Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Looking for a cleaner/quicker way to evaluate the integral of a quadratic Looking for a clean/quick way to evaluate
$$\int_{2-\sqrt{3}}^{2+\sqrt{3}}{ \left(x^2-4x+1\right)\textrm{d}x}=\left.\frac{x^3}{3}-2x^2+x \right|_{2-\sqrt{3}}^{2+\sqrt{3}}$$
So I evaluated all of this out leaving this expression as it was and ther... | Notice that $x^2 - 4x + 1 = (x - 2)^2 - 3$, so a substitution $y = x - 2$ yields
\begin{align*}
\int_{2-\sqrt{3}}^{2+\sqrt{3}} (x^2-4x+1)dx = \int_{-\sqrt{3}}^{\sqrt{3}}(y^2-3)dy = \left[\frac{y^3}{3} - 3y\right]_{-\sqrt{3}}^{\sqrt{3}} = 2\sqrt{3}-6\sqrt{3} = -4\sqrt{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I ... | Substitute $$x = \sin \theta$$ The expression is $$x + \sqrt{-x^2 + 1} = \sin \theta + |\cos \theta|$$
We want to find the max. value of $\sin \theta + |\cos \theta|$. Now, two cases: $\cos \theta < 0$ and $\cos \theta \geq 0$.
$$\textbf{Case 1:} \cos \theta < 0$$
We need to find the max. value for $\sin \theta - \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is this interval notation for the solution of an inequality problem correct? Is this notationally correct?
$$
\begin{align*}
5 - x^2 &< -2, \\[1ex]
5 - x^2 + (-5) &< -2 + (-5), \\[1ex]
-x^2 &< -7, \\[1ex]
(-x^2)(-1) &> (-7)(-1), \\[1ex]
... | I will write the final part in question as $x\in (-\infty,-\sqrt7)\cup (\sqrt 7,\infty)$ because the inequality is in $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3... | With a lot of these problems involving symmetric sums of the roots of a polynomial, there are two ways to do it. There is a long and tedious way using Vieta’s formulas and Newton sums, and a “clever” way that just involves evaluating the polynomial a couple times. I will go over the clever way for this problem, and I’l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculate $\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N$ How to calculate the limit
$$\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N?$$
By using the numerical method with Python, I guess the right answer is $-1$ but how to prove? I have no idea.
| Let $N\geq 1$. Note that
$$
\frac{{\pi ^2 }}{6} - \sum\limits_{n = 1}^N {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n(n - 1)}}} - \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} = \frac{1}{N} - \sum\limits_{n = N + 1}^\infty {\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^... | Denote $$\Delta_m(x, y, z) := (x + y + z)^m - x^m - y^m - z^m .$$
For odd $m$, $\Delta_m(x, y, y) = 0$ so $y + z$ divides $\Delta_m(x, y, z)$; by symmetry so do $z + x$ and $x + y$. Hence, $$(x + y + z)^m = x^m + y^m + z^m + (y + z)(z + x)(x + y) q_m(x, y, z)$$ for some symmetric, homogeneous polynomial $q_m$ of (even)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving by induction that $\frac{1}{n} \ge \frac{n!}{n^n}$ I have been trying to prove by induction that for all $n \ge 0, \frac{1}{n} \ge \frac{n!}{n^n} $
Here is my proof so far:
Prove that for $ n = 1, \frac{1}{n} \ge \frac{n!}{n^n} $
$ \frac{1}{1} = 1 $
$ \frac{1!}{1^1} = 1 \le 1 $
Next, assume there exists $ n \... | Just note that $$\frac{1}{n}\ge \frac{n!}{n^n} \iff n^n\ge n! \cdot n$$
$$\iff n^{n-1}\ge n!$$
Now assume that for some $n\ge 1$ the above proposition holds
$$(n+1)^n\ge (n+1)!$$
$$\iff (n+1)^{n-1}\ge n!$$
But this is trivial since $(n+1)^{n-1}\ge n^{n-1}\ge n!$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality cas... | Partial Hint :
For $x,y>0$ and $x,y\leq 1$ such that $\left(y\right)^{2}+\left(x-\frac{1}{5}\right)^{2}\geq\frac{2}{50}$ we have the inequality :
$$\sqrt{\frac{x^{2}}{y^{2}+x^{2}}}-\left(f\left(x\right)-f\left(y\right)\right)-\frac{3}{4}\le0$$
Where :
$$f\left(x\right)=x\sqrt{\frac{1}{x^{2}+1}}$$
Summing gives the ineq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove that the equation has solutions in two different intervals [-1,1] and [1,2] $a,b,c \in \mathbb{R}$ and $a,b,c$ positive numbers
$\frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=1$
Show that the equation has a solution in the interval $[-1,1]$ and a solution in the interval $[1,2]$.
$\frac{(a+b)\cdot x+a-b}{x^2-1}+\... | Hint: $$\lim_{x \to -1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
$$\lim_{x \to 1-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$
$$\lim_{x \to 1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
and
$$\lim_{x \to 2-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$.
By continuity, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A number theory question that is probably wrong Prove that $a^3-b^3 = 2011$ has no integer solutions.
I think the question is wrong as
$a^3-b^3 = 2011$
$(a-b)(a^2+ab+b^2) = 2011$
As $2011$ is prime so the only factors $2011$ has are $1$ and $2011$ itself.
So if $a-b = 1$ and $a^2+ab+b^2 = 2011$ , then we can say that
$... | Well, you seem to be doing a very good job up until when you realize that
one must have $(a-b)(a^2+ab+b^2)= 2011$.
You then come to the key observation $a-b$ must be a divisor of $2011$.
At this point I would consider the problem "almost solved", it seems like now we must only do some case checking and everything will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$? This question came in the Dhaka University admission exam 2006-7
Q) The value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$ is -
(a) $0$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $2\pi$
My attempt:
$$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$$
$$=\tan^{-1}\frac{1+2+3-1\cdot2\cdot3}{1-1\cdot2-2\cdot... | There's a bit of a discrepancy between the page title and your actual question.
Conventionally, $\tan^{-1}(0)$ is defined as 0.
However, the individual terms $\tan^{-1} 1$, $\tan^{-1} 2$, and $\tan^{-1} 3$ are all positive numbers, so their sum can't be zero.
*
*$\tan^{-1} 1 = \frac{\pi}{4}$
*$\tan^{-1} 2$ and $\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluation of $\int\sqrt{1+2\sin(\theta)^2\cos(\theta)^2}\mathrm{d}\theta~~\text{where}~~\theta\in\left[0,{\pi\over4}\right]$ $$\begin{align}
I&:=\int \sqrt{1+2\sin(\theta)^2\cos(\theta)^2} \mathrm{d} \theta ~~~\text{where}~~~~\theta\in\left[0,{\pi\over4}\right]\\
\cos(\theta)^2&=1-\sin(\theta)^2\\
I&=\int\sqrt{1+2\sin... | $$I=\int_0^{\frac \pi 4}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\theta=\frac{1}{2}\int_0^{\frac \pi 2} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\frac{1}{2}E\left(-\frac{1}{2}\right)$$ where appears the complete elliptic integral of the second kind.
If it had been
$$J=\int_0^{t}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluate $\int \frac {x+3}{\sqrt{5-4x-2x^2}} dx$ Question:
The question is about evaluating the following indefinite integral:-
$$\int \frac {x+3}{\sqrt{5-4x-2x^2}}dx$$
My Work:
We know the numerator can be expressed in the form of :
$$ x+3 = A \frac {d}{dx}(5-4x-2x^2) + B$$
So we get ,
$$ x+3 = A(-4-4x) +B $$
Equating... | Another method using Completing the square and Integration by substitution.
