Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\frac{n}{3} \le \sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2} \le \sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1}$.
Given $0 < y < 1 < n \in \mathbb Z^+$ and $n$ positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $\displaystyle \sum_{i = 1}^nx_i = n$.
Prove that $$\large \frac{n}{3... | The left inequality.
We need to prove that
$$\sum_{i=1}^n\left(\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{i=1}^n\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\geq0$$ or $$\sum_{i=1}^n\left(\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Irrational equation high school I've been trying to solve this one without success... can anybody help me?
The result should be $x=\frac{17}{16}$ and it's correct, I've already checked.
This is the equation:
$$\frac{1}{\sqrt {x+2} - \sqrt x}+\frac{2}{\sqrt {x+4} + \sqrt x}=2$$
Thanks in advance!
| Rationalize the denominators.
$$\frac{1}{\sqrt{x+2} - \sqrt{x}} = \frac{\sqrt{x+2} + \sqrt{x}}{2},$$
and
$$\frac{2}{\sqrt{x+4} + \sqrt{x}} = \frac{2\sqrt{x+4} - 2\sqrt{x}}{4},$$
and so we find
$$\sqrt{x+2} + \sqrt{x+4} = 4.$$
Thus,
$$\sqrt{x+4} = 4 - \sqrt{x+2},$$
and so we find that
$$x+4 = x + 18 - 8\sqrt{x+2},$$
$$8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach :
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^... | The following steps are not allowed
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \color{red}{=\lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} }= \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} =\\ \color{red}{= \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} }$$
since $\lim_{x\to0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of product of rational functions How can I prove that the sequence $a_{n} = \frac{4}{5} * \frac{104}{105} * ... * \frac{50n^{2} - 50n + 4}{50n^{2} - 50n + 5}$ converges or not (where $a_{1} = \frac{4}{5}$, $a_{2} = \frac{4}{5} * \frac{104}{105}$, and $a_{3}= \frac{4}{5} * \frac{104}{105} * \frac{304}{305}$)... | Hint: Your sequence is
$$a_N=\prod_{n=0}^N\frac{50n^2-50n+4}{50n^2-50n+5}=\prod_{n=0}^N\left(1-\frac1{50n^2-50n+5}\right).$$
Each additional term in the product causes it to decrease in value since the $a_n$ are positive and the term is less than $1$. Now can you think of a way to use monotone convergence theorem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Direct telescopic proof for sum of $1^2+2^2+...+n^2$ I was teaching $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ for the first I did direct telescopic proof like below
$$\sum_{k=1}^{n}k=\sum_{k=1}^{n}k(\frac{k+1}{2}-\frac{k-1}{2})=\sum_{k=1}^{n}(\frac{k(k+1)}{2}-\frac{k(k-1}{2}))=\\\s... | Noting that
$$ (n+1)^2-n^2=2n+1, (n+1)^3-n^3=3n^2+3n+1 $$
one has
$$ n=\frac12\bigg[(n+1)^2-n^2\bigg]-\frac12,n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13. $$
So
$$ n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13=\frac13\bigg[(n+1)^3-n^3\bigg]-\frac12\bigg[(n+1)^2-n^2\bigg]+\frac16 $$
and hence
\begin{eqnarray}
\sum_{k=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find the the least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real The least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real.
let $f(x) =x^4-4x^3 -8x^2 +a=0$
$ f'(x) = 4x^3 -12x^2 -16x $
Put $f'(x) = 0 $ we get x =0, -1, 4. How to p... | Let $f(x)=x^4-4x^3-8x^2+a$, $$f'(x)=4x^3-12x^2-16x=0 \Rightarrow x=4,-1,0$$
Next $f''(x)=12x(x-2)-16.$ so $f(x)$ has local minima at at $x=4,-1$ and loca max at $x=0.$ Hence $f_{min}=a-128, a-3$, $f_{max}=a$. For four real rootss: $f_{min}<0$ and
$f_{max}>0.$ Thus, for four real roots $$a \in( 0,3).$$
Both the integers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that t... | Note that the induction hypotesis reduces to show that
$$ \frac{1}{\sqrt{2k+1}} \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}} \implies\sqrt{(2k+1)(2k+3)}\le 2k+2$$
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will b... | You can use this result: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
Since $T_n$ is of degree $4$ then its sum will be of degree $5$.
So let $S(n)=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5$.
Then you can proceed by solving the system obtained calculating the first $6$ terms $S(1)$ to $S(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Surface tangent plane Determine a plane that is tangent to the surface $x ^{2}$ + $3y ^ 2$ + $2z ^ 2$ = $\frac{11}{6}$ and parallel to the plane x + y + z = 10
$$\nabla f (x, y, z) = (2x, 6y, 4z)$$
Quotation vector $X + Y + Z = 10 \implies (1,1,1)$
$$(2x, 6y, 4z) = K (1,1,1)$$
for $K = 1$
*
*$2x = 1 x = \frac{1}{2}$... | I think there is an error for $x$ and $y$, I obtain $K=\pm 2$ and
*
*$x=\pm1$
*$y=\pm \frac13$
*$z=\pm \frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Perfect numbers (prime number theory) Calculate x and y, so that $2 ^ x. 3 ^ y$ be a perfect number.
I tried to produce using this formula of the sum of the divisors of a number: Let $a ^ x, b ^ y,$ and $c ^ z$ prime factors of a number: $\frac{ a^{x+1} -1}{ a-1}. \frac{b^{y-1} -1}{b-1}. \frac{c^{c +1} -1}{c-1}$
It re... | An even number is perfect if and only if it is of the form $2^{n-1}(2^n-1)$ for some $n$ with $2^n-1$ being prime. And $2^n-1=3^m$ is prime only for $n=2$ and $m=1$. So we obtain $2^1\cdot 3^1=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$ Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$
First, we have
$$ n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 $$
for all positive integers $n$. (Since $\frac{\sin x}{x} < 1 \forall x > 0$.) Therefore,
$$ \sum_{n=1}^{\infty}\le... | We have that
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + O\left(\frac1{n^5}\right)$$
$$n\sin \frac{1}{n} = 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)$$
$$\log \left(n \sin \frac {1}{n}\right)=\log \left( 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)\right)=-\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$.
Then i calculated the area with
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next.
| If $AB=x$, then $AC=x-2,BC=x+1$.
Now use the cosine formula: $$x^2=(x-2)^2+(x+1)^2-(x-2)(x+1)$$ Expanding we get $x=7$. So the longest side is 8 and the height $AC\sin60^o=5\sqrt3/2$. Hence the area is $10\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| If $a\geq b$ then $a^3\geq b^3$ and $a^5\geq b^5$ so $(a^3-b^3)(a^5-b^5)\geq 0$. Otherwise $a^3-b^3$ is negative, and so is $a^5-b^5$, so their product is positive, and the inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
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Calculating $\iiint_E\sqrt{3x^2+3z^2}\,dV$
Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.
I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $... | Observe that the solid is symmetric with respect to the y-axis. So, it is efficient to integrate using cylindrical coordinates along the y-direction. Rewrite the original integral as
$$\iiint_E\sqrt{3x^2+3z^2}\,dV =\int_0^8 dy \int_0^{2\pi}d\theta\int_0^{\sqrt{y/2}} \sqrt{3}r\>rdr$$
$$=2\pi\sqrt 3 \int_0^8 dy \frac13 \... | {
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"timestamp": "2023-03-29T00:00:00",
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Numeric sequence Beginning with $2$ and $7$, the sequence $2, 7, 1, 4,7 ,4, 2 , 8,$... is constructed by multiplying successive pairs of its members and adjoining the result as next one or two members of the sequence, depending on whether the product is a one- or a two-digit number. Prove that the digit 6 appears an in... | (Too long to be a comment)
To slightly demystify how to approach this problem, one idea is to find a never-ending sequence, which then hopefully repeats. There is nothing obvious to do, so let's start naively:
Starting with 2, 7:
2, 7, ...
