Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Looking for closed-forms of $\int_0^{\pi/4}\ln^2(\sin x)\,dx$ and $\int_0^{\pi/4}\ln^2(\cos x)\,dx$ A few days ago, I posted the following problems
Prove that
\begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt]
-\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}... | Following the same approach as in this answer,
$$ \begin{align} &\int_{0}^{\pi/4} \log^{2} (2 \sin x) \ dx = \int_{0}^{\pi/4} \log^{2}(2) \ dx + 2 \log 2 \int_{0}^{\pi/4}\log(\sin x) \ dx + \int_{0}^{\pi /4}\log^{2}(\sin x) \ dx \\ &= \frac{\pi}{4} \log^{2}(2) - \log (2) \left(G + \frac{\pi}{2} \log (2) \right) + \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 6,
"answer_id": 4
} |
The diophantine equation $a^2+ab-b^2=0$ I first tried with brute force with $-1000 \leq a,b \leq 1000$ but found no solution.
But then a simple argument showed me that there was no solution. Not only in the integers, but even for the rationals. What was that argument?
| And why is it impossible to solve the equation? $a^2+ab-b^2=0$
$$a=\frac{-b\pm\sqrt{b^2+4b^2}}{2}=b\frac{-1\pm\sqrt{5}}{2}$$
you can see the whole decision no.
If $a^3+ab-b^2=0$ would have acted similarly. $b^2-ab-a^3=0$
$$b=\frac{a\pm\sqrt{a^2+4a^3}}{2}=a\frac{1\pm\sqrt{4a+1}}{2}=a\frac{1\pm{k}}{2}$$
$a$ - to choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Integrate $\int\frac{dx}{x\sqrt{x^2+1}}$ I would like to ask for some help regarding the following indefinite integral, tried integration by parts and trigonometric substitution which both brought me to $\int\frac{\sec\theta}{\tan\theta}d\theta$, and from this point it is messy to integrate by parts, any help would be ... | $\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}$
$\displaystyle z^{2}=x^{2}+1\Rightarrow 2zdz=2xdx\Rightarrow xzdz=x^{2}dx=
(z^{2}-1)dx$
$\displaystyle\Rightarrow \frac{dz}{z^{2}-1}=\frac{dx}{xz}$
$\displaystyle\int\frac{dx}{x\sqrt{x^{2}+1}}=\int\frac{dz}{z^{2}-1}=\frac{1}{2}\int(\frac{1}{z-1}-\frac{1}{z+1})dz$
$\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Let $p \geq 2$. Prove that if $2^p-1$ is prime then $p$ must also be prime. Would the following be a valid proof?
Let $r$ and $s$ be positive integers, then the polynomial $x^{rs}-1=(x^r -1)(x^{s(r-1)}+x^{s(r-2)}+\cdots+x^r+1)$.
So if $p$ is composite (say $rs$ with $1<s<p$), then $2^p-1$ is also composite (because it ... | By contrapositive, if $p$ is not prime, then $2^p-1$ is not prime. Since $p$ is not prime, is composite, so $p=nr$ with $1<r<p, 1<n<p$. Now $2^p-1=(2^r)^n-1$, use the change $x=2^r$ we get $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1).$$
Clearly $(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1)>1.$ Since $x=2^r \wedge r>1, 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/920991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum $1+\cos (x)+\cos (2x)+\cos (3x)+....+\cos (n-1)x$ By considering the geometric series $1+z+z^{2}+...+z^{n-1}$ where $z=\cos(\theta)+i\sin(\theta)$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+...+\cos(n-1)\theta$ = ${1-\cos(\theta)+\cos(n-1)\theta-\cos(n\theta)}\over {2-2\cos(\theta)}$
I've tried... | $$M=cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}M=\\\frac{2sin(\frac{\theta}{2})}{2sin(\frac{\theta}{2})}cos(\theta)+cos(2\theta)+cos(3\theta) +....+cos((n-1)\theta)=\\
\frac{(sin(\frac{3\theta}{2})-sin(\frac{1\theta}{2}))+(sin(\frac{5\theta}{2})-s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y... | Note we also have $x^2+y^2+z^2=2$ (expand $(x+y+z)^2=4$).
Note $y^2+z^2 \geq 2yz=2-2x(y+z)=2-2x(2-x)$, so
$x^2+y^2+z^2 \geq 3x^2-4x+2$
But $3x^2-4x+2 >2$ for $x>\frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have... | The quadratic (not quartic) equation of a real number $x$ referres to the monic quadratic polynomial that annihilates $x$. So, the equation $ax^2+bx+c=0$ should only be verified on $x = x_1$ and $x= x_2$.
As expected, making $x = x_1$ you should attain a zero regardless of $b$, you should do that calculation again.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Finding maxima values of a multivariate function
Find the maximal value of the function for $a=24.3$, $b=41.5$:
$$f(x,y)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$$
Using the second derivative test for partial derivatives, I find the critical point in terms of $a$ and $b$ by taking partial derivatives of $x$ and $... | Another way is to consider equivalently the maximum of
$$\frac{f^2}{a^2b^2}=\frac{x^2}{a^2} \frac{y^2}{b^2} \left(1-\frac{x^2}{a^2} -\frac{y^2}{b^2} \right)$$
which is the product of three positive terms with constant sum, so each term must be equal to (in this case) $\frac13$ at maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Known exact values of the $\operatorname{Li}_3$ function We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.
Known real analytic values for $\operatorname{Li}_3$:
*
*$\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$
*$\operatorname{Li}_3(0)=0$
*$\operatorname{Li}_3\left( \frac{1}{2} \right)... | It appears that
$$
\Im\left[\mathrm{Li}_3\left((-1)^{\frac{1}{2^n}}\right) \right]=\frac{(2^n-1)(2^{n+1}-1)}{3\cdot2^{3n+2}}\pi^3
$$
but that's by experimentation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
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show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
Prove $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$
Things I have done: after trying many ways and failing, I reached the fact that$\left(\frac{1}{2}\times\frac{3}{4}\time... | Note that $$\frac{n-1}{n}\gt\frac{n-2}{n-1}$$
for $n\gt 1$. So, we have $$\begin{align}\left(\frac 12\times \frac 34\times \cdots\times \frac{99}{100}\right)^2&\gt \left(\frac 12\times \frac 34\times\frac 56\times \cdots\times \frac{99}{100}\right)\left(\color{red}{\frac 12}\times \frac 23\times \frac{4}{5}\times\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to ... | Use
$$\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C.$$
Note that $$\int\frac{2}{x+2}dx=2\int\frac{(x+2)'}{x+2}dx$$
$$\int\frac{1-2x}{x^2+1}dx=-\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx=-\int\frac{(x^2+1)'}{x^2+1}dx+\int\frac{1}{x^2+1}dx.$$
Here, set $x=\tan\theta$ for $$\int\frac{1}{x^2+1}dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $ I'm looking for a closed form of this integral.
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$
where $\operatorname{Li}_2$ is the dilogarithm function.
A numerical approximation of it is
$$ I \appr... | My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.
First, from Landen's identity of the following form
$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 7,
"answer_id": 1
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Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| $$ \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + \dots \right)^2 = \left( 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 + O(x^4) \right)^2 =$$
$$= 1+ \dfrac{1}{4}x^2 + \dfrac{1}{64}x^4 + \dfrac{1}{256} x^6+1-\dfrac{1}{4}x^2 - \dfrac{1}{8}x^3 + \dfrac{1}{8}x^3 +\dfrac{1}{16}x^4 - \dfrac{1}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Rubiks Cube Landing on Red A Rubik’s cube in which each side is painted one of six colors (white, orange, red, blue, green and yellow).