We have,
$$\begin{align}I &= \int \dfrac{x + 3}{\sqrt{5 - 4x - 2x^2}} dx\\& = \int \frac{x+3}{\sqrt{5 - 2(x^2 +2x)}}dx\\& = \int \frac{x+3}{\sqrt{5 - 2[(x + 1)^2 - 1]}}dx\\& = \int \frac{x+3}{\sqrt{7 - 2(x + 1)^2}}dx\end{align}$$
Let $(x+1) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
Aft... | Hint:
$$A^5(A^3-5A^2+7A-3I)+A(A^3-5A^2+7A-3I)+A^2+A+I=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume of $y=x^2+1; y=-x^2+2x+5; x=0; x=3$ about $x$ axis (Shell Method). I was working on this exercise for an assignment. However, I get stuck in the following part.
$
y=-x^2+2x+5
$
Complete the square
$y=-(x^2-2x)+5$
$(b/2)^2=(-2/2)^2=1$
$y=-(x^2-2x+1-1)+5$
$y=-(x^2-2x+1)+5+1$
$y=-(x-1)^2+6$
$y-6=-(x-1)^2$
... | Note that in $[0,3]$, $-x^2+2x+5\geq x^2+1$ for $x\in [0,2]$. Therefore, by the Washer Method, the volume is given by
$$V=\pi\int_0^2((-x^2+2x+5)^2-(x^2+1)^2)\,dx+\pi\int_2^3((x^2+1)^2-(-x^2+2x+5)^2)\,dx=\frac{277\pi}{3}$$
Using the Shell Method is a bit more complicated because we have to split the evaluation into 4 p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \ma... | Because the integral is indefinite.
If you differentiate the first form,
$${d2\arcsin{(\sqrt{x})} \over dx}=2{1 \over \sqrt{1-(\sqrt{x})^2}}\cdot {1 \over 2\sqrt{x}}={1 \over \sqrt{x(1-x)}}$$
As for the second one,
$${d\arcsin{(2x-1)} \over dx}={1 \over \sqrt{1-(2x-1)^2}}\cdot 2={1 \over \sqrt{x(1-x)}}$$
Therefore, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$
Attempt:
Notice that $a=b=c=d=1$ is a solution.
Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negati... | This isn't a full answer, but noting that $$2a+2b+2c=6d$$ first subtract this from the second equation to obtain $$(a-1)^2+(b-1)^2+(c-1)^2=4(d-1)^2$$
Then set $A=a-1, B=b-1, C=c-1, D=d-1$ to get the system$$A^2+B^2+C^2=4D^2$$
and $$A+B+C=3D$$
So for given $D$ the solutions lie on a sphere of radius $2D$ centred at the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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To show $\int_{0}^{\infty} \frac{x}{1+x^6\sin^2x}\,dx$ is convergent
Show that $\displaystyle \int\limits_{0}^{\infty} \frac{x}{1+x^6\sin^2x}\,dx$ is convergent
Since we need to check if the given improper integral is convergent, I considered Abels/Dirichlet test for convergence, but for that, although I got monotone... | You can proceed as in this answer:
Note that $\left| \sin x \right| \geq \frac{2}{\pi}|x|$ for $|x| \leq \frac{\pi}{2} $. Using this, we get:
$$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a \sin^2 x}
\leq \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a\bigl(\frac{2}{\pi}x\bigr)^2}
\leq \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating integral $\int_{-\infty}^\infty \frac{x}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$ and one more similar I'm trying to evaluate two integrals:
$$f(m,c) = \int_{-\infty}^\infty \frac{x}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$
and
$$g(m,c) = \int_{-\infty}^\infty \frac{1}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$
where $c,m\in\... | If $m$ is "small" you could expand the exponential term as
$$e^{-(x-m)^2}=e^{-x^2}\,\sum_{n=0}^\infty P_n(x)\, m^n$$ where the first polynomials are
$$\left(
\begin{array}{cc}
n & P_n(x) \\
0 & 1 \\
1 & 2 x \\
2 & 2 x^2-1 \\
3 & \frac{4 x^3}{3}-2 x \\
4 & \frac{2 x^4}{3}-2 x^2+\frac{1}{2} \\
5 & \frac{4 x^5}{15}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to generalize this equation more? This is one of my discoveries, and my question is: can it be more general?
Let $x, y, z$ be three arbitrary complex numbers and set
$x_1=3 x^2+y^2$,
$x_2=1-z$,
$x_3=1+z$,
$x_4=3 x^2+y^2+4 x y z$,
$x_5=3 x^2+y^2-4 x y z$,
$x_6=3 x^2+y^2+z (3 x^2+2 x y-y^2)$,
$x_7=3 x^2+y^... | This is very nice. It relies to the Tarry-Escott problem, on which there is a vast literature going back more than 100 years. I do not know how much you know of this. I suggest you consult Dickson, the History of the Theory of Numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tr... | From the recursive definition, we have the identity
$$\sum_{n=5}^\infty f(n) x^n = \sum_{n=5}^\infty \frac{n-4}n f(n-1) x^n + \sum_{n=5}^\infty x^n$$
With $G(x) = \sum\limits_{n=4}^\infty f(n)x^n$, for $|x|<1$ we have
$$\begin{align*}
G(x) - x^4 &= \sum_{n=5}^\infty f(n-1) x^n - 4 \sum_{n=5}^\infty \frac{f(n-1)}n x^n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum value of $M$ with given constraints.
Let $a, b \in \mathbb{R}$ such that $n^{4a-\log_5{n^2}} \leq 25^{40-b^2}$ for all positive real number $n$. Find the maximum value of $M=a^2+b^2+a-3b$.
What I have tried:
Taking the logarithm base 5 of both sides from the inequality, we get $(4a-2\log_5{n})\log_5n... | I want to contribute my solution using geometry :D
As Cesareo told above, our problem now is to find the maximum value of $f(a,b)=a^2+b^2+a-3b$ knowing that $a^2+b^2 \leq 40$.
Geometrically, $\sqrt{a^2+b^2}$ is the distance between the point $M(a,b)$ and the origin $O(0,0)$, so the condition $a^2+b^2 \leq 40$ means tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$
Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?
I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cd... | It is a telescoping product. Everything on the RHS cancels in pairs, except the $1-x$ and $1-x^2$, which appear in the denominator but not the numerator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Conditional Variance of PDF A random vector $(X,Y)$ has a continuous distribution with a density function $$f(x,y)=\begin{cases}c⋅x & \text{when }0 ≤ x ≤ 2, \max\{0,1−x\} ≤ y ≤2−x\\ 0& \text{otherwise}\end{cases}$$ where $c > 0$ is a constant. Find variance of a $Y$ conditioned on $X = 1.5$, $Var(Y |X = 1.5)$.
I found ... | You have the region being the area trapped between the triangles $B$ and $A$ . Let it be denoted by $S$ .
You have $$\iint_{S}f(x,y)\,dxdy=1$$
$$\iint_{S}f(x,y)\,dxdy=\iint_{B}f(x,y)\,dxdy-\iint_{A}f(x,y)\,dxdy=\frac{8c}{6}-\frac{c}{6}=\frac{7c}{6}$$.
Hence $c=\frac{6}{7}$ as you correctly calculated.
Now the marginal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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For a triangle, prove that $\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2} $
Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contes... | Note that $AE=s-a, BE=s-b, BG=s-b, GC=s-c, CF=s-c, FA=s-a$. The expression under the square root becomes
$$\sqrt{\frac{1}{(s-a)(s-b)}+\frac{1}{(s-b)(s-c)}+\frac{1}{(s-a)(s-c)}}$$
By Heron's formula, $K=\sqrt{s(s-a)(s-b)(s-c)}\implies \frac{1}{(s-b)(s-c)}=\frac{s(s-a)}{K^2}$, etc. and so the expression under square root... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff ... | An easier proof for me is the following:
Firstly, $\frac a{1+b}+\frac b{1+a}\geq \frac{b}{1+a}>0,\ $ as numerator and denominator of $\frac{b}{1+a}$ are both positive.