1, 4, ...
4, ...
At this point, we cannot continue.
So, let's start with 2, 7, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Help solve recurrence : f(x) = 2f(x-1) + x While trying to analyze the time complexity of the heapify algorithm ,
I came up with the following recurrence for it's time complexity in terms of heap height:
T(h) = 2T(h-1) + h
Could you help me approach this recurrence?
| You can write it as a matrix equation:
$$\begin{pmatrix}1\\ T_h\end{pmatrix}=\begin{pmatrix}1&0\\ h&2\end{pmatrix}{1\choose T_{h-1}}=\begin{pmatrix}1&0\\ h&2\end{pmatrix}\begin{pmatrix}1&0\\ h-1&2\end{pmatrix}{1\choose T_{h-2}}=\\
\begin{pmatrix}1&0\\ (1+2)h-2&2^2\end{pmatrix}\begin{pmatrix}1&0\\ h-2&2\end{pmatrix}{1\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But ... | Intuitively, all those big numbers at the right are under so many square root signs they don't matter much. I would think it "obvious" that $3$ is larger. I made a spreadsheet (love copy down) and found the value to be about $2.0903$. For comparison, I changed all the numbers after $9$ to $0$ and the value only chan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coefficient of $ x^6 $in the given expansion. Find the coefficient of $x^6$ in $\left[(1+x)(1+x^2)^2(1+x^3)^3 \cdots (1+x^n)^n\right]$.
Expansion
$$\left[\left(1+\binom11x \right)\left(1+\binom21x + \binom22x^2 \right)\left(1 +\binom31x+\binom32x^2+\binom33x^3 \right)\cdots \left(1 + \binom n1 x + \binom n2x^... | Taken mod $x^7$ (meaning we drop anything with a power of $x$ greater than $6$), we have
$$(1+x)(1+x^2)^2(1+x^3)^3\cdots\equiv(1+x)(1+2x^2+x^4)(1+3x^3+3x^6)(1+4x^4)(1+5x^5)(1+6x^6)\\
\equiv(1+x+5x^5+5x^6)(1+2x^2+5x^4+8x^6)(1+3x^3+9x^6)\\
$$
where in the second step we've paired $(1+x)$ with $1+5x^5$, $(1+2x^2+x^4)$ wit... | {
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Sigma notation breakdown What does the sum below evaluate to?
$$\sum_{i=1}^n 4^i*i $$
I know the rule for a sum of a geometric series:
$$\sum_{i=1}^n a\cdot r^{i-1}=\frac{a(r^n-1)}{r-1} $$
But I cant seeem to apply to this formula because its to the power of i
and there is also the *i factor
| The formula $3S= 1+ n4^{n+1} - \sum_{i=0}^{n}4^i$ (where $S=\sum_{i=1}^n 4^i\cdot i$) from @Random-15's answer simplifies to
$$S=\frac{(3n-1)4^{n+1}+4}9$$ which does produce correct answers $4, 36$ and $228$ for $n=1,2,3,$ respectively.
Now, knowing the formula let me try to prove it by induction.
When $n=1$ we have ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
My attempt:-
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$
Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$
and now squaring this ... | $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is a root of $x^4 - 22 x^2 - 48 x - 23$.
By the rational root theorem, $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is either irrational or an integer.
But
$$
1.4 < \sqrt 2 < 1.5
\\
1.7 < \sqrt 3 < 1.8
\\
2.4 < \sqrt 6 < 2.5
\\
$$
imply
$$
-0.9 < \sqrt{6}-\sqrt{2}-\sqrt{3} <-0.6
$$
and so $\sqrt{6}-\sqrt{... | {
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"url": "https://math.stackexchange.com/questions/3385302",
"timestamp": "2023-03-29T00:00:00",
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f... | Since the argument and output are linear, we can try to extract $2x-3$ from $4x-2$ through arithmetic means.
\begin{align}
f(\color{red}{2x-3})&=4x-2\\
&=4x\color{blue}{-6+6}-2\\
&=2(\color{red}{2x-3})+4
\end{align}
Therefore, $f(x)=2x+4$ as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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Inequality involving Harmonic mean Let $a, b \ge 0 $ and $x,y > 1$ Show $\frac{1}{1/x+1/y}(a+b) \le \max(ax,by)$
If $a = b = 0$ then this is clear so assume not both are zero.
This seems to be related to the harmonic mean but I am not quite getting it.
$\frac{1}{1/x+1/y}(a+b) = \frac{xy}{x+y}(a+b) = \frac{1}{2}H(x,y)(... | In fact, the conditions for $x,y,a,b$ can be loosened:
For all $x,y > 0$ and $a,b \in \mathbb{R}$, we have
$$\frac{1}{1/x+1/y}(a+b)=\dfrac{y\cdot ax+x\cdot by}{x+y} \le \dfrac{y\cdot \max(ax,by)+x\cdot \max(ax,by)}{x+y} =\max(ax,by).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreci... | The function
$$
f(z)=\frac{m/z}{(z/a)^m-1}
$$
has residue $1$ where $z^m=a^m$ and residue $-m$ at $0$. For this question, we want $m=2n$ and $a^{2n}=-e^{i2ny}$. Furthermore,
$$
|z|=1\implies\csc^2(\arg(z))=-\frac{4z^2}{(z-1)^2(z+1)^2}
$$
Thus,
$$\newcommand{\Res}{\operatorname*{Res}}\newcommand{\sech}{\operatorname{se... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to prove $\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}$$\frac{1}{(n+1)n^{1/p}}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{n}{n^{1/p}}=\frac{1}{n^{1/p}}-\frac{n}{(n+1)n^{1/p}},$$
Can we go on frome here?
| I find the proper conclusion should be
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}\color{red}\leq p,$$
which is because, if $p=1$,then
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}=\sum_{n=1}^{\infty}\frac{1}{(n+1){n}}=1=p.$$
Here is my proof I just figured out. First, we have
\begin{align*}
a_n:&=\frac{1}{(n+1)n^{1... | {
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"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\b... | Using the reduced equations , we get
$$ \color{ red} {x = 3k+2 \quad \quad {\text(1)}}$$
Substituting it into the next gives us :
$$\color {#F0A}{3k+2\equiv 3\mod 4 \implies k \equiv 3 \mod 4 \implies k = 4j +3 \quad \quad \text { (2)}}$$
From $(1)$ and $(2)$ we get :
$$\color{ blue}{x = 12j + 11}$$
Substituting it in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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A term that is not bounded by 1. $f_{n}(x)= \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ I think when $0 < x <1$ this expression is bounded by 1, am I correct? if so how can I prove it?