Suppose each side of the Rubik’s cube consists of only one color, if the Rubik’s cube is tossed 6 times what is the probability that the cube will land on the red side at least 4 times... | Add up the following:
*
*The probability that it lands on the red side exactly $4$ times is $\dbinom{6}{4}\cdot\dfrac{5^2}{6^6}=\dfrac{375}{46656}$
*The probability that it lands on the red side exactly $5$ times is $\dbinom{6}{5}\cdot\dfrac{5^1}{6^6}=\dfrac{30}{46656}$
*The probability that it lands on the red si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck her... | Hint: Let be $u = x^2$, then $du = 2x\,dx$, and so you got
$$ \int\,x^3\sqrt{x^2+4}\,dx = \dfrac{1}{2}\int\,u\sqrt{u+4}\,du $$
and the let be $t = u+4$, so $u = t-4$, $dt = du$, and
$$ \dfrac{1}{2}\int\,u\sqrt{u+4}\,du = \dfrac{1}{2}\int\,(t-4)\sqrt{t}\,dt $$
you can calculate this integral by yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer wo... | Substitute $x=1/t,$ then our expression becomes $$\lim_{t \to 0} \frac{\sqrt{1+at^2+t^4}-\sqrt{1+bt^2+t^4}}{t^2}$$
Note that here we can apply the L'Hospitals rule.
$$\frac{\frac{1}{2}(1+at^2+t^4)^{-1/2}(2at+4t^3)}{2t} \to \frac {a}{2}$$
as $t \to 0.$
Therefore it easy to see that required limit is $$\frac{a-b}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to na... | Consider the expression
$$\frac{1}{x}+\frac{1}{b}+\frac{1}{c} = \frac{1}{x+b+c}.$$
This can be simplified to
$$(x+b+c)(x(b+c)+bc) - x(bc)=0.$$
This is a second degree polynomial in $x$. If we set $x=-b$, then this satisfies the given equation, hence $x+b$ is a factor of this polynomial. By symmetry $x+c$ is also a fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/943529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Equivalent of $ x(x+1)(x+2)\cdots(x+n)$? Assume $x>0$. Is there an equivalent for this quantity $$ x(x+1)(x+2)\cdots(x+n)$$ as $n$ tends to $+\infty$?
I've tried to write $$x(x+1)(x+2)\cdots(x+n)=x^{n+1}\left(1+\frac 1x\right)\left(1+\frac 2x\right)\cdots\left(1+\frac nx\right)$$ I don't know if I'm on the right track.... | I showed many years ago that
the closest $n$-th power
of an integer
to $x(x+1)...(x+n-1)$
for integer $x$
is $(x+[(n-1)/2])^n$
whenever $x$ is large enough
compared to $n$.
A sufficient condition is,
with $u = x+[(n-1)/2]$,
$u \geq (n^2-n)/4$
for even $n$
and
$u \geq (n^2+3n-4)/4$
for odd $n$.
Some more precise results... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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The maximum possible value of $ (xv - yu)^2 $ over the surface ... The maximum possible value of $ (xv - yu)^2 $ over the surface given by the equations
$ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $ is :
I solved it and my answer comes out to be $9$ but the correct answer is $36$.
Here is what I did :
Put $ x = r_1\cos\thet... | Another way :
$$(xv-yu)^2=(x^2+y^2)(u^2+v^2)-(xu+yv)^2=36-(xu+yv)^2\le 36.$$
The equality is attained when $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right).$
P.S. When $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right)$, the followings hold :
$$xu+yv=0,\ \ \ x^2+y^2=4,\ \ \ u^2+v^2=9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the second derivative of a function (implicit differentiation) I am not sure if I got the correct answer or not. It was a homework from my textbook, and it does not have answers for even number questions...
Original Function: $x^2y-4x=5$
$\frac{dy}{dx}$ = $\frac{4-2xy}{x^2}$
It seems I got to finding the dy/d... | i have also
$y'=\frac{4-2xy}{x^2}$
and by the quotient rule
$y''=\frac{(-2y-2xy')x^2-(4-2xy)2x}{x^4}$
plugging $y''$ in the equation above we obtain
$y''=\frac{\left(-2y-2x\left(\frac{4-2xy}{x^2}\right)\right)x^2-(4-2xy)2x}{x^4}$
simplifying this we get
$y''=\frac{6xy-16}{x^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
proportion of primes in a polynomial sequence It is conjectured (Bunyakovsky) that when $P(x)$ is a polynomial from $\mathbb{Z}[X]$, irreducible, with positive leading coefficient and so that the integers $P(n)$ , $n\gt0$ do not share a common factor different from $1$, then there are infinitely many primes in the seq... | Chinese remainder theorem.
Pick $R(x)$ with positive coefficients so that $$\begin{align}
R(x)&\equiv x^9(x-2)\pmod 3\\
R(x)&\equiv x^7(x-1)(x-2)(x-3)\pmod 5\\
R(x)&\equiv x^5(x-1)(x-2)(x-4)(x-5)(x-6)\pmod 7\\
R(x)&\equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)\\&\equiv x^{10}-1\pmod {11}
\end{align}
$$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
... | We have $2\sin x\cos2x=\sin2x=2\sin x\cos x\iff2\sin x(\cos2x-\cos x)=0$
$\sin x=0\implies x=m\pi\ \ \ \ (1)$ where $m$ is any integer
else $\cos2x-\cos x=0\iff\cos2x=\cos x\implies2x=2r\pi\pm x$ where $r$ is any integer
Taking the '+' sign, $x=r\pi$ which is same as $(1)$
Taking the '-' sign, $2x=2r\pi-x\iff x=\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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how many solutions to non-linear simultaneous equations I'm doing a Lagrange multiplier optimization problem, and I wound up with the following simultaneous equations:
$2x + 1 -2\lambda x = 0$
$4y-2 \lambda y = 0$
$6z-2 \lambda z = 0$
$-x^2 - y^2 - z^2 + 1 = 0$
I am asked to find ALL solutions, and I managed to find 2 ... | Since the 2nd and 3rd equation are contradictory when both y and z are not zero, I must try several scenarios where just y, just z, and both y and z are equal to 0. For the y=z=0 case:
Solve equation 1 for x:
$x = \frac{-1}{2-2\lambda}$
plug into equation 4, solve for lambda:
$\lambda = \frac{8 \pm \sqrt{64-48}}{8}$
Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate by partial fraction decomposition $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$
Here's what I have so far...
$$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$
$$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$
$$\mathrm A(x^2+2x+5) + \... | (Just working on the last integral, but not checking the partial fractions.) Complete the square:
$$\int \frac{dx}{x^2+2x+5} = \int \frac{dx}{(x+1)^2 + 4}$$
Now, substitute $x+1 = 2\tan(t)$. So, $dx = 2\sec^2 t\;dt$. Thus:
$$\begin{align}
\int \frac{dx}{(x+1)^2 + 4} &= \int\frac{2\sec^2(t)dt}{4(\tan^2t + 1)} \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Compute $\sum_{n=0}^\infty \frac{(n+1)^2}{n!}$ This is what I've done so far:
$$\sum_{n=0}^\infty\frac{(n+1)^2}{n!} = 1 +4 +4.5 + \frac 83 + \frac{25}{24} +\frac{3}{10} + \frac {49}{720} +\dots$$
I know I need to manipulate $\frac{(n+1)^2}{n!}=\frac{(n+1)(n+1)}{n(n-1)(n-2)}=$
| Using the standard Maclaurin series for $e^x$, we find that
$$xe^x=\sum_0^\infty
\frac{x^{n+1}}{n!}.$$
Differentiating we get
$$xe^x+e^x=\sum_0^\infty (n+1)\frac{x^n}{n!}.$$
Multiply by $x$, and differentiate again. We get
$$x^2e^x+3xe^x+e^x=\sum_0^\infty (n+1)^2\frac{x^n}{n!}.$$
Finally, set $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving quadratic equations in the field $F_5$ Let $y = x^2 + 2x + 2 = 0$.
