Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is the value of $a$ so that this condition holds? Let $f(x) \colon= x-x^2$, $g(x) \colon= ax$. Determine the value of $a$ so that the region above the graph of $g$ and below the graph of $f$ has area equal to $9/2$.
Here $f(x) - g(x) = (1-a)x - x^2 = x((1-a) - x) = 0$ if and only if $x = 0$ or $x = 1-a$.
So the ... | You should have gotten $\displaystyle\int_{0}^{1-a}[(1-a)x-x^2]\,dx = \left[\dfrac{1-a}{2}x^2-\dfrac{1}{3}x^3\right]_{0}^{1-a} = \dfrac{(1-a)^3}{6}$.
Then, you have $\dfrac{(1-a)^3}{6} = \dfrac{9}{2} \leadsto (1-a)^3 = 27$ which should be easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
| $$x^2+4x+3=0$$
$$\Delta=4^2-4 \cdot 1 \cdot 3=4$$
$$x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-4 \pm 2}{2}=-3 \text{ or } -1$$
$$x^2+4x+3=(x+3)(x+1)$$
$$\frac{2x}{x^2+4x+3}=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}=\frac{A(x+1)+B(x+3)}{(x+3)(x+1)}=\frac{(A+B)x+(A+3B)}{(x+3)(x+1)}$$
Since,$\frac{2x}{(x+3)(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
| one approach is to pair off the integers in the harmonic sum as $n+k$ and $2n-k$. so if $n$ is even we have:
$$
S_n = \sum_{k=0}^{\frac{n}2-1}\left(\frac1{n+k} + \frac1{2n-k} \right) + \frac2{3n}
$$
now
$$\frac1{n+k} + \frac1{2n-k} = \frac{3n}{(n+k)(2n-k)} \ge \frac{4n}{3n^2} = \frac4{3n}
$$
and there are $\frac{n}2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| Since $a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff (a^2-3a)+(b^2-3b)+(c^2-3c)+abc\ge-5 \iff$
$\;\;\;\;\;(a-\frac{3}{2})^2+(b-\frac{3}{2})^2+(c-\frac{3}{2})^2+abc\ge\frac{7}{4},$
it suffices to show that $f(x,y,z)=(x-\frac{3}{2})^2+(y-\frac{3}{2})^2+(z-\frac{3}{2})^2+xyz$ has
a minimum value of $\frac{7}{4}$
for $x\ge0, y\ge0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$
Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$
I tried by taking $x^2$ out of the root by taking it common.
i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in nume... | $$
\displaylines{
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {1 + x^2 } - x}}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)x^2 } - x}}{x} \cr
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{x^2 }}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Closed form of a sum of binomial coefficients? I have the following function:
$T_n(d)=\sum\limits_{k=\frac{n-d}{2}}^{\lceil \frac{n}{2} \rceil}{k\choose \frac{n-d}{2}}$
${n \choose 2k}$, where $n,d\in \mathbb{N}^0$, and $n,d$ have the same parity.
Looking at the sequences for various $d$, it seems that the formula ... | Another useful apparently classic technique uses basic complex
variables.
Suppose we seek to evaluate
$$\sum_{k\ge m} {k\choose m} {2m+d\choose 2k}
= \sum_{k\ge m} {k\choose m} {2m+d\choose 2m+d-2k}$$
where the second binomial coefficient serves to delimit the upper
bound of the range of $k$ which is fini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}$ How would one find the integral of $\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}$ with respect to $x$?
I already know the antiderivative (from Wolfram Alpha), but I don't know how to integrate this function with pencil and paper.
Is there maybe a clever variable substitution?
| HINT:
I would say that
$\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}=\frac{2+5x^3}{2(x^3+1)}\frac{x^3+1}{\sqrt{x^3+1}}=\frac{2+5x^3}{2\sqrt{x^3+1}}=\frac{2(x^3+1)+3x^3}{2\sqrt{x^3+1}}=\frac{x^3+1}{\sqrt{x^3+1}}+\frac{3x^3}{2\sqrt{x^3+1}}=$
$=\sqrt{x^3+1}+\frac{3x^3}{2\sqrt{x^3+1}}=1\cdot\sqrt{x^3+1}+x\cdot\frac{3x^2}{2\sqrt{x^3+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Conditional extreme value of a function Let $x,y,z$ be the positive real numbers, if $x^2+y^2+z^2=1$, then how can we find the minimal value of this function $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.
| I will show that,
$$ \frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y} \ge \sqrt{3}$$
This is equivalent to,
$$ (\frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y})^2 \ge 3$$
Which is same as,
$$ \frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} + 2(x^2+y^2+z^2) \ge 3 = 3(x^2+y^2+z^2) $$
So it suffice to prove that,
$$ \frac{x... | {
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"url": "https://math.stackexchange.com/questions/892749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$
*
*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$
*$(x+3)(x-3) = -\sqrt{2-x}$
*$(x+3)^2(x-3)^2 = 2-x$
*$x^4-18x^2+81 = 2-x$
*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please... | One way to solve (or approximate) such an equation (or to simply check how many solutions exist) is to graph each side of the equation: $$x^2 - 9 = -\sqrt{2-x}$$
*
*$y_1 = x^2-9$
*$y_2 = -\sqrt{2-x}$
Then look for when $y_1 = y_2$: if there are solutions to this equality, they will appear as points where $y_1$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}... | An element of order $2$ is a product of $k$, say, disjoint 2-cycles.
*
*For $k=1$, there are $\frac{n(n-1)}{2^1\cdot 1!}$ elements of order two.
*For $k=2$, there are $\frac{n(n-1)(n-2)(n-3)}{2^2\cdot 2!}$ elements of order two.
*For $k=3$, there are $\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{2^3\cdot 3!}$ elements of ord... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| WOLG:$a\ge b\ge c$
we have
$$2b(a+c)^2-(a+b)(b+c)(a+c)=(a+c)(a-b)(b-c)\ge 0$$
so
$$\dfrac{8abc}{(a+b)(b+c)(a+c)}\ge\dfrac{4ac}{(a+c)^2}$$
so we only prove
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$
since
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}- 2=\dfrac{(a^2+c^2
-ab-bc)^2}{(a+c)^2(ab+bc+ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Solution of $\frac{d^2y}{dx^2} - \frac{H(x) y}{b} = H(-x)$ Does the equation
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(x)$$
have a solution where $H(x)$ is the Heaviside step function and $b$ and $c$ are constant?
Update: What about the second step function be $H(-x)$:
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-... | I am not sure wether this answer is correct or not, but as far as the step function is only a sign, I brought it out of the integration.
Please note that, for finding the constants, you need the boundary values and compare the value of the function at the point $x=0$.
$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895420",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True
if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true.
$\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space... | This is a "brutal force" solution with the help of WolframAlpha. Hopefully someone else will give a more elegant proof later.
Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$
We can suppose, w.l.o.g, that $x\leq y \leq z$. Thus by the rearrangement inqeuality, $2x + 3y +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How many ways can 5 dice produce a total of 20?
How many ways can $5$ dice produce a total of $20$?
I set up the equation $x_1+x_2+x_3+x_4+x_5 = 20$. The total possible number of combinations is $\binom{19}4$. From there I subtracted the number of possibilities where $1$ of the variables is greater than $6$, which I ... | As in this answer, we can approach this question using either generating functions or inclusion-exclusion. Instead of counting the number of ways for $5$ numbers from $1$ to $6$ to sum to $20$, we will count the number of ways for $5$ numbers from $0$ to $5$ to sum to $15$ (then add $1$ to each of the $5$ numbers).
Ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Simplifying $\sqrt[3]{a\pm\sqrt{b}}$ Let
$$x=\sqrt{a\pm\sqrt{b}}$$
We know that
$$x=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
But, what about cubic root?
Let
$$y=\sqrt[3]{a\pm\sqrt{b}}$$
Is there any formula to find $c$ and $d$ such that $c,d\in\mathbb{Q}$ and $c\pm\sqrt{d}=y$ if $c$ and $d$ e... | There isn't a formula for $\sqrt[3]{a\pm\sqrt{b}}$ The best method that I know is simplifying $\sqrt{b}$ (if possible) and assuming that it can be denested into $x+y\sqrt{b}$.
More generally, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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When is $f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$ decreasing $\forall\; x$? When is $$f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$$ decreasing $\forall\; x$?
Diffrentiating:
$$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$
If $f$ is decreasing, $f'<0$:
$$5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3<... | In order the function be always decreasing, the requirement is that $$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$ be negative for any value of $x$. Since $x^4 \geq 0$, then $$\frac{\sqrt{p+4}}{1-p}-1 \lt 0$$ which implies either $p \gt 1$ or $-4 \lt p \lt \frac{1}{2} \left(3-\sqrt{21}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897969",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd = k$ and $ad+bc = l$ for fixed $k,l$? Suppose, for now, that all numbers are real numbers. Let us fix numbers $k,l$.
Then can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd =k$, $ad+bc = l$ for some $k$ and $l$?
What h... | For real numbers, yes. One infinite family of solutions is for instance
$$
(a,b,c,d) = \left(r\frac{l+k}{2}, r\frac{l-k}{2}, \frac{1}{r}, \frac{1}{r}\right),
$$
for $r \neq 0$.
For integers, if $k = l = 0$, then $(a,b,c,d) = (n,n,0,0)$ gives infinite number of solutions. If $k \neq 0$ or $l \neq 0$,
$$
0 <k^2+l^2 = (ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Intersection of circle and ellipse I'm looking for the points of intersection of a circle
$x^2 + y^2 = r^2$ ($r$ is known, origin is $(0,0)$)
and an ellipse
$(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1$ ($a,b,x_0,y_0$ are known).
Actually i do only need the Angles $\varphi$ at which the circle with radius $r$ is intersecti... | Hint:$$x^2 + y^2 = r^2\implies x^2 + y^2 - r^2=0$$ and $$(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1\implies (x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1=0$$
So you're solving $$x^2 + y^2 - r^2=(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1$$
But the solution will be long, however if $x_0=y_0=0$ then $$y=\pm\frac{\sqrt{a^2r^2+a^2x^2+a... | {
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"url": "https://math.stackexchange.com/questions/898313",
"timestamp": "2023-03-29T00:00:00",
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Prove $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$
if $a,b,c$ are positive real numbers,Prove:$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$
Things I have done so far: I know the fact that $$a^3+b^3+c^3\geq\frac{1}{2}[ab(a+b)+bc(b+c)+ca(c+a)]$$
However i tried to reach the problem inequality,but I was not succesful.
Source of problem: schoo... | By aM-GM:
$$\frac{2a^3+b^3}3\ge(a^6b^3)^{\frac13}=a^2b$$
$$\frac{2b^3+c^3}3\ge(b^6c^3)^{\frac13}=b^2c$$
$$\frac{2c^3+a^3}3\ge(c^6a^3)^{\frac13}=c^2a$$
Add them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/900973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
| $$\begin{align}S&=3+3\cdot 4+3\cdot 4^2+3\cdot 4^3+...+3\cdot 4^{\log_2 n-1}\\
&=3[1+4+4^2+4^3+...+4^{\log_2 n-1}]\\
4S&=3[\quad\;\; 4+4^2+4^3+...+4^{\log_2n-1}+4^{\log_2 n}]
\quad \quad \text{as $n=1,2,4,8,...,2^m,...$}
\end{align}$$
Subtracting:
$$\begin{align}
3S&=3[4^{\log_2n}-1]\\
&=3[2^{2\log_2n}-1]\\
&=3[(2^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Unique least square solutions There is a theorem in my book that states: If $A$ is $m\times n$, then the equation $Ax = b$ has a unique least square solution for each $b$ in $\mathbb{R}^m$.
But can we find a counter-example to this by providing a matrix $A$ and vector $b$ such that $A^TAx = A^Tb$ produces a general so... | Your theorem statement is incomplete. Requirements have been omitted.
To amplify the insights of @Troy Woo, given a matrix $\mathbf{A}\in\mathbb{C}^{m \times n}$, a solution vector $x\in\mathbb{C}^{n}$, and a data vector $b\in\mathbb{C}^{m}$ such that $b\notin\mathcal{N}(\mathbf{A}^{*})$, and where $n\in\mathbb{N}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Two sequences defined by recurrence relations satisfy $x_n/y_n<\sqrt{7}$ for all $n$ Let $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq 1}$ be two sequences such that:
$$x_{n+1}=x_n^2+1 \quad \text{ and } \quad y_{n+1}=x_n y_n$$
with $x_1=2$ and $y_1=1$
Prove that for all $n$
$$\dfrac{x_n}{y_n} < \sqrt{7}.$$
Can I have any he... | Given that:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_n y_n}$$
we have:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n y_n}+\frac{1}{x_{n-1} y_{n-1}}+\ldots+\frac{1}{x_1 y_1}+\frac{x_1}{y_1}$$
or just:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n\cdot\ldots\cdot x_1}+\frac{1}{x_{n-1}\cdot\ldots\cdot x_1}+\ldots+\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of positive integers whose digits are all $1$, $3$, or $4$, and add up to $2k$, is a perfect square I have been stuck on this question for a pretty long time. My teacher says that we should find a small pattern, but I can't find one. Can anyone give me a hand?
Let $b_n$ be the number of positive integers ... | First note that since $F_n=F_{n-1}+F_{n-2}$, we have
$$
\begin{align}
F_n^2
&=F_{n-1}^2+F_{n-2}^2+2F_{n-1}F_{n-2}\\
&=2F_{n-1}^2+2F_{n-2}^2-(F_{n-1}-F_{n-2})^2\\
&=2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2\tag{1}
\end{align}
$$
a recursion for the squares of the Fibonacci numbers.
The generating function for the count of numbers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to find that triple integral? How to find the triple integral of $$ \frac{(z-z_0)z}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}$$ over the sphere $ \{(x,y,z):x^2+y^2+z^2 \le 1 \} $ under the assumption $x_0^2+y_0^2+z_0^2 \le 1?$ Its physical interpretation suggests
the integral can be expressed through elementary funct... | Let $\vec{p} = (x,y,z)$ and $\vec{p}_0 = (x_0,y_0,z_0)$. Let $(r,\theta,\phi)$ and $(r_0,\theta_0,\phi_0)$ be their spherical polar coordinates. More precisely,
$$\begin{array}{lll}
\vec{p} &= (x,y,z) &= r(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)\\
\vec{p}_0 &= (x_0,y_0,z_0) &= r_0(\sin\theta_0\cos\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $a_{1}\;a_{2},a_{3}$ are the Roots of cubic eq. , Then $1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right)$
If $a_{1}\;a_{2},a_{3}$ are three real values of $a$ which satisfy the equation $$\displaystyle \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx-\frac{4a}{\pi-2}\int_{0}^{1}x\cdot \cos xdx = 2.$$ Then value of $\displ... | If you can get the cubic into the form $C_3a^3+C_2a^2+C_1a+C_0 = 0$, then the roots satisfy
$a_1+a_2+a_3 = -\dfrac{C_2}{C_3}$
$a_1a_2+a_2a_3+a_3a_1 = \dfrac{C_1}{C_3}$
Thus, $a_1^2+a_2^2+a_3^2 = (a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1) = \dfrac{C_2^2-2C_1C_3}{C_3^2}$.