My trials:
At $x = 1,$ the following fraction $\{ \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ becomes $\{\frac{n^3 }{ 1 + n^4}\}$ which is clearly les... | If you differentiate $f_n$ for $n \geq 1$ you have that the function attains its maximum at $x=\frac{\sqrt{3}}{n^2}$ on $[0,1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$
How to evaluate the following challenging sum:
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+i)?$$
where $H_{n}$ is the harmonic number, $\ope... | From here we got
$$X=2\sum_{n=0}^\infty\frac{H_{n/2}-H_n+\ln2}{(2n+1)^3}=\frac{\pi^2}{4}G-\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx$$
Using the identity in (4)
$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=\frac{\pi/2-x}{2\cos x}$$
multiply both sides by $x^2$ then integrate from $x=0$ to $\pi/2$ we get
\begin{align}... | {
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"source": "stackexchange",
"question_score": "3",
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$\int \frac{dx}{1-\sin x+\cos x}$ Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$
I tried:
$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos ... | My hint for your question:
$$\int \frac{dx}{1-\sin x+\cos x}$$
If I apply the $t-$substitution $t=\tan(\frac{x}{2})$ you will have $$\int \frac{dx}{1-\sin x+\cos x}=\int \frac{1}{-u+1}du$$
Using another $y-$substitution for last integral of the type $p=-y+1$ you will have:
$$\int \frac{1}{-u+1}du=\int -\frac{1}{p}dp$$
... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Do the coefficients of these polynomials alternate in sign? Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=... | I reckon that
$$p_{2n-1}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+1-k}k x^{n-1-k}$$
and
$$p_{2n}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+2-k}k x^{n-1-k}$$
and that these can be proved by induction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}... | Since $y=x\exp\frac{\pm2\pi i}{3}$,$$\frac{x^n+y^n}{(x+y)^n}=\frac{1+\exp\frac{\pm2\pi i n}{3}}{(1+\exp\frac{\pm2\pi i}{3})^n}=\frac{2\exp\frac{\pm\pi i n}{3}\cos\frac{\pi n}{3}}{(2\exp\frac{\pm\pi i}{3}\cos\frac{\pi}{3})^n}=2\cos\frac{\pi n}{3}.$$In the case $n=2013$, this simplifies to $-2$ because $n/3$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Determine the residuals of $f(z)=\frac{\log(1+z)}{\cos z-1}$ Determine the residuals of $f$
$f$ is given by
$$
f(z)=\frac{\log(1+z)}{\cos z-1}, \quad z \in \mathbb{C}-(-\infty, -1]
$$
where $\log z$ is the principal logarithm in the cutplane $\mathbb{C}_\pi:=\mathbb{C}-\{z \in \mathbb{R} \mid z \leq 0\}$.
Determining t... | I'll do it when $k=1$. The general case is similar.
You have\begin{align}\frac{\log(1+z)}{\cos(z)-1}&=\frac{\log(1+2\pi+z-2\pi)}{\cos(z-2\pi)-1}\\&=\frac{\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots}{-\frac{(z-2\pi)^2}{2!}+\frac{(z-2\pi)^4}{4!}-\cdots}\\&=\frac1{(z-2\pi)^2}\times\frac{\log(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404219",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ is bounded and has monotonicity?
How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ with $(a_n)_{n\in\mathbb{N}}$ is bounded and has monotonicity?
Boundedness: I observed, that $a_1<a_2>a_3>a_4>a_5>\dots$ That's why I assume, that there is a bound $... | Your proof is correct and for $n\leq 2$ you can simply say that $a_1=1<a_2=2$.
With a little effort we can show more.
For $n\geq 2$: $0<a_{n+1}<a_n$ because
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n+1)!}\cdot \frac{n!}{n^2}
=\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}\leq \frac{1}{2}+\frac{1}{4}=\frac{3}{4}<1.$$
Since $a_... | {
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"source": "stackexchange",
"question_score": "3",
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$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \... | The final result is correct, but in my opinion your approach is too complicated and you should avoid the "inverse of the inverse" part. More simply, by using the same standard limits as $x\to 0$,
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{\overbrace{x^{10}-3x}^{\to 0}+\overbrace{\frac{\sin x}{x}}^{\to 1}}{\under... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac... | Beware: overkill incoming. We may use the AM-GM inequality for producing tight bounds for $\sqrt[n]{n}$. In particular, by noticing that
$$ n = 1\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n}{n-1} $$
we have
$$ \sqrt[n]{n}=\text{GM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)<\text{AM}\left(1,1+1,... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D... | Perhaps the neatest approach is to square $\frac{1}{x^2-4}=\frac14\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$ to give$$\frac{1}{(x^2-4)^2}=\frac{1}{16}\left(\frac{1}{(x-2)^2}+\frac{1}{(x+2)^2}-\frac{2}{x^2-4}\right)=\frac{1}{32}\left(\frac{2}{(x-2)^2}+\frac{2}{(x+2)^2}+\frac{1}{x+2}-\frac{1}{x-2}\right).$$Now you can int... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Number Theory (Polynomials) Find The Remainder? A polynomial $f(x) = x^{50} + 3x^{49} + 3x + 12$ when divided by $x - a$, it leaves remainder $3$ & when its quotient is further divided by $x - b$ it leaves remainder $5$, also when $f(x)$ is divided by $x^2 - ( b + a)x + ab$ the remainder is $x + 6$. Find $a$?
| $P=x^{50}+3x^{49}+3x+12=x^{49}(x+3)+3(x+3)+3=k(x-a)+3$
So the quotient k is:
$k=\frac{(x^{49}+3)(x+3)}{x-a}$
⇒$\frac{(x^{49}+3)(x+3)}{x-a}=k_1(x-b)+5$
⇒$(x^{49}+3)(x+3)=k_1(x-a)(x-b)+5(x-a)$
⇒$p=(x^{49}+3)(x+3)+3=k_1(x-a)(x-b)+5(x-a)+3$
Therefore:
$5(x-a)+3=x+6$ ⇒ $a=\frac{4x-3}{5}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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What is the smallest multiple of $3^{1000}$, which has only 1’s in the digits?(Like 1, 11, 111 etc) I think the answer is $3^{1000}$ ones,
and I want to prove it with induction.
$3^0 | 1$ and $1$ is the smallest.
$ 3^1 | 111 $; it’s the smallest too.
Let’s say it's true for $ 3^k | 111...111 $ ($3^k$ ones). How c... | If $m$ is coprime to $3$ then the g.c.d. of $10^m-1$ and $10^3-1$ is $10-1=9$. In particular, $9$ but not $27$ is a factor of $10^m-1$.
If $n=3^km$ where $m$ is coprime to $3$ then $$\frac {10^{3^km}-1}{10^{3^{k-1}m}-1}=1+10^{3^{k-1}m}+10^{3^{k-1}(2m)}\equiv3\pmod9$$
Repeating this process we obtain
$10^{3^km}-1=\fr... | {
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Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction.
Please, explain how to remove the radical from the denominator. Thanks, for your hel... | Call the fraction by $r$.
$$r:=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}\frac{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
$$=\frac{8\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{2-\sqrt{\frac{5+\sqrt{5}}{2}}}=\frac{8\left(\sqrt{2-\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420241",
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"source": "stackexchange",
"question_score": "1",
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Given a defined region, write down the integral and then evaluate
Let the region be defined by $x \geq 0$, $x^2 + y^2 \leq 2$, and $x^2 + y^2 \geq 1$. Write down the integral over the region in each of the two possible orders of $f(x,y) = x^2$. Evaluate both integrals.
I was wondering before I try to evaluate the int... | None of your integrals has correct bounds. For the first one, since the integral w.r.t. $y$ is inside, $x$ should not depend on $y$ so the bounds of the outer integrals should have no $y$ at all. Similarly, for the second one, you have the integral w.r.t. $x$ inside; therefore the outer integral should have no $x$ ap... | {
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"source": "stackexchange",
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Prove that following sequence is bounded I have to prove that $x_n=\frac{1-n}{\sqrt{2+n^2}}$ sequence is bounded. I've solved some easier ones, but these seems to be kinda different. I have to find actual bounds of the sequence.
| Lower bound:
$$ \frac{1-n}{\sqrt{2+n^2}} = \frac{1}{\sqrt{2+n^2}} - \frac{n}{\sqrt{2+n^2}} \geq \frac{1}{\sqrt{2+n^2}} - \frac{n}{\sqrt{n^2}} = \frac{1}{\sqrt{2+n^2}} - 1 > -1$$
Upper bound:
Since $\sqrt{2+n^2} > 0$ we get
$$\frac{1-n}{\sqrt{2+n^2}} \leq \frac{1}{\sqrt{2+n^2}} < \frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421978",
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"source": "stackexchange",
"question_score": "1",
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$a^2b^2+a^2c^2+b^2c^2\leq 3$ Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$
Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.
My idea:
I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.
Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$
I have to show that $$y^2-2xz\leq 3$$
I tried to prove it with the sign of trinom, but it ... | Remark: I just realize that my Fact 1 is the same at Atticus's solution (he gave an elegant solution).
My Fact 1 is derived from the optimization theory (the method of Lagrange Multipliers, penalty functions, etc.).
We want to minimize $f(a,b,c) = 3 - (a^2b^2 + b^2c^2 + c^2a^2)$ subject to $g(a,b,c) = (a^2-a+1)(b^2-b+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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show this inequality $\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)$ let $x,y,z>0$, and such
$xyz=1$,show that
$$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)\tag{1}$$
I have konwn use C-S we have
$$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}... | The hint.
By the $uvw$'s technique prove that:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}.$$
After this by $uvw$ prove that:
$$\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}\geq\sqrt[3]{13(x^3+y^3+z^3)-12xyz}.$$
A proof of the second inequality.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $... | {
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"source": "stackexchange",
"question_score": "2",
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The sum $\sum^{n}_{k=0} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$ I was trying to evaluate:
$$\sum_{k=0}^{n} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$$
And in particular, when $x=1$.
I already have $2$ proofs for the $x=1$ case, (it is $2$) by counting subsets of a $2n-1$ size set and by splitting the term into a teles... | (I started an answer some hours ago, but had to stop, well, it is time to finish and submit.) Let $y=1/x$, (if $x$ is not zero,) and let us compute closer for some integer $n>0$
$$
\begin{aligned}
\frac{\binom nk}{\binom {2n-1}k}
&=\frac{n!}{k!(n-k)!}\cdot\frac{k!(2n-k-1)!}{(2n-1)!}
\\
&=\frac{n!(n-1)!}{(2n-1)!}\cdot\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$... | The domain of $$f(x)=\sin^{-1}(\sqrt{x^2+x+1})$$ is where you get $$0\le x^2+x+1\le 1$$ which is the interval $[-1,0]$
Note that $f(-1/2)=\pi/3$ is the minimum value and $f(-1)=f(0)= \pi/2$ is the maximum.
Thus the range is the interval $[\pi/3, \pi/2]$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 3
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Show that $\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$ Show that $\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$ for $k\geq6$ and even. I tried using $(1+2)^k=\sum_{i=0}^k{{k}\choose{i}}2^i$, the first half has the same binomials as the second and larger powers of two. Also tried induction, but the step is proving tha... | Here's an inductive approach. Let $a_n = \sum_{i \leq n} \binom{2n}{i} 2^i$ (including $i = 0$ for convenience). Then since $\binom{2n+2}{i} = \binom{2n}{i-2} + 2\binom{2n}{i-1} + \binom{2n}{i}$, we have
\begin{align*}
a_{n+1}
&= \sum_{i \leq n+1} \binom{2n+2}{i} 2^i \\
&= \sum_{i \leq n+1} \left(\binom{2n}{i-2} 2^i +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Comparing ranks of block matrices This seems simple, but I don't really know how to prove it, so I'd be thankful for any help.
For the following block matrix:
$A$, $B$, and $C$ are all $n\times n$ matrices. Entries of $A$, $B$, $C$ are arbitrary, $0$ is a zero matrix.
$$\operatorname{rk}\left(\left[\matrix{A&0\\0&B}\ri... | Let $A = \begin{bmatrix} \alpha_1 & \cdots & \alpha_n\end{bmatrix}$, $B = \begin{bmatrix} \beta_1 & \cdots & \beta_n\end{bmatrix}$, $C = \begin{bmatrix} \gamma_1 & \cdots & \gamma_n\end{bmatrix}$, where $\alpha_i, \beta_i, \gamma_i$ are column vectors of $A, B, C$ respectively.
Consider any linearly independent column... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to approach this? If $2m^3 - 8m^2 + 8n^3 - 12n^2 -10 \equiv 0 \mod 10$, then $8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 0 \mod 10$ Assuming
$$2m^3 - 8m^2 + 8n^3 - 12n^2 -10 \equiv 0 \mod 10$$
Prove
$$8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 0 \mod 10$$
I tried the following:
$$8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 6m... | Let $$A=2m^3 - 8m^2 + 8n^3 - 12n^2 -10 $$ and $$B= 8m^3 - 12m^2 + 2n^3 - 8n^2 - 10$$
Note that $$A+B = 10 m^3-20m^2+10n^3-20n^2-20 \equiv 0 , \mod (10)$$
So if one of $A$ or $B$ is a multiple of $10$ so is the other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving a summation by induction How can I prove $$\sum_{i=2}^{n} \frac{1}{i^2-i}\lt1$$ by induction on $n$?
So far:
If $m$ is a natural number such that $m\ge2$, let $P(m)$ be the statement:
$$\sum_{i=2}^{m} \frac{1}{i^2-i}\lt1$$
We will prove $P(m)$ by induction on $m$.
Base Case:
P(2) is the statement:
$$\sum_{i=2}^... | Well, since $\frac 1{(n+1)^2 -n} > 0$ and knowing $\sum\limits_{k=2}^n \frac 1{k^2 - k} = M < 1$, that isn't enough to prove $M + \frac 1{(n+1)^2 -n}< 1$.
We have to prove something a little stronger that $\sum\limits_{k=2}^n \frac 1{k^2 - k} $ is not just $< 1$ but less then $1-v_n$ for some $v_n > \frac 1{(n+1)^2 - n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Monotonicity of function averages Please let me know if you know an answer to this problem. May be you could provide a reference to some publication on this topic?
Let $f(x)$ be a real-valued strictly convex function on $[0, 1]$. For any integer $k$ between $0$ and some positive integer $n$ let $x_k =k/n$. Consider th... | $g(n-1) \geq g(n) \Leftrightarrow (n+1) \sum f( \frac{i}{n-1}) \leq n \sum f( \frac{i}{n})$.
This follows by summing up the following inequlities:
$\begin{array} { l l l l l l l }
& & n \times f(0) & = & n \times f(0) &=& n \times f(0), \\
1 \times f(\frac{0}{n-1}) &+& (n-1) \times f( \frac{1}{n-1} ) & \geq & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find all triples of non-negative real numbers $(a,b,c)$ Find all triples of non-negative real numbers $(a,b,c)$ such that:
$a^2+ab=c$
$b^2+bc=a$
$c^2+ca=b$.
This question was problem number of 3 in the RMO(India) Olympiad in 2019 held on $10^{th}$ November. link .
My attempt-
Assume $a \geq b\geq c$,
$\therefore a^2 ... | Your method is not full because the system is cyclic and not symmetric.
You can not assume that $a\geq b\geq c$.
We can solve our system by the following way.
We have: if $a=0$ so $c=0$ and from here $b=0$, which gives a solution $(0,0,0)$.
Thus, it's enough to solve our system for $abc\neq0$ and sice
$$a(a+b)=c,$$
$$b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $n$ distinct items, distribute them to $m$ distinct bins such that no bin is empty One solution can be solved by using inclusion-exclusion principle:
Let $A$ be the group that all items exist and to compute $A'$ then reduce it from the universe.
Then I thought about another way to solve it but I think it's not ac... | Let's look at the case $n = 5$ and $m = 3$. We will solve the problem in two ways, then examine why your approach does not work.
Method 1: We use the Inclusion-Exclusion Principle.
There are three choices for each of the five objects, so the objects can be distributed in $3^5$ ways. From these, we must subtract thos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with paramater Find all values of $a$ such that the inequality
$$\frac{a-\left(\log_3x+2\sqrt6\log_x3-5\right)}{\left(3\cos\sqrt{x-9}-4\right)-a}\le0$$
has no solution.
