Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $x=2+i$, $gcd(a,b,c)=1$, and $ax^4+bx^3+cx^2+bx+a=0$, then what is $|c|$?
Suppose
$$a(2+i)^4 + b(2+i)^3 + c(2+i)^2 + b(2+i) + a = 0,$$
where $a,b,c$ are integers whose greatest common divisor is $1$.
Determine $|c|$.
So I first simplified the exponents and combined like terms.
I received $$a(-6+24i)+b(4+12i)... | A polynomial of this symmetric form can be reduced in degree by the following trick. Divide by $x^2$ and get:
$\begin{align}
a (x^2 + x^{-2}) + b (x + x^{-1}) + c
&= a (x + 1/x)^2 + b (x + 1/x) + c - 2 a
\end{align}$
The substitution $y = x + 1/x$ finishes this off.
This works because the powers of $x + 1/x$ turn out... | {
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"url": "https://math.stackexchange.com/questions/1596576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proof by induction that $5|11^n-6$ for all positive integers $n$
Prove by induction that $5|11^n-6$ for all positive integers $n$
Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).... | Let $x=11^n-6$ then $$11^{n+1}-6=11(x+6)-6=11x+60$$ and this later is divisible by $5$ if $x$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Has this equation appeared before? I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is... | This provides the explicit polynomial in $x$ (for those curious), though I'm not aware if the equation has appeared in the mathematical literature. We get rid of the square roots by multiplying out the $8$ sign changes,
$$\prod^8 \left(\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}\pm\sqrt{\frac{a+c+x}{b}}\pm\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor an... | We may suppose that $a\geq b>0$. Hence dividing by $b^9$ and putting $x=a/b$, we have to show that for $x\geq 1$, we have $(x^7+1)(x^2+1)\geq (x^5+1)(x^4+1)$ or
$$\frac{x^7+1}{x^5+1}\geq \frac{x^4+1}{x^2+1}$$
For $x\geq 1$ fixed, put $\displaystyle f(u)=\frac{x^2u+1}{u+1}$. It is easy to see that $f$ is increasing on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}... | I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...
Using AM-GM we can show that:
$$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$
$$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$
We can also show that:
$$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$
We can therefore infer that:
$$a^2+b^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then... | Converting your expression so that all that is cubed is positive:
$$(z-x)^3 - (z-y)^3 - (y-x)^3$$
$$(A+B)^3-A^3-B^3$$
with $A,B$ positive. Since cubing is concave up with $0^3=0$, this expression must be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Umbilics on the ellipsoid Show that, if p, q and r are distinct positive numbers, there are
exactly four umbilics on the ellipsoid $$\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1$$
What happens if $p$, $q$ and $r$ are not distinct?
$$$$
To show that I considered the parametrization $\sigma (u,v)=(p\cos u\sin v, ... | Consider a parametrization $f$ :
$$ x=a\cos\ t \cos\ s,\ y=b\cos\ t\sin\ s,\ z=c\sin\ t $$
Define $$ A:=\cos\ t,\ B=\sin\ t,\ C=\cos\ s,\ D=\sin\ s $$
Then $$
f_t=(-aBC,-b BD, c A ) ,$$ $$ f_s=(-aAD, b AC, 0 )
$$
$$ f_{tt} =(-aAC,-b AD,-cB) $$
$$ f_{st} = (aB D, -bBC,0) $$
$$ f_{ss} = (-aAC,-bA D,0 ) $$
Hence we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of ... | By AM-GM $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tangent line Exercise. $f (x) = 5e^{−(x−2)^2}$ . Find the coordinates of points where the hill is the most steep. I have this exercise, look easy but I don't know where to start, I think that I need a extra function or value to continue with the calculus.
Imagine that you are riding over a hill having its profile give... | Notice, the slope of the tangent to the curve: $f(x)=5e^{-(x-2)^2}$ is given as
$$f'(x)=\frac{d}{dx}(f(x))=\frac{d}{dx}(5e^{-(x-2)^2})$$
$$=5e^{-(x-2)^2}(-2(x-2))$$$$=-10(x-2)e^{-(x-2)^2}$$
$$f''(x)=-10(x-2)\frac{d}{dx}(e^{-(x-2)^2})-10e^{-(x-2)^2}\frac{d}{dx}(x-2)$$
$$=20(x-2)^2e^{-(x-2)^2}-10e^{-(x-2)^2}$$
$$f'''(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Least squares problem: am I solving it correctly? So I have this question:
In $\mathbb R^3$ with inner product calculate all the least square solutions, and choose the one with shorter length, of the system:
$ x + y + z = 1 $
$ x + z = 0 $
$ y = 0 $
My attempt: So I applied the formula $ A^T A x = A^T b $ with ... | The reduced row echelon form is
$$
\begin{align}
\mathbf{A} &\mapsto \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
There is a ran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int (\tan x)^{1/ 6} \,\text{d}x$? How do I compute the following integral
$$
I=\int (\tan x)^{1/ 6} \,\text{d}x
$$
| Given you said you "won't even try" to solve this integral, try Mathematica:
$$\frac{-2 \left(\sqrt{3}-1\right) \tan ^{-1}\left(\frac{-2 \sqrt{2} \sqrt[6]{\tan
(x)}+\sqrt{3}-1}{1+\sqrt{3}}\right)+4 \tan ^{-1}\left(1-\sqrt{2} \sqrt[6]{\tan
(x)}\right)-4 \tan ^{-1}\left(\sqrt{2} \sqrt[6]{\tan (x)}+1\right)+2
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric contour integral I cannot figure out what I'm doing wrong:
$$\int_0^{2\pi} \frac{1}{a+b\sin\theta} d\theta\quad a>b>0$$
$$\int_{|z|=1} \frac{1}{a+\frac{b}{2i}(z-z^{-1})} \frac{dz}{iz}$$
$$\int_{|z|=1} \frac{2i}{2ia+b(z-z^{-1})} \frac{dz}{iz}$$
$$\int_{|z|=1} \frac{2}{2iza+bz^2-b} dz$$
$$2iz_0a+bz_0^2-b=0... | The integral $I=\int_0^{2\pi}\frac{1}{a+b\sin \theta}\,d\theta$ is simply $2\pi i$ times $2/b$ times the residue of $z^2+i 2(a/b) z-1$ at $z=i\left(-\frac ab +\sqrt{(a/b)^2-1}\right)$. Thus, we have
$$I=(2\pi i)\frac2b \frac{1}{2i\sqrt{(a/b)^2-1}}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}... | By Cauchy-Schwarz,
$(LHS)\le \sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}+\sqrt{2(ab+bc+ca)2(xy+yz+zx)}=f(a,b,c,x,y,z)$
However, $$f(a,b,c,x,y,z)^2\le (a^2+b^2+c^2+2ab+2bc+2ca)(x^2+y^2+z^2+2xy+2yz+2zx)=(RHS)^2(\because Cauchy)$$
Therefore, $(LHS)^2\le (RHS)^2$. Our proof is done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Finding Laurent's series of a function I am trying express the function $$f(z)=\frac{z^3+2}{(z-1)(z-2)}$$ like a Laurent's series in each ring centering in $0$, but I do not now how could I express it, in first I said that $$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
Ok, now, I see two posibilities: $A\eq... | You have to find possible Laurent series expansion for each case !
