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Proof involving polynomial roots From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit like they've been plucked out of thin air, in that I don't understand their motivation. So I'd be interested to see some alternate solutions/extra explanation.
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $a^4+a^3-1=0$ and $b^4+b^3-1=0$, but was unsuccessful.
|
I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $\,\ldots$
Following up on OP's approach, let $\,ab=p\,$ and, for symmetry, $\,a+b=s\,$. To begin with:
$$
\begin{cases}
\begin{align}
a^4 + a^3 - 1 &= 0 \\
b^4 + b^3 - 1 &= 0
\end{align}
\end{cases} \tag{1}
$$
Subtracting the two equations in $\,(1)\,$, and using that $\,a \ne b\,$:
$$\require{cancel}
\begin{align}
a^4-b^4+a^3-b^3 = 0 \;\;&\implies\;\; \cancel{(a-b)}(a+b)(a^2+b^2) + \cancel{(a-b)}(a^2+ab+b^2) = 0 \\
&\implies\;\; s(s^2-2p) + s^2 - p = 0 \\
&\implies\;\; s^3 + s^2 - (2s+1)p = 0 \tag{2}
\end{align}
$$
Adding the two equations in $\,(1)\,$:
$$
\begin{align}
a^4+b^4+a^3+b^3 - 2 = 0 \;\;&\implies\;\; \big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2\big) \\
&\quad\quad\quad\quad+\big((a+b)^3-3ab(a+b)\big)-2=0 \\
&\implies\;\; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \\
&\implies\;\; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 \tag{3}
\end{align}
$$
At this point, the equation satisfied by $\,p\,$ can be derived by eliminating $\,s\,$ between $\,(2)\,$ and $\,(3)\,$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $\,s^4+s^3=s \cdot (2s+1)p\,$ from $\,(2)\,$ into $\,(3)\,$:
$$
\begin{align}
s(2s+1)p -4ps^2-3ps+2p^2-2=0 \;\;&\implies\;\; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \\
&\implies\;\; ps^2+ps-p^2+1 = 0 \tag{4}
\end{align}
$$
Eliminating once again $\,s(s+1)\,$ between $\,(2)\,$ and $\,(4)\,$ by $\,s \cdot (4) - p \cdot (2)\,$ gives:
$$
\begin{align}
0 &= s \cdot \big(\cancel{ps(s+1)}-p^2+1\big) - p \cdot \big(\cancel{s^2(s+1)} - (2s+1)p\big) = (p^2+1)s+p^2 \tag{5}
\end{align}
$$
Then, substituting $\,s = \frac{-p^2}{p^2+1}\,$ back in $\,(4)\,$ gives in the end the sextic $\,p^6 + p^4 + p^3 - p^2 - 1 = 0\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?
| It seems easier to use $\log_{12}$ rather than $\log$.
$\log_{12}(36)=\log_{12}(3)+1=k$, and
$$2\log_{12}(2)+\log_{12}(3)=\log_{12}(12)=1,$$ so $2\log_{12}(2)=2-k$,
or $$\log_{12}(2)=1-\frac{k}{2}$$
Thus
$$\log_{24}(48)= \frac{\log_{12}(48)}{\log_{12}(24)}=\frac{2\log_{12}(2)+1}{\log_{12}(2)+1}$$
$$=\frac{2-k+1}{2-\frac{k}{2}}=\frac{6-2k}{4-k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find eigenvalues & eigenvectors for an integral. Can anyone please explain me how to solve it?
Find the nonzero eigenvalues and the corresponding eigenvectors:
$T:[-1,1]\rightarrow[-1,1]$
$$T((f(x))=\int_{-1}^1(x^2 y + y^2 x) f(y) \, dy$$
| Since $$Tf\left(x\right)=x^2\int_{-1}^{1} yf\left(y\right)dy+x\int_{-1}^{1} y^2f\left(y\right)dy$$
we have that $Tf\left(x\right)=ax^2+bx$ for some $a,b$. Hence, if $f\left(x\right)$ is an eigenvector it must take this form. Plugging this in we get $$Tf\left(x\right)=x^2\int_{-1}^{1} y\left(ay^2+by\right)dy+x\int_{-1}^{1} y^2\left(ay^2+by\right)dy$$
$$=x^2\int_{-1}^{1} \left(ay^3+by^2\right)dy+x\int_{-1}^{1} \left(ay^4+by^3\right)dy$$
$$=bx^2\int_{-1}^{1} \left(y^2\right)dy+ax\int_{-1}^{1} \left(y^4\right)dy$$
$$=\frac {2} {3} bx^2+\frac {2} {5} ax$$
and if this is an eigenvector with eigenvalue $\lambda$ then we find $\lambda a=\frac {2} {3} b$ and $\lambda b=\frac {2} {5} a$. Solving this for $\lambda$ and $a,b$ we find the eigenvalues and the eigenvectors in my other post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| Checking other answers:
Note that $\frac{x^2-1}{x}\ge 0;\frac{x-1}{x}\ge 0$ and $x>0$ imply $x>1$.
If $x^2=x+1 \ \ (1)$, then:
$$ x^2-1=x \Rightarrow \frac{x-1}{x}=\frac{1}{x+1},\\
{\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x \ \ \ \ \ \ \ \ \ \ (2) \ \ \ \ \ \Rightarrow \\
{\sqrt {(x+1)- 1\over x}} + {\sqrt{1\over x+1}} = \sqrt{x+1} \Rightarrow \\
1+\frac1{\sqrt{x+1}}=\sqrt{x+1} \Rightarrow \\
\sqrt{x+1}+1=x+1 \Rightarrow \\
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as,
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
After further simplification I got,
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$
Taking $-2^{99}$ common I got,
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So,
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
| $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= $$
$$2^{2x}-2^x + 2^{2x-2}-2^{x-1} + .... + 2^{2x-198}-2^{x-99}= $$
$$(2^{2x}+ 2^{2x-2} .... + 2^{2x-198})-(2^x +2^{x-1} + .... + 2^{x-99})= $$
$$2^{2x-198}(2^{198}+ 2^{196} .... + 2^{0})-2^{x-99}(2^{99} +2^{98} + .... + 2^{0})= $$
$$2^{2x-198}{2^{200}-1\over 2^2-1} -2^{x-99}{2^{100}-1\over 2-1}=0 $$
If we cancel this with $2^{x-99}(2^{100}-1)$ we get
$$2^{x-99}{2^{100}+1\over 3} -1=0 $$
So $$2^{x-99} = {3\over 2^{100}+1}$$ and thus $$x= 99+\log_{2} {3\over 2^{100}+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$ Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$
There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$
By trigonometry, we know the ratio of the triangle, that is the opposite to adjacent is $2t : 1-t^2$, and hence it follows the hypotenuse is $\sqrt{(2t)^2+(1-t^2)^2} = \sqrt{(t^2+1)^2} = (t^2+1)$
My question is, can the above square rooted answer be $-(t^2+1)$ also? Why do we reject the negative answer?
| Because $$\frac{2t}{1+t^2}=\frac{2\tan\frac{\theta}{2}}{\frac{1}{\cos^2\frac{\theta}{2}}}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\sin\theta.$$
You can use your formula, but we need to write before:
$$\frac{1-t^2}{1+t^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=\cos\theta.$$
Id est,
$$\sin\theta=\tan\theta\cos\theta=\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}=\frac{2t}{1+t^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confusing problem determining the exact formula for area of 'complex' shape Draw a square with length n on each side, and then draw arcs of 1/4 circle along the DIAGONAL LINES inside this square. So we have $4$ arcs of 1/4 circle inside this square. My question is, What is the exact formula for area of the region at the center of this square?. I don't know any methods to find this, and I gave up. I want to draw it here using latex but I don't know how to do it, I'm new here. I will be amazed if someone can solve this problem quickly.I wish the formula only contain n, constant pi, rational numbers,sin/cos/tan/log/in, roots of any degree.
| In case this is your diagram:
Think of the area in the middle as:
You have the twice the total area of the areas of two sectors, $FEC$ and $BEG$, less areas of $\triangle PGB$ and $\triangle FQC$, less the area of $\triangle PEQ$. Since the polygons are symmetric, you just have to solve the area of one each:
$$\frac12 A=2 A_{FEC}-2A_{\triangle FQC}-A_{\triangle PEQ}\tag{1}$$
WLOG, consider your square to lie on the $xy$ -plane, centered at $(0,0)$. Thus, your vertices will be a combination of the points $\left(\pm\frac n2,\pm\frac n2\right)$.
Consider the intersection between the two circles centered at $B,C$. Using the formula for circles, with $r=n$, we know they intersect at:
$$E=\left(0,\frac{1}{2}\left(\sqrt{3}n-n\right)\right)$$
The circles intersect the $x$-axis at the following points:
$$F=\left(-\frac{1}{2}\left(\sqrt{3}n-n\right),0\right)\\
G=\left(\frac{1}{2}\left(\sqrt{3}n-n\right),0\right)$$
For $O=(0,0)$, it's very clear that $FO=EO$, therefore $FE=FO\sqrt2$, and using the cosine law, we get that:
$$\left(\frac{n-\sqrt{3} n}{\sqrt{2}}\right)^2=-2 n^2 \cos (\angle FCE)+n^2+n^2\\
\angle FCE=\frac\pi6=30^\circ$$
Therefore, $$A_{FEC}=\pi n^2 \cdot \frac{30}{360}=\frac{\pi n^2}{12}$$
To find $A_{FQC}$, consider the line $EC:=\frac{1}{2}\left(\sqrt{3}-1\right)n-\sqrt{3}x$, which intersects the $x$-axis at $Q=\left(\frac{\left(\sqrt{3}-1\right)n}{2\sqrt{3}},0\right)$. Since we already know $F$, then $FQ=\frac{n}{\sqrt{3}}$. Using the distance formula tells us that $QC=\frac{n}{\sqrt{3}}$ thus $A_ {FQC} $ is :
$$A_{FQC}=\frac12\cdot n\cdot \frac{n}{\sqrt{3}}\cdot \sin(30^\circ)=\frac{n^2}{4 \sqrt{3}}$$
To find $A_{\triangle PEQ}$, notice that $PQ=2QO=\frac{\left(\sqrt{3}-1\right) n}{\sqrt{3}}$. Solving for the length of $EQ$, tells us that $\triangle PEQ$ is equilateral, therefore:
$$A_{\triangle PEQ}=\frac12 \left(\frac{\left(\sqrt{3}-1\right) n}{\sqrt{3}}\right)^2\cdot \sin \frac \pi3=\frac{1}{6} \left(2 \sqrt{3}-3\right) n^2$$
Going back to $(1)$, we now get that the area of the figure in the middle is:
$$A=\frac{1}{3} \left(-3 \sqrt{3}+\pi +3\right) n^2 \tag{2}$$
Consider the circle centered at $B:=\frac{1}{2}\left(\sqrt{3n^2-4nx-4x^2}-n\right)$, for $n=2$, the area in question, for $G=\frac{1}{2} \left(2 \sqrt{3}-2\right)$, should be:
$$A=4\int_0^G \frac{1}{2} \left(\sqrt{-4 x^2-8 x+12}-2\right) \, dx$$
Which evaluates to:
$$A=4\cdot\left(\frac{1}{3} \left(-3 \sqrt{3}+\pi +3\right)\right)$$
which is exactly what we get from $(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{(1+(xi)^2)}}\\
&=-x^{-1}+\arctan{xi}+C
\end{split}
Is this correct?
Thanks in advance!
| Note that, if $f(x)=\arctan(xi)$, then
$$
f'(x)=i\frac{1}{1+(xi)^2}=\frac{i}{1-x^2}
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-\frac{1}{x}-i\arctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
\frac{1}{x^2(1-x)^2}=
\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x-1}+\frac{d}{x+1}
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
\int\frac{1}{x^2(1-x^2)}\,dx=
\int\left(
\frac{1}{x^2}-\frac{1}{2}\frac{1}{x-1}+\frac{1}{2}\frac{1}{x+1}
\right)\,dx=-\frac{1}{x}+\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|+c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding integer solutions to $ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $ I was browsing through facebook and came across this image:
I was wondering if we can find more examples where this happens?
I guess this reduces to finding integer solutions for the equation
$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for integers a,b,c
Or can we even further extend to when they are all distinct that is finding solutions to
$$ \frac{a^3+b^3}{c^3+d^3} = \frac{a+b}{c+d} $$ for integers a,b,c
I don't really have that much knowledge in the number theory area so I have come here
| Let $S$ be the set of allowed values of $a$, $b$, and $c$ ($S=\mathbb{Z}$ in this OP's setting, but $S$ can be something else like $\mathbb{Q}$, $\mathbb{Q}_{>0}$, $\mathbb{R}$, or even $\mathbb{F}_p$, where $p$ is a prime natural number). If $b=-a$, then $c$ can be any number not equal to $-a$. That is, $(a,b,c)=(a,-a,c)$ with $c\neq -a$ is always a solution. From now on, we assume that $b\neq-a$.
Then, as invidid found, $$\frac{a^2-ab+b^2}{a^2-ac+c^2}=\left(\frac{a^3+b^3}{a^3+c^3}\right)\left(\frac{a+b}{a+c}\right)^{-1}=1\,.$$
Hence, $a^2-ab+b^2=a^2-ac+c^2$, or
$$(b-c)(b+c-a)=0\,.$$
That is, $b=c$ or $a=b+c$.
This concludes that all solutions $(a,b,c)\in S^3$ takes the form $(a,-a,c)$ with $c\neq -a$, $(a,b,b)$ with $b\neq -a$, and $(b+c,b,c)$ with $c\neq -\frac{b}{2}$. You can check that these indeed are solutions. The two solutions $(a,b,c)=(5,2,3)$ and $(a,b,c)=(579,123, 456)$ that you found are of the form $(a,b,c)=(b+c,b,c)$.
| {
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Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$ Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$
Things I have done so far:
$$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$
$$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\frac{1}{2(a-1)^2+7-2a}\leq \frac{1}{7-2a}$$
Then, I tried to use the UCT to solve this problem.
I created the new inequality:
$$\frac{1}{7-2a}\leq \frac{1}{5}+m.(a-1)(*)$$
with $0<a<3$. I needed to find "m" which make (*) always true.After that, I found $$m=\frac{2}{25}$$
However,
$$(*)\Leftrightarrow \frac{1}{7-2a}\leq \frac{1}{5}+\frac{2}{25}.(a-1)\Leftrightarrow 4(a-1)^{2}\leq 0$$
which is wrong with any $$a\in \mathbb{R} $$
Can you show me what my mistake is?
I hope you can have "smart" way to solve this problem.
Sorry, I am not good at English.
| $$\sum_{cyc}\dfrac{1}{(b+c)^2+a^2} \le \dfrac{2}{25}(a+b+c)+\dfrac{9}{25}=\dfrac{3}{5}$$
$$\dfrac{1}{(b+c)^2+a^2}=\dfrac{1}{(3-a)^2+a^2}\le \dfrac{2}{25}a+\dfrac{3}{25}\ , (a\ge 0)$$
| {
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"url": "https://math.stackexchange.com/questions/2874847",
"timestamp": "2023-03-29T00:00:00",
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Why does $L$ have to be lower triangular in the LU factorization? I was studying LU factorization, when I didn't understand a particular phrase, or rather, how it works or why it works.
During LU factorization L is said to be a lower triangular matrix and U is said to be an upper triangular matrix, for A=LU.
All my book says is: "Suppose A can be reduced to an echelon form U using only row replacements that add a
multiple of one row to another row below it. In this case, there exist unit lower triangular
elementary matrices $E_1, E_2..... E_p$ such that
$$
E_p....E_1A=U.
$$
Then
$$
A = (E_p...E_1)^{-1}U=LU,
$$
where
$$
L=(E_p...E_1)^{-1}.
$$
All the book says is the unit lower triangular matrix exists not mentioning why it exists or how it exists. A substantial amount of theory missing here. Could anyone please explain this?
| There is a slightly different LU decomposition called the Cholesky decomposition that creates two upper triangular matrices. The reason primarily is how it constructs the matrices. Gaussian elimination is simple to illustrate and I will show you. I recommend Trefethan and Bau personally.
Gaussian Elimination without Pivoting is as follows. The primary purpose of Gaussian elimination, if you follow this, is to find $\ell_{jk}$ which zeros out the row below. That is why it is the ratio of the two rows and then you subtract them. This continues on and on. Why does it have to be lower triangular? Because that is an identity matrix and it is only entering below the diagonal.
Suppose that
$$ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ A = LU $$
$$ U =A, L=I$$
$$ k=1,m=3,j=2$$
$$\ell_{21} = \frac{u_{21}}{u_{11}} = \frac{a_{21}}{a_{11}} = 3 $$
$$ u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3} $$
Then we're going to subtract 3 times the 1st row from the 2nd row
$$ \begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix} $$
Updating each of them
$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
$$k=1,j=3,m=3 $$
$$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2 $$
$$ u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3} $$
Then we add two times the first row to the third row
$$ \begin{bmatrix} -2 & 2 & 7 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 & 1& 1 \end{bmatrix} = \begin{bmatrix}0 & 4 & 9 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} $$
$$ k=2, j=3,m=3 $$
$$ \ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$
We're subtracting out little blocks
$$ u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3} $$
$$ \begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix} $$
Updating
$$ U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix} $$
It now terminates
$$ A = LU $$
$$ \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U} $$
A more specific reason is the generalization of how this works
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. If it's convergent with limit $a$, then $a^2=7a$, so either $a=0$ or $a=7$. We show that the sequence is monotonically increasing and bounded above by $7$ so its limit cannot be equal $0$, thus it is $7$.
Claim. $a_n < 7$ for all $n$.
Proof: Induction on $n$. The base is clear. Assume $a_{n-1} < 7$. This is equivalent to saying that $7a_{n-1} < 49$, which happens iff $\sqrt {7a_{n-1}} < 7$. Then $a_n=\sqrt{7a_{n-1}} < 7$.
It remains to show $a_n$ is monotonically increasing.
Consider
$a_n/a_{n-1}=\sqrt{7/a_{n-1}}$. We prove that this is greater than $1$. It will follow that $a_n > a_{n-1}$.
Since $a_n < 7$, $7/a_{n-1}> 1$, whence $a_n/a_{n-1} > 1$. Thus the sequence is increasing and bounded above, so it's convergent. It's limit cannot be zero, so it must be $7$.
