-{\cos\left({2a\over x}\rig... | The value of $a$ is irrelevant, the $a$-parameter can be removed through a suitable substitution, and by replacing $x$ with $\frac{1}{x}$ the problems boils down to computing
$$ \int_{0}^{+\infty}\frac{\sin^2(x)}{(4x+1/x)^2}\,dx=\frac{1}{16}\int_{0}^{+\infty}\frac{1-\cos(x)}{(x+1/x)^2}\,dx =\frac{1}{16}\left[\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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For how many primes $p$ is $p^2 + 2$ is also prime? For how many primes $p$ is $p^2 + 2$ also prime?
I'm not sure but I think you can consider every prime number as $(3k+1)$ or $(3k+2)$
| We will not consider $p=2$ because $2^2+2=6$ which is obviouly not prime.
Most definitely $p\neq n^2 + 1$ for all odd $p$ because if we assume otherwise, we want $$(n^2+1)^2+2=n^4+2n^2+3$$ to be prime. Therefore $3\nmid n$. (The only prime divisible by $3$ is $3$ itself, so $n=0$ is a solution, but note that $0^2+1=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
... | There's possibly a better way to see what's going on; instead of manipulating what you want to show, proceed from the left-hand side:
\begin{align}
1^2+2^2+\dots+(k-1)^2+k^2
&<\frac{k^3}{3}+k^2 && \text{(induction hypothesis)}\\[4px]
&=\frac{1}{3}(k^3+3k^2) && \text{(because we want $1/3$)}\\[4px]
&=\frac{1}{3}(k^3+3k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Lagrange Multiplier: Distance to the Origin Find the points on the curve $x^2+xy+y^2=2$ that are closest to the origin.
Is there a way to use Lagrange multipliers to answer this question?
| Without using calculus,
$$8=(2x+y)^2+3y^2$$
WLOG $\sqrt3y=2\sqrt2\sin t, 2x+y=2\sqrt2\cos t\iff x=?$
So, we need to minimize $$\left(\dfrac{2\sqrt2\sin t}{\sqrt3}\right)^2+\left(\dfrac{2\sqrt2\cos t-\dfrac{2\sqrt2\sin t}{\sqrt3}}{2}\right)^2$$
$$=\dfrac{8\sin^2t}3+\dfrac{8(\sqrt3\cos t-\sin t)^2}{12}$$
$$=\dfrac{8\sin^... | {
"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": 1
} |
Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it. Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.
I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M... | Another possibility: apply two times Cesàro-Stolz.
$$
\lim_{n\to\infty}\frac{2n^2 + 10n + 5}{n^2} =
\lim_{n\to\infty}\frac{((2(n+1)^2 + 10(n+1) + 5) - (2n^2 + 10n + 5)}{(n+1)^2 - n^2} =
\lim_{n\to\infty}\frac{4n + 12}{2n + 1} =
\lim_{n\to\infty}\frac{(4(n+1) + 12) -(4n + 12)}{(2(n+1) + 1) - (2n + 1)} =
\lim_{n\to\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Finding the number of multiple of $3$ in the coefficients of $x(x+1)(x+2)\cdots (x+239)$ Suppose
$$
x(x+1)(x+2)\cdots(x+239)=\sum_{n=1}^{240}a_nx^n
$$
What's the total number of $a_n$ which is exactly the multiple of $3$?
I've calculated using Mathematica and got the answer is $160$, but I don't know how to solve it u... | Some variant based on Robert Z's answer not requiring evaluating $\binom{80}{k}\pmod 3$, even though it is the same at final.
Note that $(x-1)^2(x+1)^2=(x^2-1)^2=x^4-2x^2+1\equiv 1+x^2+x^4\pmod 3$
Then we also have a pattern for the cube of such expressions:
*
*$(1+x^2+x^4)^3\equiv 1+x^6+x^{12}\pmod 3$
*$(1+x^6+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$ I've been working on this interesting problem for a while already, and here it is:
In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.
I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart... |
Let $|AB|=7=c$,
$|AC|=15=b$,
$|AM|=10=m_a$,
$|BM|=|MC|=\tfrac{a}2=x$.
Then by Stewart’s theorem for $\triangle ABC$
\begin{align}
b^2x+c^2x&=2x(m_a^2+x^2)
,\\
x&=\sqrt{\tfrac12(b^2+c^2)-m_a^2}
=\sqrt{\tfrac12(49+225)-100}
=\sqrt{37}
,\\
a&=2x=2\sqrt{37}
.
\end{align}
And the area
\begin{align}
S&=\tfrac14\sqrt{4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
inequality of series, showing their difference is less than a number given it is monotonically increasing Given $a_1=1$ $$a_{n+1} =
\frac{2+2a_n}{2+a_n} \text{is bounded: } 1\leq a_n \leq 2, \quad a_{n+1}-a_n=\frac{2(a_n-a_{n-1})}{(2+a_n)(2+a_{n-1})}$$
Prove $$|a_{n+1}-a_n|\leq \frac{1}3\left(\frac{2}9\right)^{n-1}... | You have done most of the work already. Let $b_n=|a_{n+1}-a_n|$ for $n\in \mathbb{N} ^*$.
Then, you proved that $\forall n\in \mathbb{N}^*, b_n\leqslant \frac{2} {9} b_{n-1}$. You can prove $b_n \leqslant b_1 \left( \frac{2} {9} \right)^{n-1}$ using induction since $b_n$ is always positive.
Goal:
Prove that, $\forall n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
... | There is a misprint in the announcement. The $(1,3)$ element of $A$ should be $+1$ rather than $-1$ in order for $A$ to be consistent with the calculations that follow in your posted question.
So let $A=\begin{bmatrix} 3&0&1\\-1&4&-3\\-1&0&5 \end{bmatrix}$ and set
$N=A-4I= \begin{bmatrix} -1&0&1\\-1&0&-3\\-1&0&1 \end{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
sequences above 0,1,2 start&end with 0 with no 2 same successive digits find the number of sequences of length n above {0,1,2} that start and end with 0 and without 2 successive digits that are equal (00,11,22).
