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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$
Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274....)
this series numerically checked without any problem, but I need the proving. Any help?
|
$$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$
so
$$\log(x)=-\log\frac{1}{x}=-\log(1-\frac{x-1}{x})=\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n$$
And
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
so
$$x=\frac{1}{1-\frac{x-1}{x}}=\sum_{n=0}^\infty (\frac{x-1}{x})^n$$
and thus
$$x\log(x)=(\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n)\cdot \sum_{m=0}^\infty (\frac{x-1}{x})^m = \sum_{n=1}^\infty a_n(\frac{x-1}{x})^n$$
where
$$a_n=\sum_{k=0}^{n-1} \frac{1}{n-k}=1+\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}=\frac{S_n}{n!}$$
So the result is
$$x\log(x)=\sum_{n=1}^\infty \frac{S_n}{n!}(\frac{x-1}{x})^n$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1130883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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|
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\begin{align}
& \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\
& \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\
& e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\
& e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\
& e^{-\infty} = 0\\
\end{align}
$$
Edit #1:
Continuing after the mistake of the $\tan(x)$:
$$\begin{align}
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\
\end{align}
$$
|
Take the logarithm of your limit then use L'Hopital and Taylor series. Let the expression in the limit be $L$:
$$\begin{align}
\ln L&=\ln\left[ \left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \right] \\[2ex]
&= \frac{\ln\left(\frac{\sin(x)}{x}\right)}{x^2} \\[2ex]
&\to\frac{\frac{x}{\sin x}\cdot \frac{x\cos x-\sin x}{x^2}}{2x}\qquad \text{(L'Hopital)}\\[2ex]
&= \frac{x\cos x-\sin x}{2x^2\sin x} \\[2ex]
&\to\frac{x(1-\frac{x^2}2)-(x-\frac{x^3}6)}{2x^2(x)}\qquad \text{(Taylor series)} \\[2ex]
&= \frac{-\frac 13x^3}{2x^3} \\[2ex]
&= -\frac 16
\end{align}$$
Therefore $L\to e^{-1/6}$. (I left a few minor gaps for you to fill in.)
If you don't like Taylor series, we can finish the whole thing with L'Hopital, though this way takes more steps.
$$\begin{align}
\ln L&\to\frac{x\cos x-\sin x}{2x^2\sin x} \qquad\text{(from above, before Taylor series)}\\[2ex]
&\to\frac{x\cdot -\sin x +\cos x-\cos x}{2x^2\cos x + 4x\sin x} \qquad\text{(L'Hopital)}\\[2ex]
&= \frac{-\sin x}{2x\cos x + 4\sin x} \\[2ex]
&\to\frac{-\cos x}{2x\cdot -\sin x+2\cos x+4\cos x} \qquad\text{(L'Hopital)}\\[2ex]
&= \frac{-\cos x}{-2x\sin x+6\cos x} \\[2ex]
&\to\frac{-1}{0+6} \\[2ex]
&= -\frac 16
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integral proof $I_n=\int\frac{x^n}{\sqrt{x^2+a}} \, dx$ If $$I_n=\int\frac{x^n}{\sqrt{x^2+a}}\,dx$$ prove that $$I_n=\frac{1}{n}x^{n-1}\sqrt{x^2+a}-\frac{n-1}{n}aI_{n-2}$$
I tried
$$I_n=\int\frac{x\times x^{n-1}}{\sqrt{x^2+a}} \,dx =\int(\sqrt{x^2+a})' \times x^{n-1} \,dx \\ =x^{n-1}\sqrt{x^2+a} -(n-1)\int x^{n-2}\sqrt{x^2+a} \, dx$$
I cant any further. Any hint?
|
Note that
\begin{align}
\int\frac{x^n}{\sqrt{x^2+a}}\,dx
= &\int \frac{x^{n-1}}{n(x^2+a)^{\frac{n-1}2}}\ d\left[(x^2+a)^{\frac n2}\right]
\end{align}
Then, integrate by parts to immediately establish
$$I_n= \frac{x^{n-1}}{n}\sqrt{x^2+a}-\frac{(n-1)a}{n}I_{n-2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1132007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that $ABCD$ has an incircle if and only if $\frac {1}{PE} + \frac {1}{PG} = \frac {1}{PF} +\frac {1}{PH}$ Let $ABCD$ be a convex quadrilateral.Let the diagonals $AC$ and $BD$ intersect in $P$.Let $PE,PF,PG,PH$ be the altitudes from $P$ onto the sides $AB,BC,CD$ and $DA$ respectively.Show that $ABCD$ has an incircle if and only if
$\frac {1}{PE} + \frac {1}{PG} = \frac {1}{PF} +\frac {1}{PH}$
|
Source: http://www.artofproblemsolving.com/Forum/blog.php?u=166324&b=103587
Let me denote the distances $PE,PF,PG,PH$ as $h_1,h_2,h_3,h_4$. Let $AB=a,BC=b,CD,c,DA=d$
$\frac{h_1}{h_4}=\frac{d(C,AB)}{d(C,AD)}=\frac{b \sin B}{c \sin D}$
$\frac{h_1}{h_2}=\frac{d(D,AB)}{d(D,BC)}=\frac{d \sin A}{c \sin C}$
$\frac{h_1}{h_3}=\frac{h_1}{h_2} \cdot \frac{h_2}{h_3} = \frac{d \sin A}{c \sin C} \cdot \frac{d(A,BC)}{d(A,CD)}=\frac{d \sin A*a \ sin B}{c \sin C*d \sin D}$
Therefore, we get
$\frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}$
$\iff 1+\frac{h_1}{h_3}=\frac{h_1}{h_2}+\frac{h_1}{h_4}$
$\iff 1+\frac{d \sin A \cdot a \sin B}{c \sin C \cdot d \sin D}=\frac{d \sin A}{c \sin C}+\frac{b \sin B}{c \sin D}$
$\iff a \sin A \sin B+ c \sin C \sin D=b \sin B \sin C+ d \sin D \sin A (\star)$
For the 'if' part, let r be the inradius of ABCD. Then, we get
$a=r \left( \cot \frac{A}{2} + \cot \frac{B}{2} \right) = r \frac{\sin \frac{A+B}{2}}{\sin \frac{A}{2} \sin \frac{B}{2}}$
$
\implies a \sin A \sin B = 4r \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{A+B}{2}$
$=4r \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C+D}{2}$
$= 4r \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \cos \frac{D}{2} \left( \tan \frac{C}{2} + \tan \frac{D}{2} \right)$
$= 4r \gamma \left( \tan \frac{C}{2} + \tan \frac{D}{2} \right)$
$where \gamma=\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \cos \frac{D}{2}$.
Analogously, $b \sin B \sin C = 4r \gamma \left( \tan \frac{D}{2} + \tan \frac{A}{2} \right)$
$c \sin C \sin D = 4r \gamma \left( \tan \frac{A}{2} + \tan \frac{B}{2} \right)$
$d \sin D \sin A = 4r \gamma \left( \tan \frac{B}{2} + \tan \frac{C}{2} \right)$
It is now easily seen that both sides of equation $(\star)$ are equal to $4r \gamma \left( \tan \frac{A}{2} + \tan \frac {B}{2} + \tan \frac{C}{2} + \tan \frac{D}{2} \right)$, hence the proof of the 'if' part is complete.
For the 'only if' part. Suppose first that in ABCD, both pairs of opposite sides are parallel. Then, the hypothesis, $a \sin A \sin B+ c \sin C \sin D=b \sin B \sin C+ d \sin D \sin A$ reduces to $a+c=b+d$ so ABCD becomes a rhombus and it obviously has an incircle. So we'll assume that ABCD has atleast one pair of nonparallel opposite sides, WLOG, they are AD and BC. Assume that ABCD is not tangential. There exists a circle $\Gamma$ touching segments $DA,AB,BC$. Because of our assumption, this circle does not touch side CD. Construct a parallel to CD which is tangent to the circle, and meeting BC,AD at C',D' respectively. Let BC' = b', C'D'=c' and D'A=d'. Since quadrilateral ABC'D' is tangential, we have
$a \sin A \sin B+ c' \sin C \sin D=b' \sin B \sin C+ d' \sin D \sin A$
Along with the equation,
$a \sin A \sin B+ c \sin C \sin D=b \sin B \sin C+ d \sin D \sin A$
we get
$(CD - C'D') \sin C \sin D = C'C \sin B \sin C + D'D \sin D \sin A$
If $\Gamma$ intersects the segment CD, then note that $C'C$ and $D'D$ are negative, while $CD-C'D' $is positive, contradiction. If $\Gamma$ does not intersect the segment $CD$, then $C'C$ and $D'DD'D$ are positive, while $CD - C'D'$ is negative, again a contradiction. Thus, we conclude that $\Gamma$ is tangent to $CD$ and hence, $ABCD$ is tangential.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1133077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula:
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!
|
First notice that $\dfrac{7\pi }{8}=\pi-\dfrac{\pi}{8}$ and also $\dfrac{5\pi }{8}=\pi-\dfrac{3\pi}{8}$.
Substituting these values in the original expression, you'll get:
$$\left(1+\cos {\pi \over 8}\right)\left(1+\cos {3\pi \over 8}\right)\left(1-\cos {3\pi \over 8}\right)\left(1-\cos {\pi \over 8}\right)$$
Using the obvious algebraic identity, you get
$$ \left(1-\cos^2 {\pi \over 8}\right)\left(1-\cos^2 {3\pi \over 8}\right)$$
Which is nothing but:
$$\sin^2{\pi \over 8}.sin^2{3\pi \over 8}$$
Now finally notice that $ \dfrac{3\pi}{8}=\dfrac{\pi}{2}-\dfrac{\pi}{8}$ and use the identity $\sin2\theta=2\sin\theta \cos\theta$ to get the simple form
$$\frac{1}{4}\sin^2{\pi \over 4}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws.
|
$$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\dfrac{(x+1)^{x+2}-(x+1)^{x+1}}{(x+2)^{x+2}}$$
$$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\Big(1-\dfrac{1}{x+2}\Big)^{x+2}\Big(\dfrac{x+2}{x+1}\Big)-\Big(1-\dfrac{1}{x+2}\Big)^{x+1}\dfrac{1}{x+2}\to e^{-1}+0$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
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|
Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that... Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that:
$(i)~~0\le 3ab\le 1$
$(ii)~~b\ge \sqrt3$.
My attempt:
I could solve the first part by Vieta's theorem. But, I am stuck on the second part. Please help. Thank you.
|
Let $x, y, z > 0$ be the three roots. Then, $x+y+z = xyz = \dfrac1a$ and $xy+yz+zx = \dfrac{b}a$
$(i),\quad $ Clearly, $a, b > 0$. Also $(x+y+z)^2 \ge 3(xy+yz+zx) \implies \dfrac1{a^2} \ge 3\dfrac{b}a \implies 1 \ge 3ab$.
For $(ii),\quad (xy+yz+zx)^2 \ge 3xyz(x+y+z) \implies \dfrac{b^2}{a^2} \ge 3\dfrac{1}{a^2} \implies b \ge \sqrt3$.
P.S. In case the second inequality used is not familiar, you can show that it is equivalent to the following rearrangement:
$$(xy)^2+(yz)^2+(zx)^2 \ge (xy)(yz)+(yz)(zx)+(zx)(xy)$$
|
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"url": "https://math.stackexchange.com/questions/1136109",
"timestamp": "2023-03-29T00:00:00",
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|
Having problems finding $x$ in terms of $a$ and $b$. I have attempted this question but have not found a solution. I am currently stuck. Hints on how I may go further would be helpful. Thank You in advance.
The Question:
$$\frac{a^2b}{x^2} + \left(1+\frac{b}{x}\right)a = 2b+ \frac{a^2}{x}$$
What I have done so far that I believe is correct:
$$\frac{a^2b}{x^2} + \frac{ab}{x} + a = 2b+ \frac{a^2}{x}$$
If you would like to see other work that I have attempted on this question I can also place it up.
|
Take the equation:
$$\frac{a^2b}{x^2} + \frac{ab}{x} + a = 2b+ \frac{a^2}{x}$$
Multiply each side by $x^2$:
$$a^2b+abx+ax^2=2bx^2+a^2x$$
Subtract $2bx^2+a^2x$ from each side:
$$a^2b+abx+ax^2-2bx^2-a^2x=0$$
Rearrange it as a quadratic equation:
$$(a-2b)x^2+(ab-a^2)x+(a^2b)=0$$
And solve as you would any quadratic equation:
$$x_{1,2}=\frac{-(ab-a^2)\pm\sqrt{(ab-a^2)^2-4(a-2b)(a^2b)}}{2(a-2b)}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$
$(4x^4+4x^3+6x^2+6x+6x^4+12x^3-48x^2+3x^2+6x-24)$
$=10x^4+16x^3+39x^2+6x-18$
|
Let $y=(2x^3+3x)(x-2)(x+4)$. Then if we are not too fussy about whether this makes sense in the reals, we get
$$\ln y=\ln(2x^3+3x)+\ln(x-2)+\ln(x+4).$$
Differentiating, we get
$$\frac{1}{y}\frac{dy}{dx}=\frac{6x^2+3x}{2x^3+3x}+\frac{1}{x-2}+\frac{1}{x+4},$$
and now we know $\frac{dy}{dx}$.
In this problem, the implicit differentiation approach does not give a significicant advantage over the plain Product Rule. However, it does become useful if the problem is a little more complicated.
|
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|
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me that:
$$\frac{x}{(x-1)^2} = \frac{1}{(x-1)} + \frac{1}{(x-1)^2}$$
How to intuitively think about this partial fraction expansion? I've seen some examples but suddenly these conter intuitive examples opo out and I get confused. I can check that his is true but I couldn't find this expansion by myself
Then:
$$\int \frac{x}{(x-1)^2} dx = \int \frac{1}{(x-1)} + \frac{1}{(x-1)^2}dx = \ln (x-1) + (x-1)^{-1}$$
Then:
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+\frac{x}{(x-1)^2}dx = \frac{x^2}{2} + 2x + 3[\ln(x-1)+(x-1)^{-1}]$$ but wolfram alpha gives another answer. What I did wrong?
|
change of variable gets you there faster. let $u = x - 1, x = 1 + u.$ then
$\begin{align}\int\dfrac{x^3 + 2}{(x-1)^2}\,dx &= \int\frac{(1+u)^3+2}{u^2} \,du\\
&=\int\frac{3+3u+3u^2+u^3}{u^2}\,du\\
& = \int3u^{-2}+3u^{-1}+ 3+u \,du\\
&=-\frac{3}{u} + 3\ln u + 3u + \frac{u^2}{2} +C
\end{align}$
note that this substitution works on any rational function of the form $\dfrac{p(x)}{(ax+b)^k}$ where $p$ is polynomial in $x.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that $m^2+n^2=1$, $p^2+q^2=1$, and $mp+nq=0$, how much is $mn+pq$? without using
$(m,n)\perp(p,q)$
|
The answer is obtained using trigonometry.
The condition $x^2 + y^2 = 1$ implies $x = \cos\theta$ and $y = \sin\theta$. Write
$$ m = \cos \alpha \ , \quad n = \sin\alpha $$
$$ p = \cos \beta \ , \quad q = \sin\beta $$
Then
$$ mp + nq = \cos\alpha cos\beta + \sin\alpha \sin \beta = \cos(\alpha - \beta) $$
$$ mn + pq = \cos\alpha \sin\alpha + \cos\beta \sin \beta $$
You are given that $\cos(\alpha-\beta) = mp + nq = 0$ and therefor $\alpha - \beta = \pm \pi/2$ (since $\cos(\pm \pi/2)=0$). Then $\alpha = \pi/2 + \beta$ and
Then
$$
mn + pq = \cos(\pi/2 + \beta) \sin(\pi/2 + \beta) + \cos\beta \sin \beta = \sin(-\beta) \cos(-\beta) + \cos\beta \sin \beta = -\sin\beta\cos\beta + \cos\beta \sin \beta = 0 $$
|
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"timestamp": "2023-03-29T00:00:00",
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}
|
finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-10 x-7 x^2+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2 x-3 x^2+x^4}}$$.
But failed to solve $f'(x)=0$ for finding $f(x)_{max}$.
I would be glad for your help.
Thanks.
|
Hint:
Consider the points $A=(4,5)$ and $B=(1,2)$ and a point $P$ on the line $x=y$ parallel to the line AB. We have for the distances $PA$, $PB$ $AB$ the triangle inequality
$$PA-PB \le AB$$ with equality if an only if $A$,$B$, $P$ lie on a line in this order. This will only happen for points $P$ chosen on $x=y$ at $-\infty$. The supremum value is $AB = 3\sqrt{2}$, the maximum is not achieved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
When is $2^x+3^y$ a perfect square?
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?
I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is $2^x+3^y$ is of the form $6k+1$ and so on .. but couldn't succeed. Can anyone help?
|
Let $2^x+3^y=k^2$.
mod $3$ gives $x=2m$ for some $m\in\mathbb Z^+$, since $(-1)^x\equiv k^2\pmod {3}$ and since $2$ (i.e. $-1$) is not a quadratic residue mod $3$.
More generally, $\left(\frac{-1}{p}\right)=1\iff p\equiv 1\pmod {4}$, where $p$ is an odd prime and $\left(\frac{a}{b}\right)$ is the Legendre symbol.
Thus $4^m+3^y=k^2$.
So $(-1)^y\equiv k^2\pmod {4}$, but $\left(\frac{-1}{4}\right)=-1$. So $y=2n$ for some $n\in\mathbb Z^+$.
Thus $(2^{m})^2+(3^n)^2=k^2$.
Hence $(2^m,3^n,k)$ is a Pythagorean triple. Moreover, it is primitive, because $(2^m,3^n)=1$, which means:
$$\begin{cases}2^m=2ab\\3^n=a^2-b^2\\k=a^2+b^2\end{cases},$$
where exactly one of $a,b$ is odd.
This implies $(a,b)=(2^i,2^j)$ for some $i,j\in\mathbb Z^+$.
But exactly one of $a,b$ is odd (as I said before), and also $a>b\iff 2^i>2^j$.