My work:
I used $\frac bc< 0$ if only if $bc< 0$ and considered case $a=0, b\not=0$ but I don't know how to get answer.
| First of all, we have to have
$$x\ge 9\tag1$$
Under $(1)$, we have, by AM-GM inequality,
$$\begin{align}&\left(\log_3x+2\sqrt6\log_x3-5\right)-\left(3\cos\sqrt{x-9}-4\right)\\\\&=\log_3x+\frac{2\sqrt 6}{\log_3x}-3\cos\sqrt{x-9}-1
\\\\&\ge 2\sqrt{2\sqrt 6}-3-1\gt 0\end{align}$$
from which we see that
$$\left(\log_3x+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Systems of equations over reals
Solve over reals
$$2x^2+xy+x-3y+3=0$$
$$3xy+y^2-2x+6y-6=0$$
Adding or subtracting these equations doesn't help, and completing the square also doesn't work. I'm considering substituting $x+y$ and $xy$, but it also doesn't show much promise. Thanks!
| $$2x^2+xy+x-3y+3=0\tag{1}$$
$$3xy+y^2-2x+6y-6=0\tag{2}$$
Take 2$\times$(1)+(2) to get,
$$4x^2+5xy+y^2=0 \implies (4x+y)(x+y)=0$$
which yields $y=-x$ and $y=-4x$. Plug them into (1)
$$x^2+4x+3=0\implies x=-1,\>x=-3$$
$$2x^2-13x-3=0\implies x=\frac{13\pm\sqrt{193}}{4}$$
Then, evaluated the corresponding $y$'s to obtain t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strategy for expanding the function $e^{e^x-1}$ into Maclaurin series I am trying to find a way to expand the function $e^{e^x-1}$ into Maclaurin series but I don't know how to. I don't want to use the cumbersome Maclaurin formula of taking successive higher derivative.
Do you know how to nicely expand this function in... | The exponent is
\begin{eqnarray*}
e^x-1 = x+\frac{x^2}{2}+ \frac{x^3}{6}+\cdots.
\end{eqnarray*}
Now just plug this in the power series for the exponential function
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = 1&+& \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right) \\
&+& \frac{1}{2} \left(x+ \frac{x^2}{2}+ \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Divisibility Induction $17|18^{(5n+1)}+13^{(4n+1)}+3$ Prove that by induction
$17|18^{(5n+1)}+13^{(4n+1)}+3$ for all n∈N.
So far I'm stuck on the proof for n=k+1:
When n=k+1:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$= (18^{5k+1}+13^{4k+1}+3)+[(18^5-1)18^{5k+1}+(13^4-1)13^{4k+1}]$$
From the assumption n=k, I can prove the first ... | For completeness, here is another way to achieve the same result:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$=18^5 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+3$$
$$=13^4 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$ I'm trying to solve the following problem:
Find the coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$.
Can I use the multinomial theorem to solve it? I'm unsure how to start..
Thanks!
| Definition:
$[x^n]f(x)~$ means the coefficient of $~x^n~$ of the Taylor expansion of $~f(x)~$ at $~x=0~$ .
Calculation:
$\displaystyle [x^{10}]x^3(x^2-3x^3-1)^6 = [x^{7}](x^2-3x^3-1)^6 = [x^{7}]\sum\limits_{k=0}^6 {\binom 6 k}x^{2k}(3x-1)^k = $
$\displaystyle =\sum\limits_{k=0}^6 {\binom 6 k}[x^{7-2k}](3x-1)^k =\sum\... | {
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"url": "https://math.stackexchange.com/questions/3436779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Simple formula for determinant of circulant matrix built on an arithmetic sequence. Let $a$ be an arithmetic sequence:
$$a_i=a_1+\lambda(i-1),\tag1$$
and consider a $n\times n$ circulant matrix $M_{n}(a)$ "built" on rotational shifts of the sequence $a$, i.e. with elements:
$$M_{ij}=a_{1+(j−i)\operatorname{mod}n}.\tag2... | The determinant of circulant matrix $M$ is given by the formula:
$$\det M = \prod_{\ell=0}^{n-1} f(\omega^\ell)$$
where $\omega = e^{\frac{2\pi i}{n}}$ is a primitive $n^{th}$ root of unity and
$$\begin{align}
f(x) &= \sum_{k=0}^{n-1} a_{k+1} x^k
= \sum_{k=0}^{n-1} (a_1 + \lambda k)x^k\\
&= \left(a_1 + \lambda x\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute the possibilities for $\gcd (2a^4 + 2a^2+3,2a^5+2a^3+a^2+a+3)$
Given that $a$ is an integer, compute the possibilities for $\gcd (2a^4 + 2a^2+3,2a^5+2a^3+a^2+a+3).$
Here's my work (I'm not entirely sure if it's correct).
Just to make sure we're on the same page, we say that for integers $a,b,$ $d=\gcd(a,b)$ i... | Trusting on your calculations, it is enough to find the possible values of $\gcd(48-a, -201)$. Now, assume $f(a):=48-a, f:\mathbb{Z}\to\mathbb{Z}$. Since, $f$ is monic linear polynomial it is surjective. Therefore, possibilities of $\gcd(48-a, -201)$ will be all positive divisors of $201$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\t... | We can actually utilize $x,y$ are positive integers
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac1y<\tan^{-1}3$$
As $\tan^{-1}x$ is an increasing function
$$\implies x<3$$
So, $x$ can be $1$ or $2$
Check which ones make $y$ positive integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$ I have to find the following limit:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}}$$
This is what I tried:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} =
\lim\limits_{n \to \i... | $$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k+1}{k}\dfrac{k+1}{k+2}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \frac{ \dfrac{k+1}{k}}{\dfrac{k+2}{k+1}}}$$
$$ = \lim\limits_{n \to \infty} \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$. Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$.
| Let $n$ be any odd positive integer. Then:
$$(2^n)^3 = 2^{3n} = 2 \times 2^{3n-1}= 2 \times (2^{(3n-1)/2})^2$$
The requirement that $n$ is odd ensures that $(3n-1)/2$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Given the parameter $a \in \mathbb{R}$ such that the given sequence $(x_n)_{n \ge 1}$ is bounded, find the limit of the sequence $x_n$. I am given the following sequence $(x_n)_{n \ge 1}$:
$x_n = \bigg ( 1 + \dfrac{1}{3}
+ \dfrac{1}{5} + ... + \dfrac{1}{2n-1}\bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + ... | \begin{align*}
x_{n}=\left(1-\left(a+\dfrac{1}{2}\right)\right)\left(\sum_{k=1}^{n}\dfrac{1}{k}\right)+\left(\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\right),
\end{align*}
and we have
\begin{align*}
\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\rightarrow\int_{0}^{1}\dfrac{1}{1+x}dx=\log 2,
\end{align*}
so this forces $1-(a+1/2)=0$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum_{k=0}^{n} 2^{k} {n \choose k}{n-k \choose \lfloor{\frac{n-k}{2}}\rfloor} = {2n+1 \choose n}$ The problem is the following: prove that $\sum_{k=0}^{n} 2^{k} {n \choose k}{n-k \choose \lfloor{\frac{n-k}{2}}\rfloor} = {2n+1 \choose n}$
I was trying to prove some concrete combinatorial examples to make the... | We start with
$$\sum_{k=0}^n 2^k {n\choose k}
{n-k\choose \lfloor \frac{n-k}{2} \rfloor}
= \sum_{k=0}^n 2^{n-k} {n\choose k}
{k\choose \lfloor k/2 \rfloor}
\\ =
\sum_{k=0}^{\lfloor n/2\rfloor} 2^{n-2k} {n\choose 2k}
{2k\choose k}
+ \sum_{k=0}^{\lfloor (n-1)/2\rfloor} 2^{n-2k-1} {n\choose 2k+1}
{2k+1\choose k}.$$
Now fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\frac {1+x}{1-x}\right)$ Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$
My Attempt:
$$y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$$
Put $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \... | $$D^{n}y= D^{n-1} Dy=D^{n-1}\frac{1}{1+x^2}= D^{n-1} \frac{1}{2i} \left( \frac{1}{x-i}- \frac{1}{x+i} \right)$$
The nth derivative of $(x+b)^{-1}$ is givwn as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$
Using this
$$D^n y= (-1)^{(n-1)} (n-1)! ~ \Im (x-i)^{-n-2}=(-1)^{n-1} (n-1)!~ (x^2+1)^{(-n-2)/2} \sin[(n+2) \tan^{-1}(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find all complex solutions $z^{10} + 2z^5 + 2 = 0$. So I’m working on this equation $z^{10} + 2z^5 + 2 = 0$ to find all complex solutions, and I think I managed to solve it, but I can’t find solution manual for it, since it is really old exam task. The thing that makes me uncomfortable with my solution is that, shouldn... | solve for $$w^2 + 2w + 2=0$$
which gives $$w_{1,2} = -1 \pm i = \sqrt{2} e^{i(\pi \pm \frac{\pi}{4})}$$
Now you got two equations to solve
\begin{align}
z_1^5 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4})} \\
z_2^5 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4})}
\end{align}
which gives
\begin{align}
z_1 &= \sqrt{2} e^{i(\pi + \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find GCD between these two equations? I've been trying for a couple of hours to find a way to prove the next claim:
$$\gcd(n-1,a) = \gcd(\frac{(n^a-1)}{n-1},n-1)$$
I have already proven that
$$\frac{(n^a-1)}{n-1}$$ is always divisible by $(a-1)$.