Find and solve a recursice equation.
I would be thankful for a clue or an approach to this.
|
We define a recurrence relation as follows. Let
*
*$a_k$ denote the number of valid strings with $k$-th symbol $0$.
*$b_k$ denote the number of valid strings with $k$-th symbol $1$.
*$c_k$ denote the number of valid strings with $k$-th symbol $2$.
Since valid strings start with $0$ we have
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$. Find its 10th term. Given this sequence, find its 10th term and its exact limit.
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
I've been stuck in this question forever. I can't find any relation between them. The answer for the 10th term is $\frac{5... | $$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
$$\frac{1}{2},\frac{5}{(2+1)},\frac{11}{(2+1+5)},\frac{27}{(2+1+5+11)},...$$
$$\frac{1}{2},\frac{(2+3)}{3},\frac{(3+8)}{8},\frac{(8+19)}{19},...$$
In this way I would define the series as follows:
$$\text{ for } a_1=\frac{x_1}{y_1} = \frac{1}{2}$$
$$a_n := \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Question about functional equations
Let $F(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $$F(x) + F\left(\frac{x-1}x\right) = 1+x$$ Find $F(x)$ satisfying these conditions.
Write $F(x)$ as a rational function with expanded polynomials i... | It's just a simplification error. Let $g(x)=\frac{x-1}{x}$. Then, note
$$
g(x)=1-\frac{1}{x},\quad g(g(x))=\frac{1}{1-x},\quad g(g(g(x)))=x.
$$
And so you have the system:
\begin{align*}
1+x&=F(x)+F(g(x)),\\
1+g(x)&=F(g(x))+F(g(g(x))),\\
1+g(g(x))&=F(g(g(x)))+F(x)
\end{align*}
from which
\begin{align*}
F(x)&=[F(x)+F(g(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many different strings of length $9$ containing only the letters a, b, and c have exactly two a's or exactly three b's? The question as stated in the title is how many different strings of length $9$ containing only the letters a, b, and c have exactly two a's or exactly three b's?
I came up with the idea that ther... | This would be a good case for inclusion/exclusion.
Number of ways with exactly $2$ $A$ will $2^7*{9\choose 2}$. (This includes those with $0,1,2,3,4,5$ $B$s.... so this includes those with $2$ $A$ and $3$ $B$s).
Number of ways with exactly $3$ $B$s will be $2^6*{9\choose 3}$ (This includes those with $0,1,2,3,4$ $A$s..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given:
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof th... | I faced same problem here. We know that series approximated around $ x= 0$ and it converged but how do we know that it converge to $\sin x $ ?
As limited knowledge in math. I prefered to start with Euler's formular see more in wikipedia
$$e^{i\theta} =\cos \theta + i\sin \theta $$
Which can be proof as shown in wiki's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Solution of system of equations involving $x_{1},x_{2},x_{3}$
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$
Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_... | option 1:
$x_1 = x_2 = x_3$
$ax^2 + bx + c = x\\
ax^2 + (b-1)x + c = 0\\
x= \frac {-(b-1) \pm \sqrt {(b-1)^2 - 4ac}}{2a}\\
$
option 2:
$x_1 \ne x_2 \ne x_3$
Find a polynomial that intersects:
$(x_1,x_2),(x_2,x_3),(x_3,x_1)\\
p(x) = x_2 + (x-x_1)\frac {x_3-x_2}{x_2-x_1} + (x-x_1)(x-x_2)\left(\frac {x_1 - x_2}{(x_3-x_2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.
| The log series hints at finding the function
$$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{3n+1}}{3n+1} $$
when $x=1$. Taking the derivative, we find
$$ f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-x^3)^n = \frac{1}{1+x^3} $$
Therefore
$$ f(1) = \int_0^1 \frac{1}{1+t^3} dt $$
You can solve the inte... | {
"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": 0
} |
$\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?
| $\sqrt[5]{2^4\sqrt[3]{16}} =$
$\sqrt[5\cdot 3]{2^{4\cdot 3}\color{blue}{16}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}} = \sqrt[15]{2^{\color{blue}{16}}} = 2^{\frac{\not 1\not 5\color{red}{16}}{\not{\color{blue}{1\not 6}}\color{red}{15}}}$
But in my opinion your method seems a littl... | {
"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": 3
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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... | Let $p_i$ denote the number of ways we can obtain $i$ tosses.Obviously $2\le i\le (n+1)$
To find $p_i$ -
1.First find the number of ways we can put $(i-1)$ balls in the bins such that no two go into the same bin
Number of ways = $^np_{i-1}$
2.Put the next ball in one of the bins that already has a ball
Number of way... | {
"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": 3
} |
What's the Probability of a drunk man open a door with $n$ possible keys? I have this following problem in my problem set and I would like to check if my work is in the right directio.
A drunk man with $n$ keys wants to open his door and tries the keys at random. Exactly one key will open the door. Find the mean numbe... | Yes, you are correct. Let $X$ be the number of trials. Expectation value $E[X]$ is given by the standard formula:
$$E[X] = \sum\limits_{k=1}^{\infty} k \cdot P(X=k),$$
where $P(X=k)$ is the probability that in the $k$-th trial the guy opened the door.