Secondly,
$$\frac a{1+b}+\frac b{1+a} = \frac{a(1+a)+b(1+b)}{(1+b)(1+a)} = \frac{a+b+a^2+b^2}{a+b+ab+1}. $$
So if we show that $ab+1 \geq a^2 + b^2, $ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the general solutions of the PDEs below...
a)$3u_x-4u_y = x^2$
b)$u_x - 4u_y + u = x + y + 1$
I'm starting in PDEs, I saw two resolution methods, but I think I didn't understand them right.
a) $\frac{dx}{3}= \frac{dy}{-4}=\frac{dz}{x^2}$
$\frac{dx}{3}= \frac{dy}{-4} \to -4x + c = 3y \to c_1= 4x+3y$
$\frac{dy}{... | a)
$$3u_x-4u_y=x^2$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{3}=\frac{dy}{-4}=\frac{du}{x^2}$$
A first characteristic equation comes from solving $\frac{dx}{3}=\frac{dy}{-4}$ :
$$4x+3y=c_1$$
A second characteristic equation comes from solving $\frac{dx}{3}=\frac{du}{x^2}$ :
$$u-x^3=c_2$$
The solution of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $... | Define:
$$u^4=8x^4,~v^4=27y^4,~w=64z^4$$
$$f=u^4+v^4+w^4$$
Constraint becomes: $$8^{-1/4}u+27^{-1/4}v+64^{-1/4}w=13/4$$
So we have:
$$a=8^{-1/4},~~b=27^{-1/4},~~c=64^{-1/4}$$
Now you can use C-S inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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$\binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$ There is a "direct" way to prove this equality.
$\displaystyle \binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$
I am trying to skip the induction. Maybe there is a rule or formula that will help me.
Thank you
| another way
Let's rewrite part of the RHS in terms of gamma
$$
\frac{1}{n}\left( \begin{array}{c}
2n - 2 \\ n - 1 \\ \end{array} \right)
= \frac{{\Gamma \left( {2n - 1} \right)}}
{{n\Gamma \left( n \right)\Gamma \left( n \right)}}
= \frac{{\Gamma \left( {2n - 1} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the definite integral of $\frac{\log x}{1+ x^3}$ The problem is apparently simple:
Find $\displaystyle{\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x }$.
However, I have not been able to get around it. The usual "big circle" strategy is useless since we don't have parity. With substitution I was able to... | Continue with
\begin{align}
\int_0^\infty&\frac{\log x}{1+x^3}{d}x
=
\int_0^1\frac{(1-x)\log x}{1+x^3}{d}x \\
&=\int_0^1\frac{\log x}{1+x}
- \underset{x^3\to x}{\frac{x^2\log x}{1+x^3}}\ {dx}
= \frac89\int_0^1\frac{\log x}{1+x}{d}x
=-\frac{2\pi^2}{27}
\end{align}
$\int_0^1\frac{\log x}{1+x}{d}x=-\frac{\pi^2}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)?
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$
Hello!
I was solving this problem, and here is my approach:
Inside the square root $x^2$ overpowers $2x$ as $x \to... | Let us try to eliminate or cancel the square term & the root term :
$\lim_{x \to \infty} {\sqrt{x^2 + 2x} - x} = \lim_{x \to \infty} {\sqrt{x^2 + 2x + 1 -1} - x} = \lim_{x \to \infty} {\sqrt{(x + 1)^2 -1} - x}$
We can now see that the square term will Over-Power the Constant.
We then get a way to eliminate or cancel th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$...
Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$ classify the singularity $ z=a.$
a)$f(z)=\frac{z^2+1}{z(z+1)}, a=0.$
b)$f(z)=\... | You don't need that formula.
a) For each $z\in\Bbb C$ with $|z|<1$, you have\begin{align}\frac{z^2+1}{z+1}&=z-1+\frac2{z+1}\\&=z-1+2\sum_{n=0}^\infty(-1)^nz^n\\&=1-z+2\sum_{n=2}^\infty(-1)^nz^n.\end{align}Therefore, if $z\ne0$,$$\frac{z^2+1}{z(z+1)}=\frac1z-1+2\sum_{n=1}^\infty(-1)^{n+1}z^n.$$So, $0$ is a simple pole.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$ From R. D. Sharma's Objective Mathematics,
Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$
My attempt:
By A.M.-G.M. inequality,
$$\begin{align}
(x+y)^2... | Note that,
If $x,y\to 0$ with $z\to 1$, we have
$$\inf \left\{xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\mid x+y+z=1 \wedge x>0\wedge y>0\wedge z>0\right\}=0$$
As for obtaining the inequality $A≥4xyz$, you can easily obtain this result using the Cauchy-Schwarz inequality:
$$(xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2)\left(\frac 1{xy}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests... | $$\frac{1}{1+t^2}= 1 - \frac{t^2}{1+t^2}$$ implies
$$\frac{1}{1+t^2}=1-t^2\left(1 - \frac{t^2}{1+t^2}\right)= 1 -t^2 + \frac{t^4}{1+t^2}$$ and continue in this manner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43... | Problem: Let $x_i \ge 0, i= 1, 2, \cdots, n$. Prove that
$$\sum_{i=1}^n \frac{x_i}{S - x_i} \ge \frac{n}{n - 1}$$
where $S = \sum_{j=1}^n x_j$.
Proof.