which means I can set define
$$\frac{(n^a-1)}{n-1}=\frac{c(n-1)}... | You can expand $n^a - 1$ as a difference of $a$th powers:
\begin{equation*}
n^a - 1 = (n - 1)(n^{a - 1} + n^{a - 2} + \dotsb + 1)
\end{equation*}
so
\begin{equation*}
\frac{n^a - 1}{n - 1} = n^{a - 1} + n^{a - 2} + \dotsb + 1
\end{equation*}
But since
\begin{align*}
n^{a - 1} + n^{a - 2} + \dotsb + 1 &\equiv
\under... | {
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"url": "https://math.stackexchange.com/questions/3454779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we h... | If you are trying to use GM<=AM, I suppose you have $x,y,z\geq 0$.
Check this link QM>=AM>=GM>=HM
First Step
Using HM<=GM:
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq \sqrt[3]{xyz} \implies xyz\geq \frac{3^3}{1}=27$$
Second Step
Using the AM<=QM:
$$ \frac{x+y+z}{3} \leq \sqrt{\frac{x^2+y^2+z^2}{3}}\implies \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\be... | Finally solved it.
Thanks to all for looking into this problem, special thanks to @Andrei for pointing out the exact mistake.
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\dfrac{a\lambda}{b}... | {
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"url": "https://math.stackexchange.com/questions/3456535",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$ twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom... |
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}}
&=\sum_{k=0}^{18}\binom{18}{k}\binom{20}{k}\tag{1}\\
&=\sum_{k=0}^{18}\binom{18}{18-k}\binom{20}{k}\tag{2}\\
&\,\,\color{blue}{=\binom{38}{18}=33\,578\,000\,610}\tag{3}
\end{align*}
Comment:
*
*In (1) we omit the index $k=19$, since... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(... | I think that this has to be solved numerically using Newton method. Graphing for $0 \leq x \leq 2\pi$, we can notice, beside the trivial roots $x=0$ and $x=2\pi$, solutions close to $1.0$, $2.7$, $3.8$, $4.9$ and $5.6$.
Using these as starting points, Newton iterates are
$$\left(
\begin{array}{cc}
0 & 1.00000 \\
1 &... | {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Minimize $x + y + z$ subject to $x^2 + y^2 + z^2 \le 1$ and $x \ge 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + y + z \\
\text{s.t} & \quad x^2 + y^2 + z^2 &&\le 1 \\
& \quad x &&\ge 0
\end{align*}$$
The linear independence constraint constraint qualificat... | Your ultimate solution is wrong because $x$ is not $\ge 0$, but the other solution where $x = 0$ is correct. Also there's a lot of work there and I'm not sure how much of it is necessary. (It's all correct except for the solution you have with $x < 0$.) I guess my point is we can just assume that $x^2 + y^2 + z^2 = 1$ ... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Minimize $x + 2y + 3z$ subject to $x^2 + y^2 + z^2 = 1$ and $x + y + z \le 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + 2y + 3z \\
\text{s.t} & \quad x^2 + y^2 + z^2 && = 1 \\
& \quad x + y + z && \le 0
\end{align*}$$
The linear independence constraint qua... | Let $k = x + 2y + 3z$ and observe that it represents a plane with $k$, $\frac k2$ and $\frac k3$ as the intercepts with the $x$, $y$ and $z$ axes, respectively. So, $k$ is at the minimum if the intercepts have the largest negative values, corresponding to the plane tangential to the sphere $x^2 + y^2 + z^2=1$ in the ne... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot... | I get
$(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}
$.
I use
$\ln(1+x)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n}
$.
$\begin{array}\\
f(x)
&=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}(\dfrac1{n+1}-\dfrac1{n+2})\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+1}-\sum_{n=0}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sides of non-right triangle
Given a non-right triangle with vertex $C = 91^{\circ}, \
\overline{AC} =39.9 \ cm, \ \overline{BC} = 32.6 \ cm$, find
a) $\overline{AB}$
b) $m\measuredangle A$
c) $m\measuredangle B$
I was able to find the length of $\overline{AB}$, which I labeled $c$
\begin{align*} c=& \ \sqrt{32.6^2+... | Side length $c \approx 51.9$ is ok.
The triangle side lengths are close to the right triangle of sides $ 30,40,50$ in proportion.
$$ \angle A ( \approx 38.96^{\circ})+ \angle B^{\circ} + 91 ^{\circ} = 180^{\circ}\rightarrow \angle B\approx 50.04 ^{\circ} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$x,y,z>0$, prove:$\frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$ If $x,y,z>0$, prove:$$ \frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$$ I tried to use classical i... | Since $$\sqrt[4]{\frac{x^4+y^4}{2}}\leq\sqrt{x^2-xy+y^2}$$ it's just $(x-y)^4\geq0,$ it's enough to prove that
$$\sum_{cyc}\frac{z}{x+y+\sqrt{x^2-xy+y^2}}\geq1$$ or
$$\sum_{cyc}\frac{z(x+y-\sqrt{x^2-xy+y^2})}{3xy}\geq1$$ or
$$\sum_{cyc}(x^2y+x^2z-xyz)\geq\sum_{cyc}z^2\sqrt{x^2-xy+y^2}$$ or
$$\sum_{cyc}\left(\frac{x^2y+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof explanation of $\prod_{k=2}^n \big(1- \frac{2}{k(k+1)} \big) = \frac{1}{3} \big(1+\frac{2}{n} \big), n \geq 2$ I want to understand the proof of
$$\prod_{k=2}^n \big(1- \frac{2}{k(k+1)} \big) = \frac{1}{3} \big(1+\frac{2}{n} \big), n \geq 2$$
via induction.
$$\text{Base case: } n=2 \Rightarrow\prod_{k=2}^2\big(1-... | In the first step we are using the induction hypotesis that is
$$\prod_{k=2}^n \left(1- \frac{2}{k(k+1)} \right) = \frac{1}{3} \left(1+\frac{2}{n} \right)$$
the second one is
$$= \frac{(n+1)(n+2)-2}{(n+1)\color{red}{(n+2)}} \cdot \frac{1}{3} \cdot \frac{\color{red}{n+2}}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and s... | Yes, add and subtract $1$. You will get
$$\left[
\left(
1+\frac{\sqrt{x^2+2x-3}-x-2}{x+2}
\right)^{\frac{x+2}{\sqrt{x^2+2x-3}-x-2}}
\right]^{\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}}$$
The part inside the $\left[...\right]$ tends to $e$.