In the first case, we are returning the key back. If $k$-th trial w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding parameters of an ellipse in terms of Semi-Latus Rectum and Directrix. I am trying to solve for all the parameters of an ellipse in terms of the Semi-Latus Rectum, $\ell$, and Directrix, $x$. This is for some equation tables I am making so I am looking for the most simplistic expression. In nearly all the cases ... | There's one bug in the first equation though it's unaffected after squaring both sides.
\begin{align}
x &= \frac{a^2}{c} \\
a &= \sqrt{cx} \\
\ell &= \frac{b^2}{a} \\
&= \frac{a^2-c^2}{a} \\
&= \frac{cx-c^2}{\sqrt{cx}} \\
\ell \sqrt{x} &= \sqrt{c} \, \color{red}{(x-c)} \\
0 &= c\sqrt{c}-x\sqrt{c}+\ell \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The inverse image of an interval Someone help me to find $f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)$ where $$f(x)=\begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0\end{cases}$$
I know that
$f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)=\{y\in \mathbb{R}, f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\vare... | Hint:Based on the graph of $f(x)$
$${{f}^{-1}}\left( \left] a,b \right[ \right)=\left\{ \begin{align}
& \phi \ \quad \quad \quad \quad \quad \quad \quad \quad ,if\ b\le 0\ \\
& \left] -\infty ,0 \right[\quad \quad \quad \quad \quad \quad ,if\ b\le 1\ AND\ a<0 \\
& \phi \quad \quad \quad \quad \quad \quad \quad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
The correct answer is $\dfrac13$ while... | With replacement
As you basically note $x^3+y^3 \equiv x+y \pmod 3$, which follows from Fermat's Little Theorem.
We note that $2004 \equiv 0 \pmod 3$, so there's exactly $2004/3=668$ numbers equivalent to $i \pmod 3$, for all $i \in \{0,1,2\}$.
So, no matter what $x$ value is randomly chosen, there are $2004/3$ out of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Minimum of $xy+yz+xz=k$ Given that the sum of x,y,z is 3 find the minimum of xy from the relation$$xy+yz+xz=k$$ Is there anything wrong with my solution since someone said the correct answer differs? $$xy=k-z(x+y)<=>xy=k-z(3-z)=k+z^2-3z=k+\left(z-\frac{3}{2}\right)^2-\frac{9}{4}$$ So $xy\ge k-\frac{9}{4}$ Can anyone g... | With Lagrange multipliers the problem can be formulated as
$$
L(x,y,z,\lambda_1,\lambda_2) = x y + \lambda_1(x+y+z-3)+\lambda_2(xy+yz+xz-k)
$$
The stationary points are the solutions for
$$
\lambda_1 + y + \lambda_2 (y + z) = 0\\
\lambda_1 + x + \lambda_2 (x + z) = 0\\
\lambda_1 + \lambda_2 (x + y) = 0\\
x + y + z = 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What do polynomials solve for? Roots? I had a question in which I’ve been hung up over on. I understand if we had a graph x-y plane and we drew points that intercept the $x$-axis at $2$ and $3$, we would write a quadratic equation that satisfies our condition as $(x+3)(x+2)$, or $x^2+5x+6$, and set this equal to $0$ us... | The graph of a function $y=f(x)$ intersects the $x$-axis when $y=0$, that is when $f(x)=0$.
Now if a factor $(x-a)$ appears in your polynomial, then the polynomial is zero when $x-a = 0$, which occurs precisely when $x=a$. Therefore
$$\begin{align*}
(x-a)(x-b) = 0 & \Leftrightarrow x-a=0 \text{ or } x-b = 0\\
& \Leftri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Application of AM-GM Inequality
If $a,b,c >0$ so that $a+b+c=27$ then what is the maximum value of $(a^2)(b^3)(c^4)$?
I tried the AM-GM inequality but the product term has different powers of $a,b,c$. So how to go about it?
| Hint:
$$a+b+c=27 \iff \frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}=27.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the derivative at (1,2)
$$f(x) = x^2 \sqrt{5 - x^2}$$
Find the derivative at $(1, 2)$.
\begin{align}
\frac{d}{dx} \left[ x^2 \sqrt{5 - x^2} \right] & = \frac{d}{dx} \left[ x^2 (5 - x^2)^{1/2} \right] \\
& = x^2 \frac12 (5 - x^2)^{-1/2}(-2x) + (5 - x^2)^{1/2}(2x)
\end{align}
The equation is formed using the produ... | The derivative actually reads $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x)=x^2$ and $v(x)=\sqrt{5-x^2}$.
For $u(x)$, we have $u'(x)=2x$.
For $v(x)$, we can rewrite $v(x)=f(g(x))$ where $f(t)=\sqrt{t}$ and $g(x)=5-x^2$, so that $v'(x)=g'(x)f'(g(x))=-\frac{2x}{2\sqrt{5-x^2}}$.
Hence $f'(x)=2x\sqrt{5-x^2}-(x^2)(-\frac{2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
How would I obtain the square root of this multinomial? I was doing some problems from a book I found on finding the square root of a polynomial expression. I came across this problem:
$$\frac{a^4}{64}+\frac{a^3}{8}-a+1$$
I utilised the method outlined here, and obtained the following result
$$\frac{a^4}{64}+\frac{a^3}... | We can start small and improve a bit at a time. We can begin with $\frac{a^2}{8}$ which does give the top term when squared. Next,
$$ \left( \frac{a^2}{8} + B a \right)^2 = \frac{a^4}{64} + B \frac{a^3}{4} + B^2 a^2. $$ To get the $a^3/8$ we take $B = 1/2.$ So far, we have
$$ \left( \frac{a^2}{8} + \frac{a}{2} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the ... | Solve for $a$ in the second equation. Plug that into the first. Then solve for $b$. You get:
$b=A\pm \sqrt{A^2-B^2}$.