Solve the equation in the field $F_5$.
So I used the common $b^2 - 4ac$ formula and got that $x$ is either $-1/2$ or $-3/2$ but I'm not sure if this is in the field...
| We have $a=1$, $b=2$, $c=2$, so $b^2-4ac=4-8=-4=1$. Moreover $2^{-1}=3$, so
$$
x=2^{-1}(-2\pm1)=3(-2\pm1)
$$
which means $x=3(-3}=-9=1$ or $x=3(-1)=-3=2$. The formula
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
indeed means
$$
x=(2a)^{-1}(-b\pm\sqrt{b^2-4ac})
$$
Check:
$$
1^2+2\cdot1+1=0,\qquad 2^2+2\cdot2+2=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \... | $$n^4+6n^3+11n^2+6n=n(n^3+6n^2+11n+6)=n(n+1)(n^2+5n+6)$$
$$=n(n+1)(n+2)(n+3)$$ which is a Product of $4$ consecutive integers
Now see The product of n consecutive integers is divisible by n factorial
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Matrix Problem about commutative multiplication The problem is:
"Find all $2\times 2$ matrices A that have the property that for any $2\times 2$ matrix B, AB = BA."
Given hint: "The given equation must hold for all B. Try matrices B that have lots of zero entries."
I tried bashing by letting $A=\begin{pmatrix}a & b\\c... | $$\begin{pmatrix}a & b\\c & d\end{pmatrix} \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}a & 0\\c & 0\end{pmatrix}$$
$$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix} \begin{pmatrix}a & b\\c & d\end{pmatrix}=\begin{pmatrix}a & b\\0 & 0\end{pmatrix}$$
So, you get $b=c=0.$
$$\begin{pmatrix}a & b\\c & d\end{pmatrix} \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of the locus of a point which has its sum of distance from (0,3) and (0,-3) equal to 8. I know the answer but due to various expressions under root, I'm unable to reach there.
Is there something which I'm missing to make the solution easier? By the way the answer given is $$\frac{x^2}{16} + \frac{y^2... | We start from
$$\sqrt{x^2+(y+3)^2}+\sqrt{x^2+(y-3)^2}=8.$$
Bring $\sqrt{x^2+(y-3)^2}$ to the other side and square. We get
$$x^2+(y+3)^2=64-16\sqrt{x^2+(y-3)^2}+x^2+(y-3)^2.$$
A little algebra brings us to
$$12y-64=-16\sqrt{x^2+(y-3)^2}.$$
Minor cancellation gives
$$4\sqrt{x^2+(y-3)^2}=16-3y.$$
Square. We get some can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $ How do we find $$\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $$
NB: It is not mandatory that $ax^2+bx+c$ has only a single root
| Hint:
$\int\dfrac{x\sin\sqrt{ax^2+bx+c}}{ax^2+bx+c}dx$
$=\int\dfrac{x}{ax^2+bx+c}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(ax^2+bx+c)^{n+\frac{1}{2}}}{(2n+1)!}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx(ax^2+bx+c)^{n-\frac{1}{2}}}{(2n+1)!}dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Compute $\displaystyle \lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$ $$\lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$$
Can I multiply the top and bottom by $3x$? If not, why not. I am having a hard time understanding when it is appropriate to multiply the numerator and denominator by th... | \begin{align*}\lim_{x\rightarrow -3}\ \frac{\frac{1}{3} + \frac{1}{x}}{3+x} & = \lim_{x\rightarrow -3}\ \frac{x+3}{3x(3+x)} \\&= \lim_{x\rightarrow -3}\ \frac{1}{3x} = \frac{-1}{9}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find transformation matrix which converts matrix to simple standard form I have a matrix A$$ \left( \begin{array}{ccc}
0 & 1 \\
a^2 & 0\\
\end{array} \right) $$
Using eigen values, I convert it into simple standard form B:
$$\left( \begin{array}{ccc}
a & 0 \\
0 & -a\\
\end{array} \right) $$
How can I find the t... | *
*find it's eigenvalues using $|\lambda I -A| = 0$ equation, you must find them $a$ and $-a$.
Hint: $\lambda^2-a^2 = (\lambda-a)(\lambda+a)$
*find their corresponding eigenvectors $e_1$ and $e_2$, using the criteria of eigen-value-vector
$\lambda_1 e_1 = Ae_1 $
$$a\left(\!
\begin{array}{c}
x \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Computing the limit of an expression I have the following question. Determine the limit of the following expression.
$\lim_{x \rightarrow 0} \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4}$.
My attempt to this question is the following. Let the function $f(x) = \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4} = \frac{\cos(x... | we get by simplifying your term
$\frac{(1+x^2)\cos(x\sqrt{2})-1)}{x^4(1+x^2)}$ and the taylor expansion for $x=0$ is $-\frac{5}{6}+O(x^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| More precisely, $\sqrt{a^2+4} = a + 2/a + O(1/a^3)$ as $a \to \infty$. In fact
it is easy to see that $$(a+2/a)^2 > a^2 + 4 > (a+2/a - 2/a^3)^2 \ \text{for} \ a > 1$$
so
$$a + \dfrac{2}{a} > \sqrt{a^2 + 4} > a + \dfrac{2}{a} - \dfrac{2}{a^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\s... | $$(\sqrt{3x}+\sqrt{2x})^2=17^2=289$$
$$(\sqrt{3x}+\sqrt{2x})^2 = 3x + 2x + 2\sqrt{6}x= (5+2\sqrt{6})x=289,$$
hence
$$x=\frac{289}{(5+2\sqrt{6})} = \frac{289 (5-2\sqrt{6})}{5^2-(2\sqrt{6})^2}
=289 (5-2\sqrt{6})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding the limit of ${x+\sqrt{x^2-4x+1}}$ at negative infinity I'm having trouble finding the following limit:
\begin{equation}
\lim_{x \to -\infty} {x+\sqrt{x^2-4x+1}}
\end{equation}
I tried to simplify it in many ways but couldn't get it to a form where I could evaluate the limit. How should I go about modifying thi... | $\displaystyle \lim_{x\to-\infty}x+\sqrt{x^2-4x+1}=\lim_{x\to-\infty}\frac{(x+\sqrt{x^2-4x+1})(x-\sqrt{x^2-4x+1})}{x-\sqrt{x^2-4x+1}}= \\ =\displaystyle \lim_{x\to-\infty}\frac{x^2-x^2+4x-1}{x-\sqrt{x^2-4x+1}}=\lim_{x\to-\infty}\frac{4-\frac{1}{x}}{1-\frac{|x|}{x}\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}=\frac{4}{2}=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Algorithm to find the coefficient of GCD linear combination? One of the properties of the GCD of two integers is that it can be written as the linear combination of the two, is there an algorithm that can be used to find the coefficients of this linear combination?