Case I : $0<|z|<1$
Case II : $1<|z|<2$
Case III : $2<|z|$
Case I : $0<|z|<1$
$$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
$$\Rightarrow f(z)=(z^3+2)\left[\frac{1}{1-z} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
For all $|z|<1$ , $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfi... | Since you have simplified the given equation as $(x-2)^2+(y+1)^2=2^2$, then taking the parameter $\theta$, put $x=2+2\cos\theta$ and $y=-1+2\sin\theta$.
As $\theta$ varies, you can clearly see that the point $(x,y)=(2+2\cos\theta,-1+2\sin\theta)$ is on the circle $(x-2)^2+(y+1)^2=2^2.$
Now use this parametrization in $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$. The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4).$ Find 3 more solutions in positige integers. [hint. look for solutions of the form $(a,b,c) = (xz, yz, z^2)$
Attempt:
So I tried to use the hint in relation to the triple that they g... | Here is an
infinite number of solutions:
For any positive integer $c$,
$x = (1+c^{3})$,
$y
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
Check:
$x^3+y^3
=(1+c^3)^3+(c(1+c^3))^3
=(1+c^3)^3(1+c^3)
=(1+c^3)^4
$
and
$z^2
=((1+c^{3})^2)^2
=(1+c^{3})^4
$.
This is gotten from the following,
which is a re-creation
of some algebra of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | Let $\sin^2 \theta =u$ and $\cos^2 \theta =v.$ Then $u+v=1.$ The expression exists only when $0<u<1$ and $0<v<1,$ when it is $$(u+v)+\left(\frac {1}{u}+\frac {1}{v}\right)+\left(\frac {u}{v}+\frac {v}{u}\right)=$$ $$=1+\frac {u+v}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Prove that $\left | \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \right | < \frac{1}{8}.$
Let $a,b,$ and $c$ be the lengths of the sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.$$
The best idea I had was to expand the fractions to get something... | I'll prove a stronger inequality.
Let $a,b,$ and $c$ be the lengths-sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{22}$$
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence,
$$\sum_{cyc}\frac{a-b}{a+b}=\frac{\sum\limits_{cyc}(a-b)(a+c)(b+c)}{\prod\limits_{cyc}(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equi... | For any triangle, we have
$$\tan(A) + \tan(B) + \tan(C) = \tan(A) \tan(B) \tan(C)$$
This means we have
$$\tan(A) \tan(B) \tan(C) = 6$$
Hence, we have $\tan(C) = 3$. Hence, we have
$$\tan(A) + \tan(B) = 3 \text{ and }\tan(A) \tan(B) = 2$$
This gives us that $\tan(A) = 2$ and $\tan(B) = 1$. Hence, the triangle is an acut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}
$$
The answer is given to be 2 in the textbook.
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solve... | Another way : change variable $x=1+y$; so $$A=\frac{\sqrt x - 1}{ \sqrt[3] x - 1}=\frac{\sqrt{1+y} - 1}{ \sqrt[3] {1+y} - 1}$$ Now, using the fact that, close to $y=0$ (using the generalized binomial theorem as lulu commented) $$(1+y)^a=1+a y+\frac{1}{2}a \left(a-1\right) y^2+O\left(y^3\right)$$ which makes $$A=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) +... | An other way
$$\sum_{k=1}^n(6k-3)=6\sum_{k=1}^n k-3n=6\cdot\frac{n(n+1)}{2}-3n=3n^2.$$
The only proof you need to do (by induction if you need to use induction) is that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Series expansion of $\frac{1}{\sqrt{x^3-1}}$ near $x \to 1^{+}$ How can I arrive at a series expansion for $$\frac{1}{\sqrt{x^3-1}}$$ at $x \to 1^{+}$? Experimentation with WolframAlpha shows that all expansions of things like $$\frac{1}{\sqrt{x^y - 1}}$$ have $$\frac{1}{\sqrt{y}\sqrt{x-1}}$$ as the first term, which I... | I like to expand around zero.
So, in
$\frac{1}{\sqrt{x^3-1}}
$,
let $x = 1+y$.
Then
$\begin{array}\\
x^3-1
&=(1+y)^3-1\\
&=1+3y+3y^2+y^3-1\\
&=3y+3y^2+y^3\\
&=y(3+3y+y^2)\\
\end{array}
$
so
$\begin{array}\\
\frac{1}{\sqrt{x^3-1}}
&=\frac{1}{\sqrt{y(3+3y+y^2)}}\\
&=\frac1{\sqrt{y}}\frac{1}{\sqrt{3+3y+y^2}}\\
&=\frac1{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over... | Here's another way to solve the integral.
$$\int_{0}^{\pi}\theta \ln\tan\frac{\theta}{2}\mathrm{d}\theta = 4\int_{0}^{\frac{\pi}{2}}x\ln\tan{x}\mathrm{d}x = 4\left(\int_{0}^{\frac{\pi}{2}}x\ln\sin{x}\mathrm{d}x-\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x\right)$$
We solve the later integral first. But before that, r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ... | An other way is to write
$$\int\frac{x+3}{\sqrt{x+2}}dx=\int\frac{x+2+1}{\sqrt{x+2}}dx$$
$$=\int\frac{x+2}{\sqrt{x+2}}dx+\int\frac{1}{\sqrt{x+2}}dx=\int\sqrt{x+2}dx+\int\frac{1}{\sqrt{x+2}}dx$$
$$=I_1+I_2.$$
In $I_1$ we put the change of variable $u=x+2,\ du=dx$ and in $I_2$ we put $w=\sqrt{x+2},\ dw=\frac{1}{2\sqrt{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
| HINT: To do it without linear algebra, expand both sides and subtract like terms to leave
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\;.$$
Notice that you can rearrange this as
$$(a^2y^2-2abxy+b^2x^2)+(a^2z^2-2acxz+c^2x^2)+(b^2z^2-2bcyz+c^2y^2)=0\;,$$
or
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\;.$$
*
*Wh... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we h... | The general solution to $\alpha x^2+\beta x+\gamma=0$ is $x=\frac{-\beta\pm \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}$.
Here, after factoring out the common $b$, you get $\alpha = b+3, \beta=-(3b-1),\gamma=2b^2$. In particular, you need $(3b-1)^2-4(b+3)(2b^2)=-8b^3-15b^2-6b+1$ to be a perfect square to even get a rational... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2})... | Look at what happens when we complete the square:
$$\begin{array}{lll}
ax^2+bx+c &=& \frac{1}{4a}(4a^2x^2 + 4abx + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) +b^2-b^2+ 4ac)\\
&=& \frac{1}{4a}((2ax+b)^2-(b^2 - 4ac))\\
&=& \frac{1}{4a}((2ax+b)^2-\bigg(\sqrt{b^2 - 4ac}\bigg)^2)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angl... | By complex numbers:
By the complex definition of the trigonometric functions, setting $z=e^{ix}$,
$$3\frac{z+z^{-1}}{2i}+4\frac{z+z^{-1}}2=2.$$
Multiplying by $z$ and rearranging,
$$\left(2+i\frac32\right)z^2-2z+\left(2-i\frac32\right)=0.$$
Then solving the quadratic equation
$$z=\frac{(8+6i)\pm\sqrt{21}(3-4i)}{25}.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Modeling with Markov Chains and one-step analysis
I have set up the following model:
Let $X_n$ be the number of heads in the $n$-th toss and $P(X_0=0)=1$. I can calculate the transition matrix $P$. Define
$$
T=\min\{n\geq 0\mid X_n=5\}.
$$
Then $P(X=1)=P(X_{T-1}=4)$. Noting that $X_n$ is a Markov chain and defining
$... | It might simplify slightly to say we just require the chain to reach state $4$ at any point, starting at $0$, and we can forget about the variable $T$. So re-define $u_i$ as
$$u_i = P(X_n = 4 \text{ for some $n$}\mid X_0=i).$$
Our transition probability matrix:
$$
\begin{matrix}
\qquad 0 \qquad & 1 \qquad & 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$ $$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$
This result seems to me digitally correct?