Thus $j=0$ so that $(a,b)=(2^i,1)=(2^{m-1},1)$.
Then $4^{m-1}=3^n+1\iff (2^{m-1}-1)(2^{m-1}+1)=3^n$ so that $2^{m-1}-1$ and $2^{m-1}+1$ are powers of $3$, and because their difference is $2$, it follows that $2^{m-1}-1=1\iff m=2\iff x=4$, and $n=1\iff y=2$, which leads us to the only solution $(x,y)=(4,2)$, which after checking works.
The following is an alternative continuation of my solution using Zsigmondy's theorem.
We have $4^{m-1}=3^n+1$ ($n$ is odd (check $\pmod 4$ and note that $m-1\ge 1$)), but Zsigmondy's theorem implies:
$$\exists p\in\mathbb P (p\mid a^n+b^n\wedge p\not\mid a+b), \forall n\ge 2, a>b$$ (with the exception of $(a,b,n)=(2,1,3)$, which won't work here, since in our case $(a,b,n)=(3,1,n)$) so that $a^n+b^n$ has more than one prime divisor for any $n\ge 2, a>b$.
Thus $n\ge 2$ and $3>1$ lead to a contradiction, which means $n=1\iff y=2$ and $m=2\iff x=4$.
This leads to $(x,y)=(4,2)$ as the only solution, which after checking works.
Yet another way I could've proceeded after seeing that $4^{m-1}=3^n+1$ is Catalan's conjecture (now proved), which tells us that $$\begin{cases}x^a-y^b=1\\x,a,y,b\in\mathbb Z_{>1}\end{cases}\iff (x,a,y,b)=(3,2,2,3)$$
Continuation of André Nicolas solution using Zsigmondy's theorem.
$3^y=2^{s+1}+1$.
Notice that $y\ge 1$ so that $s+1\ge 1$. Applying Zsigmondy's theorem leads to a contradiction whenever $s+1\ge 2$ and $2>1$ except for the exception $s+1=3$, which means either $s+1=1\iff (x,y)=(0,1)$, which doesn't work, or $s+1=3\iff s=2$, giving us the only solution $(x,y)=(4,2)$, which after checking works.
This could've also been solved using Catalan's conjecture (now proved), which I introduced above.
But, of course, the easiest route here is to note that (we have $y\ge 1$ and thus $s+1\ge 1$, but $s+1\neq 1$, so $s+1\ge 2$. Now assume $s+1\ge 2$) $\pmod{4}$ leads to $y=2c$ for some $c\in\mathbb Z^+$ so that $(3^c-1)(3^c+1)=2^{s+1}$ and thus $3^c-1=2\iff c=1\iff y=2$ and thus $s+1=3\iff s=2\iff x=4$.
It follows that $(x,y)=(4,2)$ is the only possible solution and after checking it it works.
|
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|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
|
Consider $$I=\int\sqrt{\frac{x}{1+x}}dx.$$ We proced by the change of variable $u=\sqrt{1+x}$, $du=\frac{1}{2\sqrt{1+x}}$dx and $x=u^2-1,$ which gives $$I=2\int \sqrt{u^2-1}\ du=u\sqrt{u^2-1}-\log\left(u+\sqrt{u^2-1}\right)+C,$$ see https://owlcation.com/stem/How-to-Integrate-Sqrtx2-1-and-Sqrt1-x2 for more informations. The calculations ends by replacing value of u by $\sqrt{x^2-1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
reduction of order method for 3rd order DE Use the Reduction of Order method to solve $$(2-x)y'''+(2x-3)y''-xy'+y=0$$ (such that $x<2$) using $u(x)=e^x$.
How do you use the method for $3$rd order... I have seen it been used on $2$nd but not higher. Please help.
|
Don't forget $y=x$ is also a particular solution of the ODE.
Let $y=xu$ ,
Then $y'=xu'+u$
$y''=xu''+u'+u'=xu''+2u'$
$y'''=xu'''+u''+2u''=xu'''+3u''$
$\therefore(2-x)(xu'''+3u'')+(2x-3)(xu''+2u')-x(xu'+u)+xu=0$
$x(2-x)u'''+3(2-x)u''+x(2x-3)u''+2(2x-3)u'-x^2u'-xu+xu=0$
$x(2-x)u'''+2(x^2-3x+3)u''-(x^2-4x+6)u'=0$
Let $v=u'$ ,
Then $v'=u''$
$v''=u'''$
$\therefore x(2-x)v''+2(x^2-3x+3)v'-(x^2-4x+6)v=0$
Note that $v=e^x$ is a particular solution of the ODE.
Let $v=e^xw$ ,
Then $v'=e^xw'+e^xw$
$v''=e^xw''+e^xw'+e^xw'+e^xw=e^xw''+2e^xw'+e^xw$
$\therefore x(2-x)(e^xw''+2e^xw'+e^xw)+2(x^2-3x+3)(e^xw'+e^xw)-(x^2-4x+6)e^xw=0$
$x(2-x)(w''+2w'+w)+2(x^2-3x+3)(w'+w)-(x^2-4x+6)w=0$
$x(2-x)w''+2x(2-x)w'+x(2-x)w+2(x^2-3x+3)w'+2(x^2-3x+3)w-(x^2-4x+6)w=0$
$x(2-x)w''-2(x-3)w'=0$
$x(2-x)w''=2(x-3)w'$
$\dfrac{w''}{w'}=\dfrac{2(x-3)}{x(2-x)}$
$\int\dfrac{w''}{w'}=\int\dfrac{2(x-3)}{x(2-x)}dx$
$\int\dfrac{w''}{w'}=-\int\left(\dfrac{2}{x}+\dfrac{6}{2-x}\right)dx$
$\ln w'=-2\ln x-6\ln(2-x)+c_1$
$\ln w'=-\ln(x^2(2-x)^6)+c_1$
$w'=-\dfrac{C_1}{x^2(2-x)^6}$
|
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|
Find two numbers whose sum is 20 and LCM is 24 With some guess work I found the answers to be 8 and 12.
But is there any general formula for this?
Note that the question is asked to my nephew who is at 4th grade.
|
A general algorithm
If $a+b = n$ and $\mathrm{lcm}(a,b) = c$ we let $\gcd(a,b) = d$ to get
$$ab = cd\\
n = a+b$$
Solving $b = n-a$ gives us
$$a(n-a) = cd \\
\Leftrightarrow a^2-na + cd = 0 \\
\Leftrightarrow a = \frac n2 \pm \sqrt{\frac{n^2}4 - cd}$$
So for even $n$ we must find $d$ such that $\frac{n^2}4 - cd$ is a perfect square (since $c>0$ this will amount to a finite number of possibilities). The $\pm$ is irrelevant because $b$ will take the alternate value.
If $n$ is odd, $n^2 - 4cd$ must be a perfect square instead and we obtain an analogous formula:
$$a = \frac12 (n \pm \sqrt{n^2 - 4cd})$$
In both cases, $d$ can range between $1$ and $\left\lfloor \frac{n^2}{4c} \right\rfloor$
Your case thus allows $1\le d\le \lfloor\frac{400}{96}\rfloor = 4$ and $100-24d$ must be a perfect square. $d=4$ yields $4 = 2^2$ so
$$a = 10 + \sqrt{4} = 12; \quad b = 10 - \sqrt4 = 8$$
|
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|
Another method of proving $x^3+y^3=z^3$ has no integral solutions? The equation $$ax^2+by^2=z^3$$ has the following parametrization:
$$y=q(3ap^2-bq^2)$$
$$x=p(ap^2-3bq^2)$$
$$z=ap^2+bq^2$$ can we deduct from that the Diophantine equation $$x^3+y^3=z^3$$ has no nontrivial solutions by choosing $x=a$ and $y=b$?
|
Your parametrization is almost right, you flipped the second factors for $x$, $y$.
In fact, from
$$\sqrt{a}\, x + i \sqrt{b}\, y = (\sqrt{a}\, p + i \sqrt{b}\, q)^3$$
we have with
\begin{eqnarray}
x &=& p (a p^2 - 3 b q^2) \\
y &=& q( 3 a p^2 - b q^2) \\
z &=& ap^2 + b q^2
\end{eqnarray}
($\pm$ if you wish) the equality
$$a x^2 + b y^2 = (a p^2 + b q^2)^3= z^3$$
While this parametrization provides solutions to the equation $ax^2 + b y^2 = z^3$ it is not clear apriori that all solutions are of this form. Let's just assume furthermore $a=b=1$. Is it clear that all the solutions of the equation
$x^2 + y^2 = z^3$ are of the form above, for some $p$ and $q$ ( and $a=b=1$)? In other words, if $N(x+iy)$ is a cube in $\mathbb{Z}$ then $x+iy$ is itself a cube in $\mathbb{Z}[i]$ ?( the converse, explained above, is clear) OK, knowing a bit of arithmetic for the Gaussian integers this can be proved. Is it true in general for $a$, $b$ (positive) integers? It probably has to do with the arithmetic of some quadratic fields. Maybe so.
Assuming it it so, your reasoning is :
From $x^3 + y^3 = z^3$ I get $x \cdot x^2 + y \cdot y^2 = z^3 $ and so for some $p$ and $q$ we have
\begin{eqnarray}
x &=& p (x p^2 - 3 y q^2) \\
y &=& q( 3 x p^2 - y q^2) \\
z &=& x p^2 + y q^2
\end{eqnarray}
All right, I suppose you will work with this equation.. Hm...plausible. I vaguely remember seeing something like that in the book of Hardy and Wright..
|
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|
How to factorize $n(n+1)(n+2)(n+3)+1$? How to factorize $n(n+1)(n+2)(n+3)+1$ ?
It's turned into the lowest power of $n$ ever possible but the question wants me to factorize it.
How can I do that ?
|
Set $\dfrac{n+n+1+n+2+n+3}4=m\iff 2n+3=2m$
$\implies n(n+1)(n+2)(n+3)+1=\dfrac{2n(2n+2)(2n+4)(2n+6)}{2^4}+1$
$=\dfrac{(2m-3)(2m-1)(2m+1)(2m+3)}{2^4}+1$
$=\dfrac{(4m^2-9)(4m^2-1)}{2^4}+1$
$=\dfrac{(4m^2)^2-2\cdot4m^2\cdot5+5^2}{2^4}=\dfrac{(4m^2-5)^2}{(2^2)^2}$
|
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|
prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x - 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)
|
Using sum of angles identity,
\begin{align*}
\cos x\cos (x-60^{\circ})&\cos(x+60^{\circ})\\
& = \cos x(\cos x\cos 60^{\circ}+\sin x\sin 60^{\circ})(\cos x\cos 60^{\circ}-\sin x\sin 60^{\circ})\\ \\
& = \cos x(\cos^2x\cos^260^{\circ}-\sin^2x\sin^260^{\circ})\\\\
& = \cos x\left(\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x\right)\\\\
& = \frac{1}{4}(\cos^3x-3\cos x(1-\cos^2x))\\\\
& = \frac{1}{4}(4\cos^3x-3\cos x)
\end{align*}
|
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|
Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2) or 5 = 2(2) + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 3) = 2F(r + 1) + F(r)4) Then F(k + 4) = F(k + 2) + F(k + 3) = 5) 2F(k) + F(k - 1) + 2F(k + 1) + F(k) = 6) 2[F(k) + F(k + 1)] + [F(k - 1) + F(k)] = 7) 2F(k + 2) + F(k + 1)My first question here is how do I know how many values of n to test for? Here they chose two.My next question is how did they get from line 3 to line 4? I understand how the statement is correct but why is this chosen? I also understand that I need to prove it's true for all values of r because if I do that it implies that it is true for k + 1. Is it just to find a relation to F(r + 3) on line 3? If that was the case why not just have F(k + 3) = F(k + 2) + F(k + 1)?My final question about this is how did they get from line 4 to 5?The second equation I want to prove is:F(n + 6) = 4F(n + 3) + F(n) for n ≥ 1I'm able to prove n = 1 and n = 2 is true but I get stuck on going from what would be line 3 - 4 on this problem. As this is my problem for homework the answer is not in the back of the book.Now that I've gotten the help I just want to update this with the proof for my second equation (I haven't gotten the formatting down yet so bear with me):F(n + 6) = 4F(n + 3) + F(n)1) n = 1: F(7) = 4F(4) + F(1) or 13 = 12 + 1, true.2) n = 2: F(8) = 4F(5) + F(2) or 21 = 20 + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 6) = 4F(r + 3) + F(r)4) Then F(k + 7) = 4F(k + 4) + F(k + 1) =5) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 4) + F(k + 1) =6) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 3) + F(k + 2) F(k + 1) =7) F(k + 4) + F(k + 4) + F(k + 4) +F(k + 3) + F(k + 3) =8) F(k + 5) + F(k + 5) + F(k + 4) =9) F(k + 6) + F(k + 5) =10) F(k + 7)
|
the following relation on the Fibonacci numbers is sometimes useful: (for $n \ge k \ge 0$):
$$
F(n+k) = \sum_{j=0}^k \binom{k}{j}F(n-j)
$$
this gives:
$$
F(n+6) = \sum_{j=0}^3 \binom{3}{j}F(n+3-j) \\
= F(n+3) +\color{blue}{3F(n+2)+3F(n+1)} +F(n) \\
=4 F(n+3)+F(n)
$$
|
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|
Inequality in Hanoi Open Mathematics Competition $\frac{1}{bc} + \frac{1}{ab} + \frac{1}{ca} \geq 1$.
Prove $\frac{a}{bc} + \frac{c}{ab} + \frac{b}{ca} \geq 1$, where
$a,b,c$ are positive real numbers
|
Let $x=\frac{1}{bc},y=\frac{1}{ab},z=\frac{1}{ca}$. The question becomes given $x+y+z\geq1$ , prove $\frac{x\sqrt{x}}{\sqrt{yz}}+\frac{y\sqrt{y}}{\sqrt{xz}}+\frac{z\sqrt{z}}{\sqrt{xy}}=\frac{x^2+y^2+z^2}{\sqrt{xyz}}\geq 1$.
From Cauchy-Schwarz $3(x^2+y^2+z^2)\geq(x+y+z)^2$
From AM-GM $x^2+y^2+z^2\geq 3(xyz)^{2/3}$
$$\frac{x^2+y^2+z^2}{\sqrt{xyz}}=\frac{(x^2+y^2+z^2)^{3/4}(x^2+y^2+z^2)^{1/4}}{\sqrt{xyz}}\geq\frac{3^{3/4}\sqrt{xyz}3^{-1/4}\sqrt{x+y+z}}{\sqrt{xyz}}\geq \sqrt{3}\geq 1$$
|
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|
Proving $\sum_{n =1,3,5..}^{\infty }\frac{4k \ \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$ Proving
$$\sum_{n =1,3,5..}^{\infty }\frac{4k \sin^2\left(\frac{n}{k}\right)}{n^2}=\pi$$
Where $k$ any number greater than $0$
I tried to prove it by using the Fourier series but I couldnt find any form likes the above formula . Any helps. Thanks
|
Using (in the interval $(0,\pi)$
) $$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(n\frac{x}{2}\right)^{2}}{n^{2}}\,\,(1)$$
we have$$\frac{2\pi x-x^{2}}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{x}{2}\right)^{2}}{\left(2n-1\right)^{2}}+\frac{1}{4}\underset{n\geq1}{\sum}\frac{\sin\left(nx\right)^{2}}{n^{2}}$$
and using $(1)$
again we get$$\frac{1}{4}\underset{n\geq1}{\sum}\frac{\sin\left(nx\right)^{2}}{n^{2}}=\frac{\pi x-x^{2}}{8}$$
so$$\frac{\pi x}{8}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{x}{2}\right)^{2}}{\left(2n-1\right)^{2}}$$
now put $$x=\frac{2}{k}$$
and we have$$\frac{\pi}{4k}=\underset{n\geq1}{\sum}\frac{\sin\left(\left(2n-1\right)\frac{1}{k}\right)^{2}}{\left(2n-1\right)^{2}}$$
as wanted.
|
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|
Complete the character table of group of order $21$ You are given the incomplete character table of a group $G$ with order $21$ which has $5$ conjugacy classes, $C_1,\dots,C_5$, which have sizes $1,7,7,3,3$.
$$ \begin{array}{|c|c|c|c|c|}
\hline
& C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline
& & & & & \\ \hline
& & && & \\ \hline
\chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline
& & & & & \\ \hline
\chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline
\end{array}
$$
Complete the character table.
Im guessing that $\chi_0$ has to be the trivial representation so we get that
$$ \begin{array}{|c|c|c|c|c|}
\hline
& C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline
\chi_0 & 1 &1 & 1& 1&1 \\ \hline
& 1 & && & \\ \hline
\chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline
& 3 & & & & \\ \hline
\chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline
\end{array}
$$
and as $21=1+1+1+9+9$.
Im sure you have to obtain something from the fact we have 3rd and 7th roots of unity which correspond to the sizes of the conjugacy classes but I cannot see what I am meant to glimmer from this.
Hints only please.
|
The remaining two characters are algebraic conjugates of $\chi_2$ and $\chi_4$. You get them by replacing $\zeta_3$ by $\zeta_3^2$, and $\zeta_7$ by $\zeta_7^3$, respectively.
|
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|
Euler Mascheroni Constant is Zero $$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y}
$$
Let $x = \frac{y - 1}{n - 1}$, or $y = (n-1)x + 1$. Then, $dy = (n - 1) dx$.
$$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_0^1 \frac{n - 1}{(n - 1) x + 1} dx = \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx
$$
$$
\gamma = \lim_{n \rightarrow \infty} \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\lim_{n \rightarrow \infty} \frac{1 - x^n}{1 - x} - \lim_{n \rightarrow \infty} \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\frac{1}{1 - x} - \frac{1}{x}) dx
$$
(The first limit relies on the fact that $0 < x < 1$, so that $\lim_{n \rightarrow \infty} x^n = 0$. The second is solved via. L'Hospital's rule.)
Let $t = 1 - x$. Then, $dt = -dx$.
$$
\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x} = - \int_1^0 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = \int_0^1 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = 0
$$
Evidently something's wrong with this, but I genuinely can't find it. The only part I find even fishy is how the limit is split in two. Any ideas?
|
You obviously took a wrong turn along the way to conclude
$$\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x},$$
since each integral is improper and divergent. The limit/integral switch cannot be justified by uniform convergence, monotone convergence, etc.