| The answer of @LonelyMathematician is perfect, but I confess that I have always found this way of describing the action just too confusing. Let’s step back a bit and look at what’s going on. In each step, you start with (dividend, divisor) and get a new pair, (newdividend, newdivisor), where the new dividend is the old... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. Please advice how to prove that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. By this I mean there exists no positive real $B$ such that for any natural $n$ $$-B <\sum\limits_{i=1}^n \cos \sqrt{i} < B$$
UPD: it looks like the sum is not bounded (it follow... | Given any smooth function $f(x)$ defined over $(1 - \epsilon,\infty)$, we can rewrite
its partial sum over $\mathbb{Z}_{+}$ as a Riemann-Stieltjes integral:
$$\sum_{k=1}^n f(k) = \int_{1^{-}}^{n^{+}} f(x) d\lfloor x \rfloor \tag{*1}$$
Let $\;B_n(x)\;$ be the $n^{th}$ Bernoulli polynomial and $\;P_n(x) = B_n(\{x\})\;$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
One Step in an Integration I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but ... | Let $$I(a)= \int \frac{1}{a+x^2} = \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}})$$
Then,
$$I'(a) = \int \frac{-1}{(a+x^2)^2} = d/da( \frac{1}{\sqrt{a}}\arctan(\frac{x}{\sqrt{a}})) $$
Find the expression in RHS and then put $a=u^2$. And you are done! :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result when I tes... | (as per David H's suggestion to use Integration By Parts)
Hypothesis: The two definite integrals $D_1$ and $D_2$ have the same value, i.e. $D_1 = D_2$.
Using integration by parts
$$
\int uv' \,d\theta= uv-\int u'v \,d\theta
$$
let us define
$$
u = \frac{1}{(1-a\cos\theta)^3}
\qquad
v' = \cos\theta
$$
So ($u'$ obtained... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Proving a palindromic integer with an even number of digits is divisible by 11 I'm in an introductory course for discrete math so I'm a novice at English proofs. I'm not sure if my reasoning here is valid or if I'm using modular arithmetic correctly. Specifically the line I marked with $(**)$. I would appreciate any fe... | You have some off-by-one errors, but you have the right idea. Note that:
\begin{align*}
p
&= (x_1 + x_2 10 + \cdots + x_n 10^{n-1}) + (x_{n} 10^n + x_{n-1} 10^{n+1} + \cdots + x_1 10^{2n - 1}) \\
&= \sum_{k=1}^n x_k 10^{k-1} + \sum_{k=1}^n x_k 10^{2n-k} \\
&= \sum_{k=1}^n x_k10^{k-1}(1 + 10^{2n - 2k - 1}) \\
&\equiv \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Evaluating Limit - $(1-\cos(x^2))/(x^3\sin(x))$ How would you go about evaluating the following limit as $x$ approaches $0$?
$$\lim_{x\to 0}\frac{1-\cos(x^2)}{x^3\sin(x)}$$
| $$\frac{1-\cos(x^2)}{x^3\sin(x)}=\frac{2\sin^2\left(\frac{x^2}{2}\right)}{x^3\sin(x)}=\frac{1}{2}\left(\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\right)^2\frac{x}{\sin(x)}$$
then use $\displaystyle \lim_{x\to 0}\frac{x}{\sin(x)}=1$, you get your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving $y^2 - yx - y + x = 0$ for $y$? I solved this equation for $y$ by inspection and confirmed it with Wolfram Alpha -
$y^2 - yx - y + x = 0$
I got the values $y = 1$ and $y = x$
However I was wondering is there a formal method for solving it? I expressed it as a polynomial -
$y^2 + (-1 - x)y + x = 0$
and used the ... | $$y^2-(x+1)y+x=0$$
First method - using the quadratic formula:
$$y_{1,2}=\frac{x+1\pm\sqrt{(x+1)^2-4x}}{2}=\frac{x+1\pm\sqrt{(x-1)^2}}{2}=\frac{(x+1)\pm(x-1)}{2}$$
Thus the solutions are $y=x$ or $y=1$.
Second method - taking common factors:
$$y^2-yx-y+x=y(y-x)-1(y-x)=(y-1)(y-x)=0$$
Thus the solutions are $y=x$ or $y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find Max Of $M=|z^{3}-z+2|$ Give a complex number $z$: $|z|=1$
Find Max $M=|z^{3}-z+2|$
Could someone help me solve this ?
| Let $z=x+yi$, hence $$\displaystyle \small \left|z^3-z+2\right|=\left|(x+yi)^3-(x+yi)+2\right|=\left|x^3+3x^2yi-3xy^2-y^3i-x-yi+2\right|= \\ \small =\left|(x^3-3xy^2-x+2)+i(3x^2y-y^3-y)\right|=\left|(x^3-3xy^2-x+2)+iy(3x^2-y^2-1)\right|$$
It is given that $|z|=1 \ \mathrm{thus\,} \ x^2+y^2=1 \rightarrow y^2=1-x^2$.
Set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.
| Expand $\left(a + \frac{1}{a}\right)^4$ by using foil, repeatedly using the famous result $(a+b)^2 = a^2 + 2ab + b^2$ (which is best in this case) or as I've done, you can use the binomial theorem assisted by pascal's triangle:
$$\left(a + \frac1{a}\right)^4 = \left(\sqrt{3}\right)^4\\
\\ a^4 + 4\cdot a^3\frac{1}{a} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
| Divide both sides of your inequality by (a+b+c), so you now have
$$bc+ac+ab\geq\frac {9abc}{(a+b+c)}$$
Now divide both sides by a so you get
$$\frac{(bc)}{a} + c +b \ge\frac {9bc}{(a+b+c)}$$
Next divide by b
$$\frac c a + \frac a b + 1 \ge \frac {9c}{(a+b+c)}$$
Finally, divide by c
$$\frac 1 a + \frac 1 b + \frac 1 c \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove this limit of $x^4$ using $\epsilon - \delta$ $\lim_{x\to a} x^4 = a^4$ using epsilon and delta.
We know $|x^4 - a^4| < \epsilon \space \text{such that} \space |x - a| < \delta$
So we have to find a $\delta$ for, which it works.
$|x^4 - a^4| = |x^2 + a^2||x-a||x+a|$
$|x^2 + a^2||x-a||x+a| < \epsilon \space \tex... | Aside from a couple of typos, it looks pretty good! Here's a cleaned up version of your proof.
Given any $\epsilon > 0$, let $\delta = \min\left(1, \dfrac{\epsilon}{(1 + 2|a|)(1 + 2|a| + 2a^2)}\right) > 0$. Then if $0 < |x - a| < \delta$, notice that:
\begin{align*}
|x^4 - a^4|
&= |x - a| \cdot |x + a| \cdot |x^2 + a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Determinant of a block matrix times the determinant of the inverse of a block Let $A$ be an $n\times n$ invertible matrix. Let $a$ be a number in $\mathbb{F}$, let $\alpha$ be a row $n$-tuple of numbers from $\mathbb{F}$ and let $\beta$ be a column $n$-tuple of numbers from $\mathbb{F}$. Show that $$
|A|^{-1}\left|\beg... | You can use another decomposition.