Can we prove that the equality is exact?
| We have:
$$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$
hence:
$$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1... | You can proceed in this way:
$$\tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}}=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}=\tan \frac{B}{2}$$
And hence comparing, we can write that $A=n\pi + \frac{B}{2}$ where $n$ is any integer.
So we can say that $2A=2n\pi + B \Rightarrow \tan 2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x... | Let the parametric equation of the tangent be
$$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$ where $\theta$ is unknown.
Plug in the equation of the ellipse to get
$$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\\
=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+2\left(\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the smallest positive value taken by $a^3+b^3+c^3-3abc$
Find the smallest positive value taken by $a^3+b^3+c^3-3abc$ for positive integers $a,b,c$. Find all integers $a,b,c$ which give the smallest value.
Since it is generally hard to find the minimum of a multivariate polynomial, I tried factoring it at first. ... | $$\quad F=a^3+b^3+c^3-3 a b c=(a+b+c)(a^2+b^2+c^2- a b -b c- c a)=$$ $$= \frac {1}{2}(a+b+c)(\;[a-b]^2+[b-c]^2+[c-a]^2\;).$$
If $a=b=c$ then $F=0.$
If $a,b,c$ are not all equal then $a+b+c\geq 1+1+2=4,$ and also at least two of $|a-b|, |b-c|, |c-a|$ are non-zero, giving $(a-b)^2+(b-c)^2+(c-a)^2\geq 1^2+1^2+0^2=2 .$
Th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I find the difference between the gradients of two lines represented by an equation I want to find the difference between the gradients (or slopes?) of two lines. The equation of the lines is $$x^2(\tan^2 \theta+\cos^2 \theta)-2xy\tan\theta+y^2 \sin^2 \theta=0$$
I have assumed the gradients are $m_1$ and $m_2$.
... | I'm going to guess wildly at what you actually mean because you do not write what you actually mean. I expect that if you clarify your question and it turns out I have guessed wrongly, others will vote this answer into oblivion, as they should...
Say we have two lines in point-slope form: $$\begin{align}
y &= m_1 ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20.$ Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$
Let $n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$ be the positive integers such that $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20$
Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$ is
$\bf{My\... | Here is an elementary argument.
The minimum possible $n_1$ is $1$.
Since the sum is $20$, dividing $20$ by $5$ we get $4$, which means $n_1$ cannot be $4$, since the numbers are strictly increasing.
If $n_1=3$, increasing each following number by $1$, we get $3+4+5+6+7=25$.
So the maximum possible $n_1$ is $2$. With... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: L... | Hint: Note that $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | Here's the symmetric polynomial approach.
$a^n+b^n+c^n$ is a symmetric homogeneous polynomial of degree $n$. So it can be expresses as a linear combination of polynomials $s_1^is_2^js_3^k$ where $i+2j+3k=n$, and $s_1=a+b+c,$ $s_2=ab+bc+ac,$ $s_3=abc$ are the elementary symmetric polynomials.
Now, $s_1=a+b+c=0$ implies... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$
If $x,y,z$ are positive real number number, Then minimum value of $\displaystyle \frac{x^4+y^4+z^2}{xyz}$
$\bf{My\; Try::}$ Given $x,y,z>0.$ So Using $\bf{A.M\geq G.M\;,}$ We get
$$\displaystyle x^4+y^4\geq 2x^2y^2$$ and then ... | We have, with AM-GM, that \begin{align}
x^4+y^4+z^2&=4\cdot\frac{x^4+y^4+\tfrac12z^2+\tfrac12z^2}{4}\\
&\geq 4\sqrt[4]{x^4y^4z^4\cdot\tfrac14}\\
&=xyz\cdot 2\sqrt{2}
\end{align}
So that $$\frac{x^4+y^4+z^2}{xyz}\geq 2\sqrt{2}$$
With equality iff $x^4=y^4=\frac12z^2$, so for example $x=y=1$ and $z=\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Proving that a series converges I am given the series $\Sigma_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$, where $a_n \geq 0 \forall n$ , and $\Sigma_1^{\infty} a_n $ converges.
I was advised to expand $|\sqrt{a_n} - \frac{1}{n}|^2 $ . Doing this gives me expressions like $-\frac{2 \sqrt{a}}{n}+a+\frac{1}{n^2}$ , but I don'... | Writing $(x-y)^2 = x^2 - 2xy + y^2$, we see that
$$ xy \leq \frac{1}{2} \left( x^2 + y^2 \right). $$
This implies that
$$ \frac{\sqrt{a_n}}{n} \leq \frac{1}{2} \left( a_n + \frac{1}{n^2} \right). $$
Since both $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge, the series $\sum_{n=1}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662941",
"timestamp": "2023-03-29T00:00:00",
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| $2^8 + 2^{11} + 2^n = 2^8(1 + 8 + 2^{n-8})=2^8(9 + 2^{n-8})$
Therefore, $9 + 2^{n-8}$ has to be a perfect square.
Clearly, $9 + 16 = 25$ is a perfect square.
So, $2^{n-8} = 2^4$ giving, $$n = 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Given two sets, is there a formula to know the quantity of injective function realationships? For example.
\begin{align*}
&A\quad =\quad \{ 1,2\} \\
&B\quad =\quad \{ a,b\} \\
&\texttt{Then 2 injective functions from } A\rightarrow B \\
&\left\{ (1,a),(2,b) \right\} \quad \{ (1,b),(2,a)\} \\ \\
&A\quad =\quad \{ 1,... | Let $A$ and $B$ be nonempty, finite sets such that $A$ has at most as many elements as $B$. Let $m$ be the number of elements in $A$ and let $n$ be the number of elements in $B$. WLOG assume that $A = \{ 1, 2, \ldots, m \}$ and $B = \{1, 2, \ldots, n \}$. Then there are precisely $\binom{n}{m} \cdot m!$ many injections... | {
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"timestamp": "2023-03-29T00:00:00",
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How does this factoring work? $$ (z^2 - 2i ) = (z -1 -i)(z + 1 +i) $$
I see if you multiply out the right-hand side, you obtain the left-hand side, but how does one know to factor like that or this?
$$ (z^2 − 3iz − 3 + i) = (z − 1 − i)(z + 1 − 2i) $$
| The first one
is using
$z^2-a^2
=(z-a)(z+a)
$
where
$a = \sqrt{2i}
=1+i
$
since
$(1+i)^2
=1+2i-1 = 2i
$.
The second one
just uses the quadratic formula,
which works for
complex as well as real coefficients,
to solve
$z^2 − 3iz − 3 + i
= 0
$.
If the roots are
$u$ and $v$,
then
$z^2 − 3iz − 3 + i
= (z-u)(z-v)
$.
(This wa... | {
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If $b$ is an odd composite number and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$? (Note: An improved version of this question has been cross-posted to MO.)
Let $\sigma(X)$ be the sum of the divisors of $X$. For example, $\sigma(2) = 1 + 2 = 3$, and $\sigma(4) = 1 ... | The only thing that I can say is that if $b=3^k$ for $k>1$ then $q=2$.