Is this a correct proof?
| It all boils down to showing that $$0<a_n<7\implies a_n<\sqrt{7a_n}=a_{n+1}<7,$$ which is fairly obvious (geometric average).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying expression: $1 - \frac{1}{ (1 + a) / (1 - a)}$ Correct Answer is $a$
My attempt:
$$1 - \frac{1}{ (1 + a) / (1 - a)}= 1 - \frac{1 - a}{1 + a} $$
multiply $(1 + a)$ on num and dem for the first term.
$$=\frac{1 + a}{1 + a} - \frac{1 - a}{1 + a}$$
combine and subtract.
$$=\frac{2a}{1 + a}$$
How does this simplify to $a$?
| $$1 - \dfrac{1}{\dfrac{1 + a}{1 - a}} = 1 - \dfrac{1 - a}{1 + a} = \dfrac{1 + a}{1 + a} - \dfrac{1 - a}{1 + a} = \dfrac{(1 + a) - (1 - a)}{1 + a} = \dfrac{2a}{1 + a} \ne a;$$
I agree with user582949; our OP is right at the answer he/she was given is wrong.
| {
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Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy=0$
Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy$
This is not one of the standard forms and neither is it an exact form. How do I go about doing this question?
| $$
(3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0
$$
$$
\implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$}
$$
We know that:
$$
d\left(\frac{x^2}{y}\right) = \frac{2xydx - x^2dy}{y^2}
$$
Substituting this back into $(1)$:
$$
\implies 3x^2y^4dx + 2x^3y^3dy + d\left(\frac{x^2}{y}\right)y^2 = 0
$$
Assuming $y \not= 0$:
$$
\implies 3x^2y^2dx + 2x^3ydy + d\left(\frac{x^2}{y}\right) = 0 \tag{$2$}
$$
Also:
$$
d(x^3y^2) = 3x^2y^2dx + 2x^3ydy
$$
Substituting this into $(2)$:
$$
\implies d(x^3y^2) + d\left(\frac{x^2}{y}\right) = 0\\
\implies x^3y^2 + \frac{x^2}{y} = c
$$
I don't know if this is formally legal in mathematics (I only have high school mathematics under my belt), but this is how I would have done it.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please correct me If I have gone wrong anywhere?
| Using the Mean Value Theorem, the following statement holds:
For $f(z) = 2z\tan^{-1}{z}$
$$f'(x+\xi)=\frac{2(x+2)\tan^{-1}{(x+2)}-2x\tan^{-1}{x}}{x+2-x}$$
For some $\xi \in (0,2)$
The left hand side simplifies to:
$$f'(x+\xi)=2\tan^{-1}{(x+\xi)}+\frac{2(x+\xi)}{1+(x+\xi)^2}$$
The right hand side simplifies to the limit you seek.
By taking the limit of the left hand side, the limit is clearly $\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Need some help finding the equation (nth term) of a sequence The sequence goes like this
Sequence = 1 , 4 , 14 , 58 , 292
Each number is being multiplied by (n+1) and then 2 is added to them.
Should i proceed to find the equation from
$ 2(n+1)^n $
this? . Thank you very much in advance!
| An exponential isn't quite right here, since you are multiplying by a different amount each step. At the first step, $n+1=2$, while at the second step, $n+1=3$, etc. It seems like factorials will be a much better fit for this sequence, since they have the same sort of "multiply by an increasing sequence of numbers behavior. Let's just play around a bit with the first few terms:
$$
1 = 1
$$
$$
4 = 2(1) + 2
$$
$$
14 = 3(2(1) + 2) + 2
$$
$$
58 = 4(3(2(1) + 2) + 2) + 2
$$
Let's take 58 as out example, and expand it out.
We get: $58 = 1(4\cdot3\cdot2) + 2(4\cdot3) + 2(4) + 2$
If we take a bigger example, like 292, we get: $292 = 1(5\cdot4\cdot3\cdot2) + 2(5\cdot4\cdot3) + 2(5\cdot4) + 2(5) + 2$. This expression is just the same as the previous one, but we've multiplied all the terms by 5, and then added another 2 onto the end. I'm noticing that all of the terms here begin with a 2, except for the first one. So let's just rewrite things so the first term also starts with a 2, to make it a bit neater.
$$
292 = 2(5\cdot4\cdot3\cdot2) + 2(5\cdot4\cdot3) + 2(5\cdot4) + 2(5) + 2 - 5!
$$
Here, I've both added and subtracted $5!$, and combined the positive one into the first term. Now we have: $
292 = 2X - 5!
$ where $X = (5\cdot4\cdot3\cdot2) + (5\cdot4\cdot3) + (5\cdot4) + (5) + 1$
Now, if we can just manage to find $X$, we'll be done. Let's rewrite $X$ as the following:
$$
X = \frac{5!}{1!} + \frac{5!}{2!} + \frac{5!}{3!} + \frac{5!}{4!} + \frac{5!}{5!}
$$
I'm noticing the Taylor series for $e$ in that series of fractions. The series for $e$ is $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + ...$.
Therefore: $5!(e-1) = X + \frac{5!}{6!} + \frac{5!}{7!} + ...$
Notice that that the terms $\frac{5!}{6!} + \frac{5!}{7!} + ...$ are equal to $\frac{1}{6} + \frac{1}{42} + ...$. In other words, their sum is less than $1$. So we could legitimately write the following expression for $X$:
$$
X = \left\lfloor{5!(e-1)}\right\rfloor
$$
Now we did all this for the specific example of the $5^{th}$ term, 292, but we can also make a general formula out of this by replacing the 5s with $n$s.
For $X$, we have: $X = \left\lfloor{n!(e-1)}\right\rfloor$.
So my proposed formula is:
$$
t_n = 2\left\lfloor{n!(e-1)}\right\rfloor - n!
$$
Obviously you still need to prove that this works, and taking the floor to find $X$ might fail for small $n$, meaning you'll need to add some special cases. But this gives you some idea of how one might go about finding a candidate formula in the first place.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple limit with asymptotic approach. Where's the error? Simply calculus question about a limit.
I don't understand why I'm wrong, I have to calculate
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}
$$
Using asymptotics, limits and De l'Hospital rule I would write these passages...
$$ = \lim_{x \rightarrow 0} \frac{x \, (2 - \sqrt[3]{8 - x^2})}{x^3/2}
= \lim_{x \rightarrow 0} \frac{\frac{2}{3}\frac{\sqrt[3]{8 - x^2}}{8-x^2}x}{x}
= \frac{1}{6}
$$
But the answers should be $\frac{5}{6}$. Thank you for your help.
| You are just replacing $\sin x$ by $x$ and $\cos \sqrt{x^3}$ with $1-(x^3/2)$. Both these replacements are wrong for the very simple reason that $\sin x\neq x$ and $\cos \sqrt{x^3}\neq 1-(x^3/2)$ unless $x=0$.
The right approach is to use the standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =1,\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{1}$$ Using above limit and L'Hospital's Rule it can be easily proved that $$\lim_{x\to 0}\frac{x-\sin x} {x^3}=\frac {1}{6}\tag{2}$$ Another limit which is needed here is $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{3}$$ We can evaluate the limit in question using limits $(1), (2)$ and $(3)$ as follows
\begin{align}
L&=\lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{1-\cos\sqrt{x^3}}\notag\\
&= \lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{x^3}\cdot\frac{x^3}{1-\cos\sqrt{x^3}}\notag\\
&= 2\lim_{x\to 0}\frac{2x-2\sin x+\sin x(2-\sqrt[3]{8-x^2}) } {x^3}\text{ (using (1))}\notag\\
&=2\left(\frac{1}{3}+\lim_{x\to 0}\frac{\sin x} {x} \cdot\frac{2-\sqrt[3]{8-x^2}}{x^2}\right)\text{ (using (2))}\notag\\
&= \frac{2}{3}+2\lim_{t\to 8}\frac{8^{1/3}-t^{1/3}}{8-t} \text{ (putting }t=8-x^2) \notag\\
&= \frac{2}{3}+2\cdot\frac{1}{3}\cdot 8^{-2/3}\text{ (using (3))}\notag\\
&=\frac{5}{6}\notag
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find two $2\times2$ real matrix $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$ Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$
Tried to write the matrices as
$$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g&h}$$
and solve $(A+B)^{-1}=A^{-1}+B^{-1}$, But make it too complex. Any more convenient ways?
| Using the formula from
Inverse of the sum of matrices
we can take, for arbitrary $a,b$ with $a+b\neq 0$,
$$
A=\begin{pmatrix} -1 & 1 \cr 0 & 2 \end{pmatrix},\quad
B=\begin{pmatrix} a & b \cr \frac{2(a^2-a+1)}{a+b} & \frac{-2(b-ab+1)}{a+b} \end{pmatrix}.
$$
All of $A$, $B$ and $A+B$ have determinant $-2$. It is easy to check that $(A+B)^{-1}=A^{-1}+B^{-1}$. So your way is not too "complex".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How does $\frac{x + \frac{1}{2}}{\frac{1}{x} + 2} = \frac{x}2$? How does simplifying $\dfrac{x+\frac12}{\frac1x+2}=\dfrac{x}2$
Plugging and chugging seems to prove this, but I don’t understand the algebra behind it.
How would you simplify $\dfrac{x+\frac12}{\frac1x+2}$ to get $\dfrac{x}2$?
I tried multiplying the expression by $x/x$, but that left me with $\frac{x^2+\frac{x}2}{1+2x}$
Is the fact that $\frac1x$ and $2$ are the reciprocals of $x$ and $\frac12$ respectively of any significance?
Thanks!
| If $x \neq \frac{-1}{2}$, then $\frac{(x+1/2)}{(1/x+2)} = \frac{x^2+x/2}{1+2x} = \frac{2x^2+x}{2+4x}$. This is multiplying by 2 on top and on the bottom. Note that we can write $x = \frac{2x+4x^2}{2+4x}$. Therefore $\frac{x}{2} = \frac{x+2x^2}{2+4x}$, and the expressions are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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$112x\equiv392\pmod{91}$ has $7$ solutions? True or false? $112x\equiv392\pmod{91}$ has exactly $7$ solutions.
A necessary and sufficient condition for a linear congruence equation to have solution says that $\gcd(112,91)\mid392$, and the solutions are $\gcd(112,91)=7$. Since $7\mid392$ then the statement is true.
Is it correct? Am I supposed to find the general solution $(x=10+91k,\;0\leq k<7)$?
Thanks!
| Since $91=13\cdot 7$ we have that
$$112x\equiv392\pmod{91} \iff 21x\equiv 28\pmod{91}$$
is equivalent to
$$21x\equiv 28\pmod{7} \iff 0\equiv 0 \pmod{7}$$
$$21x\equiv 28\pmod{13} \iff 8x\equiv 2\pmod{13}\iff 4x\equiv 1\pmod{13}$$
and for the latter since $10\cdot4-3\cdot13=1$ we have
$$10\cdot 4x\equiv 10\cdot 1\pmod{13}\iff x\equiv 10\pmod{13}$$
therefore the $7$ positive solutions less than $91$ are
$$x\in\{10,23,36,49,62,75,88\}$$
| {
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"url": "https://math.stackexchange.com/questions/2894452",
"timestamp": "2023-03-29T00:00:00",
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Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$ By the Binomial expansion for integer powers,
$$ (A+B)^n = \sum_{k=0}^n {n\choose{k}} A^{n-k} B^{k}$$
(I'm assuming $A,B\ge 0$) and so we get the easy estimate $A^n + B^n \le (A+B)^n$ for any positive integer $n$.
Now what happens when we take $n$ to be non-integer? Well if $0<n\le1$, then we have the reverse inequality,
$$(A+B)^n \le A^n + B^n ,\qquad {(0<n\le 1)} $$
What about if $n>1$ is a non-integer? Can we still say $A^n + B^n \le (A+B)^n$? I'm pretty sure the answer is yes, but I wanted to check with you guys. Here is my quick argument. Let $n = p + \alpha$ where $p \in \mathbb{N}$ and $0<\alpha<1$.
\begin{align} A^n+B^n &= A^\alpha A^p + B^\alpha B^p \\ &\le \text{max}(A^\alpha,B^\alpha) (A^p+B^p) \\ &\le \text{max}(A^\alpha,B^\alpha) (A+B)^p \\ &\le (A+B)^\alpha (A+B)^p \\ &= (A+B)^n\end{align}
| Alt. hint: $\;\displaystyle \left(\frac{A}{A+B}\right)^n + \left(\frac{B}{A+B}\right)^n \le \frac{A}{A+B}+\frac{B}{A+B}=1\,$ when $\,n \ge 1\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}}}{x+\sqrt x^2 \sqrt \frac{x+\frac{1}{x}}{x+1}} \\
= {} & \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt x^2 \sqrt \frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}} = \frac{x+1}{x+ x \sqrt 1} = \frac{x+1}{2x} \to \frac {1}{2}
\end{align}
I used this for symplifying:
*
*${(x-\sqrt \frac{x^3+x}{x+1})} \cdot {{(x+\sqrt \frac{x^3+x}{x+1})}} = x^2 -\frac{x^3+x}{x+1}=> (a-b)(a+b) = a^2-b^2$.
*$\sqrt \frac{x^3+x}{x+1} = \sqrt {x^2\cdot\frac{({x+\frac{1}{x}})}{x+1}} = \sqrt x^2 \sqrt\frac{x+\frac{1}{x}}{x+1}$.
I didn't write everything formally correct but hopefully you'll get the idea. The problem I see is that sometimes I'm applying $\lim\limits_{x \rightarrow \infty}$ to just the lower part of a fraction, when I probably have to apply it to both parts of the fraction? I'm talking about this part especially: $\lim\limits_{x \rightarrow \infty}\frac{x+1}{1-\frac{1}{x}} \sim x+1$.
As always very grateful for any comments/help. Cheers.
| another solu tion
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt \frac{x^3+x}{x+1}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ x^2 - x +2 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ (x - 1/2)^2+7/4 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x- x +\frac12 )) = \frac {1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim _{x\to \infty \:}|\frac{x+1}{x-2}|^{\sqrt{x^2-4}}$ I need help solving the following tough limit , please.
$$\lim _{x\to \infty} \left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}$$
| Since eventually $x-2>0$ we have
$$\lim _{x\to \infty \:}\left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}=\lim _{x\to \infty \:}\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}$$
then we have
$$\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}=\left(\frac{x-2+3}{x-2}\right)^{\sqrt{x^2-4}}=\left[\left(1+\frac{3}{x-2}\right)^{\frac{x-2}3}\right]^{\frac{3\sqrt{x^2-4}}{x-2}}\to e^3$$
indeed by $y=\frac{x-2}3 \to \infty$
$$\left(1+\frac{3}{x-2}\right)^{\frac{x-2}3}=\left(1+\frac{1}{y}\right)^{y}\to e$$
and
$$\frac{3\sqrt{x^2-4}}{x-2}=3\cdot \frac x x \frac{\sqrt{1-4/x^2}}{1-2/x}\to 3\cdot 1\cdot 1 =3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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p is prime. Show that there is at most one $(m,n) \in \mathbb Z^2$ with $0I've been trying to solve this question without great success.
I've found that the problem is correct for $p \equiv 1 \mod 4$.
Thank you,
| Let $p=2$, then $p=1^2+1^2$. So assume $p$ is an odd prime s.t. $p=a^2+b^2$, hence neither $a$ nor $b=0$. Now $a$ has a multiplicative inverse $c$ modulo $p$: $ac\equiv1\pmod{p}$.
$$(ac)^2+(bc)^2\equiv1+(bc)^2\equiv0\pmod{p}\implies (bc)^2\equiv-1\pmod{p}$$
and so $-1$ is a quadratic residue modulo $p$. Let $bc\equiv d$ then
$$d^2+1\equiv0\pmod{p}\tag{1}$$
Now $\gcd(d,p)=1$ so there exists a unique solution to the congruence
$$dx\equiv y\pmod{p}\tag{2}$$
where $0<|x|<\sqrt{p}$ and $0<|y|<\sqrt{p}$. To see this let $m=\lfloor\sqrt{p}\rfloor$, and consider the $(m+1)^2$ numbers $jd+k$, where $1\le j,k\le m+1$. Since $(m+1)^2>p$, at least two of the numbers are in the same residue class, say
$$x_1d+y_1\equiv x_2d+y_2\pmod{p}\implies (x_1-x_2)d\equiv (y_2-y_1)\pmod{p}$$
where $x_1\neq x_2$ or $y_1\neq y_2$. Since if one of the terms is zero, so is the other, then neither is, and we have $x=x_1-x_2$, $y=y_2-y_1$.
Multiply $(1)$ by $x^2$ and use $(2)$ to give
$$(d^2+1)x^2=(dx)^2+x^2\equiv y^2+x^2\equiv0\pmod{p}\tag{3}$$
Now $0<x^2+y^2<2p$, implying $x^2+y^2=p$.
To discover which primes can be written as the sums of two squares we have that if $\gcd(a,p)=1$ then Euler's criterion gives
$$(-a|p)\equiv(-a)^{(p-1)/2}\equiv(-1)^{(p-1)/2}a^{(p-1)/2}\equiv(-1)^{(p-1)/2}(a|p)\pmod{p}$$
from which $(-a|p)=(-1)^{(p-1)/2}(a|p)$ (here $(a|p)$ is the Legendre symbol). This gives the results:
If $p=4n+1$, $(p-1)/2$ is even, so
$$(-a|p)= (a|p)$$
If $p=4n+3$, $(p-1)/2$ is odd, so
$$(-a|p)=-(a|p)$$
This shows an odd prime $p$ can be written as the sum of two squares if and only if it is of the form $4n+1$ (for then $(-1|p)=(1|p)=1$, and we need $-1$ a quadratic residue modulo $p$).
We now show these are the only numbers that can be represented in this way.
By the Brahmagupta-Fibonacci Identity has it that if two numbers can be written as the sum of two squares then so can their product:
$$(a^2+b^2)(d^2+c^2)=(ad+bc)^2+(ac-bd)^2$$
Now if $N=m^2k$ where $k$ is square free, then $N$ can be written as the sum of two squares if and only if $k$ contains no factors of the form $4n+3$. If $k$ has no such factors we can write $k=f^2+g^2$, and so $m^2k=(mf)^2+(mg)^2$ is the sum of two squares.