So, all you have to do is evaluate the following i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A double sum involving the Riemann zeta function Evaluate the sum $S=\sum_{k=2}^{\infty} \frac{\zeta (k)-1}{k+1}$, where $\zeta (s)$ denotes the Riemann zeta function.
The sum is equal to $\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{(k+1)n^k}$, then switching the order (since the summand converges uniformly) gives... | Remember Stirling's formula
$$\sum_{m=1}^n \log m = \log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{n}\right).$$
Then we have
$$\begin{align}
\sum_{n=2}^K \left(-n\log\left(1 - \tfrac{1}{n}\right) - 1 - \tfrac{1}{2n}\right)
&= 1-K +\tfrac{1}{2} - \tfrac{1}{2}H_K - \sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove logarithmic inequality with greatest integer function. $\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$
How to show this? I tried using $\left \lfloor k \right \rfloor \leq k$, but I'... | hint:
$\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1 =\left \lfloor (n\log_2 (n+1)^2)+\log_2 (n+1)^2 \right \rfloor +1\ge \left \lfloor (n\log_2 (n+1)^2) \right \rfloor +\left \lfloor \log_2 (n+1)^2\right \rfloor+1$
now prove:
$\left\lfloor (n\log_2 (n+1)^2) \right \rfloor \ge \left\lfloor (n\log_2 n^2) \right ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
| The following solution is based on a suggestion of lab bhattacharjee:
$\displaystyle\int \frac{\sqrt{1-k^2\sin^2 x}}{\sin x} dx=\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin^{2} x}\cdot\sin x dx=\int\frac{\sqrt{1-k^2(1-\cos^{2}x)}}{1-\cos^{2}x}\cdot\sin x dx$.
Now let $t=\cos x$, $dt=-\sin x dx$ to get $\displaystyle -\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| Suppose that $x \neq 1$. We wish to show by inducting on $n$ that $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$.
Then $1 = \frac{x-1}{x-1}$ so that the formula holds for $n = 1$.
Suppose that we have $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$. We must show that $\sum\limits_{k=0}^{n}x^{k} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Calculate Integrals $ \int \sqrt{\sec 2x-1}\;dx$ and $ \int \sqrt{\sec 2x+1}\;dx$ Calculation of Integral of
$$\displaystyle \int \sqrt{\sec 2x-1}\;dx,\>\>\>\>\>\displaystyle \int \sqrt{\sec 2x+1}\;dx$$
$\bf{My\; Solution}::$ For $(a)::$
Let $$\displaystyle I = \int \sqrt{\sec 2x-1}\;dx = \int \sqrt{\frac{1-\cos 2x}{\c... | Integrate as follows
\begin{align}
&\int \sqrt{\sec 2x-1}\;dx \\=&
\int \frac{\sqrt2\sin x}{\sqrt{\cos 2x}}dx=- \int \frac{d(\sqrt2\cos x)}{\sqrt{2\cos^2 x -1}}dx
= -\cosh^{-1}(\sqrt2 \cos x)+C\\
\\
&\int \sqrt{\sec 2x+1}\;dx \\=&
\int \frac{\sqrt2\cos x}{\sqrt{\cos 2x}}dx=\int \frac{d(\sqrt2\sin x)}{\sqrt{1-2\sin^2 x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the correct way to compare to algebraic quantities? If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other?
For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} ... | $$\begin{align}\frac{2ds}{s^2-c^2}-\frac{2d}{\sqrt{s^2-c^2}}&=\frac{2d}{\sqrt{s^2-c^2}}\left(\frac{s}{\sqrt{s^2-c^2}}-1\right)\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s-\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}}\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-(s^2-c^2)}{\sqrt{s^2-c^2}(s+\sqrt{s^2-c^2})}\\&=\frac{2dc^2}{(s^2-c^2)(s+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int\frac{x^3}{(x^3+1)^2}dx$ any ideas on how I could continue this integral
$$\int\frac{x^3}{(x^3+1)^2}dx$$
I am half way done, by entirely using fraction decomposition
$\int\frac{x^3}{(x^3+1)^2}dx=\int\frac{1}{x^3+1}dx-\int\frac{1}{(x^3+1)^2}dx$
$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$ from whi... | Hint: Try integrating by parts first, with $u=x$ and $dv={x^2\over(x^3+1)^2}dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have... | From your equation:
$−\frac{b}{2}x^2+bx+b=0$
notice that it's equal to $b*(−\frac{1}{2}x^2+x+1)=0$
Multiplying both sides of the equation by -2 we get:
$b*(x^2-2x-2)=0$
So the solutions to this become either $b=0$
or $x^2-2x-2=0$
Which imply $x=1 \pm \sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\... | well, notice that $$\frac{x^2-2}{x^2-\sqrt{2}x-4} = \frac{(x-\sqrt{2})(x+\sqrt{2})}{(x-\sqrt{2})(x+2\sqrt{2})}$$ which gives the same answer as yours.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
| Note that
$$5\times 10\times 15\times 20\times 25=5\times(2\cdot 5)\times (3\cdot 5)\times (2^2\cdot 5)\times 5^2=2^3\cdot 3\cdot 5^6$$
and that
$$10^6=2^6\cdot 5^6=\frac{8\cdot 5\cdot 10\cdot 15\cdot 20\cdot 25}{3}=8\cdot 5\cdot 10\cdot 5\cdot 20\cdot 25.$$
So, in mod $10$, we have
$$\begin{align}\frac{28!}{10^6}&\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Calculate number of integers less than n fitting the form 6n±1 Of course the approximation is n/3, but I am looking for a way to get the number of integers, not an approximation.
| There are several sequences that count the amount of $6k\pm1$ less than or equal to $n$.
Here is a specific sequence:
$$A_n=2\cdot\left\lfloor\frac{n+1}{6}\right\rfloor-\frac{(n\bmod6)^4}{24}+\frac{5(n\bmod6)^3}{12}-\frac{35(n\bmod6)^2}{24}+\frac{25(n\bmod6)}{12}$$
Here is the general solution:
*
*Divide by $3$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Rational Expression Simplification The function:
$$f(x) = \frac{3x - 4}{x^2 - 2x}$$
is simplified to:
$$f(x) = \frac{2}{x} + \frac{1}{x - 2}$$
How? And in what way?
| We would like to write $\dfrac{3x-4}{x^2-2x} = \dfrac{3x-4}{x(x-2)}$ in the form $\dfrac{A}{x} + \dfrac{B}{x-2}$ for some constants $A,B$.
Now, we just need to solve for $A,B$. First, let's get rid of fractions:
$\dfrac{A}{x} + \dfrac{B}{x-2} = \dfrac{3x-4}{x(x-2)}$
$A(x-2)+Bx = 3x-4$
This equality must be true for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the circle which passes through two points
Find the equation of a circle which passes through $(4,-3)$ and $(-3,-4)$ with radius $5$.