$$
\begin{pmatrix}
1 & -\alpha A^{-1} \\
0 & A^{-1}
\end{pmatrix}
\begin{pmatrix}
a & \alpha \\
\beta & A
\end{pmatrix}
=
\begin{pmatrix}
a - \alpha A^{-1} \beta & 0 \\
A^{-1} \beta & I
\end{pmatrix}
$$
Taking determinants on both sides gives
$$
|A|^{-1}
\begin{vmatrix}
a & \alpha \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we express the following ordinary generating function? I wish to express the following power series
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that $0< m \le n$
| In fact, the polynomial part $$\sum_{k=0}^n \binom{n-k}{m} x^k= \sum_{k=0}^{n-m} \binom{n-k}{m} x^k + \sum_{k=n-m+1}^n \binom{n-k}{m} x^k$$
$$=\sum_{k=0}^{n-m} \binom{n-k}{m} x^k$$
$$ =x^{n} \sum_{k=0}^{n-m} \binom{n-k}{m} x^{k-n} $$ $$=x^{n-m} [\sum_{i=0}^{\infty} \binom{m+i}{m} \frac{1}{x^i} - \sum_{i=n-m+1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of series with ^(n+1)
I want to check, if this series is convergent or not:
$$\sum\limits_{n=1}^\infty \frac{n^n(n!)}{(2n)!}$$
I tried with the ratio test:
$$\frac{\frac{(n+1)^{n+1}(n+1)!}{(2n+2)!}}{\frac{n^n*n!}{(2n)!} } =\frac{(n+1)^{n+1}(n+1)!(2n)!}{(2n+2)!(n!)n^n}=\frac{(n+1)^{n+1}(n+1)}{n^n(2n+2)(2... | \begin{align}
\dfrac{n^n n!}{(2n)!} &= \dfrac{n^n}{(n+1)(n+2) \cdots 2n} \\
&= \prod_{k=1}^n \dfrac{n}{n+k} \leq \prod_{k= \lfloor n/2\rfloor + 1}^n \dfrac{n}{n+k} \\
&\leq \left(\dfrac{n}{n + n/2}\right)^{n- \lfloor n/2\rfloor } = \left(\frac{2}{3}\right)^{n- \lfloor n/2\rfloor }\\ &\leq \left(\dfrac{2}{3}\right)^{n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
| Troubled by the quotient rule? Then it might be helpful to rewrite your question:
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{1 + 4\cos x}{2\sqrt{x+ 4\sin x}}\right) = \frac{1}{2} \left(\frac{\mathrm d}{\mathrm dx}(1 + 4\cos x)(x + 4\sin x)^{-1/2}\right)$$
Now, just use the product rule accompanied by the chain rule.
Try... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Equation $(a+b)^a=a^b$ How can we find the positive integer solutions to $(a+b)^a=a^b$?
Since $a+b>a$, it is necessary that $a<b$, otherwise the left-hand side is less than the right-hand side. So let $b=a+x$. The equation turns into $(2a+x)^a=a^{a+x}$, or equivalently, $\left(2+\dfrac{x}{a}\right)^a=a^x$.
| Your equation means that $(2+\frac xa)^a$ is an integer. This is only possible if $2+\frac xa$ is itself an integer. In particular, $x$ is divisible by $a$. Therefore we may write $b=ka$ for some integer $k\geq2$. Using this in the original equation, we obtain $$((k+1)a)^a = a^{ka}$$ or equivalently $$(k+1)^a = a^{(k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$
Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then:
$$
\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\tex... | Hint
As Integrator said, multiply.
Another simple way : consider the function $$e^{a x}=(2+b x)(\frac{1}{2}+\frac{x^2}{4}-cx^3)$$ Use the Taylor series for the lhs (just as you did) and expand the rhs. Now, identify as many terms as you can. You will then get the following equations $$ a-\frac{b}{2}=0$$ $$a^2-1=0$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ solve the differential equation.
$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$
The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.
Thanks in advance.
| This question is a bit of a nightmare for an entrance exam. Anyway, first observe that $$d(x^2 + y^2) = 2xdx + 2ydy$$ This suggests that using a variable $u = x^2 + y^2$ is useful, as we also have the RHS (right hand side)
$$RHS = \sqrt{\frac{a^2 - u}{u}}$$
The denominator on the LHS (left hand side) is slightly tricke... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Fibonacci divisibility Prove that the following holds: $3|F_n$ if and only if $4|n$
Base case for $n=1$:
$F_1$=1, so $F_1$ is not divisible by 3 and 1 is not divisble by 4.
So the proposition holds for $k=1$
Continue for n=2, n=3 and n=4.
Inductive step
Assume $3|F(4k)$. We now have to prove that the proposition also ... | Along the lines off mookid:
Lemma: In mod 3 arithmetic, for all $k \in \Bbb{N}$,
$$
\begin{array}{c}
F_{8k-7} \equiv 1 \\
F_{8k-6} \equiv 1 \\
F_{8k-5} \equiv 2 \\
F_{8k-4} \equiv 0 \\
F_{8k-3} \equiv 2 \\
F_{8k-2} \equiv 2 \\
F_{8k-1} \equiv 1 \\
F_{8k} \equiv 0 \\
\end{array}
$$
Proof: The basis is trivially esta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality with five variables Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$
Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is... | A proof for even $n$.
Let $a_i>0$, $a_{n+1}=a_1$ and $n$ is an even natural number. Prove that:
$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
Proof.
By C-S and AM-GM $\sum\limits_{i=1}^n\frac{a_i}{a_i+a_{i+1}}=\sum\limits_{k=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 2
} |
How find this inequality is $a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac)$
let $x,y,z\ge 0$ ,Assmue that
$$a^3+b^3+c^3+3xabc+y(a+b+c)\ge z(ab+bc+ac),\forall a,b,c\ge 0$$ if and only if
$$z^2\le \min{(16y,4y(1+x))}$$
My idea： $$\Longrightarrow $$
let $a=b=0,c>0$,then we have
$$c^3+yc\ge 0\Longrightarrow c(c^2+y)\ge ... | The necessary condition:
$a=b=c$ gives $a^2(1+x)-za+y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 4(1+x)y$
Further, $a=b, c = 0$ gives $2a^2-za+2y \ge 0, \: \forall a \in \mathbb R \implies z^2 \le 16y$, so together, we have $$z^2 \le 4y\;\min(4, 1+x).$$
The sufficient condition:
An easy way would be to use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $P(x)$ is a poly. of least degree and local Max. at $x=1$ and Local Min. at $x=3,$ Then $P'(0)$ is If $P(x)$ is a polynomial of least degree which has a local Maxima at $x=1$ and Local Minima at
$x=3.$ If $P(1)=6$ and $P(3)=2$. Then $P'(0)=$
$\bf{My\; Try::}$ Given function has one Maxima and one Minima So $P(x)$ mu... | Thanks Friends ,I have got it
My Solution:: Let $P'(x) = A\cdot (x-1)\cdot (x-3)\;,$ Then Integrate both side w. r. to $x\;,$ We Get
$\displaystyle P(x) = A\left[\frac{x^3}{3}-2x^2+3x\right]+\mathcal C\;,$ Now for calculation of $A$ and $\mathcal C\;,$ Put $x=1$ and $x=3.$
So $\displaystyle P(1) = \frac{4A}{3}+\mathcal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ Finding the closed form of:
$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$
where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$
It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{... | Different approach:
By Cauchy product we have
$$\operatorname{Li}^2_2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$
take $x=1/2$ and rearrange the terms to get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}=\frac12\operatorname{Li}^2_2\left(\frac12\right)+3\operatorname{Li}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim_{x\to-2}\frac{x+8}{x+3}=6 $ using $\epsilon-\delta$
Prove:
$$\lim_{x\to-2}\frac{x+8}{x+3}=6 $$
I started with:
$ |\frac{x+8}{x+3} - 6| < \epsilon => |\frac{x+8-6x-18}{x+3} | < \epsilon => |\frac{-5x-10}{x+3} | < \epsilon =>| \frac{-5(x+2)}{x+3} | < \epsilon $
From the definition of limit we know that:
$ ... | First let us restrict $x$ such that $\vert x+2 \vert < 1/2$, i.e., $-5/2<x<-3/2$.