Conversely if $N=a^2+b^2=m^2k$, and $\gcd(a,b)=d$ then $d^2\mid m^2$. Let $a=a_1d$, $b=b_1d$. Then
$$a_1^2+b_1^2\equiv0\pmod{k}\tag{4}$$
Now $\gcd(a_1,b_1)=1$, so if an odd prime $p\mid k$ then one of $a_1$, $b_1$ is prime to $k$. Say $\gcd(a_1,k)=1$. Then there exists a multiplicative inverse $c$ s.t.
$$ca_1\equiv1\pmod{p}$$
and so by $(4)$
$$(a_1c)^2+(b_1c)^2\equiv1+(b_1c)^2\equiv0\pmod{p}\implies (b_1c)^2\equiv-1\pmod{p}$$
implying $-1$ is a quadratic residue modulo $p$. Hence $p$ is of the form $4n+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.
| Set
$$ I = \int \frac{\sin x}{\sin x - \cos x} dx = \int 1 + \frac{\cos x}{\sin x - \cos x} dx$$
Therefore:
$$ 2I = \int 1 + \frac{\sin x +\cos x}{\sin x - \cos x} dx $$
$$ 2I = x + \log(\sin x - \cos x) + C$$
$$ I = \frac{x}{2} + \frac{1}{2} \log(\sin x - \cos x) + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
$\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
Clearly, we can see that $$T_r=\frac{n-r+1}{r(r+1)(r+2)}$$
Now, somehow, we have to make this telescoping.
But we do not have $n$ as a factor of the denominator, so we have to multiply the denominator by $n$ and after our manipulation, multiply the whole sum by $n$. But that too did not work.
PLease help
| We have that
$$T_r=\frac{n-r+1}{r(r+1)(r+2)}=\frac{n+1}{r(r+1)(r+2)}-\frac{r}{r(r+1)(r+2)}=$$$$=(n+1)\left(\frac{1}{2r}+\frac{1}{2(r+2)}-\frac{1}{r+1}\right)+\frac1{r+2}-\frac1{r+1}$$
and by telescoping we can see that the first term gives
$$(n+1)\left(\frac12+\frac14+\frac12\cdot 2\sum_{r=3}^n\left(\frac1r\right)+\frac12\frac1{n+1}+\frac12\frac1{n+2}-\sum_{r=2}^n\left(\frac1r\right)-\frac1{n+1}\right)$$
$$(n+1)\left(\frac{1}{4} -\frac12\frac1{n+1}+\frac12\frac1{n+2}\right)$$
$$(n+1)\left(\frac{(n+1)(n+2)-2(n+2)+2(n+1)}{4(n+1)(n+2)}\right)$$
$$\frac{n^2+3n}{4(n+2)}$$
and the second one
$$\frac1{n+2}-\frac12$$
therefore
$$\sum_{r=1}^n T_r=\frac{n^2+3n}{4(n+2)}+\frac1{n+2}-\frac12=\frac{n^2+3n+4-2(n+2)}{4(n+2)}=\frac{n^2+n}{4(n+2)}$$
| {
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Induction step in proof that $\binom{s}{s} + \binom{s + 1}{s} + \cdots + \binom{n}{s} = \binom{n + 1}{s + 1}$ Prove by induction that (binomial theorem)
$$
\binom{s}{s} + \binom{s+1}{s} + \dotsb + \binom{n}{s} = \binom{n+1}{s+1}
$$
for all $s$ and all $n>s$.
I used base case $s=0$, and I got my base case to work. However when I go to my induction step, I can’t seem to get it to work.
| Alternatively, note that:
$${n\choose m}={n\choose n-m}; \ \text{so}:\\
\begin{align}&\binom{s}{s} + \binom{s+1}{s} + \binom{s+2}{s} +\binom{s+3}{s} +\dotsb + \ \ \ \binom{n}{s} \ \ \ \ = \binom{n+1}{s+1} \iff \\
&\color{red}{\binom{s}{0} + \binom{s+1}{1}} + \color{blue}{\binom{s+2}{2}} +\color{green}{\binom{s+3}{3}} +\dotsb + \mathbf{\color{purple}{\binom{n}{n-s}}} = \binom{n+1}{n-s}.\end{align}$$
Now we get:
$$\begin{align}\color{red}{\binom{s}{0} + \binom{s+1}{1}}&=\binom{s+2}{1};\\
\binom{s+2}{1} + \color{blue}{\binom{s+2}{2}}&=\binom{s+3}{2};\\
\binom{s+3}{2} + \color{green}{\binom{s+3}{3}}&=\binom{s+4}{3};\\
\vdots\\
\binom{n}{n-s-1} + \mathbf{\color{purple}{\binom{n}{n-s}}}&=\binom{n+1}{n-s} \ \ \ \ \ \ \ \ \ (\text{inductive hypothesis});\\
\binom{n+1}{n-s} + \binom{n+1}{n-s+1}&=\binom{n+2}{n-s+1} \ \ (\text{inductive step}).\end{align}$$
| {
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prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
| By definition of the limit
$$f(x)-3=\frac{x^3-1}{x-1}-3=x^2+x-2=(x-1)^2+3(x-1)$$
\begin{align}
|f(x)-3|
&= |(x-1)^2+3(x-1)| \\
&= |x-1|^2+3|x-1| \\
&<\delta^2+3\delta \\
&= \varepsilon
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve polynomial equation with real parameter Solve the equation $x^4-(2m+1)x^3+(m-1)x^2+(2m^2+1)x+m=0,$ where $m$ is a real parameter.
My work: So far I've been able to factor the polynomial to $(-x^2+x+m)(-x^2+2mx+1)=0$. Then after using the quadratic formula with each of the factors I'm here: $x=\frac{-1 \pm \sqrt{1+4m}}{-2}$ and $x=\frac{-2m \pm \sqrt{4m^2+4}}{-2}$
| Take the negatives out. It will look tidier.
So:
$$(x^2-x-m)(x^2-2mx-1)=0$$
Then:
$$x=\frac{1\pm\sqrt{1-4m}}{2}=\frac 12\pm\frac{\sqrt{1-4m}}{2}$$
And:
$$x=\frac{2m\pm\sqrt{4m^2+4}}{2}=m\pm\sqrt{m^2+1}$$
You can't do anything more here.
| {
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $
Why can we just add $ 2 - x ^ { 2 } $ in the numerator?
Step 1 to step 2, as well as step 3 to step 4 is clear to me.
| You multiply by $1$ in a fancy fashion:
$$\sqrt{2-x^2}=\sqrt{2-x^2}\cdot 1=\sqrt{2-x^2}\cdot \frac{\sqrt{2-x^2}}{\sqrt{2-x^2}}=\frac{(\sqrt{2-x^2})^2}{\sqrt{2-x^2}}=\frac{2-x^2}{\sqrt{2-x^2}}$$
This of course only holds if $2-x^2\neq 0$, as otherwise also $\sqrt{2-x^2}=0$ and the last expression(and the intermediate ones) would be ill-defined.
Note, that his holds for any $a\in (0,\infty)$ by the same method as above, i.e.
$$\sqrt{a}=\sqrt{a}\cdot 1=\sqrt{a}\cdot\frac{\sqrt{a}}{\sqrt{a}}=\frac{(\sqrt{a})^2}{\sqrt{a}}=\frac{a}{\sqrt{a}}$$
| {
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How can I determine general formula of this sequence? I am trying to find general formula of the sequence $(x_n)$ defined by
$$x_1=1, \quad x_{n+1}=\dfrac{7x_n + 5}{x_n + 3}, \quad \forall n>1.$$
I tried
put $y_n = x_n + 3$, then $y_1=4$ and
$$\quad y_{n+1}=\dfrac{7(y_n-3) + 5}{y_n }=7 - \dfrac{16}{y_n}, \quad \forall n>1.$$
From here, I can't solve it. How can I determine general formula of above sequence?
With Mathematica, I found $x_n = \dfrac{5\cdot 4^n-8}{4^n+8}$. I want to know a method to solve problem, than have a given formula.
| The chracteristic equation of the given sequence is
$$y=\dfrac{7y+5}{y+3} \Leftrightarrow y_1 = 5 \lor y_2 = -1.$$
Let us consider the sequence
$$b_n = \dfrac{x_n-y_1}{x_n - y_2}=\dfrac{x_n-5}{x_n+1}.$$
We note that
$$b_{n+1}=\dfrac{x_{n+1}-5}{x_{n+1}+1}=\dfrac{\dfrac{7x_n+5}{x_n+3}-5}{\dfrac{7x_n+5}{x_n+3} + 1}=\dfrac{2x_n-10}{8x_n + 8} = \frac{1}{4} \cdot \dfrac{x_n-5}{x_n +1 }=\frac{1}{4} b_n.$$
Therefore $(b_n)$ is a geometric progression, with ratio $\dfrac{1}{4}$ and the first term
$$b_1 = \frac{x_1-5}{x_1+1}=\frac{1-5}{1+1}=\frac{-4}{2}=-2.$$
Therefore
$$b_{n+1}=b_1 \cdot \left (\dfrac{1}{4}\right)^n= - 2 \cdot \left (\dfrac{1}{4}\right)^n,$$
or equaivalently,
$$\dfrac{x_{n+1}-5}{x_{n+1}+1}=- 2 \cdot \left (\dfrac{1}{4}\right)^n \Leftrightarrow x_{n+1} = \frac{5\cdot 4^n - 2}{4^n +2}\Leftrightarrow x_n = \frac{5\cdot 4^n - 8}{4^n + 8}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what would the value of determinant of a matrix be if a specific entry changed?
What will the value of determinant of matrix $A=\pmatrix{1&3&4\\5&2&a\\6&-2&3}$ be change if we change $a$ to $a+2$.
This is an easy problem because $|A|=-127+20a$ and if we did that changing, we would get $-87+20a$. My question here is:
Can we do this request by doing another way? Where does that $+40$ come from regarding the whole entries of the matrix? Thanks
| The property of the matrix (in general, for any size $n\times n$):
$$\begin{vmatrix}
a+b&c+d \\ e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ e&f
\end{vmatrix}+\begin{vmatrix}
b&d \\ e&f
\end{vmatrix};\\
\begin{vmatrix}
a+b&c \\ d+e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ d&f
\end{vmatrix}+\begin{vmatrix}
b&c \\ e&f
\end{vmatrix};$$
Hence:
$$\begin{vmatrix}
1&3&4 \\ 5&2&(a+2) \\6&-2&3
\end{vmatrix}
=\begin{vmatrix}
1&3&c+(4-c) \\ 5&2&a+2 \ \ \ \ \ \ \ \ \ \ \\6&-2&d+(3-d)
\end{vmatrix}=\\
\begin{vmatrix}
1&3&c \\ 5&2&a \\6&-2&d
\end{vmatrix}+\begin{vmatrix}
1&3&4-c \\ 5&2&2 \\6&-2&3-d
\end{vmatrix}
$$
For easy calculation of determinant you can create as many zeros as possible (especially in rows/columns).
| {
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Calculate $\int \frac{x^4+1}{x^{12}-1} dx$ So I found this problem:
Calculate $$\int \frac{x^4+1}{x^{12}-1} dx$$ where $x\in(1, +\infty)$
and I don't have any ideea how to solve it. I tried to write $$x^{12}-1=(x^6+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)$$ but with $x^4+1$ I had no idea what to do, an obvious thing would be to add some terms (an $x^5$ would be helpful) and then discard them, but again I couldn't do anything. What should I do?
| You can use that $$x^4+1=(x^2+1)^2-2x^2=…$$
You will get $$\frac{x^4+1}{x^{12}-1}=\frac{-2 x-1}{12 \left(x^2+x+1\right)}-\frac{1}{3 \left(x^2+1\right)}+\frac{2 x-1}{12
\left(x^2-x+1\right)}+\frac{-x^2-1}{6 \left(x^4-x^2+1\right)}+\frac{1}{6
(x-1)}-\frac{1}{6 (x+1)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 \ x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 \ -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x \ -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
| The Euclidean algorithm steps are
\begin{align*}
x^3+1&=x (x^2+1) + (-x+1)\\
x^2+1&=-x (-x+1)+ (x+1)\\
-x+1&=-(x+1)+2
\end{align*}
So the GCD is $1$.
We can go back up
\begin{align*}
2 &= (-x+1) +(x+1)\\
&= (-x+1) + (x^2+1)+x (-x+1)\\
&= (-x+1) (x+1)+(x^2+1)\\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
\end{align*}
and so $1 = \frac{1}{2} [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
| {
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$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum
$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$$
I have tried to make this series in the form $1 + n x + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (n+2)x^3}{3!} + \cdots$. But I failed to do so. Can anyone please help me
| Let $a_n=\frac{1}{2}\prod_{k=1}^{n}(3k-1)$ and $b_n=\prod_{k=1}^{n}(6k)$. Then
$$ \frac{a_n}{b_n} = \frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\Gamma(n+1)\Gamma\left(\frac{2}{3}\right)}=\frac{\pi}{2^n \sqrt{3}} B\left(n+\frac{2}{3},\frac{1}{3}\right)=\frac{\pi}{\sqrt{3}}\int_{0}^{1}\frac{x^{n-1/3}(1-x)^{-2/3}}{2^n}\,dx $$
hence
$$ \sum_{n\geq 1}\frac{a_n}{b_n} = \frac{\pi}{\sqrt{3}}\int_{0}^{1}\frac{x^{2/3}}{(2-x)(1-x)^{2/3}}\,dx\stackrel{\frac{x}{1-x}\mapsto z}{=}\frac{\pi}{\sqrt{3}}\int_{0}^{+\infty}\frac{z^{2/3}}{(z+1)(z+2)}\,dz. $$
By setting $z=v^3$ and applying partial fraction decomposition we get
$$ \sum_{n\geq 1}\frac{a_n}{b_n}=\color{red}{\frac{2^{2/3}-1}{2}}. $$
The same can be proved by considering the Maclaurin series of $(1-x)^{-2/3}$ evaluated at $x=\frac{1}{2}$.
| {
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Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)^{n}$
So this takes care of the first terms, but then I just need to show
$\sum_{n = 3}^{\infty} (1 - x)^{n} = (1 - x)^{3}/x.$
Or, maybe I'm approaching this completely incorrectly.
| Alternatively:
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{(1-(1-x))(1 + (1 - x) + (1 - x)^2)}{1-(1-x)} + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{1^3-(1-x)^3}{x} + \frac{(1 - x)^3}{x},$$
where it was used:
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
| {
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} |
Prob. 7, Sec. 6.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: A necessary and sufficient condition for the existence of $\big(|f|\big)^\prime$ Here is Prob. 7, Sec. 6.1, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Suppose that $f \colon \mathbb{R} \to \mathbb{R}$ is differentiable at $c$ and that $f(c) = 0$. Show that $g(x) \colon= \lvert f(x) \rvert$ is differentialbe at $c$ if and only if $f^\prime(c)=0$.
My Attempt:
Suppose that $f^\prime(c)=0$.
Then, given a real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) - f(c) }{ x -c } \right\rvert = \left\lvert \frac{ f(x) }{ x -c } \right\rvert < \varepsilon $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$
Therefore for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$ we find that
$$ \left\lvert \frac{g(x) - g(c) }{x-c} - 0 \right\rvert = \left\lvert \frac{ \lvert f(x) \rvert - \lvert f(c) \rvert }{ x-c} \right\rvert = \left\lvert \frac{ \lvert f(x) \rvert }{ x-c} \right\rvert = \frac{ \lvert f(x) \rvert }{ \lvert x-c \rvert } = \left\lvert \frac{ f(x) }{ x -c } \right\rvert < \varepsilon. $$
Since $\varepsilon > 0$ was arbitrary, it follows that $g$ is differentiable at $c$ and that
$$ g^\prime(c) = 0. $$
Conversely, suppose that $f^\prime(c) \neq 0$. Then either $f^\prime(c) < 0$ or $f^\prime(c) > 0$.
Case 1. If $f^\prime(c) > 0$, then for $\varepsilon \colon= f^\prime(c)/2$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) - f(c) }{ x -c } - f^\prime(c) \right\rvert < f^\prime(c)/2 $$
or
$$ 0 < \frac{f^\prime(c)}{2} = f^\prime(c) - f^\prime(c)/2 < \frac{ f(x) - f(c) }{ x -c } < f^\prime(c) + f^\prime(c)/2 $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$
But $f(c) = 0$. Thus for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$
we have
$$ \frac{ f(x) }{x-c} > \frac{f^\prime(c)}{2} > 0, \tag{1} $$
which implies that, for all $x \in \mathbb{R}$,
$$ f(x) \ \begin{cases} > 0 \ \mbox{ if } & c < x < c + \delta, \\ < 0 \ \mbox{ if } & c-\delta < x < c. \end{cases} $$
So from (1) it follows that
$$ \frac{g(x) - g(c) }{x-c} = \frac{ \lvert f(x) \rvert }{ x-c} = \begin{cases} \frac{f(x)}{x-c} & \mbox{ if } \ c < x < c+\delta, \\ -\frac{ f(x)}{x-c} & \mbox{ if } \ c-\delta < x < c. \end{cases} \tag{2}$$
Moreover, from (2) we can also conclude that
$$ \lim_{x \to c+} \frac{g(x) - g(c)}{x-c} = \lim_{x \to c+} \frac{ f(x) }{x-c} = \lim_{x \to c+} \frac{ f(x) - f(c) }{x-c} = f^\prime(c), $$
and
$$ \lim_{x \to c-} \frac{g(x) - g(c)}{x-c} = \lim_{x \to c-}\left(- \frac{ f(x) }{x-c} \right) = \lim_{x \to c-} \left( - \frac{ f(x) - f(c) }{x-c}\right) = - \lim_{x \to c-} \frac{ f(x) - f(c) }{x-c}=- f^\prime(c). $$
Thus if $f^\prime(c) > 0$, then
$$ \lim_{x \to c+} \frac{g(x) - g(c)}{x-c} \neq \lim_{x \to c-} \frac{g(x) - g(c)}{x-c}, $$
and so $g^\prime(c)$ does not exist. Therefore if $g$ is differentiable at $c$, then we must have $f^\prime(c) \not> 0$.