As an example where the switch is valid, change variables in the first integral with $x = (1-y/n)$ to get
$$H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y} \\= \int_0^n \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n \frac{dy}{y}\\=\int_0^1 \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n \left(1- \frac{y}{n}\right)^n\frac{dy}{y}.$$
Now you can justify a switch of limit and integral using LDCT to obtain
$$\gamma = \lim_{n \to \infty}(H_n - \ln n)= \int_0^1 \left(1 - e^{-y}\right)\frac{dy}{y}-\int_1^\infty e^{-y}\frac{dy}{y},$$
where the improper integrals converge.
Further manipulation leads to the well-known result
$$\gamma = \int_0^1 \left( \frac{1}{1-x}+ \frac{1}{\ln x}\right) \, dx.$$
Here the improper integral converges, as the singular behavior of $1/(1-x)$ is offset by that of $1/\ln x$ near $x=1$.
|
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|
Find the last digit of $77777^{77777}$ Find the last digit of $$77777^{77777}$$
I got a pattern going for $77777^n$ for $n=1, 2, ....$ to be:
$$7, 9, 3, 1$$ for $n = 1, 2, 3, 4$ respectively.
The idea is:
$$77777^{77777} \pmod{10}$$
I see that:
$$77777^n \equiv 77777^{n + 4} \pmod{10}$$
Using the spotted pattern, but letting $n = 77777$ doesnt help at all.
Please give hints only...!
|
$\varphi(10)
= \varphi(2 \times 5) = \varphi(2) \times \varphi(5) = (2-1)(5-1) = 4
$
Since $\gcd(7,10) = 1,$ then $7^4 \equiv 1 \pmod{10}$
Note that $77777 \equiv 777 \times 100 + 76 + 1 \equiv 1 \pmod 4$
So
\begin{align}
77777^{77777} \pmod{10}
&\equiv 7^{77777} \pmod{10}\\
&\equiv 7^{4(\text{something})+1} \pmod{10}\\
&\equiv (7^4)^\text{something} \times 7^1 \pmod{10}\\
&\equiv 1^\text{something} \times 7 \pmod{10}\\
&\equiv 7 \pmod{10}
\end{align}
|
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|
Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
|
if you want to find just $C$ not other two, then look at the behavior of $$ \frac{3x^2 + 17x}{(x+1)(x-2)(x+4)} = \frac{3*(-4)^2+17*(-4)}{(-4+1)(-4-2)(x+4)} +\cdots=-\frac{10/9}{(x-4)}+\cdots$$ therefore $$C = -\frac{10}9. $$
|
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|
Find the equation of a circle, given a point on it and a point where it is tangent to a given line The given question is:
Find the equation of the circle that passes through point $(-3,-4)$ and touches the line $x-y+7=0$ at the point $(-5,2).$
What I did was:
Took the given points $(-5,2)$ and $(-3,-4)$ as the diameter of the circle and derived the equation using the 'formula for the equation of a circle when the end points of diameter are given' :-
$(x+5)(x+3) + (y+4)(y-2) = x^2+y^2+15x+2y+7 =0 .$
Then I affixed the equation of the line derived by the given points with a parameter so that I get the equation of any circle which would pass through the given points $(-5,2)$ and $(-3,-4) $:-
$$x^2+y^2+15x+2y+7+k(3x+y+13) = 0 \tag1 $$
I substituted the mid points $(-15-3k/2,-2-k/2)$ of (1) above equation to the line perpendicular to the given tangent because the mid points lie on that $(x+y+3=0) $ line. Then
$$(-15-3k/2) + (-2-k/2) + 3 = 0
= -15-3k-2-k+6 = 0
= -11 = 4k
= k = -11/4$$
But when I substitute this value of 'k' to the equation (1) I get the wrong answer. So I tried using another method namely: finding the mid points of the required circle using simultaneous equations of the lines of perpendicular to the given tangent and the perpendicular of the line thru points (-5,2) and (-3,-4) and. That worked for me.
What I need to know
If the first method is inapplicable to the given question or what I am doing wrong there. Thanks in advance.
|
the mid point of the chord connecting the points $(-5, 2)$ and $(-3, -4)$ is $(-4, -1)$ and the chord has slope $\frac{-4-2}{-3 + 5} = -3.$ the parametric equation of a point on the bisector of this chord is $$x = -4 + 3t, y= -1 + t$$ the slope of the line connection this point and point of contact $(-5, 2)$ is $$\frac{3-t}{-1-3t} $$ must be orthogonal to the tangent so must equal $-1.$ that gives $t = 1/2$ and the center of the circle is $$(-5/2, -1/2)\text{ and the radius is }\sqrt{(-5/2)^2 + (5/2)^2} = \frac{5\sqrt 2}2.$$
|
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|
Integral $\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$ How do I determine the value of this integral?
$$\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$$
Plugging in Euler's identity gives
$$\int_{-\infty}^{\infty} \frac{i\sin^3x}{x^2}dx + \int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$
and since $\dfrac{i\sin^3x}{x^2}$ is an odd function, all that is left is
$$\int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$
at which point I am stuck.
I feel I am not even going the right direction, anybody willing to help?
Thanks in advance.
|
The integral can be evaluate by converting it to a double integral first.
Let
$$I = \int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = 2 \int_0^\infty \frac{\sin^2 x \cos x}{x^2} \, dx.$$
Noting that
$$\sin^2 x \cos x = \frac{1}{4} (\cos x - \cos 3x),$$
yields
$$I = \frac{1}{2} \int_0^\infty \frac{\cos x - \cos 3x}{x^2} \, dx.$$
Now, by observing that
$$\int_1^3 \sin (tx) \, dt = \frac{\cos x - \cos 3x}{x},$$
the integral for $I$ can be rewritten as
$$I = \frac{1}{2} \int_0^\infty \int_1^3 \frac{\sin (xt)}{x} \, dt \, dx,$$
or
$$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin (xt)}{x} \, dx \, dt,$$
after the order of integration has been changed.
Enforcing a substitution of $x \mapsto x/t$ leads to
$$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin x}{x} \, dx \, dt.$$
Making use of the well-known result of
$$\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2},$$
one has
$$I = \frac{\pi}{4} \int_1^3 dt,$$
or
$$\int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = \frac{\pi}{2},$$
as expected.
|
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|
Showing that $f_n$ is the number closest to $\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n$ I want to prove that the the $n$th Fibonacci number $f_n$ is the integer closest to $\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n$. What would be a rigorous way to go about this? I assume I'll have to use a recurrence relation and maybe Binet's Forumla. I just don't know how to go about tying it all together, so any advice on how to proceed would be much appreciated.
|
Copy-paste from Wikipedia:
Like every sequence defined by a linear recurrence with constant coefficients, the Fibonacci numbers have a closed-form solution. It has become known as Binet's formula, even though it was already known by Abraham de Moivre:
$$F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi^n}{\sqrt 5}$$
where
$$\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\cdots\,$$
is the golden ratio, and
$$\psi = \frac{1 - \sqrt{5}}{2} = 1 - \varphi = - {1 \over \varphi} \approx -0.61803\,39887\cdots$$
Since $\psi = -\frac{1}{\varphi}$, this formula can also be written as
$$F_n = \frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}$$
To see this, note that $\phi$ and $\psi$ are both solutions of the equations
$$x^2 = x + 1,\, x^n = x^{n-1} + x^{n-2},\,$$
so the powers of φ and ψ satisfy the Fibonacci recursion. In other words
$$\varphi^n = \varphi^{n-1} + \varphi^{n-2}\, $$
and
$$\psi^n = \psi^{n-1} + \psi^{n-2}\, .$$
It follows that for any values $a$ and $b$, the sequence defined by
$$U_n=a \varphi^n + b \psi^n\,$$
satisfies the same recurrence
$$U_n=a \varphi^{n-1} + b \psi^{n-1} + a \varphi^{n-2} + b \psi^{n-2} = U_{n-1} + U_{n-2}.\,$$
If $a$ and $b$ are chosen so that $U_0=0$ and $U_1=1$ then the resulting sequence $U_n$ must be the Fibonacci sequence. This is the same as requiring $a$ and $b$ satisfy the system of equations:
$$\left\{\begin{array}{l} a + b = 0\\ \varphi a + \psi b = 1\end{array}\right.$$
which has solution
$$a = \frac{1}{\varphi-\psi} = \frac{1}{\sqrt 5},\, b = -a$$
producing the required formula.
Computation by rounding
Since
$$\frac{|\psi|^n}{\sqrt 5} < \frac{1}{2}$$
for all $n \geq 0$, the number $F_n$ is the closest integer to
$$\frac{\varphi^n}{\sqrt 5}\, .$$
Therefore it can be found by rounding, or in terms of the floor function:
$$F_n=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor,\ n \geq 0.$$
Or the nearest integer function:
$$F_n=\bigg[\frac{\varphi^n}{\sqrt 5}\bigg],\ n \geq 0.$$
Similarly, if we already know that the number $F > 1$ is a Fibonacci number, we can determine its index within the sequence by
$$n(F) = \bigg\lfloor \log_\varphi \left(F\cdot\sqrt{5} + \frac{1}{2}\right) \bigg\rfloor$$
|
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|
how does this expression cancel out How does $$\frac{2t-1-i}{2t^2-2t+1}=\frac{1+i}{-1+(1+i)t}$$ I just can't see how this works...
I typed the LHS in WFA and it gave the RHS but I don't know how anything can cancel.
|
By using quadratic formula we get the roots of $2t^2-2t+1$, they are $\frac{1+i}{2}$ and $\frac{1-i}{2}$, then, we can factor out $2t^2-2t+1$ as
\begin{align*}
2t^2-2t+1&=2\left(t-\frac{1+i}{2}\right)\left(t-\frac{1-i}{2}\right)\\
&=(2t-1-i)\left(t-\frac{1-i}{2}\right)
\end{align*}
Hence
\begin{align*}
\frac{2t-1-i}{2t^2-2t+1}&=\frac{2t-1-i}{(2t-1-i)\left(t-\frac{1-i}{2}\right)}\\
&=\frac{1}{t-\frac{1-i}{2}}\cdot\frac{1+i}{1+i}\\
&=\frac{1+i}{(1+i)t-\frac{(1-i)(1+i)}{2}}\\
&=\frac{1+i}{(1+i)t-1}
\end{align*}
|
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|
Partial fraction decomposition of $\frac{x^3+x+2}{x(x^2+1)^2}$ Help me solve this
$$\dfrac{x^3+x+2}{x(x^2+1)^2}$$
It looked like a simple one, but became complicated in my hands because i tried it like this:
$$\dfrac{x^3+x+2}{x(x^2+1)^2}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2} $$
multiply all sides by $x(x^2+1)^2$ to get
$x^3+x+2= A(x^2+1)+Bx(x^3+x)+C(x^2+1)+Dx^2+Ex$
group like terms:
$x^3+x+2=Bx^4+ Ax^2+Bx^2+Cx^2+Dx^2+Ex+A+C$
$x^3+x+2=Bx^4+ (A+B+C+D)x^2+Ex+A+C$
The rest seems like am in the wrong path..because I think $x^4$ seems to be misplaced.. any idea?
|
You're working with an incorrect equation.
After multiplying both sides by $x(x^2 + 1)^2$, you should have $$\begin{align} x^3+x+2 & = A(x^2+1)^2+Bx(x^3+x)+C(x^3+x)+Dx^2+Ex\\
&= Ax^4+ 2Ax^2 + A +Bx^4 + Bx^2 +Cx^3 + Cx +Dx^2 + E x
\end{align}$$
|
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|
Introduction to probability Dice We roll a fair die repeatedly until we see the number four appear and then we stop.
What is the probability that we needed an even number of die rolls?
For this problem I said that it would be the Sum of (5/36)*(25/36)^j and I got an answer of 5/9? Does that seem correct.
Thanks
|
We want the probability of having an even number of rolls. That means we want an odd number of non-fours, and then a four. So, we could have 1, 3, 5, 7, etc. non-fours, followed by a four. Given that the die is fair, we will assume that $P(4) = \frac{1}{6}$ and that the rolls are independent, so we can multiply when appropriate. If I am not mistaken, this becomes:
$$
\begin{align}
&\frac{1}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i + 1}\\
&=\frac{1}{6}\times\sum_{i=0}^\infty \frac{5}{6}\left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\left(\frac{5}{6}\right)^2\right)^i\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{25}{36}\right)^i\\
&=\frac{5}{36}\times\frac{1}{1 - \frac{25}{36}}\\
&=\frac{5}{36}\times \frac{36}{11} = \frac{5}{11}
\end{align}
$$
|
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|
How to solve this limit, normal methods result in zero. Online sources say otherwise. in one part of my homework, we are asked to solve this limit:
$$
\lim_{x\to\infty}\sqrt{\frac{x^3}{x-3}} - x
$$
The result should be $3/2$, but when I try it, I always get $0$ instead.
Please, do not post the full solution, I would only like to be pointed in the right direction.
ps:
Right now I got to:
$$
\lim_{x\to\infty}\frac{x\sqrt{x^2-3x} - x^2 + 3x}{x-3}
$$
EDIT:
Thanks everyone, now I understand it :) . Marking as solved!
|
We have
\begin{align} \sqrt{\frac{x^3}{x - 3}} - x &= x\sqrt{\frac{x}{x - 3}} - x\\
&= x\left(\sqrt{\frac{1}{1 - \frac{3}{x}}} - 1\right)\\
&= x\left(\frac{1}{\sqrt{1 - \frac{3}{x}}} - 1\right)\\
&= \frac{x}{\sqrt{1 - \frac{3}{x}}}\left(1 - \sqrt{1 - \frac{3}{x}}\right)\\
&= \frac{x}{\sqrt{1 - \frac{3}{x}}}\frac{\frac{3}{x}}{1 + \sqrt{1 - \frac{3}{x}}}\\
&= \frac{3}{\sqrt{1 - \frac{3}{x}}} \frac{1}{1 + \sqrt{1 -\frac{3}{x}}}
\end{align}
Since $3/x \to 0$ as $x\to \infty$, the last expression tends to $$\frac{3}{1}\cdot\frac{1}{2} = \frac{3}{2}$$
Hence, your limit is $3/2$.
|
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|
Expansion of cumulant transform
Verify the following expansion for a cumulant generating function of a random variable $X$.
\begin{align}
\kappa(t) & = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\rho_3\sigma^3t^3 + \frac{1}{24}\rho_4\sigma^4t^4 + \cdots \\
& = \mu t + \frac{1}{2}\sigma^2t^2+\frac{1}{6}\kappa_3 t^3 + \frac{1}{24}\kappa_4 t^4 + \cdots, \tag{$*$}
\end{align}
where $\mu=\mathbb EX$ and $\rho_r=\frac{\kappa_r}{\sigma^r}$ is the standardized cumulant.
By applying Taylor expansion twice, I got:
\begin{align}
\kappa(t) & := \log\mathbb Ee^{tX} = \log\mathbb E\left( 1+tX+\frac{t^2X^2}{2} + \frac{t^3X^3}{3!} + \frac{t^4X^4}{4!} + \cdots\right) \\
& = \log \left( 1+t\mathbb EX + \frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{3!} + \frac{t^4\mathbb EX^4}{4!} + \cdots \right) \\
& = t\mathbb EX + \frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{3!} + \frac{t^4\mathbb EX^4}{4!} + \cdots \\
& =\mu t+\frac{t^2\mathbb EX^2}{2} + \frac{t^3\mathbb EX^3}{6} + \frac{t^4\mathbb EX^4}{24} + \cdots \tag{$**$}
\end{align}
It is clear that $(*)$ and $(**)$ are not the same since cumulant and moment are not equal. Where did I get wrong, please? Thank you!
|
The Taylor expansion of $\log(1+x)$ is $x-x^2/2+x^3/3-\cdots$. Therefore up to second order we get
$$
\newcommand{\EE}{\mathbb{E}}
\newcommand{\VV}{\mathbb{V}}
\log\left(1+t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right) \\ =
\left(t\EE[X]+\frac{t^2}{2}\EE[X^2]+O(t^3)\right)-\frac{1}{2}\left(t\EE[X]+O(t^2)\right)^2 \\ =
t\EE[x] + \frac{t^2}{2}\EE[X^2]-\frac{t^2}{2}\EE[X]^2 + O(t^3) \\ =
t\EE[x] + \frac{t^2}{2}\VV[X] + O(t^3).
$$
|
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Product of Sums: Show that the following is a Polynomial by converting it into standard form. $$\prod_{k=0}^n (1+x^{2^k})$$
The given expression simplifies to $(1+x)(1 + x^2)...(1 + x^{2^n})$
I am not able to proceed further. How do I express this in Summation form?
|
We have
$$\prod_{k = 0}^n (1 + x^{2^k}) = \prod_{k = 0}^n \frac{1 - (x^{2^k})^2}{1 - x^{2^k}} = \prod_{k = 0}^n \frac{1 - x^{2^{k+1}}}{1 - x^{2^k}} = \frac{1 - x^{2^{n+1}}}{1 - x},$$
and
$$\frac{1 - x^{2^{n+1}}}{1 - x} = 1 + x + x^2 + \cdots + x^{2^{n+1} - 1}.$$
Thus
$$\prod_{k = 0}^n (1 + x^{2^k}) = 1 + x + x^2 + \cdots + x^{2^{n+1} - 1}.$$
|
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Evaluate $\int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta$
Evaluate the following \begin{align*} \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \mathrm d\rho \mathrm d\phi \mathrm d\theta \end{align*}
Attempt at solution: We have \begin{align*} 5 \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) \ d\phi \Big[\frac{8}{3} - \frac{\sec^3 \phi}{3} \Big] \\ = \frac{40}{3} \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) \ d\phi - \int_{0}^{\pi/3} \sin(\phi) \sec^3 (\phi) \ d\phi \\ = \frac{40}{3} \int_{0}^{\pi} d\theta \big[-\cos(\pi/3) +1 \big] - \int_{0}^{\pi/3} \tan(\phi) \sec^2 (\phi) d\phi \end{align*}
The expression to the left of the minus sign becomes $\frac{40}{3}\int_{0}^{\pi} d\theta \frac{1}{2} = \frac{40 \pi}{6}$. For the expression on the right side we let $\tan (\phi) = u$. If we adapt the integration bounds, that integral then simplifies to $\int_{0}^{\sqrt3} u \ du$., which equals $3/2$.
