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Evaluation of $\int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
Evaluation of $\displaystyle \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
My Try:: Let $\displaystyle I = \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$
So $\displaystyle I = -\ln\left(\cos x+\sqrt{\cos 2x}\right)\cdot \cot x+\int \frac{1}{\left(\cos x+\sqrt{\cos 2x}\right)}\cdot \left(-\sin x-\frac{1}{2\sqrt{\cos 2x}}\cdot 2\cdot \sin 2x\right)\cdot \cot xdx$
Now How Can I solve after that
Help me
Thanks
|
As you have done, we first use integration by parts to get
$$\begin{align}
I &= \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx
\\&= \int \ln(\cos x+\sqrt{\cos 2x})\csc^2 (x)dx
\\&=-\cot (x)\ln(\cos x+\sqrt{\cos 2x})+\int \left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x)dx
\end{align}$$
Now we just need to deal with the remaining integral which we will call $J$. First note that $$(\cos x+\sqrt{\cos 2x})(\cos x-\sqrt{\cos 2x})=\cos^2x-\cos2x=\cos^2x-(\cos^2x-\sin^2x)=\sin^2x$$
Hence
$$\begin{align}
J&=\int \left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x)dx
\\&=\int\left(\frac{\cos x-\sqrt{\cos 2x}}{\cos x-\sqrt{\cos 2x}}\right)\left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \right)\cot (x) dx
\\&=\int\left(\frac{\cos x-\sqrt{\cos 2x}}{\sin^2x}\right) \left(-\sin x-\frac{2\sin x\cos x}{\sqrt{\cos 2x}} \right)\cot (x) dx
\\&=\int\left(\frac{-\sin x\cos x+\sin x\sqrt{\cos 2x}-\frac{2\sin x\cos^2 x}{\sqrt{\cos2x}}+2\sin x\cos x}{\sin^2x}\right)\cot (x) dx
\\&=\int\frac{-\cos^2 x+\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}+2\cos^2 x}{\sin^2x}dx
\\&=\int\left(-\frac{\sin^2x-1}{\sin^2x}+\frac{\sin^2x-1}{\sin^2x}\right)+\frac{\cos^2x +\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}}{\sin^2x}dx
\\&=-x-\cot x+\int\frac{\cos x\sqrt{\cos 2x}-\frac{2\cos^3 x}{\sqrt{\cos2x}}}{\sin^2x}dx
\\&=-x-\cot x+\int\frac{\cos x\cos 2x-2\cos^3 x}{\sin^2x\sqrt{\cos2x}}dx
\\&=-x-\cot x+\int\frac{\cos x(1-2\sin^2x)-2\cos^3 x}{\sin^2x\sqrt{\cos2x}}dx
\\&=-x-\cot x+\int\frac{-2\cos x\sin^2x+\cos x(1-2\cos^2 x)}{\sin^2x\sqrt{\cos2x}}dx
\\&=-x-\cot x+\int\frac{-\sin2x\sin x-\cos x\cos2x}{\sin^2x\sqrt{\cos2x}}dx
\\&=-x-\cot x+\int\frac{\frac{-\sin2x}{\sqrt{\cos2x}}\sin x-\cos x\sqrt{\cos2x}}{\sin^2x}dx
\\&=-x-\cot x+\frac{\sqrt{\cos 2x}}{\sin x}+C
\end{align}$$
Hence
$$\begin{align}I&=-\cot (x)\ln(\cos x+\sqrt{\cos 2x})-x-\cot x+\frac{\sqrt{\cos 2x}}{\sin x}+C
\\&=\frac{\sqrt{\cos 2x}}{\sin x}-x-\cot (x)(1+\ln(\cos x+\sqrt{\cos 2x})+C
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor series convergence for sin x a. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!},$
b. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots-\frac{x^{4k-1}}{(4k-1)!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots+\frac{x^{4k+1}}{(4k+1)!},$
c. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots = \sin x.$
Hello!
I have been trying to prove these, but only had luck with the first two. I do get that you have to use part b. to solve c., but I have no idea how to connect them.
|
For a) and by the Maclaurin formula for the sine function there's $\theta\in(0,1)$ such that
$$\sin x=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin^{(4)}(\theta x)=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin(\theta x)>x-\frac{x^3}{3!},\; \forall x\in(0,\frac\pi2)$$
and
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^6}{6!}\sin^{(6)}(\theta x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^6}{6!}\sin(\theta x)\\<x-\frac{x^3}{3!}+\frac{x^5}{5!},\; \forall x\in(0,\frac\pi2)$$
so we deduce the result. Can you generalize this to find b)?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True
if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true.
$\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$
Additional info: The Proof should be by contradiction.we can use Cauchy , AM-GM and other simple inequalities.
Things I have done so far: I don't have a complete idea for this Problem.I just think
that for starting step I should prove at least one of those inequalities are true.
So, I assume that $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}< 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$
.Summing these
inequalities gives us: $$\frac{11}{a}+\frac{11}{b}+\frac{11}{c}<18$$
using Cauchy and $a+b+c\geq abc$ I can write:$$ab+bc+ac\geq9$$
So I can rewrite Previous inequality as: $$\frac{99}{abc}<18$$
And I stuck here.
UPDATE
Thanks to user169478 help, we proved that at least one of these inequalities is true.So the remaining is to prove that if one of these 3 is true then the another one is true. So any hint for starting this part is appreciated.
As it was Proved that at least one of those 3 inequalities is true, I assume that $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ is true. Now we suppose that $\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6$ and $\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$.So we can say $$\frac{8}{b}+\frac{5}{c}+\frac{9}{a}<12$$
We have $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ So we can re write last inequality as $$\frac{7}{a}+\frac{5}{b}-\frac{1}{c}<6$$
and I stuck at proving this inequality is false.
|
Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$.
So we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$
Assuming that all three inequalities are false, like you did, we obtain $11(x+y+z)<18$. But
$$(x+y+z)^2\geq3(xy+yz+xz)
\Rightarrow x+y+z\geq\sqrt{3}
\Rightarrow 11(x+y+z)\geq11\sqrt{3}>18$$
We have a contradiction.
Now we suppose only one of them is true. Assume that $2z+3x+6y\geq6$ is true. Then
$$2x+3y+6z<6
\Rightarrow4x^2+9y^2+36z^2+12xy+36yz+24xz<36 \tag{*}$$
Since $xy+yz+xz\geq1$ and using $*$ ,We obtain
$$4x^2+9y^2+36z^2+12xy+36yz+24xz<36(xy+yz+xz)$$
So
$$4x^2+9y^2+36z^2-24xy-12xz<0 \tag{1}$$
And using the same process, from $2y+3z+6x<6$ We obtain
$$4y^2+9z^2+36x^2-24yz-12xy<0 \tag{2}$$
Sum up $(1)$ and $(2)$ , We get
$$40x^2+13y^2+45z^2-36xy-12xz-24yz<0$$
which is equivalent to
$$9(2x-y)^2+4(y-3z)^2+(3z-2x)^2<0$$
We have a contradiction again. So we can conclude that at least two of the inequalities are true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integer solutions of the factorial equation $(x!+1)(y!+1)=(x+y)!$ The problem is: are there solutions for the next equation?
$$(x!+1)(y!+1)=(x+y)!$$
with $x,y\in\mathbb{N}$.
My solution:
$\left(x!+1\right)\cdot \left(y!+1\right) = \left(x+y\right)!$
$x!y!+x!+y!+1= \left(x+y\right)!$
$\displaystyle\frac{x!y!+x!+y!+1}{x!y!}= \displaystyle\frac{\left(x+y\right)!}{x!y!}$
$1+\displaystyle\frac{1}{y!}+\displaystyle\frac{1}{x!}+\displaystyle\frac{1}{x!y!}=\displaystyle\binom{x+y}{x}$
how $1+\displaystyle\frac{1}{y!}+\displaystyle\frac{1}{x!}+\displaystyle\frac{1}{x!y!}\leq{4}$, then $\displaystyle\binom{x+y}{x}\leq{4}$, If $\displaystyle\binom{x+y}{x}\leq{4}$, is neccesary that $x,y\leq{4}$.
Therefore I can check a possible finite set of solutions $\{(x,y)|x,y\leq{4}\}$.
Is correct my proof?
There are other form?
|
The pairs $(1,2)$ and $(2,1)$ are the only solutions. To see this assume that $x>1$ and $y>1$. Then $x!$, $y!$ and $(x+y)!$ are even. But $(x!+1)(y!+1)$ is odd, which is a contradiction. Thus, $x=1$ or $y=1$, and since the problem is symmetric in $x$ and $y$ we assume that $y=1$. Then $2(x!+1)=(x+1)!$, which is equivalent to $x!+2=x\cdot x!$. Thus, $x!|2$, and this shows that $x=1$ or $x=2$. But $x=1$ is not possible. Thus, $x=2$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
|
If $\log $ is 2-base logarithm, you have:
$\;\;\; 3+4 \cdot 3+4^2 \cdot 3+ \cdots + 4^{\log n -1} \cdot 3
\\ = 4-1+4(4-1)+4^2(4-1)+4^3(4-1)+\cdots+4^{\log n -1} \cdot (4-1)
\\ = 4-1+4^2-4+4^3-4^2+\cdots+4^{\log n}-4^{\log n-1}
\\ = 4^{\log n}-1
\\ = 2^{2\log n}-1
\\ = n^2-1$
$\textbf{Edit}$ In sigma notation :
$$\sum_{i=1}^{\log n}3\cdot 4^{i-1}=\sum_{i=1}^{\log n}(4-1)\cdot 4^{i-1}=\sum_{i=1}^{\log n}4^{i}-4^{i-1}=\sum_{i=1}^{\log n}4^{i}-\sum_{i=1}^{\log n}4^{i-1}=4^{\log n}-1$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/901379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane. I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equations :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
$$a_3x+b_3y+c_3=0$$
Then the area of the triangle that these lines will enclose is given by the magnitude of :
$$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$
Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.
What I'm wondering is, where did this come from? And why isn't it famous? Earlier we had to calculate areas by finding the vertices and all but this does it in a minute or so and thus deserves more familiarity.
|
equations in the reduced form
$y=ax+b$
$y=cx+d$
$y=ex+f$
solving by the Cramer rule we get the A, B and C points
calculating the area of the ABC triangle
$ΔABC=\frac{1}{2}\left| \begin{array}{} \frac{b-d}{c-a} & \frac{bc-ad}{c-a} & 1 \\ \frac{d-f}{e-c} & \frac{de-cf}{e-c} & 1 \\ \frac{f-b}{a-e} & \frac{af-be}{a-e} & 1 \\ \end{array} \right| $
$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left| \begin{array}{} b-d & bc-ad & c-a \\ d-f & de-ef & e-c \\ f-b & af-be & a-e \\ \end{array} \right| $
$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left [ a(d-f)+c(f-b)+e(b-d) \right ] ^{2}$
$ΔABC=\frac{1}{2}\frac{\left|\begin{array}{} a & b & 1 \\ c & d & 1\\ e & f & 1 \\ \end{array} \right| ^{2}}{\left| \begin{array}{} a & a^{2} & 1 \\ c & c^2 & 1 \\ e & e^2 & 1 \\ \end{array} \right| }$
$a=\frac{-a_1}{b_1}$
$b=\frac{-c_1}{b_1}$
$c=\frac{-a_2}{b_2}$
$d=\frac{-c_2}{b_2}$
$e=\frac{-a_3}{b_3}$
$f=\frac{-c_3}{b_3}$
$ΔABC=|\frac{1}{2}\frac{\left| \begin{array}{} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \\ \end{array} \right| ^2}{\left| \begin{array}{} a_1^2 & b_1^2 & a_1b_1 \\ a_2^2 & b_2^2 & a_2b_2 \\ a_3^2 & b_3^2 & a_3b_3 \\ \end{array} \right| }|$
$ΔABC=|\frac{1}{2}\frac{Δ^2}{δ}|$
|
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|
Minimizing the expression $(1+1/x)(1+m/y)$ over positive reals such that $mx+y=1$ Let $x$ and $y$ be positive real numbers such that $mx+y=1$. Find the positive $m$ such that the minimum of:
$$\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right).$$
is $81$.
I have tried expanding, using Cauchy-Schwarz inequality, but nothing helped.
|
Since $x=\frac{1}{m}(1-y)$, we have:
$$\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=\left(1+\frac{m}{1-y}\right)\left(1+\frac{m}{y}\right),\tag{1}$$
and since $f(x)=1+\frac{m}{x}$ is a log-convex function, the minimum of the RHS of $(1)$ over $(0,1)$ occurs when $y=\frac{1}{2}$, and the value of such a minimum is just $(1+2m)^2$. The answer is so $m=4$.
In order to avoid log-convexity, we can just write down the RHS of $(1)$ in the following form:
$$ g(y) = 1+\frac{m+m^2}{y(1-y)}$$
and notice that $y(1-y)\leq \frac{1}{4}$ due to the AM-GM inequality, from which:
$$ f(y) \geq 1+4m+4m^2 = (2m+1)^2 $$
with equality occurring only when $y=(1-y)$, i.e. $y=\frac{1}{2}$.
|
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|
Prove $a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$ if $abc=1$
if $a,b,c$ are positive real numbers that $abc=1 $,Prove:$$a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$$
Additional info: We should only use AM-GM and Cauchy inequalities.
Things I have done so far: for $a^2b+b^2c+c^2a $ minimum I can say: $$a^4b^2+b^2c^3a + b^2c^3a \ge 3 \sqrt{a^6b^6c^6}=3$$
So like this I can conclude that: $$(a^2b+b^2c+c^2a)^2=\sum \limits_{cyc}a^4b^2+\sum \limits_{cyc}b^2c^3a + \sum \limits_{cyc}b^2c^3a \ge 27$$
So $a^2b+b^2c+c^2a \ge 3\sqrt 3$. So my idea is to show that $3 \ge \sqrt{a^2 + b^2 + c^2}$. And I stuck here.
|
Expand $(a^2b+b^2c+c^2a)^2$ and use the fact that $abc=1$ to reduce it and then factorise to get,
$$ a^2(a^2b^2+2c) + b^2(b^2c^2+2a) + c^2(c^2a^2+2b) $$
Using the fact $abc=1$ again,
$$ a^2\left(\frac{1}{c^2}+2c\right) + b^2\left(\frac{1}{a^2}+2a\right) + c^2\left(\frac{1}{b^2}+2b\right) $$
and then it's easy to show that $\frac{1}{x^2}+2x\geq 3$ for all $x>0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating the limits $\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$ and $\lim_{(x,y)\to(\infty,8)}(1+\frac{1}{3x})^\frac{x^2}{x+y}$ I got the following problem:
Evaluate the following limits or show that it does not exist:
$$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
and
$$\lim_{(x,y)\to(\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}$$
I tried for an hour and half evaluating each of those limits but I failed and I got nothing useful to share.
Some hints will be appreciated.
Thanks.
|
Alternative solutions:
Consider the limit on all lines passing through the origin, i.e.,
$y=ax$. $$\lim_{(x,y)\to (\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}=\lim_{x\to\infty}\frac{x\left[2-a\right]}{x^2\left[1-a+a^2\right]}=\lim_{x\to\infty}\frac{\left[2-a\right]}{x\left[1-a+a^2\right]}=0, $$ independently of the choice of $a$.
For second question, note that
$$\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}=e^{\frac{x^2}{x+y}\ln\left(1+\frac{1}{3x}\right) }.$$
Once that numerator and denominator are going to the infinity we can applied the L'Hopital rule. So
\begin{eqnarray}
\lim_{(x,y)\to (\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}&=&\lim_{x\to\infty}e^{\frac{x^2}{x+8}\ln\left(1+\frac{1}{3x}\right) } \\
&=&\lim_{x\to\infty}e^{\left[2x\ln\left(1+\frac{1}{3x}\right) +x^2\frac{1}{1+\frac{1}{3x}}\frac{-1}{3x^2}\right]}\\
&=&e^{-\frac{1}{3}}\lim_{x\to \infty}\left(1+\frac{1}{3x}\right)^{2x}
\end{eqnarray}
Make $2x=y$. So $3x=(3/2)y$ and $y\to\infty$ as $x\to\infty$. Therewith
$$
\lim_{x\to \infty}\left(1+\frac{1}{3x}\right)^{2x}=\lim_{y\to\infty}\left(1+\frac{2}{3}\frac{1}{y}\right)^{y}=e^{\frac{2}{3}},\,\,\text{(for defition).}
$$
therefore
$$
\lim_{(x,y)\to (\infty,8)}\left(1+\frac{1}{3x}\right)^\frac{x^2}{x+y}=e^{-\frac{1}{3}}e^{\frac{2}{3}}=e^{\frac{1}{3}}.
$$
|
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|
Finding the points where a circle intersects an axis A circle has the equation:
x²+y²+4x-2y-11 = 0
What would be the coordinates of the points where the circle intersects with the y-axis and how would you calculate it?
|
$x^2+y^2+4x-2y-11=0$
The points of intersection with the $y$ axis are when $x=0$. Thus, plug in $x=0$ and solve.
$$0^2+y^2+4(0)-2y-11=0$$
$$y^2-2y-11=0$$
Use the quadratic formula: $\large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1, b=-2, c=-11$
$$\frac{-(-2)\pm \sqrt{4-(-44)}}{2}$$
$$\frac{2\pm \sqrt{48}}{2}$$
$\sqrt{48}=\sqrt{8\cdot 6}=\sqrt{4\cdot 2\cdot 3\cdot 2}=4\sqrt{3}$
$$\frac{2\pm 4\sqrt{3}}{2}=1\pm 2\sqrt{3}$$
So, your coordinates for intersection with the $y$-axis are: $(0,1+ 2\sqrt{3})$ and $(0,1- 2\sqrt{3})$
|
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|
Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?
$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$
I've found that it cannot be denested into $a+b\sqrt{2}+c\sqrt{5}+d\sqrt{10}$ by squaring and equating coefficients. Resulting system didn't have rational solutions.
Maybe it can be denested to something as sum of $\sqrt{a+b\sqrt{c}}$ or $\sqrt{a+b\sqrt{c}+d\sqrt{e}}$ or even $\sqrt[4]{a+b\sqrt{c}}$?
For example, following can be denested:
$$\sqrt{8+2 \sqrt{2}-2 \sqrt{6}}=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}$$
$$\sqrt{110-60 \sqrt{3}+46 \sqrt{5}-28 \sqrt{15}}=2\sqrt{15+6\sqrt{5}}-\sqrt{50+22\sqrt{5}}$$
$$\sqrt{48+12 \sqrt{2}+16 \sqrt{5}-12 \sqrt{6}+4\sqrt{10}-4\sqrt{30}}=$$
$$=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}+\sqrt{30+15\sqrt{2}}-\sqrt{10-5\sqrt{2}}$$
|
Try factoring $2\sqrt{10}$ as $2\sqrt{5}\sqrt{2}$, then the expression under the radical can be factored.
|
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|
Solving these two equations simultaneously I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..