Case 2. If $f^\prime(c) < 0$, then for $\varepsilon \colon= -f^\prime(c)/2 > 0$, we can find a real number $\delta > 0$ such that
$$ \left\lvert \frac{ f(x) - f(c) }{ x -c } - f^\prime(c) \right\rvert < - f^\prime(c)/2 $$
or
$$ \frac{3f^\prime(c)}{2} = f^\prime(c) - \frac{-f^\prime(c)}{2} < \frac{ f(x) - f(c) }{ x -c } < f^\prime(c) + \frac{-f^\prime(c)}{2} = \frac{ f^\prime(c)}{2} < 0 $$
for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta. $$
But $f(c) = 0$. Thus for all $x \in \mathbb{R}$ which satisfy
$$ 0 < \lvert x-c \rvert < \delta, $$
we have
$$ \frac{ f(x) }{x-c} < \frac{f^\prime(c)}{2} < 0, \tag{3} $$
which implies that, for all $x \in \mathbb{R}$,
$$ f(x) \ \begin{cases} < 0 \ \mbox{ if } & c < x < c + \delta, \\ > 0 \ \mbox{ if } & c-\delta < x < c. \end{cases} $$
So from (3) it follows that
$$ \frac{g(x) - g(c) }{x-c} = \frac{ \lvert f(x) \rvert }{ x-c} = \begin{cases} -\frac{f(x)}{x-c} & \mbox{ if } \ c < x < c+\delta, \\ \frac{ f(x)}{x-c} & \mbox{ if } \ c-\delta < x < c. \end{cases} \tag{4}$$
Moreover, from (4) we can also conclude that
$$ \lim_{x \to c-} \frac{g(x) - g(c)}{x-c} = \lim_{x \to c-} \frac{ f(x) }{x-c} = \lim_{x \to c-} \frac{ f(x) - f(c) }{x-c} = f^\prime(c), $$
and
$$ \lim_{x \to c+} \frac{g(x) - g(c)}{x-c} = \lim_{x \to c+}\left(- \frac{ f(x) }{x-c} \right) = \lim_{x \to c+} \left( - \frac{ f(x) - f(c) }{x-c}\right) = - \lim_{x \to c+} \frac{ f(x) - f(c) }{x-c}=- f^\prime(c). $$
Thus if $f^\prime(c) < 0$, then
$$ \lim_{x \to c+} \frac{g(x) - g(c)}{x-c} \neq \lim_{x \to c-} \frac{g(x) - g(c)}{x-c}, $$
and so $g^\prime(c)$ does not exist. Therefore if $g$ is differentiable at $c$, then we must have $f^\prime(c) \not< 0$.
From the above two cases, we can conclude that if $g$ is differentiable at $c$, then we must have $f^\prime(c)=0$.
Is this proof correct? If so, then is the presentation clear and rigorous enough too? If not, then where are the issues as far as accuracy, rigor, or clarity of the argument go?
| Of course everything you did is correct. But such a simple statement deserves a shorter proof, which then also shows what's going on here.
Assume that $c=0$, $f(0)=0$, and $f'(0)=a$. Then
$$m(x):={f(x)\over x}\to a \quad (x\to0)\ ,$$
and therefore
$${\bigl|f(x)\bigr|\over x}={\rm sgn}(x)\,\bigl|m(x)\bigr|\to\left\{\eqalign{|a|\quad&(x\to0+) \cr -|a|\quad&(x\to0-)\cr}\right.\quad.$$
Here the two one-sided limits coincide iff $a=0$.
| {
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"question_score": "4",
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} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
|
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
No, since $\frac10$ is not defined.
*
*Does $\infty\cdot\frac{1}{2}$ equal to $\infty$?
There are the "extended real numbers" $\mathbb{R}\cup\{\pm\infty\}$
Where you can define this.
Note, that $\infty$ is in general not a number. So you can not do some calculations with it.
You can check out: https://en.wikipedia.org/wiki/Extended_real_number_line
For more informations.
*
*In conclusion, what is the limit of the sequence
above? $\infty?$
Yes, the limit is $\infty$. Or in other words the sequence does not converge.
| {
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"url": "https://math.stackexchange.com/questions/2914728",
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"question_score": "10",
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"answer_id": 3
} |
Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.
Base case is easy, n=1.
$$
\frac{1}{2^1}<2
$$
Induction case we assume that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2
$$
Then we get to fun old induction. How do I show that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ?
$$
| In that case a possible trick to use induction is to prove the stronger condition
$$\frac{1}{2} + \frac{2}{2^2} + \ldots +\frac{n-1}{2^{n-1}} + \frac{n}{2^{n}} <2-\frac{n+1}{2^{n-1}}<2$$
and the induction step becomes
$$\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} + \frac{n+1}{2^{n+1}}\stackrel{Ind. Hyp.}<2-\frac{n+1}{2^{n-1}}+ \frac{n+1}{2^{n+1}}\stackrel{?}<2-\frac{n+2}{2^{n}}$$
and the last inequality holds indeed
$$2-\frac{n+1}{2^{n-1}}+ \frac{n+1}{2^{n+1}}\stackrel{?}<2-\frac{n+2}{2^{n}}$$
$$\frac{n+1}{2^{n-1}}- \frac{n+1}{2^{n+1}}\stackrel{?}>\frac{n+2}{2^{n}}$$
$$4(n+1)-(n+1)\stackrel{?}>
2(n+2)$$
$$3n+3\stackrel{?}>2n+4$$
wich is true for $n>1$.
| {
"language": "en",
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"answer_id": 4
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Closed form solution for logarithmic inequality We have the following inequality that was very hard to be solved in a closed form. Yet, someone solved it this way, and I can't get to fully understand what is done.
eq: $\sqrt{x-2\sqrt{x-1}} + \sqrt{x+2\sqrt{x-1}} +log_2(x-1)=0$
Solution:
$x\in Df \iff x \geq 1$
We have: $|\sqrt{x-1} -1|+|\sqrt{x-1} +1|+log_2(x-1)=0$
$|\sqrt{x-1} -1|= \{\sqrt{x-1} -1 \ if\ \sqrt{x-1} -1 \geq0\ and\ \sqrt{x-1} -1 \geq0 \Rightarrow x\geq2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \{\ 1 - \sqrt{x-1} \ if\ \sqrt{x-1} -1 \leq0\ and\ \sqrt{x-1} -1 \leq0 \Rightarrow 1\leq x\leq2$
$|\sqrt{x-1} +1|= \{\sqrt{x-1} +1 \ if\ \sqrt{x-1} +1 \geq0\ and\ \sqrt{x-1} +1 \geq0,\ \forall x\geq1$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \{\ -\sqrt{x-1}-1 \ if\ \sqrt{x-1} +1 \leq0\ and\ \sqrt{x-1} +1 \leq0 \Rightarrow x\in \emptyset$
Then, the solution moves to have two intervals for study, (1,2) and (2, +oo[, for which we find the solution to be x = 5/4 on the interval (1,2) and no solution in the second interval. My question is what is the logic of the use of absolute value and replacing x with 1 as you can notice?
| Suppose you have, instead,
$$
\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}+\log_2x=0
$$
Then the substitution with the absolute values would be sensible, but there's a better method.
Let $y=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}$; then $y>0$ and
$$
y^2=2x+2\sqrt{x^2-4(x-1)}=2x+2\sqrt{(x-2)^2}
$$
However, $\log_2x$ should be negative, because $y>0$, which implies $1<x<2$. Hence $\sqrt{(x-2)^2}=2-x$ and we have $y^2=4$, so $y=2$. Thus the solution is $\log_2(x-1)=-2$, hence $x-1=1/4$ and therefore
$$
x=\frac{5}{4}
$$
The equation in your question has a solution, but it's strictly between $1.2$ and $1.25=5/4$.
| {
"language": "en",
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"question_score": "1",
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"answer_id": 1
} |
The number of positive integral solutions of $abc =$ $30$ is
The number of positive integral solutions of $abc =$ $30$ is
My attempt
Factors of $30$ are $3 \cdot 5 \cdot 2$.
Therefore $a$ will have three choices, similarly $b$ & $c$ will have $2$ choices & $1$ choice.
So,it's answer must be $3 \cdot 2 \cdot 1$
But answer in my textbook is given is $3 \cdot 3 \cdot 3$.
.It states that"$2$ can be assigned to either $a$ or $b$ or $c$". Similarly, each of $3$ and $5$ can be assigned in $3$ ways. Thus, the number of solutions is $3 \cdot 3 \cdot 3$
According to me, if $a$ is already assigned to $2$ than how can other letters ($b$ or $c$) have same value as that of $a$.
Please explain me where am I wrong?
| The texbook is right.
Start with $1\cdot1\cdot1$ and assign $2$, say to $b$ and get $1\cdot2\cdot1$.
Now assign $3$, say to $a$ and get $3\cdot2\cdot1$.
Then assign $5$ to $a$ again, giving $15\cdot2\cdot1$.
On every stage, you have three choices.
| {
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"answer_id": 1
} |
Calculate$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$
Calculate $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$
First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t-2}+t^2}}dt$$ and use the fact that $$2(\sqrt{e^{t-2}+t^2})'=\frac{1}{\sqrt{e^{t-2}+t^2}} \cdot(e^{t-2}+2t)$$ Using integration by parts we get that we have to calculate (excluding some terms we know) $$\int\limits_{0}^{2} \sqrt{e^{t-2}+t^2}\cdot \frac{6e^{t-2}-2te^{t-2}+8}{(e^{t-2}+2t)^2}dt$$ which is uglier then the initial problem. Do you have any idea how to solve the problem?
| Long method
One possible method is the following. Using
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^n \, x^n}{4^n} &= \frac{1}{\sqrt{1 + x}} \\
\gamma(s,x) &= \int_{0}^{x} e^{-t} \, t^{s-1} \, dt
\end{align}
then:
\begin{align}
I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\
&= \int_{-2}^{0} \frac{x \, e^{-x/2} \, dx}{\sqrt{1 + \left(\frac{(x+2)^2}{e^{x}}\right)}} \\
&= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{-2}^{0} e^{-(n+1/2) x} \, x \, (x+2)^{2n} \, dx \\
&= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^n} \, e^{2n+1} \, \int_{0}^{2} e^{-(n+1/2) t} \, (t-2) \, t^{2n} \, dt \hspace{5mm} t \to x + 2 \\
&= \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n}}{4^{n}} \, e^{2n+1} \, \left[ \left(\frac{2}{2n+1}\right)^{2n+2} \, \gamma(2n+2, 2n+1) - 2 \, \left(\frac{2}{2n+1}\right)^{2n+1} \, \gamma(2n+1, 2n+1) \right] \\
&= 4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, \left[ \gamma(2n+2, 2n+1) - (2n+1) \, \gamma(2n+1, 2n+1) \right] \\
&= - 4 \, \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, (2n+1)^{2n+1} \, e^{-(2n+1)} \\
&= -4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n}}{(2n+1)} \\
&= - \, \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{0}^{2} y^{2n} \, dy \\
&= -2 \, \int_{0}^{2} \frac{dy}{\sqrt{1+y^2}} \\
&= -2 \, \sinh^{-1}(2).
\end{align}
Short method
\begin{align}
I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\
&= -2 \int_{-2}^{0} \frac{- (x/2) \, e^{-x/2} \, dx}{\sqrt{1 + ((x+2) \, e^{-x/2})^2}} \\
\end{align}
Let $u = (x+2) \, e^{-x/2}$, $du = -(x/2) \, e^{-x/2} \, dx$, then
\begin{align}
I &= -2 \int_{-2}^{0} \frac{du}{\sqrt{1+u^2}} = -2 \left[\sinh^{-1}(u)\right]_{-2}^{0} \\
&= -2 \, \sinh^{-1}(2).
\end{align}
It is of interest to note that
$$\int_{-2}^{\infty} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} = 0.$$
| {
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Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1$$
Since $$\binom{1}{0} = \binom{1}{1} = 1$$
it's true.
Now suppose that is true with $n$ if is true with $n + 1$
Then, multiply both sides by two
$$2(2^n) = 2(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n})$$
$$2^{n+1} = 2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n}$$
$$2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n} = \binom{n}{0} + \binom{n}{0} + \binom{n}{1} + \binom{n}{1} + ... + \binom{n}{n} + \binom{n}{n} $$
The first term have two equal term, then, you sum the last one with the first one of the next term, and you'll get this
$$\binom{n}{0} + \binom{n}{1} +... + \binom{n}{n-1} + \binom{n}{n}$$
If we use this equation (Already proved)
$$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $$
Of course, we'll have two term without sum, one $\binom{n}{0}$ and $\binom{n}{n}$
We can write these two term like this
$$\binom{n}{0} = \binom{n}{n} = \binom{n+1}{0} = \binom{n+1}{n+1}$$
Then, we get
$$2^{n+1} = \binom{n+1}{0} + \binom{n+1}{1} + ... + \binom {n+1}{n+1}$$
And it's already proved. Note: Just if we take $0! = 1$
I have to prove these too.
$\sum_{k}^n \binom{n}{m} = 2^{n-1}$ If $m$ is even. And
$\sum_{j}^n \binom{n}{j} = 2^{n-1}$ If $j$ is odd. Then, I just said that
If
$$\sum_{m}^n \binom{n}{m} + \sum_{j}^n \binom{n}{j} = \sum_{k=0}^n \binom{n}{k} $$
Then
$$2^{n - 1} + 2^{n - 1} = 2^n$$
Which is true, then, I already prove this. And I have a last one.
$$\sum_{i=0}^n (-1)^i\binom{n}{i} = 0$$
if n is odd. Then
$$\binom{n}{0} - \binom{n}{1} + ... + \binom{n}{n-1} - \binom{n}{n} = 0$$
And that's can be solve knowing that $$\binom{n}{k} = \binom{n}{n - k}$$
And if n is even
$$\binom{n}{0} - \binom{n}{1} + ... - \binom{n}{n-1} + \binom{n}{n} = 0$$
That means that every negative term if when n is odd, then, we can use our two last prove to prove it
If
$$\sum_{m}^n \binom{n}{m} - \sum_{j}^n \binom{n}{j} = 0$$
Then
$$2^{n-1} - 2^{n-1} = 0$$
Which is true.
And that's it, I want to know if my proves are fine and are rigorous too and what is the meaning of every combinatorics prove .
I want to know too better approaches to prove these (Or forms more intuitive)
| here is another approach:
Consider a set with $n$ objects. There are $\binom nk$ ways to choose $k$ of them. That way gives $\sum_{i=0}^n\binom ni$ ways to choose objects from the set. On the other hand, we can choose to choose each item or not.That gives $2^n$. And of course they are equal, since they both represent the number of ways to choose some objects from the set.
The second and the third one are actually the same (can you see why?). Using Pascal's triangle $\sum_{2|i}^n\binom ni=\sum_{2\nmid i}^n\binom ni$ since they both equal to the sum of the line above them. (which is $2^{n-1}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Convergence of $\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$? $a_n=\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$
I think that $\sqrt[n]{1^n+2^n+...+(2n)^n}\rightarrow 2n+1$
So $a_n=\frac{2n+1}{2n}\rightarrow 1$
Any ideas if that's correct? And if so how do I prove it?
| We have
\begin{align}
a_n=&
\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}
\\
=&\frac{1}{2n}\sqrt[n]{(2n)^n\left[\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1\right]}
\\
=&\sqrt[n]{\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1}
\\
=&\sqrt[n]{\left(\frac{1}{2n}\right)^{n}+\left(\frac{2}{2n}\right)^{n}+...+\left(1-\frac{1}{2n}\right)^{n}+1}
\end{align}
For $ n $ big enough we have
$$
0\leq \left(\frac{1}{2n}\right)^{n}\leq 1, \quad 0\leq \left(\frac{2}{2n}\right)^{n} \leq 1,\ldots \qquad \ldots, 0\leq \left(1-\frac{1}{2n}\right)^{n}\leq 1.
$$
It is easy to see that
$$
\sqrt[n]{0+0+\ldots+0+1}\leq a_n \leq \sqrt[n\,]{\underbrace{1+1+\ldots+1+1}_{n \mbox{ times }}}
$$
or
$$
1\leq a_n\leq \sqrt[n]{n}
$$
Once $\sqrt[n]{n}\to 1$, thus $a_n\to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
Here's a picture, since I'm not good with MathJax.
| I would rather approach on a very different way...I want to solve for y.
So,here we are given,
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
Now,assume that $g$ is our variable and $x$ and $y$ are constant.then equalling the coefficients we can say that,
$$\dfrac{y}{x}=\dfrac{y}{y^2} ~~and~~\dfrac{2y^2}{x}=\dfrac{2x}{y^2}$$
All of these come to an end,where
$$y^2=x$$
Hence,$$y=\pm\sqrt{x}$$
If you want to solve for x, then you should follow leucippus solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sort those 3 logarithmic values without using calculator I found this problem interesting, namely we are given three values: $$\log_{1/3}{27}, \log_{1/5}{4}, \log_{1/2}{5}$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $\log_{1/3}{27} = \log_{1/3}{3^3} = \log_{1/3}{(\frac{1}{3})^{-3}} = -3\\ \log_{1/5}{4} = \log_{1/5}{2^2} = 2 \cdot \log_{1/5}{2} = 2 \cdot \log_{5^{-1}}{2} = -2 \cdot \log_{5}{2}$
Since $\log_{5}{2} < 1, \text{then: } -2\cdot \log_{5}{2} > -2 $
And the third value: $\log_{1/2}{5} = \log_{2^{-1}}{5} = -\log_{2}{5}$
Since $2 < \log_{2}{5} < 3, \text{then: } -2 > -\log_{2}{5} > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$\log_{1/3}{27}, \log_{1/2}{5}, \log_{1/5}{4}$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
| For clarity, we take the inverses of the bases, which will reverse the order ($\log_{1/x}y=-\log_x y$).
We have
$$\log_3{27}=3,$$
$$\log_54<1,$$
because $4<5^1$ and
$$1<\log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$\log_54<1<\log_25<3=\log_327$$
and
$$\log_{1/5}4>\log_{1/2}5>\log_{1/3}27.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Modular Multiplicative Inverses I'm in a cryptography and struggling with understanding how the Euclidean Algorithm necessarily works for finding multiplicative inverses. We haven't actually covered the algorithm yet, just a way to brute force finding the inverses but I like working with an algorithm instead of just guess and checking so I learned it myself. In the equation 1/5 mod 13 I understand that this is equivalent to 5x = 1 mod 13 however when going through the algorithm I get the answer to be -5 instead of 8 which it is supposed to be. Can anyone shed some light on where I'm going wrong? Here's my work.
(a) 1/5 mod 13
1/5 mod 13 ≡ 5-1 mod 13 ≡ 5x = 1 (mod 13)
13 = 2(5) + 3
5 = 1(3) + 2
3 = 1(2) + 1
GCD of 13 and 5 is 1.
