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How do you solve this trig/geometry question? In a quadrilateral $ABCD$, if
$\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right) + \sin\left(\frac{C+D}2\right)\cos\left(\frac{C-D}2\right) = 2$
then $\sin\left(\frac A 2\right) \sin\left(\frac B 2\right) \sin\left(\frac C 2\right) \sin\left(\frac D 2\right) = $
a) $\frac 1 8 $
b) $\frac1 4$
c) $\frac {1}{2\sqrt2}$
d) $\frac 1 2$
|
For $w$, $x$, $y$, $z$ between $0$ and $\pi$, the relation
$$\sin w \;\cos x \;+\; \sin y \; \cos z \;=\; 2$$
holds if and only if each term ---and each factor of each term--- is itself $1$. Specifically,
$$w = y = \frac{\pi}{2} \quad\text{ and }\quad x = z = 0$$
Now, note that any two angles of a quadrilateral have a sum and (absolute) difference between $0$ and $2\pi$ (because the sum of all angles is $2\pi$); thus, they have a half-sum and half-difference between $0$ and $\pi$.
Therefore, if for $\square ABCD$ we define
$$w := \frac{A+B}{2} \quad x := \frac{|A-B|}{2} \quad y := \frac{C+D}{2} \quad z := \frac{|C-D|}{2}$$
then the above applies to $w$, $x$, $y$, $z$ and allows us to conclude that $A=B=C=D = \pi/2$, so that
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\sin\frac{D}{2} = \sin^4\frac{\pi}{4} =\left(\frac{\sqrt{2}}{2}\right)^4 = \frac{1}{4}$$
|
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|
What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.
|
Multiplying
$$\frac{\sqrt{x^2+8}-3}{x+1}$$
by
$$\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}\ \ (=1)$$
gives you
$$\begin{align}\lim_{x\to -1}\frac{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}{(x+1)(\sqrt{x^2+8}+3)}&=\lim_{x\to -1}\frac{\color{red}{(x+1)}(x-1)}{\color{red}{(x+1)}(\sqrt{x^2+8}+3)}\\&=\lim_{x\to -1}\frac{x-1}{\sqrt{x^2+8}+3}\\&=\frac{-2}{\sqrt 9+3}\\& =-\frac 13.\end{align}$$
|
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|
Verify Demorgan's Law Algebraically If $\overline X \equiv \text { not }X$,
De Morgan's Laws are stated as:
*
*$ \overline{(A + B)}= \overline A\cdot \overline B$
*$ \overline{(A\cdot B)} = \overline A + \overline B$
Verify the above laws algebraically.
I can prove this using truth tables and logic gates but algebraically, I don't know any intuitive way to prove it.
$$
\begin{array}{ |c|c| }
\hline
{\text{Axioms}}\\
\hline
\text{Property of 0} & X + 0 = X \space ;\space X\cdot0 = 0 \\
\text{Property of 1} & X + 1 = 1 \space;\space X\cdot1 = X\\
\text{Idempotence Law} & X + X = X \space;\space X\cdot X = X\\
\text{Involution Law} & \overline{\overline X} = X\\
\text{Complementarity Law} & X + \overline X = 1 \space;\space X\cdot\overline X = 0\\
\text{Commutative Law} & X+Y = Y+X \space;\space X\cdot Y = Y\cdot X\\
\text{Associative Law} & (X+Y)+Z = X+(Y+Z) \space;\space (X\cdot Y)\cdot Z = X\cdot (Y\cdot Z)\\
\text{Distributive Law} & X(Y+Z) = XY + XZ \space;\space X + YZ = (X+Y)(X+Z)\\
\text{Absorption Law} & X + XY = X \space;\space X(X+Y) = X\\
\text{Other (3rd Distributive)} & X + \overline XY = X+Y\\
\hline
\end{array}
$$
|
By Complementarity Law,
$$P + \overline P = 1 \space\text{ and }\space P \cdot \overline P = 0$$
(Note: I shall only be using $P + \overline P = 1$ as its dual is automatically true)
First Law::
DeMorgan's $1^{\text{st}}$ law states $\overline{X+Y} = \overline X \cdot \overline Y$
It is sufficient to prove that $(X + Y) + \overline X \cdot \overline Y = 1$
$$
\begin{align}
\text{LHS} &= Y + (X + \overline X \cdot \overline Y)\\
&= Y + X + \overline Y\\
&= (Y+\overline Y) + X\\
&= 1 + X \\&= 1 = \text{RHS}
\end{align}
$$
Second Law:: DeMorgan's $2^{\text{nd}}$ Law states that $\overline{X\cdot Y} = \overline X + \overline Y$
It is sufficient to prove that $X\cdot Y + (\overline X + \overline Y) = 1$
$$
\begin{align}
\text{LHS} &= \overline Y + (\overline X + \overline{\overline X}\cdot Y)\\
&= \overline Y + (\overline X + Y)\\
&= (Y + \overline Y) + \overline X\\
&= 1 + \overline X\\
&= 1 = \text{RHS}
\end{align}
$$
Hence, DeMorgan's Laws are verified algebraically.