Plug in for $A=10x+y$ and $B=10y+x$ and expand. You get:
$b=10x+y\pm 3\sqrt{11(x^2-y^2)}$
The only multiples of 11 that are perfect squares are even powers of 11 times even powers of other primes. So, e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
A question about exponential function/equation. I'm solving the following exponential equation
$$4^{x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$$
(My attempt is below)
\begin{align}4^{x}-3^{x-\frac{1}{2}}&=3^{x+\frac{1}{2}}-2^{2x-1}\\\\
4^{x}+2^{2x-1}&=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}\\\\
2^{2x}+2^{2x}\cdot2^{-... | Exponential functions are strictly increasing. At $x<3/2$ we have that $3^{x-\frac32} < 2^{2x-3}$ and at $x>3/2$ we have that $3^{x-\frac32} > 2^{2x-3}$. Using these observations We can say that $3/2$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$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}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Chole... |
I have failed to prove it using 1
Note:
$$\begin{pmatrix}a&b&c&d&e\end{pmatrix}\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}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\
\begin{pmatrix}4a+2b&2a+5b+2c&2b+5c+2d&2c+5d+2t&2d+5... | {
"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": 1
} |
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)) & =\... | Let arcsinh$(x)=y\implies2x=2\sinh(y)=e^y-e^{-y}$
$\implies(e^y+e^{-y})^2=(e^y-e^{-y})^2+4e^y\cdot e^{-y}=4(x^2+1)$
and tanh$(y)=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}=?$
| {
"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": 0
} |
Find the volume of the region common to the interiors of the cylinders First cylinder $x^2+y^2=4$
Second cyclinder $x^2+z^2=4$
My progress so far
$$V=8\int_0^2\int_0^\sqrt{4-y^2}\int_0^\sqrt{4-x^2}dzdxdy$$
I know I can substitute $x$ with $2\sin(\theta)$ in $\int\sqrt{2-x^2}$ but if I do that then I came across $sin(4\... | You may use a triple integral to find the volume.
$$ V= 8\int _0^2 \int _0^{\sqrt {4-x^2}}\int _0^{\sqrt {4-x^2}}dzdydx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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_... | Hint. The function
$$
z \mapsto \frac{1}{\sin(z)}-\frac{1}{z}
$$ is regular and odd near $0$ thus the residue of
$$
f(z) = \frac{1}{z^3 \sin(z)}
$$is equal to zero.
| {
"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": 2
} |
Finding every solution for equation of complex numbers I need to find every solution for:
$\ z^{3} + 3i \overline z = 0 $
So I tried was just to compare imaginary and complex part of $\ z^{3} $ and $\ 3i\overline z$
Ill spare you the alegbra, here is the result:
$$\ a^{3} - 3ab^{2} + i(3a^{2}b-b^{3}) = -3b -3ai \\a^... | Write your equation in the form: $$z(z^2+3i)=0$$
and then you will get: $$x^2-y^2+i(2xy+3)=0$$ or $$z=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\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 \... | You're right; but it depends on what the full question is; if one is asked to compute
$$
\int_{0}^{1}\arcsin\frac{2x}{1+x^2}\,dx
$$
then just the antiderivative $2\arctan x$ is sufficient. Not if one wants to compute an integral involving points outside the interval $[-1,1]$.
Here's a shorter way to get at your result.... | {
"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": 0
} |
If $x^2-bx+c=0$ has real roots, then prove that both are greater than $1$ when $c+1>b>2$.
If $ x^2-bx+c=0$ has real roots, prove that both roots are greater than $1$, when $c+1>b>2$.
Working
I tried to prove the given inequality by taking roots greater than $1$.
Let $\alpha$, $\beta$ be the roots of the quadratic ... | The roots of
$x^2-bx+c=0$
are
$x
=\dfrac{b\pm\sqrt{b^2-4c}}{2}
$.
If
$c+1 > b > 2$,
the smallest root is
$\dfrac{b-\sqrt{b^2-4c}}{2}
$
and
$b^2 > 4c$
so
we want
$b-\sqrt{b^2-4c}
\gt 2$
or
$(b-2)^2
\gt b^2-4c
$
or
$b^2-4b+4
\gt b^2-4c
$
or
$c+1 > b$
which we are given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Confused about finding LCM of three numbers I need to find the LCM of three numbers:
*
*$x^2-36$
*$2x^2-12x$
*$x^2-6x$
I factor:
*
*$x^2-36=(x+6)(x-6)$
*$2x^2-12x=2x(x-6)$
*$x^2-6x=x(x-6)$
Then I multiply all factors. Every factor that appears in all three expressions I multiply only once. $(2x)(x)(x-6)(x... | Note that $x$ is already contained in $2x$ thus $2\cdot x\cdot (x+6)\cdot (x-6)$ is the correct answer.
Indeed the different factors which appear are
*
*$2$
*$x$
*$(x-6)$
*$(x+6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
which of the following statement is TRue ?..... let $f : R \rightarrow R$ be a continious and nonnegative function.
which of the following statement is TRue ?
a) if there exist $ c \in (0,1)$ such that $f(c) = 100$ then $\int_{0}^{1} f(x) dx \ge \frac {1}{2}.$
b)$\int_{0}^{1} f(x) dx > \frac {1}{2}$.then $f(c) > \... | In the same vein as GNU Supporter's answer, but not as elementary (and neat :D)
Consider another "peak function", namely
$$f(x) = 100\ e^{-a(x-\frac{1}{2})^2}\ .$$
Clearly $f(\frac{1}{2}) = 100.$ We can compute
\begin{align*}
\int_0^1 100\ e^{-a(x-1/2)^2} \ dx
&= 100 \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-a x^2} \ dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int x\sin^2 (x) dx$ Integrate $\int x\sin^2 (x) dx$
My attempt:
$$=\int x\sin^2 (x) dx\\
=x^2\sin^2 (x) - \int 2\sin (x)\cos (x)x^2 dx\\
=x^2\sin^2 (x) - \int \sin (2x) x^2 dx.$$
| Here is a method using tabular integration, where we use $\sin^2 x = \frac{1}{2} (1 - \cos 2x)$:
$$\begin{array}{c|c} D & I \\ \hline \color{red}{x} & \frac{1}{2} - \frac{1}{2} \cos(2x) \\ \hline \color{blue}{1} & \color{red}{\frac{1}{2}x - \frac{1}{4} \sin(2x)} \\ \hline 0 & \color{blue}{\frac{1}{4}x^2 + \frac{1}{8} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}... | $$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
Substitute $z=y''$
$$z'+z=x^2+2x+2$$
$$(z-x^2-2)'+(z-x^2-2)=0$$
$$v'=-v$$
Where $v=z-x^2-2$
$$\implies \ln|v|=-x+K \implies v=K_1e^{-x} $$
$$\implies z=x^2+2+K_1e^{-x}$$
Integrate twice to get y
$$\boxed{y=K_1e^{-x}+K_2x+K_3+x^2+\frac {x^4}{12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ I am tring to evaluate
$$I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$$
The first thing I did was to notice that
$$\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)$$
So I integrated by parts in order to get
$$I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x... | There is no need to evaluate the integral to answer the original question.