Since the desired inequality is homogeneous, assume that $\sum_{j=1}^n x_j = 1$. We need to prove that
$$\sum_{i=1}^n \frac{x_i}{1 - x_i} \ge \frac{n}{n - 1}.$$
We have
... | {
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"source": "stackexchange",
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"answer_count": 6,
"answer_id": 5
} |
Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$
$\color{red}{\textbf{Problem:}}$
Let, $x_i>0,1\le i\le n$, then
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$
$\color{red}{\textbf{Proof:}}$
Using AM-GM inequality we have,
\begin{alig... | OP's work is correct however $x_j>0$ needs to be mentioned. Here is another approach:
Let $s=\sum_{j=1}^{n} x_j,$ then
$$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad 0<x_j<s.$$
So by Jensen's inequality
$$\frac{1}{n}\sum_{j=1}^{n} f(x_j) \ge f(\frac{s}{n}) \implies \sum_{j=1}^{n} \frac{x_j}{s-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculating eigenvalues of a matrix I have this quadratic matrix:
\begin{align*}
A = \begin{bmatrix}
0 & -1 & -2\\
-1 & 0 & -2\\
-2 & -2 & -3
\end{bmatrix} \implies A - \lambda I = \begin{bmatrix}
0 - \lambda & -1 & -2\\
-1 & 0 - \lambda & -2\\
-2 & -2 & -3 - \lambda
\end{bmatrix}
\end{align*}
After forming the charact... |
$$p(\lambda)=-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$
Proceed from your work,
$$\begin{align}
p(\lambda)&=-\lambda^3-5\lambda^2+2\lambda^2+9\lambda-5\\
\\
&=-\lambda^2(\lambda+5)+(2\lambda-1)(\lambda+5)\\
\\
&=(\lambda+5)(-\lambda^2+2\lambda-1)\\
\\
&=-(\lambda+5)(\lambda^2-2\lambda+1)\\
\\
&=-(\lambda+5)(\lambda-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
| You can write:
$$\sqrt{x^2+p^2}-p=p\cdot\left(\sqrt{1+\frac{x^2}{p^2}}-1\right)$$
And:
$$\sqrt{x^2+q^2}-q=q\cdot\left(\sqrt{1+\frac{x^2}{q^2}}-1\right)$$
We know that $(1+t)^\alpha-1 \,\, \sim\,\, \alpha \cdot t$ when $x\to 0$; notice that in our cases $t=\frac{x^2}{p^2}$ for the numerator and $t =\frac{x^2}{q^2}$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ Question:
Let $x<0$ then $\cos^{-1}\sqrt{1-x^2}=$
(1) $\pi-\cos^{-1}x$
(2) $-\sin^{-1}x$
(3) $\pi-\sin^{-1}x$
(4) $\sin^{-1}x$
My approach is as follow
Given $x < 0,{\cos ^{ - 1}}\sqrt {1 - {x^2}} $
Let $x = \sin \theta $, therefore ${\cos ^{ - 1}}\sqrt {1 - {{\... | Because $x=\sin \theta <0$ and $\cos \theta = \sqrt{1-x^2} >0$, angle $\theta$ is in the fourth quadrant. Thus, $\cos^{-1} |\cos \theta|=-\theta$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{({x}^{2} + {b}^{2}) {({x}^{2} + {a}^{2})}^{2}} = \frac{\pi (2 a + b)}{2 {a}^{3} b {(a + b)}^{2}}$ I am in chapter 6 of Whittaker and Watson, and I am currently confused attempting to find the mistake I made in computing the following integral. The problem asks me to pr... | Actually,\begin{align}\lim_{z\to ai}\frac{\mathrm d}{\mathrm dz}\frac1{(z^2+b^2)(z+ai)^2}&=\lim_{z\to ai}\frac{-2z(z+ai)^2-2(z^2+b^2)(z+ai)}{(z^2+b^2)^2(z+ai)^{\color{red}4}}\\&=\frac{i \left(3 a^2-b^2\right)}{4 a^3 (a-b)^2 (a+b)^2}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A deceptively difficult linear recurrence problem Consider the sequence $b_0 = 0$, $b_1 = 1$, $b_{n+2} = 2b_{n+1} - 3b_n$ for all $n\geq 0$. Prove that the only positive integers $m$ with $b_m = \pm 1$ are $m = 1$ and $m = 3$. (In fact, I expect that $b_m = -1$ never occurs but not proving this is not an issue for the ... | We will work in the ring
$$ R=\Bbb Z[a]\ ,\qquad\text{ where } a=\sqrt{-2}\ .
$$
Then the characteristic equation of the given linear recursion is
$\lambda^2 -2\lambda +3=0$, it has the roots $(1+a)$ and $(1-a)$, both in $R$.
The general formula for $b_n$ is then:
$$
b_n =\frac 1{2a}\Big[\ (1+a)^n -(1-a)^n \ \Big]\ ,
$... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ..... | Yes, this is correct; there's nothing of note to mention otherwise that I can think of.
I'm choosing to answer like this and mark it as Community Wiki since I have nothing further to add, but don't want the question to end up in the unanswered queue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Find the intervals to which $p$ belong for these $4$ conditions
Suppose there is a quadratic equation $$f(y)=y^2-(p+1)y+p^2+p-8=0$$ then find the intervals in which $p$ should belong to fulfill the following conditions
$1.$ both the roots are less than $2$
$2.$ one root is greater than $2$ and the other one is greate... | Part 3.
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{(p+1)^2 - 4(p^2 + p - 8)} ~\right] \implies $$
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{(p^2 + 2p + 1) - 4(p^2 + p - 8)} ~\right] \implies $$
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{-3p^2 -2p + 33} ~\right].$$
As the OP (i.e. original poster) has already determined,
$$-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Integration $\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$ Here is the integral:
$$\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$$
This is an elliptic integral, with such an easy result, maybe some clever substitutions or integrating methods can solve it, but so ... | I would like to solve it directly.
For $|t|$ large enough, let
$$
\tag{1}J\left(t\right)=\int_{0}^{\infty}\frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x/t^2)} }\text{d}x.
$$
Firstly, apply the substitution $y=xt$, and
$$
(1)={t^{2/3}}\int_{0}^{\infty}\frac{2y-t}{y^{2/3}\sqrt{(y+t)((y+t)^3-y)} }\text{d}y.
$$
Secondly, let $z=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How to derive the polar equation of a circle.
I have learnt that the standard equation of a circle whose centre is at (a,b) is
$$(x-a)^2 + (y-b)^2 = c^2
$$
I am trying to derive the polar equation of this circle but I am unfortunately stuck.
$$\begin{align}(x-a)^2+(y-b)^2&=c^2\\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=c^... | You got off to a good start: Just plug in $x = r \cos\theta$ and $y = r \sin\theta$ in the standard rectangular-coordinate equation $(x-a)^2 + (y-b)^2 = c^2$
$$(r \cos\theta-a)^2 + (r \sin\theta-b)^2 = c^2$$
Or, if you need the equation expressed as a polynomial of $r$,
$$r^2 \cos^2\theta - 2ar\cos\theta + a^2 + r^2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Determinant of block matrices. $$
X = \begin{pmatrix}
1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\
1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\
b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\
b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\
b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\
b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6}
\end{pm... | The blocks in the Schur complement need not be contiguous. In your case, the diagonal blocks corresponds to the submatrices
$$\begin{aligned}
A &=\begin{bmatrix}X_{11} & X_{1n} \\ X_{n1} & X_{nn}\end{bmatrix}
=\begin{bmatrix}1+b_1 & \frac{1}{a_6} \\ b_6& 1 - \frac{a_5}{a_6}\end{bmatrix}
\\
D&=\begin{bmatrix}X_{2,2} & ⋯... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\li... | HINT
I would start with noticing that:
\begin{align*}
\lim_{x\to\infty}(\sqrt{4x^{2} + 2x + 1} - (ax + b)) & = \lim_{x\to\infty}\frac{(4x^{2} + 2x + 1) - (ax + b)^{2}}{\sqrt{4x^{2} + 2x + 1} + (ax + b)}\\\\
& = \lim_{x\to\infty}\frac{(4 - a^{2})x^{2} + (2 - 2ab)x + 1 - b^{2}}{\sqrt{4x^{2} + 2x + 1} + ax + b}\\\\
& = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
What happens with indexes in this proof? I am trying to understand a proof we did in our discrete math class.
Theorem: For $n \geq 0,$ $$x^{\overline{n}} = \sum_k s(n,k) x^k,$$
where $x^\overline{n} = x (x+1)\cdots (x+n-1)$ and $s(n,k)$ is the Stirling number of the first kind.
Proof:
We prove the theorem by induction.... | We obtain using the recurrence relation
\begin{align*}
\begin{bmatrix}n\\k\end{bmatrix}&=\begin{bmatrix}n-1\\k-1\end{bmatrix}+(n-1)\begin{bmatrix}n-1\\k\end{bmatrix}\qquad\qquad n\geq k>0\tag{1}\\
\begin{bmatrix}0\\0\end{bmatrix}&=1\qquad\begin{bmatrix}n\\0\end{bmatrix}=\begin{bmatrix}0\\n\end{bmatrix}=0\qquad\qquad\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving that $x^n=a$, for $n>0$ an odd natural number, has exactly one real root In my school book, I read this theorem
Let $n>0$ is an odd natural number (or an odd positive integer), then the equation $$x^n=a$$ has exactly one real root.
But, the book doesn't provide a proof, only tells $x=\sqrt [n]a$.