Up to $10^8$ there are no other values of $b$ that make $q$ prime.
| {
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Taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$ I am a bit confused about taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$.
$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2 y^2)}{dx} $$
Edit: Incorrect step
$$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 10xy^2 + 5x^22y\frac{dy}{dx}$$
$$= 3... | An expression $f(x,y)=g(x,y)$ is implicitly differentiated by
$$ \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial x} x' =\frac{\partial g}{\partial y} y' + \frac{\partial g}{\partial x} x'$$
but with $x'=1$ in your case.
What I get is
$$ \left(6 y (x^2+y^2)^2 \right) y' + \left( 6 x (x^2 + y^2)^2 \right) 1... | {
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How to calculate $\operatorname{taxicab}(3,8,2)$ (sum of 8 cubes in two different ways) Could someone explain me how I can calculate $\operatorname{taxicab}(3,8,2)$?
$\operatorname{taxicab}(3,8,2)$ is the smallest natural number that can be written in $2$ different ways as a sum of $8$ powers $3$.
For instance:
\begin... | $$\eqalign{132=1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 (or 1+1+1+1+1+1+1+125) =1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 (or 1+8+8+8+8+8+27+64)}$$
This can be found by a recursive computation.
Let $N(x,m)$ be the number of different ways to write $x$ as the (unordered) sum of $m$ positive cubes. Then $N(x,m) ... | {
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"url": "https://math.stackexchange.com/questions/1672292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$ Compute $$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$$
$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$
$k - 2 = t , k = t + 2$
$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^{t+2}}{3^{2t+1}} - \binom {n}{n} \frac{2^n}{3^{2n}}$$
Let's simplify thi... | Let $k-2=u$
$$=3\sum_{u=-1}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^{u+2}=3\left(\dfrac2{3^2}\right)^2\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u$$
Now,
$$\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u=\left(1+\dfrac2{3^2}\right)^n-\binom nn\left(\dfrac2{3^2}\right)^n=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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recursive function: non-recursive form possible? Can the following recursive function be converted to a non-recursive form?
$$f(x,c,\ell)=\frac{c-c^\ell}{1-c}+(c-2)\sum \limits_{k=1}^{\ell-1}f(x,c,k)$$
$$f(x,c,1)=c$$
$$c= \text{constant}$$
$$\ell=\text{length}$$
If so, where do I start?
| I use $n$ instead of $\ell$, hope you do not mind. Let's prove by (strong) induction that, for $n \geq 2$, it holds $f(c,n) = c^n - c(c-1)^{n-2}$. For $n=2$ it is clear, since $f(c,2) = c+(c-2)c = c^2-c$. If the assumption holds for every $2 \leq j \leq n$, then
\begin{align*}
f(c,n+1) &= \frac{c^{n+1}-c}{c-1} + (c-2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | Alternative Approach:
Let $x=\tan^2{\theta}$,$dx=2\tan{\theta} \sec^2{\theta} d\theta$
\begin{align}
I&=\int{\frac{\tan{\theta}\cdot 2\tan{\theta}\sec^2{\theta} d\theta}{ \sec^4{\theta} }}\\&=2\int{\sin^2{\theta}}d\theta\\&=\int{1-\cos{(2\theta)}}d\theta\\&=\theta-\frac12 \sin{(2\theta)}+C\\&=\arctan{\sqrt x}-\frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Integral inequality $\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx$? If $f: [0, 1] \to \mathbb{R}$ is a convex and integrable function with $f(0) = 0$, does it necessarily follow that$$\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx?$$
| For $0\le x\le \frac{1}{2}$, by convexity one has $$f\left(\frac{1}{2}\right)\le \frac{1}{2}\left[f\left(\frac{1}{2} + x\right) + f\left(\frac{1}{2} -x\right)\right].$$ Also, $$\int_0^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_{\frac{1}{2}}^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_0^{\frac{1}{2}}f\left(x+\frac{1}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to prove the trigonometric identity $\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$ $$\frac{1-\sin\varphi}{1+\sin\varphi}$$
I have no idea how to start this, please help. This problem is essentially supposed to help me solve the proof of
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi -... | If we solve the Pythagorean identity
$$\sin^2\varphi + \cos^2\varphi = 1$$ for $\cos^2\varphi$
we obtain
$$\cos^2\varphi = 1 - \sin^2\varphi$$
Since
$$\sec\varphi = \frac{1}{\cos\varphi}$$
and
$$\tan\varphi = \frac{\sin\varphi}{\cos\varphi}$$
we need to obtain $\cos^2\varphi$ in the denominator. We can do this by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Trying to find volume by rotating a region..getting a negative value I'm working on a review assignment for volumes of solids obtained by rotating a region bounded by given curves around a certain line (using washers or shells) and seem to be getting a negative answer for this one problem...
Find volume of solid obtai... | First of all, let us have a look at the region. The blue line is $y=-2$, and the region is clearly visible.
So we will have $x$ ranging over $1$ and the $x$ of the rightmost intersection of the curves. Let us find the intersections: $x-y=3$ or $y=x-3$, so subbing into $y^2=x-1$ we get $(x-3)^2=x-1$, or $x^2-6x+9=x-1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve the following system of equations in real $x$, $y$ solve for real $x,y$
$$2^{x^2+y}+2^{x+y^2}=8 \tag{1}$$
$$\sqrt{x}+\sqrt{y}=2 \tag{2}$$
Trivially $x=y=1$ Now Equation $(1)$ can be written as
$$2^{x^2+(\sqrt{y})^2}+2^{x+(\sqrt{y})^4}=8$$ so we get
$$2^{x^2+(2-\sqrt{x})^2}+2^{x+(2-\sqrt{x})^4}=8$$ so
$$2^{x^2+4+x... | We will show that the solution $(x,y)=(1,1)$ is unique.
$$2^{x^2+y}+2^{x+y^2}=8 \tag{1}\\ln(2^{x^2+y}+2^{x+y^2})=ln8⇒$$
We assume that $x^2+y>y^2+x$, (similarly you can assume that $x^2+y>y^2+x$ and reach the same result) and it follows that:
$$ln(2^{x^2+y}+2^{x+y^2})=\Big(ln2^{x^2+y}\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Sum of two squares and the rearrangement Let $a,b,c,d$ be four nonzero integers such that $ab=cd$. If $a^2+b^2\ne c^2+d^2$, then what is the minimal value of $|(a^2+b^2)-(c^2+d^2)|$? Surely it must be bigger than or equal to $1$, but I guess it is strictly bigger than $1$. How can I prove(or disprove) this conjecture?
| You can write $|a^2+b^2-c^2-d^2|=|a^2+2ab+b^2-c^2-2cd-d^2|=|(a+b)^2-(c+d)^2|$ and note that the difference of squares is never $1$ unless they are $0,1$. To have them $0,1$ we would need $c=-d, a=1-b$ but then we would need $-c^2=b-b^2$ or $b(b-1)=c^2$, which requires $c=0$
Added: WOLOG we can ask $a+b \gt c+d$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and... | Set $u=x^7$ and $du=7x^6dx$
$$=\frac 1 7 \int u \sqrt{u+1}du$$
Set $\nu=u+1$ and $d\nu=du$
$$\begin{align}
\frac 1 7\int(\nu-1)\sqrt{\nu}d\nu&=\frac 1 7\int\nu ^{3/2}d\nu -\frac 1 7\int\sqrt{\nu}d\nu\\\\
&=\frac{2}{35}(x^7+1)^{5/2}-\frac{2}{21}(x^7+1)^{3/2}+\mathcal C\\\\&=\color{red}{\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $p>3$ a prime number then $\binom {p-1}{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} 4^{p-1} \pmod {p^3}$ Here is one of Morley's theorem in number theory.