This means we have $1/2<x+3<3/2$, i.e., $\dfrac23 < \dfrac1{x+3} < 2$
Hence, we have
$$\left \vert \dfrac{-5(x+2)}{x+3}\right \vert = 5 \left \vert \dfrac{x+2}{x+3}\right \vert < 10 \vert x+2 \vert$$
Hence, choose your $\delta = \min(1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The inequality $ abc\geq(a-b+c)(a+b-c)(b+c-a)$ holds for $a,b,c\geq 0$
What is the proof that:
$$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$
I tried :
we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$
then
$$a\ b\ c\geq(a-b+c)(a+b... | Let $a\geq b\geq c$.
Hence, $$abc-(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=$$
$$=(a-b)(a(a-c)-b(b-c))\geq0$$
because $a\geq b\geq0$ and $a-c\geq b-c\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the zeroes in a function I have come across a problem on my trigonometry homework where we need to find the zeros of a function without the use of a calculator.
The Equation:
Given one of the zeros is $x=5$
$f(x) = x^3+2x^2-23x-60$
Is there a more efficient way to go about this problem than simply factor it ou... | Yes if one of the roots is $5$ it means $x-5$ is factor of the given polynomial.Dividing the polynomial by $x-5$ we get the quadratic equation $x^2+7x+12$. Now the roots of this quadratic equation are $\frac{-7+(49-4*12)^{\frac{1}{2}}}{2}$ and $\frac{-7-(49-4*12)^{\frac{1}{2}}}{2}$. Thus the remaining two roots are $-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The Cosine and an Enigmatic Parabola In the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$, the cosine function superficially resembles an inverted parabola of the form $-ax^2+1$:
I wanted to know more and computed the $L^2$ norm (surrounding square root omitted for readability):
$$
\int_{-\pi/2}^{\pi/2} \,(\cos(x)-(ax^2+1... | These rather unexpected and large coefficients arise, in the calculation of the integral and of the successive minimization, only if we search an expression for $a$ where the numerator has no fractional terms and no "collected" factors: in fact, in this case we necessarily have to perform some multiplications that lead... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$ Can one help finding this limit
$$\lim_{x \to 0}\frac{x^{3}-\sin^{2}x\tan x}{\tan(\sin x) - \sin (\tan x)}$$
L'Hospital's rule is permited.
(Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$)
| Hint. Recall that, for $x$ near $0$, you have
$$\tan x = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\mathcal{O}(x^9)$$
$$ \sin x = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{17x^7}{720}+\mathcal{O}(x^9)$$ to obtain (using only relevant terms) $$\tan(\sin x) = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Derive Formula for Sine Half Angle I got stack with another problem (From the book: The Forgotten Art of Spherical Trigonometry):
Not having OC as 1 always creates me complicated formulas to define the rest. Once again looking for the smart track that I am missing.
| For ease of notation, let $\beta = \alpha/2$. Note that in the figure, $\triangle ABO \cong \triangle CBO$. Hence $$AO = CO = \cos \beta, \quad AB = CB = \sin \beta.$$ Since $DO/CO = \cos \alpha$, it follows that $DO = \cos \alpha \cos \beta$. Similarly, $CD/CO = \sin \alpha$, hence $CD = \sin \alpha \cos \beta$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute: $\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}$ Help me please to compute the limit of:
$
\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdot \left ( 1-\frac{2}{4} \right ) ^{\frac{4}{n}}\cdot \left ( 1-\frac{2}... | Let
$$f_n = \left ( 1-\dfrac{2}{3} \right ) ^{\dfrac{3}{n}}\cdot \left ( 1-\dfrac{2}{4} \right ) ^{\dfrac{4}{n}}\cdot \left ( 1-\dfrac{2}{5} \right ) ^{\dfrac{5}{n}}\cdot \cdot \cdot \left ( 1-\dfrac{2}{n+2} \right ) ^{\dfrac{n+2}{n}} $$
We have
$$\ln(f_n) = \sum_{k=1}^n \dfrac{k+2}{n}\ln\left(1-\dfrac2{k+2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ How can I prove that
$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$$
| The sum is equal to
$$\frac1{16} \sum_{n=0}^{\infty} \frac{(2 n+1)!}{2^{3 n} n !(n+1)!} = \frac1{16} \sum_{n=0}^{\infty} \frac{2 n+1}{n+1} \frac1{2^{3 n}} \binom{2 n}{n} $$
The sum may be written as
$$\frac1{8} \sum_{n=0}^{\infty} \frac1{2^{3 n}} \binom{2 n}{n} - \frac1{16}\sum_{n=0}^{\infty} \frac1{n+1} \frac1{2^{3 n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proving this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ by induction Where $n \in \mathbb{N}$
and
$$
F_n = \begin{cases}
0 & \text{ if } n = 0 \\
1 & \text{ if } n = 1 \\
F_{n-1} + F_{n-2} & \text{ if } n > 1
\end{cases}
$$
This is basically describing the famous Fibonacci sequence.
If we... | Your proof is completely incorrect. You need to prove why $$F_{n+2} F_{n} - F_{n+1}^2 = -(F_{n+1} F_{n-1} - F_{n}^2)$$ to prove that $F_{n+2} F_{n} - F_{n+1}^2 = (-1)^{n+1}$.
You can check the base case. For the inductive step, we have
\begin{align}
F_{n+2} F_{n} - F_{n+1}^2 & = (F_{n+1}+F_{n})F_{n} - F_{n+1}^2 & \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How does the recursion relation work in the solution to this differential equation (using series)? Sorry for the vague title but it would not let me post the first step and last step of this equation (too many characters!).
How does $$\dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} = \... | Use the fact that $\Gamma{(x+1)} = x \Gamma{(x)} $:
$$\begin{align}\Gamma{\left (n+\frac{2}{3} \right )} &= \left (n-\frac{1}{3} \right ) \Gamma{\left (n-\frac{1}{3} \right )}\\ &= \left (n-\frac{1}{3} \right ) \left (n-\frac{4}{3} \right )\Gamma{\left (n-\frac{4}{3} \right )}\\ &=\cdots\\&=\left (n-\frac{1}{3} \right ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Help finding the residue of $1/(z^8+1)$ Help finding the residue of $1/(z^8+1)$
I'm integrating over $\{ Re^{it} | 0 \leq t \leq \pi \}$, and I found 4 simple poles at $z_0=e^{in\pi/8}$ where $n = 0,...,3$ and I'm trying to calculate $res(1/(z^8+1),z_0)$
calculating this: $$\lim_{z\to z_0} (z-z_0)f = \lim_{z\to z_0}\f... | You don't need to factor $z^8+1$. The function $f(z)=\frac{1}{z^{8}+1}$ has 8 simple poles $z_{k}$, $k\in\left\{ 0,1,2\ldots ,7\right\} $, which are the zeros of the equation $z^{8}+1=0\Leftrightarrow z^{8}=-1$, i.e. the complex numbers $z_{k}=e^{i\left( \pi +2k\pi \right) /8}=e^{i\left(2k+1 \right)\pi /8}$ (and not $z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.
| The sum may be written as
$$\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1\cdot 4 \cdots (3k-2)}{5^k k!} $$
Now, based on a suggestion from @Robert Israel:
$$(1+t)^{-1/3} = 1 -\frac1{1!}\frac13 t + \frac{1}{2!}\left ( -\frac13 \right ) \left ( -\frac{4}{3} \right )t^2 - \cdots$$
so that
$$1-(1+t)^{-1/3} = \frac1{1!}\frac13 t -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
| Using Rearrangement inequality for $a,a^2$ and $b,b^2$
$$a^3+b^3\gt a^2b+b^2a$$
Similarly,
$$b^3+c^3\gt b^2c+c^2b$$
and
$$c^3+a^3\gt c^2a+a^2c$$
Adding these three, we have
Using sing AM-GM, we have
$$2(a^3+b^3+c^3)> a^2b+a^2c+b^2c+b^2a+c^2a+c^2b$$
$$\dfrac{a^2b+a^2c+b^2c+b^2a+c^2a+c^2b}{6}\ge\sqrt[6]{a^6b^6c^6}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$ I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?