How can I pr... | If $n$ is even then $x^n=a$ has two real roots, $x=\pm\sqrt[n]{a}$, since $x^n=\left(x^{(n/2)}\right)^2$ is always positive, but solutions are restricted to $a\geq0$. If $n$ is odd then $x^n=a$ does not have the negative root $x=-\sqrt[n]{a}$ since $(-\sqrt[n]{a})^n=(-1)^n (\sqrt[n]{a})^n=-a\neq a$. It may also have so... | {
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"url": "https://math.stackexchange.com/questions/4531652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible ... | The trick here is that the coefficients are palindromic. Let $$f(x)=x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1,$$ and let $$g(x)=f(x)/x^3=(x^3+1/x^3) - (x^2+1/x^2) +4(x+1/x)-4.$$ We can rewrite this in terms of $w=x+1/x$. We note that $w^2=(x+1/x)^2=x^2+1/x^2+2$, and $w^3=x^3+1/x^3 + 3(x+1/x)$, and so $x^2+1/x^2=w^2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531994",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
A Gaussian elimination problem on Matrices Considering these linear equations :-
$-3x_1 - 3x_2 + 9x_3 = 12$
$2x_1 + 2x_2 - 4x_3 = -2$
$ -2x_2 - 4x_3 = -8$
Now to solve this problem I used the Gaussian elimination method.
I converted into an augmented matrix of the same
$$ \begin{bmatrix} -3 & -3 & 9 & 12 \\ 2 & 2 & -4 ... | You went wrong in the second step (where you should also write $[R_{2}-2R_{1}\to R_{2}]$) and again in the last step.
With problems like this, you may want to consider leaving it be for a moment and coming back some time later. That makes it easier to spot your own mistakes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substitut... |
We might also think of the first two equations as describing the intersections of a line with intercepts $ \ (0 \ , \ 1) \ \ $ and $ \ (1 \ , \ 0 ) \ $ and the circle of radius $ \ \sqrt2 \ $ centered on the origin. These intersections are symmetric about the line $ \ y \ = \ x \ \ . $
Since the axis intercepts are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x... | We'll assume $x_1>0,$ since the left hand side is undefined if $x_1=0.$ But if ww treat the first term as zero when $x_1=0,$ we can just use that the right side is decreasing on $n.$
Try induction on $n.$ When $n=1,$ you have $x_1=1,$ the inequality becomes $1\geq 1.$
If true for $n,$ take $x_1,\dots,x_{n+1}.$
Since $x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How to prove you found ALL limit points: $x_n = (-1)^n + 1/n + 2\sin(\frac{n\pi}{2})$ How does one find all limit points? How do you prove you're not missing a limit point? For example, if $x_1,x_2,\cdots$ converges to $x$ and $y_1,y_2,\cdots$ converges to $y$, and we consider the shuffled sequence, obvious limit point... | We have that, starting from $n=4k$
*
*$x_{4k} = (-1)^{4k} + \frac1{4k} + 2\sin\left(\frac{4k\pi}{2}\right)=1+\frac1{4k}+0=1+\frac1{4k}$
*$x_{4k+1} = (-1)^{4k+1} + \frac1{4k+1} + 2\sin\left(\frac{4k\pi}{2}+\frac \pi 2\right)=-1+\frac1{4k+1}+2=1+\frac1{4k+1}$
*$x_{4k+2} = (-1)^{4k+2} + \frac1{4k+2} + 2\sin\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all real values of x and y such that $\left(x+yi\right)^3$ is real and greater than $8$, represent these values in the xy-plane My solution
$\left(x+yi\right)^3>8$
$\left(x^3-3xy^2-8\right)+\left(3x^2y-y^3\right)i>0$
and now I'm stuck with these equations
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
| We need the cube is real then
$$3x^2y-y^3 =y(3x^2-y^2)=0$$
(and not $3x^2y-y^3>0$) then we have three cases
*
*$y=0 \land x\neq 0 \implies x^3-8>0 \implies \begin{cases}x>2\\y=0 \end{cases}$
*$y\neq 0 \land 3x^2=y^2\neq 0 \implies x^3-9x^3-8>0 \implies \begin{cases}x<-1\\y=\pm \sqrt 3 x \end{cases}$
*$y=0 \land x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Pairs $(x,y)$ such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$. How many possible pairs of integers $x$ and $y$ exist such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$?
Since $y| 3x-1$, there exists some $k$ such that $3x-1 = ky$.
On the other hand since $x| 3y-1$ we have $3y-1 = l... | Multiplying the two divisibility relations together gives
$$xy\mid(3x-1)(3y-1)=9xy-3(x+y)+1\implies xy\mid3(x+y)-1$$
Now neither $x$ nor $y$ can be zero, since the original relations would then force $3x-1=3y-1=0$ which is impossible for integral $x,y$, so the quotient $\frac{3(x+y)-1}{xy}$ exists and is an integer wit... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact ... | Alternative approach:
Start out differently: Let
$$A = \frac{1-x}{1+x}, ~B = A^{(1/2)}.$$
Then
$$\frac{d [B]}{dx} = \frac{1}{2} \times {A}^{-(1/2)} \times \frac{dA}{dx}. \tag1 $$
and
$$\frac{1}{2} \times {A}^{-(1/2)} = \frac{1}{2} \times \sqrt{\frac{1+x}{1-x}}.\tag2 $$
So, what remains is to compute $~\displaystyle \fr... | {
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"url": "https://math.stackexchange.com/questions/4542228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:
Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$
I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{... | The universe is the set of real numbers.
$\forall x\exists y<1:3x^2y+y-x=0$
$\iff\forall x\exists y<1:3x^2y+y=x$
$\iff\forall x\exists y<1:y(3x^2+1)=x$
$\iff\forall x\exists y<1:y=\frac{x}{3x^2+1}$, because we know that $3x^2+1\not=0$
$\iff\forall x\exists y(y<1\land y=\frac{x}{3x^2+1})$
$\iff\forall x\exists y(y=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\frac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$ as much as possible
Let $\alpha\in\mathbb{R}$ be so that $\cos\alpha\neq 1$. Simplify as much as possible $\dfrac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$. To clarify, if for a finite number of values of $\alpha,$ the ... | Let $x = 2022$ and $y = 2022 + \alpha$. Observe that using sum-to-product identities
\begin{align*}
\frac{\sin(x) + \sin(y)}{\cos(x) - \cos(y)} &= \frac{2\sin\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}{-2\sin\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}\\
&=\cot\left (\frac{y - x}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Check if this series is convergent or not I've been going crazy for a long time in determining if this series converges or diverges. Most likely it converges.
$\displaystyle \sum_{n=1}^\infty \left(\frac{n}{2}\, \sin\frac{1}{n}\right)^\frac{n^2+1}{n+2}$
I am stuck in the necessary condition.
I tried to solve in this wa... | Using the ratio test
$$a_n
=
\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}\implies
\log(a_n)=\frac{n^2+1}{n+2}\log\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)$$
For large $n$, compose Taylor expansion
$$\log(a_n)=\frac{n^2+1}{n+2}\bigg[-\log (2)-\frac{1}{6 n^2}+O\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15... | If $n$ must be an integer: HINT: First, $n^2+15n+122$ is even for all integers $n$. Then $n^2+15n+122$ will be divisble by $6$ iff $n^2+15n+122$ is a muliple of $3$. So what integers $n \pmod 3$ is $n^2+15n+122$ a multiple of $3$?