So the final answer is $\frac{40 \pi}{6} - 3/2$. This is what I answered, and the online learning platform (where I encountered this integral) said it was wrong. So where did I go wrong?
Any help would be appreciated.
Edit: I'll try again from second step. \begin{align*} \frac{5}{3} \Big[ \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} 8 \sin(\phi) d\phi - \int_{0}^{\pi/3} \sin(\phi) \sec^3 (\phi) \Big] \\ =\frac{40}{3} \int_{0}^{\pi} d\theta \int_{0}^{\pi/3} \sin(\phi) d\phi - \frac{5}{3} \int_{0}^{\pi/3} \tan(\phi) \sec^2 (\phi) d\phi \\ = \frac{40 \pi}{6} - (\frac{5}{3} \cdot \frac{3}{2}) = \frac{40 \pi - 15}{6}
\end{align*}
|
$\displaystyle \int_{0}^{\pi} \int_{0}^{\pi/3} \int_{\sec \phi}^{2} 5\rho^2 \sin(\phi) \ d\rho \ d\phi \ d\theta=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8-\sec^3\phi)\sin\phi\;d\phi d\theta$
$\displaystyle=\frac{5}{3}\int_0^{\pi}\int_0^{\pi/3}(8\sin\phi-\tan\phi\sec^2\phi)\;d\phi d\theta=\frac{5}{3}\pi\left[-8\cos\phi-\frac{1}{2}\tan^2\phi\right]_0^{\pi/3}$.
(Now use the values you have above to finish evaluating.)
|
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|
Proving that $ \int_{0}^{\pi/2} \frac{\mathrm{d}{x}}{\sqrt{a^{2} {\cos^{2}}(x) + b^{2} {\sin^{2}}(x)}} = \frac{\pi}{2 \cdot \text{AGM}(a,b)} $. I know Neumann’s solution of this famous definite integral that is totally based on substitution, but is there any solution using complex analysis? Assuming that $ a > b $, show that
$$
\int_{0}^{\pi/2} \frac{\mathrm{d}{x}}{\sqrt{a^{2} {\cos^{2}}(x) + b^{2} {\sin^{2}}(x)}}
= \frac{\pi}{2 \cdot \text{AGM}(a,b)},
$$
where $ \text{AGM}(a,b) $ is the arithmetic-geometric mean of $ a $ and $ b $, which satisfies the following functional equation:
$$
\text{AGM}(a,b) = \text{AGM} \! \left( \frac{a + b}{2},\sqrt{a b} \right).
$$
|
Let
$$I(a,b)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{a^2\sin^2 x+b^2\cos^2 x}}=\int_{0}^{1}\frac{dt}{\sqrt{(1-t^2)(a^2 t^2 + b^2(1-t^2))}}.\tag{1}$$
Obviously $I(a,a)=\frac{\pi}{2a}$, hence in order to prove our identity we just need to prove that:
$$ I(a,b)=I\left(\sqrt{ab},\frac{a+b}{2}\right).\tag{2}$$
By replacing $b\tan x$ with $z$, we have:
$$ I(a,b) = \int_{0}^{\infty}\frac{dz}{\sqrt{(z^2+a^2)(z^2+b^2)}}.\tag{3}$$
By Lagrange's identity for multiplying the sums of two squares,
$$ (z^2+a^2)(z^2+b^2) = (z^2-ab)^2+(a+b)^2 z^2, \tag{4} $$
hence, with the substitution $z=\frac{1}{2}\left(u-\frac{ab}{u}\right)$, $(3)$ turns into
$$ \int_{0}^{\infty}\frac{du}{\sqrt{\left(u^2+\frac{1}{4}(a+b)^2\right)(u^2+ab)}}=I\left(\sqrt{ab},\frac{a+b}{2}\right),$$
so we're done, since $I(a,b)$ is a continuous function with respect to its parameters and the iterates of the map $\varphi:(a,b)\to\left(\sqrt{ab},\frac{a+b}{2}\right)$ converge towards $\left(\text{AGM}(a,b),\text{AGM}(a,b)\right)$ giving:
$$ I(a,b)=\frac{\pi}{2\cdot\text{AGM}(a,b)} $$
as wanted.
|
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|
Show the series is convergent and find the limit Sequence of real numbers $a_n$ defined recursively with $a_1=1/2$ and $a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$ for all $n \geq 1$.
Show that $\sum_{n=1}^\infty a_n$ is convergent and find its limit.
I have tried to convert the recursive form to explicit form but it's too difficult.
|
Define the following sequence $b_n = \frac{1}{a_n}$. Then
$$b_{n+1} =
\frac{a_n^2-a_n+1}{a_n^2} = 1 - \frac{1}{a_n}+ \frac{1}{a_n^2} = b_n^2-b_n+1
$$
so the sequence $b_n$ is an increasing sequence of natural numbers whose first terms are
$$2, \ 3, \ 7, \ 43, \ 1807, \ \dots$$
Partial sums have a curious pattern: if you denote $s_n = a_1 + \cdots + a_n$ you have
*
*$s_1 = \frac{1}{2} = 1 - \frac{1}{b_1} = 1- \frac{1}{b_2-1}$
*$s_2 = \frac{5}{6} = 1- \frac{1}{b_1b_2} = 1- \frac{1}{b_3-1}$
*$s_3 = \frac{41}{42} = 1- \frac{1}{b_1b_2b_3} = 1- \frac{1}{b_4-1}$
*$s_4 = \frac{1805}{1806} = 1- \frac{1}{b_1b_2b_3b_4} = 1- \frac{1}{b_5-1}$
So it seems that the sum of the series is $1$ and the partial sums are $s_n=1- a_1 \cdots a_n$
EDIT: Wikipedia says that this is called Sylvester Sequence. See http://en.wikipedia.org/wiki/Sylvester%27s_sequence
|
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|
Bounds for $\log(1-x)$ I would like to show the following
$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequality on the left, the difference is $\frac{x^2}{2}-\left(\frac{x^3}{3}+\frac{x^4}{4} + \cdots\right)$. How do I show this is nonnegative?
|
$\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \frac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \frac{x^k}{k}\\
&=x+x^2\sum_{k=2}^{\infty} \frac{x^{k-2}}{k}\\
&=x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{k+2}\\
&\le x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{2}\\
&\le x+\frac{x^2}{2}\sum_{k=0}^{\infty} x^{k}\\
&= x+\frac{x^2}{2}\frac1{1-x}\\
&= x+\frac{x^2}{2(1-x)}\\
&\le x+x^2 \quad \text{if $0 \le x \le \frac12$}\\
\end{array}
$
|
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|
quadratic Gauss sum over a power of 2 Is there a general formula for the generalized quadratic Gauss sum defined by
$$
G(a,b,c)=\frac{1}{c}\sum_{n=0}^{c-1}e\left(\frac{an^2+bn}{c}\right)
$$
where $e(x)=\exp(2\pi ix)$ and $c$ is a power of 2?
|
You could probably use the generalized quadratic Gauss sum reciprocity formula, defined for integers $a, b, c$ satisfying $ac \neq 0$ and $ac+b$ even. In this case, we set $\displaystyle S(a,b,c) = \sum_{x=0}^{|c|-1} e\left(\frac{ax^2+bx}{2c}\right)$, and the theorem states
\begin{eqnarray*}
S(a,b,c) = \left|\frac{c}a\right| e\left(\frac{sgn(ac)-(b^2/ac)}8\right)S(-c,-b,a).
\end{eqnarray*}
See Gauss and Jacobi Sums by Berndt, Evans and Williams.
Observe that we can complete the square $ax^2+bx = a(x+\frac{b}{2a})^2-\frac{b^2}{4a}$, so if $2 \mid b$ and $a$ is odd we have $ax^2+bx \equiv a(x+a^{-1}\frac{b}2)^2 - a^{-1}(\frac{b}2)^2 \mod c$. Under these assumptions, we have that
\begin{align*}
G(a,b,c) &= \sum_{x=0}^{|c|-1} e\left(\frac{a(x+a^{-1}\frac{b}2)^2}c\right) e \left(\frac{- a^{-1}(\frac{b}2)^2}c\right)\\
&= e \left(\frac{- a^{-1}(\frac{b}2)^2}c\right) G(a;|c|)
\end{align*}
where $G(a;|c|) = (1+\imath^a)\left(\frac{|c|}a\right) \sqrt{|c|}$ is the quadratic Gauss sum when $c = 2^r$ for some integer $r > 1$.
|
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|
Find the derivative of $\arccos\frac{b+a\cos x}{a+b\cos x}$ Find the derivative of $\arccos\dfrac{b+a\cos x}{a+b\cos x}$
is there a smart way to find this derivative
i tried by the conventional chain rule way, and it got very complicated
|
It is not too complicated if one goes through the chain rule of derivatives.
$$\begin{split}\arccos \left( {\frac{{b + a\cos x}}{{a + b\cos x}}} \right) &= \arccos \left( u \right) = F\left( {u\left( x \right)} \right)\\
u &= \frac{{b + a\cos x}}{{a + b\cos x}}\\
\frac{{dF}}{{dx}} &= \frac{{dF}}{{du}}\frac{{du}}{{dx}}
\end{split}
$$
$$\frac{{dF}}{{du}} = - \frac{1}{{\sqrt {1 - {u^2}} }} = - \frac{1}{{\sqrt {1 - {{\left( {\frac{{b + a\cos x}}{{a + b\cos x}}} \right)}^2}} }}
$$
$$
\begin{split}
\frac{{du}}{{dx}} &= \frac{{\left( { - a\sin x} \right)\left( {a + b\cos x} \right) - \left( {b + a\cos x} \right)\left( { - b\sin x} \right)}}{{{{\left( {a + b\cos x} \right)}^2}}}\\
&= \frac{1}{{{{\left( {a + b\cos x} \right)}^2}}}\left[ { - {a^2}\sin x - ab\sin x\cos x + {b^2}sinx + ab\sin x\cos x} \right]\\ &= \frac{1}{{{{\left( {a + b\cos x} \right)}^2}}}\left[ { - {a^2}\sin x + {b^2}sinx} \right]\\
&= \frac{{\left( {{b^2} - {a^2}} \right)\sin x}}{{{{\left( {a + b\cos x} \right)}^2}}}\end{split}
$$
$$
\begin{split}
\frac{{dF}}{{dx}} = \frac{{dF}}{{du}}\frac{{du}}{{dx}} &= - \frac{1}{{\sqrt {1 - {{\left( {\frac{{b + a\cos x}}{{a + b\cos x}}} \right)}^2}} }}\frac{{\left( {{b^2} - {a^2}} \right)\sin x}}{{{{\left( {a + b\cos x} \right)}^2}}}\\ &= - \frac{1}{{\sqrt {{{\left( {a + b\cos x} \right)}^2} - {{\left( {b + a\cos x} \right)}^2}} }}\frac{{\left( {{b^2} - {a^2}} \right)\sin x}}{{\left|{a + b\cos x}\right|}} \\&= {\mathop{\rm sgn}} \left( {\sin x} \right)\frac{{\sqrt {{a^2} - {b^2}} }}{{\left| {a + b\cos x} \right|}}
\end{split}
$$
|
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How to prove that $x^9-9x^7+27x^5-30x^3+9x-1$ is irreducible in $\mathbb{Q}[x]$? The problem says:
Given the irreducible polynomial $x^3-3x-1$ with root $2\cos(\pi/9)$, prove that $2\cos(\pi/27)$ is a root of a monic irreducible polynomial of degree 9 over $\mathbb{Q}$, and hence $[\mathbb{Q}(2\cos(\pi/27)):\mathbb{Q}]=9.$
To solve it, I substituted the equality $2\cos(\pi/9)=8\cos^3(\pi/27)-6\cos(\pi/27)$ on the equation $x^3-3x-1=0$ to obtain that $2\cos(\pi/27)$ is a root of the monic polynomial $x^9-9x^7+27x^5-30x^3+9x-1$. But now, how I prove that $x^9-9x^7+27x^5-30x^3+9x-1$ is irreducible over $\mathbb{Q}$?
|
The computation is a little horrendous (I used WolframAlpha), but you can show that
$$(x+1)^9-9(x+1)^7+27(x+1)^5-30(x+1)^3+9(x+1)-1$$
expands to be
$$x^9+9 x^8+27 x^7+21 x^6-36 x^5-54 x^4+9 x^3+27 x^2-3$$
which is irreducible by Eisenstein's criterion with $p=3$, which means that
$$x^9-9x^7+27x^5-30x^3+9x-1$$
is also irreducible.
EDIT:
As Michalis points out, we can avoid the computation just by considering the polynomial modulo $3$, and then considering the constant coefficient modulo $9$. Since we have coefficients which are multiples of $3$, $P(x+1)\pmod{3}$ becomes
$$(x+1)^9-1\pmod3\equiv x^9+1-1\pmod3\equiv x^9\pmod3$$
Hence all the coefficients of the expanded polynomial are divisible by $3$ except for the first. Now we consider the constant coefficient modulo $9$: This is just
$$1-9+27-30+9-1\pmod9=1-0+0-3+0-1\pmod9\equiv-3\pmod9$$
So $9$ doesn't divide the final coefficient, and we have satisfied all the conditions necessary to use Eisenstein's criterion as above.
|
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|
sums of squares and Pythagorean Triples I'm told that if $r^2 + s^2 = z^2$ then $(r+s)^2 + (r-s)^2 = 2z^2$, which is obvious. But i'm trying to show that every integer solution to $x^2 + y^2 = 2z^2$ arises this way from a Pythagorean Triple (r,s,z) and then go on to find the general solution of $x^2 + y^2 = 2z^2$ where $gcd(x,y,z)=1$.
I started by saying that $a,b \in \mathbb{Z}$ was an integer to solution to $x^2 + y^2 = 2z^2$, then $a^2 + b^2 = 2z^2$ and then trying to show that there exist some $a_1 , b_1$ such that $a = a_1 + b_1$, $b=a_1 - b_1$, but i haven't seem to have gotten anywhere. I find myself going in circles. Any tips?
|
From $a^2+b^2=2z^2,$ assuming $a\ge b,$ if you want $a=a_1+b_1,\ b=a_1-b_1$ then $a_1=(a+b)/2$ and $b_1=(a-b)/2.$ Here you need then to require that $a,b$ have the same parity to be able to do this manipulation.
Note: If one already has $a^2+b^2=2z^2$ then the requirement mentioned above, that the parity of $a,b$ are the same, is implied (an odd square plus an even square would be odd, not $2z^2.$)
An example: $1^2+7^2=2\cdot 5^2,$ and then $(7+1)/2=4,\ (7-1)/2)=3.$ This then goes with $4^2+3^2=5^2.$ So I can see your method will generate Pythagorean triples. SOme care may need to be taken to make sure one ends up with primitive triples.
Added explanation: Starting from $$x^2+y^2=2z^2\tag{*}$$ in which we may assume $\gcd(x,y,z)=1$ and $x \ge y$ for definiteness, we have $x,y$ odd and so may define $r=(x+y)/2,\ s=(x-y)/2.$ Then $$r^2+s^2=(1/2)(x^2+y^2)=(1/2)(2z^2)=z^2.$$
So we do have the solution to $(*)$ arising from a Pythagorean triple $(r,s,z)$ as required.
|
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writing $pq$ as a sum of squares for primes $p,q$ Let $p$ and $q$ be distinct primes congruent to $1$ mod $4$. How many ways are there to write $pq$ as a sum of squares?
I know that any prime $p\equiv 1\pmod 4$ can be written uniquely as a sum of squares and $pq\equiv 1\pmod 4$ but $pq$ is not a prime so I'm stuck.
|
Since $n$ can be written as a sum of two squares in a number of ways that depends on how many ways there are to split $n$ as:
$$ n = z\bar{z} = (a-bi)(a+bi) $$
over $\mathbb{Z}[i]$, and the latter is an Euclidean domain hence a UFD, it follows that for any $n\in\mathbb{N}$:
$$\#\{(a,b)\in\mathbb{Z}^2:a^2+b^2= n\} = 4(\chi_4*1)(n) = 4\left(d_1(n)-d_3(n)\right)$$
where $d_1(n)$ is the number of divisors of $n$ of the form $4k+1$ and $d_1(n)$ is the number of divisors of $n$ of the form $4k+3$. If $p$ and $q$ are primes of the form $4k+1$, it follows that:
$$\#\{(a,b)\in\mathbb{N}^2:a^2+b^2=pq\} = \color{red}{4}.$$
These representations can be recovered from Lagrange's identity:
$$ (a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ac+bd)^2.$$
For instance, let we assume $p=13$ and $q=41$. Then:
$$ p = 2^2 + 3^2,\qquad q=4^2+5^2$$
and $pq=13\cdot 41=533$ can be written as a sum of two squares in the following ways:
$$ pq = 2^2 + 23^2 = 7^2 + 22^2$$
since $2\cdot 4+3\cdot 5=23$ and $2\cdot 5+3\cdot 4=22$.
|
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|
$n^2(n^2-1)(n^2-4)$ is always divisible by 360 $(n>2,n\in \mathbb{N})$ How does one prove that $n^2(n^2-1)(n^2-4)$ is always divisible by 360? $(n>2,n\in \mathbb{N})$
I explain my own way:
You can factorize it and get $n^2(n-1)(n+1)(n-2)(n+2)$.
Then change the condition $(n>2,n\in \mathbb{N})$ into $(n>0,n\in \mathbb{N})$ that is actually equal to $(n\in \mathbb{N})$.
Now the statement changes into :
$$n(n+1)(n+2)^2(n+3)(n+4)$$
Then I factorized 360 and got $3^2 \cdot 2^3 \cdot 5$.
I don't know how to prove the expression is equal to $3^2 \cdot 2^3 \cdot5$.