I tried putting the $x$ and $y$ into the equation $(x-h)^2 + (y-k)^2 = r^2$, but then I don't know how to continue.
| Since we have
$$(4-h)^2+(-3-k)^2=5^2\iff h^2-8h+16+k^2+6k+9=25\tag1$$
$$(-3-h)^2+(-4-k)^2=5^2\iff h^2+6h+9+k^2+8k+16=25$$
subtracting the latter from the former gives you
$$(-8-6)h+7+(6-8)k-7=0\iff k=-7h.$$
Then, use $(1)$ to get $h=0,1$. So, the answer is the followings :
$$x^2+y^2=25,\ \ (x-1)^2+(y+7)^2=25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $... | $3^n\geq2n^2+3n$
(1) $n=4$
$81\geq44$
(2) Assume true for $n=k$ ,Therfore we have :
$3^k\geq2k^2+3k$
(3) Then for $n=k+1$
$3^{k+1}\geq2(k+1)^2+3k+3$
Simplifying this gives:
$3(3^k)\geq2k^2+7k+5$
Now to prove this ^^^ you can subsititute what we assumed in (2) to show this inequality stands.
$3^k\geq2k^2+3k$
So..
$3(2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \fr... | We have $$\lim_{x\to0}(1+\sin x)^{\dfrac1x}\cdot\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}$$
$$\lim_{x\to0}(1+\sin x)^{\dfrac1x}=\left(\lim_{x\to0}(1+\sin x)^{\frac1{\sin x}}\right)^{\lim_{x\to0}\dfrac{\sin x}x}=e^1$$
$$\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}=\left(\lim_{x\to0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygam... | Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.
We can break up the integral into
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 5,
"answer_id": 1
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finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to ... | Well,
$$\int\frac{2}{x+2} dx= 2\log(x+2)$$
So you have that part.
But for:
$$\int\frac{1-2x}{x^2 + 1} dx$$
You must further decompose this fraction into partial fractions:
$$\frac{1 - 2x}{x^2 + 1} = \frac{1}{x^2 + 1} - \frac{2x}{x^2 + 1}$$
So this integral becomes:$$\int\frac{1 - 2x}{x^2 + 1} dx= \int\frac{1}{x^2 + 1}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
| Another way to find this integral would be to use
$\displaystyle\int\frac{y^2-x^2}{(y^2+x^2)^2}dy=\int\bigg(\frac{2y^2}{(x^2+y^2)^2}-\frac{1}{y^2+x^2}\bigg)dy$;
Using integration by parts for the first term with $u=y, dv=\frac{2y}{(x^2+y^2)^2}dy$ gives
$\bigg(\displaystyle-\frac{y}{y^2+x^2}+\int\frac{1}{y^2+x^2}dy\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find whether vector w belongs in the span $$v_1=[1,0,1,2]$$
$$v_2 = [0,1,1,3]$$
$$v_3 = [2,1,3,7]$$
$$w = [1,2,3,4]$$
We are supposed to determine if $w$ is in $\operatorname{span}(v_1,v_2,v_3)$.
| The row reduced form:
$$
\begin{align}
\mathbf{A} &\to \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 1 \\
1 & 1 & 3 \\
2 & 3 & 7 \\
\end{array}
\right]
%
&\to
%
\left[
\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
In $\triangle ABC$, $D$ is a point on side $BC$ that $\angle BAD = \angle CAD =\angle ABC$. If $BD=1$ and $DC=2$, what would be the length of $AB$?
In $\triangle ABC$, $D$ is a point on side $\overline{BC}$ that $\angle BAD = \angle CAD =\angle ABC$. If $\overline{BD}=1$ and $\overline{DC}=2$, what would be the length... | Let $\alpha=\angle ABC$.
Since $AD$ is the bisector of $\angle CAB$, we have
$$AB:AC=BD:CD=1:2\Rightarrow AC=2AB.$$
So, since
$$\angle ACB=180^\circ-3\alpha\Rightarrow \sin(\angle ACB)=\sin(3\alpha)=3\sin \alpha-4\sin^3\alpha,$$we have with $\sin\alpha\not=0$, by the law of sines,
$$\begin{align}\frac{AB}{\sin (\angle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is this proof by induction for a sum of odd squares correct? Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$
Proof by induction
-Base case: when $n = 1$
$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$
$1=1$ hence statement holds for $n = 1$
-Inductive step
assume $n = k$ is true
then $1^2 + 3^2 + 5^2 + ... | After mentioning the case when $n=1$, the way you are writing looks strange to me.
Also, note that you don't need to expand them.
Assuming that $1^2+3^2+\cdots +(2k-1)^2=\frac{k(2k-1)(2k+1)}{3}$ holds, we have
$$\begin{align}1^2+3^2+\cdots +(2k-1)^2+(2(k+1)-1)^2&=\frac{k(2k-1)(2k+1)}{3}+(2k+1)^2\\&=\frac{2k+1}{3}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factor $(x^5+1) (x^5-1) $ I have this:
$ (x^5+1) (x^5-1) $
And I don't know how to continue factor.
Geogebra's Factor says:
$(x+1)(x-1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$
| Hint:
$$x^n-1=(x - 1)[x^{n - 1} + x^{n - 2} + ... + x^2 + x + 1] $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}d... | Other answers offer alternative approaches to integrating form the beginning. If you are looking to find where your steps went astray, it starts when you are using: $$2(1-\cos(2a))=4\sin^2(4a)$$ but the correct identity is: $$2(1-\cos(2a))=4\sin^2(a)$$ Just check both sides with $a=\pi/4$ and you'll believe it. And si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 1
} |
Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Expr... |
Given the picture, let $x$ be the side of the equilateral triangle.
We have:
$$ 6 = \frac{x}{2}\cot\arctan\frac{4}{3}+\frac{x}{2}\cot\frac{\pi}{6},$$
or:
$$ 6 = \frac{3x}{8}+\frac{\sqrt{3}\,x}{2},$$
so:
$$ 48 = x(4\sqrt{3}+3) $$
and:
$$ 48(4\sqrt{3}-3) = 39 x,$$
so $k=\color{red}{\frac{16}{13}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is the series of the function 3 / ( 1- x^4) I know that $f(x) = \frac{1}{ 1-x } = \sum_{n=1}^\infty x^n$. We can find that $g(x) = \frac{1}{ 1-x^4 } = \sum_{n=1}^\infty (x^4)^n = \sum_{n=1}^\infty x^{4n}$. Does the sum converge? what is the convergence radius?
| The series $ 1+t+t^2+\cdots$ has radius of convergence $1$, so converges if $|t|\lt 1$ and diverges if $|t|\gt 1$. Thus our series $1+x^4+x^8+\cdots$ converges if $x^4\lt 1$, and diverges if $x^4\gt 1$. So we have convergence if $|x|\lt 1$, divergence if $|x|\gt 1$.
Remark: Your sums should start at $n=0$, not at $n=1... | {
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"timestamp": "2023-03-29T00:00:00",
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Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| When you square the right-hand side, you will get another (infinite) polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots$ . Now go through the coefficients $a_i$ and ask yourself which terms in the product $\big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big) \big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big)$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim_{n\to\infty}\frac{2n}{n+2}=2$ Prove $$\lim_{n\to\infty} \frac{2n}{n+2} = 2$$
I know for all $\epsilon > 0$ there exists some $k\in\mathbb{N}$ for all $n > K$ and $\left|\frac{2n}{n+2} - 0\right| < \epsilon$.
Proof:
Fix any $\epsilon > 0$ choose $k \in \mathbb{N}$ where there exists $???? < K$
I am getting s... | $\textbf{Proof:}$
Let $\epsilon >0$ be given.
$\textbf{RTP:}$ $\exists N_\epsilon \in \mathbb{N} \ni |\frac{2n}{n+2} -2|<\epsilon \ \forall \ n \geq N_\epsilon $.
$\textbf{ROUGH WORK:}$
We know that $|\frac{2n}{n+2}-2| = |\frac{-4}{n+2}| =\frac{4}{n+2} \forall n\in \mathbb{N}$.