Then compute the limit of the exponent
$$\begin{align}\frac{(\sqrt{x^2+2x-3}-x-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472821",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is there a closed form for this transcendental equation? Wondering if there’s a closed form for $\exp(\frac{1}{\log(x)})=\frac{x}{e}$?
Numerically solving yields $x \approx .539$
| $\exp \left (\dfrac{1}{\ln x} \right ) = \dfrac{x}{e}; \tag 1$
$\dfrac{1}{\ln x} = \ln \left ( \dfrac{x}{e} \right ) = \ln x - \ln e = \ln x - 1; \tag 2$
$1 = (\ln x)^2 - \ln x; \tag 3$
$(\ln x)^2 - \ln x - 1 = 0; \tag 4$
quadratic formula:
$\ln x = \dfrac{1 \pm \sqrt 5}{2}; \tag 5$
$x = \exp \left ( \dfrac{1 \pm \sqrt... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
If $x_1, x_2, x_3, x_4$ are the roots of $x^4-x^3+2x^2+3x+1$, find $\frac{\sum x_i^3+\sum{(x_ix_jx_k)^3}}{\sum{(x_ix_j)^3}-(x_1x_2x_3x_4)^3}$ If $x_1, x_2, x_3, x_4$ are the roots of
$x^4-x^3+2x^2+3x+1$
then find
$$\frac{\sum x_i^3+\sum{x_i^3x_j^3x_k^3}}{\sum{x_i^3x_j^3}-x_1^3x_2^3x_3^3x_4^3}$$
My attempt: I have trie... | Hint: Let $P(x)=x^4-x^3+2x^2+3x+1$, and let $\omega\ne1$ be a third root of unity. Then consider $Q(x^3)=P(x)P(\omega x)P(\omega^2 x)$. Convince yourself that $Q(x)$ has roots $x_i^3$ and using Vieta you can read off the terms you need from the coefficients of $Q$
Working that out, $Q(x)=x^4+14x^3+50x^2+6x+1$, so we... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra (Matrixes with powers) Let
$$
M \colon= \left[ \begin{matrix}
2 & -1 \\
2 & 5
\end{matrix} \right]
$$
Find formulas for the entries of $M^n$, where $n$ is a positive integer.
$$ M^n = \text{?} $$
| I think simple diagonilization makes sense here for $M$ has two different eigenvalues, $\{3,4\}$. Two corresponding eigenvectors are $[1,-1]^\top$ and $[1,-2]^\top$. Setting
\begin{align}
P=\begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}
\end{align}
can be expressed $M$ as $M= P\Delta P^{-1}$ where $\Delta=\operatorname... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inverse of a matrix using bases Consider the bases
$(4,2)$ , $(5,3)$ and $(1,0)$,$(0,1)$ of $F^2$. The Matrix of
$$
M(I,\bigg(\Big(4,2),(5,3)),((1,0),(0,1)\Big)\bigg) = \Bigg(\begin{matrix}4 &5 \\2 &3 \end{matrix}\Bigg)
$$
The matrix above was computed using the procedure
$I(4,2) = 4(1,0) + 2(0,1)$ and
$I(5,3) = 5(... | You calculated the first matrix correctly by writing $(4,2)$ and $(5,3)$ as linear combinations of $(1,0)$ and $(0,1)$. You should calculate the inverse matrix by writing $(1,0)$ and $(0,1)$ as linear combinations of $(4,2)$ and $(5,3)$:
$$(1,0) = \frac{3}{2}(4,2) + (-1) (5,3)$$
$$(0,1) = -\frac{5}{2}(4,2) + 2 (5,3)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $\lim_{x\to\infty} f(x)$, where $f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}$
Determine $\lim_{x\to\infty} f(x)$ where
$$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$
I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos an... | We have that
$$\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}=\left[\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}}\right]^{(x^3+3x)\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}}$$
and
$$\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}} \to e$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find factor $f(x)$ conmmon to two quartic equation Let $f(x) = x^2 + bx + c$, where $b, c ∈ R$. If f(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x + 5$, then find $f(x)$
My approach ,on dividing both quartic equation by $f(x)$ remainder is zero, but not getting the answer
| Hint: $$ x^4+6x^2+25 = (x^2+3)^2+16 $$ $$= (x^2+3+4i)(x^2+3-4i)$$
So, the roots of the first functions are$$ \pm 1\pm 2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}$. The presented solution of the following problem on Art of Problem Solving using Jensen's inequality is wrong, since the function $f(x):=\sqrt[3]{1+\frac{2t}{x}} $ is convex rather than concave. How would one prove this inequality, correctly?
Let $a, b ,c ... | TLDR
A standard computer-assisted (but rigorous) proof is to use Lagrange multipliers method together with interval arithmetic libraries such as IntervalRoots.jl.
We are optimizing within a compact set in $\mathbb R^3$ as shown below
So there exist maximum points, either in the interior, or on the boundaries.
We can ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the maximum value of $\frac{7x+2y}{2x+2}+\frac{3x+8y}{2y+2}$ while $0≤x,y≤1$
Find maximum value of $\left(\dfrac{7x+2y}{2x+2}+\dfrac{3x+8y}{2y+2}\right)$ for $0≤x,y≤1$
My approach was to use A.M-G.M inequality or cauchy shbert inequality, but I failed.
| $$
f(x,y) = \frac{7x+2y}{2x+2} + \frac{3x+8y}{2y+2}
=
\left( \frac 72 - \frac{7-2y}{2x+2} \right)+
\left( 4 - \frac{8 - 3x}{2y+2} \right)
$$
is strictly increasing in both variables on $[0, 1]^2$, therefore
$$
f(x, y) \le f(1,1) = 5
$$
with equality exactly for $(x, y) = (1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the range of $y$ in a DE
Consider the equation $$y' = y^2 - y - 2 = (y+1)(y-2).$$ If $y(10) = 0$, find the range of $y(t)$ for $t>10$. That is, find the best $A$ and $B$ such that $A<y(t)<B$ for $t>10$.
From integration by parts, and using $y(10) = 0$, I get the equation $$2e^{3t-30} = \frac{|y-2|}{|y+1|}.$$
Le... | $$y'=(y+1)(y-2) \implies \int \frac{dy}{(y+1)(y-2)}= \int dx+C$$ $$ \implies \int \frac{1}{3} \left ( \frac{1}{y-2}-\frac{1}{y+1} \right) dy=x+c \implies \ln\frac{(y-2) }{(y+1)}=(3x+3C)$$ $$ \implies \frac{y-2}{y+1}=e^{3x+3C}>0 \implies y<-1 ~~or ~~ y>2.$$
Hence the range is $y \in (-\infty, -1) \cup y\in (2,\infty)$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Issue about the generating function
A rabbit initially stands at the position $0$, and repeatedly
jumps on the real line. In each jump, the rabbit can jump to any
position corresponds to an integer but it cannot stand still. Let
$N(a)$ be the number of ways to jump with a total distance of
$2019$ and stop at the posit... |
Question: What is $1$ in $F$ since rabbit must jump at least once? Shouldn't it start with $T$ and not with $1$?
We are free to choose $F$ either starting with $1$ or starting with $T$. A value $1$ represents no jump at all and will simply be ignored when doing the analysis. Its benefit is to have a slightly simpler ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{... | In real life the indefinite integral is usually given via Kepler's angle:
$$\sin\psi=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$
For $0<e<1$. So
$$\cos^2\psi=\frac{1+2e\cos x+e^2\cos^2-\sin^2 x+e^2\sin^2x}{\left(1+e\cos x\right)^2}=\frac{\left(\cos x+e\right)^2}{\left(1+e\cos x\right)^2}$$
Since we want a small positive $x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proof for any positive integer n the following equality hold $$\sum_{k=1}^n (-1)^{k+1} \binom {n}{k} \frac{1}{k}= 1 + \frac{1}{2} + ... + \frac{1}{n}$$
How do I prove this by induction?