The original question is a multiple choice question so ruling out every option but the right one leads to the right answer of course. One sees that the integrand is positive almost everywhere. Moreover the arctangent function is increasing, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the Eigenvalues and its Eigen vectors. Consider the $ \ n \times n \ $ matrix $$ \begin{pmatrix} a & -1 & & & & \\ -1 & a & -1 & & & \\ & -1 & a & -1 & & & \\ & & -1 & a & -1 & \\ & & ..... & .... & .... \\ & & && a & -1 \\ & & & &-1 & a \end{pmatrix} $$
Find the Eigenvalues and its Eigen vectors.
Also d... | First question: No, it appears you've made a computational error. I'm getting eigenvalues that follow in the example.
Given a matrix $A,$ then one computes eigenvectors with eigenvalue $\lambda$ by computing a basis for $$\ker(A-\lambda I_n).$$
Example:
$$A=\begin{pmatrix} a & -1 &0 \\ -1 & a & -1 \\0& -1 & a \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
${3^n\choose k}$ is divisible by $3$? How can I prove that ${3^n\choose k}$ is divisible by $3$ for all positive integer values of $n$? (where $k$ is any positive integer smaller than $3^n$) Can you use induction? Thanks.
| Looking at the fractional expansion of the binomial coefficient, you can cancel out factors of 3 in the $(3n+1)^\text{th}$ value in the numerator with the $(3n)^\text{th}$ value in the denominator, for example:
$${3^3 \choose k} = \begin{array} {c}
27 & 26 & 25 & \color{red}{24} & 23 & 22 & \color{blue}{21} & 20 & 19 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possib... | The first thing that screams at me is $3|x^3 + x^3$ so
$x^3 \equiv -y^3 \mod 3$ so $x\equiv -y \mod 3$
Let $x \equiv i \mod 3$ and $y \equiv -i \mod 3$. If $x = 3k + i$ and $y = 3j -i$ then $\frac {x^3 + y^3}{3} = 9(k^3 + j^k) + 9(k^2i-j^2i)+3(ki -ji) + \frac {i^3 - i^3}3 \equiv 0 \mod 3$
So $xy + \frac {x^3 + y^3}3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding equation with roots $\cos(2k+1)\pi/9$ and using Vieta's formula's Question
From the equation who's roots are $\cos\frac{\pi}9,\cos\frac{3\pi}9,\cos\frac{5\pi}9,\cos\frac{7\pi}9$ and hence prove
a) $8\cos\frac{\pi}9\cos\frac{5\pi}9\cos\frac{7\pi}9=1=8\cos\frac{\pi}9\cos\frac{2\pi}9\cos\frac{4\pi}9$
b) $\sec... | $$\implies x^4-3x^2+1=x(x^2-2)$$
$$\implies(x^4-3x^2+1)^2=x^2(x^2-2)^2$$
Set $\sec^2\dfrac{(2k+1)\pi}9=\dfrac4{x^2}=v$(say)
On replacement & simplification
$$ v^4-v^3(24+16)+v^2(144+32+64)+v(\cdots)+1=0$$
whose roots are $\sec^2\dfrac{(2k+1)\pi}9,k=0,1,2,3$
$$\sum_{k=1}^4\left(\sec^2\dfrac{(2k+1)\pi}9\right)^2$$
$$=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"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
$$\mathbf{F}'(r)=\int_{0}^{\pi}{-2\cos x + 2r \over1-2r\cos x +r^2}dx$$... | If you continue $$A={r+ru^2+u^2-1 \over 1+u^2-2r+2ru^2+r^2}{1 \over 1+u^2}=\frac{(r+1) u^2+(r-1) } {(1+u^2)((r+1)^2 u^2+(r-1)^2)} $$ For the time being, let $t=u^2$ to get
$$A=\frac{(r+1) t+(r-1) } {(1+t)((r+1)^2 t+(r-1)^2)} $$
being and partial fraction decomposition leads to
$$A=\frac{1}{2 r}\frac{1}{
\left(t+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove: $\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$ Prove: $$\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$$
for $x,\,y\in \left [ 1,\,2 \right ]$
My unsuccessful try (with my... | An alternate way to your substitution: $x\in [1, 2] \implies (x-1)(x-2)\leqslant 0 \implies x^2\leqslant 3x-2$. Similarly $y^2\leqslant 3y-2$. Using these in the LHS, we get
$$LHS \geqslant \frac{x+2y}{3(x+y+1)}+\frac{y+2x}{3(x+y+1)}+\frac1{4(x+y-1)}$$
Simplifying and letting $x+y=t$, we are trying to find the minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it ... | Another way to see the solution is by using
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})\tag{1}$$
leading to
$$1^n-3^n-6^n+8^n=(8^n-3^n)-(6^n-1^n)=\\
(8-3)(8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}+3^{n-1})-\\
(6-1)(6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}+1^{n-1})=...$$
we can see that $8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Finding limit for infinite quantities.