My idea is to begin in $\mathbb{Z}/p\mathbb{Z}$ :
$\binom {p-1}{\frac{p-1}{2}} = \frac{(p-1)!}{(\frac{p-1}{2})!(\frac{p-1}{2})!}=\frac{(p-1)!}{(-1)^{\frac{p-1}{2}}... | Just a partial answer, for now. By Wolstenholme's theorem we have $\binom{2p-1}{p-1}\equiv 1\pmod{p^3}$, and
$$\begin{eqnarray*} \binom{2p-1}{p-1} &=& \frac{\color{red}{(2p-1)\cdot (2p-3)\cdot\ldots\cdot(p+2)}}{\color{blue}{(p-2)\cdot(p-4)\cdot\ldots\cdot 1}}\cdot\binom{p-1}{\frac{p-1}{2}}\\ &=&(-1)^{\frac{p-1}{2}}\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to prove $\forall x,y\in\mathbb{R} : x^3+x^2y=y^2+xy \Leftrightarrow y=x^2\lor y=-x?$ Let $x,y\in\mathbb{R}$
Assume $x^3+x^2y=y^2+xy$
Then $x^2(x+y)=y(x+y)$
Then either $(x+y)=0$ or $(x+y)\ne0$
Assume $x+y=0$
Then $y=-x$
Assume $(x+y)\ne0$
Then $y=x^2$
Then $x^3+x^2y=y^2+xy \Rightarrow y=x^2\lor y=-x$
Now assume $... | This is not feedback on your proof, but instead an alternative proof, which seems a lot shorter and simpler.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which... | For small $x$ and $n>4$,
$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots+\sqrt{x^{n}}}}}=
x^{n/2^{n}}+\frac{1}{2}x^{1-n/2^{n}}+\frac{1}{8}x^{2-n/2^{n}}+\ldots$$
Thus $$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}} \approx
1+\frac{x}{2}+\frac{x^{2}}{8}$$
For large $x$,
$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}}=
\sqrt{2x}+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 2
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Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2... | Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 2
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How to evaluate the integral $\int_0^{2\pi}\mathrm{d}\theta e^{ia\cos(\theta-\theta_1)}\cos^2(\theta-\theta_2)$ I have an integral:
$$
\int_{0}^{2\pi}
\mathrm{e}^{\mathrm{i}a\cos\left(\theta - \theta_{1}\,\right)}
\,\,\,\cos^{2}\left(\theta - \theta_{2}\right)\,
\mathrm{d}\theta,
$$
where $a, \theta_1$ and $\theta_2$ a... | Let $\theta \mapsto \theta + \theta_1$, so that the integral becomes
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} \, \left [ 1+\cos{(2 (\theta+\theta_1-\theta_2))} \right ]$$
The first term produces
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} $$
Let $z=e^{i \theta}$; then the integral is
$$-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where:
$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
and $x$, $y$, $z$ are angles of a triangle.
I tried to use sine law but got nothing.
| Unless $x, y, z$ represent the angles of a triangle, there'll be infinitely many solutions.
If three sides of a triangle are given, it's better to use cosine law:
\begin{align*}
\cos x &= \frac{3^{2}+2^{2}-4^{2}}{2(3)(2)} \\
\cos y &= \frac{4^{2}+2^{2}-3^{2}}{2(4)(2)} \\
\cos z &= \frac{3^{2}+4^{2}-2^{2}}{2(3)(4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2B+\cos^2C-1=\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
$$x^2+2x\cos B\cos C+\cos^2B+\cos^2C-1=0$$
$$\iff x^2+x\{\cos(B+C)+\cos(B-C)\}+\cos(B+C)\cos(B-C)=0$$
Now use $y^2+(a+b)y+ab=(y+a)(y+b)$ and $\cos(B+C)=\cos(\pi-A)=-\cos A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find x for $\sqrt{(5x-1)}+\sqrt{(x-1)}=2$ Solve:
$$\sqrt{(5x-1)}+\sqrt{(x-1)}=2$$
When $x=1$, we get the following equation to equal to $2$
I've been trying to solve this problem but when I square both sides and simplify I end up with:
$$x^2+6x+2=0$$ and of course $x=1$ cannot be a solution. So im not sure what im doin... | Yes, squaring both sides gives:
$$ (5x - 1) + 2\sqrt{5x-1}\sqrt{x-1} + (x-1) = 4 $$
Then:
$$ (5x - 1) + 2\sqrt{(5x-1)(x-1)} + (x-1) = 4 $$
Which simplifies to:
$$ (5x - 1) + 2\sqrt{(5x^2-6x+1)} + (x-1) = 4 $$
And thus:
$$ \sqrt{(5x^2-6x+1)} = 3 - 3x $$
Now squaring both sides again:
$$ 5x^2-6x+1 = 9- 18x +9x^2 $$
Wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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how to show that $\tan^2 y=-\csc 2x$ If $\csc y=\sin x -\cos x$ how to show that $\tan^2 y=-\csc 2x$
Can anyone explain to me? What identity I should use?
| Question: Show $tan^2(y) = -csc(2x)$, given that $csc(y) = sin(x) - cos(x)$.
Identities/Knowledge:
*
*$csc(\theta) = \frac {1}{sin(\theta)}$
*$sin(2\theta) = 2sin(\theta)cos(\theta)$
*$cos^2(\theta) + sin^2(\theta) = 1$
*$tan(\theta) = \frac {1}{csc(\theta)cos(\theta)}$
Working Out:
Lets' begin with what the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Radius of convergence - Complex plane A example in my textbook explain that $x^2-2x+2=0$ has the solutions $x = 1± i$. The distance between $x=0$ and $x = 1± i$ is $\sqrt{2}$ in the complex plane. So the radius of convergence of the Taylor series for $(x^2-2x+2)^{-1}$ is $\sqrt{2}$ around $x=0$.
I looked for why this ... | Hint: Compute the partial fraction decomposition of $\frac{1}{x^2-2x+2}$ and consider the radius of convergence of the power series for $\frac{1}{1-x}$.
More precisely, we can write
$$
\frac{1}{x^2-2x+2}=\frac{A}{x-(1+i)}+\frac{B}{x-(1-i)}.
$$
Using standard techniques from calculus, one can find the constants $A$ and ... | {
"language": "en",
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"source": "stackexchange",
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Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $ , What is the exact value of $\sin(3x)$?
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $
What is the exact value of $\sin(3x)$?
What I have done:
Given $\tan(x) = 2\sqrt2 $ , I drew a right angled triangle and found the hypotenuse to be $3$... | Your strategy is correct, but you failed to take into account that $x \in \left[\pi, \frac{3\pi}{2}\right] \implies \sin x < 0$.
As for an alternative method:
We use the identity
$$\sin(3x) = 3\cos^2x\sin x - \sin^3x$$
Using the identity $\sec^2x = \tan^2x + 1$ yields
$$\sec^2x = (2\sqrt{2})^2 + 1 = 8 + 1 = 9 \impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| Hint: $x^4+x^2-2$ can be factored as $(x-1)(x+1)(x^2+2)$. then find $A,B,C$, and $D$ which are constants, such that
$$
\frac{x^2}{x^4+x^2-2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+2}
$$
and integrate termwise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the expected value of the largest of the three dice rolls?