$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$
Thanks in advance.
| Here I derive the rhs by simplification.
Use $\displaystyle \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ to write
$$\begin{align}\frac{\cos x - \sin x +1}{\cos x + \sin x -1}&=\frac{\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
calculation $f(x)$ from given expression $$f(x)+xf(-x)=x-2$$
what is $f(x$)?
I try to solve this problem, but I don't know how to remove $f(-x)$ or converting it to $f(x)$.
| After solving such task as in the above answers, you should always test it: you put $\frac{x^2 +3x - 2}{1+x^2}$ instead of $f(x)$ into the given equation:
$$\frac{x^2 +3x - 2}{1+x^2} + x\frac{x^2 -3x - 2}{1+x^2} =$$
$$= \frac{x^2 + 3x - 2 +x^3 - 3x^2 - 2x}{1+x^2} =$$
$$= \frac{x^3 - 2x^2 + x - 2}{1+x^2} =$$
$$= \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to find $\lim\limits_{x \to 0} \frac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$? How to find $\;\;\lim\limits_{x \to 0} \dfrac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$ ?
| We have
$$f(x) = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} \dfrac{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}$$
Hence,
$$f(x) = \dfrac{\sin(2x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{1+\tan(x) - 1 + \tan(x)} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the value of $a+b+c$?
What is the value of $a+b+c$?
if $$a^4+b^4+c^4=32$$
$$a^5+b^5+c^5=186$$
$$a^6+b^6+c^6=803$$
How to approach this kind of problem. Any help.
UPDATE: Thank you all for answers. Now I realize there is no integer solution. But is there any real number solution? I am curious to know.
| The lowest possible values for fourth powers of integers are $$0^4 = 0 \qquad 1^4 = (-1)^4 = 1 \qquad 2^4 = (-2)^4 = 16$$
By inspection, it becomes very clear that two of $a,b,c$ must have absolute value $2$, and one must be zero, in order for your first equation to hold.
Without loss of generality, suppose $c=0$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\int_0 ^{\pi}\left (\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$ How would you evaluate the integral
$$\int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$$
The answer from Wolfram is $0$.
Would you use a substitu... | Using the triple angle formula for sine and the power reduction formula for the square of cosine, we have:
$$\begin{align}
\sin{\left(3\theta\right)}
&=3\sin{\theta}-4\sin^3{\theta}\\
&=\sin{\theta}\left(3-4\sin^2{\theta}\right)\\
&=\sin{\theta}\left(4\cos^2{\theta}-1\right)\\
&=\sin{\theta}\left[4\left(\frac{1+\cos{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Indefinite integral with sector of ellipse An ellipse is given by the following equation:
$$
152 x^2 - 300 x y + 150 y^2 - 42 x + 40 y + 3 = 0
$$
After solving for the midpoint we have:
$$
152 (x-1/2)^2 - 300 (x-1/2) (y-11/30) + 150 (y-11/30)^2 = 1/6
$$
Introducing polar coordinates:
$$
x = 1/2 + r \cos(\theta) \quad ;... | The integral is elementary.
$$
I = \int \frac{d \theta}
{a \cos^2(\theta) - b \cos(\theta) \sin(\theta) + c \sin^2(\theta)}=
\int \frac{d \tan(\theta)}
{a - b \tan(\theta) + c \tan^2(\theta)}
$$
Let $u=\tan(\theta)$ , then:
$$
I = \int \frac{d u}{c u^2 - b u + a} =
\int \frac{d u/c}{\left[ u-b/(2c) \right]^2+a/c - \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\ph... | \begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\
&= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\
&= 4\sin^2(\theta + \phi) \\
&= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\
&= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 4
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Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ M... | You could have made it a little easier here:
$$
\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} = (n+1)^2 \left( \frac{n^2}{4}+\frac{4(n+1)}{4} \right) = \frac{(n+1)^2(n^2+4n+4)}{4} = \frac{((n+1)(n+2))^2}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Convergence test for the series $\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$
Determine convergence of the series
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n}}$$
My proof: using comparison test I have
$$\frac{\sqrt{n^2+1}-n}{\sqrt{n}} = \sqrt{n+\frac{1}{n}} - \sqrt{n} \to 0$$ for large $n$.
Th... | Since $$\sqrt {n^2 + 1} < n + \frac {1} {2n}$$ we have $$\frac {\sqrt {n^2 + 1} - n} {\sqrt n} < \frac {1} {2n \sqrt n}.$$ So the series is dominated by $$\sum_{n = 1}^{\infty} \frac {1} {2n \sqrt n} = \frac {1} {2} \zeta \left(\frac {3} {2} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrating trigonometric functions. I need a little help with solving the following two integrals. I am not able to approach these two problems correctly. Little hints will suffice. Thanks for your help!
*
*$$\int\frac{\tan^2x\sec^2x}{1+ \tan^6x}\,dx$$
*$$\int (\sin x \cos x)^{1/3}\,dx$$
| $1.$
$\int\dfrac{\tan^2x\sec^2x}{1+\tan^6x}dx$
$=\int\dfrac{\tan^2x}{1+\tan^6x}d(\tan x)$
$=\int\dfrac{1}{3(1+\tan^6x)}d(\tan^3x)$
$=\dfrac{\tan^{-1}\tan^3x}{3}+C$
$2.$
$\int(\sin x\cos x)^\frac{1}{3}dx=\int\dfrac{\sin^\frac{1}{3}2x}{2^\frac{1}{3}}dx$
Let $u=\sin^\frac{1}{3}2x$ ,
Then $x=\dfrac{\sin^{-1}u^3}{2}$
$dx=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}... | You differentiated $\sqrt{x^2-a^2}$ incorrectly. It looks like you forgot to apply the chain rule.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
| Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = \frac{1}{2}\int\frac{\sec x(\sec x+\tan x)+\sec x(\sec x-\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
$$\displaystyle = \frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx+\frac{1}{2}\int\frac{\sec x\cdot (\sec x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 5
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Evaluation of $2$ limit problems Evaluation of $(a)\;\;\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
and $(b)\;\; \displaystyle \lim_{x\rightarrow 0^{-}}\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+...............+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+... | Assuming $\;\lfloor x\rfloor\;$ is the floor function, we get that for $\;x\;$ very close to zero from the left,
$$\;\forall\,k\in\Bbb N\;,\;\;\lfloor x^k\rfloor=\begin{cases}-1&,\;\;k\;\;\text{is odd}\\{}\\\;\;\,0&,\;\;k\;\;\text{is even}\end{cases}\;$$ , and then:
$$\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+\ldot... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$ For any natural number $n$ ,let $S(n)$ denote the sum of the digits of $n$.Find the number of all 3 digit numbers $n$ such that $S(S(n))=2$
| First, by the nice answer/hint provided by mfl, the only way that $S(S(n)) = 2$ is if $a+b+c = 2$ or $11$ or $20$.
Next, suppose that $a$, $b$, and $c$ are digits with $a+b+c=2$ or $11$ or $20$. Then since,
$$100a+10b+c+9 = 100a + 10(b+1) + (c-1),$$
the new digits are either $a$, $b+1$, and $c-1$ (when $c>0$) or $a$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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A couple of definite integrals related to Stieltjes constants In a (great) paper "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" by Iaroslav V. Blagouchine, the following integral representation of the first Stieltjes constant $\ga... | Using contour integration, I am able to evaluate the first integral in terms of a slightly different infinite series which may or may not be able to be evaluated in closed form.