Now we need $\frac{4}{n+2} < \epsilon \... | {
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"timestamp": "2023-03-29T00:00:00",
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If $y = a\sin{x} + b\cos{x} +C$ then find maxima and minima for $y$. I was able to solve it till
$$y = \sqrt{(a^2 + b^2)}\sin(\alpha + x) + C.$$
But I don't know how to find maxima and minima from here.
If $C = 0$ then maxima & minima equals the amplitude of the sine curve but when $C$ is non-zero then?
I need help f... | The old way:
$$y'=a\cos x-b\sin x=0\iff\tan x=\frac ab\iff x=\arctan\frac ab+k\pi.$$
Then
$$\sin\theta=\pm\frac{a}{\sqrt{a^2+b^2}},\\\cos\theta=\pm\frac{b}{\sqrt{a^2+b^2}},$$ and
$$y=\pm\frac{a^2+b^2}{\sqrt{a^2+b^2}}+c=\pm\sqrt{a^2+b^2}+c.$$
The maximum obviously corresponds to the plus sign.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\limits^{\infty}_{0} \frac{x^\alpha}{(1+x^2)^2}dx, \ -1 < \alpha<3.$ Evaluate $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx, \ -1 < \alpha<3.$
May I verify if my solution is correct? Thank you.
Consider $\gamma_1:=\{x:-\dfrac{1}{r}\leq x \leq-r\}, \ \gamma_2:=\{re^{it}: \pi\le... | I am providing an alternative route towards the evaluation of this integral.
Consider a change of variable $x^2=u$ then the integral can be expressed as
$$\int^{\infty}_{0}\frac{x^{\alpha}}{(1+x^2)^2}\,dx=\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{2(1+u)^2}\,du=\frac{1}{2}\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{(1+u)^2}... | {
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"source": "stackexchange",
"question_score": "3",
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Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}... | $\dfrac{8}{\sqrt{x-2}} = 8\cdot (x-2)^{-1/2}$.
Now you can use $\dfrac{d}{dx}(x^n) = n x^{n-1}$, where $n = -1/2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\frac{ 1 }{ 1010 \times 2016} + \frac{ 1 }{ 1012 \times 2014} + \frac{ 1 }{ 1014 \times 2012} + \cdots + \frac{ 1 }{ 2016 \times 1010} = ?$ $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = ?
$$
My attempt so far :
$... | The answers already showed that the expression cannot be more simplified. However, it could be quite accurately approximated since $$S_{m,n}=\sum_{i=m}^n \frac{1}{i}=H_n-H_{m-1}$$ Now, consider that both $m$ and $n$ are large numbers; we can use asymptotic expansions and, at the fourth order, $$S_{m,n}\approx \log \lef... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \... | If we write the polynomial as a linear combination of combinatorial polynomials, then the polynomial is divisible by $24$ for all integer arguments if and only if all of the coefficients are divisible by $24$.
$$
n^4+6n^3+11n^2+6n=24\binom{n}{4}+72\binom{n}{3}+72\binom{n}{2}+24\binom{n}{1}
$$
and so the polynomial is d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Can someone explain how to solve linear inequality?
Can someone explain how to solve this linear inequality?
| Multiply the both sides by $x^2-9=(x-3)(x+3)$.
But note that you need to have two cases as $x^2-9\gt 0$ or $x^2-9\lt 0$.
(Case 1)
If $x^2-9\gt 0\iff x\lt -3\ \text{or}\ x\gt 3$, then
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\le\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\le 3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating product of exponent and polynomial In a probability theory problem, I need to solve an inequality over $n\in\mathbb{N}$ which can be expressed in a general form like this:
$a^n ( b_1 n^2 + b_2 n + b_3) \leq c$ where $a, b_1, b_2, b_3, c$ are real numbers.
What is the standart way (or a good way) to tackle t... | Let's clean up a bit
$$ 0.8^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq 0.1 $$
$$ \left(\frac{8}{10}\right)^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{10^n}\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{8}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\righ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to show that $a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2} $ I want show the following $$a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2},\ s,\ t\in{\mathbb Q} $$
How can we prove this ?
[Add] Someone implies that we mus... | Set $a=\cos\theta$ and $b=\sin\theta$ rational. Then,
Set $s=1$ and $t=\tan(\frac{\theta}{2})$ then $a=\frac{s^2-t^2}{s^2+t^2}$ and $b=\frac{2st}{s^2+t^2}$.
To show that $t$ is rational, remark that $\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$.
Q.E.D.
And actually it works for all $a,b\in\mathbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x... | 1) $$\bigg( \frac{-3}{x^{5/2}} \bigg)'=-3x^\frac{-5}{2}=\frac{15}{2}
x^\frac{-7}{2}$$
2) $$ \bigg(\frac{2x^2+3}{(x^3-4)^3}\bigg)' =\bigg( (2x^2+3
)(x^3-4)^{-3}\bigg)' $$ $$= (4x)(x^3-4)^{-3} + (2x^2+3
)(-3)(x^3-4)^{-4}(3x^2) $$$$= (x^3-4)^{-3} [ 4x -9x^2(2x^2+3 )(x^3-4)^{-1}]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplification of a series so that it converges to a given function I am trying to rearrange the series
$ \frac{1}{1-z} - \frac{(1-a)z}{(1-z)^2} + \frac{(1-a)^2z^2}{(1-z)^3} - \cdots$
In such a way that I can show it converges to
$\frac{1}{1-az} $
What I have so far
Let $ w = \frac{z}{1-z} $, we can then writ... | Even though it has already been answered, I might aswell post my already typed solution:
Your expresion is
$$\frac{1}{1-z} \cdot \sum_{n=0}^\infty \left(-\frac{1-a}{1-z}z\right)^n$$
Using the geometric series, this is
$$\frac{1}{1-z} \cdot \frac{1}{1+\frac{1-a}{1-z}z}$$
which is
$$\frac{1}{1-z} \cdot \frac{1-z}{1-z+(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/969471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conditional expectations of $E(X+Y|z)$ Given: $$f(x,y,z) = \frac23 (x+y+z), \,\,\, 0<x<1,\,\,\, 0<y<1,\,\,0<z<1$$ zero elsewhere.I was instructed to determine the cumulative df of $x,y,z$. Here is my answer $$F (x,y,z) = \frac {xyz (x+xy+z)} {3} $$ Another problem that I can not answer is this one: Find the conditiona... | The first thing to notice is that you have mis-interpreted the first question. There is no common notion of the c.d.f. of multiple fvariables; what your professor or book wanted is the c.d.f. of $x$ which is just like the c.d.f. of $y$ or of $z$.
Thus for a value $x = X$,
$$
F(X) = \frac{2}{3}\int_{x=0}^{X} dx\int_{y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Log integrals III The integral
\begin{align}
J_{m} = \int_{0}^{1} \frac{t^{m}}{1+t} \, \ln(1+t) \, dt
\end{align}
has the general form
\begin{align}
J_{m} = (-1)^{m} \left[ A_{m} - B_{m} \, \ln(2) + C_{m} \, \ln^{2}(2) \right]
\end{align}.
Is there a general form for the coefficients $A_{m}$, $B_{m}$, and $C_{m}$ ?
| The generating function is, according to Maple,
$$ \sum_{m=0}^\infty J_m x^m = \int_0^1 \dfrac{\ln(1+t)}{(1+t)(1-xt)}\; dt
= (1+x)^{-1} \left(\dfrac{1}{2} \ln(2)^2 + \ln \left(\dfrac{x}{1+x}\right) \ln(1-x) - \text{dilog} \left(\dfrac{x}{1+x}\right) + \text{dilog} \left(\dfrac{2x}{1+x}\right) \right)$$
but I don't kno... | {
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"url": "https://math.stackexchange.com/questions/974157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hard Mathematical Induction I have a mathematical induction question and I know what I need to do just not how to do it.