You can tell working $\pmod 3$, that $n^2+15n+122$ is divisible by $3$ for every integer ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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System of first order differential equation. Consider the linear system $y’=Ay+h$ where
$$
A=\begin{bmatrix}
1 & 1\\
4 & -2
\end{bmatrix}
$$and $$h=\begin{bmatrix}
3t+1\\
2t+5
\end{bmatrix}$$ Suppose $y(t)$ is a solution such that $$\lim_{t\to\infty}\frac{y(t)}{t}=k\in\mathbb R^2$$ What is the value of $k?$
$1$. $\begi... | We are given the linear system
$$y'(t) = Ay(t) + h(t) =\begin{bmatrix}1 & 1\\4 & -2\end{bmatrix}y(t) + \begin{bmatrix}3t+1\\2t+5\end{bmatrix}$$
I will use Example $2$ of these notes as a guide.
We find the eigenvalues and eigenvectors as
$$\lambda_1 = -3, v_1 = \begin{bmatrix} -1\\4 \end{bmatrix}\\ \lambda_2 = 2, v_2 =... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Complex contour integral using Cauchy integral formula I am trying to get integral in contour $C=\left \{ 3\cos(t)+2i\sin(t) : 0\leq t\leq 2\pi \right \}$
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz$$
What I tried was using partial fractions and integrate them separately
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz =\oint_{C}... | Using the Residue at Infinity
You need not evaluate the residues at $-1$ and $1$. Rather, note that the Residue at $\infty$ is given by
$$\begin{align}
\text{Res}\left(\frac{z}{(z+1)(z-1)^2},z=\infty\right)&=\text{Res}\left(-\frac1{z^2}\frac{1/z}{(1/z+1)(1/z-1)^2},z=0\right)\\\\
&=-\text{Res}\left(\frac{1}{(z+1)(z-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Finding the closed formula of a generating function for flower selection Find the generating function for the number of ways to select $n$ flowers from a boundless supply of dandelion, tulip, rose and blossoms if we have at least one flower of each type, the number of dandelions is less than 4, the number of tulips is ... | Yes, the desired generating function is the product
$$\frac{x^9 (1-x^3)}{(1 - x)^2(1-x^2) (1-x^5)}=\frac{x^9 (1+x+x^2)}{(1+x) (1 - x)^3 (1+x+x^2+x^3+x^4)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$
Let $$A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$$
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$.
I already got these Eigenvalues $D=\begin{bmatrix}0&0&0\\0&0&0\\0&0&3\end{bmatrix}$and Eigenvectors $P=\beg... | I try to give my input from a methodical approach. From your above computations, we can see that:
For $\lambda = 0$, these eigenvectors $\{v_1,v_2\}$ span the eigenspace $E_{\lambda=0}$:
\begin{Bmatrix}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}
\end{Bmatrix}
and for $\lambda ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{... | You have\begin{align}-4&=0-4\\&=4A-B+C-4(A+B)\\&=-5B+C\end{align}and\begin{align}-4&=0-4\\&=4A-6B-3C-4(A+B)\\&=-10B-3C.\end{align}And now, you have\begin{align}4&=(-2)\times(-4)-4\\&=(-2)(-5B+C)-10B-3C\\&=-5C,\end{align}and therefore $C=-\frac45$. Now, since $-5B+C=-4$, you get that $B=\frac{16}{25}$. And, since $A+B=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
\int_a^bf(x)dx = (b-a)\int_... | Noticing that $(x-3)(5-x)=1 -(x-4)^2$, we use the substitution $x-4=\sin \theta$ to transform the integral into
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(\sin \theta+4)(\sin \theta+5)}{\cos^{\frac 2 3} \theta} \cos \theta d \theta \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\sum_{i=0}^{n/2} (-1)^i \frac{n}{n-i} {n-i \choose i} = 2\cos(\frac{\pi n}{3})$ I was investigating the Girard-Waring identity, specifically for two variables:
$$x_1^n + x_2^n = \sum_{i=0}^{\frac{n}{2}} (-1)^i \frac{n}{n-i} {n-i \choose i}(x_1+x_2)^{n-2i}(x_1x_2)^i $$
This lead to me considering the foll... | I saw that one binomial coefficient, so that got me thinking about Fibonacci polynomials, and what do you know I find that this is an evaulation of a Lucas polynomial
$$L_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor}\frac{n}{n-k}\binom{n-k}{k}x^{n-2k}$$
By the Wikipedia page, $L_n(x) = \alpha^n(x) + \beta^n(x)$ where $\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ by induction I need to prove that
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$$ by induction.
The base case was fine, and after that we have the induction hypothesis:
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k}{2^k}=2-\frac{k+2}... | Some simple mistake in your algebra. $-2(k+2)+k+1=-((k+1)+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Change of variables for an ODE Question: Use the substitution $u = 2 \sqrt{x}$ to show that the equation:
$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-(1-x)y=0$$
becomes:
$$u^2\frac{d^2y}{du^2}+u\frac{dy}{du}-(u^2-u)y = 0$$
My attempt:
$$x^2\left[\frac{d^2y}{du^2}\left(\frac{du}{dx} \right)^2+\frac{dy}{du}\left(\frac{d^2u}{dx... | You must write everything in terms of $u$.
First derivative is easy, noting that $x=\frac{u^2}{4}$:
$$\frac{dy}{dx}=\frac{\frac{dy}{du}}{\frac{dx}{du}}=\frac{\frac{dy}{du}}{\frac{u}{2}}=\frac{2}{u}\frac{dy}{du}$$
Second derivative is difficult and key step:
$$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{du^2}-\frac{d^2x}{du^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Find the all positive integer solutions to: $x^3-x^2+x=3y^3$ Number theory problem:
Find the all positive integer solutions to:
$$x^3-x^2+x=3y^3$$
Here are my attempts:
$$x(x^2-x+1)=3y^3$$
$$x(x+1)(x^2-x+1)=3y^3(x+1)$$
$$x(x^3+1)=3(x+1)y^3$$
$$x^4+x^2=3(x+1)y^3$$
I can not see how can I proceed.
| As hinted in the LMFDB database (see my comments under the question), this elliptic curve is birationally equivalent to the Fermat cubic. In addition to the transformation described in the comments, I needed to bring it into the short Weierstrass form $r^2=s^3-432$, and then use the transformation given here. A bit of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1 \cos^{-1} x\ dx$ by first finding $\frac{d}{dx}(x\cos^{-1} x)$ Question
Evaluate $$\int_0^1 \cos^{-1} x\ dx$$ by first finding the value of $$\frac{d}{dx}(x\cos^{-1} x).$$
My Working
As the question said to evaluate $$\frac{d}{dx}(x\cos^{-1} x),$$ I used the product rule to differentiate. We first ha... | Integration by parts gives
$$
\begin{aligned}
\int_0^1 \cos ^{-1} x d x &=\left[x \cos ^{-1} x\right]_0^1+\int_0^1 \frac{x}{\sqrt{1-x^2}} d x \\
&=-\frac{1}{2} \int_0^1u^{-\frac{1}{2}} d u, \quad (\textrm{ where }u=1-x^2)\\
&=-\frac{1}{2} \cdot 2\left[u^{\frac{1}{2}}\right]_1^0 \\
&=1
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Verification of solution of contest problem related to algebra Given that $m, n$ are positive integers such that $mn<2023$ and $|n^2-mn-m^2|=1$, find the maximum value of $mn$.
This was problem 3 on the preliminary IMO exam where I live. My answer was $6$. However, there is space for $4$ numbers on the answer sheet, in... | If anything, here's a counter-example and its derivation:
$$|21^2-21\cdot 13-13^2|=1$$
$$21\cdot 13=273\gt 6\quad$$
derivation: one of the pell equations in $\quad |m^2-mn-n^2|=1\quad$is:
$$(m+2n)^2-5m^2=4,\quad subs\quad N=m+2n,\quad\text{consider}\quad N^2-5m^2=1\quad (9-4\sqrt5)(9+4\sqrt 5)=1\implies (9N+20m)^2-5(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Symmetric point of a point wrt a line I am trying to solve the following question.