Find $$\lim_{x\rightarrow 0}{\color{red}{x}} \cdot\bigg(\dfrac{1}{1+x^4}+\dfrac{1}{1+(2x)^4}+\dfrac{1}{1+(3x)^4}\cdots\bigg)$$
As terms are written infinitely my intuitions doesn't let to give answer as $0$. So tried calculated that weird sum something like,
$T_n=\dfrac{1}{1+(nx... | Making $n = \frac{1}{x}$
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^4}\equiv\int_0^1\frac{d\xi}{1+\xi^4} = \frac{\pi +2 \coth ^{-1}\left(\sqrt{2}\right)}{4 \sqrt{2}}
$$
NOTE
For integration purposes
$$
\frac{1}{1+\xi^4} = \frac{a_1\xi+b_1}{\xi^2+\sqrt 2 \xi + 1}+\frac{a_2 \xi+ b_2}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Proof involving generating function The following is part of a proof that the number of ways of associating a product with $n$ terms (different ways of inserting parentheses) is $$
a_1 = 1,\ a_n = \frac{1}{n} \binom{2n-2}{n-1},$$
and the relationship$$
a_{n+1} = a_1a_n + a_2a_{n-1} + a_3a_{n-2} + \cdots + a_na_1$$
is a... | Let
$$f_n = \frac{1}{n} \binom{2n-2}{n-1} = \frac{1}{2n - 1} \binom{2n-1}{n}$$
\begin{align}
f(x) &= \sum_{n=1}^\infty f_nx^n\tag{1}\\
&= \sum_{n=1}^\infty \frac{1}{2n - 1} \binom{2n-1}{n}x^n\\
&= \sum_{n=1}^\infty \frac{(2n-2)!}{n!(n-1)!}x^n\\
&= \sum_{n=1}^\infty \frac{1}{n} \binom{2n-2}{n-1}x^n\tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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... | $$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{5-2e^{ix}-2e^{-ix}}=\frac{e^{ix}}{5e^{ix}-2e^{2ix}-2}
$$
Let $X=e^{ix}$ then
$$
-2X^2+5X-2=-\left(2X-1\right)\left(X-2\right)
$$
Now we do a partial decomposition
$$-\frac{1}{\left(2X-1\right)\left(X-2\right)}=\frac{2}{3}\frac{1}{2X-1}-\frac{1}{3}\frac{1}{X-2}
$$
So far we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also. For some real $x$, let $y=x +\frac1x$, with $y\in \mathbb{Z}$.
$y=x +\frac1x\implies x^2 -xy +1 =0\implies x= \frac{y\pm \sqrt{y^2 -4}}2$, so $x+\frac1x = \frac{y\pm \sqrt{y^2 -4}}2+ \frac2{y\pm \sqrt{y^2 -4}}$
This implies: $ y ... | Let $A_n = (x + 1/x)^n$.
Then $A_n$ is a palindromic polynomial with integer coefficients.
Let $Y_k = x^{k} + 1/x^{k}$.
Then $Y_{k+1} = A_{k+1} - \sum_{j = 1}^k c_k Y_k$ where $c$ is the list of integer coefficients necessary to make the construction.
So if $x$ is such that $A_1$ (and thus $Y_1$) is an integer, then al... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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}+... | No, it's not correct.
$3d+6=0$ and $3e-4d-1=0$ gives $d=-2$ and $e=-\frac73$.
From $D_n=4D_{n-1}$, we get $D_n=D_0 4^n$ with $D_0 = C_0 - 0d - e = 2 + \frac73 = \frac{13}{3}$, that is, $D_n=\frac{13}{3} 4^n$.
This gives $C_n = D_n -2n -\frac73 = \frac{13}{3} 4^n -2n -\frac73 = \frac13 (13 \cdot 4^n - 6 n - 7)$.
Your ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Arithmetic doubt while studying limits of sequences While studying limits of sequences, I came across these expressions.
$\left|\frac{-5}{n+2}\right|<\delta \iff\frac{5}{n+2}<\delta \iff n+2>\frac{1}{\delta }$
$n\in \mathbb N$
$\delta\in \mathbb R$ and $\delta >0$
I'm struggling to understand how it went from the secon... | To treat that as an equal sign just makes it dead wrong. What they mean is implies the next. (Actually; we must assume $\delta > 0$ and that $n + 2 > 0$. I presume those are conditions stated in the source that you omitted? That $n\in \mathbb N$?)
(Note: these implications only work one way.)
$\left|\frac{-5}{n+2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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integration by parts $\int \frac{x^{2}+4x}{x+2}\,dx$ $$\int \frac{x^{2}+4x}{x+2}\,dx$$
I have written it in this form:
$$\int \frac{(x+2)^{2}-4}{x+2}\,dx$$: on this stage I try to do integration by parts, which gets me to :
$$\frac{x^{2}}{2}+2x-4\ln\left | {x+2} \right |$$
but it's wrong for some reason. The right answ... | Note that we have $$\frac{x^2+4x+4}2-4\ln|x+2|+C=\frac{x^2}2+2x\color{red}{+2}-4\ln|x+2|+C$$ and this is the same as your expression if we let $C_1=C+2$!
A quicker way: $$\int \frac{(x+2)^{2}-4}{x+2}\,dx=\int\left[x+2-\frac4{x+2}\right]\,dx=\frac{x^2}2+2x-4\ln|x+2|\color{blue}{+C}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Is element $5 + 2i$ irreducible in the ring $\mathbb{Z}[i]$? I want to prove that $5 + 2i$ is an irreducible in the ring $\mathbb{Z}[i]$. I was wondering how to do so without the use of the norm defined on this ring.
Let's suppose that it is reducible, i.e. $5 + 2i = (a + bi)(c + di)$ for some $a, b, c, d \in \mathbb{Z... | Assume $5+2i = (a+ib)(c+id) = (ac - bd) + i(ad + bc)$.