You toss a fair die three times. What is the expected value of the largest of the three outcomes?
My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{... | Let X denote the largest value, then:
*
*$P(X=1)=\frac{ 1}{216}$
*$P(X=2)=\frac{ 7}{216}$
*$P(X=3)=\frac{19}{216}$
*$P(X=4)=\frac{37}{216}$
*$P(X=5)=\frac{61}{216}$
*$P(X=6)=\frac{91}{216}$
Hence the expected value is:
$$1\cdot\frac{1}{216}+2\cdot\frac{7}{216}+3\cdot\frac{19}{216}+4\cdot\frac{37}{216}+5\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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The right way to cancel out the terms in the following telescoping series So how do I cancel and simplify the terms in the following telescopic series.
Been at it for hours, cant seem to figure it out.
$\sum\limits_{k = 1}^n \frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)} $
Any help would be deeply appreciated.
P.S: I... | We have:
$$
a_n=\sum\limits_{k = 1}^n \left(\frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)}\right)= a_{n-1}+ \left(\frac{1}{2(n+1)} -\frac{1}{n+2}+\frac{1}{2(n+3)}\right)=a_{n-1}+\frac{1}{(n+1)(n+2)(n+3)}
$$
Does this solve your problem?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$
My Approach : I would assume that we can prove by induction.
Base case $n=2$.
$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$
$$n^2 =2^2 = 4.$$
Assume for $n\ge 2$, $n\ge 2$, $2{n \c... | We can also give this a combinatorial interpretation:
*
*$\binom{n}{2}$ is the number of unordered pairs of distinct elements of $\underline{n}$
*So $2\binom{n}{2}$ is the number of ordered pairs of distinct elements of $\underline{n}$
*So $2\binom{n}{2}+\binom{n}{1}$ is the number of ordered pairs of $\underline{... | {
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"question_score": "5",
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Number of common roots of $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ The equations $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ have
*
*exactly one common root;
*no common root;
*exactly three common roots;
*exactly two common roots.
I factored the first equation. I think the root... | $$x^3+2x^2+2x+1=(x+1)(x^2+x+1)$$
The roots are $-1,\omega,$ and $\omega^2$, where $\omega,\omega^2$ are non real cube roots of unity.
Substituting in the other equation,
$$(-1)^{200}+(-1)^{130}+1=1\ne0$$
$$\omega^{200}+\omega^{130}+1=\omega^2+\omega+1=0$$
$$(\omega^2)^{200}+(\omega^2)^{130}+1=\omega+\omega^2+1=0$$
Thus... | {
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"source": "stackexchange",
"question_score": "4",
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Prove the sequence converge and find its limit Consider the following sequence:
$a_{n+2} = \frac{a_{n+1}+a_{n}}{2}$, $a_{1}$ and $a_{2}$ are given.
Write $a_{n}$ as a function of $a_{1}$ and $a_{2}$ and show that its limit is $\frac{a1 + 2a_{2}}{3}$
I think I am loosing myself on algebra here. I can't even do the fir... | $A = \pmatrix{1/2&1/2\\1&0}$
$\pmatrix{a_{n}\\a_{n-1}}=A\pmatrix{a_{n-1}\\a_{n-2}}$
$\pmatrix{a_{n+2}\\a_{n+1}}=A^n\pmatrix{a_2\\a_1}$
$A = PDP^{-1}; A^n =PD^nP^{-1}$
$A = \pmatrix{1&1\\1&-2}\pmatrix{1&0\\0&-1/2}\pmatrix{2/3&1/3\\1/3&-1/3}$
$A^n = \pmatrix{1&1\\1&-2}\pmatrix{1&0\\0&-2^{-n}}\pmatrix{2/3&1/3\\1/3&-1/3}$
... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Sum of the Powers of $2$ Suppose I have a sequence consisting of the first, say, $8$ consecutive powers of $2$ also including $1$: $1,2,4,8,16,32,64,128$. Why is it that for example, $1 + 2 + 4 = 7$ is $1$ less than the next term in the series, $8$? Even if one was to try, for instance, $524,288 + 1,048,576$ ($2^{19}$ ... | If you think about the list of the positive integers written in base 2, it goes like this:
$$\begin{align*}
1&=1\cdot 2^0 &=1_2\\
2&=1\cdot 2^1 + 0\cdot 2^0 &=\textbf{10}_2\\
3&=1\cdot 2^1+1\cdot 2^0 &=11_2\\
4&=1\cdot 2^2+0\cdot 2^1+0\cdot 2^0 &=\textbf{100}_2\\
5&=1\cdot 2^2+0\cdot 2^1+1\cdot 2^0 &=101_2\\
6&=1\cdot ... | {
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"answer_id": 3
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What's wrong with my infinite series expansion for $\log(x)$? Here, log is natural log.
Looking at $f(x)=\frac{1}{x}$, I tried to put $f(x)$ in the form $\frac{a}{1-r}$ that an infinite geometric series $\sum_{n=0}^\infty (a \cdot r^n)$ converges to when $\mid r \mid < 1$.
That gave me $f(x) = \frac{1}{1-(1-x)}$, so th... | Your problem is from expanding all the terms in parentheses; while it's not wrong, it's obfuscating the point, which is to end up with an expression involving powers of $x - 1$.
So don't expand. Get
$$C + x - \frac{(1 - x)^2}{2} - \frac{(1 - x)^3}{3} - ...$$
I'll leave it to you to check that $C = -1$, so the series is... | {
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Solve the following trigonometric inequalities I'm solving the following inequality:
$$\cos(2x+\frac \pi 3) \ge - \frac 1 2 \text{ with } -\frac \pi 2<x<\frac \pi 2$$
I found the following range of solutions:
$$\frac 4 3 \pi \leq2x+\frac \pi 3 \leq \frac 2 3 \pi$$
That I simplified with
$$\frac \pi 3 \leq 2x\leq\pi \ri... | Let $\theta = 2x + \dfrac{\pi}{3}$.
Observe that the requirement that $x$ satisfies the inequalities
$$-\frac{\pi}{2} < x < \frac{\pi}{2}$$
implies that
$$\theta = 2x + \frac{\pi}{3} > 2\left(-\frac{\pi}{2}\right) + \frac{\pi}{3} = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \tag{1}$$
and that
$$\theta = 2x + \frac{\pi}{3... | {
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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am n... | You can also use Cauchy-Schwarz:
$$\left(x(2-x)+y(2-y)+z(2-z)\right)\left(\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\right)\ge (x+y+z)^2=1.$$
Furthermore,
$$\begin{aligned}x(2-x)+y(2-y)+z(2-z) &= 2(x+y+z)-(x^2+y^2+z^2)\\
&\le 2-\frac{(x+y+z)^2}{3}=\frac53.