1 = 3 – 2(1)
= 3 – (5 – 3(1)) = 3 - 5 + 3 = 2(3) - 5
= 2(13 - 2(5)) - 5
= 2(13) – 4(5) – 5
= 2(13) -5(5)
-> mod 13 on both sides
1 mod 13 = 0 – 5(5) mod 13
1 = -5(5) mod 13
x = -5
| You are right indeed by Euclidean algorithm we have
*
*$13=5\cdot 2+\color{red}3$
*$5=\color{red}3\cdot 1+\color{red}2$
*$3=\color{red}2\cdot 1+1$
and therefore starting from the last one
$$1=3-2=3-(5-3)=2\cdot 13-2\cdot 5\cdot 2-5 \implies 2\cdot 13-5\cdot 5=1 $$$$\implies -5\cdot 5=1-2\cdot 13$$
and then
$$5x\equiv 1 \pmod {13} \implies x\equiv -5 \equiv 8 \pmod {13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Induction proof of a series Suppose we have the series
$$f(n) = \sum_{i=1}^n \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}.$$
As a hint it was said that this series "telescopes". I observed the pattern to be $f(n)=\frac{n^2}{n^2 +1}$. I wish to prove this via induction. The base case holds, since $f(1)=\frac{1}{2}$ which comes out of both the definition of $f(n)$ and my proposed formula. Now I wish to prove that for some arbitrary $k$ we assume $f(k)=\frac{k^2}{k^2+1}$, then I wish to prove that $f(k+1)=\frac{(k+1)^2}{(k+1)^2+1} (=\frac{k^2 +2k +1}{k^2 +2k +2})$.
I wanted to start out by noticing:
$$f(k+1) = \sum_{i=1}^{k+1} \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}= \\f (k)+ \frac{2(k+1)-1}{(k+1)^4 - 2 (k+1)^3 + 3 (k+1)^2 - 2 (k+1) + 2}$$
This eventually simplifies to
$$ f(n+1)= \frac{k^2}{k^2 +1} + \frac{2k+1}{k^4 +2k^3+3k^2+2k+2}$$
| For completeness sake I want to finish my own proof, without telescoping and simply by induction as I intended to:
Now notice that:
$$(k^2+1) (k^2 +2k +2) =k^4 +2k^3 +3k^2+2k +2$$
We can thus write:
$$f(k+1)=\frac{k^2}{k^2 +1}+ \frac{2k+1}{(k^2+1)(k^2+2k+2)} = \\ \frac{k^2 (k^2+2k+2)+ (2k+1)}{(k^2+1)(k^2+2k+2)}=\frac{k^4+2k^3+2k^2+2k+1}{(k^2+1)(k^2+2k+2)}$$
Finally, observe that:
$$(k^2+1)(k^2+2k+1)=k^4+2k^3+2k^2+2k+1 $$
Such that we can write:
$$ f(k+1)=\frac{(k^2+1)(k^2+2k+1)}{(k^2+1)(k^2+2k+2)}=\frac{k^2+2k+1}{k^2+2k+2}=\frac{(k+1)^2}{(k+1)^2+1}$$
If the formula holds for arbitrary k, it also holds for $k+1$, by the principle of mathematical induction, it holds for all $n$ $\square$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the integral $\leq 1/\sqrt{5} \leq 1$.
Correct?
Unsure how to progress from here
| Hint: $$\frac{1}{\sqrt{a^3+4}} \le \frac{1}{\sqrt{x^3+4}} \le \frac{1}{\sqrt{0^3+4}}$$
for $0 \le x \le a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by inspection)?
I am currently using the addition of cosine formula, but it is too slow. I'm thinking perhaps a graphic method can be used?
| Let $\,a=\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}=\dfrac{1+i}{\sqrt{2}}\,$ and $\,b=\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}=\dfrac{-1+i\sqrt{3}}{2}\,$, so:
$$
ab = \dfrac{(1+i)(-1+i\sqrt{3})}{2 \sqrt{2}}=\dfrac{\left(-1-\sqrt{3}\right)+i\left(-1+\sqrt{3}\right)}{2 \sqrt{2}}
$$
Then $\,\cos\left(\dfrac{\pi}{4} + \dfrac{2 \pi}{3}\right) = \operatorname{Re}(ab) = \dfrac{-1-\sqrt{3}}{2\sqrt{2}}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Implication of equations Just an interesting question I saw online. :)
Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
| If $a,b,c$ are all positive then we have by Am-Gm:
$$\dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} \geq 3\sqrt[3]{\dfrac{ab}{c^2} \cdot \dfrac{bc}{a^2} \cdot \dfrac{ca}{b^2}} = 3$$
With eqaulity iff $ \dfrac{ab}{c^2} = \dfrac{bc}{a^2} = \dfrac{ca}{b^2}$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0\implies k=-m \;\;\;{\rm or} \;\;\;k^2-km+m^2=3m$$
We would like to know what is the value of $${(x+y)(xy+x+y+1)\over xy}-2$$
i.e. $$E:={k(m+k+1)\over m}-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2={k^2+m^2+(k-m)^2\over 2}\geq 0$, then $m\geq 0$ but this is a contradiction since $m<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On the eigenvalue of a particular kind of matrix Let $x+y=1$. Consider the matrix $A=\begin{pmatrix} x^5+y^5&5x^5&10x^5&10x^5&5x^5\\5y^5&x^5+y^5 &5x^5&10x^5&10x^5 \\10y^5&5y^5&x^5+y^5&5x^5&10x^5\\10y^5&10y^5&5y^5&x^5+y^5&5x^5\\5y^5&10y^5&10y^5&5y^5&x^5+y^5 \end{pmatrix} $
Is $1$ an eigenvalue of $A$ ?
It is not obvious to me if $1$ is an eigenvalue or not (the row sums of $A$ are not all equal ... if they were equal then $1$ would obviously be an eigenvalue ).
Please help
| If $x = \frac{g}{g+h}$ and $y = \frac{h}{g+h},$ then
$$
\left(
\begin{array}{c}
g^4 \\
g^3 h \\
g^2 h^2 \\
g h^3 \\
h^4
\end{array}
\right)
$$
is an eigenvector with eigenvalue $1.$ So is the multiple
$$
\left(
\begin{array}{c}
x^4 \\
x^3 y \\
x^2 y^2 \\
x y^3 \\
y^4
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's Find limit of
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$
I started by defining $\ x = y + \frac{\pi}{4} $
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$
This is exactly the same question where the solution suggested also used $\ x = y + \pi/4 $ yet I couldn't understand this equality: $$\ \tan(x) = \tan(y + \pi/4) = \frac{\tan y -1 }{1- \tan y} $$ and then how is that exactly equal to limit of $$\ \lim_{y \to 0}\frac{1}{y} \left( 1-\frac{\tan y +1}{1 - \tan y}\right)$$
| As an alternative without derivatives, let $x=\frac{\pi}{4}-y$ then
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4}=\lim_{y \to 0} \frac{\tan \left(\frac{\pi}{4}-y\right)-1}{-y}=\lim_{y \to 0} \frac{1-\frac{\cos \left(2y\right)}{1+\sin \left(2y\right)}}{y}=\lim_{y \to 0} \frac{1-\cos \left(2y\right)+\sin \left(2y\right)}{y(1+\sin \left(2y\right))}=$$
$$=\lim_{y \to 0} \frac{1}{1+\sin \left(2y\right)}\left(4y\frac{1-\cos \left(2y\right)}{4y^2}+2\frac{\sin \left(2y\right)}{2y}\right)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are... If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are...
I've been trying to find the limit as usual:
$\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}$ and I still get zero in the denominator, so I don't know how to go from here...
| $\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}$ so $b$ must be 1 and $c$ must be $1/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website
| $$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin^2(u)}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$
$$\int \frac{-(1 - \cos(u))(1 + \cos(u)) }{1-\cos(u)}du=-\int (1 + \cos(u)) du=-u-\sin u+C$$
Can you replace the value of $u$ by using $\cos u =x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Calculate $\int_0^1{x·\lceil1/x\rceil dx}$ I am trying to calculate following integral:
$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$
I tried usual change t=1/x but not able to further advance.
Thanks!
| $$
\begin{aligned}
\int_0^1x\cdot\left\lceil \frac{1}{x}\right\rceil\; dx
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
\underbrace{\left\lceil \frac{1}{x}\right\rceil}_{\in (k,k+1)\text{ a.e}} \;dx
\\
&=
\sum_{k\ge 1}
\int_{1/(k+1)}^{1/k}x\cdot
(k+1)\;dx
\\
&=
\sum_{k\ge 1}
(k+1)
\left[\ \frac 12 x^2\ \right]_{1/(k+1)}^{1/k}
\\
&=
\frac 12
\sum_{k\ge 1}
(k+1)
\left[\ \frac 1{k^2}-\frac 1{(k+1)^2}\ \right]_{1/(k+1)}^{1/k}
\\
&=
\frac 12
\sum_{k\ge 1}
\frac {2k+1}{k^2(k+1)}
\\
&=
\frac 12
\sum_{k\ge 1}
\frac k{k^2(k+1)}
+
\frac 12
\sum_{k\ge 1}
\frac {k+1}{k^2(k+1)}
\\
&=
\frac 12
\sum_{k\ge 1}
\left[
\frac 1k-\frac 1{k+1}\right]
+
\frac 12
\sum_{k\ge 1}
\frac 1{k^2}
\\
&=
\frac 12\cdot 1+\frac 12\cdot\frac{\pi^2}6\ .
\end{aligned}
$$
Computer check, here PARI/GP:
? intnum( x=1.e-12, 1, x*ceil(1/x), 8 )
%12 = 1.3224146303918600341384604269228003273
? 1/2 + Pi^2/12
%13 = 1.3224670334241132182362075833230125946
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$
Simplify
$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$$
Found in a book with tag "Moscow 1982", the stated answer is $1+\sqrt[4]{5}$. Used all tricks that I know but without success. The answer appears to be correct, checked in Wolfram Alpha.
Hints and answers welcomed. Sorry if this is a duplicate.
| $$\begin{align}\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125})(\sqrt{1+\sqrt[4]{5}})}}\cdots \cdots(1)\\&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{5^3})+(4\cdot \sqrt[4]{5}-3\sqrt{5}+2\sqrt[4]{5^3}-5)}}\\&= \frac{2\cdot\sqrt{1+\sqrt[4]{5}}}{\sqrt{5^{1/4}-5^{1/2}+5^{3/4}-1}}\\&=\frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{(5^{1/4}-5^{1/2}+5^{3/4}-1)+5^{1/4}(5^{1/4}-5^{1/2}+5^{3/4}-1)}}\\&= \frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{4}}=1+\sqrt[4]{5}.\end{align} $$
at $(1)$ and 4th line multiplied the denominetor and numerator by $\sqrt{1+\sqrt[4]{5}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime
Let $p>3$ be a prime number. How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$?
When $p=3$, you also have $p+1$ as a divisor of $\left(p^2+1\right)^2$.
|
Proposition. For a positive integer $p$, $\left(p^2+1\right)^2$ is divisible by some positive integer $d\equiv 1\pmod{p}$ such that $$d\notin\left\{1,p^2+1,\left(p^2+1\right)^2\right\}$$ if and only if
$$p=t_s(b)\text{ for some integers }b\geq 1\text{ and }s\geq 2\,,$$
where
$$t_j(b):=\frac{\left(\frac{\left(b^2+2\right)+b\,\sqrt{b^2+4}}{2}\right)^j-\left(\frac{\left(b^2+2\right)-b\sqrt{b^2+4}}{2}\right)^j}{\sqrt{b^2+4}}$$
for all positive integers $b$ and $j$. Additionally, if $p$ is in this form $t_s(b)$, then there are at least two positive integers not of the form $1$, $p^2+1$, and $\left(p^2+1\right)^2$ that divide $\left(p^2+1\right)^2$, which are
$$t_s(b)\,t_{s-1}(b)+1\text{ and }\left(b^2+2\right)\,\left(t_s(b)\right)^2-t_s(b)\,t_{s-1}(b)+1\,.$$
Finally, the only prime natural number $p$ with this property is $p=3$.
Suppose that $p$ and $d$ are positive integers such that $d\equiv 1\pmod{p}$ and $d$ divides $\left(p^2+1\right)^2$. Without loss of generality, we may assume that $d\leq p^2+1$; otherwise, replace $d$ by $\dfrac{\left(p^2+1\right)^2}{d}$. Because $d\equiv 1\pmod{p}$, we have
$$d=pk+1\text{ for some integer }k\text{ such that }0\leq k\leq p\,.$$
As $d\mid \left(p^2+1\right)^2$, we get that $d=pk+1$ divides
$$\left(p^2+1\right)^2+(pk+1)(-2p^2+pk-1)=p^2\,\left(p-k\right)^2\,.$$
Since $\gcd\left(pk+1,p^2\right)=1$, it follows that
$$\frac{(p-k)^2}{pk+1}\in\mathbb{Z}\,.$$
Of course, $k=0$ and $k=p$ are possible solutions, but they lead to trivial divisors $d=1$ and $d=p^2+1$.
From Zvi's answer in this link, all integral solutions $(p,k)$ with $p>k>0$ take the form
$$(p,k)=\big(t_{j+1}(b),t_j(b)\big)\text{ for some positive integers }b\text{ and }j\,,$$
where $\left\{t_j(b)\right\}_{j\in\mathbb{Z}_{\geq 0}}$ is a sequence defined by the initial conditions $t_0(b):=0$ and $t_1(b):=b$, along with the recurrence
$$t_{j}(b):=\left(b^2+2\right)\,t_{j-1}(b)-t_{j-2}(b)\text{ for all integers }j\geq 2\,.$$
For a given positive intger $b$, note that
$$t_j(b)=\frac{\left(\frac{\left(b^2+2\right)+b\,\sqrt{b^2+4}}{2}\right)^j-\left(\frac{\left(b^2+2\right)-b\sqrt{b^2+4}}{2}\right)^j}{\sqrt{b^2+4}}\text{ for all }j=0,1,2,\ldots\,.$$
From here, we conclude that all positive integers $p$ for which $\left(p^2+1\right)^2$ is divisible by some positive integer $d\equiv 1\pmod{p}$ not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$ must satisfy
$$p=t_{s}(b)\text{ for some integers }b>0\text{ and }s>1\,.$$
Amongst $1$ to $100$, positive integers $p$ in the said form are
$$t_2(1)=3\,,\,\,t_3(1)=8\,,\,\,t_2(2)=12\,,\,\,t_4(1)=21\,,\,\,t_2(3)=33\,,$$
$$t_5(1)=55\,,\,\,t_3(2)=70\,,\text{ and }t_4(1)=72\,.$$
For example, with $p:=t_2(2)=12$, we have $\left(p^2+1\right)^2=145^2$, which is divisible by $$d:=5^2=25\equiv 1\pmod{p}\,,$$ which is not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$. In particular, if $p$ is a prime natural number, then we must have
$$p=t_2(1)=3\,,$$
where $\left(p^2+1\right)^2=10^2=100$ has a divisor $d:=4\equiv 1\pmod{p}$ not of the form $1$, $p^2+1$, or $\left(p^2+1\right)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Understanding how $x^2+2x+y^2-6y-15=0$ becomes $(x+1)^2+(y-3)^2=25$
I am looking at how this equation was simplified, $15$ was pushed to the other side and $10$ I believe was added by adding $1$ to the first bracket equation and $9$ to the other part.
The next part where to get $x² + 2x + 1$ to $(x + 1)^2$ there $2x$ seems to be dropped off and the same for $6y$.
| Yes, you’re right. The equation was simplified by completing the square. Remember that $(m+n)^2 = m^2+2mn+n^2$ and vice versa.
$$x^2+\frac{b}{a}x = -c \implies x^2+\biggr(\frac{b}{2a}\biggr)^2+\frac{b}{a}x = -c+\biggr(\frac{b}{2a}\biggr)^2$$
$$\implies \biggr(x+\frac{b}{2a}\biggr)^2 = -c+\biggr(\frac{b}{2a}\biggr)^2 \implies \biggr(x+\frac{b}{2a}\biggr)^2-\biggr(\frac{b}{2a}\biggr)^2= -c$$
This process was done for both $x$ and $y$.
$$x^2+2x+y^2-6y-15 = 0$$
Break it down into two parts. Complete the square for both $x$ and $y$.
$$(x^2+2x)+(y^2-6y)-15 = 0$$
$$(x^2 + 1^2+2x)-1^2+(y^2+3^2-6y)-3^2-15 = 0$$
$$(x+1)^2-1+(y-3)^2-9-15 = 0 \implies (x+1)^2+(y-3)^2-25 = 0$$ $$\implies (x+1)^2+(y-3)^2 = 25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing that the inequality $\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}$ stands.
If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that
$$\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}.$$
I have already proven the second inequality, but couldn't the first one. I managed to show that $\dfrac{1}{n+1} \leq u_n$, and I wrote a recursive formula by using the $\displaystyle \int_{0}^{1} \dfrac{t^n}{\sqrt{t+1}} dt$ integral, it got worse. Seems to be some easier way.
| $\sqrt{t+1}$ is approximately constant on $(0,1)$, hence the following application of Cauchy-Schwarz
$$ I_n=\int_{0}^{1}t^n\sqrt{1+t}\,dt\leq \sqrt{\int_{0}^{1}t^{n}\,dt\int_{0}^{1}t^n(1+t)\,dt}=\color{red}{\frac{1}{n+1}\sqrt{\frac{2n+3}{n+2}}}<\frac{\sqrt{2}}{n+1}$$
is expected to provide a tight upper bound. Indeed,
$$ \frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}t^n\left(\sqrt{2}-\sqrt{1+t}\right)\,dt = \int_{0}^{1}(1-t)^n\left(\sqrt{2}-\sqrt{2-t}\right)\,dt $$
leads to
$$\begin{eqnarray*}\frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}\frac{t(1-t)^n}{\sqrt{2}+\sqrt{2-t}}\,dt&\leq&\frac{1}{\sqrt{2}+1}\int_{0}^{1}t(1-t)^n\,dt\\&=&\frac{1}{\sqrt{2}+1}\cdot\frac{1}{(n+1)(n+2)}\\&<&\frac{1}{2(n+1)^2}.\end{eqnarray*}$$
Actually we have
$$ I_n = \frac{\sqrt{2}}{n+1}+\frac{C_n}{(n+1)^2} $$
where $C_n\in\left[\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}+1}\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A proof on multinomial roots If $x_1,x_2,...,x_{n-1},x_n$ be the roots of the equation $$1 + x + x^2 + ... + x^n = 0$$ and $y_1,y_2,...,y_{n},y_{n+1}$ be those of equation $$1 + x + x^2 + ... + x^{n+1} = 0$$ show that $$(1-x_1)(1-x_2)...(1-x_n)=2\left[\frac{1}{1-y_1}+\frac{1}{1-y_2}+...+\frac{1}{1-y_{n+1}}\right]$$
I couldn't get any clue to solve the question. Any suggestion will be very helpful.
| The hint:
$$x^{n+1}-1=(x-1)(x-x_1)(x-x_2)...(x-x_n),$$ which gives
$$x^n+x^{n-1}+...+x+1=(x-x_1)(x-x_2)...(x-x_n),$$ which for $x=1$ gives
$$n+1=(1-x_1)(1-x_2)...(1-x_n).$$
By the similar way you can calculate the second expression.