$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$
$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$
|
Drop subscripts of $x$ and $y$.
Put $u=x+2, v=y-5$.
Then the equations become
$$(u-2)^2+(v+5)^2=(7\sqrt{2})^2 \qquad \cdots (1)$$
and
$$\sqrt{25+u^2}+\sqrt{4+v^2}=7\sqrt{2}\qquad \cdots (2)$$
Squaring $(2)$ equals $(1)$, i.e.
$$\begin{align}(25+u^2)+(4+v^2)+2\sqrt{(25+u^2)(4+v^2)}&=(u^2-4u+4)+(v^2+10v+25)\\
\sqrt{(25+u^2)(4+v^2)}&=-2u+5v\\
\end{align}$$
Squaring:
$$\begin{align}
100+4u^2+25v^2+u^2v^2&=4u^2-20uv+25v^2\\
(uv)^2+20uv+100&=0\\
(uv+10)^2&=0\\
uv&=-10 \Rightarrow v=-\frac{10}u\end{align}$$
Substituting back into $(1)$:
$$\begin{align}
(u-2)^2+(-\frac {10}u+5)^2&=98\\
\end{align}$$
Solving numerically gives
$$\begin{align}u&=-5, -2.6893, \quad \ \; 0.6752,\ 11.0142\\
v=-\frac {10}u &=\quad 2,\ \ 3.7184, \; -14.8104, -0.9079\\
x=u-2&=-7, -4.6893, \ -1.3248, \ \ \ 9.0142\\
y=v+5&=\ \ 7,\quad 8.7184, \ -9.8104, \ \ \ 4.0921
\end{align}$$
Checking by substitution shows that only the first two sets of numbers are valid, hence solution is
$$(x,y)=(-7,7), (-4.6893, 8.7184)\qquad \blacksquare$$
|
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|
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.
|
Observe that $\displaystyle\frac{13\pi}{24}-\frac\pi2=\pi\dfrac{(13-12)}{24}=\frac\pi{24}$
and $\displaystyle\frac{19\pi}{24}-\frac\pi2=\pi\dfrac{(19-12)}{24}=\frac{7\pi}{24}$
So, $\displaystyle\sin\frac{13\pi}{24}=\sin\left(\frac\pi2+\frac\pi{24}\right)=\cos\frac\pi{24}$ and $\displaystyle\sin\frac{19\pi}{24}=\sin\left(\frac\pi2+\frac{7\pi}{24}\right)=\cos\frac{7\pi}{24}$
So, we need the four degree equation whose roots are $\displaystyle\sin\frac{7\pi}{24},\cos\frac{7\pi}{24};\sin\frac{\pi}{24},\cos\frac{\pi}{24}$
Now the equation whose roots are $\displaystyle\sin\frac{\pi}{24},\cos\frac{\pi}{24}$ is $$t^2-\left(\sin\frac{\pi}{24}+\cos\frac{\pi}{24}\right)t+\sin\frac{\pi}{24}\cos\frac{\pi}{24}=0$$
Now $\displaystyle\sin\frac{\pi}{24}\cos\frac{\pi}{24}=\frac{\sin\dfrac\pi{12}}2$ and $\displaystyle\dfrac\pi{12}=\frac\pi4-\frac\pi6$
Again, $\displaystyle\sin\frac{\pi}{24}+\cos\frac{\pi}{24}=+\sqrt{1+\sin\dfrac\pi{12}}$
The same method should be applied to find the equation whose roots are $\displaystyle\sin\frac{7\pi}{24},\cos\frac{7\pi}{24}$
|
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|
Finding substitution in the integral $\int{\frac{2+3x}{3-2x}}dx$ In a problem sheet I found the integral $$\int{\frac{2+3x}{3-2x}}dx.$$
In the solution the substitution $z=3-2x$ is given which yields $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2}dz$. We have
$$\int{\frac{2+3x}{3-2x}}dx = \int{\frac{2+3\left(\frac{3-z}{2}\right)}{z}}\left(-\frac{1}{2}dz\right)=\cdots$$
How do I recognise these kind of substitutions and how do I choose them?
After reading the line $z=3-2x \Rightarrow x=\frac{3-z}{2} \Rightarrow dx=-\frac{1}{2}dz$. I was able to solve the integral. My main problem is, how do I recognise from the integrand that I have to substitute and solve for the free variable $x$. Is there a trick to remember?
Dmoreno's hint which I was at first not able to follow but after some thinking made me realise that equations like $\frac{a+bx}{c-dx}$ can be alternated like:
$$\frac{a+bx}{c-dx} = \left(\frac{a+bx}{c-dx} + \frac{b}{d}\right) - \frac{b}{d} = \frac{ad+bc}{d(c-dx)} - \frac{b}{d},$$
therefore $$\int{\frac{a+bx}{c-dx}}dx = \frac{ad+bc}{d}\cdot\int{\frac{1}{c-dx}}dx - \frac{b}{d}\cdot \int dx,$$
which made my problem much easier.
|
You were on the good way $$I =\int{\frac{2+3x}{3-2x}}dx$$ Change variable using $3-2x=z$, so $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2} dz$ (which is what you obtained). Replace and simplify as much as you can; you should easily arrive to $$I=\int \Big(\frac{3}{4}-\frac{13}{4 z}\Big)dz=\frac{3}{4}\int dz-\frac{13}{4}\int\frac{dz}{z}$$
I am sure that you can take from here.
|
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|
How to find all values of $z$ at which $\sum\limits_{n=1}^{\infty} \frac{1}{n^2} \exp(\frac{nz}{z-2})$ converges? Could anyone advise me on how to find all $z$ such that $\begin{align} \sum^{\infty}_{n=1} \dfrac{1}{n^2} \end{align}\text{exp}\left(\dfrac{nz}{z-2}\right)$ converges ? Does it suffice to find all $z$ such that $ \begin{align}\left|\text{exp}\left(\dfrac{nz}{z-2}\right)\right|\leq 1 \ ?\end{align}$
Hints will suffice, thank you.
|
The radius of convergence of $\sum \frac{1}{n^2} x^n$ is $1$. Plug in $x = \exp(\frac{z}{z-2})$, so that the series converge iff
$$|\exp(\frac{z}{z-2})| < 1 \iff \Re\left( \frac{z}{z-2} \right) < 0.$$
By writing down $z = a+bi$, where $a,b \in \mathbb{R}$, you get (let's assume $z \neq 2$ all the time):
$$\begin{align}
\Re \frac{a+bi}{a-2+bi} < 0 & \iff \Re\frac{(a+bi)(a-2-bi)}{(a-2)^2+b^2} < 0 \\
& \iff \frac{a^2-2a+b^2}{(a-2)^2 + b^2} < 0 \\
& \iff a^2-2a+b^2 < 0.
\end{align}$$
So that's the region of convergence, with the caveat that $(a,b) \neq (2,0)$. (There remains the problem at the boundary)
|
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|
How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
|
Maple says:
$$
\int \!{\frac {\sqrt {1-{k}^{2} \sin^2 \left( x \right)
}}{\sin \left( x \right) }}{dx}
= \frac{1}{2}\left[k\ln \left( 2 \right) +k
\ln \left( k \right) -k\ln \left( -2\, \cos^2 \left( x \right)
{k}^{2}+2\,\cos \left( x \right)k \sqrt { \cos^2
\left( x \right){k}^{2}-{k}^{2}+1}+{k}^{2}-1 \right) -
{\rm atanh} \left({\frac { \cos^2 \left( x \right)
{k}^{2}+ \cos^2 \left( x \right)-{k}^{
2}+1}{2\cos \left( x \right) \sqrt { \cos^2 \left( x \right)
{k}^{2}-{k}^{2}+1}}} \right) \right]
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^{{\pi}/{2}} \log(1+\cos x)\, dx$ Find the value of $\displaystyle \int_0^{{\pi}/{2}} \log(1+\cos x)\ dx$
I tried to put $1+ \cos x = 2 \cos^2 \frac{x}{2} $, but I am unable to proceed further.
I think the following integral can be helpful:
$\displaystyle \int_0^{{\pi}/{2}} \log(\cos x)\ dx =-\frac{\pi}{2} \log2 $.
|
\begin{align}
\color{red}{\int^{\pi/2}_0\log(1+\cos{x}){\rm d}x}
&=\int^{\pi/2}_0\frac{x\sin{x}}{1+\cos{x}}{\rm d}x\tag1\\
&=\int^{\pi/2}_0 x\tan{\frac{x}{2}} \ {\rm d}x\tag2\\
&=4\int^{\pi/4}_0x\tan{x} \ {\rm d}x\tag3\\
&=8\sum^\infty_{n=1}(-1)^{n-1}\int^{\pi/4}_0x\sin(2nx) \ {\rm d}x\tag4\\
&=2\sum^\infty_{n=1}(-1)^{n-1}\frac{\sin\left(\frac{n\pi}{2}\right)}{n^2}-\pi\sum^\infty_{n=1}(-1)^{n-1}\frac{\cos\left(\frac{n\pi}{2}\right)}{n}\\
&=2\sum^\infty_{n=0}\frac{\sin\left(\frac{(2n+1)\pi}{2}\right)}{(2n+1)^2}-\color{grey}{2\sum^\infty_{n=1}\frac{\sin\left(n\pi\right)}{4n^2}}\\
&-\color{grey}{\pi\sum^\infty_{n=0}\frac{\cos\left(\frac{(2n+1)\pi}{2}\right)}{2n+1}}+\pi\sum^\infty_{n=1}\frac{\cos\left(n\pi\right)}{2n}\tag5\\
&=2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}+\frac{\pi}{2}\sum^\infty_{n=1}\frac{(-1)^n}{n}\tag6\\
&\large{\color{red}{=2G-\frac{\pi}{2}\log{2}}}\tag7\\
\end{align}
Explanation:
$(1)$: Integrate by parts
$(2)$: Trigonometric identities
$(3)$: Substitute $x\mapsto \frac{x}{2}$
$(4)$: Use the Fourier series of $\tan{x}$
$(5)$: Split the sum into odd and even terms. It is obvious that the grey sums $=0$
$(6)$: $\sin\left(\frac{(2n+1)\pi}{2}\right)=\cos{n\pi}=(-1)^n$ for integer $n$. This can be verified easily.
$(7)$: The first sum is Catalan's constant. The second is $-\log(1+1)$. (Taylor's series)
|
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|
Integral of $\ln(x)\operatorname{sech}(x)$ How can I prove that:
$$\int_{0}^{\infty}\ln(x)\,\operatorname{sech}(x)\,dx=\int_{0}^{\infty}\frac{2\ln(x)}{e^x+e^{-x}}\,dx\\=\pi\ln2+\frac{3}{2}\pi\ln(\pi)-2\pi\ln\!\Gamma(1/4)\approx-0.5208856126\!\dots$$
I haven't really tried much of anything worth mentioning; I've had basically no experience with $\ln\!\Gamma$.
|
You can get the value of the integral you're interested in from the integral $$I(a) =\int_{0}^{\infty} \frac{\ln (1+\frac{x^{2}}{a^{2}})}{\cosh x} \, dx, \quad a>0.$$
Notice that $\lim_{a \to \infty} I(a) = 0$.
Differentiating under the integral sign, we get $$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{2a}{(a^{2}+x^{2})\cosh x} \, dx - \frac{2}{a} \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx \\ &= \int_{0}^{\infty} \frac{2}{(1+u^{2})\cosh (au)} \, du - \frac{\pi}{a}. \end{align}$$
From the answers to this question, we know that $$\int_{0}^{\infty} \frac{2}{(1+u^{2}) \cosh (au)} \, du= \psi\left(\frac{3}{4}+ \frac{a}{2 \pi} \right) - \psi \left(\frac{1}{4} + \frac{a}{2 \pi} \right). $$
Therefore, $$ \begin{align} I(a) &= 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right) \right] - \pi \ln (a)+ C \\ &= 2 \pi \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] +C. \end{align}$$
Letting $ a \to \infty$, we get $$ 0 = 2 \pi \lim_{a \to \infty} \log \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] + C.$$
Using Stirling's approximation formula for the gamma function, we see that $$ \frac{\Gamma(x+\frac{1}{2})}{\Gamma(x)} \sim \frac{\sqrt{\frac{2\pi}{x+1/2}} \left(\frac{x+1/2}{e} \right)^{x+1/2}}{\sqrt{\frac{2\pi}{x}} \left(\frac{x}{e} \right)^{x}} =\sqrt{x} \left(1+ \frac{1}{2x} \right)^{x} e^{-1/2} \sim \sqrt{x} .$$
Therefore, $$ \lim_{a \to \infty} \ln \left[\frac{\Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right)}{\sqrt{a} \ \Gamma \left(\frac{1}{4} + \frac{a}{2 \pi} \right)} \right] = \lim_{a \to \infty} \ln \ \frac{\sqrt{\frac{1}{4} + \frac{a}{2 \pi}}}{\sqrt{a}} = \lim_{a \to \infty} \ln \sqrt{\frac{1}{4a}+\frac{1}{2 \pi}} = - \frac{\ln (2 \pi)}{2}, $$
which implies $$C= \pi \ln (2 \pi).$$
So we have $$I(a) = 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right] - \pi \ln (a) + \pi \ln (2 \pi).$$
But since $$ \int_{0}^{\infty} \frac{\ln (a^{2}+x^{2})}{\cosh x} \, dx = I(a) + \ln(a^{2}) \int_{0}^{\infty} \frac{dx}{\cosh x} \, dx = I(a) + \pi \ln (a),$$ it follows that $$2 \int_{0}^{\infty} \frac{\ln x}{\cosh x} \, dx = \lim_{a \to 0^{+}} 2 \pi \left[\ln \Gamma \left(\frac{3}{4} + \frac{a}{2 \pi} \right) - \ln \Gamma \left(\frac{1}{4} + \frac{a}{2\pi} \right) \right] + \pi \ln (2 \pi). $$
The final step is to apply the reflection formula for the gamma function.
If we had started with the integral $\int_{0}^{\infty} \frac{\ln(a^{2}+x^{2})}{\cosh x} \, dx $, we wouldn't have had a known initial condition.
|
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|
Cards in box - probability a given type is picked last I came out with a probability question which I find difficult to solve. I hope some kind souls can provide me with some ideas.
There is a box with four different types of cards, namely A, B, C, D. There are 7 A, 4 B, 3 C and 2 D. One starts to pick cards from the box. The card picked out is not put back into the box. I would like to calculate the probability for certain type the cards that is to be picked last.
For example, if the sequence of cards picked goes like AABABC, then D is identified instantly as the card to be picked last.
Can anyone provide a non-exhaustive method of calculating the probability of certain type of a cards to be picked last? Thank you!
Furthermore, it would be very nice of you to provide a generalized formula of evaluation.
|
The cards in the box are: 7 of type A, 4 of B, 3 of C and 2 of D.
Let, for example, $A=1,B=2,C=3$ represent the event of encountering the type A first, type B second, type C third, and type D last. (We don't have to write the last, it's implicit.) One such example is to draw cards in order $\mathbf A,A,\mathbf B, A, \mathbf C,B,A,C,\mathbf D, A...$
Clearly the probability of encountering A first is : $\mathsf P(A=1) =a/(a+b+c+d) =7/16$
Given that, the probability of encountering B second is: $\mathsf P(B=2 \mid A=1)=b/(b+c+d)= 4/9$
And likewise, $\mathsf P(C=3\mid A=1,B=2) = c/(c+d) = 3/5$
So $$\begin{align}
\mathsf P(A\!=\!1,B\!=\!2,C\!=\!3) &= \frac{abc}{(a+b+c+d)(b+c+d)(c+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(4+3+2)(3+2)}
\\
\mathsf P(A\!=\!1,B\!=\!3,C\!=\!2) &= \frac{abc}{(a+b+c+d)(b+c+d)(b+d)} &=\frac{7\times 4\times 3}{(7+4+3+2)(4+3+2)(4+2)}
\\
\mathsf P(A\!=\!2,B\!=\!1,C\!=\!3) &= \frac{abc}{(a+b+c+d)(a+c+d)(c+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+3+2)(3+2)}
\\
\mathsf P(A\!=\!2,B\!=\!3,C\!=\!1) &= \frac{abc}{(a+b+c+d)(a+b+d)(b+d)} &=\frac{7\times 4\times 3}{(7+4+3+2)(7+4+2)(4+2)}
\\
\mathsf P(A\!=\!3,B\!=\!1,C\!=\!2) &= \frac{abc}{(a+b+c+d)(a+c+d)(a+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+3+2)(7+2)}
\\
\mathsf P(A\!=\!3,B\!=\!2,C\!=\!1) &= \frac{abc}{(a+b+c+d)(a+b+d)(a+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+4+2)(7+2)}
\end{align}$$
Then $\mathsf P(D=4)$ is the sum of these six.
$\begin{align}\mathsf P(D=4) & = \frac{7\times 4\times 3}{16}\times(\frac 1 {9\times 5}+\frac 1{ 12\times 5}+\frac 1{9 \times 6}+\frac 1{12 \times 9}+\frac 1{13\times 6}+\frac 1{13\times 9})
\\ & =\frac{721}{1560}\end{align}$
And so forth.
|
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|
Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
|
*
*Show $3^{1-1}=\dfrac{3^1-1}{2}$
*Assume $\sum\limits_{k=0}^{n-1}3^k=\dfrac{3^n-1}{2}$
*Prove $\sum\limits_{k=0}^{n}3^k=\dfrac{3^{n+1}-1}{2}$:
*
*$\sum\limits_{k=0}^{n}3^k=(\sum\limits_{k=0}^{n-1}3^k)+3^n$
*$(\sum\limits_{k=0}^{n-1}3^k)+3^n=\dfrac{3^n-1}{2}+3^n$
*$\dfrac{3^n-1}{2}+3^n=\dfrac{3^n-1+2\cdot3^n}{2}$
*$\dfrac{3^n-1+2\cdot3^n}{2}=\dfrac{3^n+2\cdot3^n-1}{2}$
*$\dfrac{3^n+2\cdot3^n-1}{2}=\dfrac{3\cdot3^n-1}{2}$
*$\dfrac{3\cdot3^n-1}{2}=\dfrac{3^{n+1}-1}{2}$
Note that the induction-step is applied only at the second bullet.
|
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|
What's the sum of this series? I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?
Thanks!
|
It is well known that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
Thus, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \sum_{m=1}^{\infty}\dfrac{1}{(2m)^2} = \dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$.
Hence, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} = \sum_{n=1}^{\infty}\dfrac{1}{n^2} - \sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$.