My attempt: let $n = 1$, then $(-1)^2 \binom{1}{k} =1=\frac{1}{1}$
Assuming $n$ is true, how do I prove the $n + 1$ case?
| The induction strategy is: suppose that the result holds for $n\geq 1$, and we want to show that it holds for $n+1$. The solution I came up with uses a few classic results about binomial coefficients (see here if you want). Try using the recursion formula $$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$ and then the "abs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue ... | By C-S
$$\sum_{cyc}\frac{a}{7a+b}=\frac{3}{7}+\sum_{cyc}\left(\frac{a}{7a+b}-\frac{1}{7}\right)=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b}{7a+b}=$$
$$=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
Id est, it's enough to prove that
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?
| Put $y = f(x)$. Taking natural logarithm on both sides of the equation,
$$\ln(y) = \ln(x-3) + \ln(x-2) + \ln(x-1) + \ln(x) - \ln(x+1) - \ln(x+2) - \ln(x+3)$$
$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x-3} + \frac{1}{x-2} + \frac{1}{x-1} + \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} - \frac{1}{x+3}$$
Now, multiplying both s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it ou... | So we have $c(a+b)=ab$. Now \begin{eqnarray}a^2+b^2+c^2 &=&\underbrace{a^2+\color{red}{2ab}+b^2}\color{red}{-2c(a+b)}+c^2\\
&=&(a+b)^2-2c(a+b)+c^2\\
&=& (a+b-c)^2
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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How many real solutions $(x,y)$ for $x + y^2 = y^3 $ and $y + x^2 = x^3$ I think the answer is that there are $3$ real solutions.
*
*When $x = y$
*When $x = -y$
If $x\neq y$ is there a real solution? If so, how do we find the real solutions?
| Substituting $y=x^3-x^2$ into the other equation gives
\begin{eqnarray*}
x(x^8-3x^7+3x^6-2x^5+2x^4-x^3-1)=0 \\
x(x^2-x-1)( x^6-2x^5 +2x^4 -2x^3 +2x^2-x+1) =0 \\
x(x^2-x-1) \left( (x^4+x^2)(x-1)^2+ \frac{1}{4} (2x-1)^2 + \frac{3}{4} \right) =0. \\
\end{eqnarray*}
So you are indeed correct: there are $3$ real solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integration using spherical coordinates I'm trying to evaluate this triple integral:
$$\iiint_{V}(x^2+y^2+z^2-1)\,dx\,dy\,dz$$ where $V=\{(x,y,z)\in\mathbb{R}^3,\ x^2+y^2+z^2\le2,\ x^2+y^2\le z\}$.
After applying the spherical coordinates, I find that $V'=\begin{cases} 0\le \rho\le \sqrt2 \\ \cfrac{5\pi}{4}\le \theta\l... | We have $x = \rho \sin(\theta)\cos(\phi) , y = \rho \sin(\theta)\sin(\phi) , z = \rho \cos(\theta)$, thus the first limit translates into $0\le \rho \le \sqrt{2}$ and the second one turns to $\rho^2\sin^2(\theta)\le\rho\cos(\theta)$ which means $\rho \le \frac{\cos(\theta)}{\sin^2(\theta)}$ and the $\phi$ is unconditio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find all naturals $n>1$, such that the value of the sum $2^2 +3^2 +4^2 +\cdots+n^2$ equals $p^k$ where $p$ is a prime and $k$ is natural I simplified the sum using the formula of the sum of the squares
$$\frac{(n-1)(2 n^2 + 5n+6)}{6}=p^k$$
Moreover $\gcd(n-1,2n^2+5n+6)=1$ or $13$
I need help to complete the solution.
| You have done most of the work. If the gcd is 1, then $n-1$ must be 1,2,3 or 6. It is easy to check that $n=2,3,4,7$ are solutions (giving $k=2,p=2$ and $k=1$ and $p=13,29,139$).
Otherwise, we must have $p=13$ and $k>1$. We cannot have $13^2|n-1$, because if $13^2|2n^2+5n+6$, then $13$ is not the gcd, so $2n^2+5n+6<6\c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find minimum value $E(x, y) = x^2 + y^2 -6x -10y$, where $x^2 + y^2 - 2y \le 0$ I am given the expression:
$$E(x, y) = x^2 + y^2 -6x -10y$$
And I have to find the minimum value of $E(x, y)$ for $(x, y) \in D$ where:
$$D = \{ (x,y) \in \mathbb{R}^2 \hspace{0.25cm} | \hspace{0.25cm} x^2 + y^2 -2y \le 0 \}$$
Doing the fol... | For $(x,y)=\left(\frac{3}{5},\frac{9}{5}\right)$ we'll get a value $-18$.
We'll prove that it's a minimal value.
Indeed, it's enough to prove that:
$$x^2+y^2-6x-10y\geq-18+4(2y-x^2-y^2)$$ or
$$5x^2-6x+5y^2-18y+18\geq0$$ or
$$5\left(x-\frac{3}{5}\right)^2+5\left(y-\frac{9}{5}\right)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.
Here ... | You can use the fact that
$$\begin{align}\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}&\Leftrightarrow \sin x + \cos x = \dfrac{\pi}{3}\\
&\Leftrightarrow \cos x = \dfrac{\pi}{3}-\sin x.\end{align}$$
Since $\sin (2x) = 2\sin x \cos x$, we have $\sin (2x) = 2\cdot\dfrac{\frac{2\pi}{3}+ \frac{2}{3}\sqrt{18-\pi^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2... | The function to find an antiderivative of can be rewritten as
$$
f(x)=\begin{cases}
-\pi-2\arctan x & x<-1 \\[6px]
2\arctan x & -1\le x\le 1 \\[6px]
\pi-2\arctan x & x>1
\end{cases}
$$
It's sufficient to see that the derivative is
$$
f'(x)=\dfrac{1-x^2}{|1-x^2|}\dfrac{2}{1+x^2}
$$
Thus an antiderivative is
$$
F(x)=\int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int_{0}^{\frac{\pi}{6}}\frac{\sqrt{3\cos2x-1}}{\cos x}dx$ $$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3\cos2x-1}}{\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3(1-\tan^2x)-(1+\tan^2x)}}{\sec x\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\sqrt{2-4\tan^2x}\quad dx$$
$$\sqrt{2}\int_{0}^{\frac{\pi}{6}}\sqrt{1-2\tan^2x... | I tried $\sin x=u$, then $\sqrt3u=\sin\theta$, and finally $v=\tan\theta$ to get
$$\begin{align}\int_0^{\pi/6}\frac{\sqrt{3\cos2x-1}}{\cos x}dx&=\sqrt2\int_0^{1/2}\frac{\sqrt{1-3u^2}}{1-u^2}du\\
&=\sqrt6\int_0^{\pi/3}\frac{\cos^2\theta}{3-\sin^2\theta}d\theta\\
&=\sqrt6\int_0^{\sqrt3}\frac{dv}{(1+v^2)(3+2v^2)}\\
&=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Derivation of the graph of $r=\frac{1}{\sin(\theta)}$ When playing around with desmos, I found a very interesting function:
$r=\frac{1}{\sin(\theta)}$
The graph of this function is a straight line with constant value $y=1$
I tried to prove this, however I failed:
Assuming $r=\sqrt {x^2+y^2}$ and $\theta = \arctan(\frac... | I found the error myself! It lies within my very first step:
$\frac{1}{\sin(\arctan(\frac{y}{x}))}$ is not equal to $\frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$ but rather to $\frac{1}{\frac{\frac{y}{x}}{\sqrt{1+\frac{y^2}{x^2}}}}$
From there, the derivation is straightforward:
$\sqrt{x^2+y^2} = \frac{\sqrt{1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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