The question is:
Prove the equality of:
$$(1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3$$
Base case:
$$(1 + 2)^2 = 1^3 + 2^3\\
(3)^2 = 1 + 8\\
9 = 9$$
and I understand I have to get the sides to eq... | Let the statement hold for $n \in \mathbb{N}$. For $n+1$, by applying the binomial formula $(a+b)^2=a^2+2ab+b^2$ you have that $$\begin{align*}\left(\underbrace{1+2+\ldots+n}_{=a}+\underbrace{(n+1)}_{=b}\right)^2&=(1+2+\ldots+n)^2+2(1+2+\ldots+n)(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+2\frac{(n+1)n}{2}(n+1)+(n+1)^2=\\&=1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/974788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to integrate by parts, without changing variable How can I solve:
$$\int \frac{x^2}{\sqrt{1-x^2}}\;dx$$
without changing variables, by parts?
| $$\int\frac{x^2}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int\frac{x}{\sqrt{1-x^2}}\,d(1-x^2)=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx\Big)=..$$\begin{align}&...=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{1-x^2}{\sqrt{1-x^2}})\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| For positive $a$:
$$
\sqrt{a^2 + 4} = a\sqrt{1+\frac{4}{a^2}}
$$
and
$$\lim_{a\to\infty}\sqrt{1+\frac{4}{a^2}} = 1.
$$
So, the Taylor series of $\sqrt{1+x}$ around $0$ (when $x\to 0$) is:
$$
\sqrt{1+x} = 1 + \left.\frac{1}{2\sqrt{1+x}}\right\rvert_{x=0} x + O(x^2) = 1+ \frac{x}{2} + O(x^2)
$$
So, when $1<<a$ (it's impo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how to solve $a\sin x+b\cos x$ Let's solve:
$\sqrt{3}\sin x - \cos x=2$
The left hand side may be expressed as $R\sin(x+ \phi)$
We know that $R=\sqrt{3+1}=2$
We also know that $\tan \phi= \frac{-1}{\sqrt{3}}$
The solution to $\tan \phi=\frac{-1}{\sqrt{3}}$ has many solutions, for example, -30, 150, 330 degrees etc.
Whi... | $$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x=1$$
$$\sin\left(x -\frac{\pi}{6}\right) = 1$$
$$\sin\left(x -\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right)$$
$$ x = n\pi + (-1)^n \frac{\pi}{2} +\frac{\pi}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all posi... | If $x=2k$, then
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+2}{4}\right\rfloor=\left\lfloor k^2+k+\frac 12\right\rfloor=k^2+k.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+1}{4}\right\rfloor=\left\lfloor k^2+k+\frac 14\right\rfloor=k^2+k.$$
If $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How do you solve this fraction?
What is the value of:
$$ \frac{2}{3} - \frac{1}{16} $$
My answer: $ \dfrac{1}{48} $ .
I believe that it's incorrect.
Three does not divide into $16$, so I cross multiplied. What am I doing wrong?
| \begin{align*}
\frac{2}{3}+\frac{-1}{16}&=\frac{2}{3}-\frac{1}{16} \\
&=\left(\frac{16}{16}\right)\times\frac{2}{3}-\frac{1}{16}\times\left(\frac{3}{3}\right) \\
&=\frac{32}{48}-\frac{3}{48} \\
&=\frac{29}{48}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Stuck with modular arithmetic problem using multiplication property I have the following problem:
Given $k\geq 1$, find $h$ such that
$$2^h \frac{4^k-1}{3}-1 \equiv 0 ~(\text{mod}~3).$$
This is my attempt using the invariance of multiplication:
$$2^h \frac{4^k-1}{3} \equiv 1 ~(\text{mod}~3) \Rightarrow 2^h \frac{4^k-... | $$
\begin{align}
\frac{(1+3)^k-1}{3}
&=\frac{\left[\binom{k}{0}3^0+\binom{k}{1}3^1+3^2\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\right]-1}{3}\\
&=\binom{k}{1}+3\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\\
&\equiv\binom{k}{1}\pmod3
\end{align}
$$
Therefore,
$$
2^h\frac{(1+3)^k-1}{3}-1\equiv0\pmod3
$$
is the same as
$$
2^hk\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ I am trying to evaluate:
$$\int_0^{\pi/12} \ln(\tan x)\,dx$$
I think the integral is quite simple but I am having a hard time evaluating it. I started with the result:
$$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$
where $G$ is the Catalan's constant. With the change of variables $x\... | An alternative "elementary" method.
Consider,
\begin{align*}
K&=\int_0^1 \frac{\arctan\left(\frac{x}{1-x^2}\right)}{x}\,dx\\
\end{align*}
Perform the change of variable $x=\tan\left(\frac{t}{2}\right) $,
\begin{align*}
K&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(\frac{1}{2}\tan t\right)}{\sin t}\,dt
\end{align*}
Défin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 3
} |
How to find the derivative of $f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$? Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:
$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$
would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$... | given $f(x)=(x^3-4*x+6)\ln(x^4-6x^2+9)$ we find by the product and chaine rule
$f'(x)=(3x^2-4)\ln(x^4-6x^2+9)+(x^3-4x+6)\frac{4x^3-12x}{x^4-6x^2+9}$
simplifying this in a few minutes
simplifying this i got
$f'(x)=\frac{4 x^4-16 x^2+3 x^4 \ln \left(x^4-6 x^2+9\right)-13 x^2 \ln \left(x^4-6
x^2+9\right)+12 \ln \left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying radicals I am stuck in the following puzzle and couldn't find a way to approach this.
$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$
Please help.
| $$
\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{(\sqrt{5}+3)^2}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{5} + 3}} = \sqrt{5 + \sqrt{5} + \sqrt{6+2\sqrt{5}}} = \sqrt{5 + \sqrt{5} + \sqrt{(\sqrt{5}+1)^2}} = \sqrt{5 + \sqrt{5} + \sqrt{5}+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Recognising that $\sum_{n=0}^\infty \frac{a^2-b^2(2n+1)^2}{(a^2+b^2(2n+1)^2)^2}=-\frac{\pi^2\mathrm{sech}^2\left(\frac{a\pi}{2b}\right)}{8b^2}$ So I know from Mathematica that:
$$\sum_{n=0}^\infty \frac{a^2-b^2(2n+1)^2}{(a^2+b^2(2n+1)^2)^2}=-\frac{\pi^2\mathrm{sech}^2\left(\frac{a\pi}{2b}\right)}{8b^2}$$
I am wondering... | As Semiclassical noted, you look at
$$ \sum_{n=0}^\infty \dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2} $$
Expand the summand in partial fractions:
$$ \dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2} ={\frac {2 {r}^{2}}{ \left( r^2 + (2n+1)^2 \right) ^{2}}} - \dfrac{1}{ r^2 + (2n+1)^2 }$$
First deal with the term on the right:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty} x \cdot \cos(x^3) dx$ convergence $$\int_{0}^{\infty} x \cdot \cos(x^3) dx$$
I only want to prove, that this integral converges, I don't need to calculate the exact value.
I don't know what to do with the cosinus, I can't get rid of it.