Finally, $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2} = \displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} - \displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{12}$.
|
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|
Find $\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)$ I am having a tough time with these TYPE of problems
looking forward ideas, All ideas will be appreciated
|
Multiply numerator and denominator by the conjugate of the "numerator": $\sqrt{x^2 + x+1} + \sqrt{x^2 -x}$, to get a difference of squares in the numerator.
$$\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)\cdot \frac{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}$$
$$ = \lim_{x\to \infty} \frac{x^2 + x + 1 -(x^2 -x)}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}
$$ $$ = \lim_{x\to \infty} \frac{2x+1}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}$$
Now, divide numerator and denominator by $x$:
$$ = \lim_{x\to \infty} \frac{2+\frac 1x}{\sqrt{1+\frac 1x +\frac 1{x^2}} + \sqrt{1 -\frac{1}{x}}}= \frac 22 = 1$$
So whenever you encounter limits of the form $$\lim_{x\to \infty} \left(\sqrt{f(x)} - \sqrt{g(x)}\right) \leadsto \infty - \infty,$$ "multiply by $1$". That is, multiply by $\dfrac{\sqrt{f(x)} + \sqrt{g(x)}}{\sqrt{f(x)} + \sqrt{g(x)}}.\;$ You'll likely be using this strategy in a whole host of situations.
|
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|
Find $\frac{dy}{dx}$ for $x=2\theta+sin2\theta$ and $y=1-cos2\theta$
The parametric equations of a curve are
$$x=2\theta+\sin2\theta,\:y=1-\cos2\theta.$$
Show that $\frac{dy}{dx}=\tan\theta$.
I can use the chain rule to get
$$\frac{dx}{d\theta}=2+2\cos2\theta$$
$$\frac{dy}{d\theta}=2\sin2\theta$$
$$\frac{dy}{dx}=\frac{dy}{d\theta}\div\frac{dx}{d\theta}$$
$$=\frac{2\sin2\theta}{2+2\cos2\theta},$$
but I'm not sure how to get to a final proof.
|
You can see that $$\frac{2\sin 2 \theta}{2+2 \cos 2 \theta} = \frac{\sin 2 \theta}{1+ \cos 2 \theta} \\ = \frac{2 \cos \theta \sin \theta}{2 \cos^2 \theta} = \tan \theta$$
I used the identities $\cos 2\theta = 2 \cos^2 \theta - 1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$
|
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|
prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:
let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that
$$a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$$
I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\sum\limits_{cyc}a^3b^3=27v^6-27uv^2w^3+3w^6$, we see that our inequality is equivalent to $f(w^3)\geq0$,
where $f$ is a concave function, which says that it's enough to prove our inequality
for an extremal value of $w^3$, which happence in the following cases.
*
*$w^3=0$.
Let $c=0$. Thus, we need to prove that $$a^2+b^2\geq2a^3b^3,$$
which is AM-GM: $a^2+b^2\geq2ab$ and it's enough to prove that $ab\leq1$,
which follows from our condition: $2=a+b\geq2\sqrt{ab}$;
*$b=a$ and $c=2-2a$, where $0\leq a\leq1$.
In this case we need to prove that
$$2a^2+4(1-a)^2\geq2(a^6+2a^3(2-2a)^3+4a^4(2-2a)^2)$$ or
$$(1-a)(a^5+17a^4-15a^3+a^2-2a+2)\geq0,$$
which is true.
Done!
|
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|
Calculate Integrals $ \int \sqrt{\sec 2x-1}\;dx$ and $ \int \sqrt{\sec 2x+1}\;dx$ Calculation of Integral of
$$\displaystyle \int \sqrt{\sec 2x-1}\;dx,\>\>\>\>\>\displaystyle \int \sqrt{\sec 2x+1}\;dx$$
$\bf{My\; Solution}::$ For $(a)::$
Let $$\displaystyle I = \int \sqrt{\sec 2x-1}\;dx = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}}\;dx $$
$$\displaystyle = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}\times \frac{1+\cos 2x}{1+\cos 2x}}dx = \int \frac{\sin (2x)}{\sqrt{\cos 2x\cdot (1+\cos 2x)}}dx$$
Now Let $\cos (2x) = t\;,$ Then $\displaystyle \sin (2x)dx = -2dt$
So Integral $$\displaystyle I =-2\int\frac{1}{\sqrt{t\cdot (1+t)}}dt=-2\int\frac{1}{\sqrt{(t+\frac{1}{2})^2+(\frac{1}{2})^2}}dt$$
So $\displaystyle I = -2\cdot \ln \left|\left(t+\frac{1}{2}\right)+\sqrt{t^2+t}\right|+\mathcal{C} = -2\cdot \ln \left|\left(\cos 2x+\frac{1}{2}\right)+\sqrt{\cos^2(2x)+\cos (2x)}\right|+\mathcal{C}$
Is There is any Method other then that,
If yes then plz explain here
Thanks
|
Why not just perform this on computer?
$\int \sqrt{\sec{2 x}+1} dx = -8 \cos^4\left(\frac{x}{2}\right) \sqrt{\frac{-\left(1+\sqrt{2}\right) \sec
(x)+\sqrt{2}+2}{\sec (x)+1}} \left(\left(1+\sqrt{2}\right) \sec (x)+\sqrt{2}+2\right)
\sqrt{\frac{1}{\sec (x)+1}-\sqrt{2}+1} \sqrt{\frac{3-2 \sqrt{2}}{\sec (x)+1}-5
\sqrt{2}+7} \sqrt{\frac{3-2 \sqrt{2}}{\sec (x)+1}+\sqrt{2}-1} \sec (2 x) \sqrt{\sec
(2 x)+1} \left(F\left(\sin ^{-1}\left(\left(1+\sqrt{2}\right) \tan
\left(\frac{x}{2}\right)\right)|17-12 \sqrt{2}\right)+2 \Pi \left(-3+2 \sqrt{2};-\sin
^{-1}\left(\left(1+\sqrt{2}\right) \tan \left(\frac{x}{2}\right)\right)|17-12
\sqrt{2}\right)\right)$
|
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|
Why do the $\sec$ & $\cosh$ substitutions give different antiderivatives for $\frac1{\sqrt{x^2-a^2}} \,?$ Suppose we want to find
$$\int \frac{1}{\sqrt{x^2-a^2}}\,\mathrm dx.$$
Trigonometric substitution: $$=\ln \left| x+\sqrt{x^2-a^2} \right|$$
Hyperbolic substitution: $$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln\left({x+\sqrt{x^2-a^2}}\right)$$
How are they the same when one has the absolute value and one doesn't? Clearly, the first answer is defined for all $|x|>a,$ whereas the second one is defined only for $x>a.$
|
Suppose we want to find
$$\int \frac{1}{\sqrt{x^2-a^2}}\,\mathrm dx.$$
Trigonometric substitution: $$=\ln \left| x+\sqrt{x^2-a^2} \right|$$
Hyperbolic substitution:
$$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln\left({x+\sqrt{x^2-a^2}}\right)$$
the fact that trig substitution gives a more COMPLETE solution than hyperbolic substitutions bug me.
Noting that antiderivatives exist on $(-\infty,-a)\cup(a,\infty)$ and using the hyperbolic substitution $\displaystyle \theta=\operatorname{arccosh} \left(\frac{|x|}a\right):$
\begin{align}\int \frac{1}{\sqrt{x^2-a^2}}\,\mathrm
dx&=\begin{cases}\displaystyle-\operatorname{arccosh}\,\left(-\frac
x{a}\right)+C_1, &&&x<-a; \\
\displaystyle\mathrm{arccosh}\,\left(\frac x{a}\right)+C_2, &&&
x>a \end{cases}
\\&=\begin{cases}\displaystyle-\left[\ln\left(\sqrt{x^2-a^2}-x\right)-\ln (a)\right]+C_1, &&&x<-a; \\
\displaystyle\left[\ln\left(x+\sqrt{x^2-a^2}\right)-\ln (a)\right]+C_2, &&&
x>a \end{cases}
\\&=\begin{cases}-\ln \left(\sqrt{x^2-a^2}-x\right)+C_1+\ln(a^2), &&&x<-a; \\\ln \left| x+\sqrt{x^2-a^2}\right|+C_3, &&&x>a \end{cases}
\\&=\begin{cases}\ln \left( \frac{a^2}{\sqrt{x^2-a^2}-x}\right)+C_1,&&&x<-a; \\\ln \left| x+\sqrt{x^2-a^2}\right|+C_3, &&&x>a \end{cases}
\\&=\begin{cases}\ln \left( -\left(\sqrt{x^2-a^2}+x\right)\right)+C_1,&&&x<-a; \\\ln \left| x+\sqrt{x^2-a^2}\right|+C_3, &&&x>a \end{cases}
\\&=\begin{cases}\ln \left| x+\sqrt{x^2-a^2}\right|+C,&&&x<-a; \\\ln \left| x+\sqrt{x^2-a^2}\right|+D, &&&x>a, \end{cases}\end{align} which is the same result as the trigonometric substitution.
|
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|
Differentiation help (lots of square roots) help finding the first derivative of this question:
$y=\sqrt{ 1 + \sqrt{ 1+\sqrt{1+8x}}}$
I get confused, is it meant to be done implicitly or is it just a really long chain rule?
|
It seems like it is a long chain rule indeed:
\begin{align}
\frac{dy}{dx}&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1+8x}}}} \cdot \frac d{dx} \left(1+\sqrt{1+\sqrt{1+8x}} \right) \\
&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1+8x}}}}\cdot \frac 1{2\sqrt{1+\sqrt{1+8x}}} \cdot \frac d{dx} \left(1+\sqrt{1+8x} \right) \\
&= \cdots
\end{align}
Perhaps you can finish the rest?
|
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|
Solve $\int (4x+2)\sqrt{x^2+x+1}\,dx$ Trying to solve this for a while now, so far I was able to come up without a proper answer.
Problem : $\displaystyle \int (4x+2)\sqrt{x^2+x+1}\,dx$.
I tried to take two common from $(4x+2)$ and also to take $(x+1)^2 - x$ from the root, but wasn't able to come up with something to take for substitution. A hint in the right direction would be highly appreciated
Edit :
I forgot to mention this. As this is part of the integration by substitution exercise it'd be highly appreciated if you could provide the hint in that direction.
|
I'm extremely new to integration and my methods are sloppy but I'd like to try out this question. The following isn't an answer but a demonstration that any substitution can do the job as long as it is done right (Although, a fundamental zen of calculus dictates that the rightest way is the fastest way, I think any path is better than no path)
I'm going to try my luck with $u = 4x+2 \implies \frac{du}{dx} = 4 \implies dx= \frac{1}{4} du$
Also, by the above assumption, $x = \frac{u-2}{4}$
$$
\require{cancel}
\begin{align}
&\int (4x+2)\sqrt{x^2+x+1}\,dx \\
&= \int{u\cdot\sqrt{\left(\frac{u-2}{4}\right)^2 + \frac{u-2}{4}+ 1}}\cdot\frac{1}{4} du\\
&= \frac{1}{4} \int{u}\cdot\sqrt{\frac{u^2 - 4u + 4 +4(u-2) + 4^2}{4^2}}\, du\\
&= \frac{1}{4} \int{ u \cdot \frac{1}{4}\cdot \sqrt{u^2 \cancel{- 4u} + 4 \cancel{+ 4u} - 8 +16}}\, du\\
&=\frac{1}{16} \int{u\cdot\sqrt{u^2 + 12}}\, du
\end{align}
$$
Mhh, we need to substitute again inorder to proceed. Let's continue this with
$$v =\sqrt{ u^2 + 12 }
\implies \frac{dv}{du} = \frac{2u}{2\sqrt{u^2 + 12}} = \frac{u}{v}
\implies du = \frac{v}{u}\, dv$$
Continuing,
$$
\begin{align}
&\frac{1}{16} \int{u\cdot\sqrt{u^2 + 12}}\, du \\
&= \frac{1}{16} \int{u\cdot v \cdot \frac{v}{u}\, dv} \\
&= \frac{1}{16}\int{v^2}\, dv\\
&= \frac{1}{16}\frac{v^{2+1}}{(2+1)} + C\\
&= \frac{v^3}{48} + C
\end{align}$$
Putting $v$ interms of $x$, you'll probably get the right answer.
$\dots$ Maybe taking $u=x^2+x+1$ would have been a better substitution.
Well, practice makes perfect! :D
|
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|
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y)=(y-1)^2$$
this equality shows that $x$ and $z$ have same sign(both of them are positive or negative).like this we can conclude that $xy=(x-1)^2$ and $yz=(x-1)^2$.So all variables are positive or negative.If all of them are negative then $x+y+z \neq 2$ so all of $x,y,z$ are non negative. I don't know what to do for showing that none of $x,y,z$ will be be bigger than $\frac{4}{3}$.
|
You have
$$
1=xy+z(x+y)=xy+(x+y)(2-(x+y))=xy+x(2-x)+2y-y^2,
$$
so that
$$
0=y^2+(x-2)y+x^2-2x+1=\bigg(y+\frac{x-2}{2}\bigg)^2+\frac{3}{4}x^2-x
$$
and hence $x-\frac{3}{4}x^2 =-\bigg(y+\frac{x-2}{2}\bigg)^2 \leq 0$. So we have
$\frac{3x}{4}(\frac{4}{3}-x) \leq 0$, i.e. $x(x-\frac{4}{3}) \geq 0$, hence
$x\in [0,\frac{3}{4}]$. By symmetry, the same holds for $y$ or $z$.
|
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|
What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$? What does this number equal if it goes on forever?
$$\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\frac{\frac{17}{18}}{\frac{19}{20}}}{\frac{\frac{21}{22}}{\frac{23}{24}}}}{\frac{\frac{\frac{25}{26}}{\frac{27}{28}}}{\frac{\frac{29}{30}}{\frac{31}{32}}}}}$$
$$.$$$$.$$$$.$$
Treat the fraction in blocks, one divides what is under it, then that pair is divided by the next pair, those four are divided by the next four. ETC
Edit. I'm reformatting as I can't read the original... aged eyes I guess ;-( but in deference to almagest's comment I am leaving the original too.
$$\frac{\frac{\frac{1}{2}\Big/\frac{3}{4}}{\frac{5}{6}\Big/\frac{7}{8}}\Bigg/
\frac{\frac{9}{10}\Big/\frac{11}{12}}{\frac{13}{14}\Big/\frac{15}{16}}}
{\frac{\frac{17}{18}\Big/\frac{19}{20}}{\frac{21}{22}\Big/\frac{23}{24}}\Bigg/
\frac{\frac{25}{26}\Big/\frac{27}{28}}{\frac{29}{30}\Big/\frac{31}{32}}}
\cdots$$
|
SPOILER. Computation of the expression out to $20$ levels suggests strongly that the answer is...
... don't rollover unless you really want to see it...
$\displaystyle\frac{1}{\sqrt2}$
|
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|
Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why?
The final answer is: $x^2-2x-2$.
|
Here's how I would do the "calculate and check" steps:
$$\begin{align}(x-x_1)(x-x_2)&=0\\
x_1&=1+\sqrt 3\\
x_2&=1-\sqrt 3\\
x^2-(x_1+x_2)x+x_1x_2&=0\\
x_1+x_2&=2\\
x_1x_2&=1^2-\sqrt 3^2=-2\\
x^2-2x-2&=0\end{align}$$
Now, taking the form that you created, we have
$$-\frac b2x^2+bx+b = 0$$
Plugging in $x=x_1=1+\sqrt 3$ results in
$$\begin{align}-\frac b2(1^2+2\sqrt 3+3)+b(1+\sqrt 3)+b&=0\tag 1\\
-\frac b2(4+2\sqrt 3)+2b+b\sqrt 3&=0\\
-2b-b\sqrt 3+2b+b\sqrt3 &= 0\\
0 &= 0\end{align}$$
Note that using $x=x_2=1-\sqrt 3$ simply places a couple negative signs so we end up with $(1^2-2\sqrt 3+3)$ and $(1-\sqrt 3)$ in $(1)$.
This check step should always result in $0=0$ when the equation and other calculations are done correctly.
|
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|
What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$? What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$?
I know that $\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle$ or equivalently $\nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$.
Eventually, I am going to evaluate this gradient at a point and then dot it with some given unit vector to find a directional derivative. I think maybe what is troubling me is the radical in $f$. Will I have to use the chain rule in order to find the three partial derivatives?
I tried this:
$$\frac{\partial f}{\partial x} = \frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{1}{2}}2x$$
$$= \frac{x}{\sqrt{x^2 + y^2 + z^2}}$$
That result looks ugly to me. However, if I did that partial derivative correctly, then I think I know what I am doing.
|
So you're asking about what is $\nabla_{\vec{v}}f$ where $f = \sqrt{x^2 + y^2 + z^2}$. First, lets organize the answer:
$\nabla_{\vec{v}}f =
\begin{aligned}
\begin{bmatrix}
\frac{\partial f}{\partial x} \\
\frac{\partial f}{\partial y} \\
\frac{\partial f}{\partial z}
\end{bmatrix}
\end{aligned}$ $\cdot \, \vec{v}$
Now, lets solve for
$\frac{\partial f}{ \partial x}$.
$$\frac{\partial}{\partial x}\left(\sqrt{x^2 + \overbrace{y^2 + z^2}^{\mathrm{a\ constant}}}\right)$$
Now, lets start, also just for the sake of space, I'll make this substitution $y^2 + z^2 =c$, so we have
$$
\frac{\partial f }{\partial x}
= \frac{d}{dx}\sqrt{x^2 + c}
$$
so $g(x) = \sqrt{x},\, h(x) = x^2 + c$
I am going to skip some steps.
$$g'(x) = \frac{1}{2\sqrt{x}}, \qquad h'(x) = 2x$$
$$\frac{d}{dx} = g'(h(x)) \bullet h'(x) \rightarrow \frac{4x}{\sqrt{x^2 \, + \, c }} \rightarrow \mathbf{\frac{x}{\sqrt{x^2 + y^2 + z^2}}}$$
Then, $$\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \, \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \, \frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}.$$
Okay, so now let's find a directional derivative! But first, let's go to the gradient. say $\vec{v} =
\begin{aligned}
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
\end{aligned}$
$$\begin{aligned}
\begin{bmatrix}
\frac{x}{\sqrt{x^2 + y^2 + z^2}} \\
\frac{y}{\sqrt{x^2 + y^2 + z^2}} \\
\frac{z}{\sqrt{x^2 + y^2 + z^2}}
\end{bmatrix}
\cdot
\begin{bmatrix}
a \\
b \\
c \\
\end{bmatrix}
\end{aligned} =
\frac{xa \,+ \, yb \, + \, zc}{\sqrt{x^2 \, +\, y^2 \, + \, z^2}}$$
I hope I helped you out on this page.
|
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|
Trigonometric equation $\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$ I want to solve a trigonometric equation below:
$$\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$$
What is the value of $x$ for $x>0$
Thank you for your help.
|
Note $\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}$, and $\tan$ is periodic with period $\pi$, so this means that for some $n\in\mathbb{Z}$
$$\frac{\sqrt{3}x}{2}=n\pi-\frac{\pi}{3}\implies x=\frac{2\pi}{\sqrt{3}}\left(n-\frac{1}{3}\right).$$
Since we require $x>0$, this is equivalent to saying $n-\frac{1}{3}>0$, and since $n$ is an integer, we must have $n>0$. So in conclusion all possible solutions are of the form:
$$x=\frac{2\pi}{\sqrt{3}}\left(n-\frac{1}{3}\right),\quad n\in\mathbb{N}.$$
|
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|
Using Fermat's Little Theorem to find remainders My professor does not want us to use mods.