Who can help me solve it?
|
$$n^2(n^2-1)(n^2-4)=(n-2)(n-1)n(n+1)(n+2)\cdot n \\
=(n-2)(n-1)n(n+1)(n+2)(n+3)-3(n-2)(n-1)n(n+1)(n+2)$$
And
$$(n-2)(n-1)n(n+1)(n+2)(n+3)=720 \cdot \binom{n+3}{6}$$
$$3(n-2)(n-1)n(n+1)(n+2)=3\cdot 120 \binom{n+2}{5}=360 \binom{n+2}{5}$$
|
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|
$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit solved using only rules of algebra of limits.
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do not fully address my question.
A challenging limit problem for the level of student who knows that:
$$\begin{align*}
\lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\
\lim\limits_{x\to -\infty} e^x&=0\tag2\\
\lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\
\lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\
\lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5
\end{align*}$$
|
$$\left(\frac{e^x-1}x\right)^2=\frac{e^{2x}-2e^x+1}{x^2}=2^2\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}.$$
Then, taking the limit,
$$1=4L-2L.$$
UPDATE: the same approach can be used for the next order,
$$\left(\frac{e^x-1}x\right)^3=\frac{e^{3x}-3e^{2x}+3e^x-1}{x^3}=3^3f(3x)-3\cdot2^3(2x)+3f(x),$$
where $f(x)=\dfrac{e^x-1-x-\dfrac{x^2}2}{x^3},$
and
$$1=27L-24L+3L.$$
|
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|
Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
|
A quick check shows that $x = 1$ is a root of $x^4 - 4x + 3$. By synthetic division we find $x^4 - 4x + 3 = (x - 1)(x^3 + x^2 + x - 3)$. By similar reasoning, $x^3 + x^2 + x - 3 = (x - 1)(x^2 + 2x + 3)$. So $x^4 - 4x + 3 = (x - 1)^2(x^2 + 2x + 3)$. Now you have the setup to do partial fraction decomposition. The result is
$$\frac{-2x^2 + 6x + 8}{x^4 - 4x + 3} = \frac{2}{(x - 1)^2} - \frac{1}{x - 1} + \frac{x - 1}{x^2 + 2x + 3}.$$
So
\begin{align}
\int \frac{-2x^2 + 6x + 8}{x^4 - 4x + 3}\, dx &= \int \frac{2\, dx}{(x - 1)^2} - \int \frac{dx}{x - 1} + \int \frac{x - 1}{x^2 + 2x + 3}\, dx\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \int \frac{x - 1}{(x + 1)^2 + 2}\, dx\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \int \frac{u - 2}{u^2 + 2}\, du \quad (u = x + 1)\\
&= - \frac{2}{x - 1} - \ln|x - 1| + \int \frac{u}{u^2 + 2} - \int \frac{2}{u^2 + 2}\, du\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \frac{1}{2}\ln|u^2 + 2| - \frac{2}{\sqrt{2}}\arctan \frac{u}{\sqrt{2}} + C\\
&= -\frac{2}{x - 1} - \ln|x - 1| + \frac{1}{2}\ln[(x + 1)^2 + 2] - \sqrt{2}\arctan \frac{x+1}{\sqrt{2}} + C.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to solve equations of the form $(a+x)^2 + (b+x)^2 +(c+x)^2 + (d+x)^2-e=0$ I have an equation of the form $(a+x)^2 + (b+x)^2 +(c+x)^2 + (d+x)^2-e=0$ where $a$, $b$, $c$, $d$ and $e$ are known.
My question is how would I derive $x$.
|
HINT:
$(a+x)^2+(b+x)^2+(c+x)^2+(d+x)^2-e=0\implies$
$\color{red}{4}x^2+\color{green}{2(a+b+c+d)}x+\color{blue}{a^2+b^2+c^2+d^2-e}=0\implies$
$x_{1,2}=\frac{-\color{green}{2(a+b+c+d)}\pm\sqrt{(\color{green}{2(a+b+c+d)})^2-4\cdot\color{red}{4}\cdot(\color{blue}{a^2+b^2+c^2+d^2-e})}}{2\cdot\color{red}{4}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1183480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
A sum involving a ratio of two binomial factors. Let $a\ge 0$, $a_1\ge 0$ ,$b \ge 0$ and $b_1\ge0$ be real numbers subject to $1+b+a_1-b_1-a >0$. Let $m$ be a positive integer. Then using methods similar to those in Another sum involving binomial coefficients. I have shown that the following identity holds:
\begin{eqnarray}
\sum\limits_{i=0}^{m-1}\frac{\binom{i+a}{b}}{\binom{i+a_1}{b_1}} &=&
\frac{\binom{a+m}{b+1}}{\binom{a_1+m}{b_1}}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+ a +m \\ 2+b & 1+a_1+m \end{array}; 1
\right]\\
&-&
\frac{\binom{a}{b+1}}{\binom{a_1}{b_1}}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+ a \\ 2+b & 1+a_1 \end{array}; 1
\right]
\end{eqnarray}
Now, what is the asymptotic behaviour of this sum when $m \rightarrow \infty$ ?
|
We need to analyze a certain rational function of $m$. We decompose that rational function into simple fractions and we have:
\begin{equation}
\frac{(1+a+m)^{(n)}}{(1+a_1+m)^{(n)}} = 1 + \sum\limits_{l=1}^n \frac{{\mathcal A}_l}{m+a_1+l}
\end{equation}
where ${\mathcal A}_l = (a-a_1-l+1)^{(n)} (-1)^{l-1}/((l-1)!(n-l)!)$. Therefore we have:
\begin{equation}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=
F_{2,1} \left[
\begin{array}{rr} 1 & b_1 \\ 2+b\end{array};1
\right]
+
\sum\limits_{l=1}^\infty \frac{(-1)^{l-1}}{(l-1)!}
\left.
\frac{d^l}{d x^l}
F_{2,1} \left[
\begin{array}{rr} b_1 & a-\theta-l+1 \\ 2+b\end{array};x
\right]
\right|_{x=1} \cdot \frac{1}{m+\theta+l}
\end{equation}
We use the ``generalized Gauss' theorem'':
\begin{equation}
\left.
\frac{d^l}{d x^l}
F_{2,1}\left[
\begin{array}{rr} a & b \\ c \end{array};x
\right]
\right|_{x=1} = \frac{a^{(l)} b^{(l)} (-1)^l}{(1+a+b-c)^{(l)}} \cdot \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
\end{equation}
and we easily get:
\begin{eqnarray}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=\\
\frac{1+b}{1+b-b_1} +
\frac{(a-a_1) b_1}{1+a_1+m} \frac{\Gamma(2+b) \Gamma(1-a+a_1+b-b_1)}{\Gamma(2+b-b_1) \Gamma(2-a+a_1+b) } \cdot \\
F_{3,2}\left[
\begin{array}{rrr} 1-a+a_1 & 1+b_1 & 1+a_1+m \\ 2-a+a_1+b & 2+a_1+m\end{array};1
\right]
\end{eqnarray}
The first term on the right hand side is the large $m$ limit and the second term is of the order of $O(1/m)$ . Repeating the above procedure $p$ times we obtain a following large-$m$ expansion of the hypergeometric function:
\begin{eqnarray}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=\\
\frac{1+b}{1+b-b_1} +
\sum\limits_{l=1}^p \frac{(a-a_1)_{(l)} b_1^{(l)} (1+b)}{(1+a_1+m)^{(l)} (1+b-b_1)_{(l+1)}} +
\frac{(a-a_1)_{(p+1)} b_1^{(p+1)}}{(1+a_1+m)^{(p+1)}}
\frac{\Gamma(2+b) \Gamma(1-a+a_1+b-b_1)}{\Gamma(2+b-b_1) \Gamma(p+2-a+a_1+b)}\cdot \\
F_{3,2}\left[
\begin{array}{rrr} p+1-a+a_1 & p+1+b_1 & p+1+a_1+m\\ p+2-a+a_1+b & p+2+a_1+m\end{array}; 1
\right]
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How am I supposed to know that $\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$? I'm currently reading through the solution to a problem that involves finding generating functions. In some of the intermediary steps, it is written that
$$\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$$
without any justification.
Is there some kind of reasoning behind this or is this just one of those things I have to accept?
|
$$\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots\,\,\,\,\,\forall|x|\lt1$$ Then $$\dfrac{1}{(1-x)^2}=(1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots)$$ $x^k$ in the expansion of right hand side is produced by $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Hence $$\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\cdots.$$ Similarly we can find $\dfrac{1}{(1-x)^3}$ and so on.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How many bags would need to be bought to have all the red noses? There are 9 types of Red Noses, for comic relief this year:
Each one is sold in an opaque packet, so it is lucky dip which one you will get.
Assuming there is the same amount of each type (${1\over9}$th). On average, how many bags would need to be bought for someone to get all 9 of them?
|
I'm going assume that there are, effectively, infinite bags of noses—if we only had 9 bags... well, you get it. ;)
Say we have $n-1$ of the $9$ unique noses. The probability of getting a new style of nose on the next bag is $p_n = 1 - \frac{n-1}{9}$. Then, on average, it takes $\frac{1}{p_n}$ bags to get the next unique nose. (i.e. - Say we have $0$ unique noses. Then, $n = 1$ and $p_n = 1 - \frac{0}{1} = 1$ and it will take only one bag to get a new nose. Similarly, if we have $4$ nose-styles, $n = 5$ and it will take $1 - \frac{4}{9} = \frac{9}{5}$ bags to get a $5^{\text{th}}$ nose.)
The probabilities here are independant. That is, the chance of getting the $5^{\text{th}}$ new nose on the next bag doesn't depend on the chance of getting the $4^{\text{th}}$ one (once you have the $4^{\text{th}}$ nose, the probability of getting the $5^{\text{th}}$ will always be $\frac{1}{p_5}$). Thus, we can just sum the average of each $p_n$ until $n = 9$.
So, we have:
$$\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \frac{1}{p_4} + \frac{1}{p_5} + \frac{1}{p_6} + \frac{1}{p_7} + \frac{1}{p_8} + \frac{1}{p_9} + = \\
1 + \frac{9}{8} + \frac{9}{7} + \frac{9}{6} + \frac{9}{5} + \frac{9}{4} + \frac{9}{3} + \frac{9}{2} + \frac{9}{1} = \\
9 \sum_{n=1}^{9} \frac{1}{n} = \frac{9 \times 7129}{2520} = 25.46$$
That is, on average, it'd take about 25 bags to get all nine noses.
|
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|
Find trig derivative of $y=4x(7x+\cot{7x})^6$ Find trig derivative of $y=4x(7x+\cot{7x})^6$.
I got $y'= 4(7x+\cot{7x})^6 + 168x(7x+\cot{7x})^5 (\cot^2 {7x})$ but I'm not sure I did it right.
Your help is appreciated (:
|
use the logarithmic derivative. here is how it works.
let $$y = 4x(7x + \cot 7x)^6, \ln y = \ln 4 + \ln x + 6 \ln(7x \sin 7x + \cos 7x) - 6 \ln \sin 7x $$ taking the derivative, we get
$$\begin{align}\frac 1y\frac{dy}{dx} &=\frac 1x+6\left(\frac{7\sin 7x+49x\cos7 x-7\sin7 x}{7x\sin 7x + \cos 7x}- \frac{7\cos 7x}{\sin 7x}\right) \\
&=\frac 1x+42\left(\frac{(\sin 7x+7x\cos 7x-\sin 7x)\sin 7x - \cos 7x(7x \sin 7x+\cos 7x)}{7x\sin 7x + \cos 7x}\right)\\
&=\frac 1x-\frac{ 42\cos^2 7x}{7x\sin 7x + \cos 7x}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1191475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding the sum of $\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}$
Find the sum of the series and for which values of $x$ does it converge:
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n} $$
My attempt:
$$\begin{align}&S_n=(x+1)^2+(x+1)^3/3+...+(x+1)^{n+1}/3^{n} \\
-(x+1)^2&S_n=-(x+1)^3/3-...-(x+1)^{n+2}/3^{n} \\
&S_n=\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2} \end{align}$$
And for $-4<x<2$ we have:
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2}=\frac{(x+1)^2}{1-(x+1)^2}$$
Which is the sum of the series for $-4<x<2$.
The sum looks a bit odd, is that alright?
Note: no integrals or Taylor or Zeta.
|
The series is of the form
\begin{align}
f(t) = \sum_{n=0}^{\infty} t^{n} = \frac{1}{1-t}.
\end{align}
When $t = (x+1)/3$ the value becomes
\begin{align}
f\left(\frac{x+1}{3}\right) = \sum_{n=0}^{\infty} \frac{(x+1)^{n}}{3^{n}} = \frac{3}{2-x}.
\end{align}
Now multiply by $(x+1)^{2}$ to obtain
\begin{align}
\sum_{n=0}^{\infty} \frac{(x+1)^{n+2}}{3^{n}} = \frac{3(x+1)^{2}}{2-x}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
|
$$ \lim \limits_{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1}=\lim \limits_{x \to \infty }\dfrac{(\sqrt{x^2 +x} - \sqrt{x^2 +1})(\sqrt{x^2 +x} + \sqrt{x^2 +1})}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{x-1}{\sqrt{x^2+x}+\sqrt{x^2+1}}=\lim \limits_{x \to \infty } \dfrac{1-\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x^2}}}=\dfrac{1}{2} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Why General Leibniz rule and Newton's Binomial are so similar? The binomial expansion:
$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$
The General Leibniz rule (used as a generalization of the product rule for derivatives):
$$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$
Both formulas can be obtained simply by induction; Newton's binomial also has a combinatorial proof (here's the relevant wikipedia page).
It's striking how these formulas are similar; is there a possible connection between them?
I was thinking that maybe the General Leibniz rule could be obtained using a combinatorial argument as well (hence the binomial coefficients)...
|
Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....
$$
Taylor series expansion of $f(x+h)g(x+h)$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....
$$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....)
$$
If you order $h^n$, it will give you binom expansion
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$
Because it has exactly the same coefficient of
$$
=(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)}
$$
$$
e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+....
$$
If you order $h^n$ of $e^{hx}e^{hy}$ and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.
$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n \frac{x^i y^{n-i}}{i! (n-i)!} $$
$$(x+y)^n=\sum_{i=0}^n \frac{n! x^i y^{n-i}}{i! (n-i)!} $$
$$(x+y)^n=\sum_{i=0}^n {n \choose i} x^i y^{n-i} $$
Thus it is also true
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$
|
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|
calculating a limit of sequence Can some one help to show that for $a>0$
$$ \lim_{n\to \infty} \frac{a^\frac{1}{n}}{n+1}+\frac{a^\frac{2}{n}}{n+\frac{1}{2}}+...+\frac{a^\frac{n}{n}}{n+\frac{1}{n}}=\frac{a-1}{\log a}$$
My tries : I was not able to use riemann sum for calculating.
For the special case $a=1$ the RHS has limit $1$ and for the LHS for $a=1$ we can write
$$\frac{n}{n+1} \le \frac{1}{n+1}+\frac{1}{n+\frac{1}{2}}+...+\frac{1}{n+\frac{1}{n}} \le \frac{n}{n+\frac{1}{n}} $$ so the limit of LHS is $1$ . For arbitrary $a$ I have failed to answer.
|
Let
$$S(n) = \frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + \frac{1}{2}} + \cdots + \frac{a^{\frac{n}{n}}}{n + \frac{1}{n}}.$$
Then
$$\frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + 1} + \cdots + \frac{a^{\frac{n}{n}}}{n + 1} < S(n) < \frac{a^{\frac{1}{n}}}{n} + \frac{a^{\frac{2}{n}}}{n} +\cdots + \frac{a^{\frac{n}{n}}}{n}$$
that is,
$$\frac{n}{n+1}\cdot \frac{a^{\frac{1}{n}} + a^{\frac{2}{n}} + \cdots + a^{\frac{n}{n}}}{n} < S(n) < \frac{a^{\frac{1}{n}} + a^{\frac{2}{n}}+\cdots + a^{\frac{n}{n}}}{n}. \tag{*}$$
The left- and right-most sides of $(*)$ converge to
$$\int_0^1 a^x\, dx = \frac{a^x}{\log a}\bigg|_{x = 0}^1 = \frac{a - 1}{\log a}.$$
Hence, by the squeeze theorem,
$$\lim_{n\to \infty} S(n) = \frac{a - 1}{\log a}.$$
|
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|
Find a formula for $\sin(3a)$ and use to calculate $\sin(π/3)$ and $\cos(π/3)$? Problem:
Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$.
My attempt:
\begin{align}
\sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\
&= \sin(a + a)\cos(a) + \cos(a + a)\sin(a) \\
&= [\sin(a)\cos(a) + \cos(a)\sin(a)]\cos(a) + [\cos(a)\cos(a) - \sin(a)\sin(a)]\sin(a).
\end{align}
It can then be simplified to
$$2\sin(a)\cos^2(a) + \sin(a)\cos^2(a) - \sin^3(a) = 3\sin(a)\cos^2(a) - \sin^3(a).$$
My question is this: How am I supposed to use this formula to find $\sin(π/3)$ and $\cos(π/3)$?
|
$$\sin 3a = \sin (2a+a) = \sin(2a)\cos(a)+ \cos(2a)\sin(a)= 2\sin(a)\cos^2(a)+\sin(a)(1-2\sin^2(a))=2\sin(a)(1-\sin^2(a))+\sin(a)(1-2\sin^2(a)) =3\sin(a)-4\sin^3(a) $$
So...
$$\sin(\pi) = 0 = 3\sin(\dfrac{\pi}{3})-4\sin^3(\dfrac{\pi}{3})\implies \dfrac{3}{4} =\sin^2(\dfrac{\pi}{3})\implies \sin(\dfrac{\pi}{3})=\dfrac{\sqrt{3}}{2}$$
Using $cos^2(a)+\sin^2(a)=1$
$$\cos^2(\dfrac{\pi}{3})+\dfrac{3}{4} = 1\implies \cos^2(\dfrac{\pi}{3}) =\dfrac{1}{4}\implies \cos(\dfrac{\pi}{3}) =\dfrac{1}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help integrate $\frac{1}{x^3+x^8}$ Could you help me to integrate
$$
\int{\frac{dx}{x^3+x^8}}
$$
I've tried partial fraction decomposition but got $(x^4-x^3+x^2-x+1)$ as the last term when factored the denominator.