Notice that
$$(a-ib)(c-id) = (ac - bd) - i(ad + bc) = 5-2i$$
Multiply these two equalities:
$$29 = 5^2 + 2^2 = (5+2i)(5-2i) = (a+ib)(c+id)(a-ib)(c-id) = (a^2+b^2)(c^2+d^2)$$
Since $29$ is prime, this implies $a^2+b^2 = 1$ or $c^2+d^2 = 1$. In the first case we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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... | The polynomial $x^4+6x^3+11x^2+6x+1$ has symmetric coefficents - more precisely, it's called a palindromic polynomial:
$$p(x)=x^4+ax^3+bx^2+ax+1$$
The goal is to factor $p(x)$ (and show that it factors to a square of some expression). Let's start by dividing by $x^2$ and refactoring:
$$\frac{p(x)}{x^2} = q(x) = x^2+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 6
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The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
(A) $2 \cos 10$
(B) $2 \cos 20$
(C) $2 \cos 40$
(D) $2 \cos 80$
I was dumb enough to square the expression to reach $x^8-4x^6+4x^4+2-x=0$ which is clearly a dead end ;-;
| Using the trigonometric identity, $$\cos2x=2\cos^2x-1=1-2\sin^2x$$
$$2\cos2x+2=4\cos^2x$$
We are looking for an angle that allows for the relationship $\sin x=\cos(\frac{\pi}{2}-x)$ in the final step. This is because, working from the inside out, the first gives a cosine, the second gives a sine, so in order for the la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluating trivial integral with linear algebra $$\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c} = \frac{\pi}{\sqrt{\det(A)}}$$
where $A = \begin{bmatrix}a&\frac{b}{2}\\\frac{b}{2}&c\end{bmatrix}$
The connection to matrix quadratic form is given via the equivalence:
$$ax^2 + bx + c \equiv \begin{bmatrix}x&1\end{bm... | Note: I am relatively new here, so if I didn't interpret your actual question correctly, sorry in advance.
The integral in question is
$$I=\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c}.$$
Completing the square for the denominator yields
$$I=\frac{1}{a}\int_{-\infty}^{\infty} \frac{\text{d}x}{(x+\frac{b}{2a})^2+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Solving a congruence of the form $a^x = b \pmod m$ without indices or primitive roots Consider $9^x\equiv 7 \mod 19$. So $9^x\equiv 26 \equiv 45$, $9^{x-1} \equiv 5 \equiv 24 \equiv 43 \equiv 62 \equiv 81$, so $x=3$, and $19 \mid 722$.
But what I really want to solve is $12^x\equiv 17 \mod 25$. Using the same method, $... | There are several possible answers to your question.
*
*You don't do the "same thing", since you are breaking the left hand side in two powers. In fact, it is not true that $a^xb^y\equiv a^zb^t \pmod n$ implies that $x=z$ and $y=t$, even not modulo the respective order $\pmod n$, or even if they are both "primes" (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\binom{n}{k}$ for $n>k^{100}$ This paper I'm reading says this
"We used the fact that $4k^2n^{k-1}<4(k+1)!\binom{n-1}{k-1}$ for $n>k^{100}$." The only condition here is that $k\geq3$ and $n$ can be sufficiently large. However, this seems incorrect to me.
| Note that
$$\begin{align}\frac{(k-1)!{n-1\choose k-1}}{n^{k-1}}
&=\frac{(n-1)!}{n^{k-1}(n-k)!}\\
&>\frac{(n-k+1)^{k-1}}{n^{k-1}}\\
&=\left(1-\frac{k-1}{n}\right)^{k-1}\\
&>1-\frac{(k-1)^2}n \end{align}$$(where we first use that $\frac{(n-1)!}{(n-k)!}$ consists of $k-1$ factors $\ge n-k+1$ and in the last step use Bern... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$... | Consider the function for $x>1,n>1$
$$f(n)=n^x-(n-1)^x$$
$$f'(n)=xn^{x-1}-x(n-1)^{x-1}=x(n-1)^{x-1}[(1+\frac{1}{n-1})^{x-1}-1]>0$$
So $f(n)$ is increasing for $x>1,n>1$.
Now rewrite the equation
$$5^x+7^x+11^x=6^x+8^x+9^x$$
$$\color{red}{(11^x-10^x)}+\color{blue}{(10^x-9^x)}=\color{red}{(8^x-7^x)}+\color{blue}{(6^x-5^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solving $a! + b! = 2^n$
$a! + b! = 2^n$, find $a,b,n$
My obseravtions thus far:
*
*If $a>4$, $b<4$, otherwise $a! + b! = 0 \mod 10$
*If $a=1$, $b=1$
If we try all combination with the numbers 2 to 4, we get following solutions:
(1,1,1), (2,2,2), (2,3,3),(3,2,3)
Are there more solutions and how to find them?
| I guess I would do: wolog $b < a$ so $a! + b! = b!([a-b]! + 1) = 2^n$.
If $b \ge 3$ then $3|b!([a-b]! + 1)=2^n$ but that's impossible. So $b \le 2$.
Also the only factor of $2^n$ are powers of $2$ so $[a-b]! + 1 = 2^m$ so $[a-b]! = 2^m -1$ which is an odd number if $m > 0$. But if $[a-b] \ge 2$ then $[a-b]!$ is even.... | {
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"url": "https://math.stackexchange.com/questions/2840866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
| Given $f(x) = ax^2+bx+c\geq 0\forall x \in \mathbb{R}$
Now put $x=-2,$ We get $f(-2)\geq 4a-2b+c\geq 0\Rightarrow 2a+c\geq 2(b-a)$
So $\displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}\geq 1+2=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, tha... | $$\frac{4x^2+8x+13}{6(x+1)} = \frac{4(x+1)^2+9}{6(x+1)} \ge \frac{2\sqrt{4(x+1)^2\cdot 9}}{6(x+1)} = \frac{2\cdot2\cdot3}{6} =\frac{12}{6} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using De Moivre's to find roots of a polynomial a) Use De Moivre's theorem to express $\frac{\sin 8\theta}{\sin\theta \cos\theta}$ as a polynomial in$ s$, where $s=\sin\theta$
b) Hence solve the equation $x^6-6x^4+10x^2-4=0$
I've been able to do the first question and I worked out the answer to be $8(1-10s^2+24s^4-16s^... | Solution
$$x^6-6x^4+10x^2-4=(x^2-2)(x^4-4x^2+2)=(x^2-2)[(x^2-2)^2-2]=0$$
Thus, let $x^2-2=0$ or $x^4-4x^2+2=0$. We obtain $$x=\pm\sqrt{2},$$or$$x=\pm\sqrt{2\pm\sqrt{2}}.$$
| {
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"url": "https://math.stackexchange.com/questions/2847248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can these two equations be solved by elimination? In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.