Use Fermat's Little Theorem to find the remainder of $12^{7641}$ divided by $7$.
I have $7 \mid \left(\,12^{7} - 12\,\right)$ and that $12^{7651}=\left(\,12^{7}\,\right)^{1093}$
but I'm basically stuck because a remainder of $12^{1093}$ makes no sense.
|
By Fermat's Little Theorem we know that $12^6 = 1 \pmod 7$
*
*Now $7641 =6 \times 1273 + 3$. So that $(12^6)^{1273} = 1^{1273} = 1 \pmod 7$
*hence $12^{7638} = 1 \pmod 7$
*hence $12^{7638} \times 12^3 = 12^3 \pmod 7$
*hence $12^{7641} = 12^3 \pmod 7$
Now $12 = 2^2 3$. Hence $12^3 = 2^6 \times 3^3$
*
*now $2^3 = 1 \pmod 7$
*hence $(2^3)^2 = 1^2 = 1 \pmod 7$ and $3^3 = 6 \pmod 7$
*now we know that $a = b \pmod p$ and $c = d \pmod p$
*then $a \times c = b \times d \pmod p$
*hence $2^6 \times 3^3 = 1 \times 6 \pmod 7$
*hence $12^3 = 6 \pmod 7$
*hence $12^{7641} = 12^3 = 6 \pmod 7$
so the remainder is 6.
|
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|
Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\frac{w^2}{2}\text{ and if }x \to \sqrt{2} \text{ then }w\to 2.$$
Is that correct?
|
That is correct, although you could do the same without the change of variables, and with some courage to manipulate the $\sqrt{2}$
$\lim\limits_{x\to \sqrt{2}} \frac{x^2 - 2 }{ x^2 + \sqrt{2}x - 4}= \lim\limits_{x\to \sqrt{2}} \frac{ ( x- \sqrt{2} ) ( x + \sqrt{2} ) }{ (x - \sqrt{2} ) ( x + 2 \sqrt{2} ) } = \lim\limits_{x\to \sqrt{2}} \frac{x + \sqrt{2}}{ x + 2\sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{2}} = \frac{2}{3}$
As I said, is just a question of courage, but irt may save you some time factoring the polynomials.
|
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|
Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
|
Perhaps a small reduction in the calculations in other answers. First, we use the standard technique for finding the power of a prime which divides a factorial:
$$\Bigl\lfloor\frac{28}{5}\Bigr\rfloor+\Bigl\lfloor\frac{28}{25}\Bigr\rfloor
=5+1=6$$
so $5^6$ is a factor of $28!$ (and $5^7$ is not), and likewise
$$\Bigl\lfloor\frac{28}{2}\Bigr\rfloor+\Bigl\lfloor\frac{28}{4}\Bigr\rfloor
+\Bigl\lfloor\frac{28}{8}\Bigr\rfloor
+\Bigl\lfloor\frac{28}{16}\Bigr\rfloor=14+7+3+1=25\ ,$$
so $2^{25}$ is a fcator of $28!$. Therefore $28!$ ends with $6$ zeros and we need the last digit of $28!/10^6$. Simplifying modulo $10$, we have
$$2^\equiv2^5\equiv2^9\equiv2^{13}\equiv2^{17}\ ,\quad 3^4\equiv1\ ,\quad
7^4\equiv1\ ,\quad 9^2\equiv1\ ;$$
cancelling $2$s and $5$s and reducing the resuts modulo $10$ gives
$$\eqalign{28!/10^6
&\equiv2^{19}1.1.3.1.1.3.7.1.9.1.1.3.3.7.3.1.7.9.9.1.1.1.3.3.1.3.7.7\cr
&\equiv2^3.3^8.7^5.9^3\cr
&\equiv8.1.7.9\cr
&\equiv4\ .\cr}$$
|
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|
Find $ \int \frac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$ Find $$\int \dfrac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$$
I cannot figure out how start this problem, can anyone explain
|
$$
\begin{aligned}
\int\frac{\sin^2 x}{1 + \sin^2 x}\,\mathrm{d}x&=\int\frac{1+\sin^2x - 1}{1+\sin^2 x}\,\mathrm{d}x\\
&=x - \int\frac{\mathrm{d}x}{1+\sin^2 x}\\
&=x+\int\frac{-\csc^2 x}{\csc^2 x + 1}\,\mathrm{d}x\\
&=x+\underbrace{\int\frac{-\csc^2 x}{2+\cot^2 x}\,\mathrm{d}x}_{t=\cot x\implies\mathrm{d}t=-\csc^2x\,\mathrm{d}x}\\
&=x+\int\frac{\mathrm{d}t}{2+t^2}\\
&=x+\frac{1}{\sqrt{2}}\arctan\!\left(\frac{t}{\sqrt{2}}\right)+C\\
&=x+\frac{1}{\sqrt{2}}\arctan\!\left(\frac{\cot x}{\sqrt{2}}\right)+C.
\end{aligned}
$$
|
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|
Simplifying an expression written as the sum of three fractions Specifically, I don't know what to do first given the following expression:
$$
\frac{4x - 2}{6} - \frac{2 - x}{4} + \frac{x + 3}{3}
$$
So I think of it as $\frac 16(4x-2) - \frac 14(2-x) + \frac 13(x+3)$?
That just gives me more fractions.
|
Find the common denominator first, noting that the least common multiple of $6, 4, 3$ is $12: $12 = 2\times 6 = 3\times 4 = 4\times 3$. This way, you can write the entire expression as one fraction:
$$\frac{4x - 2}{6} - \frac{2 - x}{4} + \frac{x + 3}{3} = \frac{2(4x-2) - 3(2-x) + 4(x+3)}{12}$$
Now expand the factors in the numerator on the right-hand side, and then simplify (add [or subtract] the factors of $x$, and do the same for the constants. Doing this will give you a final result of the form $\dfrac{ax + b}{12}$, where $a, b$ are nonzero constants.
|
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|
Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$
x^3 + 1 = 2\sqrt[3]{{2x - 1}}
$$
I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here?
|
For $x$ - real
$$2\cdot(2x-1)^\frac13=x^3+1$$
Let $y= (2x-1)^\frac13$
Therefore, $$y^3=2x-1$$
$$x=\frac{(y^3+1)}{2}$$
Then you get
$$2\cdot y=\left(\frac{y^3+1}2\right)^3+1$$
Resolve it for $y$, and then replace find the $x$
Resolving for $y$:
*
*multiply both sides by 8 and you get
$$16\cdot y=(y^3+1)^3+8$$
$$16\cdot y = y^9+ 3\cdot y^6+3\cdot y^3 +9$$
$$(y-1)(y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 )(y^2 + y - 1)=0$$
Then we get:
$y-1 = 0 \implies y=1$
$y^2 +y - 1=0 \implies y = \frac{1}{2}(-1\pm \sqrt5)$
$y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 =0\implies$ No roots
|
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|
Symmetric inequality for a rational function of three variables If $x,y,z$ are positive real numbers such that $xyz \geqslant 1$ prove: $$\dfrac{x^3+y^3}{x^2+xy+y^2}+\dfrac{y^3+z^3}{y^2+yz+z^2}+\dfrac{x^3+z^3}{x^2+xz+z^2} \geqslant 2$$
I have tried with Hölder's inequality, but it is not working. Can you help, please?
|
Since:
$$\frac23x^3+\frac13y^3\ge x^y\text{ or }\frac23y^3+\frac13x^3\ge y^2x$$
So:
$$\large x^3+y^3\ge\frac13x^3+\frac23x^2y+\frac23xy^2+\frac13y^3=\frac13(x+y)(x^2+xy+y^2)$$
Hence:
$$\large \sum_{cyc}\frac{x^3+y^3}{x^2+xy+y^2}\ge\sum_{cyc}\frac{x+y}{3}=\frac23(x+y+z)\ge2\sqrt[3]{xyz}=2$$
|
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|
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
Prove $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$
Things I have done: after trying many ways and failing, I reached the fact that$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{100}{101}\right)=\frac{1}{101}$
So $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true. it remains to show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$.
I'm thinking of applying my approach on proving this part. something like this. $$\frac{1}{225}<\frac{1}{x}=\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right) \times B<\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2$$
And another thing I'm curious about it,is there a way to approximate value of $\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}$ ?
|
Don't you already have the answer?
$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2>\frac{1}{2}\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{98}{99}\right)=\frac{1}{200}$
|
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|
Solving a recurrence relation of second order I have a pattern, which goes: $x_n =2(x_{n-1}-x_{n-2})+x_{n-1}$ and this pattern holds for all $n \ge 2$. I also know that $x_0 = 1 \ and \ x_1 = 5.$
$x_2 = 2(x_1-x_0)+x_1$
$\begin{align}
x_3 = 2(x_2-x_1) + x_2 &= 2(2(x_1-x_0)+x_1)+x_2\\
&=2^2x_1-2^2x_0+2x_ 1+x_2
\end{align}$
$\begin{align}
x_4 = 2(x_3-x_2)+x_3 &= 2(2^2x_1-2^2x_0+2x_1+x_2-(2(x_1-x_0)+x_1)+x_2)+x_3 \\
&=2^3x_1-2^3x_0+2^2x_1+2x_2-(2x_1-2x_0+x_1)+x_2)+x_3 \\
&=2^3x_1-2^3x_0+2^2x_1+2x_2 - 2x_1 +2x_0-x_1+x_2+x_3 \\
&=2^2x_1-2^2x_0+2x_1+2^2x_2+x_3
\end{align}$
(The formula below is no longer valid, I noticed since last edit)
$[x_n = 2^{n-1}(x_1-x_0) + \sum^{n-1}_{i=0}(2^{n-2-i}x_i)]$
Well, here I'm confused; there's no pattern to be found anymore, or am I just blind?
|
Here is another answer which uses a trick special to this problem. Note that the recurrence may be written as $x_n-x_{n-1}=2(x_{n-1}-x_{n-2})$ for $n\geq 2$. So the difference in terms is doubling at each step, and if we iterate this we come to $x_n-x_{n-1}=2^{n-1}(x_1-x_0)=2^{n+1}$ since $x_1-x_0=4$. So we can solve the recurrence $x_n=x_{n-1}+2^{n+1}$ instead (valid for $n\geq 1$). But this iterates to
\begin{align}
x_n
&=x_{n-1}+2^{n+1}\\
&=x_{n-2}+2^n+2^{n+1}\\
&\cdots\\
&=x_0+2^2+2^3+\cdots +2^{n+1}\\
&=1+4\cdot(2^{n}-1)\\
&=4\cdot 2^{n+2}-3
\end{align}
which is the same answer as in the other solutions.
|
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|
Proving $\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi }{2-x}\right)$ How to solve the following question?
If $n$ is an integer, show that
\begin{eqnarray} \left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi}{2}-x\right) \end{eqnarray}
where x is a real number such that $\left(1+\sin x\right)^2+\cos^2x>0$.
My work
I have no idea how to continue...
Thank you for your help.
|
$$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$
$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right)$$
Apply de Movire's Theorem assuming $1+\sin x\ne0$
as $\sin x+1=0\implies 1+\sin x-i\cos x=0$
|
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|
finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to integrate. I tried to split them into two $\displaystyle\int\frac{2}{x+2
}dx$ and $\displaystyle\int\frac{1-2x}{x^2+1}dx$ but I don't know how to do after.
|
Hint:
Let $u=x+2$ for
\begin{equation}
\int\frac{2}{x+2}dx
\end{equation}
Split the latter integral into two parts
\begin{equation}
\int\frac{1-2x}{x^2+1}dx=\int\frac{1}{x^2+1}dx-\int\frac{2x}{x^2+1}dx
\end{equation}
then let $x=\tan\theta$ also use identity $\sec^2\theta=1+\tan^2\theta$ for
\begin{equation}
\int\frac{1}{x^2+1}dx
\end{equation}
and
let $v=x^2+1$ for
\begin{equation}
\int\frac{2x}{x^2+1}dx
\end{equation}
|
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|
How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
|
$$
\int\frac{y^2-x^2}{(y^2+x^2)^2}dy
$$
\begin{align}
y & = x\tan\theta \\
dy & = x\sec^2\theta\,d\theta \\
y^2+x^2 & = x^2(\tan^2\theta+1) = x^2\sec^2\theta \\
y^2-x^2 & = x^2(\tan^2\theta-1) \\
& \phantom{=}\text{etc.}
\end{align}
Usually with $\displaystyle\int \Big(\cdots\cdots\text{something involving }(y^2+\text{constant}^2)\cdots\cdots\Big)\,dy$, this will work.
|
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|
Sum of series with cosines I need to prove this:
$$
\sum_{n = 1}^{\infty}{8 \over \left(\,2n - 1\,\right)^{2}\pi^{2}}\,
\sin\left(\,\left[\,2n - 1\,\right]\,{\pi x \over 2}\,\right)
\sin\left(\,\left[\,2n - 1\,\right]\,{\pi z \over 2}\,\right) = \min\left\{x, z\right\}
$$
I got this:
$$ \frac{8}{\pi ^ 2} \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} \cos((2n -1)\frac{\pi (x - z)}{2}) - \frac{8}{\pi ^ 2} \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} \cos((2n -1)\frac{\pi (x + z)}{2})$$
But now I'm completely stuck. Can anybody please help me?
Thanks a lot!!
|
Big Hint:
Integrating the negative of
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(nx)}{n}
&=\mathrm{Im}\left(\sum_{k=1}^\infty\frac{e^{inx}}{n}\right)\\
&=-\mathrm{Im}\left(\log(1-e^{ix})\right)\\
&=\frac{x}{2|x|}(\pi-|x|)\tag{1}
\end{align}
$$
we get
$$
\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}=\frac{2\pi^2-6\pi|x|+3x^2}{12}\tag{2}
$$
and subtracting $\frac14$ of $(2)$ at $2x$, which is the even terms of $(2)$,
$$
\sum_{n=1}^\infty\frac{\cos(2nx)}{4n^2}=\frac{\pi^2-6\pi|x|+6x^2}{24}\tag{3}
$$
we get
$$
\sum_{n=1}^\infty\frac{\cos((2n-1)x)}{(2n-1)^2}=\frac{\pi^2-2\pi|x|}8\tag{4}
$$
for $x\in(-\pi,\pi)$.
Next, recall that $\min(x,y)=\dfrac{x+y-|x-y|}2$.
|
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|
Sums $\sum_{n=1}^{N}\sqrt{4n+1}$ I need to find sum of the first N terms of the sequence whose nth term is as follow :
T(n)= $\sqrt{4*n+1}$
So the sequence is : $\sqrt{5}$,$\sqrt{9}$,$\sqrt{13}$,$\sqrt{17}$,$\sqrt{21}$......
Please help how to find it for a given N.
|
There is an answer which is not simple at all since $$\sum_{i=1}^n \sqrt{4i+1}=2 \left(\zeta \left(-\frac{1}{2},\frac{5}{4}\right)-\zeta
\left(-\frac{1}{2},n+\frac{5}{4}\right)\right)$$ in which appears Hurwitz $\zeta$ function.
However, you could notice that $$2\sqrt{i}<\sqrt{4i+1}<2\sqrt{i+1}$$ and so $$2H_{n}^{\left(-\frac{1}{2}\right)}<\sum_{i=1}^n \sqrt{4i+1}<2H_{n+1}^{\left(-\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers.
For large values of $m$, an approximation is $$H_{m}^{\left(-\frac{1}{2}\right)}=\frac{2 m^{3/2}}{3}+\frac{\sqrt{m}}{2}+\zeta
\left(-\frac{1}{2}\right)+O\left(\left(\frac{1}{m}\right)^{3/2}\right)$$
For $n=10$, the exact summation leads to $46.1629$ while the expression on the left is $44.9102$ and the one on the right is $51.5447$
Added later
The approximation can be improved writing $$\sqrt{4i+1}=2 \sqrt{i}+\frac{\sqrt{\frac{1}{i}}}{4}-\frac{1}{64}
\left(\frac{1}{i}\right)^{3/2}+O\left(\left(\frac{1}{i}\right)^{5/2}\right)$$ and expanding again the harmonic numbers. As a result comes the approximation $$\sum_{i=1}^n \sqrt{4i+1}\approx \frac{4 n^{3/2}}{3}+\frac{3 \sqrt{n}}{2}+\left(2 \zeta \left(-\frac{1}{2}\right)+\frac{\zeta
\left(\frac{1}{2}\right)}{4}-\frac{\zeta \left(\frac{3}{2}\right)}{64}\right)+O\left(\left(\frac{1}{n}\right)\right)$$ which does not seem to be too bad $(46.0854$ for $n=10)$.
For $n=1000$, the value of the sum is $42210.32452$ while the approximation gives $42210.31462$.
|
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|
How to factor $(x^5+1) (x^5-1) $ I have this:
$ (x^5+1) (x^5-1) $
And I don't know how to continue factor.
Geogebra's Factor says:
$(x+1)(x-1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$
|
Since $-1$ is a root of $x^5+1$, we know that it is divisible by $x+1$. Likewise, since $1$ is a root of $x^5-1$, we know that it is divisible by $x-1$. You can use polynomial long division to obtain the other factors.
|
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|
Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now:
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:
$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$
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Too long for a comment, perhaps the following approach would help.