Thank you.
|
The best way to get the answer given by Wolfram Alpha is to write:
$$\frac{1}{x^3+x^8} = \frac{1}{x^3} -\frac{x^2}{x^5+1}$$
Then solve:
$$\frac{x^2}{1+x^5}=\frac{a}{x+1} + \frac{bx+c}{x^2-2\cos(\pi/5)x+1}+\frac{dx+e}{x^2-2\cos(3\pi/5)x+1}$$
Since $\sin\pi/5$ and $\cos\pi/5$ are in terms of $\sqrt{5}$, you can rewrite the answer in terms of $\sin(\pi/5),\cos(\pi/5),\sin(3\pi/5),\cos(3\pi/5)$. It still won't be prtty.
|
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"url": "https://math.stackexchange.com/questions/1199047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why Does $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ sum to $ (1-(1-p)^{n+1}) $? I was browsing around when I found this question: Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial.
I understood the solution until I hit this portion where $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ becomes $ (1-(1-p)^{n+1}) $...
Any help would be appreciated.
Thanks!
|
By using $t = k + 1$:
$$ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} = \sum\limits_{t = 1}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1}$$
$$ = \sum\limits_{t = 0}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{n-t +1} - (1 - p)^{n+1} $$
$$=\sum\limits_{t = 0}^{n+1} \begin{pmatrix} n+1 \\ t \end{pmatrix} p^{t} (1-p)^{(n+1)-t } -(1 - p)^{n+1} = (p + 1 - p)^{n+1} - (1-p)^{n+1} = 1 - (1 - p)^{n+1}$$
(Using the binomial theorem)
|
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|
If $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , $\exists 1< aIf $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , then is it true that either $2^n|x-1$ or $2^n|y-1$ ? I have only been able to observe that both $x,y$ are odd . Please help
EDIT : As is seen from an answer : the original claim does not hold ; so I ask does there exist $a>1 , a<n$ such that $2^a|x-1$ or $2^a|y-1$ ?
|
Let $x=2^{a+p}c+1,y=2^ad+1$ where $p\ge0$ and $c,d$ are add
$2^n+1=xy=2^{2a+p}cd+2^{a+p}c+2^ad+1$
$\iff2^n=2^a(2^{a+p}cd+2^pc+d)$
$\iff2^{n-a}-[2^{a+p}cd+2^pc]=d$ which is odd
If $p>0,2^{n-a}-[2^{a+p}cd+2^pc]$ is even
$\implies p=0\implies$ the highest power of $2$ that divides $x-1$ and $y-1$ will be the same
|
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|
Implicit line equation Trying to understand parametric and implicit line equations but I'm at a complete halt now. I have a line that goes through P(0,1) and Q(3,2), I need to find the implicit equation N((x,y)- Z) = 0 such that Z is a point on the line and N is a vector perpendicular to the line. Please point me in the right direction. Thank you.
|
Given the points:
$$\begin{gathered}
P(0,1) \hfill \\
Q(3,2) \hfill \\
\end{gathered}$$
Build the vector pointing from $P$ to $Q$:
$$\overrightarrow {PQ} = Q - P = \left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$$
Set parametrization:
$$\left( {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right) = P + t\overrightarrow {PQ} = \left( {\begin{array}{*{20}{c}}
0 \\
1
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$$
Observe that
$$N = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
3
\end{array}} \right)$$
is orthogonal for $\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$. With:
$$\left( {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
0 \\
1
\end{array}} \right) = t\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$$
we get from dot product:
$$N \cdot \left( {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right) - N \cdot \left( {\begin{array}{*{20}{c}}
0 \\
1
\end{array}} \right) = t \cdot N \cdot \left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right)$$
so:
$$ - x + 3y - 3 = 0$$
Now we set:
$$f(x,y) = - x + 3y - 3$$
and verify:
$$\begin{gathered}
f(P) = - 0 + 3 - 3 = 0 \hfill \\
f(Q) = - 3 + 6 - 3 = 0 \hfill \\
\end{gathered} $$
Function $f$ implicitly defines a line.
|
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|
Proving bounds of a harmonic series Let $p>1$. Prove that the series:
$\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^p}$
is between $\frac {1}{2}$ and 1.
Any help is appreciated.
Just a challenge problem I was presented and curious on the solution
Thanks
|
Since $\sum_{n = 1}^\infty (-1)^{n+1}/n^p$ is absolutely convergent for $p > 1$, we may rearrange the terms without affecting the sum. Now
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^p} = \left(1 - \frac{1}{2^p}\right) + \left(\frac{1}{3^p} - \frac{1}{4^p}\right) + \cdots$$
and each term in parentheses is positive, so the sum of the series is greater than $1 - 1/2^p$, which is greater than $1/2$. Since
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^p} = 1 - \left(\frac{1}{2^p} - \frac{1}{3^p}\right) - \left(\frac{1}{4^p} - \frac{1}{5^p}\right) - \cdots$$
and each term in parentheses is positive, it follows that the sum of the series does not exceed $1$. So $\sum_{n = 1}^\infty (-1)^{n+1}/n^p$ lies between $1/2$ and $1$.
|
{
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|
A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
|
assuming $$x\geq0 \land y\geq0$$
$$\dfrac{\sqrt{x}+\sqrt{y}}{2} \leq \sqrt{\dfrac{x+y}{2}}$$
Squaring:
$$\dfrac{x+y+2\sqrt{xy}}{4} \leq \dfrac{x+y}{2} \implies \dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4}$$
Squaring again
$$\dfrac{\sqrt{xy}}{2} \leq \dfrac{x+y}{4} \implies \dfrac{xy}{4} \leq \dfrac{x^2+2xy+y^2}{16}\implies$$
$$0 \leq \dfrac{x^2-2xy+y^2}{16} \implies 0 \leq \dfrac{(x-y)^2}{4}$$
But we now that $a^2\geq 0 \forall a \in \mathbb{R}$ so it's true...
|
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|
Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6
|
First, show that this is true for $n=1$:
$1^3+11=6\cdot2$
Second, assume that this is true for $n$:
$n^3+11n=6k$
Third, prove that this is true for $n+1$:
$(n+1)^3+11(n+1)=$
$n^3+3n^2+14n+12=$
$\color{red}{n^3+11n}+3n^2+3n+12=$
$\color{red}{6k}+3n^2+3n+12=$
$6k+3n(n+1)+12=$
*
*$2 |n\implies6|3n \implies6|3n(n+1)\implies3n(n+1)=6m$
*$2\not|n\implies2|n+1\implies6|3n(n+1)\implies3n(n+1)=6m$
$6k+6m+12=$
$6(k+m+2)$
Please note that the assumption is used only in the part marked red.
Alternatively, consider the following options:
*
*$n\equiv0\pmod6 \implies n^3+11n\equiv 0+ 0\equiv6\cdot 0\equiv0\pmod6$
*$n\equiv1\pmod6 \implies n^3+11n\equiv 1+11\equiv6\cdot 2\equiv0\pmod6$
*$n\equiv2\pmod6 \implies n^3+11n\equiv 8+22\equiv6\cdot 5\equiv0\pmod6$
*$n\equiv3\pmod6 \implies n^3+11n\equiv 27+33\equiv6\cdot10\equiv0\pmod6$
*$n\equiv4\pmod6 \implies n^3+11n\equiv 64+44\equiv6\cdot18\equiv0\pmod6$
*$n\equiv5\pmod6 \implies n^3+11n\equiv125+55\equiv6\cdot30\equiv0\pmod6$
|
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|
If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to find out relation but that will be long and tedious process , I want shorter approach to this problem . Btw I was given following options for this question
A) $a^2+c^2+ab=0$
B) $a^2-c^2+ab=0$
C) $a^2-c^2-ab=0$
D) $ap^2+bp+c=0$
Maybe we can relate something by looking at options?
|
Just observe that the product of two roots of the quadratic is $1$. So the third root has to be $\frac{-c}{a}$. Now substitute this root instead of $x$ in the cubic.
|
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|
the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$
or, as another example if a = 0.01 .$$\frac{n (n + 0.01) (2 n + 0.01) }{6 \cdot 0.01} = 0.01^2 + 0.02^2 + \cdots + n^2 .$$
Now if we follow the same logic, I suppose if the difference between the closest numbers becomes smallest possible (a = 0.0...1), we will obtain
$$ \frac{n (n + 0.0..1) (2 n + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \ (0.0..1 ^ 2 + 0.0..2 ^ 2 .. + n ^ 2) * {0.0..1}$$ If we take that, a = 0.0...1 and, b = $\frac {n}{0.0...1}$ and we rearrange the formula we will get this:
$$\frac{n ^ 3}{3} = 1^2a^3 + 2^2a^3 + \dots+b^2 a^3$$
following the same logic we prove that the formula holds for each degree (it is easy to check), so that we could generalize, If we take that degree call m, we will get
$$\frac{n ^ m}{m} = 1^{m-1}a^m + 2^{m-1}a^m + \dots+b^{m-1} a^m$$
so the question is this expression can be interpreted as a Riemann sum, and why?
|
The quantity on the right of your last equation can be expressed as
$$ R(n, \epsilon) =
\sum_{k=1}^{n/\epsilon} k^2 \epsilon^3 $$ where $\epsilon = 0.00\ldots 01$ approaches zero.
Yore conclusion is correct in the following sense:
$$\lim_{\epsilon \to 0^+} \frac{R(n, \epsilon)}{n^3/3} = 1 $$
|
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|
Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?
|
The probability we pick an odd number first is $\frac{3}{5}$. Then the probability of another odd number being chosen from the remaining numbers is $\frac{2}{4}=\frac{1}{2}$. Put together, the probability is $\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$.
The probability we pick an even number first is $\frac{2}{5}$. Then the probability of an odd number being chosen from the remaining numbers is $\frac{3}{4}$. Put together, the probability is $\frac{2}{5}\times\frac{3}{4}=\frac{3}{10}$.
The probability that the second number being chosen is odd is the sum of these two probabilities. So $\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$ is the correct answer.
As others have noted, picking randomly could also be thought of as shuffling the numbers like a deck of cards, and then choosing top to bottom. Of course, there's a $\frac{3}{5}$ chance that the second "card" would be an odd number, but it's nice to see that the conditional probability works out as well.
|
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|
Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
|
Although the substitution $y=x+2$ works well, I would first try $y=x+3$, which avoids having to expand the highest power (quartic) term and leads to $$y^4+y^3-2y^2-y+1=0$$This yields two roots $y=\pm 1$ to the rational root theorem. Or alternatively to the observation that $$y^4+y^3-2y^2-y+1=(y^4+y^3-y^2)-(y^2+y-1)=(y^2-1)(y^2+y-1)$$
A second thought would be to spot the even powers $(x+1)^2, (x+3)^4$, and to see that the minimum value of the sum of these for integer $x$ is $2$, which would lead to spotting the root $x=-2$.
|
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How to find all positive integers $m,n$ such that $3^m+4^n$ is a perfect square? How to find all positive integers $m$, $n$ such that $3^m+4^n$ is a perfect square? I have found $m=n=2$ is a solution, but cannot find any other and cannot prove whether there is any other solution or not. Please help. Thanks in advance.
|
Since $3=-1$ modulo $4$ and $3^2=1$ modulo $4$ the sum $3^m+4^n$ can only be a square when $m$ is even. Therefore we want that
$$(3^j)^2+ (2^n)^2$$
is a square. Using the formula for Pythagorean triples we see that necessarily
$$3^j=p^2-q^2,\qquad 2^n=2pq\ .$$
It follows that $p=2^r$, $q=2^s$ with $r\geq s\geq0$ and therefore
$$3^j=2^{2r}-2^{2s}=2^{2s}(2^{2(r-s)}-1)\ .$$
This implies $s=0$ and $3^j=(2^r+1)(2^r-1)$. Here not both factors on the right hand side can be nontrivial powers of $3$; whence $r=1$ and then $j=1$.
It follows that $3^2+4^2=25$ is the only solution to the problem.
|
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Maximum and minimum value of $f(z)= |1+z|+|1-z+z^2|$ Let $f(z)= |1+z|+|1-z+z^2|$. Then what is the maximum and minimum value of $f(z)$, if $|z|=1$, where $z$ is a complex number.
Its very difficult to go through the process when we put $z=x+iy$ in the function and then quantify the maximum value. Is there any other method to find out. Any suggestion would be appreciated.
|
$$\begin{align}z = e^{2it}, f &= |1 + z| + |1-z+z^2|\\ &= |1+z| + \frac{|1 + z^3|}{|1+z|}\\
& = 2\cos t+ \frac{\cos 3t}{\cos t} \\
&= 2|\cos t|+|4\cos^2 t - 3|\\
& = 2|x| + |4x^2-3|, x = \cos t \end{align}$$
we can break $$f(x) = 2|x| + |4x^2 - 3| = \begin{cases} 2x+3-4x^2 & if\, 0 < x < \sqrt 3/2\\ 2x + 4x^2 - 3 & if \, \sqrt 3/2< x < 1\end{cases}$$
$$ \text{the maximum value is } f(1/4) = 3.25 \text{ and the minimum value is } f(\sqrt 3/2) = \sqrt 3.$$
|
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Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2(n-1)+1)(n-1)$=$2n^2-3n+1$
for $k=n$
$a_n$= $\sum\limits_{k=1}^{2n} {(-1)^n \cdot n^2}$+$a_{n-1}$
$a_n$=${(-1)^n \cdot n^2}$+$2n^2-3n+1$
But that last equation seems somehow wrong to get $(2n+1)\cdot n$
|
The equality is true for $n=1$, because
$$
(-1)^1\cdot 1^2+(-1)^2\cdot 2^2=-1+4=3
$$
and
$$
(2\cdot 1+1)\cdot 1=3.
$$
Now suppose the assert is true for $a_{n-1}$. Then
\begin{align}
a_{n}&=\sum_{k=1}^{2n}(-1)^k\cdot k^2\\
&=\biggl(\sum_{k=1}^{2n-2}(-1)^k\cdot k^2\biggr)+
(-1)^{2n-1}\cdot(2n-1)^2+(-1)^{2n}(2n)^2\\
&=a_{n-1}+(-1)^{2n-1}\cdot(2n-1)^2+(-1)^{2n}(2n)^2\\
&=(2(n-1)+1)(n-1)-(2n-1)^2+(2n)^2
\end{align}
(the last equality by the induction hypothesis).
You have to verify that the last expression equals
$$
(2n+1)\cdot n
$$
which is easy algebra.
You seem to have forgotten a summand.
|
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|
Proof that $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$ $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$
my work:
I assumed $m = da$ , $n = db$ for $a,b \in \mathbb{Z}$.
Now, $2^m - 1$ = $2^{da} - 1$ = $(2^d)^a - 1$ = $x^a - 1$ where $x = 2^d$.
similarly
$2^n - 1$ = $x^b - 1$
Now,
using $x^a - 1 = (x - 1)(x^{a-1} + x^{a - 2} + .... + x + 1)$ and $x^b - 1 = (x - 1)(x^{b - 1} + x^{b - 2} + ..... + x + 1)$
clearly $(x - 1)$ is a factor common in both but how to show that
$(x^{a-1} + x^{a - 2} + .... + x + 1)$ and $(x^{b-1} + x^{b - 2} + .... + x + 1)$ can't have a common factor
|
If $a=b$ this means you have $m=n=d$ so this is trivial.
Assume now that $a\neq b$, well you can exchange both so that we can always assume $a>b$ :
$$x^{a-b}(x^{b-1}+...+x+1)=x^{a-1}+...+x^{a-b+1}+x^{a-b} $$
Hence :
$$x^{a-1}+...+x+1-x^{a-b}(x^{b-1}+...+x+1)=x^{a-b-1}+...+x+1 $$
It implies that $\gcd(x^{a-1}+...+x+1,x^{b-1}+...+1)$ divides $\gcd(x^{a-b-1}+...+x+1,x^{b-1}+...+1)$. Furthermore $\gcd(a-b,b)=\gcd(a,b)=1$. This is the heridity of an induction on $N$ :
$$\text{ for all integers } 1\leq a,b\leq N \text{ with } \gcd(a,b)=1 \text{ we have : } $$
$$\gcd(x^{a-1}+...+x+1,x^{b-1}+...+1)=1$$
|
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|
How to factor $\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}$ I'm stuck with the following:
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}$
My idea was/is the following:
$\frac{3^3a^6b^9}{5^3}-\frac{c^{12}}{8^2}$
Trouble is that I don't know where to go from here. If I go via LCD for $5^3$ and $8^2$ I'll get 8000 in the denominator and high numbers in the numerator, which I think isn't the point here with factorization. Ideally I'd get something like $(3a^3b^3)^2$ in the first numerator but for the whole expression, of course. Maybe I'm looking at it for too long, so I've lost perspective, but usual helping tools like wolfram and symbolab aren't of much help.
|
Use
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}
=\left(\frac{3a^2b^3}{5}\right)^3-\left(\frac{c^4}{4}\right)^3$.
Now,
let $x=\frac{3a^2b^3}{5}$
and
$y=\frac{c^4}{4}$
and use that factorization to get
$[\frac{3a^2b^3}{5}-\frac{c^4}{4}][\frac{9a^4b^6}{25}+\frac{3a^2b^3c^4}{20}+\frac{c^8}{16}]$
|
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|
Integrate with substitution Use the substitution $x=2\sin(\theta)$ to find the value of
$$\int_0^{\sqrt{3}} \frac{1}{(4+x^2)^{3/2}}dx$$
I got to $$\int_{0}^{\pi/3} \frac{2\cos(\theta)}{(4+4\cos^2(\theta))^{3/2}}d\theta.$$
However, I don't know how to further integrate this!
Thanks!
|
For now let's perform the indefinite integral: $$\int \frac{1}{(4+x^2)^{3/2}}dx.$$
If we use the substitution $x=2\tan(\theta)$ this gives $dx=2\sec^2(\theta)d\theta$.