Indeed, $\frac{1}{y_1-1}$, $\frac{1}{y_2-1}$,...,$\frac{1}{y_{n+1}-1}$ they are roots of the equation:
$$\frac{\left(\frac{1}{x}+1\right)^{n+2}-1}{\frac{1}{x}+1-1}=0$$ or
$$(1+x)^{n+2}-x^{n+2}=0$$ or
$$(n+2)x^{n+1}+\frac{(n+2)(n+1)}{2}x^n+...=0$$ or
$$x^{n+1}+\frac{n+1}{2}x^n+...=0,$$ which gives
$$\frac{1}{y_1-1}+\frac{1}{y_2-1}+...+\frac{1}{y_{n+1}-1}=-\frac{n+1}{2}$$ or
$$\frac{1}{1-y_1}+\frac{1}{1-y_2}+...+\frac{1}{1-y_{n+1}}=\frac{n+1}{2}$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$ So I got this problem:
Determine the following limit value:
$$\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$$
What I tried is:
$\large{\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}\cdot{\sqrt{x^2+16}+4\over\sqrt{x^2+16}+4}=\\\lim_\limits{x\to0}{({\sqrt{x^2+1}-1})(\sqrt{x^2+16}+4)\over x^2+16-16}}$
From this point I just get everything messy, and can't get anything out of it, so I believe this is not the way to solve this. Doing it from a table of values gives 4, but I should solve this without the table.
| Hint:
$$f(a)=\lim_{x\to0}\dfrac{\sqrt{x^2+a^2}-a}{x^2}=\lim...\dfrac{x^2+a^2-a^2}{x^2(\sqrt{x^2+a^2}+a)}=\dfrac1{\sqrt{a^2}+a}$$
For $a>0,$ $$f(a)=\dfrac1{2a}$$
$$\dfrac{f(1)}{f(4)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ According to De Moivre's theorem, we get:
$$z^n= \cos{\left(n\arctan{\frac{4}{3}}\right)} + i\sin{\left(n\arctan{\frac{4}{3}}\right)}$$
Now, if for $n\in \mathbb{N}: z^n=1$, then the imaginary part of the expression above must be zero, therefore:
$$\sin{\left(n\arctan{\frac{4}{3}}\right)}=0 \iff n\arctan{\frac{4}{3}} = k\pi, \ \ \ k \in \mathbb{N}$$
And we get that for $z$ to be a root of unity for some natural number $n$, $n$ must be in the form:
$$n = \frac{k\pi}{\arctan{\frac{4}{3}}}, \ \ \ k \in \mathbb{N}$$
On the other hand, for $z^n=1$ it must be that:
$$\cos{\left(n\arctan{\frac{4}{3}}\right)} = 1 \iff n\arctan{\frac{4}{3}} = 2l\pi, \ \ \ l \in \mathbb{N}$$
and thus
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
By comparing those two forms of $n$, it must be the case that $k=2l$ and for $n$ to satisfy $z^n = 1$. What follows is that $n$ should be in the form
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
But, at the same time, $n$ must be a natural number. Should I prove now that such $n$ cannot even be a rational number, let alone a natural one? Or how should I approach finishing this proof?
| $$z=\frac{3+4i}5=\frac{(1+2i)^2}5=\frac{1+2i}{1-2i}.$$
The ring $\Bbb Z[i]$ is a UFD and $1+2i$ and $1-2i$ are non-associate
primes therein. So in $\Bbb Z[i]$, $(1+2i)^n$ is never divisible by $1-2i$
so that
$$z^n=\frac{(1+2i)^n}{(1-2i)^n}$$
is always in lowest terms and cannot cancel to equal $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What does $\lim \inf _ { n \rightarrow \infty }$ mean? I'm new into Mathematical Analysis, and my textbook says:
Consider the series: $\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac {
1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 2 ^ { 3
} } + \frac { 1 } { 3 ^ { 3 } } + \frac { 1 } { 2 ^ { 4 } } + \frac {
1 } { 3 ^ { 4 } } + \cdots$
for which
$\liminf _ { n \rightarrow \infty } \frac { a _ { n + 1 } } { a _ { n
} } = \lim _ { n \rightarrow \infty } \left( \frac { 2 } { 3 } \right)
^ { n } = 0$
$\liminf _ { n \rightarrow \infty } \sqrt [ n ] { a _ { n } } = \lim _
{ n \rightarrow \infty } \sqrt [ 2 n ] { \frac { 1 } { 3 ^ { n } } } =
\frac { 1 } { \sqrt { 3 } }$
$\limsup _ { n \rightarrow \infty } \sqrt [ n ] { a _ { n } } = \lim _
{ n \rightarrow \infty } \sqrt [ 2 n ] { \frac { 1 } { 2 ^ { n } } } =
\frac { 1 } { \sqrt { 2 } }$
$\limsup _ { n \rightarrow \infty } \frac { a _ { n + 1 } } { a _ { n
} } = \lim _ { n \rightarrow \infty } \frac { 1 } { 2 } \left( \frac {
3 } { 2 } \right) ^ { n } = + \infty$
My question is: What does $\lim \inf _ { n \rightarrow \infty }$ mean?
And why $\lim \inf _ { n \rightarrow \infty } \frac { a _ { n + 1 } } { a _ { n > } } = \lim _ { n \rightarrow \infty } \left( \frac { 2 } { 3 } \right)$?
| You can refer to your notes or to the wiki page Limit superior and limit inferior for the definition.
It is just a generalization for the limit of a sequence, in particular we consider
$$I_n=\inf\{a_n:k\ge n\}$$
and since $I_n$ is an increasing sequence the limit always exist (finite or infinite) and then we can consider and define
$$\liminf_{n\to \infty} a_n =\lim_{n\to \infty} I_n $$
To show that for a given sequence $a_n$ we have $\liminf_{n\to \infty} a_n =L$ we need to show two fact
*
*$a_n \ge b_n \quad b_n\to L$
*find a sub-sequence $a_{i_n}$ such that $a_{i_n}\to L$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4 + y^2}$ Prove or disprove each of the follow function has limits $x \to a$ by the definition
$$\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4 + y^2}$$
Let $y = mx^2$
$$\frac{x^2y}{x^4 + y^2}=\frac{x^4m}{x^4(1+m^2)} = \frac{m}{1+m^2}$$
Hence, $$\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4+y^2} = \frac{m}{1+m^2}$$
Therefore, since it depends on the value of m, the limit does not exist
Correct?
| Yes that correct, indeed we can consider for example
*
*for $x=0,\, y\neq 0 \implies \frac{x^2y}{x^4 + y^2}=0$
*for $x=t,\,y=t^2, \,t\to 0 \implies \frac{x^2y}{x^4 + y^2}=\frac{t^4}{t^4 + t^4}\to \frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $4\alpha^2–5\beta^2+6\alpha+1=0$.Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. If $4\alpha^2–5\beta^2+6\alpha+1=0$. Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. I tried to solve this question by taking a General equation of circle and then substituting the values but could not proceed further I took the line as a tangent and try to prove it by equating the radius with perpendicular distance of the line from the assumed centre.
| We have $\beta(\alpha)=\pm\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}$ so you want to find the envelope of the family $f_\alpha(x,y)=x\alpha+y\beta(\alpha)+1=0$.
In other words, $F(x,y,\alpha):=x\alpha\pm y\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}+1=0$, $\dfrac{\partial}{\partial\alpha}F(x,y,\alpha)=0$,
i.e.,
$$
\left\{
\begin{aligned}
x\alpha\pm y\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}+1&=0\\
x\pm y\dfrac{4\alpha+3}{\sqrt{4\alpha^2+6\alpha+1}\sqrt{5}} &=0
\end{aligned}
\right.
$$
which gives
$$
x=\frac{4\alpha+3}{3\alpha+1}, y=\mp\frac{\sqrt{5}\sqrt{4\alpha^2+6\alpha+1}}{3\alpha+1}
$$
and eliminating $\alpha$ gives $x^2+y^2-6x+4=0$, from which you can read off the centre and radius of the circle.
| {
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"timestamp": "2023-03-29T00:00:00",
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?
| Let $$f(x)=\sum^\infty_{k=1} \frac{(-1)^k}{k+1}x^k$$
Then, $G(x)=e^f$.
To expand $G$ in Taylor series up to the cube term, we have to compute up to the third derivative of $G$.
This isn’t too difficult.
$$G(0)=1$$
$$G’(0)=f’(0)=-\frac12$$
$$G’’(0)=f’’(0)+f’^2(0)=2\cdot \frac13+\frac14$$
$$G’’’(0)=f’’’(0)+2f’(0)f’’(0)+f’’(0)f’(0)+f’^3(0)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
| $$x_n=\frac{(n^3+1)^2-(n^2-1)^3}{\sqrt[3]{(n^3+1)^6}+\sqrt[3]{(n^3+1)^5}\sqrt{n^2-1}+...}\rightarrow0,$$
which says that $\{x_n\}$ is bounded.
| {
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How to find the average value of $f(x,y)$ on $R=\{(x,y): -1\le x\le1,0\le y\le3\}$ Given $R=\{(x,y): -1\le x\le1,0\le y\le3\}$
For the function $f(x,y)$ evaluate $\int\int_R f(x,y)dA$ and find the average value of $f(x,y$ on $R$
$f(x,y)=x^3+xy^3-3y$
My Try:
$=\int_{-1}^{1}\int_0^3x^3+xy^3-3ydydx$
$=\int_{-1}^{1}(x^3y+\dfrac{xy^4}{4}-\dfrac{3y^2}{2})\bigg|_0^3dx$
$=\int_{-1}^{1}3x^3+\dfrac{81x}{4}-\dfrac{27}{2}dx$
$=\dfrac{3x^4}{4}+\dfrac{81x^2}{8}-\dfrac{27x}{2}\bigg|_{-1}^{1}$
$=-\dfrac{54}{2}$
$=-27$
Is my above answer correct?
How to find the average value of $f(x,y)$ on $R$?
| So, not quite!
First, we need to know how exactly the average value of a two variable function $f(x,y)$ is computed. Typically, the following equation is utilized to compute the average value of a function $f(x,y)$ over a rectangle $R$:
$$f_{ave} = \frac{1}{\text{Area}(R)} \iint_R f(x,y) dxdy$$
Since we're given a rectangle $R$ with side lengths $2$ and $3$, we find that $\text{Area(R)}$ = 6.
So, our equation above becomes:
$$f_{ave} = \frac{1}{6} \iint_R f(x,y) dxdy$$
Then, we need only simply compute $\iint_R f(x,y) dxdy$ which you have done correctly:
$$\iint_R f(x,y) dxdy = \int_{-1}^{1} \int_{0}^{3} (x^3 + 3xy^3 -3y)dydx = -27$$
So, your answer will simply be:
$f_{ave} = \frac{1}{6} (-27) = -\frac{27}{6}$
| {
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Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R=0$.
For $p \equiv 3 \mod 4$, how can we find $R$? Does $R$ depend on the value of $p$?
| Supposing $$(x-1)(x-2)...(x-(p-1))\equiv x^{p-1}\pm1\pmod{p}$$
The left-hand side is $$(x^2-1^2)(x^2-2^2)...(x^2-({p-1\over2})^2)\pmod p$$
Replace $x^2$ by $y$; and then replace $y$ by $-1$. The left-hand side is now a square-root of your product $\pmod p$, give or take a sign. The right-hand side is $1\pm1$.
Can you fill in the details?
| {
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Proving the given inequalities Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.Thanks in advance.
| First of all, you should have $\geq $ or $\leq$ instead of $>$ and $<$ in your inequalities. In each of them, the equality occurs if and only if $a=b=c$.
(i) Use the Weighted AM-GM Inequality to show that
$$\frac{bc+ca+ab}{a+b+c}=\frac{b}{a+b+c}c+\frac{c}{a+b+c}a+\frac{a}{a+b+c}b\geq c^{\frac{b}{a+b+c}}a^{\frac{c}{a+b+c}}b^{\frac{a}{a+b+c}}\,.$$
Similarly,
$$\frac{bc+ca+ab}{a+b+c}=\frac{c}{a+b+c}b+\frac{a}{a+b+c}c+\frac{b}{a+b+c}a\geq b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}}a^{\frac{b}{a+b+c}}\,.$$
Multiplying these two inequalities to get
$$\left(\frac{bc+ca+ab}{a+b+c}\right)^2\geq \left((bc)^a(ca)^b(ab)^c\right)^{\frac{1}{a+b+c}}\,,$$
which is equivalent to the required inequality.
(ii) For the inequality on the right, note that
$$\frac{a^2+b^2+c^2}{a+b+c}=\frac{a}{a+b+c}a+\frac{b}{a+b+c}b+\frac{c}{a+b+c}c\geq a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}}=\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}$$
by the Weighted AM-GM Inequality. Thus,
$$\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c}\geq a^ab^bc^c$$
as desired.
For the inequality on the left, observe that
$$\frac{3}{a+b+c}=\frac{a}{a+b+c}\left(\frac{1}{a}\right)+\frac{b}{a+b+c}\left(\frac1b\right)+\frac{c}{a+b+c}\left(\frac1c\right)\,.$$
Thus, by the Weighted AM-GM Inequality,
$$\frac{3}{a+b+c} \geq \left(\frac1a\right)^{\frac{a}{a+b+c}}\left(\frac1b\right)^{\frac{b}{a+b+c}}\left(\frac{1}{c}\right)^{\frac{c}{a+b+c}}=\frac{1}{\left(a^ab^bc^c\right)^{\frac{1}{a+b+c}}}\,.$$
This is equivalent to the inequality to be proven.
| {
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Prove following inequality Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$
What I tried was to use AM-GM for the left side of this inequality, what I got was
$3(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 3$ and $(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 1$ or just $\frac{8abc}{(a+b)(b+c)(c+a)}≥ 1$, but this isn't true.
| Hint:
WLOG you may set $a+b+c=3$ and show instead that for $x\in[0,3]$,
$$\left(\frac{2x}{3-x}\right)^{2/3}\geqslant x$$
But this is equivalent to the obvious $x^2(4-x)(x-1)^2\geqslant0$.
| {
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Question about the dimension of elements in quotient ring Here is the question:
Let $\def\Z{\mathbb Z}\Z[X]$ be the polynomial ring over $\Z$ with variable $X$. Let $f(X) = X^3+X^2+2X+2$. Let $I$ be the ideal of $\Z[X]$ generated by $f(x)$ and $5$. Put $A =\Z[X]/I$.
*
*Find the number of elements $a$ in $A$ such that $a^2=1$
*Find the number of elements $b$ in $A$ such that $b^{18}=1$
I try to use the properties of multiplication and addition on quotient ring and I almost solved the question 1 very straightly by enumeration. Then I find that enumeration is useless for question 2, I wonder whether there is a common solution such that I can analyze the dimensions of the elements in quotient ring and the dimension of the quotient ring especially in the situation that the original ring is infinite ring. Thank you for your help.
| Note that $X^3+X^2+2X+2 = (X+1)(X^2+2)$. So we have, by the Chinese reminder theorem:
$$
\Bbb Z_5[X]/(X^3+X^2+2X+2) \cong \Bbb Z_5[X]/(X+1)\times \Bbb Z_5[X]/(X^2+2)
$$
We can find the number of solutions to $a^2 = 1$ and $b^{18} = 1$ in each of those separately. This is helpful because each of those two factors in the product to the right are fields (one is isomorphic to $\Bbb F_5 = \Bbb Z_5$, and the other one is $\Bbb F_{25}$).
For $a^2 = 1$, which becomes $(a+1)(a-1) = 0$. Using that fields are integral domains (a product is $0$ iff one of the factors is zero) we see that $(\pm 1, \pm1)$ (where the two $\pm$ are independent) are the only four solutions. These correspond to $\pm 1$ and $\pm(X^2 + 3)$ in the original ring.
$b^{18} = 1$ is a bit more tricky. Let's take $\Bbb Z_5[X]/(X+1)$ first. By Fermat's little theorem, $b^{18} = b^2$, so the same two solutions as above are the only two solutions.
For $\Bbb Z_5[X]/(X^2+2)$, the corresponding version of Fermat's little theorem says that as long as $b\neq 0$ (and $0$ isn't a solution anyways, so we don't lose anything by assuming this), we have $b^{24} = 1$. If we want to have $b^{18} = 1$, then this implies that we also need $b^6 = 1$. Thus we get
$$
0 = b^6-1 = (b^3-1)(b^3+1)\\
= (b-1)(b^2+b+1)(b+1)(b^2-b+1)
$$
We are interested in just finding the number of solutions to $b^6-1 = 0$, and there is theory which lets us do that rather quickly. I have done the actual calculation too, and you may skip to that if there is too much unfamiliar theory here.
$\Bbb Z_5[X]/(X^2+2)$ is a quadratic extension of the finite field $\Bbb Z_5$, and it must therefore contain the roots of any quadratic polynomial, including $b^2\pm b+1$. We know that there cannot be repeated roots because $b^6-1$ and its derivative $b^5$ are coprime polynomials. Thus there are six distinct solutions.
So that means $b^6 = 1$ has six solutions in $\Bbb Z_5[X]/(X^2+2)$. Combined with the two solutions from $\Bbb Z_5[X]/(X+1)$, this gives $2\cdot 6 = 12$ solutions in our original ring.