Express the inner term of square root as follows
\begin{align}
\frac{\sin(x-a)}{\sin(x+a)}&=\frac{\sin(x)\cos(a)-\cos(x)\sin(a)}{\sin(x)\cos(a)+\cos(x)\sin(a)}\qquad\Rightarrow\qquad\text{divide by}\,\cos(x)\cos(a)\\
&=\frac{\tan(x)-\tan(a)}{\tan(x)+\tan(a)}
\end{align}
Let $t^2=\tan(x)+\tan(a)$, then
\begin{align}
\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\,dx&=\int\sqrt{\frac{\tan(x)-\tan(a)}{\tan(x)+\tan(a)}}\,dx\\
&=2\int\frac{\sqrt{t^2-2\tan(a)}}{1+(t^2-\tan(a))^2}\,dt\\
&=2\int\frac{\sqrt{1-2\frac{\tan(a)}{t^2}}}{\frac{1}{t^4}+\left(1-\frac{\tan(a)}{t^2}\right)^2}\,\frac{dt}{t^3}\qquad\Rightarrow\qquad\text{let}\,u=\frac{\tan(a)}{t^2}\\
&=-\int\frac{\sqrt{1-2u}}{\frac{u^2}{\tan^2(a)}+\left(1-u\right)^2}\,\frac{du}{\tan(a)}\\
&=-\int\frac{\tan(a)\sqrt{1-2u}}{u^2\sec^2(a)-2u\tan^2(a)+\tan^2(a)}\,du\\
&=-\frac{1}{\tan(a)}\int\frac{\sqrt{1-2u}}{u^2\csc^2(a)-2u+1}\,du\\
\end{align}
|
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|
Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
|
You can put $t=\frac 12 x-\frac 18 x^2 + \frac 1{16}x^3$ then, you get:
$$\sqrt{\sqrt{1+x}}=\sqrt{1+t}=1+ \frac 12 t--\frac 18 [t^2]_3 + \frac 1{16}[t^3]_3 ... $$
where $[P(x)]_3$ for a polynom $$P(x)=a_0+a_1 x +a_2x^2+a_3x^3+a_4x^4 ... +a_n x^n $$ is:
$$[P(x)]_3=a_0+a_1x+a_2x^2+a_3x^3.$$
For example , since: $t=x(\frac 12 -\frac 18 x + \frac 1{16}x^2)$ we have :
$$t^2=x^2(\frac 12 -\frac 18 x + \frac 1{16}x^2)^2$$ then : $$[t^2]_3=x^2(\frac 14 -\frac 18 x)$$
|
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|
Integral $\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$ Integrate:
$$
\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}
$$
|
$$I=\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\int_0^\pi \frac{(\pi-x)\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$$
$$2I=\pi\int_0^\pi \frac{\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\pi\int_0^\pi \frac{\sec^2x\,\operatorname dx}{a^2+b^2\tan^2x}$$
$$\frac{2I}{\pi}=\int_0^{\pi}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}=\int_0^{\pi/2}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}+\int_{\pi/2}^{\pi}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}$$
$$\frac{2I}{\pi}=\frac1{ab}\arctan\left(\frac{b\tan x}{a}\right)|_0^{\pi/2}+|_{\pi/2}^{\pi}$$
$$I=\ldots\ldots$$
|
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|
Evaluating the integral $ \int{\frac{x}{\sqrt{2x^2 + 3}}}dx $ I am trying to integrate the following:
$$
\int{\frac{x}{\sqrt{2x^2 + 3}}}dx
$$
It seems to me to be a trig substitution; however, I couldn't seem to get it into one of the three forms, i.e.,
$$\sqrt{a^2 - x^2}$$
$$\sqrt{x^2 - a^2}$$
$$\sqrt{a^2 + x^2}$$
I also tried integration by parts. If I made $u = \frac{1}{\sqrt{2x^2 + 3}}$ and $dv = x$, the next integral was more complicated, and if I made $u = x$ and $dv = \frac{1}{\sqrt{2x^2 + 3}}$, I was again unsure how to integrate the 1/sqrt term.
How do I solve this?
|
$$ \int \frac{x}{\sqrt{2x^2+3}}dx $$
Let $u=\sqrt{2x^2+3}$, then
$$ \frac{d}{dx}u=\frac{2x}{\sqrt{2x^2+3}}\Rightarrow \frac{1}{2}du=\frac{x}{\sqrt{2x^2+3}}dx $$
So now we have
$$ \frac{1}{2}\int du= \frac{1}{2}u+C = \frac{1}{2}\sqrt{2x^2+3}+C $$
|
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|
Is there any integer solutions of $3x^3+3x+7=y^3$? $3x^3+3x+7=y^3$
$x, y \in \mathbb{N}$
Having thought about it two hours, and I'm still not sure how to show there actually aren't any integer solutions.
EDIT
Another formulation of this problem: Prove that $3x^3+3x+7$ cannot be a perfect cube.
|
There are no natural solutions for $x$ and $y$.
A quick bit of important information: If $C_n$ is the $n$th non-negative cube ($0,1,8,...$), then $C_{n+1}-C_n = 3n^2+3n+1$. (This can be grasped visually by a similar method to the visual proof that the consecutive differences of consecutive squares are consecutive odd integers.)
Let $s,a$ be positive integers such that $a=s-1$ (this will make the math a bit easier later).
$y$ is clearly greater than $x$ so let $y=x+s=x+a+1$.
Subtract $x^3$ from both sides of $3x^3+3x+7=y^3$ to get $$2x^3+3x+7=y^3-x^3$$
Let us focus on expanding $y^3-x^3$. Note that this is equivalent to $$((x+a+1)^3-(x+a)^3)+((x+a)^3-(x+a-1)^3)+\cdots+((x+2)^3-(x+1)^3)+((x+1)^3-x^3)$$ Which, by expanding the differences of consecutive cubes, is equivalent to: $$3x^2+3x+1+3(x+1)^2+3(x+1)+1+3(x+2)^2+3(x+2)+1+⋯+3(x+a)^2+3(x+a)+1=$$ $$3(x^2+x+(x+1)^2+x+1+\cdots+(x+a)^2+x+a)+a+1=$$ $$3(x^2+(x+1)^2+⋯+(x+a)^2)+3(x+x+1+⋯+x+a)+a+1$$
Note $a+1$ arises since $1$ appears $a+1$ times in the expression.
We now condense the first two terms on the RHS (starting with the first): $$3(x^2+(x+1)^2+⋯+(x+a)^2)=$$ $$3(x^2+x^2+2x+1+x^2+4x+4+x^2+6x+9+⋯x^2+2ax+a^2)=$$ $$3(x^2(a+1)+xa(a+1)+\frac{a(a+1)(2a+1)}{6})$$
These coefficients are obtained by noting there are $a+1$ terms of $x^2$ (as above with $1$s), that the coefficients on the $x$ terms are the first $a$ even integers, and that the constants are the first $a$ squares.
Now we move to the second term on the RHS:
$$3(x+x+1+⋯+x+a)=$$ $$3(x(a+1)+\frac{a(a+1)}{2})$$
Again, the coefficients are found by noting the $a+1$ $x$ terms and the first $a$ positive integers.
Now, returning to $$3(x^2+(x+1)^2+⋯+(x+a)^2)+3(x+x+1+⋯+x+a)+a+1$$ we replace the first two terms with results found above to obtain: $$3(x^2(a+1)+xa(a+1)+\frac{a(a+1)(2a+1)}{6})+3(x(a+1)+\frac{a(a+1)}{2})+a+1$$
At this point it is convenient to substitute $a$ for $s-1$ which leaves us with: $$3(x^2s+xs(s-1)+\frac{s(s-1)(2s-1)}{6})+3(xs+\frac{s(s-1)}{2})+s$$
Distribute the $3$'s: $$3x^2s+3xs(s-1)+\frac{s(s-1)(2s-1)}{2}+3xs+\frac{3s(s-1)}{2}+s=$$ $$3x^2s+3xs^2+\frac{s(s-1)(2s-1)+3s(s-1)}{2}+s=$$ $$3x^2s+3xs^2+\frac{s(s-1)(2s+2)}{2}+3s=$$ $$3x^2s+3xs^2+s(s-1)(s+1)+s=$$ $$3x^2s+3xs^2+s^3$$
UPDATE This answer previously contained a mathematical error which has since been corrected, however as a result it no longer solves the problem.
|
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|
Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here
$\int \dfrac{1}{4 x^4} \dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck here please help
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As a general rule, whenever dealing with integrands containing $\sqrt{x^2+a^2}$ in their expression, the
natural substitutions are $x=a\tan t$ or $x=a\sinh u$, since, on one hand, $\sqrt{\tan^2t+1}=\dfrac1{\cos t},$
and $\tan't=\dfrac1{\cos^2t},$ transforming our integral into $\displaystyle16\int\frac{\sin^3t}{\cos^6t}dt=-16\int\frac{1-\cos^2t}{\cos^6t}d(\cos t)$
whose evaluation is trivial; and, on the other hand, $\sqrt{\sinh^2u+1}=\cosh u$, and $\sinh'u=\cosh u$
yielding $\displaystyle16\int\sinh^3u~\cosh^2u~du=16\int(\cosh^2u-1)\cosh u~d(\cosh u)$, which is also trivial to evaluate.
|
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|
At what point does normal line intersect curve second time?
At what point does the normal line to $y=-5+4x+3x^2$ at $(1,2)$ intersect the parabola a second time?
$y'=6x+4$
$m_{tangent}=6(1)+4=10$
$m_{normal}=-\dfrac{1}{10}$
$y=f'(1)(x-1)+f(1)$
$=(-1/10)(x-1)+2$
$y=-1/10+21/10 \implies$ equation of normal line
Then, set normal line = curve to find point of intersection:
$(-1/10x+21/10=-5+4x+3x^2)(-10) \rightarrow$ I multiplied by $-10$ to get rid of the $-1/10x$ fraction.
$x=60-4x-30x^2$
$-5(6x^2+x-12)=0$
Then I used the quadratic formula to find the $x$ values. I got $x= \dfrac{4}{3},-\dfrac{3}{2}$ and this is wrong.
How do I find the $x$ values and $y$ values and what am I doing wrong? Thanks.
|
Correcting some algebra, will look into the analytic geometry when I have time.
$$C_{norm}: y= \frac{21}{10}-\frac{x}{10}$$
Then you have $$\frac{21}{10}-\frac{x}{10}=-5+4x+3x^2 \iff -3x^2-\frac{41x}{10}+\frac{71}{10}=0 \iff -\frac{1}{10} ((x-1)(71+30x))=0 \iff (x-1)(71+30x)=0 \iff x=1 \; \mathrm{or} \; x=-\frac{71}{30}$$
|
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|
How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to natural logs and derivatives somehow but then i'm not even sure how that would help.
|
Note that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$$
$$\frac{ab+ac+bc}{abc}=\frac{1}{a+b+c}$$
$$(ab+ac+bc)(a+b+c)-abc=0$$
$$(a+b)(b+c)(c+a)=0$$
then $a=-b$, $a=-c$ or $b=-c$.
If $a=-b$, how $n$ is odd, then
$$a^n + b^n + c^n=(-b)^n + b^n + c^n=c^n$$ and $$(a+b+c)^n=(-b+b+c)^n=c^n$$
you continue...
|
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|
differentiating a function of a function $w=\sqrt{u^2+v^2}$ I want to find the total differential of $w$ from a given function
$w=\sqrt{u^2+v^2}\:with\:u\:=\:cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)\:and\:v\:=\:sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\right)$
to solving this problem I'm using
$\left(a\right)\:\:\:\:\frac{dw}{dp}=\frac{∂w}{∂u}\cdot \frac{du}{dp}+\frac{∂w}{∂v}\cdot \frac{dv}{dp}$
first I'm solving for the partial derivatives of $w$ with respect to $u$ and $v$ on RHS
$\left(b\right)\:\:\:\:\frac{∂w}{∂u}=\frac{u}{\sqrt{u^2+v^2}}\:\:\:\:and\:\:\:\:\frac{∂w}{∂v}=\frac{v}{\sqrt{u^2+v^2}}$
and then I'm solving for the total derivatives of $u$ and $v$ with respect to $p$ on RHS using chain rule.
$\frac{du}{dp}=\frac{d}{dx}cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)=\frac{du}{dx}\cdot \frac{dx}{dy}\cdot \frac{dy}{dp}$
let $x=ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\:$ and $y\:=tan\left(p+\frac{1}{2}\pi \right)$
$\left(1\right)\:\frac{du}{dx}=\frac{dcos\left(x\right)}{dx}=-sin\left(x\right)=-sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)$
$\left(2\right)\:\frac{dx}{dy}=\frac{dln\left(y\right)}{dy}=\frac{1}{y}=\frac{1}{tan\left(p+\frac{1}{2}\pi \right)}$
$\left(3\right)\:\frac{dy}{dp}=\frac{dtan\left(p+\frac{1}{2}\pi \right)}{dp}=sec^2\left(p+\frac{1}{2}\pi \right)$
substitute back $(1)$, $(2)$, and $(3)$ into $\frac{du}{dp}$ i get
$\left(c\right)\:\:\:\:\frac{du}{dp}=-\frac{sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{tan\left(p+\frac{1}{2}\pi \:\right)cos^2\left(p+\frac{1}{2}\pi \:\right)}$
now i want to find $\frac{dv}{dp}$
$\frac{dv}{dp}=\frac{dsin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{dp}=\frac{dv}{dx}\cdot \frac{dx}{dy}\cdot \frac{dy}{dp}$
since $\frac{dx}{dy}$ and $\frac{dy}{dp}$ have the same result as $(2)$ and $(3)$, now i only have to find $\frac{dv}{dx}$ with $v\:=\:sin\left(x\right)$ i get $\frac{dv}{dx}=cos\left(x\right)=cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)$ and then substitute them back into $\frac{dv}{dp}$ yields
$\left(d\right)\:\:\:\:\frac{dv}{dp}=\frac{cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{tan\left(p+\frac{1}{2}\pi \:\right)cos^2\left(p+\frac{1}{2}\pi \:\right)}$
finally, substitute $(b)$, $(c)$, and $(d)$ into $(a)$ i get
$\frac{dw}{dp}=\frac{cot\left(p+\frac{1}{2}\pi \:\right)\cdot sec^2\left(p+\frac{1}{2}\pi \:\right)}{\sqrt{u^2+v^2}}\left(v\cdot cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\right)-u\cdot sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\:\right)\right)\right)\right)$
this is my answer, but i'm not sure is there an error in my calculation? and how to know if my answer is correct? thank you for your time and advice, it has nothing to do with homework or school, i just have no one to ask to correct my answer.
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If we denote $$\phi := \ln \tan \left(p + \frac{\pi}{2}\right),$$ then
$$w = \sqrt{\cos^2 \phi + \sin^2 \phi} = 1,$$
and thus $$\frac{dw}{dp} = 0.$$
|
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If $y = a\sin{x} + b\cos{x} +C$ then find maxima and minima for $y$. I was able to solve it till
$$y = \sqrt{(a^2 + b^2)}\sin(\alpha + x) + C.$$
But I don't know how to find maxima and minima from here.
If $C = 0$ then maxima & minima equals the amplitude of the sine curve but when $C$ is non-zero then?
I need help from here onwards.
|
Hint:
*
*$\sin(\alpha +x) \in [-1,1]$
$$ \therefore \quad \sqrt{a^2+b^2}\sin(\alpha+x) \in \left[ - \sqrt{a^2+b^2}, \sqrt{a^2+b^2} \right]$$
$$ \quad \therefore \sqrt{a^2+b^2}\sin(\alpha+x)+C \in \left[C- \sqrt{a^2+b^2}, C+ \sqrt{a^2+b^2}\ \right] \quad ,$$ therefore the maximum of $y$ is $_____$, and the minimum of $y$ is $_____$.
|
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|
laurent series of $1/(z-5)$ about $z_o=2$ I'm calculating the Laurent Series of $f=1/(z-5)$ about the point $z_{0}=2$.
The expansion I get has no principal part and is analytic within the circle $|z-2|<3$ since there are no singularities within that disk. I write $f=1/(z-5) = 1/(z-2 -3)= 1/3[(z-2)/3-1)]$, and since $(z-2)/3<1$, we can expand it into a convergent series without containing negative powers of $z-2$.
However, I saw a book say that the answer does have a principal principal part, and in their derivation they introduce a singularity at $z_{0}=2$ . I don't see why they did that.
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See if this is it.
For $|z-2| < 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = -\frac{1}{3}\frac{1}{1-\frac{z-2}{3}} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{(z-2)^n}{3^n} = \sum_{n=0}^{\infty} \frac{(z-2)^n}{3^{n+1}}$$
For $|z-2| > 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = \frac{1}{z-2}\frac{1}{1-\frac{3}{z-2}} = \frac{1}{z-2}\sum_{n=0}^{\infty} \frac{3^n}{(z-2)^n} = \sum_{n=0}^{\infty} \frac{3^n}{(z-2)^{n+1}}$$
|
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The maximum possible value of $ (xv - yu)^2 $ over the surface ... The maximum possible value of $ (xv - yu)^2 $ over the surface given by the equations
$ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $ is :
I solved it and my answer comes out to be $9$ but the correct answer is $36$.
Here is what I did :
Put $ x = r_1\cos\theta$ and $y = r_1\sin\theta $
$ u = r_2\sin\theta $ and $ v = r_2\cos\theta$
Using $ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $
we get $ r_1 = \sqrt 2 $ and $ r_2 = \frac 3{\sqrt 2} $
Now $ (xv - yu)^2 = (r_1r_2)^2(\cos^2\theta - \sin^2\theta)^2 $
$= 9(1-2\sin^2\theta)^2$
Now this will be maximum when $ (1- 2\sin^2\theta)^2 $ is maximum which is at $ \sin^2\theta = 0,1 $ so the maximum value of the expression comes out to be $9$.
Where did I go wrong here? Or am I correct?
|
$x^2+y^2=r_1^2(\cos^2\theta+\sin^2\theta)=r_1^2$, so $r_1=2$
|
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|
Integrating $\int \frac {dx}{\sqrt{4x^{2}+1}}$ $\int \dfrac {dx}{\sqrt{4x^{2}+1}}$
I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:
$\dfrac {1}{2}\ln \left| 2\sqrt {x^2 +\dfrac {1}{4}}+2x\right|+c$
However, whatever idea I come up with to try to solve it, it seems I always obtain:
$\dfrac {1}{2}\ln \left| x + \sqrt {x^2 + 1} \right| + c$
Thanks for reading, and hopefully answering too.
Cheers
Yann
|
we have $\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx$.Now let $1+4x^{2}=u^2$, so we have $xdx=\frac{1}{4}udu$ and so $$\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx=\frac{1}{4}\int \dfrac {u}{u\frac{u^2-1}{4}}\,du=\int \dfrac {1}{u^2-1}\,du$$ also $$\int \dfrac {1}{u^2-1}\,du=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|.$$ Therefore $$\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|+C.$$
|
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|
Show that $\lim_{n\rightarrow \infty } Var(Y_{n}) = 0$. Given $Var(Y_{n}) = (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2}$.
My work: $$lim_{n\rightarrow \infty } Var(Y_{n}) = \lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = \lim_{n\rightarrow \infty } \frac{(\theta_{2}-\theta_{1})^2}{(1 + \frac{2}{n}) (n+1)^{2}} = \lim_{n\rightarrow \infty } \frac{\frac{(\theta_{2}-\theta_{1})^2}{n^2}}{(1 + \frac{2}{n}) (1 + \frac{1}{n})^{2}} = 0.$$
Is this correct?
|
Hint:
$\lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = (\theta_{2}-\theta_{1})^2\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2} $
However, $\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=0$ (Do you know why?).