I know that the integral is equal to $$\frac{1}{3} \cdot \int_{0... | Thanks, to everyone!
$$\int_{0}^{\infty} x \cdot \cos(x^3) dx = \frac{1}{3} \cdot \int_{0}^{\infty} \frac{1}{x} \cdot 3 \cdot x^2 \cdot \cos(x^3) dx$$
and because of $\int 3 \cdot x^2 \cdot \cos(x^3) dx = \sin(x^3)$ we get
$$\int_{0}^{\infty} x \cdot \cos(x^3) dx=\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex... | Step $4$ is where your mistake happens. Your original equation is
$$z^3=i$$
Then you rewrite $i=1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$ and rewrite $z = r(\cos v + i\sin v)$, meaning that $$z^3=i$$
will change into $$r^3(\cos3v + i\sin 3v) = 1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$$
What you made was you also took the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Solving $y^2 - yx - y + x = 0$ for $y$? I solved this equation for $y$ by inspection and confirmed it with Wolfram Alpha -
$y^2 - yx - y + x = 0$
I got the values $y = 1$ and $y = x$
However I was wondering is there a formal method for solving it? I expressed it as a polynomial -
$y^2 + (-1 - x)y + x = 0$
and used the ... | Well, check again.
For $y^2+(−1−x)y+x=0$ with $y$ as the quadratic variable.
The quadratic formula with $b$ as $(-1-x)$, $a$ as $1$, and $c$ as $x$ tells that the roots are
$$y_{1,2}=\displaystyle \frac{(1+x) \pm \sqrt{1+x^2+2x - 4x}}{2}=\frac{(1+x)\pm\sqrt{1+x^2-2x}}{2}=\frac{(1+x) \pm{1-x}}{2}$$
Hence the solutions a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Beautiful evaluation inequality a,b,c>0 are such that $a^2+b^2+c^2=4$ and $4(a^2+2)=(a^2+b+c)^2$. What is the biggest possible value for a+b+c?
I tried a lot of stuff like $a^2=4-b^2-c^2$. And i think it's somehow connected to QM$\geq$AM where AM is (a+b+c)/3.
| Using the first relation, the left hand side of second relation can be written as $$4(a^2+2)=(a^2+b^2+c^2)(a^2+1^2+1^2)$$ On the right hand side we can apply the Cauchy-Schwarz inequality, so that $$4(a^2+2)=(a^2+b^2+c^2)(a^2+1^2+1^2)\overset{C.S.}\ge(a^2+b+c)^2$$ But due to the second relation we know that it holds ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
sum of infinite series with telescopes I need help finding the sum of the infinite series
$$\sum_{k=1}^\infty \frac{1}{n(n+1)(n+2)}$$
I have used the partial fraction decomposition to get this as the sum of
$$\frac{-1}{k+1}+\frac{1}{2(k+2)}+\frac{1}{2k}$$
but don't know where to go from here.
Thanks!
| Hint. First observe that $$
\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)
$$ and conclude by telescoping terms.
You end up with
$$
\sum_{n=1}^{N}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(1-\frac{1}{N+1}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
what is the limit of f(n)/g(n)? I am trying to solve this problem imagine that
\begin{array}{lr}
f(n) = 2^{\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n} - \left(\dfrac{1 - \sqrt{5}}{2}\right)^{n}\right]} \\
\\
g(n) = 2^{\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n+1}}
\end{array}
and I wan... | For simplicity, I will let $\varphi$ denote the golden ratio, $\overline \varphi$ its conjugate, and $F(n)$ the $n^{th}$ Fibonacci number. That is,
$$\varphi = \frac{1 + \sqrt 5}{2} \;\;\;\;\; \overline \varphi = \frac{1 - \sqrt 5}{2} \;\;\;\;\; F(n) = \frac{\varphi^n - \overline \varphi^n}{\sqrt 5}$$
Then we have
$$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
| it is $\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ $AM-HM$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ? Problem :
If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ?
Putting the value of n=2,3,4, we get :
when n =2 ; $a_2=1+\frac{1}{1} = 2$
when n=3; $a_3 =2+\frac{1}{2}$
when n=4; $a_4 =\frac{5}{2} +\frac{2}{5}$
Now how to ge... | To get a good approximation, note that $a_n^2 = \left(a_{n-1}+\dfrac{1}{a_{n-1}}\right)^2 = a_{n-1}^2+2+\dfrac{1}{a_{n-1}^2}$
Hence, $a_{100}^2 = a_1^2 + \displaystyle\sum_{n = 2}^{100}\left(a_n^2-a_{n-1}^2\right) = 1 + \sum_{n = 2}^{100}\left(2+\dfrac{1}{a_{n-1}^2}\right) = 199 + \sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving the determinant of a tridiagonal matrix with $-1, 2, -1$ on diagonal. Let $A_n$ denote an $n \times n$ tridiagonal matrix.
$$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$
Set $D_n = \det(A_n)$
Pr... | Here is a way to compute the determinant without using induction. Multiply the matrix $A_n$ to the left by the $n\times n$ upper triangular matrix $U_n$ with all entries on and above the diagonal equal to$~1$:
$$
\begin{pmatrix}1&1&1&\ldots&1\\ 0&1&1&\ddots&\vdots\\0&0&1&\ddots&1\\
\vdots & \ddots & \ddots & \ddots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integrating a function from negative infinity to infinity i got this weird question about integration from infinity to infinity.
$$\int_{-\infty}^{\infty} \frac{1}{z^2+25}$$
First idea was to take the factor out to get $(z+5)(z-5)$ but that really did not achieve anything.
Then i tried let $x= z^2$ and the $\frac{dx}{... | I think i finally got it
let $z=5tan(x)$ hence $dz=5sec^2(x)dx$ sub this in to the integrand
$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\int_{-\infty}^{\infty} \frac{5sec^2(x)dx}{25tan^2(x)+25}$$
as sec^2(x)=1+tan^2(x)
$$\int_{-\infty}^{\infty} \frac{5(tan^2x+1)}{25(tan^2(x)+1)}= \int_{-\infty}^{\infty} \frac{1dx}{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
} |
Evaluating by real methods $\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$ $\def\Li{{\rm{Li}}}$I'm sure you guys can briefly get the result by some methods of complex analysis, but now
I'm only interested in real analysis methods of proving the result. What would you propose
for that?
\begin{align*}
\int_0^{\pi/2} \frac{x... | Using double angle formula, the integrand can be rewritten as
\begin{equation}
I=\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=\int_0^{\pi/2} \frac{2x^5}{3-\cos(2x)}\ dx
\end{equation}
Mapping the variable $2x\mapsto x$, we have
\begin{equation}
I=\frac{1}{32}\int_0^{\pi} \frac{x^5}{3-\cos x}\ dx
\end{equation}
Using iden... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
the inequality $\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$ How to show that $$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$$ for $a,b,c>0$?
I tried to prove $$\frac{a^4}{a^3+b^3}\ge \frac {5a}4+\frac{-3b}4$$
but could not continue.
Give me ideas, pleas... | Here is a solution from "Secrets in Inequalities" by Pham Kim Hung.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^2\frac{x^5}{\sqrt{x^3+6}}\,dx.$ I am stuck on the following integral:
$\displaystyle\int_0^2\dfrac{x^5}{\sqrt{x^3+6}}\,dx.$
I have no idea how one can work it out. Normally I'd try $u=x^3+6$ but this surely does not work here.
| $$\frac{x^5}{\sqrt{x^3+6}}=x^3\frac{x^2}{\sqrt{x^3+6}}$$
Set $\sqrt{x^3+6}=u\implies x^3=u^2-6,\dfrac{3x^2dx}{\sqrt{x^3+6}}=du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is the remainder on division of $z^{400} + z^{303} + 1$ by $z^4-1$? I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-1$.