Let $d$ be the
$$
\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{5},
$$
and let $P$ be the point of coordinates $(4,3,10)$. Find the coordinates of the symmetric point $P^{\prime}$ of $P$ with respect to $d$.
Book has the following solution.
L... | The equations are
$ (x_0, y_0, z_0) = (1,2,3) + t (2, 4, 5) $
$PM = M - P = (1, 2, 3) - (4, 3, 10) + t (2, 4, 5) = (-3, -1, -7) + t (2, 4, 5)$
$ PM \perp d $ , so
$ \bigg( (-3, -1, -7) + t(2, 4, 5) \bigg) \cdot (2, 4, 5) = 0 $
So that
$ -45 + 45 t = 0 $
$\Longrightarrow t = 1 $
Thus $M = (1,2,3) + (2,4,5) = (3, 6, 8) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of differentiation to find the power series and determine the sum of the series We were asked to differentiate $f(x) = \frac{1}{4-{x^2}}$ to determine the sum of the series $\sum_{n=1}^\infty \frac{n}{16^{n}}$
so far I got the power series representation which is
$$\sum_{n=0}^\infty \frac{x^{2n}}{4^{n+1}}$$
But whe... | From $$f(x)=\frac{1}{4-x^2}=\frac{1}{4}\frac{1}{1-\left(\frac{x}{2}\right)^2}=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n}$$
Now
$$f'(x)=\frac{1}{4}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n-1}=\frac{1}{4}\frac{2}{x}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n}$$
Put $x=\frac{1}{2}$ and from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How can you show the results from substituting $x=a\cos{\theta}$ and $x=a\sin{\theta}$ are the same when finding $\int{\frac{16}{\sqrt{9-x^2}}}$? How can you show the results from substituting $x=a\cos{\theta}$ and $x=a\sin{\theta}$ are the same when finding $\int{\frac{16}{\sqrt{9-x^2}}}$?
I am learning trig substitut... | If you do a simpler example of $\int \frac{1}{\sqrt{1 - x^2}} dx$, you will get $\sin^{-1}(x) + C$ and $-\cos^{-1}(x) + C$ respectively. If you draw the diagrams, you will see that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$, since it's the two angles that's not a right angle. Therefore, you see that $\sin^{-1}(x)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Arrangements of $a,a,b,b,b,c,c,c,c$ in which no two consecutive letters are the same I have the following question:
We are going to generate permutations from a,a,b,b,b,c,c,c,c. Please compute the number of permutations such that:
(a) for any consecutive 4 elements, they are not all the same;
(b) for any consecutive 3... | This answer is rather a supplement which could be used as crosscheck for manual calculations. We consider a $3$-ary alphabet built from letters $\mathcal{V}=\{a,b,c\}$. Words which do not have any consecutive equal letters are called Smirnov words. A generating function for Smirnov words is given as
\begin{align*}
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
write an explicit formula for $f(x)$ when $x$ is in $[\frac{1}{n+1},\frac{1}{n}]$
this is from spivak calculus ,i tried to find an explicit formula when $x$ is in $[\frac{1}{n+1},\frac{1}{n}]$
my attempt:
in $[\frac{1}{n+1},\frac{1}{n}]$ the function $f$ is a linear function so we can write it as $f(x)=\alpha x+ \bet... | Your attempt is right except $f$ instead of $f_{\color{red}n}$ and $x>0$(just say the function is even). I will make a few remarks in addition to the remarks made in comments :
Let $a,b\in\Bbb R$: the middle of $[a,b]$ is $\frac12(a+b)$
The middle of $[\frac{1}{n+1},\frac1n]$ is $\frac{\frac{1}{n}+\frac{1}{n+1}}{2}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y... | $$\frac23xyy'=\sqrt{x^6-y^4}+y^2.$$
$$x^3(y^2)'=3x^2\sqrt{x^6-y^4}+3x^2y^2$$
$$\dfrac {x^3(y^2)'-3x^2y^2}{x^6}=\dfrac 3 {x^4}\sqrt{x^6-y^4}$$
$$\left (\dfrac {y^2}{x^3}\right)'=\dfrac 3 {x^4}\sqrt{x^6-y^4}$$
$$\left (\dfrac {y^2}{x^3}\right)'=\dfrac 3 {x}\sqrt{1-\left (\dfrac {y^2}{x^3}\right)^2}$$
The DE is separable.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x... | Letting $x=\tan \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4} } \ln \left[\left(1+\tan ^2 \theta\right)\left(1-\tan ^2 \theta\right)\right] d \theta \\
&=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta+\ln \left(\frac{\cos^2 \theta-\sin ^2 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ ... | Observe that
\begin{align*}
x^2-9x+24&=(x-5)^2+(x-5)+4\\
6x^2-59x+149&=6(x-5)^2+(x-5)+4
\end{align*}
Say $(x-5)^2+(x-5)+4=\alpha$ (note $\alpha \geq 0$), then we have
$$\sqrt{\alpha}-\sqrt{5(x-5)^2+\alpha}=|5-x|$$
The expression on the left is $\leq 0$ and that on the right is $\geq 0$. For the equality to hold, both h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\s... | Let $s = \sqrt{n+1}$ and $t = \sqrt{n}.$ Then:
$$\mathcal{L} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{s}}{s-t} = \\
= \lim_{n \to \infty}\displaystyle\frac{1}{(s-t)}\frac{1}{s} = \lim_{n \to \infty}\displaystyle\frac{s+t}{(s-t)(s+t)}\frac{1}{s}.$$
But:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to ... | $$
p(x) = ab{c}^{2}{x}^{2}+c \left( 3\,{a}^{2}+{b}^{2} \right) x+3\,{a}^{2}-ab+{b}^{2} .
$$
Let's try to factor this with the idea of finding a factor $Ax+B$ where $A$ is a divisor of $abc^2$ and $B$ is a divisor of $3a^2-ab+b^2$. Since $3a^2-ab+b^2$ cannot be factored over the rationals, we try without factoring it. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Trig and de Moivre's theorem
A) Use de Moivre's theorem to prove that $\cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
B) Therefore deduce that $\cos(\pi/8) = \left(\frac{2 + \sqrt{2}}{4}\right)^{1/2}$
C) and write down an expression for $\cos(3\pi/8)$.
I have proved the first part of the question but I am not sure... | With $\cos(4\theta)$ on the lhs, you can substitute $\theta = \frac{\pi}{8}$ in your first equation, which gives a polynomial equation for $\cos(\frac{\pi}{8})$.
$$0 = 8\cos^4\frac{\pi}{8}- 8\cos^2\frac{\pi}{8} + 1$$
This equation is biquadratic and can be solved with the quadratic formula to show that,
$$\cos\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5... | It is clear that there are infinitely many solutions (see my comment above). We know that the identity $$(a^2+b^2+c^2+d^2)^2=(a^2-b^2-c^2-d^2)^2+(2ab)^2+(2ac)^2+(2ad)^2$$ gives all the solutions of $x^2+y^2+z^2+w^2=u^2$.