ADDED part:
$\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=$\lim_{n\rightarrow \infty }\frac{1}{(1+\frac2n)(n+1)^2}=$\lim_{n\rightarrow \infty }\frac{1}{(n+1)^2}=0$
|
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|
Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$ In order to find out the Catalan numbers from their generating function you have to evaluate the product above.
Here is what I thought:
\begin{align*}
1 \times 3 \times 5 \times...\times (2k-3) &= \frac{1 \times 2 \times 3 \times 4 \times \cdots \times (2k - 3)}{2 \times 4 \times \cdots \times (2k - 2)} \\
&= \frac{(2k - 3)!}{2^{k - 1}(k - 1)!} \\
\end{align*}
But a Mathematica session quickly proved me wrong, instead the result in the title is true. Why is that true, what mistake did I make?
|
For example, for the denominator $$2\times 4\times 6\times 8=2\cdot 1\times 2\cdot 2\times 2\cdot 3\times 2\cdot 4=2\times 2\times 2\times 2\cdot 1\times 2\times 3\times 4=2^4\cdot 4!$$
I let you adapt this hint to your exercise.
|
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|
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
\frac{(\sin 2x)}{(\cos 2x)}=2\sin x
$$
$$
\frac{(2\sin x \cos x)}{((\cos x)^2-(\sin x)^2)}=2\sin x
$$
$$
2\sin x\cos x=2\sin x\cos^2 x-2\sin^3 x
$$
$$
\sin x\cos x=\sin x(\cos^2 x-\sin^2x)
$$
$$
\cos x=\cos^2 x-\sin^2 x
$$
$$
\cos x=\cos^2 x+\cos^2x-1
$$
$$
2\cos^2x-\cos x-1=0
$$
Solving this quadratic equation gives $\cos x=1$ or $\cos x=-\frac 12$
$$
\cos x=1=\cos \theta
$$
$$
x=n2\pi
$$
$$
\cos x=-\frac 12=\cos(\frac{2\pi}3)
$$
$$
x=\pm\frac{2\pi}3+n2\pi
$$
|
with $\tan(2x)=\frac{2\cos(x)\sin(x)}{\cos(x)^2-\sin(x)^2}$ we get
$2\sin(x)\left(\frac{\cos(x)}{\cos(x)^2-\sin(x)^2}-1\right)=0$
thus we obtain
$\sin(x)=0$
or
$0=2\cos(x)^2-\cos(x)-1$
|
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|
Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution $$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$
which allows for the trig substitution of $$x = a\sec\theta$$
making the denomonator $$\sqrt{16\sec^2\theta-16}$$
$$\tan^2\theta + 1 = \sec^2\theta$$
$$\Rightarrow4\tan\theta$$
to give $$\int\frac{\frac{4\sec}{9}^3}{4\tan\theta}dx$$
Am I doing this correctly?
|
I would prefer to use integration by parts as below:
$$
\begin{aligned}
\int \frac{x^3}{\sqrt{81 x^2-16}} d x=&\frac{1}{81} \int x^2 d \sqrt{81 x^2-16} \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{1}{6565} \int \sqrt{81 x^2-16} \,d\left(81 x^2-16\right) \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{2}{19683}\left(81 x^2-16\right)^{\frac{3}{2}}+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(243 x^2-162 x^2+32\right)+C \\
= & \frac{\sqrt{81 x^2-16}}{19683}\left(81 x^2+32\right)+C
\end{aligned}
$$
|
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|
how to solve equation with cos I have this equation $\cos2x +5 \cos x + 3=0$. To solve it I rewrite $\cos2x$ to $2 \cos^{2} x- 1$ and set $\cos = t$.
I get the following equation $2t^2 - 1 +5t +3 = 0$ with that and then divide the equation with two $t^2 +\frac{5}{2} t +1 = 0$. I solve this equation and get two $t$, $t_1 = -2$ and $t_2 = - \frac {1}{2}$. $t_2$ is the valid because $t$ can't be larger than 1.
From here on I don't know how to use $t$ to solve this equation $\cos2x +5 \cos x + 3=0$.
Can anyone explain what to do next and how to solve this equation?
Thanks!!
|
It is not necessary to use substitutions when what you're working with isn't going to be too long:
$$
\cos 2x +5 \cos x + 3 = 0\\
\implies 2\cos^2 x - 1 + 5\cos x + 3 = 0\\
\implies 2\cos^2 x + 5\cos x + 2 = 0\\
\implies \cos^2 x + \frac{5}{2}\cos x + 1 = 0\\
\implies \cos^2 x + \left(2 + \frac{1}{2}\right)\cos x + 2\cdot \frac{1}{2} = 0\\
\implies (\cos x + 2)(\cos x + \frac{1}{2}) = 0
$$
Now, $\cos x \neq -2$ since $\cos x \in \left[-1,1\right]$. So we can reject that.
$$\therefore \space\cos x = - \frac{1}{2} \\
\implies \cos x = -\sin{\frac{\pi}{6}} = \cos\left(\frac{\pi}{2} + \frac{\pi}{6}\right) = \cos \left(\frac{2\pi}{3}\right)\\
\implies x = 2n\pi \pm \frac{2\pi}{3} , \space n\in \mathbb Z\\
\implies x = 2\pi \left(n \pm \frac{2}{3}\right), \space n\in \mathbb Z $$
Trivial yet Important Facts used:
*
*$x^2 + \left(\alpha + \beta\right)x + \alpha\beta = 0 \iff (x+\alpha)(x+\beta) = 0$
*$\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$
*$\cos x = \cos \theta \implies x = 2n\pi \pm \theta$
|
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|
Compute $\sum_{n=0}^\infty \frac{(n+1)^2}{n!}$ This is what I've done so far:
$$\sum_{n=0}^\infty\frac{(n+1)^2}{n!} = 1 +4 +4.5 + \frac 83 + \frac{25}{24} +\frac{3}{10} + \frac {49}{720} +\dots$$
I know I need to manipulate $\frac{(n+1)^2}{n!}=\frac{(n+1)(n+1)}{n(n-1)(n-2)}=$
|
Hint: $(n+1)^2=n^2+2n+1=n(n-1)+3n+1$
|
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|
Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
|
$f(x)=-6\sin^4 x=-6(\sin x)^4$
By Chain Rule,
$f'(x)=4\cdot-6(\sin x)^3(\sin x)'=-24(\sin x)^3\cos x=-24\sin^3x\cos x$
|
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|
Evaluate $\frac{ 1 }{ 1010 \times 2016} + \frac{ 1 }{ 1012 \times 2014} + \frac{ 1 }{ 1014 \times 2012} + \cdots + \frac{ 1 }{ 2016 \times 1010} = ?$ $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = ?
$$
My attempt so far :
$$\sum\limits_{n=0}^{503}\dfrac{1}{(1010+2n)(2016-2n)} = \dfrac{1}{6052}\sum\limits_{n=0}^{503}\left(\dfrac{1}{n+505} - \dfrac{1}{n-1008}\right)$$
It won't telescope/simplify further. I feel I am in wrong road. Any help ?
|
First, note that each term is duplicated $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = \\\dfrac{2}{ 1010 \times 2016}+ \dfrac{ 2 }{ 1012 \times 2014} + \dfrac{ 2 }{ 1014 \times 2012} + \cdots + \dfrac{ 2 }{ 1512 \times 1514}=\\\sum_{i=0}^{251}\frac 2{1513^2-(2i+1)^2}$$ Alpha gets a fraction with huge numbers that is about $0.0002289$
|
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|
polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \equiv n^4+2n^2 \pmod3 $$$$n^4+2n^2 = n^2(n^2+2)=n^2(n^2-1)=n^2(n+1)(n-1)\equiv 0 \pmod3$$
Divisble by 8: $$n^4+ 6n^3 + 11n^2+6n \equiv n(n^3-2n^2-5n-2) \pmod8 $$$$n(n^3-2n^2-5n-2) = n(n+1)(n^2-3n-2) = n(n+1)(n^2+5n+6) = n(n+1)(n+2)(n+3) = ??? \pmod8$$
So it seems that the polynomial has to be divisible by four, but I'm not sure how to show that it has to be divisible by two one more time to show that it's divisible by 8.
Any other approaches to solving the problem are welcome, as I'm preparing for an exam and would appreciate having multiple ways to tackle the problem. I think this problem was meant to be an exercise in proof by induction, but the other approach seemed more approachable.
|
We have
$$
n^4+6n^3+11n^2+6n=n(n+1)(n+2)(n+3)=24\binom{n+3}4.
$$
This settles the claim when $n\ge1$. To get it for all $n$ observe that all integers are congruent to one $>1$ modulo $24$.
|
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|
Solve $ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$ Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$
I first tried to use Eulers formula
$$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik} \right)$$
However to use the geometric formula here, I must subtract the $k=0$ term and that term is undefinted since $1/k$. I also end up with something that diverges in my calculations and since $\sin 2k$ is limited the serie should not diverge.
|
The summation is as follows:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{1}{2i} \, \sum_{n=1}^{\infty} \frac{ e^{2ai n} - e^{- 2ai n}}{n} \\
&= - \frac{1}{2i} \left( \ln(1 - e^{2ai}) - \ln(1 - e^{-2ai}) \right) \\
&= - \frac{1}{2i} \ln\left( \frac{1 - e^{2ai}}{1 - e^{- 2ai}} \right) = - \frac{1}{2i} \ln\left( - \frac{e^{ai}}{e^{-ai}} \cdot \frac{\sin(a)}{\sin(a)} \right) \\
&= - \frac{1}{2i} \ln\left( - e^{2ai} \right) = - \frac{1}{2i} \left( \ln(e^{- \pi i}) + \ln(e^{2ai}) \right) \\
&= - \frac{1}{2i} \left( - \pi i + 2ai \right) = \frac{ \pi - 2a}{2}.
\end{align}
This provides
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{ \pi - 2a}{2}.
\end{align}
|
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|
when ${\rm gcd} (a,b)=1$, what is ${\rm gcd} (a+b , a^2+b^2)$? I want to prove above statement
"what is ${\rm gcd} (a+b , a^2+b^2)$ when ${\rm gcd}(a,b) = 1$"
I've seen some proofs of it, but i couldn't find useful one.
here is one of the proof of it.
some proofs
I want prove it using method like 2nd answer. (user 9413)
but i can't understand why "if $d | 2ab$ then $d = 1$ or $2$ or $d | a$ or $d|b$"
(isn't Euclid's lemma which only holds when $d$ is prime ?)
please show me proof step by step
|
Suppose $\gcd(a,b)=1$, that is, there are $x,y$ so that $ax+by=1$. Then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(ax^3+3bx^2y)+b^2(3axy^2+by^3)
\end{align}
$$
Therefore,
$$
\begin{align}
\small2
&\small=\overbrace{(\color{#00A000}{(a^2+b^2)}+\color{#C00000}{(a+b)}(a-b))}^{\large 2a^2}(ax^3+3bx^2y)+\overbrace{(\color{#00A000}{(a^2+b^2)}-\color{#C00000}{(a+b)}(a-b))}^{\large 2b^2}(3axy^2+by^3)\\[4pt]
&\small=\color{#C00000}{(a+b)}(a-b)(ax^3+3bx^2y-3axy^2-by^3)+\color{#00A000}{(a^2+b^2)}(ax^3+3bx^2y+3axy^2+by^3)
\end{align}
$$
This means that $\gcd(a+b,a^2+b^2)\mid2$. Thus,
$$
\gcd(a+b,a^2+b^2)=\gcd(a+b,2)
$$
|
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|
How to show that $a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2} $ I want show the following $$a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2},\ s,\ t\in{\mathbb Q} $$
How can we prove this ?
[Add] Someone implies that we must use pythagorean triple :
Let $$ a=\frac{n}{m},\ b= \frac{s}{k},\ (m,n)=(s,k)=1$$
Then $$ k^2n^2+s^2m^2=m^2k^2 \Rightarrow n^2|(k^2-s^2),\ k^2|m^2
$$ so that we have $$n^2+ s^2=k^2,\ (n,s)=1,\ k=m$$ We complete the proof by the following
Proof of pythagorean triple : $$a^2+
b^2=c^2,\ (a,b)=1$$ Then which form do $a,\ b,\ c$ have ? We have $$
\frac{c+a}{b}=\frac{A}{B}=\frac{b}{c-a},\ (A,B)=1$$
So $$ (c+a)B^2=bAB=(c-a)A^2 $$
So $$ c-a=B^2t,\ c+a=A^2t$$ That is $$ b=tAB,\
c=\frac{t}{2}(A^2+B^2),\ a= \frac{t}{2}(A^2-B^2)
$$
$(a,b)=1 \Rightarrow t=2$ or $1$ If $t=1$, then $AB$ is odd. So we can derive a contradiction.
|
Consider the line through $(-1,0)$ with slope $\frac{t}{s}$. Compute the other point on the circle that the line passes through.
So, points on the line are of the form $(p,q)=(-1+\alpha s,\alpha t)$ and the points on the circle have $p^2+q^2=1$, or $1-2\alpha s + \alpha^2(s^2+t^2)=1$ or $$\alpha\left(\alpha(s^2+t^2)-2s\right)=0$$
$\alpha=0$ gives $(-1,0)$, the original point, and $\alpha=\frac{2s}{s^2+t^2}$ gives the other point. $p=-1+\frac{2s^2}{s^2+t^2}=\frac{s^2-t^2}{s^2+t^2}$ and $q=\frac{2st}{s^2+t^2}$.
Now, if $p,q$ are rational, $p\neq -1$, the line through $(p,q)$ and $(-1,0)$ has rational slope, specifically, $\frac{q}{p+1}$. (You have to treat the case $p=-1$ sepearately.)
This technique works for any quadratic and the rational points. If you have one known one rational point, $(p_0,q_0)$ you can take any pair of integers $(s,t)$ and take the line from our base rational point in the direction $(s,t)$ and get another rational point. For example, the equaion:
$$p^2+q^2=2$$
has obvious root $(-1,-1)$ and $(p,q)=(-1+\alpha s,-1+\alpha s)$ gives $1-2\alpha(s+t) + \alpha^2(s^2+t^2)=1$ or:
$$\alpha\left(\alpha(s^2+t^2)-2(s+t)\right)=0$$
or $\alpha=\frac{2(s+t)}{s^2+t^2}$ for the non-zero answer, and then:
$$(p,q)=\left(\frac{s^2+2st-t^2}{s^2+t^2},\frac{t^2+2st-s^2}{s^2+t^2}\right)$$
|
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|
A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{12}$$ using power series expansion.
Thank you.
|
Letting $u = \log(1+x)$,
$$ \begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \int_{0}^{\log 2} \frac{u^{3}}{e^{u}-1} e^{u} \ du \\ &= \int_{0}^{\log 2} \frac{u^{3}}{1-e^{-u}} \ du \\ &= \int_{0}^{\log 2} u^{3} \sum_{n=0}^{\infty} e^{-nu} \ du \\ &= \sum_{n=0}^{\infty} \int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \int_{0}^{\log 2} u^{3} \ du + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \frac{\log^{4}(2)}{4} + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du .\end{align}$$
Then integrating by parts 3 times,
$$\begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \frac{\log^{4}(2)}{4} -\sum_{n=1}^{\infty} e^{-nu} \left(\frac{6}{n^{4}} + \frac{6u}{n^{3}} + \frac{3 u^{2}}{n^{2}} + \frac{u^{3}}{n} \right)\Bigg|^{\log 2}_{0} \\ &= \frac{\log^{4}(2)}{4} - \sum_{n=1}^{\infty} \left[\frac{1}{2^{n}} \left(\frac{6}{n^{4}} + \frac{6 \log 2}{n^{3}} + \frac{3 \log^{2} (2)}{n^{2}} + \frac{\log^{3}(2)}{n}\right) - 6 \zeta(4) \right] \\ &= -\frac{3\log^{4}(2)}{4 } - 6 \text{Li}_{4} \left(\frac{1}{2} \right) - 6 \log (2) \ \text{Li}_{3} \left(\frac{1}{2} \right)-3 \log^{2}(2) \text{Li}_{2} \left(\frac{1}{2} \right) + 6 \zeta(4) \\ &\approx 0.1425141979 . \end{align}$$
The answer could of course be simplified using the known values of $\text{Li}_{2} \left(\frac{1}{2} \right)$, $\text{Li}_{3} \left( \frac{1}{2}\right) $, and $\zeta(4)$.
|
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|
How to find transformation matrix which converts matrix to simple standard form I have a matrix A$$ \left( \begin{array}{ccc}
0 & 1 \\
a^2 & 0\\
\end{array} \right) $$
Using eigen values, I convert it into simple standard form B:
$$\left( \begin{array}{ccc}
a & 0 \\
0 & -a\\
\end{array} \right) $$
How can I find the transformation matrix M which converts this to simple standard form. In other words, how to find M such that $MAM^{-1}=B$. I am actually interested in knowing the nature of the transformation. Also in general how does one find it?
|
If $a \ne 0$, the eigenvalues of $A$ are $\pm a$; this is tacitly given in the problem, by stipulating the diagonalized form of $A$, $B$, is diagonal with diagonal entries $a$, $-a$, but it is also easy to see since the characteristic polynomial of $A$ is
$\det (A - \lambda I) = (-\lambda)^2 - a^2 = \lambda^2 - a^2; \tag{1}$
the roots are clearly $\lambda = \pm a$. Since they are distinct, the corresponding eigenectors are linearly independent; we can in fact almost without effort write them down; indeed, direct calculation reveals that
$A \begin{pmatrix} 1 \\ a \end{pmatrix} = \begin{pmatrix} a \\ a^2 \end{pmatrix} = a \begin{pmatrix} 1 \\ a \end{pmatrix} \tag{2}$
and
$A \begin{pmatrix} 1 \\ -a \end{pmatrix} = \begin{pmatrix} -a \\ a^2 \end{pmatrix} = -a \begin{pmatrix} 1 \\ -a \end{pmatrix}; \tag{3}$
one can also validate the linear independence of $v_a = (1, a)^T$ and $v_{-a} = (1, -a)^T$ directly; this is an easy exercise working from the definitions so I leave it to the reader. Suppose we set up the matrix
$E = \begin{bmatrix} 1 & 1 \\ a & -a \end{bmatrix} = \begin{bmatrix} v_a & v_{-a} \end{bmatrix}; \tag{4}$
i.e., the columns of $E$ are the eigenvectors of $A$. Then it is easy to see that
$AE = \begin{bmatrix} Av_a & Av_{-a} \end{bmatrix} = \begin{bmatrix} a v_a & -a v_{-a} \end{bmatrix}. \tag{5}$
We next observe that, the columns of $E$ being linearly independent, $E$ is nonsingular, so we have $E^{-1}$ with
$E^{-1} \begin{bmatrix} v_a & v_{-a} \end{bmatrix} = \begin{bmatrix} E^{-1} v_a & E^{-1} v_{-a} \end{bmatrix} = I, \tag{6}$
i.e.,
$E^{-1} v_a = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \tag{7}$
and
$E^{-1} v_{-a} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \tag{8}$
Thus
$E^{-1}AE = E^{-1} \begin{bmatrix} Av_a & Av_{-a} \end{bmatrix} = \begin{bmatrix} a E^{-1}(v_a) & -a E^{-1}(v_{-a}) \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}. \tag{9}$
We see the matrix $E$ of eigenvectors of $A$ diagonalizes $A$ via the prescription
$A \to E^{-1}AE$; if one really needs the answer to read $MAM^{-1}$, simply set $M = E^{-1}$ so that $M^{-1} = E$ and proceed from there.