\end{aligned}$$
Thus the minimum is $3/5$ attained at $x=y=z=1/3... | {
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"answer_id": 1
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Finding the error in an evaluation of the limit $\lim_{x\to0} \frac{e^x-x-1}{x^2} $ \begin{align}
\lim_{x\to0} \frac{e^x-x-1}{x^2}
&= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\
&= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\t... | The error starts here:
$$\lim_{x\to0} \frac{e^x-x-1}{x^2} $$ $$\color{red}{\neq\lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x}
=\dots}$$
See Basic Limit Laws
Hint: Applaying L'Hopital twice gives $\frac 1 2 \lim\limits_{x\to 0} e^x$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with complex numbers involving modules Let $z_1,z_2,z_3 \in \Bbb C$ so as $|z_1|=|z_2|=|z_3|=1$ and let $a=|z_1-z_2|, b=|z_2-z_3|, c=|z_3-z_1|$. Show, using algebraic methods, that:
$$ \frac {1} {(a+b-c)^2}+ \frac {1} {(b+c-a)^2}+ \frac {1} {(a+c-b)^2} \ge 1.$$
| This solution is a bit geometric. I'll use the fact that $a,b,c$ are the sides of a triangle inscribed in a circle of radius $1$. It's clear that $$a+b+c\le 3\sqrt3.$$ Now call $a+b-c=x$, $a+c-b=y$, $b+c-a=z$. We have that $x,y,z>0$ and $x+y+z=a+b+c$. By Cauchy-Schwarz and AM-HM
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z... | {
"language": "en",
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Prove: $\csc a +\cot a = \cot\frac{a}{2}$
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$
All I have right now, from trig identities, is
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
| We start with the following identities: $\quad\sin(2a) = 2\sin a \cos a\quad$ $\quad\cos(2a) = 1-2\sin^2 a\quad$
We solve these to get the half-angle identities: $\quad \sin(a) = 2\sin \frac a2 \cos \frac a2\quad$$\quad\sin^2 \frac a2 = \frac 12 (1-\cos a)$
We now tackle the problem
$$\frac{1}{\sin a} + \frac{\cos a}{\... | {
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Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$ I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed to derive the following series... | I was able to solve this problem on my own.
Using integration by parts,
$$\begin{align*}
&\; \int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx \\
&= -2\int_0^1 \frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx-\int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx \tag{1}
\end{align*}$$
I posted the solution to both these integrals o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When is matrix $A$ diagonalizable? I got the following matrix:
$$ A =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
I need to answer when this matrix is diagonalizable.
Its characteristic polynomial is $ t(t-a)(t-1) $. So its 3 eigenvalues are 0, 1 and ... | Your matrix has at least $2$ eigenvalues, namely $0$ and $1$, and maybe a third, namely$~a$ it it is different from those two others. In the latter case we have $3$ simple roots of the characteristic polynomial, and $A$ is automatically diagonalisable. So the remaining interesting case is $a\in\{0,1\}$.
In that case $A... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Three questions about the form $X^2 \pm 3Y^2 = Z^3$ and a related lemma In Ribenboim’s Fermat’s Last Theorem for Amateurs, he gives the following lemma [Lemma 4.7, pp. 30–31].
Lemma. Let $E$ be the set of all triples $(u, v, s)$ such that $s$ is odd, $\gcd(u,v) = 1$ and $s^3 = u^2 + 3v^2$. Let $F$ be the set of all pai... | To describe the solutions of the equation. $$x^2+qy^2=z^3$$
I think best would be to describe a solution using $3$ parameters.
$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^3)^3$$
$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$
$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find the general solution to differential equation $x(x+1)^2(y'-\sqrt x)=(3x^2+4x+1)y$ Equation can be transformed to linear differential equation:
$$LHS=x^3y'+2x^2y'+xy'-x^{7/2}-2x^{5/2}-x^{3/2}$$
$$RHS=3x^2y+4xy+y$$
$$\Rightarrow y'(x^3+2x^2+x)+y(-3x^2-4x-1)=x^{7/2}+2x^{5/2}+x^{3/2}$$
After dividing by $(x^3+2x^2+x),... | What seems to be interesting is to define $$y=z\, x\,(1+x)^2$$ which makes the differential equation to be $$x (x+1)^2 \,z'-\sqrt{x}=0$$ which is separable and "quite" simple to integrate. $$z=\int \frac{\sqrt{x}}{x (x+1)^2}\,dx$$ Make $x=t^2$ to get $$z=2\int \frac{dt}{\left(1+t^2\right)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^... | Brute force says that $(\sqrt{2011}-1)^2\leq n\leq (\sqrt{2011}+1)^2$. So you can use a calculator to find the lower and upper bound integer values: $1923\leq n\leq
2101$, do there are $2101-1923+1$ integer values $n$.
Algebraically, we see that $(\sqrt{2011}-1)^2 = 2012-2\sqrt{2011}$ and $(\sqrt{2011}+1)^2=2012+2\sqr... | {
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How do I find the solution to the equation $z^2=-81i$? This question is from the Powers of Complex Numbers, Precalculus section of KhanAcademy
Find the solution to the following equation whose argument is between $90°$ and $180°$
$$z^2=-81i$$
What I understand thus far:
I am going to set $r$ and $\theta$ to be the modu... | Write $-81i$ in trigonometric form:
$$
-81i=81\cdot(-i)=81\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)
$$
so by De Moivre its square roots are
$$
9\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)
$$
and
$$
9\left(\cos\left(\frac{3\pi}{4}+\pi\right)+
i\sin\left(\frac{3\pi}{4}+\pi\right)\right)
=
9\left(\cos\fr... | {
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"source": "stackexchange",
"question_score": "4",
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Prove $\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}} = \frac{(1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)} \pi \ \ \ \forall n \in \mathbb{N}$ My attempt starts with a contour integral in the half disk, I let the radius -> infinity and so the contour integral
\begin{equation}
\int_{\gamma} \frac{dz}{(1+z^2)^{n+1}} = 2 \pi i... | $$\begin{equation}
\frac{(2n)!}{(n!)^2 2^{2n}} = \frac{(1)(2)(3)...(2n)}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2} \times \frac{1}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2}
\end{equation}$$
Now, rewrite
$$1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2=2 \cdot 4 \cdot 6 \cdot ... \cd... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that the group is abelian Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.
I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\... | Pick an arbitrary
$$g = \begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix} \in G$$
In the quotient $G/Z(G)$, $g$ is a representative of the coset
$$gZ(G) = \left\{ \begin{pmatrix}
1 & x & y+ \tilde{y} \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}: \tilde{y} \in M \right\}$$
so you can choose another (mor... | {
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In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan ... | Rephrasing Mathematics's answer:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and $\cos(A+B)=\cdots=-\cos C$
$\cos^2A+\cos^2B+\cos^2C=1+\cos^2A-\sin^2B+\cos^2C=1-2\cos A\cos B\cos C$
If $\cos A\cos B\cos C=S$
$\iff\cos^2A+\cos^2B+\cos^2C=1-2S$
let $y=\tan A\tan B\iff y-1=-\dfrac{\cos(A+B)}{\cos... | {
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"url": "https://math.stackexchange.com/questions/1724897",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes when $b_1<b_2\le 4$, where $b_1,b_2$ are the numbers of objects in box... | Here is a very painstaking approach that may help you to see exactly what’s going on.