Now we have $$x^2+y^2+z^2=m^2+n^2\iff x^2+y^2+z^2+(im)^2=n^2$$ because the identity is valid in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show that scalene triangle $\triangle ABC$ is a right-triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$ As the title suggests, this is a college entrance exam practice problem from Japan, it is as follows:
Given a scalene triangle $\triangle ABC$, prove that it is a right triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$
I found... | Here's an identity-free approach:
Writing $D$ and $E$ for the feet of the altitudes from $A$ and $B$, we have
$\require{cancel}$
$$\color{blue}{|\triangle ABE|}+\color{red}{|\triangle CBE|} \;=\; |\triangle ABC| \;=\; \color{blue}{|\triangle BAD|}+\color{red}{|\triangle CAD|} \tag1$$
$$\cancel{\color{blue}{\tfrac12c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
The sum $\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is equal to: From an exercise of a textbook of an high school of 15 years old, I have this partial sum
$$S_n=\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$$
Con... | The general term
$$\begin{align}
\frac{1}{(n+1)\sqrt{n}+ n\sqrt{n+1}} &= \frac{1}{\sqrt{n(n+1)}}\cdot\frac{1}{\sqrt{n+1}+\sqrt{n}} \\
&=\frac{1}{\sqrt{n(n+1)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\\
&=\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}
\end{align}$$
Make the sum, the result must be equal to
$$\sum_{n=1}^{99}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi $ miss solutions? So the solution to the problem is:
$$4 \tan x = 5 \sin x$$
$$4 \frac{\sin x}{\cos x} = 5 \sin x$$
$$4 \sin x = 5 \sin x \cos x$$
So, either $\sin x = 0$ or $\cos x = 4/5$. I'm fine with the logic of this. But is there... | For the same reason why solving $x = x^2$ by dividing by $x$ would lose solutions.
If you divide both sides by something, you implicitly give rise to two cases.
For $x=x^2$, if you divide by $x$, then you get $1=x$, but only if $x \ne 0$. If $x$ (what you divided by) is equal to zero, then you just divided by zero, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
substitute $x=\sqrt{y}$
$$\sqrt y\left(1+\sqrt{1-y}\right)=\sqrt{1-y}$$
divide both sides by $\sqrt{y}$
$$1+\sqrt{1-y}=\sqrt{y^{-1}-1}$$
square both sides
$$2-y+2\sqrt{1-y}=y^{-1}-1$$
isolate the square root
$$2\sqrt{1-y}=y+y^{-1}-3$$
square both sides
$$4-4y=y^2+y^{-2}-6y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead... | From $\cos(A-B)=\frac{7}{8}$, we have $\sin(\frac{A-B}{2})=\sqrt{\frac{1-\cos(A-B)}{2}}=\frac{1}{4}. {\tag 1}$
This is a nice property about interior angles: $$\sin(\frac{A-B}{2})=\frac{a-b}{c}\cos(\frac{C}{2}).$$ From this property and $(1)$ with $a=5$, $b=4$, we have $\cos(\frac C2)=\frac c4.\tag 2$
By using the half... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Find $a_n$ such that the formula is equal to $\zeta(2)$ My question start with the observation :
$$\sqrt{e}\simeq \frac{\pi^2}{6}$$
At first glance it's not really convincing but after some work I found :
$$\sqrt{e-5\left(\frac{1}{\pi}-\frac{1}{e}\right)^{2}+2\left(\frac{1}{\pi}-\frac{1}{e}\right)^{3}+9\left(\frac{1}{\... | As @Gevorg Hmayakyan says, the problem can be rephased as finding a sequence $a_n\in \mathbb{Z}$ such that
\begin{equation}
\sum_{i=1}^\infty a_i \left(\frac{1}{\pi}-\frac{1}{e}\right)^{i}=\frac{\pi^4}{36}-e
\end{equation}
To to this, I use a greedy algorithm, i.e., in each step the error is maximally reduced. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the real solutions (if any) of $x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$
Find the real solutions(if any) of $$x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$$
Note: I need only HINT:
My effort:
I let $v=x^3$, then we get
$$v^4=3v^3+4v^2+4v+2\sqrt[3]{v}+2 \Rightarrow v^4-3v^3-4v^2-4v=2(1+\sqrt[3]{v})$$
$$\begin{aligned}
& \Rightarrow v... | Let $$f(x)=x^{12}-3x^9-4x^6-4x^3-2x-2$$
By Intermediate value theorem, we have $f(1)<0, f(2)>0$, so $\exists$ atleast one root on $(1,2)$. But by Descarte's rule of signs, the number of changes in $f(x)$ is one, so only one positive root. Obviously there should be one more real root. So there exists a quadratic factor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinite... | We invoke the Fourier series for $\left|\cos x\right|$:
$$ \left| \cos x \right| = \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^n}{4n^2 - 1} \cos(2nx) $$
From this, we obtain the following series representation for the numerator of the integrand:
\begin{align*}
\left|\cos\left(x-\frac{\pi}{4}\right)\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integ... | Well, we are trying to solve:
$$\mathcal{I}\left(x\right):=\int\sqrt{x^2+x-2}\space\text{d}x\tag1$$
First, complete the square:
$$\mathcal{I}\left(x\right)=\int\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{9}{4}}\space\text{d}x\tag2$$
Substitute $\text{u}=x+\frac{1}{2}$:
$$\mathcal{I}\left(x\right)=\int\sqrt{\text{u}^2-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (... | Since $$\tan{x}+\tan\left(60^{\circ}+x\right)+\tan\left(120^{\circ}+x\right)=$$
$$=\tan{x}+\frac{\sqrt3+\tan{x}}{1-\sqrt3\tan{x}}+\frac{-\sqrt3+\tan{x}}{1+\sqrt3\tan{x}}=3\tan3x,$$ we obtain:
$$\cot10^{\circ}+\cot70^{\circ}-\cot50^{\circ}=\tan80^{\circ}+\tan20^{\circ}+\tan140^{\circ}=3\tan(3\cdot20^{\circ})=3\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure ... | Let $$S_n=\sum_{k=0}^{n} {2n \choose k}$$
and let
$$S'_n=\sum_{k=0}^{n} k {2n \choose n-k}=\sum_{k=0}^{n} (n-k) {2n \choose k}=nS_n-T_n$$
$$2^{2n}=\sum_{k=0}^{2n} {2n \choose k}=\sum_{k=0}^{n} {2n \choose k}+\sum_{k=n+1}^{2n} {2n \choose k}$$
$$\implies S_n+\sum_{k=n+1}^{2n} {2n \choose k}=2^{2n}$$
Change $k$ to $2n-j$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
A limit of a sequence that yields different answers This is the problem:
$\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}\sum^n_{k=1}\sqrt{k}$
A friend and I tried to solve it using different methods and our results are very different. Our attempts:
First attempt. Using Riemann sums
$\lim_{n\rightarrow\infty}\frac{1}{n\... | The second attempt is incorrect. You have made a mistake in your algebra as the first identity written in that part is false. You can try the following
\begin{align*}
\frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} &= \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} \times \frac{(n+1)\sqrt{n+1}+n\sqrt n }{(n+1)\sqrt{n+1}+n\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limits problem (unable to solve further) Struggling to solve this problem,
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$
My approach:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\... | You also can use Riemann-sums:
$$
\left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{6n}\right)
$$
$$
=
\left(1+\frac{n}{n+1}+\frac{n}{n+2}+\dots + \frac{n}{n+ (5n-1)}\right)\frac{1}{n}+ \frac{1}{6n}
$$
$$
=\left(1+\frac{1}{1+1/n}+\frac{1}{1+2/n}+\dots + \frac{1}{1+(5n-1)/n}\right)\frac{1}{n} + \frac{1}{6n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$... | Clearing all denominators, we have that
$$
\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1
$$
is equivalent to
$$
a^4b^4c^4d^4 = (a^4b^4 + a^4c^4 + a^4d^4 + b^4c^4 + b^4d^4 + c^4d^4) + 2(a^4 + b^4 + c^4 + d^4) + 3
$$
Applying arithmetic mean $\ge$ geometric mean to the brackets on the right gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt... | You might see that the function can be rewritten as follows:
$$
y=-\arcsin(x)\Big((1-x^2)^{\tfrac{1}{2}}\Big)'
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.