Plugging in the subsitution yields: $$\int \frac{2\sec^2(\theta)}{(4+4\tan^2(\theta))^{3/2}} d\theta=\frac{1}{4} \int \frac{\sec^2(\theta)}{\sec^3(\theta)} d\theta =\frac{1}{4} \int \frac{1}{\sec(\theta)}d\theta =\frac{1}{4} \int \cos(\theta)d\theta =\frac{1}{4} \sin(\theta)+C.$$
Since $x/2 = \tan(\theta)$, this means $$\sin(\theta) = \frac{x}{\sqrt{x^2+4}}.$$
|
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|
Easy inequality going wrong
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$
My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $x\leq-2, -5/4\leq x \leq -1$ and $-1\leq x\leq1$, x is less than or equal to zero.
But, this range i got is incorrect. Where did I go wrong then?
|
After you reached the stage
$$ (x-5)(4x+5)(x-1)(x+1)(x+2)\le 0\stackrel{\div 4}\iff$$
$$ (x+2)\left(x+\frac54\right)(x+1) (x-1)(x-5)\le0$$
and since all the factors are with odd exponent, you can apply the snake method, and get the solution
$$x<-2\;\;,\;\;\;or\;\;\;-\frac54\le x<-1\;\;,\;\;\;or\;\;\;1<x\le 5$$
|
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|
Evaluate $\int \frac{dx}{1+\sin x+\cos x}$ Evaluate $$\int \frac{1}{1+\sin x+\cos x}\:dx$$
I tried several ways but all of them didn't work
I tried to use Integration-By-Parts method but it's going to give me a more complicated integral
I also tried u-substitution but all of my choices of u didn't work
Any suggestions?
|
$$
\begin{aligned}
\int \frac{d x}{1+\sin x+\cos x}
= & \int \frac{d x}{2 \cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\\ &=\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2}}{1+\tan \frac{x}{2}} d x\\&=\int \frac{d\left(\tan \frac{x}{2}\right)}{1+\tan \frac{x}{2}}
\\&=\ln \left|1+\tan \frac{x}{2}\right|+C
\end{aligned}
$$
|
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|
How to simplify expression with Fibonacci numbers I have to simplify the expression $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n}$. I only noticed that $\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n} = \sum_{n=0}^\infty \frac{1}{10^n} \sum_{k=0}^n F_{2k}F_{n-k}$. What to do next?
|
The generating function of the Fibonacci numbers is well-known:
$$ \frac{x}{1-x-x^2} = \sum_{n=0}^\infty F_n x^n. $$
We also want to isolate only the even Fibonacci numbers:
$$ \frac{x}{1-x-x^2} + \frac{-x}{1+x-x^2} = \sum_{n=0}^\infty F_n (x^n + (-x)^n) = \sum_{n=0}^\infty 2F_{2n} x^{2n}. $$
Since
$$
\frac{x}{1-x-x^2} + \frac{-x}{1+x-x^2} =
\frac{x(1+x-x^2) -x(1-x-x^2)}{(1-x-x^2)(1+x-x^2)} =
\frac{2x^2}{(1-x^2)^2-x^2},
$$
we can conclude that
$$
\frac{x}{1-3x+x^2} = \sum_{n=0}^\infty F_{2n} x^n.
$$
It follows that
$$
\frac{x}{1-x-x^2} \frac{x}{1-3x+x^2} = \sum_{n=0}^\infty x^n \sum_{k=0}^n F_{2k} F_{n-k}.
$$
In particular, your expression equals $100/6319$.
|
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|
Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$
|
I'm afraid I can't think of any methods other than brute force. If we set $\alpha = \sqrt{2} + \sqrt[3]{5}$, then all powers of $\alpha$ lie in the span over $\Bbb Q$ of the set $\{1, \sqrt{2}, \sqrt[3]{5}, \sqrt[3]{25}, \sqrt{2}\sqrt[3]{5}, \sqrt{2}\sqrt[3]{25}\}.$ What we essentially need to show is that the set spanned by the powers of $\alpha$ then contains $6$ linearly independent elements of this span.
I start by calculating $\alpha^2 = 2 + \sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}$. You could then compute $\alpha^3, \alpha^4, \alpha^5$ similarly, and check by row reduction that the resulting set is linearly independent.
If you're doing computations by hand, it might be somewhat easier, to compute $\alpha(\alpha^2-2) = (\sqrt{2} + \sqrt[3]{5})(\sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}) = 5 + 4\sqrt[3]{5} + 3\sqrt{2}\sqrt[3]{25}$, and then similarly compute $\alpha(\alpha(\alpha^2-2)-5)$, and so forth.
|
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|
Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx$ $$I_1 =\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx= ?$$
I tried substitution: $\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) = \Xi$, but then I'm not able to do anything after the resulting integral.
Could someone help? There must be a simple way to solve this...
|
Hint. Assume $a>0,\,x+a>0$. Integrating by parts gives
$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\int x\left( \frac1{2x} \frac{\sqrt{ax}}{x+a}\right)dx \tag1 $$ and the last integral is easy to evaluate
$$\int \frac{\sqrt{x}}{x+a}\:dx=2\int \frac{u^2}{u^2+a}\:du\quad (\sqrt{x}=u). \tag2$$ Bringing $(1)$ and $(2)$ together leads to
$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\sqrt{ax}+ a\sqrt{x}\arctan \left( \sqrt{\frac{x}{a}}\right)+C.$$
|
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|
Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.
I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:
$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $
So I could simply solve: $(2+\sqrt{5}+2\sqrt{7})(a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35})=1$
Which I believe works. Is there a better way of solving this problem?
|
\begin{align}
\frac{1}{2+\sqrt{5}+2\sqrt{7}}
&=\frac{2+\sqrt{5}-2\sqrt{7}}{(2+\sqrt{5})^2-28}\\
&=\frac{2+\sqrt{5}-2\sqrt{7}}{-19+4\sqrt{5}}\\
&=\frac{(2+\sqrt{5}-\sqrt{7})(-19-4\sqrt{5})}{361-80}\\
&=\dots
\end{align}
|
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|
What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$? What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$?
Let's try a several primes greater than 3...
If $p=5$, then we have $1^2 + 2^2 + 3^2 + 4^2 = 30$, so that $30\pmod{5} = 0$
If $p=7$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2= 91$, so that $91\pmod{7} = 0$
If $p=11$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2= 385$, so that $385\pmod{11} = 0$
So it seems like we will always get 0. So for primes greater than 3, we have the sum equal to a multiple of $p$.
I know that $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1)$.Well if $p>3$, then $p-1=2a$ for some integer $a$, so that $2p-1 = 4a +1$. Then we have
$$ 1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1) = \frac16 p(2a)(4a+1)$$
If I could show that $(2a)(4a+1)$ is a multiple of 6, I'll be done, but I am having trouble showing that. Perhaps someone could provide a much needed tip, but don't figure this out for me, please.
|
HINT: You don't need $a$: consider separately the cases $p\equiv1\pmod3$ and $p\equiv2\pmod3$.
|
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|
Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$
|
Let $a>b>0$, $x=a\cdot t$ and $b\cdot t$. Then we have
$$
\lim_{t^+\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^+\to 0}\frac{a^2bt^3}{t^3(a^2\frac{1}{t} + b^3)}
=
\lim_{t^+\to 0}\frac{a^2b}{(a^2\frac{1}{t} + b^3)}=+\infty
$$
and
$$
\lim_{t^-\to 0}\frac{a^2t^2\cdot bt}{a^2t^2 + b^3t^3}
=
\lim_{t^-\to 0}\frac{a^2bt^3}{t^3(a^2\frac{1}{t} + b^3)}
=
\lim_{t^-\to 0}\frac{a^2b}{(a^2\frac{1}{t} + b^3)}=-\infty
$$
|
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|
Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& = \frac{1}{2} \frac{ \log(x)}{x^2+1} - \frac{1}{2}\int \left[(x^2+1)\frac{\frac{(x^2+1)^2}{x} - 4x(x^2+1) \log(x)}{(x^2+1)^4} \right ]dx\\
& = \frac{ \log(x)}{2(x^2+1)}-\frac{1}{2}\int \left [ \frac{x^2+1-4x^2 \log(x)}{x(x^2+1)^2} \right ] dx\\
& = \frac{ \log(x)}{2(x^2+1)} - \frac{1}{2}\int \frac{dx}{x(x^2+1)} + 2\int\frac{x \log(x)}{(x^2+1)^2}dx
\end{align}
Or this method doesn't work here, or I have done a mistake somewhere. However, I have also tried doing $u = x^2+1$ substitution, but this also didnt gave me any good results.Thank you.
|
We have
$$\int_1^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{1/x\log(1/x)}{(1+1/x^2)^2}\dfrac{-dx}{x^2} = \int_1^0 \dfrac{x\log(x)}{(1+x^2)^2} = -\int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx $$
Hence,
$$\int_0^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx + \int_1^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = 0$$
In case you are interested in $I = \displaystyle \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx$, we have
\begin{align}
I & = \int_0^1 \dfrac12\dfrac{\log(x^2)}{(1+x^2)^2} d\left(\dfrac{x^2}2\right)
\end{align}
This gives us
\begin{align}
4I & = \int_0^1 \dfrac{\log(t)}{(1+t)^2} dt = \int_0^1 \sum_{k=0}^{\infty} (-1)^k(k+1)t^k\log(t) dt = \sum_{k=0}^{\infty}(-1)^k(k+1) \int_0^1 t^k \log(t)dt\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}(k+1)}{(k+1)^2} = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{k+1} = -\log(2)
\end{align}
This gives us
$$I = \displaystyle \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx = -\dfrac{\log(2)}4$$
|
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|
What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
|
$\frac{1}{2}d(x^2+y^2) = xdx+ydy$
$d(y/x) = \frac{xdy-ydx}{x^2}$
Let $u = x^2+y^2$ and $v = \frac{y}{x}$
$LHS = \frac{1}{2}d(x^2+y^2) = \frac{1}{2}du$
$RHS = \frac{xdy-ydx}{x^2+y^2} = x^2 \frac{dv}{u} = \frac{u}{1+v^2}\frac{dv}{u} = \frac{dv}{1+v^2}$
Thus,
$$
\frac{du}{dv} = \frac{1}{1+v^2}\
$$
Now integrate and substitute for $u,v$
|
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|
How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
|
Note that for $x\not=1$,
$$(x-1)(x^5+x^4+x^3+x^2+x+1)=x^6-1$$
$$\Rightarrow \frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$$
|
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|
Floor inequality: $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$ I remember seeing the inequality $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$ somewhere which is true for all reals.
So I was wondering what's wrong with this proof?
For all reals $a,b$ with $|b|>1$ we can use the inequality to get:
$\lfloor \frac{a-1}{b}\rfloor\ge\lfloor\frac{a}{b}\rfloor+\lfloor-\frac{1}{b}\rfloor=\lfloor\frac{a}{b}\rfloor$,
since $\lfloor-\frac{1}{b}\rfloor=1$ since $|b|>1$.
Is there anything wrong?
|
$$ \left\lfloor -\dfrac{1}{b} \right\rfloor = -1 $$
so you should get
$$ \left\lfloor \dfrac{a-1}{b} \right\rfloor \ge \left\lfloor \dfrac{a}{b} \right\rfloor + \left\lfloor \dfrac{-1}{b} \right\rfloor = \left\lfloor \dfrac{a}{b} \right\rfloor - 1 $$
|
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|
Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}Using the axioms, theorem, definitions of high school algebra concerning the real numbers, then prove the following:
Given $r>0$, find a $k>0$ such that:
$$\text{for all }x, y: \sqrt{(x-2)^2+(y-1)^2}<k\implies|xy-2|<r $$
I tried with several values given to $k$ and $r$ to find the relation between them.
Suppose then $r=1$ and we choose $k$ to be $\frac{1}{2}$
Now let us check if $\forall x,y$ such that $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|xy-2|<1$
However we know that $|x-2|\leq\sqrt{(x-2)^2+(y-1)^2}$
Therefore if $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|x-2|<\frac{1}{2}$..........(1)
For the same reason $|y-1|<\frac{1}{2}$...................(2)
But, (1) implies $\frac{3}{2}<x<\frac{5}{2}$.......................(3)
Also (2) implies $\frac{1}{2}<y<\frac{3}{2}$....................(4)
And the question now is, do ALL the values of x and y satisfying (3) and (4), satisfy $|xy-2|<1$
No because for $x=\frac{16}{10}$ and $y=\frac{6}{10}$ the relation $|xy-2|<1$ is not satisfied.
And the question is, which is the proper relation between $r$ and $k$ so that the above inequality is satisfied for all the values of $x$ and $y$
|
what you are looking for are the two hyperbolas that either touch at two place or touch and go through the end point of a diameter of the semicircle. take the upper part of the semicircle $$(x-2)^2 + (y-1)^2 = k^2.$$ let the hyperbola $$xy = 2+r, \frac{dy}{dx} = -\frac y x$$ touch at $$x = 2 + \cos t, y = 1 + \sin t$$ the common tangent has the slope $-\frac{\cos t}{\sin t}.$ the constraints on $t$ are $$ (2+\cos t)(1+\sin t)= 2 + r, \frac{\cos t}{\sin t} =\frac{1+\sin t}{2+\cos t}$$ simplifying the last equation, we get $$ \cos t + 2 \sin t + \cos t \sin t= r,2\cos t-\sin t + \cos^2 t-\sin^2 t=0 \tag 1$$
here is what i will try to do. solve numerically $$2\cos t-\sin t + \cos^2 t-\sin^2 t=0$$ for $t.$ use the second equation $$r = \cos t + 2 \sin t + \cos t \sin t $$ to find $r.$
|
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|
Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$ The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$
My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer.
Is there a slicker way of doing this?
|
As you suggest, complete the square,
$$ 1+x+x^2 = (x+1/2)^2 + 3/4 $$
Now, the key is to substitute to make the square root something nice. I'll do a couple of substitutions to make it clearer. Set $y=x+1/2$, so we have
$$ \int \frac{(y-1/2)^2}{\sqrt{y^2+3/4}} \, dy $$
Now the denominator looks in the form appropriate for a hyperbolic substitution. In particular,
$$ \frac{d}{dx} \arg \sinh{x} = \frac{1}{\sqrt{1+x^2}}, $$
so setting $y=\frac{1}{2}\sqrt{3}\sinh{u}$, you find, ($dy=\frac{1}{2}\sqrt{3}\cosh{u} \, du$)
$$ \int \frac{(\frac{1}{2}\sqrt{3}\sinh{u}-\frac{1}{2})^2}{\sqrt{\frac{3}{4} (1+\sinh^2{u})}} \frac{1}{2}\sqrt{3}\cosh{u} \, du = \frac{1}{4}\int (\sqrt{3} \sinh{u}-1)^2 \, du $$
This integral is fairly elementary after using the identity
$$ \cosh{2u}=\cosh^2{u}+\sinh^2{u} = 1+2\sinh^2{u} $$
After that, you just have to invert the substitutions, for which you will also need
$$ \cosh{u} = \sqrt{1+\sinh^2{u}} $$
and
$$ \sinh{2u}=2\sinh{u}\cosh{u}. $$
|
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|
How can I integrate $\frac{1}{x^2-x-1}$? I need to find $$\int\frac{1}{x^2-x-1}dx$$ but I don't know what to do. I've thought about substitution or partial fractions but neither has worked.
|
As stated above, complete the square of the integrand to get
\begin{equation*}
\int\frac{1}{(x-\frac{1}{2})^2-\frac{5}{4}}.
\end{equation*}
Now substitute $u=x-\frac{1}{2}$ & $du=dx:$
\begin{equation*}
\int\frac{1}{u^2-\frac{5}{4}}=-\frac{4}{5}\int \frac{1}{1-\frac{4u^2}{5}}.
\end{equation*}
Substitute again using $s=\frac{2u}{\sqrt{5}}$ & $ds=\frac{2}{\sqrt{5}}du:$
\begin{equation*}
-\frac{2}{\sqrt{5}}\int\frac{1}{1-s^2}ds=-\frac{2\tanh^{-1}(s)}{\sqrt{5}}+C.
\end{equation*}
Now substitute back for $s=\frac{2u}{\sqrt{5}}$ & $u=x-\frac{1}{2}.~_{\square}$
|
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|
How prove that $0 < a_{10} - \sqrt{2}< 10^{-370}$ for $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$? Let $a_1=1,$ , $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$. How prove that
$0 < a_{10} - \sqrt{2}< 10^{-370}$?
$a_n - a_{n-1} = \frac{1}{2a_{n-1}} \left(2 - a_{n-1}^2\right) < 0 \Rightarrow a_{n-1} > \sqrt{2}$ and what next?
|
We have
$$a_{n+1} - \sqrt2 = \dfrac12\left(a_n-\sqrt2 + \dfrac2{a_n} - \sqrt2\right) = \dfrac12\left(a_n-\sqrt2 + \dfrac{2-\sqrt2a_n}{a_n}\right) = \dfrac{(a_n-\sqrt2)^2}{2a_n}$$
Clearly, $a_n > \sqrt2$ for all $n >1$. Hence, we obtain that
$$a_{n+1} -\sqrt2 \leq \dfrac{(a_n-\sqrt2)^2}{2\sqrt2} \leq \dfrac{(a_{n-1}-\sqrt2)^4}{(2\sqrt2)^{1+2}} \leq \dfrac{(a_{n-2}-\sqrt2)^8}{(2\sqrt2)^{1+2+4}} \leq \cdots \leq \dfrac{(a_2-\sqrt2)^{2^{n-1}}}{(2\sqrt2)^{2^{n-1}-1}}$$
Hence, we have
$$a_{10} - \sqrt2 \leq \dfrac{(3/2-\sqrt2)^{2^8}}{(2\sqrt2)^{2^8-1}} < 10^{-370}$$
where the last inequality can be checked using logarithms.
|
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|
Square Roots: Variables with Exponents. Alright, so let me get this straight:
$\sqrt{x^2} = |x|$
$\sqrt{x^3} = x\sqrt{x}$
$\sqrt{x^4} = x^2$
$\sqrt{x^6} = |x^3|$
Are these correct?
|
Well The first and last one are more tricky than the other two.
you know that $x^2$ is always positive and so when we take the square root it will be also positive, This is why $$\sqrt{x^2} = |x|$$
And the last one follows from the first one, But here you have to know that the an even power is always positive $x^{2n}$ is always positive for any positive integer $n$ and since $6$ is even then $x^6$ is positive and so
$$\sqrt{x^6} = |(x^6)^\frac{1}{2}| = |x^3|$$
Notice that we evaluate inside out, so we first take $x^6$ then we apply the square root that's why we get the absolute value equivalence.