Solving $b^6 = 1$: We see $b\pm 1$ in the factorisation above, so $b = \pm 1$ are two solutions. As for the other two factors, the quadratic equation works just as well here as it does for real numbers (only we have to remember we're working in a finite field, so division and square roots work a little differently):
$$
b^2+b+1 = 0\implies b = \frac{-1\pm\sqrt{1-4}}{2} = 2\pm 3\sqrt{2}
$$
Now we just need to find $\sqrt2$. By definition, we have
$$
X^2+2 = 0\\
X^2 = 3\\
4X^2 = 2\\
(2X)^2 = 2
$$
so $\sqrt2 = 2X$ (or, more precisely, $\pm\sqrt2$ represents the same two elements as $\pm2X$; we haven't decided on a square root convention, and we don't need to). Thus we have
$$
b = 2\pm3\cdot 2X = 2\pm X
$$
These are the solutions to $b^2+b+1 = 0$. Now note that $b^2-b+1 = 0$ necessarily has the same solutions with opposite sign: $b = 3\pm X$.
So, going back to our original ring, we saw that the four solutions from before, $\pm 1$ and $\pm(X^2+3)$ work. The other $8$ solutions are given by
$$
\begin{array}{|c|c|c|c|}\hline
\Bbb Z_5[X]/(X+1)&\Bbb Z_5[X]/(X^2+2)&&\Bbb Z_5[X]/(f(X))\\
\hline 1&2+X&&X+2\\
1&2-X&&X^2-X-1\\
1&3+X&&3X^2+X-1\\
1&3-X&&-X^2-X + 1\\\hline
\end{array}
$$
and their negatives.
| {
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Assuming that $|z|<1$, calculate: $\sum_{n=2}^{\infty}(n^2-3n+2){z^{n-1}}$ The answer is: $\frac{-2z^2}{(1-z)^{3}}$
When I do it, I get the answer without the minus sign.
So far, I got:
$$
\begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^n
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^n \\
&= z^3 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\
&= z^3 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}
\end{split}
$$
| \begin{split}
\sum_{n=2}^{\infty}(n^2-3n+2)z^{n-1}
&= \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\
&= z^2 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-3} \\
&= z^2 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}\\
&= z^2 \frac{d^2}{dz^2} \sum_{n=0}^{\infty}z^{n}\\
&= z^2 \frac{d^2}{dz^2} \frac{1}{1-z}\\
&= z^2 \frac{2}{(1-z)^3}
\end{split}
This proof holds for all complex $z$ with $|z|<1$ which means that, in particular, this also holds for all real $z$ with $0<z<1$. From the task setting itself, in this case the answer must be positive, since only positive terms are added. So if the result is claimed to be $ \frac{-2 z^2}{(1-z)^3}$, this is false.
| {
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Irrational equation
Solve over the real numbers: $$(x^2+x+1)^{1/3}+(2x+1)^{1/2}=2$$
I know for the second radical to be defined $x≥-0,5$ and I've attempted various methods I've solved other such equations with but to no avail; if I could write $x^2+x-7$ in terms of $2x+1$ to use a convenient notation in $x^2+x-7-6(2x+1)+12(2x+1)^{1/2}+(2x+1)(2x+1)^{1/2}=0$
I think I could solve it. I've ran it through Wolfram and the only real solution is $0$ however how the conclusion was reached I am not aware.
| If we put $$a =(x^2+x+1)^{1/3}\;\;\;{\rm and}\;\;\;b= (2x+1)^{1/2}$$
So $$a^3= x^2+x+1 \;\;\;{\rm and}\;\;\;b^2= 2x+1\;\;\;{\rm and}\;\;\; a+b=2$$
and thus $$4a^3 = 4x^2+4x+4 = (2x+1)^2+3 = b^4+3$$
and finally $$ 4(2-b)^3 = b^4+3$$ ...
$$ b^4+4b^3-24b^2+48b-29=0$$
$$ (b-1)(b^3+ \underbrace{ 5b^2-19b+29}_{f(b)})=0$$
Since the discriminat of $f$ is negative $f$ is always positive. Now $b\geq 0$ so $b^3+f(b)>0$ and thus $b=1$ is only solution i.e. $x=0$.
| {
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Congruence with Lucas sequence I want to show that for each $n \geq 2$ the following congruence holds,
$$L_{2^n} \equiv 7 \pmod{10}.$$
According to my notes,
$$L_n=\left( \frac{1+\sqrt{5}}{2}\right)^n+\left( \frac{1-\sqrt{5}}{2}\right)^n$$
and
$$L_n=F_{n-1}+F_{n+1},$$
where $F_n$ is the $n$-th Fibonacci number.
Do we use somehow the last equality in order to show the desired congruence?
| The following helps :
$$L_{2n}=L_n^2-2(-1)^n\tag1$$
(the proof is written at the end of the answer.)
This is a proof by induction.
For $n=2$, $L_{2^2}=7\equiv 7\pmod{10}.$
Suppose that $L_{2^n}=10k+7$ for some $k\in\mathbb N$.
Then, using $(1)$, we get
$$\begin{align}L_{2^{n+1}}&=L_{2^n}^2-2(-1)^{2^n}
\\\\&=(10k+7)^2-2
\\\\&=10(10k^2+14k+4)+7
\\\\&\equiv 7\pmod{10}\end{align}$$
Proof for $(1)$ :
Using that $L_{n}=\alpha^n+\beta^n$ where $\alpha=\frac{1-\sqrt 5}{2}$ and $\beta=\frac{1+\sqrt 5}{2}$, we get
$$L_{2n}-L_n^2=\alpha^{2n}+\beta^{2n}-(\alpha^n+\beta^n)^2=-2(\alpha\beta)^n=-2(-1)^n$$
| {
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Calculate the limit (Squeeze Theorem?) I have to calculate the limit of this formula as $n\to \infty$.
$$a_n = \frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$\frac{1}{\sqrt{2}}\leftarrow\frac{n}{\sqrt{2n^2}}\le\frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl) \le \frac{n}{\sqrt{n^2+n}}\to1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
| for a decreasing function such as $1/\sqrt x$ with $x$ positive, a simple picture shows
$$ \int_a^{b+1} \; f(x) \; dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b} \; f(x) \; dx $$
$$ \int_{n+1}^{2n+1} \; \frac{1}{\sqrt x} \; dx < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < \int_{n}^{2n} \; \frac{1}{\sqrt x} \; dx $$
getting there
$$ 2 \sqrt {2n+1} - 2 \sqrt {n+1} < \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2n} - 2 \sqrt {n} $$
$$ 2 \sqrt {2+\frac{1}{n}} - 2 \sqrt {1+\frac{1}{n}} < \frac{1}{\sqrt n} \sum_{k=n+1}^{2n} \frac{1}{\sqrt k} < 2 \sqrt {2} - 2 \sqrt {1} $$
| {
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+k^2 \sin^2 \theta}} \,d \theta$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1+k^2 \sin^2\theta}}
\,d \theta$
I wang to let $k=-ai \,\,\,\,\,$ ,then :$$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-a^2 \sin^2\theta}}
\,d \theta$$
But I am not sure whether the elliptic integral can be used when $k$ is complexed.Or is it right that the Taylor expansion of $(1-z)^{-\frac{1}{2}} $ an analogue of $(1-x)^{-\frac{1}{2}} $
| The following manipulations assume $k^2<1$. Since for any $x\in(-1,1)$ we have
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^n \tag{1}$$
and the central binomial coefficients also appear in
$$ \int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n}\tag{2} $$
by Maclaurin series and termwise integration it follows that
$$\begin{eqnarray*} \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1+k^2\sin^2\theta}} &=& \frac{\pi}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2(-1)^n k^{2n}=K(-k^2)\\&=&\frac{\pi}{2\,\text{AGM}(1,\sqrt{1+k^2})}\tag{3} \end{eqnarray*}$$
where the complete elliptic integral of the first kind is denoted according to Mathematica's notation (in order to avoid a proliferation of square roots, it is more practical to assume that the argument of $K$ is the elliptic modulus rather than the elliptic parameter). So your integral is given by a hypergeometric $\phantom{}_2 F_1$ function, whose numerical computation is made extremely simple by the relation to the arithmetic-geometric mean.
If $k^2\geq 1$ we may consider that
$$ \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1+k^2\sin^2\theta}} =\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1+k^2\cos^2\theta}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{(1+k^2)-k^2\sin^2\theta}}$$
equals
$$\frac{1}{\sqrt{1+k^2}} K\left(\frac{k^2}{k^2+1}\right) = \frac{\pi}{2\sqrt{1+k^2}\text{AGM}\left(1,\sqrt{\frac{1}{k^2+1}}\right)}=\color{red}{\frac{\pi}{2\,\text{AGM}(1,\sqrt{1+k^2})}},$$
so the representation through the $\text{AGM}$ is left unchanged. It gives
$$\frac{\pi}{1+\sqrt{1+k^2}}<\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1+k^2\sin^2\theta}}<\frac{\pi}{2(1+k^2)^{1/4}}.\tag{4}$$
For particular values of $k$ (like $k=1$) the middle integral has an explicit representation in terms of the $\Gamma$ function evaluated at multiples of $\frac{1}{24}$, see Wikipedia.
If $k$ takes complex values we have to be careful in managing the determinations of the involved square roots, but the mechanics stays pretty much the same.
| {
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Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
| Hint:
Numerator:
$a^3-b^3= (a-b)(a^2+ab+b^2)$.
$a=(x+5)^{1/3}$; $b=2.$
Denominator:
$c^4-d^4=(c^2-d^2)(c^2+d^2)=$
$(c-d)(c+d)(c^2+d^2);$
$c=(x-2)^{1/4};$ $d=1.$
| {
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How to calculate $\sum_{n=1}^{\infty}A_n(1/2)^{2n+2}(n+2)/(2n+2)$ This is a follow-up to a previous problem, which was answered as follows:
$$S=\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\frac{\pi}{4},$$
where $C_n$ is the $n$th Catalan number.
A related problem has arisen which asks for the following summation:
$$T=\frac{1}{2}+\sum_{n=1}^{\infty}A_n\left(\frac{1}{2}\right)^{\!2n+2}\!\!\left(\frac{n+2}{2n+2}\right),$$
where $A_n$ is integer sequence A000245 (a "Catalan sequence convolved with its shifted variant") given by:
$$A_n=3\frac{(2n)!}{(n+2)!\cdot(n-1)!}.$$
Any thoughts on how to proceed? I am wondering if $S<T$.
| Assuming that $A_n$ is defined as $\frac{3(2n)!}{(n+\color{red}{2})!(n-1)!}$ (otherwise the associated series is divergent, since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$) we have that $T$ equals
$$ \frac{1}{2}+\frac{3}{8}\sum_{n\geq 1}\frac{1}{4^n}\binom{2n}{n}\frac{n}{(n+1)^2} $$
which due to $\sum_{n\geq 0}\frac{x^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$ can be written as
$$ T = \frac{1}{2}+\frac{3}{8}\int_{0}^{1}\left(\frac{1}{\sqrt{1-x}}-1\right)(1+\log x)\,dx = \color{red}{-\frac{1}{4}+\frac{3}{2}\log 2}.$$
| {
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If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| $$\sin(x)(1+\sin^2(x))=1-\sin^2(x)$$
Square both sides & replace $\sin^2(x)$ with $1-\cos^2(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Calculate the limit $\lim \limits_{x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although $x\neq 2$ I should calculate the Limit $\lim \limits_ {x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although I noticed, that $x\neq 2$ must apply. Is the limit undefined? Otherwise, with which steps should I go on to calculate the limit?
| Given $$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{\color{red}{(x-2)}(x+4)}{x\color{red}{(x-2)}}=\lim_{x\rightarrow2}\dfrac{x+4}{x}=\lim_{x\rightarrow2}1+\dfrac4x = 3$$
OR
You could also use L'Hopital's rule
$$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{2x+2}{2x-2}=\dfrac{2(2)+2}{2(2)-2}=\dfrac{6}{2}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2} \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.
| Let $x = \tan\theta$ and make use of
\begin{align}
1 + \tan^{2}\theta &= \frac{1}{\cos^{2}\theta} \\
dx &= \frac{d\theta}{\cos^2\theta} \\
\sin(\tan^{-1}\theta) &= \frac{x}{\sqrt{1+x^2}} \\
\cos(\tan^{-1}\theta) &= \frac{1}{\sqrt{1 + x^2}}
\end{align}
to obtain:
\begin{align}
I &= \int \frac{dx}{(1+x^2)^2} \\
&= \int \cos^{2}\theta \, d\theta \\
&= \frac{1}{2} \, \int (1 + \cos(2 \theta) ) \, d\theta \\
&= \frac{1}{2} \, \left( \theta + \frac{\sin(2 \theta)}{2} \right) = \frac{1}{2} \, ( \theta + \sin\theta \, \cos\theta ) \\
&= \frac{1}{2} \, \left( \tan^{-1}x + \frac{x}{1+x^2} \right).
\end{align}
Now evaluate the integral with endpoints $a$ and $b$. If the range was $(0,1)$ then
$$\int_{0}^{1} \frac{dx}{(1+x^2)^2} = \frac{\pi + 2}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$
Prove convergence of the following sequence: $$\frac{n+3}{n^2 -3} \rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $\frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ \left|\frac{n+3}{n^2 -3}-0\right|=\frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $\frac{1}{n^2 -3}< \frac{1}{n^2 -9}$ we can thus write:
$$\frac{n+3}{n^2 -3}<\frac{n+3}{n^2 -9}=\frac{(n+3)}{(n+3)(n-3)}=\frac{1}{n-3} $$
To be able to complete this proof we want that $\frac{1}{n-3}<\epsilon$, we write $n-3>\frac{1}{\epsilon}$ or $n> \frac{1}{\epsilon} +3$. If we pick $n_0 =\lceil\frac{1}{\epsilon} +3\rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $\epsilon>0$, we let $n_0=\lceil{\frac{1}{\epsilon}+3 }\rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=\left|\frac{n+3}{n^2-3} \right|<\frac{n+3}{n^2-9}=\frac{1}{n-3}< \frac{1}{\frac{1}{\epsilon}+3-3}=\epsilon$$
And hence our sequence converges to $0$ $\square$.
Is my proof okay?
| Yup, the proof is correct.
Also, "we write $n-3>\color{blue}{\frac1{\epsilon}}$"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$ I was playing around with some integrals and noticed that some integrals of the form: $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form.
I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.
$$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$
Let's start by letting $x-1=t\,$ thus:
$$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$
With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$
$$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$
$$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$
$${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$
$$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$
Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue.
$$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$
Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$
There are more examples that I found by checking and try such as:
$$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$
$$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$
$$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$
And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.
The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$?
| The simplest approaches are sometimes the right ones. Therefore choose any arbitrarily numbers $a,b$ and $c$ and apply your algorithm. So by firstly setting $t=x-a$ we get
$$\begin{align}
I(a,b,c)=\int_a^{\infty}\frac{\arctan(x+b)}{x^2+c}dx=\int_0^{\infty}\frac{\arctan(t+a+b)}{t^2+2at+(c+a^2)}dt
\end{align}$$
Now set $\alpha=c+a^2$ and then $\displaystyle t=\frac{c+a^2}{y}$ to further get
$$\begin{align}
I(a,b,c)=\int_0^{\infty}\frac{\arctan(t+a+b)}{t^2+2at+(c+a^2)}dt&=\int_0^{\infty}\frac{\arctan\left(\frac{c+a^2}{y}+a+b\right)}{\left(\frac{c+a^2}{y}\right)^2+2a\left(\frac{c+a^2}{y}\right)+(c+a^2)}\frac{c+a^2}{y^2}dy\\
&\stackrel{y=t}{=}\int_0^{\infty}\frac{\arctan\left(\frac{c+a^2}{y}+a+b\right)}{t^2+2at+(c+a^2)}dt
\end{align}$$
Adding the first and the second form up results in using the addition theorem of the inverse tangent function. This addition looks like the following
$$\small\begin{align}
\arctan(\color{blue}{t+a+b})+\arctan\left(\color{red}{\frac{c+a^2}{y}+a+b}\right)&=\arctan\left(\frac{\color{blue}{t+a+b}+\color{red}{\frac{c+a^2}{y}+a+b}}{1-(\color{blue}{t+a+b})\left(\color{red}{\frac{c+a^2}{y}+a+b}\right)}\right)\\
&=\arctan\left(\frac{t^2+2(a+b)t+a^2+c}{-(a+b)\left(t^2+\frac{2a^2+2ab+b^2+c-1}{a+b}t+a^2+c\right)}\right)
\end{align}$$
In order to make the polynomial $t^2+2(a+b)t+a^2+c$ vanish the following condition has to be fulfilled
$$\frac{2a^2+2ab+b^2+c-1}{a+b}=2(a+b)$$
From hereon we can deduce a relation between $a,b$ and $c$ that has to be satisfied. To be precise
$$\begin{align}
\frac{2a^2+2ab+b^2+c-1}{a+b}=2(a+b)&\Leftrightarrow 2a^2+2ab+b^2+c-1=2(a+b)^2\\
&\Leftrightarrow 2a^2+2ab+b^2+c-1=2a^2+2b^2+4ab\\
&\Leftrightarrow c=b^2+2ab+1
\end{align}$$
And indeed your given case $I(\color{red}{1},\color{blue}{3},\color{green}{16})$ fulfills this relation as $\color{green}{16}=\color{blue}{3}^2+2\cdot\color{blue}{3}\cdot\color{red}{1}+1$. Another example would be $I(\color{red}{1},\color{blue}{1},\color{green}{4})$ which again turns out to work since $\color{green}{4}=\color{blue}{1}^2+2\cdot\color{blue}{1}\cdot\color{red}{1}+1$.
Therefore for your given case $I(7,13,c)$ you have to choose $c=352$ in order to let the integral be solvable via symmetry.