This technique obviously generalizes to matrices of any size $n$, as long as there are $n$ linearly independent eigenvectors, which will always be the case if the eigenvalues are distinct. Finally, we note that the case $a = 0$ is tacitly excluded by the question itself, since $A$ then becomes nilpotent and can't be diagonalized.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
|
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|
Simplification of a series so that it converges to a given function I am trying to rearrange the series
$ \frac{1}{1-z} - \frac{(1-a)z}{(1-z)^2} + \frac{(1-a)^2z^2}{(1-z)^3} - \cdots$
In such a way that I can show it converges to
$\frac{1}{1-az} $
What I have so far
Let $ w = \frac{z}{1-z} $, we can then write the series as
$ \frac{w}{z} \left ( 1 - (1-a)w + (1-a)^2 w^2 - \cdots \right ) $
Which is the taylor series expansion about $0$ of
$ \frac{w}{z} \frac{1}{1+(a-1)w} $
Which I can simplify down to
$ \frac{1}{1+(a-2)z}$ as follows
$ \frac{w}{z} \frac{1}{1+(a-1)w} = \frac{1}{1-z} \left ( \frac{1}{1 + (a-1)\frac{z}{1-z}}\right) $
$= \frac{1}{1-z} \left ( {\frac{1-z + az - z}{1-z}}\right)^{-1} $
$ = \frac{1}{1+(a-2)z} $
Obviously this is not the same as $\frac{1}{1-az} $. I would love any guidance people can give as to where I went wrong.
|
The mistake lies in this step:
$$\frac{w}{z}\left(1-(1-a)w+(1-a)^2w^2-\cdots\right)=\frac{w}{z}\cdot\frac{1}{1+(a-1)w}$$
Recall that $1+x+x^2+\cdots=\frac{1}{1-x}$. If we let $x=-(1-a)w$, then your series actually becomes:
$$\frac{w}{z}\left(1+x+x^2+\cdots\right)=\frac{w}{z}\cdot\frac{1}{1-x}=\frac{w}{z}\cdot\frac{1}{1+(1-a)w}$$
which is slightly different to what you wrote. If you continue from here, you will reach the desired answer:
\begin{align*}
\frac{w}{z}\cdot\frac{1}{1+(1-a)w} &= \frac{1}{1-z}\cdot\frac{1}{1+(1-a)\frac{z}{1-z}} \\
&= \frac{1}{(1-z)+(1-a)z} \\
&= \frac{1}{1-az}.
\end{align*}
|
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|
Determine whether $A215928(n)=G_n$. Let $G_0=1$ and $G_{n+1}=F_0G_n+F_1G_{n-1}+\cdots+F_nG_0$, where $F_n$ is the $n$th term of the Fibonacci sequence, i.e., $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$.
Let $P_0=P_1=1,\ P_2=2,$ and $P_{n+1}=2P_n+P_{n-1}$ for $n>1$.
Is $P_n=G_n$?
Edit: set $P_2=2$, thanks to Daniel R.
|
Let $G(z) = \sum\limits_{n=0}^\infty G_n z^n$
and $F(z) = \sum\limits_{n=0}^\infty F_n z^n$
where $F_n = \text{Fib}_{n+1}$ is the $(n+1)^{th}$ Fibonacci number.
In following set of recurrence relations,
$$G_{n+1} = F_0 G_n + F_1 G_{n-1} + \cdots + F_n G_0,$$
if one multiply the $n^{th}$ term by $z^n$ and sum over $n$, we get
$$\frac{G(z)-1}{z} = F(z)G(z)\quad\implies\quad
G(z) = \frac{1}{1 - zF(z)}$$
It is well known $\displaystyle\;F(z) = \frac{1}{1-z-z^2}$, this leads to
$$G(z) = \frac{1}{1 - \frac{z}{1-z-z^2}} = \frac{1-z-z^2}{1-2z-z^2} =
1 + \frac{z}{1-2z-z^2}$$
This implies
$$\begin{align}
& (1 - 2z - z^2)( G(z) - 1 ) = z\\
\iff & (1 -2z - z^2)(G_1 z + G_2 z^2 + G_3 z^3 + G_4 z^4 + \cdots ) = z\\
\iff & G_1 z + ( G_2 - 2 G_1 ) z^2 + (G_3 - 2 G_2 - G_1 ) z^3 + (G_4 - 2G_3 - G_2)z^4 + \cdots = z\\
\implies &
G_{n+2} = 2G_{n+1} + G_{n}\quad\text{ for } n \ge 1
\end{align}
$$
This matches the the recurrence relation in OEIS A215928.
By brute force, one find
$$G_0, G_1, G_2, G_3, \ldots = 1, 1, 2, 5, \ldots$$
Since $G_n$ matches $A215928(n)$ for $n = 1, 2$, above recurrence relation guarantee
$G_n = A215928(n)$ for all $n \ge 1$. It turns out $G_0 = A215928(0)$ too, but that
is more of a coincidence.
|
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|
If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$. If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.
So far, I let
$d=(a^2+b^2,a+b)$
$\implies d|(a^2+b^2-(a+b)^2)$
$\implies d|(a^2+b^2-(a^2+2ab+b^2))$
$\implies d|(-2ab)$
I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Alternatively, is there some other way of showing it?
EDIT(10:41PM):
Just noticed that
$(a^2+b^2\pm (a+b)(a-b),a+b)$
results in
$d|2a^2$ and $d|2b^2$.
$\implies d|(2a^2,2b^2)$.
$\implies d|2(a^2,b^2)$.
Now I just have to fully understand why
$(a,b)=1\implies (a^2,b^2)=1$.
I sort of intuitively understand this.
|
Lemma 1: If $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.
Proof: We will prove the contrapositive. Suppose $\gcd(a^2,b^2)>1$. Then $a^2$ and $b^2$ must have a prime common divisor, say $p$. Since $p|a^2$, Euclid's lemma implies that $p|a$. Similarly, $p|b$. Thus $p$ is a common divisor of $a,b$ and so $\gcd(a,b)>1$.
Lemma 2: If $\gcd(u,v)=1$, then $\gcd(u+v,u-v) \leq 2$.
Proof: Let $d$ be a common divisor of $u+v$ and $u-v$. Then $d|(u+v+u-v)$ and so $d|2u$. Moreover, $d|(u+v-(u-v))$ and so $d|2v$. Thus $d|\gcd(2u,2v)$. But $\gcd(2u,2v)=2\gcd(u,v)=2$. It follows that $\gcd(u+v,u-v) \leq 2$.
Given these two lemmas, we'll prove the problem statement. Since $\gcd(a,b)=1$, $\gcd(a^2,b^2)=1$ by Lemma 1. Now, by setting $u=a^2$, $v=b^2$ in Lemma 2, we can see that $\gcd(a^2+b^2,a^2-b^2) \leq 2$.
Finally, suppose $d$ is a common divisor of $a^2+b^2$ and $a+b$. Because $d|(a+b)$, $d|(a+b)(a-b)=a^2-b^2$. So $d$ is a common divisor of $a^2+b^2$ and $a^2-b^2$. It follows that $d \leq 2$, completing the proof.
|
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|
Solving the differential equation: $f(x)yy'=(y')^2-0.5$ I am trying to solve this equation:
$f(x)yy'=(y')^2-0.5$
I have already tried traditional methods...
Any ideas?
|
We have
$$
\left(\frac{y'}{y}\right)^2-\frac{y'}{y}=\frac{1}{2y^2},
$$
or
$$
\left(\frac{y'}{y}-\frac{1}{2}\right)^2=\frac{1}{4}+\frac{1}{2y^2},
$$
or
$$
\frac{y'}{y}=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{1}{2y^2}}
$$
or
$$
y'=\frac{y}{2}\pm\sqrt{\frac{y^2}{4}+\frac{1}{2}}=\frac{1}{2}\big(y\pm\sqrt{y^2+2}\big).
$$
Using now Separation of Variables the problem reduces to finding the integral
$$
\int\frac{dy}{y\pm\sqrt{y^2+2}}.
$$
|
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|
Can someone walk me through how this expression simplifies to y/x? I am just wondering how this equation comes to be: it is from an economics problem involving marginal utilities. I have my two variables, $x$ and $y$.
Intuitively, how does $$\frac{0.5\times x^{-0.5}\times y^{0.5}}{0.5\times x^{0.5}\times y^{-0.5}}= \frac{y}{x}?$$
I'm guessing that the $0.5$ and $(1/2)$ both cancel out, but I am not sure what happens next.
Thank you!
|
Let me show you. It's just the "Multiply by one" trick:
$\frac{0.5 x^{-\frac{1}{2}} y^\frac{1}{2}}{0.5 x^\frac{1}{2} y^{-\frac{1}{2}}} \cdot 1 \cdot 1 = \frac{0.5 x^{-\frac{1}{2}} y^\frac{1}{2}}{0.5 x^\frac{1}{2} y^{-\frac{1}{2}}} \cdot \frac{y^\frac{1}{2}}{y^\frac{1}{2}} \cdot \frac{x^\frac{1}{2}}{x^\frac{1}{2}}$
You know, of course, that $a^p a^q = a^{p+q}$:
$\frac{0.5 x^{-\frac{1}{2}+\frac{1}{2}} y^{\frac{1}{2}+\frac{1}{2}}}{0.5 x^{\frac{1}{2}+\frac{1}{2}} y^{-\frac{1}{2}+\frac{1}{2}}}=\frac{0.5x^0 y^1}{0.5x^1 y^0} = \frac{y}{x}$
And there you have it :)
|
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|
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
|
$$\sqrt{a^2+4}\sim a+\frac{1}{a}$$
as $a\to\infty$ if
$$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=1$$ But $$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=\lim_{a\to\infty}\frac{a\sqrt{a^2+4}}{a^2+1}=\lim_{a\to\infty}\frac{a^2\sqrt{1+4/a^2}}{a^2(1+1/a^2)}=1$$
|
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Show that $\, 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 $ How can I prove that, for $a,b \in \mathbb{Z}$ we have $$ 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 \, ? $$ Here, $\left \lfloor\,\right \rfloor$ is the floor function. I tried the following: say that $\frac{2a}{b} = x$, and $ \left \lfloor{\frac{2a}{b}}\right \rfloor = m$, with $0 \leq x - m \leq 1$. I tried the same for $ 2 \left \lfloor{\frac{2a}{b}}\right \rfloor $, and then combining the two inequalities. It did not seem to help, though.
|
$0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor \leq \frac{2*a}{b} < 2*(\left \lfloor{\frac{a}{b}}\right \rfloor + 1)$ = $2*\left \lfloor{\frac{a}{b}}\right \rfloor$ + 2
$\left \lfloor{\frac{2a}{b}}\right \rfloor$ is an integer, so being < than an other integer means being $\leq$ than this integer -1
=> $\left \lfloor{\frac{2a}{b}}\right \rfloor$ $\leq $ $2*\left \lfloor{\frac{a}{b}}\right \rfloor$ + 1
Edit:
As for the other part of the inequality:
$$\frac{2a}{b}= 2\left \lfloor{\frac{a}{b}}\right \rfloor + 2(\frac{a}{b}) $$ with $(x)$ being the fractional part of $x$.
All that's left to consider is whether $2(\frac{a}{b})$ is smaller or greater than $1$ to conclude.
|
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Proof of ${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4} = -6$ Observe:
\begin{matrix} F(n)|&{F(n)}^{4}& - {4F(n+1)}^{4}& - {19F(n+2)}^{4}&- {4F(n+3)}^{4}&{F(n+4)}^{4}& = -6\\ 1|& 1& -4& -304& -324& 625&=-6\\ 1|& 1& -64& -1539& -2500& 4096&=-6\\ 2|& 16& -324& -11875& -16384& 28561&=-6\\ 3|& 81& -2500& -77824& -114244& 194481&=-6\\ 5|& 625& -16384& -542659& -777924& 1336336&=-6\\ 8|& 4096& -114244& -3695139& -5345344& 9150625&=-6 \end{matrix}
I see that the proof is true but I cant quite grasp the pattern. I'm more interested in hints rather than a solution. Ive read through this website about Fibonomials a couple times and have a decent understanding but I cant figure how to apply it in this situation. It may not even be necessary. I've tried proving through induction but ended up with almost exactly the same problem. I think there may be a proof through some sort of recursive sequence but I don't know enough to prove something like that.
I'm more interested in hints rather than a solution.
|
This is not the most elegant solution, but its a very straight forward computation using the standard tools we use to solve recurence-relations.
Let $a_n = F_n^4$ (where $F_n$ represents the terms in your recurence relation which we don't yet know is the Fibonacci sequence) then $$a_{n+4} - 4a_{n+3} - 19a_{n+2} - 4a_{n+1} + a_n = -6$$
The characteristic polynomial is
$$x^4 - 4x^3 - 19x^2 - 4x + 1 = (x^2+3x+1)(x^2-7x+1)$$
The particular solution is $a_n = C = \frac{6}{25}$ so the full solution reads
$$a_n = \frac{6}{25} + a r_+^n + b r_-^n + c s_+^n + d s_-^n$$
where
$$r_\pm = \frac{-3 \pm \sqrt{5}}{2} = \left(\frac{1\pm \sqrt{5}}{2}\right)^2$$
$$s_\pm = \frac{7 \pm 3\sqrt{5}}{2} = r_{\pm}^2 = \left(\frac{1\pm \sqrt{5}}{2}\right)^4$$
are the roots of the characteristic polynomial. Now note that the Fibonacci numbers statisfy
$$F_n = e\left(\frac{1+\sqrt{5}}{2}\right)^n + f\left(\frac{1-\sqrt{5}}{2}\right)^n$$
Fixing the constants $a,b,c,d,e,f$ gives us $a_n = F_n^4$ where $F_n$ here is the Fibonacci sequence.
${\bf Added}:$
Defining $q_\pm = \frac{1\pm\sqrt{5}}{2}$ then $$F_n = \frac{1}{\sqrt{5}}(q_+^n - q_-^n)$$ so
$$F_n^4 = \frac{1}{25}\left((q_+^4)^n + (q^4_-)^n + 4(q_+^3 q_-)^n + 6 (q^2_- q^2_+)^n + 4(q_+q_-^3)^n\right)$$
Now we use $q_+q_- = -1$ to find
$$F_n^4 = \frac{1}{25}\left((q_+^4)^n + (q^4_-)^n - 4(q_+^2)^n - 4(q_-^2)^n + 6\right)$$
or using the notation $s_\pm,r_\pm$ defined above we have
$$F_n^4 = \frac{1}{25}\left(s^n_+ + s^{n}_{-} -4r^{n}_{+} - 4r^{n}_{-} + 6\right)$$
which gives $a=b=-\frac{4}{25}$ and $c=d=\frac{1}{25}$.
|
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|
Help in finding $\lim_{n\to\infty}\Bigl( \sum_{k=1}^{n} \frac{1}{({n \atop k})} \Bigr)^n$. I am not able to get a solution for this problem . Of finding the limit
$$\lim_{n\to\infty} \left( \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } \right)^n$$
I have tried using Mathematica and that numerically evaluates it to $7.3890560989 \cdots$
Which motivates me to think it is $e^2$ . Thanks for help
|
$$\left(\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right )^n=e^{n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}}.$$
$$\binom{n}{1}=\binom{n}{n-1}=n,\binom{n}{2}=\binom{n}{n-2}=\frac{n(n-1)}{2},\binom{n}{2}\leq\binom{n}{k}, \ k=2,\cdots,n-2,$$
$$\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}=\frac{2}{n}+\sum_{k=2}^{n-2}\frac{1}{\binom{n}{k}}<\frac{2}{n}+\frac{n-3}{\binom{n}{2}}\
\to0\ (n\to\infty).$$
$$n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}=n\log\left(1+\sum\limits_{k=1}^{n-1}\frac{1}{\binom{n}{k}}\right)
\sim n\cdot\sum\limits_{k=1}^{n-1} \frac{1}{\binom{n}{k}}.$$
So
$$\frac{n}{\binom{n}{2}}=\frac{n}{\binom{n}{n-2}}\to0,\
n\cdot\sum\limits_{k=3}^{n-3} \frac{1}{\binom{n}{k}}\leq\frac{n(n-5)}{\binom{n}{3}}\to 0\ (n\to\infty),$$
$$n\cdot\sum\limits_{k=1}^{n-1} \frac{1}{\binom{n}{k}}\to 2\ (n\to\infty).$$
$$\left(\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right )^n=e^{n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}}\to e^2\ (n\to\infty).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
|
$$\begin{align}
&1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)\\
&=n\cdot 1+(n-2)\cdot 2+(n-3)\cdot 3+\cdots +1\cdot n\\
&=\sum_{r=1}^n(n+1-r)r\\
&=\sum_{r=1}^n {n+1-r\choose 1}{r\choose 1}\\
&={n+2\choose 1+2}\\
&={n+2\choose 3}\\
&=\frac16 n(n+1)(n+2)
\end{align}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to find the derivative of $f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$? Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:
$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$
would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$?
|
Something to notice is that $$x^4 - 6x^2 + 9 = (x^2 - 3)^2$$ and consequently, $$f(x) = 2 (x^3 - 4x + 6) \log (x^2 - 3).$$ This makes differentiation slightly easier: $$f'(x) = 2\left( (3x^2 - 4)\log(x^2 - 3) + \frac{x^3-4x+6}{x^2-3} \cdot 2x \right).$$
|
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"language": "en",
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|
Finding maximum points by constrain optimization (multivariable calculus) Find the maximum value of the function $f(x,y)=x^2+y^2+2x+y$, on the closed disc (the circle together with the region inside the circle) of radius 2, centred at the origin.
What i tried
I know that i have to maximize the function
$f(x,y)=x^2+y^2+2x+y$
with a constrain of
$x^2+y^2<2$
which then give me a new function of
$L(x,y,\lambda)=x^2+y^2+2x+y+\lambda(x^2+y^2-2)=0$ which i have to maximize.