I expressed f as $f(z) = h(z)(z^4-1) + r(z)$ where $r(z)$ is a polynomial
Realising that $z^4-1 = (z^2-1)(z^2+1) = (z-1)(... | Another way:
Since
\begin{align*}
z^{400}+z^{303}+1&=z^{400}-1+(z^{300}-1)z^3 +z^3+2 \\
&=(z^4-1)(z^{396}+z^{392}+\ldots +1)+(z^4-1)(z^{296}+z^{292}+\ldots +1)z^3+z^3+2\\
\end{align*}
It follows that $z^3+2$ is the asked remainder.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I simplify $\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$? $$\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$$
I've had a go at putting everything over a common denominator in the form of:
$$\frac{\frac{\sqrt{1-x}\sqrt{1+x}+1}{\sqrt{1+x}}}{\frac{\sqrt{1-x}+1}{\sqrt{1-x}}}... | From where you left off, to remove all radicals from the denominator, we could multiply the top and bottom by $\sqrt{1+x}(1-\sqrt{1-x})$ -- the former factor to remove the factor of $\sqrt{1+x}$, and the latter being the conjugate of $1+\sqrt{1-x}$. This would result with
\begin{align*}
\frac{\sqrt{1-x}(\sqrt{1-x^2}+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Conditional probability dice problem, Two players, Adam and Eve, are throwing a die, and the first one to get a 6, will win.
Eve is throwing first, what is the probability that Adam can win?
$P(A)$ = probability Adam to win
$P(E)$ = probability Eve to win
I know the formula for conditional probability
$ P(A|B) = \frac{... | If Eve wins, it happens on either the first throw, the third throw, the fifth throw, etc...
If Eve wins on the first throw: $\frac{1}{6}$
If Eve wins on the third throw (his second throw): She missed, he missed, she won: $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$
If Eve wins on the fifth throw (her personal third th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Circle inscribed in triangle What's the radius and area of circle of max area that can be inscribed in a isoceles triangle with $2$ equal sides of length $1$? Radius formula is given, $r = \dfrac{2A}{P}$, where $A$ is area of triangle and $P$ is perimeter of triangle.
I have no idea how to do this.
I'm guessing the tri... | First assume that $\triangle ABC$ has $AB = AC = 1$, and denote $S$ = area of $\triangle ABC$, and $P$ = perimeter of the triangle, then:
$r = \dfrac{2S}{P} = \dfrac{2\cdot \dfrac{1}{2}\cdot 1\cdot 1\cdot \sin A}{1+1+2\sin A} = \dfrac{1}{2}\cdot \dfrac{\sin A}{1+\sin A} = f(A)$.
Taking derivative of $f$ w.r.t $A$ we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Large round brackets in equation I cannot recall what the large () brackets mean - Google seems to be full of links on how to create them but not what they actually are or how to resolve them.
$
\left(
\begin{array}{l}
n\\
2
\end{array}
\right)r^2
$
Can someone please clarify? In this instance n = 6 and r = 0.05 if tha... | The binomial coefficient
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
is the number of ways of selecting $k$ elements from $n$ elements when order does not matter (the number of subsets with $k$ elements in an $n$ element set).
The number $n!$, read "$n$ factorial," is defined recursively as follows:
*
*$1! = 1$
*$n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple first order differential equation I've got another first order differential that has me stumped. Please see my work and let me know where I'm going wrong:
$$y' = \frac{6x^2}{y(1+x^3)}$$
$$y'y = \frac{6x^2}{1+x^3}$$
$$ y.\frac{dy}{ dx} = \frac{6x^2}{1+x^3} $$
$$\int y dy = \int 6x^2\frac{1}{1+x^3} dx$$
$$\frac{1... | Your error is that $\displaystyle\int\dfrac{6x^2}{1+x^3}\,dx = 2\ln|1+x^3|+C$ and not $2x^3\ln|1+x^3|+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
| Looks like you will need to use the quotient rule and the chain rule to solve this. The quotient rule tells us that $$\left[\frac{1+4\cos x}{2 \sqrt{x+4\sin x}}\right]' = \frac{[1+4\cos(x)]'2 \sqrt{x+4\sin x}-\left(1+4\cos(x)\right)\left[2 \sqrt{x+4\sin(x)}\right]'}{\left(2 \sqrt{x+4\sin(x)}\right)^2}$$ and to evaluate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How find this limits with hardly form?
show that:
$$\lim_{n\to\infty}n\left[\left(\dfrac{1}{\pi}\left(\sin{\left(\dfrac{\pi}{\sqrt{n^2+1}}\right)}+
\sin{\left(\dfrac{\pi}{\sqrt{n^2+2}}\right)}+\cdots+\sin{\left(\dfrac{\pi}{\sqrt{n^2+n}}\right)}
\right)\right)^n-\dfrac{1}{\sqrt[4]{e}}\right]=-\dfrac{1}{\sqrt[4]{e}}\lef... | You seem to be on the right track and I guess we can try to get more reasonable estimates using further terms of the expansion. First we need to use $\sin x \approx x - \dfrac{x^{3}}{6}$ to get $$\begin{aligned}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(1 + \frac{i}{n^{2}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how do you solve $a^2+b^2+c^2=d^3$ let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$
| For the equation:
$$x^2+y^2+z^2=r^3$$
Will make a replacement that formula was compact.
$$c=2(q-p-s)t$$
$$d=s^2+t^2-q^2-p^2+2p(q-s)$$
$$k=p^2+t^2-q^2-s^2+2s(q-p)$$
$$n=p^2+t^2+s^2-q^2$$
$$j=p^2+s^2+t^2+q^2-2q(p+s)$$
$p,s,t,q$ - integers asked us. Then decisions can be recorded.
$$x=dn^2+2cnj-dj^2$$
$$y=cj^2+2dnj-cn^2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$ What tools would you recommend me for computing the limit below?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$
As soon as any use... | After I posted my earlier answer, I realized that this can be handled in a much simpler way.
Pick an $\epsilon\gt0$.
Most of the series is contained in a finite sum
This identity is proven in my earlier answer:
$$
\sum_{k=1}^\infty\frac1{(2k-1)^2+\pi^2}=\frac14\tanh(\pi^2/2)\tag{1}
$$
Since the series in $(1)$ converg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
Show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$ when $x>1$
If $x>1$ show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$
I know from binomial expansion that $(1+x)$ will produce a divergent series in the form of $1-x+x^2-x^3+\cdots$ but ... | Let
$$\begin{align}
f(x)&=\ln(1+x)-\ln x\\
&=\ln\left(1+\frac{1}{x}\right)
\end{align}
$$
By differenting w.r.t. $x$ and writing as sum of infinite GP
$$\begin{align}f'(x)&=\frac{1}{1+\frac1x}\cdot\left(\frac{-1}{x^2}\right)\\
&=-\frac{1}{x^2}\left(1-\frac1x+\frac{1}{x^2}-\cdots\right)\\
f'(x)&=-\frac{1}{x^2}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subtracting Quarters of Squares Equals Multiply?! Can anyone explain to me how/why this works (hopefully in mostly layman's terms)?
It seems pretty magical to me at the moment.
$${{(a+b)^2\over4} - {(a-b)^2\over4}} = a b.$$
| Expanding the squared terms gives
\begin{equation}
\frac{(a + b)^2}{4} - \frac{(a - b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} - \frac{a^2 - 2ab + b^2}{4} = \frac{4ab}{4} = ab.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 3
} |
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.