To finish the evaluation of $I(a,b,c)$, now under the restriction of $c=b^2+2ab+1$, we arrive at
$$\small\arctan\left(\frac{t^2+2(a+b)t+a^2+c}{-(a+b)\left(t^2+\frac{2a^2+2ab+b^2+c-1}{a+b}t+a^2+c\right)}\right)=\arctan\left(\frac{-1}{a+b}\right)=\pi-\arctan\left(\frac1{a+b}\right)$$
following the same argumentation as you did. So for $I(a,b,c)$ as a whole we get
$$\begin{align}
2I(a,b,c)&=\left(\pi-\arctan\left(\frac1{a+b}\right)\right)\int_0^{\infty}\frac{dt}{t^2+2at+(c+a^2)}\\
&=\left(\pi-\arctan\left(\frac1{a+b}\right)\right)\int_0^{\infty}\frac{dt}{(t+a)^2+c}\\
&=\left(\pi-\arctan\left(\frac1{a+b}\right)\right)\left[\frac1{\sqrt{c}}\arctan\left(\frac{t+a}{\sqrt{c}}\right)\right]_0^{\infty}\\
\Leftrightarrow I(a,b,c)&=\frac1{2\sqrt{c}}\left(\pi-\arctan\left(\frac1{a+b}\right)\right)\left[\frac{\pi}2-\arctan\left(\frac a{\sqrt{c}}\right)\right]
\end{align}$$
Where the final formula produces the right value for your example integral $I(1,3,16)$.
I do not claim that this deduced relation between $a,b$ and $c$ is the only one for which the integral can be evaluated via symmetry but in fact it is one possibilty.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integral $\int \frac{x^2 + 1}{x^4 + x^2 +1} dx$ Evaluate the integral $\int \frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
| It turns out that we must compute the integral in the following way:
$$I=\int\frac{x^2+1}{x^4+x^2+1}\mathrm{d}x$$
$$I=\int\frac{x^2+1}{(x^2-x+1)(x^2+x+1)}\mathrm{d}x$$
After a fraction decomposition,
$$I=\frac12\int\frac{\mathrm{d}x}{x^2+x+1}+\frac12\int\frac{\mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=\int\frac{\mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=\int\frac{\mathrm{d}x}{(x+\frac12)^2+\frac34}$$
Then the substitution $u=\frac1{\sqrt{3}}(2x+1)$ gives
$$I_1=\frac2{\sqrt{3}}\int\frac{\mathrm{d}u}{u^2+1}$$
$$I_1=\frac2{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}$$
Similarly,
$$I_2=\int\frac{\mathrm{d}x}{x^2-x+1}$$
$$I_2=\int\frac{\mathrm{d}x}{(x-\frac12)^2+\frac34}$$
And the substitution $u=x-\frac12$ carries us to
$$I_2=\frac2{\sqrt{3}}\arctan\frac{2x-1}{\sqrt{3}}$$
Plugging in:
$$I=\frac1{\sqrt{3}}\bigg(\arctan\frac{2x+1}{\sqrt{3}}+\arctan\frac{2x-1}{\sqrt{3}}\bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=\int\frac{\mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=\frac1{\sqrt{3}}\arctan\frac{x^2+1}{x\sqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a formula of a power of a matrix Part of a solution I came across of calculating the following matrix:
$$\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $M^2,M^3,M^4$ but I can seem to see the pattern. How should I approach this issue?
| Idea
$$
\\\begin{pmatrix}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{pmatrix}=\begin{pmatrix}1 & -1\\
1 & 1
\end{pmatrix}\cdot\frac{\sqrt{2}}{2}
\\\begin{pmatrix}1 & -1\\
1 & 1
\end{pmatrix}^n\cdot\Big(\frac{1}{\sqrt{2}}\Big)^n=\begin{pmatrix}-i & i\\
1 & 1
\end{pmatrix}\cdot\begin{pmatrix}(1-i)^n & 0\\
0 & (1+i)^n
\end{pmatrix}\cdot\begin{pmatrix}-i & i\\
1 & 1
\end{pmatrix}^{-1}\cdot\frac{1}{2^{\frac n 2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_{0}^{2\pi}\frac{1}{1+\sin^{2}(\theta)}d\theta$ I'm pretty sure the idea is to interpret this integral as a contour integral over a closed path in the complex plane.
$$I = \int_{0}^{2\pi}\frac{1}{1+\sin^{2}(\theta)}d\theta$$
We take $C = \{z \in \mathbb{C} : |z| = 1\}$, which is positively oriented.
$$z = e^{i\theta}$$
$$\frac{dz}{d\theta} = ie^{i\theta} = iz \rightarrow d\theta = \frac{dz}{iz} $$
$$\sin(\theta) = \frac{e^{i\theta}-e^{-i\theta}}{2i} = \frac{z-z^{-1}}{2i} $$
$$\sin^2(\theta) = \frac{z^2 - 2zz^{-1} + z^{-2}}{(2i)^2} = \frac{z^2 - 2 + z^{-2}}{-4} $$
After some manipulation:
$$I = \frac{4}{i}\oint_{C}\frac{z}{-z^4+6z^2-1}dz$$
After that one should obtain the roots of the polynomial in the denominator and solve the integral either by residues or using Cauchy's formula. However, I'm not sure how to find the roots. This was in the context of an exam, so I assume something must be wrong if the roots are too difficult to find, or if they are not "nice" numbers.
Is there a more clever way of going about this? I tried calculating the residue at infinity, but $\frac{1}{z^{2}}f(\frac{1}{z})$ yields the exact same function and it doesn't get any easier.
Thanks.
| How about using Weierstrauss substitution.
$$I=\int_0^{2\pi}\frac{1}{1+\sin^2(x)}dx$$
$t=\tan\frac{x}{2}$ so it becomes:
$$I=4\int_0^\infty\frac{1}{1+\left(\frac{2t}{1+t^2}\right)^2}\frac{dt}{1+t^2}=4\int_0^\infty\frac{1+t^2}{(1+t^2)^2+(2t)^2}dt$$
$$=4\int_0^\infty\frac{t^2+1}{t^4+6t^2+1}dt=4\int_0^\infty\frac{t^2+1}{(t^2+3-2\sqrt{2})(t^2+3+2\sqrt{2})}dt$$
now using partial fractions:
$$\frac{t^2+1}{(t^2+3-2\sqrt{2})(t^2+3+2\sqrt{2})}=\frac{At+B}{t^2+3-2\sqrt{2}}+\frac{Ct+D}{t^2+3+2\sqrt{2}}$$
giving us the simultaneous equations:
$$A+C=0$$$$B+D=1$$$$(3+2\sqrt{2})A+(3-2\sqrt{2})C=0$$$$(3+2\sqrt{2})B+(3-2\sqrt{2})D=1$$
Both $A$ and $C$ come out as $0$ so some form of $\tan$ substitution can be used to solve the two integrals that are formed.
| {
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"url": "https://math.stackexchange.com/questions/3007087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to proceed with this integral?
Let $ G:= \left\{ (x,y) \in \mathbb{R}^2 : 0 < y,\: x^2 + \frac{y^2}{9} <1\: ,\: x^2+y^2 > 1 \right\} $.
I want to calculate this integral:
$ \displaystyle\int_G x^2\,dxdy $.
I want to try with polar coordinates:
so I set $ (x,y) = (r\cos\phi,r\sin\phi)$, but
I am not sure how to get the right boundaries for $\phi $. Isn't it $ x^2 +y^2 = r^2 $ ?
Any help is very appreciated !
| Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.
Since $y>0$, we are integrating over the first two quadrants, so that $0 \leq \phi \leq \pi$.
Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
$$ r^2\cos^2\theta + \frac{r^2\sin^2\theta}{9} < 1 \quad \Rightarrow \quad r < \frac{3}{\sqrt{8\cos^2\theta+1}}. $$
Thus
\begin{align}
\int_0^\pi \int_{1}^{\frac{3}{\sqrt{8\cos^2\theta+1}}} r^2\cos^2\theta \cdot r \,\mathrm dr \, \mathrm d\theta &= \frac{81}{4} \int_0^\pi \frac{\cos^2\theta}{(8\cos^2\theta+1)^2} \, \mathrm d\theta - \frac{1}{4} \int_0^\pi \cos^2\theta \, \mathrm d\theta = \frac{\pi}{4}.
\end{align}
| {
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"source": "stackexchange",
"question_score": "3",
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Proving a point lies on a ellipse
Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
| Let $O$, the center of the ellipse, be the origin of coordinates.
Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$\frac{X-x}X=\frac{b}{a+b}$$
$$x=X-\frac{bX}{a+b}=\frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$\frac{Y-y}{Y}=\frac{a}{a+b}$$
$$y=Y-\frac{aY}{a+b}=\frac{bY}{a+b}$$
Now,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y=0$$ we get
$$12y+2\lambda+2\mu y=0 \tag{2}$$
For $$L_z=0$$ we get
$$8z+\lambda=0 \tag{3}$$
From $(1)$ and $(2)$ we get
$$x=\frac{4 \lambda}{1-2\mu}$$
$$y=\frac{8 \lambda}{6-\mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 \frac{\lambda^2}{(1-2\mu)^2}+4 \frac{\lambda^2}{(6-\mu)^2}=1 \tag{4}$$
$$\frac{4 \lambda}{1-2\mu}+\frac{16 \lambda}{6-\mu}+\frac{\lambda}{8}=4 \tag{5}$$
But its tedious to solve above equations for $\lambda$ and $\mu$
Any other approach?
| With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\\
y=0.4377195786259503\\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$
$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$
$\frac{6}{10} + \frac{5}{10}$
$\frac{11}{10}$
| $$
\cos(a+b)=\cos a\cos b-\sin a\sin b
$$
and hence
$$
\cos\left[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}\right]=\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\cos\left(\frac{\pi}{3}\right)-\sin\left(\cos^{-1}\left(\frac{3}{5}\right)\right)\sin\left(\frac{\pi}{3}\right)\\=\frac{3}{5}\cdot\frac{1}{2}-\sqrt{1-\frac{3^2}{5^2}}\cdot\frac{\sqrt{3}}{2}=\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $\int\frac{2x-3}{(x^2+x+1)^2}dx$ $\int\frac{2x-3}{(x^2+x+1)^2}dx$
$\int\frac{2x-3}{(x^2+x+1)^2}dx=\int\frac{2x+1}{(x^2+x+1)^2}dx-\int\frac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
| $$\dfrac{d\left(\dfrac{ax^2+bx+c}{x^2+x+1}\right)}{dx}=\dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0\iff a=b$
$b-c=-3\iff c=b+3$
$2(a+c)=2\iff1=a+c=b+b+3\iff b=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \frac{2}{y^{2}} - \frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$\displaystyle y = -\frac{5}{2} \pm \frac{\sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $\displaystyle y = \frac{58}{4} \pm \frac{10\sqrt{33}}{4}$
But this isn't the correct solution.
Also for $\#10$ I do the same thing:
Let $y = \sqrt n$ then $y^2 = n$
So I have $\displaystyle \frac{3}{y^2} - \frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $\displaystyle y = -\frac{3}{2}, \frac{1}{3}$
Then $n = \frac{9}{4}, \frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1)\ (x-7)^2 -13(x-7) +36=0 \qquad \qquad 4)\ 3(w/6)^2 -8(w/6) +4=0 \\
2)\ (1-3x)^2 -13(1-3x) +36=0 \qquad \qquad 5)\ 3(w^2-2)^2 -8(w^2-2) +4=0 \\
3)\ x^4 -13x^2 +36=0 \qquad \qquad 6)\ \frac{3}{p^2} -\frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
| In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $\dfrac2x -\dfrac5{\sqrt{x}}=1$ as
$$2 \left(\dfrac{1}{\sqrt x}\right)^2 - 5\left(\dfrac{1}{\sqrt x}\right)-1 = 0$$
So $$\dfrac{1}{\sqrt x} = \dfrac{5 \pm \sqrt{33}}{4}$$
We can ignore $\dfrac{1}{\sqrt x} = \dfrac{5 - \sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$\sqrt x = \dfrac{4}{5+\sqrt{33}}= \dfrac{\sqrt{33}-5}{2}$$
and $$x = \dfrac{29-5\sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
\begin{align}
3\color{red}{(w^2-2)}^2 -8\color{red}{(w^2-2)} + 4 &= 0 \\
(3\color{red}{(w^2-2)} - 2)(\color{red}{(w^2-2)} - 2) &= 0 \\
w^2-2= \dfrac 23 &\text{ or } w^2-2 = 2 \\
w^2 = \dfrac 83 &\text{ or } w^2 = 4 \\
w &\in \left\{\dfrac 23 \sqrt 6, -\dfrac 23 \sqrt 6, 2, -2 \right\}
\end{align}
You can solve the others similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find the derivative of $y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$
What is the derivative of
$$y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$$
What I did here was:
$$\begin{align}y'&=({1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x)'\\&={1\over2}\cdot4\tan^3x(\tan x)'-{1\over2}\cdot2\tan{x}(\tan x)'-{1\over\cos x}(\cos{x})'\\&={2\tan^3x\over\cos^2x}-{\tan x\over\cos^2x}+{\sin x\over\cos x}\\&={2{\sin^3x\over\cos^3x}\over\cos^2x}-{{\sin x\over\cos x}\over\cos^2x}+{\sin x\over\cos x}\\&={2\sin^3x\over\cos^5x}-{\sin x\over\cos^3x}+{\sin x\over\cos x}\\&={2\sin^3x-\sin x\cos^2x+\sin x\cos^4x\over\cos^5x}\\&={\sin x(2-\cos^2x+\cos^4x)\over\cos^5x}\end{align}$$
I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?
I made a mistake in the beginning while writing down this problem, it should be ${1\over4}\tan^4x$, I changed it out and got exact results
| You made a mistake, the first denominator should be $\cos^2 x \times \cos^3 x = \cos^5 x$ and you should be able to do more factoring then
UPDATE
Note after fixing sine power mistake in the last step you have
$$
\cos^4 x - \cos^2 x + 2\sin ^2 x
= \cos^4 x - 3\cos^2 x + 2
= \left(\cos^2 x - 1\right)\left(\cos^2 x - 2\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and
$$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$
Find the smallest possible value of
$$a^2 + b^2 + c^2+ d^2$$
I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.
Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.
| Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$
$$2ac+2ad+2bc+2bd+2ab+2cd \leq 3a^2+3b^2+3c^2+3d^2$$
So $$a^2+b^2+c^2+d^2\geq 154$$
But, unfortenuly this is not the minumum, since variables can not be all the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6}\sqrt{(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\require{cancel} \cancel{x^3} \sqrt{(9-\frac {5}{x^2})}}{\require{cancel} \cancel{x^3}(1-\frac {2}{x}+\frac{1}{x^3})}} =
\frac {3}{1} = +3$$
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
I'll be glad to get your help!
Thank you.
| Observe that if $y < 0$ then $$\sqrt{y^2} = |y| = -y.$$
Take $y=-3$ for example. In your case, $y = x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove this algebra question? If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1$$
| $$x^2+y^2+z^2=x(x+1)\implies\dfrac x{x+1}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A nice relationship between $\zeta$, $\pi$ and $e$ I just happened to see this equation today, any suggestions on how to prove it?
$$\sum_{n=1}^\infty{\frac{\zeta(2n)}{n(2n+1)4^n}}=\log{\frac{\pi}{e}}$$
| Since
\begin{align*}
\zeta(2n) &= \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!},
\end{align*}
for integers $n > 0$, the given sum $S$ is
\begin{align*}
S &= \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n}
= \sum_{n=1}^\infty (-1)^{n+1} \frac{B_{2n}}{(2n)(2n+1)}\frac{\pi^{2n}}{(2n)!}
\end{align*}
But
\begin{align*}
f(z) &= \sum_{n=1}^\infty \frac{B_n}{n(n+1)} \frac{z^{n+1}}{n!}
\end{align*}
has $\operatorname{Im} f(\pi i) = \pi S$ and
\begin{align*}
f''(z) &= \frac{1}{z}\sum_{n=1}^\infty B_n \frac{z^n}{n!} = \frac{1}{z}\left(-1 + \frac{z}{e^z - 1}\right) = -\frac{1}{z} + \frac{1}{e^z - 1}
\end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Trying to prove an equation I would like to receive some help about the next problem.
The problem:
I'm trying to prove the next equation:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = 0 \quad, n = 1, 2, ...$$
My work until now:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = \sum_{k = 0}^{n} \frac{1}{(-1)^{k}k!(n - k)!} =$$
$$= \frac{1}{(-1)^0 \cdot n!} + \frac{1}{(-1)^1 \cdot (n - 1)!} + \frac{1}{(-1)^2 \cdot 2!(n - 2)!} + \frac{1}{(-1)^3 \cdot 3!(n - 3)!} + \cdot \cdot \cdot + \frac{1}{(-1)^{n - 3} \cdot (n - 3)!3!} + \frac{1}{(-1)^{n - 2} \cdot (n - 2)!2!} + \frac{1}{(-1)^{n - 1} \cdot (n - 1)!} + \frac{1}{(-1)^n \cdot n!} \quad (1)$$
Because we have $n + 1$ summands i tried to see what happens when $n$ is even or odd number:
1) $n = 2k - 1$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k$ summands
$$(1) \iff
\frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k - 1}}\right) +
\frac{1}{(2k - 2)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 2}}\right) +
\frac{1}{2!(2k - 3)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 3}}\right) +
\cdot \cdot \cdot +
\frac{1}{(k - 1)!k!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^k}\right) = 0$$
This is equal $0$, because the difference in powers of number $-1$ is odd number, so one of them has to be $1$ and another has to be $-1$.
2) $n = 2k$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k + 1$ summands
$$(1) \iff
\frac{1}{(2k)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k}}\right) +
\frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 1}}\right) +
\frac{1}{2!(2k - 2)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 2}}\right) +
\cdot \cdot \cdot +
\frac{1}{(k - 1)!(k + 1)!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^{k + 1}}\right) +
\frac{1}{(-1)^k \cdot k!k!} \quad \Longrightarrow \quad ?$$
Here, i can't use the same princip as in the first case. Also, the last summand is the $k + 1$ -st summand, he is in the middle of this sum and doesn't have a pair.
My question:
Please, could you help me with some advice or a hint that i could use to finnish my proof of the given equation?
Thank you, for your time and your help!
| HINT
Actually you were thinking a bit to complicated. Therefore consider the series expansion of $(1-x)^n$ which is due the Binomial Theorem given by
$$\sum_{k=0}^n\binom nk (-x)^k(1)^{n-k}=\sum_{k=0}^n\binom nk (-x)^k$$
Further notice that the binomial coefficient can be written as
$$\binom nk = \frac{n!}{k!(n-k)!}$$
Putting these two together yields to
$$\sum_{k=0}^n\frac{n!}{k!(n-k)!}(-x)^k=n!\sum_{k=0}^n\frac{(-x)^k}{k!(n-k)!}$$
For $x=1$ this equals your given formula. Now look at the original term we have expanded. Can you finish it from hereon?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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