$$
\frac{3x(3x^... | If you would like to see how to get the equation by hand:
Introduce a third variable $$z = \frac{3x^2+9}{2y}$$ We have three equations
$$2yz - 3x^2 = 9\\3xz - z^3 - y = 6\\z^2 - 2x = 12$$
Multiplying the second equation by $2z$ and adding the first eliminates $y$:
$$6xz^2 - 2z^4 - 3x^2 = 12z + 9$$
Substituting for $z^2... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$ Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\... | Let $x:=n+f$. The equation is
$$f+\frac1{n+f}=1,$$
giving the solutions in $f$
$$f=\frac{\pm\sqrt{(n+1)^2-4}-n+1}2.$$
The negative sign cannot work, nor the negative $n$. Then $n\ge1$ is required, but $n=1$ yields $x=1$, which is wrong. Finally,
$$f=\frac{\sqrt{(n+1)^2-4}-n+1}2, \forall n>1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to calculate $\int \frac{\cos^2 x}{1 + \sin^2 x}dx$ My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\cos^2x = \frac{1 + \cos(2x)}{2}$
| We reduce the degree of the numerator with
$$\int\frac{\cos^2x}{1+\sin^2x}dx=\int\frac{2-(1+\sin^2x)}{1+\sin^2x}dx=2\int\frac{dx}{1+\sin^2x}-x$$
and we introduct a tangent to rationalize,
$$\int\frac{dx}{1+\sin^2x}=\int\frac{dx}{\cos^2x+2\sin^2x}=\int\frac{dx}{\cos^2x(1+2\tan^2x)}=\int\frac{dt}{1+2t^2}
\\=\frac1{\sqrt2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x... | Rewrite the equation: $x^2 - 4yx + y^2+y = 0\implies\triangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 \implies y(3y-1)=k^2$ . Observe that $\text{gcd}(y,3y-1) = 1$ since if $d = \text{gcd} \implies d \mid y, d \mid 3y-1 \implies y = md, 3y-1 = nd\implies 1 = 3y -nd= 3md - nd = d(3m-n)\implies d = 1\implies y = u^2, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Angles on a point inside a triangle Let $ABC$ be an isosceles triangle with $AB=AC$ and $∠BAC = 100$. A point $P$
inside the triangle $ABC$ satisfies that $∠CBP=35$ and $∠PCB= 30$. Find the
measure, in degrees, of angle $∠BAP$. Attached is the figure of the triangle
I tried to Angle Chase but it seemed true for all ... | Let $M$ be the midpoint of $BC.$ Let $Q$ be the intersection of $CP$ with $AM.$
Note that $\angle ABC = \frac12(180^\circ - 100^\circ) = 40^\circ$
and therefore
$$ \angle PBA = \angle ABC - \angle PBC = 40^\circ - 35^\circ = 5^\circ.$$
Since $Q$ is on the perpendicular bisector of $BC,$ triangle $\triangle BQC$
is is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For prime $p$ do we have $p^3+p^2+p+1=n^2$ infinitely often? This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?
| There are no more positive prime solution other than $p = 7$.
First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.
Let's say $p$ is an odd prime such that
$$p^3 + p^2 + p + 1 = n^2\quad\iff\quad(p^2+1)(p+1) = n^2$$
Since $\gcd(p+1,p^2+1) = \gcd(p+1,2) = 2$, we find for some $k, m > 0$,
$$\begin{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to \sqrt 2}\frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}$ I used rationalization trick where you multiply the top and bottom with opposites of given functions and thus got $\frac{2(x^2+2)}{(x^2-2)({\sqrt {3+2x} + (\sqrt2 + 1)})}$ but by then it has become too complex. The book says answer should be $\fra... | You made a mistake in the numerator.
\begin{align}
\frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}\cdot \frac{\sqrt{3+2x} + (\sqrt 2+1)}{\sqrt{3+2x} + (\sqrt 2+1)} &= \frac{3+2x - (\sqrt{2}+1)^2}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\
&= \frac{2(x-\sqrt{2})}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\
&= \frac{2}{(x+\sqrt{2})(\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(... | So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to determine two matrices are conjugate. Which of the following statements are true?
*
*The matrices $
A=\left[ {\begin{array}{cc}
1 & 1 \\
0 & 1\\
\end{array} } \right]$ and $
B=\left[ {\begin{array}{cc}
1 & 0 \\
1 & 1\\
\end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$
*The matri... | Quiver has already told you (and I am sure it is in you textbook) that two matrices, A and B, are conjugate if and only if there exist an invertible matrix, P, such that $A= P^{-1}BP$. That is equivalent to $PA= BP$.
In the first problem, $A= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $B= \begin{bmatrix}1 & 0 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximum of $\sqrt{\frac{1}{4}\cdot \sin^2(t)+\sin^2(t+\frac{\pi}{3})}$ Maximum of $\sqrt{\frac{1}{4}\cdot\sin^2(t)+\sin^2(t+\frac{\pi}{3})}$
In my opinion, the maximum of the sinus is 1, so I calculated $\sqrt{\frac{1}{4}\cdot1+1}$
This is wrong, why?
| Other people have pointed out the OP's error. Now, I am supplying a solution.
Note that
$$\sin\left(t+\frac{\pi}{3}\right)=\frac{1}{2}\sin(t)+\frac{\sqrt{3}}{2}\cos(t)\,.$$
That is,
$$\begin{align}\frac{1}{4}\sin^2(t)+\sin^2\left(t+\frac{\pi}{3}\right)&=\frac{\sin^2(t)+\big(\sin(t)+\sqrt{3}\cos(t)\big)^2}{4}\\&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy
$$
\textrm{det } X=
\begin{vmatrix}
x &2 &2 &2\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
My teacher does this by adding the three bottom rows to the top row
$$
\textrm{det } X=
\begin{vmatrix}
x+6 ... | "Subtracting a row of 2s" is NOT a valid matrix operation. But subtracting one row from another is. Perhaps what your teacher did (or meant to do) is, first subtract the third row from the second to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 2 & 2 & x & 2 \\ 2 & 2 & 2 & x \end{array}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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.