The possible values of $b_1$ are $0,1,2$, and $3$, so far starters we try
$$1+x+\frac{x^2}2+\frac{x^3}6$$
to account for $b_1$. Similarly, the possible values of $b_2$ are $1,2,3$, and $4$, so we try
$$y+\frac{y^2}2+\frac{y^3}6+\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A conjecture about traces of projections Let $M_n$ denote the space of all $n\times n$ complex matrices. Define $\tau:M_n\rightarrow \mathbb{C}$ by $$\tau(X)=\frac{1}{n}\sum_{i=1}^n x_{ii},$$ where of course $X=[x_{ij}]\in M_n$. Recall that a matrix $P\in M_n$ is called an "orthogonal projection" if $P=P^*=P^2$. Let $A... | I prove only $y\ge \frac x2(3x-1)$. We know that $A+B+C\ge 0$, so let $\lambda_1,\dots,\lambda_n$ be its eigenvalues. Then
$$
\frac 1n Tr(A+B+C)=\frac 1n \sum \lambda_i=\bar \lambda= 3x
$$
and
$$
Tr((A+B+C)^2)=\sum \lambda_i^2\ge n \bar \lambda^2=9nx^2.
$$
But
$$
(A+B+C)^2=A^2+B^2+C^2+AB+AC+BA+BC+CA+CB
$$
and so
$$
Tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorization of polynomials with degree higher than 2 I need help to factorize $x^4-x^2+16$. I have tried to take $x^4$ as $(x^2)^2$ and factorize it in the typical way of factorizing a quadratic expression but that did not help. Can someone help me to factor this and also introduce me to the procedure that i need to ... | Can be done by making perfect squares
$$
Let\ ax^2 +bx + c=0
\\Try\ to\ make\ the\ equation\ look\ in\ the\ form
\\ax^2 +2\sqrt {ca}x +c-(2\sqrt{ca}-b)x=0
\\You\ will\ see\ that\ ax^2 +2\sqrt {ca}x +c\ makes\ a \ perfect\ square
\\\therefore \quad ax^2 +2\sqrt {ca}x +c=(\sqrt ax + \sqrt c )^2
\\Thus\ the\ equation\ wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
| Let $s = \sum_{k=0}^{p}x^k$. Calculate $0 \leq a_1, a_2 < 5$, where
$$
a_1 \equiv s \pmod 2 \\
a_2 \equiv s \pmod 5
$$
$a_1$ is easy. If $x \equiv_2 0$, then $a_1 = 1$, else $a_1 \equiv_2 (1+p)$, cause it's $1 + 1^1 + 1^2 + 1^3 + \dots + 1^p$.
$a_2$ isn't harder, $x \equiv_5 0, 1$ are same. For $2, 3, 4$ you can easily... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cascading Summation $\sum_{i=1}^n\sum_{j=i}^n\sum_{k=j}^n \frac {i(j+2)(k+4)}{15} $ Evaluate
$$\sum_{i=1}^n\sum_{j=i}^n\sum_{k=j}^n \frac {i(j+2)(k+4)}{15} $$
Background
Many basic summation questions on MSE relate to a single index - it might be interesting to devise a question where the summand is a product of the ... | Our sum depends on three sums:
$$ S_1 =\!\!\!\!\sum_{1\leq i \leq j \leq k\leq n}\!\!\!ijk, \qquad S_2 = \!\!\!\!\sum_{1\leq i \leq j \leq k\leq n}\!\!\!ij, \qquad S_3 =\!\!\!\! \sum_{1\leq i \leq j \leq k\leq n}\!\!\!ik $$
that can be evaluated by using standard symmetry tricks. For instance:
$$ \left(\sum_{i=1}^{n} i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Determinant of $N \times\ N$ matrix So the question asks:
For $n \geq 2$, compute the determinant of the following matrix:
$$
B =
\begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdo... | Let $v= \begin{pmatrix} 1 \\ x \\ x^{2} \\ \cdot \\ \cdot \\ \cdot \\ x^{n-2} \\x^{n-1} \end{pmatrix}$.
Then $Bv = (x-X)v \iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\alpha}$, which are li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to prove $\sum_{i=1}^na_i> n- \frac{9}{4}$ Given that $a_1=\frac{1}{2}$ and $a_{n+1}=\sqrt[]{\frac{1}{3}a_n^{2}+\frac{2}{3}a_n}$ ,prove that
$$\sum_{i=1}^na_i> n- \frac{9}{4}$$Thanks.
| Let $f(x)=\sqrt{\frac{1}{3}(x^2+2x)}$, for $x\ge0$. Clearly $f$ is increasing and it is staightforward to check that
$$0\le x\le 1\Longrightarrow 0\le f(x)\le x\le 1$$
This alows us to prove that the sequence $(a_n)$ is increasing and bounded by $1$, so it must converge to the positive solution of the equation $f(x)=x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$ I have to find number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$. I guess it would be $[x^n](1+x+x^2 \dots)(1 + x^2 + x^4 \dots)^2$, but how to compute it? I know only that $\frac{1}{1-x} = 1+x+x^2 \dots$.
| Generating Function
The generating function is
$$
\begin{align}
\frac1{(1-x)\left(1-x^2\right)^2}
&=\frac1{(1-x)^3(1+x)^2}\\
&=\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\sum_{k=0}^\infty\binom{-2}{k}x^k\\
&=\sum_{j=0}^\infty\binom{j+2}{2}x^j\sum_{k=0}^\infty\binom{k+1}{1}(-x)^k\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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What can we do to solve the following equation with $6$ variables with some information provided?
Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3... | It is easy to see that $(a_2,\ldots,a_7)=(1,1,1,0,4,2)$ is a solution. Indeed, we have
$$
\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{4}{720}+\frac{2}{5040}=\frac{5}{7}.
$$
In which case we have $a_2+\cdots +a_7=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
Th... | For this to happen we need $|x+y+2|\geq |xy-1|$ or $x+y+2=0$.
If $x+y+2=0$ we get solutions $(a,-a-2),(-a-2,a)$.
otherwise we have $|x+y|+2\geq |xy|-1\rightarrow |x|+|y|+3\geq |x||y|$.
Notice that if $x$ or $y$ is $0$ it holds trivially.
We now find all possible values $n,m> 0$ so that $n+m+3\geq nm\iff \frac{n+3}{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Conditional Probability (dice) A die is rolled 7 times.
What is the probability that all outcomes are odd, given that the first outcome was greater than 3?
My approach: If the first outcome if > 3, then the dice rolled is either 4,5 or 6. So, we have a probability of $\frac{1}{3}$ to get a odd number.
So for the remai... | P(all outcomes are odd AND first outcome >3)=# (ways to get first outcome as 5 & remaining outcomes all odd)/n(S)=(1*3^6)/3*6^6=3^5/6^6. Regarding cardinality of the sample space, is my following logic correct? We want first outcome >3 which are 4,5,6 so there 3 possibilities for the first roll. For the remaining 6 rol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Lagrange Multiplier in 3D Find the minimum and maximum values of the function $f(x,y,z) = x+2y+3z$ where $(x,y,z)$ is on the sphere $x^2+y^2+z^2=1$ using Lagrange multiplier.
So I put them into the Lagrange form and got
$L(x,y,z,\lambda) = x+2y+3z+ \lambda(x^2+y^2+z^2-1)$
And you then get simultaneous equations
$L_x =... | $x+2y+3z\leq\sqrt {x^2+y^2+z^2}\sqrt {1+4+9} = \sqrt {14}$, by Cauchy-Schwarz, so $\max$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{1}{\sqrt{14}}$ and $\min$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = -\dfrac{1}{\sqrt{14}}$.
You should get a same answer if you solve the Lagrange equations correctly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the Min of P(x,y) Find the Minimum of the following function :
$$P(x,y) = \frac{(x-y)}{(x^4+y^4+6)}.$$
This is a math problem I found in an internet math competition but it is really complex to me !!!
| It is not difficult to see that the minimum is located at
$x<0$ and $y>0$, so by the symmetry, it suffices to consider the problem
$$\max P(x,y)\quad\mbox{s.t.}\quad x,y\geq 0.$$
Now we use the polar coordinate as below.
Let $x=r\cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and
$0\leq\theta\leq\frac{\pi}{2}$. Then
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.