Then finding critical points and equating to 0, it leads me to the system of equations
$$ 2x +2+ (2\lambda)x = 0 $$
$$ 2y +1+ (2\lambda)y = 0 $$
$$x^2+y^2=2$$
I got stuck at solving these equations and im unsure that after geeting the crticial points. Do have to substitute the critical points back to the original equation $f(x,y)$ or the equation $L(x,y,\lambda)$ to get the maximum value. Could anyone help. Thanks
|
You can rewrite the equation you have as follows $$\begin{cases}2x(1+λ)=-2\\2y(1+λ)=-1\\y^2=2-x^2\end{cases}$$ or equivalently (in order to avoid calculations with roots) as $$\begin{cases}4x^2(1+λ)^2=4\\4y^2(1+λ)^2=1\\y^2=2-x^2\end{cases}$$ Now subsituting the 3rd equation into the 2nd yields $$1=4(2-x^2)(1+λ)^2=8(1+λ)^2-4x^2(1+λ)^2=8(1+λ)^2-4$$ where the last equality is due to the first equation. Hence $$8(1+λ)^2=5 \quad \implies \quad λ=-1\pm\sqrt{\dfrac{5}{8}}$$ which leads you with straightforward calculations to the solution.
|
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|
Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$ Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
I am at a loss at how to start this problem and which contour to pick. I have been trying to get the sector with angle $2\pi/3$ to work with a bump around the pole at $e^{i2\pi/3}$ and the origin, but I am getting 5 or 6 different integrals and it is not really getting me anywhere.
|
This isn't quite complex analytic, but first denote your as $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$
Consider the double integral:
$$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{(x^2+y^3)(1+x^2)}dydx.$$
We intend to evaluate $J$ and relate $J$ to $I.$
To evaluate, $J$ we integrate with respect to $y.$ You can proceed in two ways. The long way is to integrate by partial fractions, but the short way is to let $y=ux^\frac{2}{3}$ so that $dy=x^\frac{2}{3} du.$ Then we get:
$$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x^\frac{5}{3}}{(x^2+x^2u^3)(1+x^2)}dudx=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x^\frac{-1}{3}}{(1+u^3)(1+x^2)}dudx.$$ Using the nice formula $$\int_{0}^{\infty} \frac{t^m}{1+t^n} dt=\frac{\pi}{n}\csc\left(\frac{\pi(m+1)}{n}\right),$$
we get $$J=\frac{2\pi^2}{9}.$$
Now use Fubini's Theorem on $J$ as such:
$$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{(x^2+y^3)(1+x^2)}dxdy.$$
We will need partial fractions to integrate with respect to $x.$ Omitting the details of this computation, $$\frac{x}{(x^2+y^3)(1+x^2)}=\frac{1}{y^3-1} \left(\frac{x}{x^2+1}-\frac{x}{x^2+y^3}\right).$$ Now integrating with respect to $x$, and plugging in the endpoints, we get
$$J=\int_{0}^{\infty} \frac{1}{y^3-1}\lim_{x\rightarrow \infty} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)-\frac{1}{y^3-1}\lim_{x\rightarrow 0} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)dy.$$
We get $$J=\int_{0}^{\infty} \frac{\ln(y^3)}{2(y^3-1)}dy=\int_{0}^{\infty} \frac{3\ln(y)}{2(y^3-1)}dy=\frac{3}{2}I.$$
Thus we have $$I=\frac{2}{3}J=\frac{2}{3}\left(\frac{2\pi^2}{9}\right)=\frac{4\pi^2}{27}.$$
Addendum
Consider $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$
Split the integral into two parts:
$$\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx = \int_{0}^{1} \frac{\ln(x)}{x^3-1} dx+ \int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$ We can convert $\frac{\ln(x)}{x^3-1}$ into a geometric series as such (assuming $0<x<1.$) $$ \frac{\ln(x)}{x^3-1}=-\sum_{n=0}^{\infty}\ln(x)x^{3n}.$$ If we do integration by parts, $$\int_{0}^{1} \frac{\ln(x)}{x^3-1} dx= \int_{0}^{1} -\sum_{n=0}^{\infty}\ln(x)x^{3n}dx=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^2}.$$ For,$$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx,$$ perform a $u$ substitution $x=\frac{1}{u}, dx=\frac{-1}{u^2} du,$ and convert transformed integrand into a geometric series. You will see:
$$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^2}=\sum_{n=-\infty}^{-1} \frac{1}{(3n+1)^2},$$ and the latter equality can be seen by simply writing out the terms of each summation. Now, add the two computed series together and see:
$$\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^2}= \frac{4 \pi^2}{27.}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
|
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 3\sqrt[3]{abc}\times 3\sqrt[3]{\frac{1}{abc}}=9.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/991885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Asymptotic expansion for Fresnel Integrals If you take the fresnel integrals to be $$S(x) = \int_{0}^{x}\sin \left(\frac { \pi \cdot t^2}{2} \right) dt$$ How do you find the asymptotic expansion? I know it begins with a $1/2$ but how?
|
I suppose you want an asymptotic expansion as $x\to \infty$. We start with
$$S(x) = \int_0^x \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{2} - \int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt.$$
Now, to get a handle on that integral, we substitute $u = \frac{\pi t^2}{2}$ and obtain
$$\int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{\sqrt{2\pi}} \int_{\pi x^2/2}^\infty \frac{\sin u}{\sqrt{u}}\,du.$$
Then we want an asymptotic expansion of
$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du$$
which we get via integration by parts:
\begin{align}
\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du
&= \left[-\frac{\cos u}{\sqrt{u}}\right]_y^\infty - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\
&= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\
&= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\left(\left[\frac{\sin u}{u^{3/2}}\right]_y^\infty + \frac{3}{2}\int_y^\infty \frac{\sin u}{u^{5/2}}\,du\right)\\
&= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3}{4}\int_y^\infty \frac{\sin u}{u^{5/2}}\\
&= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3\cos y}{4y^{5/2}} - \frac{15}{8} \int_y^\infty \frac{\cos u}{u^{7/2}}\,du.
\end{align}
An elementary estimate shows that the last integral is $O(y^{-5/2})$ [actually, it is $O(y^{-7/2})$, as one sees when thinking about what a further integration by parts yields], so we get the asymptotic expansion
$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du = \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} + O(y^{-5/2}),$$
and hence, inserting $y = \frac{\pi x^2}{2}$,
$$S(x) = \frac{1}{2} - \frac{\cos \frac{\pi x^2}{2}}{\pi x} - \frac{\sin \frac{\pi x^2}{2}}{\pi^2 x^3} + O(x^{-5}).$$
To get higher-order asymptotic expansions, integrate by parts more often.
|
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|
How find this value$\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^22^{3n+1}}$ show that:
$$\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\left(\frac{1}{2}\right)^{1/2}?$$
this sum is from other problem,if I solve this,then the other problem is solve it
|
Hint: what is $\frac{f^{(n)}(0)}{n!}$ when $f(x) = \sqrt{1-x}$?
Compute the first terms:
\begin{align}
f^{(0)}(x) &= \sqrt{1-x}\\
f^{(1)}(x) &= -\frac 12 (1-x)^{-1/2} \\
f^{(2)}(x) &= -\frac 14 (1-x)^{-3/2} \\
\end{align}
The exponent of $1-x$ in $f^{(n)}(x)$ is obviously $1/2 - n$. So when you derivate,
the front factor gets multiplied by $1/2 - n$ to get $f^{(n+1)}(x)$, and by $-1 $ because of the chain rule.
Hence the result is
\begin{align}f^{(n+1)}(0)
&= 1\times \left(\frac 12 - 0\right) \times \left(\frac 12 - 1\right)
\times \dots \times \left(\frac 12 - n\right) \times (-1)^{n+1} \\
&= \frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}} (-1)^n \times (-1)^{n+1}
= -\frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}}\\
&= -\frac{(2n)!}{2^{n+1} 2\times 4 \times \dots \times 2n} = -\frac{(2n)!}{2^{2n+1} n!} \end{align}
You can check that this is true for the first values:
\begin{align}
f^{(1)}(x) &= -\frac 12 = 1\times \frac 12 &\times (-1)^1 \\
f^{(2)}(x) &= -\frac 14 = 1\times \frac 12 \times \left(-\frac 12\right)
&\times (-1)^2
\end{align}
Then, you must show that the series is convergent for $x=\frac 12$ and you will get:
\begin{align}
\sqrt{1-x}
&= 1 + \sum_{n=0}^\infty \frac{f^{(n+1)}(0) }{(n+1)!}x^{n+1} \\
&= 1 - \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2 (n+1)} x^{n+1}
\end{align}
Unfortunately, when evaluated in $\frac 12$, this does not make appear the desired sum, because of the factor $n+1$.
We can get rid of it just using tern by term derivation:
\begin{align}
-\frac 12 (1-x)^{-1/2}
&=- \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2} x^n\\
(1-x)^{-1/2}
&= \sum_{n=0}^\infty \frac{(2n)!}{2^{2n} (n!)^2} x^n
\end{align}
And now we get, with $x=1/2$:
$$
\left(1-\frac 12\right)^{-1/2}
= \sum_{n=0}^\infty \frac{(2n)!}{2^{3n} (n!)^2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the Sum of the Series: $1/(x+1) + 2/(x^2 + 1) + 4/(x^4 +1) +\cdots$ $n$ terms Find the sum of $n$ terms the following series:
$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$
$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$
But after that I'm not sure as to how to proceed
|
Just another approach.
Let $x$ be a real number such that $|x|>1$. Observe that
$$
1+x^{2^{n}}=\frac{x^{2^{n+1}}-1}{x^{2^{n}}-1} \quad n=0,1,2,\ldots,
$$ giving
$$
\prod_{n=0}^{N} \left(1+x^{2^{n}}\right)=\frac{x^{2^{N+1}}-1}{x-1}
$$
Applying the logarithmic function gives
$$
\sum_{n=0}^{N}\log \left(1+x^{2^{n}}\right)=\log\left(x^{2^{N+1}}-1\right)-\log (x-1)
$$
differentiating with respect to $x$
$$
\sum_{n=0}^{N}\frac{2^{n}x^{2^{n}-1}}{x^{2^{n}}+1}=\frac{2^{N+1}x^{2^{N+1}-1}}{x^{2^{N+1}}-1}-\frac{1}{x-1}
$$ equivalently
$$
\sum_{n=0}^{N}\frac{2^{n}}{x^{2^{n}}+1}=\frac{1}{x-1}-\frac{2^{N+1}}{x^{2^{N+1}}-1}
$$
and then making $N$ tend to $+\infty$ leads to a closed form for your series.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving the determinant of a tridiagonal matrix with $-1, 2, -1$ on diagonal. Let $A_n$ denote an $n \times n$ tridiagonal matrix.
$$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$
Set $D_n = \det(A_n)$
Prove that $D_n = 2D_{n-1} - D_{n-2}$ for $n \ge 4$.
|
Expand with respect to the first column: you get
$$
D_n
=2 \times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
- (-1) \times
\begin{vmatrix}-1 & 0 & & & 0 \\
-1 & 2 & -1 & & 0 \\
& \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
$$
Now expand according to the first line the second determinant: you get
$$
=2 \times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
- (-1)^2\times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-2)}
$$
Conclusion:
$$
D_n = 2D_{n-1} - D_{n-2}
$$
|
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|
find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers I am trying to find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers:
a) $2$
b) $7$
c) $12$
d) $25$
and I am stuck on how to do this. I know that you have to set $n \geq 1$ for a) and state $n!$ is divisible by $1 = 1$ but I do not know where these numbers come from.
|
$$1!+2!+\dots 100!\\ \equiv 1! +2!+3\times 2!+ 4\times 3\times 2!...\\ \equiv 1!+0+3\times 0+ 4\times 3\times 0...\equiv 1!\pmod 2$$
Because $2!$ is congruent to $0$ $\pmod 2$. As André Nicolas pointed out in the comments, $4!\equiv 0\pmod {12}$ which means that $5!$ which is $5\times 4!$ is congruent to $0$ as well. See if you can generalize this strategy to other moduli as well.
EDIT: Lets do the $\pmod {12}$ example.
$$1!+2!+\dots 100!\\ \equiv 1! +\ldots+4!+5\times 4! + 6\times 5\times 4!+7\times 6\times 5\times 4!+...\\ \equiv 1! +\ldots+24+5\times 24 + 6\times 5\times 24+7\times 6\times 5\times 24+... \pmod {12}$$ Because $4!=24$. $24=12\times 2$ which is congruent to $0\times 2=0$ $\pmod{12}$. So... $$1! +\ldots+24+5\times 24 + 6\times 5\times 24+7\times 6\times 5\times 24+..\\ \equiv 1!+2!+3!+0+5\times 0+6\times 5\times 0\ldots\\ \equiv 1!+2!+3! \pmod{12}$$ and $1!=1,2!=2,3!=6$ so $1!+2!+3!=1+2+6=9$
|
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|
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$? How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$
for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$.
I do not know where to start. I need some idea and advice on this problem.Thanks
|
the given inequality is equivalent to
$abc+ab+ac+bc\geq 4$ (I)from the given condition we get
$abc(a+b+c)=ab+ac+bc$ (1)
this gives in our inequality (I)
$abc+abc(a+b+c)\geq 4$
since $a+b+c\geq 3$ we have in the case $abc\geq 1$:
$abc+abc(a+b+c)\geq 1+3\cdot 1=4$
now our second case:
$abc\le 1$
then we can set $a=\frac{1}{a'},b=\frac{1}{b'},c=\frac{1}{c'}$
and our condition $a'+b'+c'=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}$ is fulfilled and we can make the same like in part I.
|
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|
Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
|
Let $u(x)=\log(\sin x + 2\cos x)$, then $(\sin x + 2\cos x)'=\cos x-2\sin x$ hence $$u'(x)=\frac{\cos x-2\sin x}{\sin x + 2\cos x}.$$ More generally, for every $(a,b)$, $$(au(x)+bx)'=\frac{(a+2b)\cos x+(b-2a)\sin x}{\sin x + 2\cos x}$$ Solving for $(a+2b,b-2a)=(1,2)$ yields $(a,b)=(-\frac35,\frac45)$ hence $$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x}\,\mathrm dx=-\frac35u(x)+\frac45x+C=-\frac35\log(\sin x + 2\cos x)+\frac45x+C. $$
|
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|
Possible values of infinitely nested square root $n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$ If $$n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$$
Is it possible that $n$ is a integer for any $x=Z( \text{zahlen number})$.If yes .What is the value of $x$??
|
So, $n^2=x+n\iff n^2-x-n=0\implies n=\dfrac{1\pm\sqrt{1+4x}}2$
As $n\ge0,n=\dfrac{1+\sqrt{1+4x}}2$ So, we need $1+4x$ to be Perfect Square
As $1+4x$ is odd, $1+4x=(2m+1)^2\iff x=m^2+m$ where $m$ is any integer
|
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|
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
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Do it in order
$$\frac{\frac{d}{dx}(1+4\cos x)2\sqrt{x+4\sin x} -\frac{d}{dx}(2\sqrt{x+4\sin x})(1+4\cos x)}{4(x+4\sin x)}.$$
Solve the derivatives one by one and then replace in the big fraction. Remember to use the chain rule!
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
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Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$
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Put $y=\tan^{-1}x$. Then $\tan y=x$. So
$$1 = \frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( {\tan y} \right) = \frac{d}
{{dy}}\left( {\tan y} \right).\frac{d}{{dx}}\left( y \right) = \frac{1}{{{{\cos }^2}y}}.\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right),$$
which gives
$$\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = {\cos ^2}y = {\cos ^2}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + \tan^2 \left( {{{\tan }^{ - 1}}x} \right)}} = \frac{1}{{1 + {x^2}}}.$$
Notice: In above we used the formula ${\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }}$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1009871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
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A geometry problem involving geometric mean $ABCD$ is a quadrilateral inscribed in a circle of center $O$. Let $BD$ bisect $OC$ perpendicularly. $P$ is a point on the diagonal $AC$ such that $PC=OC$. $BP$ cuts $AD$ at $E$ and the circle circumscribing $ABCD$ at $F$.
Prove that $PF$ is the geometric mean of $EF$ and $BF$.
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This is a coordinate-based solution. Without loss of generality, you can choose your coordinate system such that the circle $\bigcirc ABCD$ becomes the unit circle and $C$ has coordinates $(1,0)$. Then, in order to simplify the formulation, we start by choosing rational coordinates for point $P$:
$$P = (2t^2,2t)/(t^2+1)$$
The coordinates are chosen in such a way that they describe any point on the circle $(x-1)^2+y^2=1$ except for the point $(2,0)$ (which corresponds to $t\to\infty$). For any point $P$ on that circle you can compute $t=y/(2-x)$.
From these coordinates, everything else can be computed:
\begin{align*}
O&=(0,0) \\
C&=(1,0) \\
\bigcirc ABCD&:\;\;x^2+y^2=1 \\
\bigcirc OBD&:\;\;(x-1)^2+y^2=1 \\
B&=(1,-\sqrt3)/2 \\
D&=(1,\sqrt3)/2 \\
P&=(2t^2,2t)/(t^2+1) \\
CP&:\;\;2t\,x+(1-t^2)\,y=2t \\
BP&:\;\;(t+\sqrt3)\,x+(1-\sqrt3t)\,y=2t \\
A&=(-t^4 + 6t^2 - 1, -4t^3 + 4t)/(t^4 + 2t^2 + 1) \\
AD&:\;\;(1-t^2-2\sqrt3t)\,x+(\sqrt3t^2 - 2t - \sqrt3)y = t^2 - 2\sqrt3t - 1 \\
E&=(-3\sqrt3t^3 + 11t^2 + \sqrt3t - 1, -3t^3 - 3\sqrt3t^2 + 9t + \sqrt3)/(4t^2 + 4) \\
F&=(t^2 + 2\sqrt3t - 1, -\sqrt3t^2 + 2t + \sqrt3)/(2t^2 + 2) \\
\lVert P-F\rVert^2&=(3t^2 - 2\sqrt3t + 1)/(t^2 + 1) \\
\lVert E-F\rVert^2&=(9t^4 - 12\sqrt3t^3 + 18t^2 - 4\sqrt3t + 1)/(4t^2 + 4) \\
\lVert B-F\rVert^2&=4/(t^2 + 1) \\
\lVert P-F\rVert^4 &= \lVert E-F\rVert^2\cdot\lVert B-F\rVert^2
= (\sqrt3t - 1)^4/(t^2+1)^2
\end{align*}
I did all the computation using projective geometry and homogeneous coordinates. So the things I write as $(x,y)/z$ were internally just homogeneous coordinate vectors $(x,y,z)$ in my computation. And of course I used a computer algebra system (sage in this case) to handle all these polynomials.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1010570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.