First notice that
$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1} &= \frac{1}{2} \int_{0}^{\infty} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
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Find the equation which has key root $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ In my last question which was Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ ,I could find the coefficients(were very easy) of fourth-degree equation, so I went to study the case which has key root $x=\sqrt{a}+\sqrt{b}... | If $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ then $(x-\sqrt{a})^2=(\sqrt{b}+\sqrt{c})^2$, that is, $\xi=\sqrt{\alpha}+\sqrt{\beta}$ with $\alpha=4ax^2$, $\beta=4bc$ and $\xi=x^2+a-b-c$.
The result in your previous question yields $\xi^4-2(\alpha+\beta)\xi^2+(\alpha-\beta)^2=0$, that is, $$(x^2+a-b-c)^4-8(ax^2+bc)(x^2+a-b-c)^2+16... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
| $$3\equiv1\pmod2\implies 1+2^n+3^n+4^n\equiv0\pmod2$$
If $n$ is odd, $1+4^n=1^n-(-4)^n$ is divisible by $1-(-4)=5$
Similarly, $2^n+3^n\equiv0\pmod5,$
For $n=4k+2,2^{4k+2}\equiv4\pmod5,3^{4k+2}\equiv-1,4^{4k+2}\equiv1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
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Solving the equation: $3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)$ Solving the equation:
$$3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)\tag{1}$$
I tried to write $(1)$ becomes
$$\sqrt{3}\sin \left(\frac{\pi}{3}-x\right)=\sin \left(\frac{\pi}{3}+2x \right)$$
Now, I have stuck :( , can you help me?
Thanks!
| Hint
$$\cos{x}(3-2\sin{x})=\sqrt{3}(2\sin{x}+1)(1-\sin{x})$$
$$\Longrightarrow \left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)^2(2\sin{x}+1)$$
(1):$$\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
| We'll show a bit more: $a^a b^b > \left( \frac{a+b}{2} \right)^{a+b} > a^b b^a.$
First we need a lemma: $P = (1+x)^{1+x} (1-x)^{1-x} > 1$ if $x < 1$.
Proof: $\ln P = (1+x)\ln (1+x) + (1-x) \ln (1-x) = x ( \ln (1+x) - \ln (1-x)) + \ln (1+x) + \ln (1-x) = 2x \left(x + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right) - 2 \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 10,
"answer_id": 4
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Linear transformation ker and image Let $\varphi\colon \mathbb{R}^4 \rightarrow \mathbb{R}^3$ be described by $\varphi(X)=AX$ where
$A=\begin{pmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{pmatrix} $
. Find base vectors of $\ker \varphi$ and $\operatorname{Im} \varphi$. In my opinion those vector... | Start with the definitions:
$$\ker\varphi:=\{X\in\mathbb R^4:\varphi(X)=0\},$$ that is, the set of all $(x_1,x_2,x_3,x_4)$ such that
$$
\begin{bmatrix}
3 & 2 & 1 & 3 \\
1 & 1 & 1 & 1 \\
2 & 1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}.
$$
Can you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
| \begin{align}J&=\int_0^\infty \frac{(x^2-1)\ln x}{1+x^4}dx\\
&=\underbrace{\int_0^\infty \frac{x^2\ln x}{1+x^4}dx}_{u=\frac{1}{x}}-\int_0^\infty \frac{\ln x}{1+x^4}dx\\
&=-2\int_0^\infty \frac{\ln x}{1+x^4}dx\\
K&=\int_0^\infty \frac{\ln x}{1+x^4}dx\\
R&=\int_0^\infty\int_0^\infty \frac{\ln(xy)}{(1+x^4)(1+y^4)}dxdy\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
How to compute the residue of $(z^2+2z+1)\sin\left(\frac{1}{1+z}\right)$ This was an example given in my notes but all it concluded was with something about an infinite principal part. How do we compute it? we have it equal to $ \left( z + 1 \right)^2 \cdot \sin \left( (z+1)^{-1} \right) = \left( z + 1 \right)^2 \cdot... | $z=-1$ is an essential singularity of the function. So we have $$(z^2+2z+1)\sin \frac{1}{1+z}=(z+1)^2\sin \frac{1}{1+z} $$ therefore $$(z+1)^2\sin \frac{1}{1+z}=(z+1)^2 \left(\frac{1}{1+z}-\frac{1}{3!(1+z)^3}+\frac{1}{5!(1+z)^5}-\cdots\right)$$ this shows that Res$[(z+1)^2\sin \frac{1}{1+z}, -1]=-\frac{1}{3!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$ Greatest value of $f(x)= (x+1)^{1/3}-(x-1)^{1/3}$ on $(0,1)$
Please guide me to solve this problem. I have differentiated it with respect to $x$ and make equal to zero, but couldn't get any point.
| By Jensen $$\left(\frac{\sqrt[3]{1+x}+\sqrt[3]{1-x}}{2}\right)^3\leq\frac{1+x+1-x}{2}=1,$$
which gives $\sqrt[3]{1+x}+\sqrt[3]{1-x}\leq2$.
The equality occurs for $x=0$.
Id est, the answer is $2$.
We used the following inequality.
For non-negatives $a$ and $b$ we have:
$$\left(\frac{a+b}{2}\right)^3\leq\frac{a^3+b^3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
| One way is to look at
$$
1+2x^2+3x^4+4x^6+5x^8+\dots
$$
as
$$
1+2t+3t^2+4t^3+5t^4+\dots
$$
where $t=x^2$. The last series is the derivative of
$$
1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t}
$$
Therefore,
$$
1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2}
$$
and
$$
1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2}
$$
Now, just multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x... | Note that if your equation includes $\sqrt{4x-1}$, this means that $4x-1\ge0$. Similarly, you are only looking for solutions among numbers that fulfill $x+2\ge0$, since the equation contains the expression $\sqrt{x+2}$.
So you can write $\sqrt{(4x-1)^2}=|4x-1|=4x-1$. But do not forget in the end include only those solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Surface Area (Integration) Find the surface area of the object by rotating $y = 4+3x^2$ about the $y$-axis, where $1 \leq x \leq 2$.
I've been using the formula:
Surface area:
Definite integral of $$2\pi y\Bigg(1+\Big(\dfrac{dx}{dy}\Big)^2\Bigg)dy$$
Limits are supposed to be in the form of $y$ for the above question.
... | Since $x=\sqrt{\frac{y-4}{3}}, \;\;\frac{dx}{dy}=\frac{1}{2}\left(\frac{y-4}{3}\right)^{-\frac{1}{2}}\cdot\frac{1}{3}\text{ and }\left(\frac{dx}{dy}\right)^2=\frac{1}{36}\frac{1}{\frac{y-4}{3}}=\frac{1}{12(y-4)}$.
Then $S=\displaystyle\int_7^{16} 2\pi\sqrt{\frac{y-4}{3}}\sqrt{1+\frac{1}{12(y-4)}}dy=2\pi\int_7^{16}\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| Hint: $\sqrt{x^2 + 1} - x$ can be written as $\sqrt{x^2 + 1} - \sqrt{x^2}$.
Generally, if you have an expression involving $\sqrt a - \sqrt b$ then it
is often useful to "expand" the expression with $\sqrt a + \sqrt b$:
$$
x \, (\sqrt{x^2 + 1} - x) = \frac {x \,(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)} {\sqrt{x^2 + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
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