The 2nd and 3rd ones are very easy.
$$\sqrt{x^3} = (x^3)^\frac{1}{2} = x^\frac{3}{2} = xx^\frac{1}{2} = x\sqrt{x}$$
and the last one follows the same way
You are right !
|
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|
Find vectors when added up equal (1, 1, 1) Question:
Let $V$ be the 2-dim subspace of $\mathbb R^3$ spanned by $(1, 2, -3)$ and $(-2, 0, 1)$. Write the vector $u = (1,1,1)$ in the form $u = v + w$, where $v$ is in $V$ and $w$ is in $V^\perp$, which is the subspace orthogonal to $V$.
What I tried:
So I found a vector that was linearly independent to the two vectors given: $(0, 0, 1)$. Then I used the Gram–Schmidt process to find a vector orthogonal to both of them. I found $(2, 5, 4)$ is orthogonal to $(1, 2, -3)$ and $(-2, 0, 1)$. However, after obtaining the unit vectors, I don't know how to proceed. No matter what vectors I add up, I do not get $u$.
My unit vectors were:
$$u_1 = (1,2,-3)/\sqrt14$$
$$u_2 = (23,-10,1)/(3\sqrt(70)$$
$$u_3 = (2,5,4)/(3\sqrt5)$$
Please tell me how to obtain the vectors so when I add $v+w$, it will equal $(1,1,1)$. Thank you!
|
You have found a basis. Let these vectors be the columns of a matrix $A$, and solve $Ax=(1,1,1)^T $.
Let $A = \begin{pmatrix} 2 & 1 & -2 \\ 5 & 2 & 0 \\ 4 & -3 & 1 \end{pmatrix}$.
Compute $$Ax = \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}.$$
Solving, we get $x = \begin{pmatrix} \frac{11}{45} \\ -\frac{1}{9} \\ -\frac{14}{45}\end{pmatrix}$.
Now, let's compute $\frac{11}{45}\begin{pmatrix}2 \\ 5 \\4\end{pmatrix} - \frac{1}{9}\begin{pmatrix} 1 \\ 2 \\ -3\end{pmatrix} - \frac{14}{45}\begin{pmatrix} -2\\0\\1\end{pmatrix}$:
$$\color{blue}{\frac{11}{45}\begin{pmatrix}2 \\ 5 \\4\end{pmatrix}} \color{red}{-\frac{1}{9}\begin{pmatrix} 1 \\ 2 \\ -3\end{pmatrix} - \frac{14}{45}\begin{pmatrix} -2\\0\\1\end{pmatrix}} = \begin{pmatrix} \frac{22}{45} - \frac{5}{45} + \frac{28}{45} \\ \frac{55}{45} - \frac{10}{45} - \frac{0}{45} \\
\frac{44}{45} + \frac{15}{45} - \frac{14}{45} \end{pmatrix} = \begin{pmatrix} \frac{45}{45} \\ \frac{45}{45} \\ \frac{45}{45} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$$
Now, the term in $\color{blue}{\textrm{blue}}$ is a vector in $V^\perp$, and the term in $\color{red}{\textrm{red}}$ is a linear combination of vectors in $V$. Since $V$ is a subspace, it is algebraically closed under linear combinations of its elements; hence, the red term is in $V$.
Therefore, let $v$ be the $\color{red}{\textrm{red}}$ term, and let $w$ be the $\color{blue}{\textrm{blue}}$ term, and we are done.
|
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|
problem based on LCM and HCF Find the least number which when divided by 7,9,11 produce 1,2,3 as reminders .
I tried a method which works only when the difference between the elements is same. But in this case 7-1=6 , 9-2=7, 11-3=8 . So that method is not working.
Please, provide the way by which I can solve the problem. Answer of the problem is 344.
|
This may be a bit longer, but it is a bit more mechanical of a solution.
To solve
$$
\begin{align}
x&\equiv1\pmod{7}\\
x&\equiv2\pmod{9}\\
x&\equiv3\pmod{11}
\end{align}
$$
we can solve three simpler problems. To solve the simpler equations, we can use the Extended Euclidean Algorithm as implemented in this answer.
$$
\begin{align}
x_7&\equiv1\pmod{7}\\
x_7&\equiv0\pmod{9}\\
x_7&\equiv0\pmod{11}
\end{align}
$$
$x\equiv0\pmod{9}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{99}$.
$$
\begin{array}{r}
&&14&7\\\hline
1&0&1&-7\\
0&1&-14&99\\
99&7&1&0
\end{array}
$$
Therefore, $x_7=99$.
$$
\begin{align}
x_9&\equiv0\pmod{7}\\
x_9&\equiv1\pmod{9}\\
x_9&\equiv0\pmod{11}
\end{align}
$$
$x\equiv0\pmod{7}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{77}$.
$$
\begin{array}{r}
&&8&1&1&4\\\hline
1&0&1&-1&2&-9\\
0&1&-8&9&-17&77\\
77&9&5&4&1&0
\end{array}
$$
Therefore, $x_9=154$.
$$
\begin{align}
x_{11}&\equiv0\pmod{7}\\
x_{11}&\equiv0\pmod{9}\\
x_{11}&\equiv1\pmod{11}
\end{align}
$$
$x\equiv0\pmod{7}$ and $x\equiv0\pmod{9}\iff x\equiv0\pmod{63}$.
$$
\begin{array}{r}
&&5&1&2&1&2\\\hline
1&0&1&-1&3&-4&11\\
0&1&-5&6&-17&23&-63\\
63&11&8&3&2&1&0
\end{array}
$$
Therefore, $x_{11}=-252$.
We can put the three simpler answers together by
$$
x\equiv x_7+2x_9+3x_{11}=-349\equiv344\pmod{693}
$$
That is
$$
\begin{bmatrix}
1\\0\\0
\end{bmatrix}
+
2\begin{bmatrix}
0\\1\\0
\end{bmatrix}
+
3\begin{bmatrix}
0\\0\\1
\end{bmatrix}
=
\begin{bmatrix}
1\\2\\3
\end{bmatrix}
\begin{array}{l}
\pmod{7}\\
\pmod{9}\\
\pmod{11}
\end{array}
$$
|
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|
Determine the degree of the extension over Q Determine the degree of the extension $Q(\sqrt{3+2 \sqrt{2}})$ over Q.
I can see that $$3+2 \sqrt{2} = (1+ \sqrt2)(1+ \sqrt2) =(1+ \sqrt2)^2$$ does that mean $$x^2 -(1+ \sqrt2)^2)$$ has a degree $2$. Is this correct
|
Example: Let $p(x) = \sqrt{1+\sqrt{2}}$.
Let $u = \sqrt{1+\sqrt{2}} \Rightarrow u^2 = (1+\sqrt{2})^2 = 2+2\sqrt{2} \Rightarrow (u^2-2)^2 = 8 \Rightarrow u^4-4u^2+4=8 \Rightarrow u^4-4u2-4=0$
Hence $u$ is the root of $x^4-4u^2-4$. Try this techniques with your adjoined root and see if you can say anything about the polynomial that it satisifies over $\mathbb{Q}$.
|
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|
Find all $c\in\mathbb Z^+$ for which $\exists a,b\in\mathbb Z^+, a\neq b$ with $a+c\mid ab$ and $b+c\mid ab$
Find all $c\in\mathbb Z^+$ for which $\exists a,b\in\mathbb Z^+, a\neq b$ with $\begin{cases}a+c\mid ab\\b+c\mid ab\end{cases}$
For those $c$, prove only finitely many $(a,b)$ exist.
My attempt:
Let $(a+c,b+c)=d$. Then $\color{RoyalBlue}{d\mid ab}$ and $\text{lcm}(a+c,b+c)\mid ab\iff \frac{(a+c)(b+c)}{d}\mid ab$.
So $d\neq 1$, because $(a+c)(b+c)\mid ab\,\Rightarrow ab\ge (a+c)(b+c)$, impossible.
$a\equiv -c\equiv b\,\Rightarrow\, a\equiv b$ mod $d$. Then $\color{RoyalBlue}{ab\equiv 0}\equiv a^2\equiv b^2\,\Rightarrow\, d\mid a^2,b^2$.
So $p\mid d\mid a+c,b+c,a^2,b^2\,\Rightarrow\, p\mid a,b,c$, and $(a,b,c)\ge 2$.
|
Your findings so far are fine, but not (yet) fully conclusive towards the main problem.
Let $$C=\{\,c\in\mathbb Z^+\mid \exists a,b\in\mathbb Z^+\colon a\ne b\land a+c| ab\land b+c| ab\,\}$$
be the set we are looking for.
For $c\ge 2$ we can let $$\tag{$\star$} a=c^3+c^2-c,\qquad b=c(a-1)=c(c+1)(c^2-1)$$Then certainly $0<a<b$ and we have
$$ab = c^2(c^2+c-1)(c+1)(c^2-1)=(a+c)(c^2+c-1)(c^2-1)$$
and
$$ ab=ac(a-1)=(b+c)(a-1).$$
This shows $c\in C$ for $c\ge 2$.
On the other hand $1\notin C$ for if $a+1\mid ab$ then $\gcd(a+1,a)=1$ implies $a+1\mid b$; likewise $b+1\mid ab$ implies $b+1\mid a$. Specifically $b\ge a+1$ and $a\ge b+1$, which is absurd. Therefore we have
$$C=\mathbb Z^+\setminus\{1\}.$$
Remains to show that for each $c\in C$ there are only finitely many $(a,b)$.
Empirically, it seems that the choice for $a,b$ made in $(\star)$ is the maximal possible choice. Showing that $\max\{a,b\}>c(c+1)(c^2-1)$ is really impossible would of course show finiteness.
|
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|
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2,$ or $6$.
I'm having a hard time with the $0,2,6$ and part but here is what I have so far.
Since $n$ is the product of two consecutive integers then one must be odd and the other must be even.
Let $m=2k+1$ represent the odd integer where $k \in \mathbb{Z}$, and let $q=2k$ represent the even integer where $k \in \mathbb{Z}$. Then
$$n=m \times q=(2k+1)(2k)=(2k)^2+2k$$
This tells me that $n$ is even. So how can I show that units digit must be $0,2,$ or $6$?
|
You can also work over all the cases of $k$ modulo $5$.
*
*If $k=5t$, then
$$(2(5t))^2+2(5t)=100t^2+10t=10(10t^2+t)$$
so our number ends with a $0$.
*If $k=5t+1$, then
$$(2(5t+1))^2+2(5t+1)=100t^2+40t+4+10t+2=10(10t^2+5t)+6$$
so our number ends with a $6$.
*If $k=5t+2$, then
$$(2(5t+2))^2+2(5t+2)=100t^2+80t+16+10t+4=10(10t^2+9t+20)$$
so our number ends with a $0$.
*If $k=5t+3$, then
$$(2(5t+3))^2+2(5t+3)=100t^2+120t+36+10t+6=10(10t^2+13t+40)+2$$
so our number ends with a $2$.
*If $k=5t+4$, then
$$(2(5t+4))^2+2(5t+4)=100t^2+160t+64+10t+8=10(10t^2+17t+70)+2$$
so our number ends with a $2$.
|
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|
How can I finish integrating $\int {\sqrt{x^2-49} \over x} $ using trig substitution? $$\int {\sqrt{x^2-49} \over x}\,dx $$
$$ x = 7\sec\theta$$
$$ dx = 7\tan\theta \sec\theta \,d\theta$$
$$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - 49} \left(\tan\theta d\theta\right)$$
$$ \int\sqrt{7^2(\sec^2\theta - 1)} (\tan\theta \,d\theta) = 7\int\sqrt{\sec^2\theta - 1} (\tan\theta \,d\theta)$$
$$ 7\int \tan^2\theta \,d\theta = 7\int \sec^2\theta - 1 \,d\theta $$
$$ 7\int \sec^2\theta - 7\int d\theta $$
$$ 7\tan\theta - 7\theta + C = 7(\tan\theta - \theta) + C$$
This makes: $$ \theta = \sec^{-1}\left(x \over 7\right)$$
And plugging back in to the indefinite integral:
$$ 7\left(\left(\sqrt{x^2-49} \over 7 \right) - \sec^{-1}\left(x \over 7 \right)\right) + C $$
My question really is, how can I evaluate $\sec^{-1}\left(x \over 7 \right)$ ?
|
This is the correct answer. Unless the original problem asked for a definite integral, in which case all you would need to do is plug in your bounds.
|
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|
Help solve for length $PQ$ how do I approach this question using simultaneous equations with trig and or pythag??? Solve for length $PQ$
Cheers bob
|
Let $PQ=x$ and $QS=y$
Then for triangle $PQS$:
$$x^2 +y^2 = 12^2$$
$$x^2+y^2 = 144$$
For triangle PTR:
$$(x-2)^2+y^2=11^2$$
$$x^2-4x+4+y^2=121$$
So adding $23$ to both sides:
$$x^2-4x+4+y^2+23=121+23$$
$$x^2-4x+27+y^2=144$$
So
$$x^2-4x+27+y^2 = x^2+y^2$$
Take $x^2+y^2$ from both sides
$$-4x+27=0$$
$$4x=27$$
$$x=\frac{27}{4}=6.75$$
$$PQ=6.75m$$
You need not solve for $y$ as all we need is $PQ$
|
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|
Represent the transformation with respect to the standard basis
Consider a linear transformation from $R^2$ to $R^2$ defined via:
$$\left(\begin{matrix} 1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 1\end{matrix} \right)$$ and $$\left(\begin{matrix} -1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 2\end{matrix} \right)$$
Represent the transformation with respect to the standard basis
It has been some time I have tried to understand this problem, and I have seen solutions to problems similar to this and I still do not understand it. Those solutions talk about images of $e_1$ and $e_2$ (which are the standard vectors of the standard basis), but I don't know what the heck they are talking about.
|
We can represent a linear map: $F:\mathbb{R}^{2} \to \mathbb{R}^{2}$ using a matrix $\mathbf{A}\in\mathbb{R}^{2\times2}$, generically, we can write:
$$\mathbf{A}=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$
Applying this to the vector $\begin{pmatrix}1 & 3\end{pmatrix}^{T}$, we have:
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}1 \\ 3\end{pmatrix}=\begin{pmatrix}a+3b \\ c + 3d\end{pmatrix} = \begin{pmatrix}3 \\ 1\end{pmatrix}$$
Similarly, applying it to the vector $\begin{pmatrix}-1 & 3\end{pmatrix}^{T}$, we have:
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}-1 \\ 3\end{pmatrix}=\begin{pmatrix}-a+3b \\ -c+3d\end{pmatrix} = \begin{pmatrix}3 \\ 2\end{pmatrix}$$
We therefore have a system of 4 linear simultaneous equations with 4 variables $a, b, c$ and $d$:
$$\begin{align*}a+3b &= 3 \\ c+3d &= 1 \\ 3b-a &= 3 \\ 3d - c &= 2\end{align*}$$
Solving these we get:
$$a=0,\quad b=1,\quad c = -\frac{1}{2},\quad d=\frac{1}{2}$$
We can thus write our transformation matrix describing the linear map, as:
$$\mathbf{A} = \frac{1}{2}\begin{pmatrix}0 & 2 \\ -1 & 1\end{pmatrix}$$
|
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|
PDE Manipulation - Calculus I need help for this question, its a lot of calculus but I'm confuse.
let $$ u= \dfrac{(x-b)^{2}+y^{2}-q^{2}}{(x-b-1)^{2}+y^{2}-q^{2}-1} $$
I need show that
$$ u_{x}^{2}+u_y^{2}= \dfrac{1}{(x-b)^{2}}\left(u-1 \right)^{2}\left( u^{2}+q^{2}(u-1)^{2}\right) $$
I begin with this
\begin{eqnarray*}u_x & =& \dfrac{2(x-b)\left[ (x-b-1)^{2}+y^{2}-q^{2}-1\right]-2(x-b-1)\left[ (x-b)^{2}+y^{2}-q^{2}\right]}{\left[(x-b-1)^{2}+y^{2}-q^{2}-1\right]^{2}}\\\\
&=& \dfrac{2(x-b)A}{A^{2}} - \dfrac{2(x-b-1)B}{A^{2}}
\end{eqnarray*}
where $A = \left[(x-b)^{2}+y^{2}-q^{2}\right]$ and $B= \left[(x-b-1)^{2}+y^{2}-q^{2}-1\right]$
so
\begin{eqnarray*}u_x & =& \dfrac{2(x-b)}{A} - \dfrac{2(x-b-1)u}{A}.
\end{eqnarray*}
I do not know how can I do "appears q" and other manipulations, I've done the square that, I've done the same for the derivative with y, do not leave the place ...
|
It is a bit easier to write
$$\begin{align}
u&=\frac{(x-b)^2+y^2-q^2}{(x-b-1)^2+y^2-q^2-1}&\\\\
&=\frac{(x-b-1)^2+y^2-q^2-1+2(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\&=1+2\frac{(x-b)}{(x-b-1)^2+y^2-q^2-1}\\\\
&=1+2\frac{s}{(s-1)^2+t^2}
\end{align}$$
where $s=x-b$ and $t^2=y^2-q^2-1$.
Using the chain rule, we have
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial s^2}$$
and
$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial t^2}\left(\frac{dt}{dy}\right)^2+\frac{\partial u}{\partial t}\frac{d^2t}{dy^2}$$
Now, using $dt/dy=y/t$, $d^2t/dy^2=-(1+q^2)/t^3$ we find
$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial t^2}\left(\frac{t^2+(1+q^2)}{t^2}\right)-\frac{(1+q^2)}{t^3}\frac{\partial u}{\partial t}$$
The rest is just persevering.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1269159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
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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.