|
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|
What exactly happens in the algebraic steps here? $$ \frac{n(n+1)}{2} + (n+1) = (n+1)(\frac{n}{2} + 1) = \frac{(n+1)(n+2)}{2} $$
I don't understand what happens from the first to the second and from the second to the third one.
|
For the first equality, $n+1$ is factored out. But it is easier if read from right to left:
$$
(n+1)\bigg(\frac{n}{2}+1\bigg) = (n+1) \cdot \frac{n}{2}+ (n+1) \cdot 1 = \frac{n(n+1)}{2}+(n+1)
$$
For the second equality, the sum $\frac{n}{2}+1$ is evaluated:
$$
(n+1)\bigg(\frac{n}{2}+1\bigg) = (n+1)\bigg(\frac{n}{2}+\frac{2}{2}\bigg) = \frac{(n+1)(n+2)}{2}
$$
|
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|
Integration of a function containing inverse trigonometric functions Q. $$\int \sin\left\{2\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)\right\}dx$$
$\implies$ $$\int \sin\left\{\sin ^{-1}\left(\frac{2\left(\sqrt{\frac{3-x}{3+x}}\right)}{1+\left(\sqrt{\frac{3-x}{3+x}}\right)^2}\right)\right\}dx$$
$\implies$ $$\int \:\frac{\left(\sqrt{3^2-x^2}\right)}{3}dx$$
$\implies$ANSWER $=$$$\frac{1}{3}\left(\left[\frac{x}{2}\sqrt{3^2-x^2}\right]+\left[\frac{9}{2}\sin ^{-1}\left(\frac{x}{3}\right)\right]\right)=\left(\left[\frac{x}{6}\sqrt{3^2-x^2}\right]+\left[\frac{3}{2}\sin ^{-1}\left(\frac{x}{3}\right)\right]\right)$$
Am I right?
|
Set $u=\arctan\sqrt{\dfrac{3-x}{3+x}}$. Then
$$
\sqrt{\dfrac{3-x}{3+x}}=\tan u
$$
and
$$
\dfrac{3-x}{3+x}=\tan^2u
$$
and, with straighforward computations,
$$
x=3\cos2u
$$
so
$$
dx=-6\sin2u\,du
$$
Thus your integral becomes
$$
\int-6\sin^{2}2u\,du
$$
that you should be able to carry over.
|
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|
Find the least value of $n$ such that $n^{25}\equiv_{83}37$ Let $n\in\mathbb{N}$. Find the least value of $n$ such that $n^{25}\equiv_{83}37$.
I concluded that $0<n<83$. Then I wrote $n^{25}$ as $\left(n^5\right)^5$ and let $t=n^5$. Now, it is hard to solve $t^5\equiv_{83}37$. How to solve this?
|
Working mod $83$, a prime number. Fermat's little theorem says:
$$a^{83} \equiv a$$ for any integer $a$ so
$$a^{m \cdot 82+1} \equiv a$$ for any $a$ and $m\ge 0$ integers
so in particular
$$37^{m \cdot 82+1} \equiv 37$$
Now try to find $m$ so that $m\cdot 82 + 1 = k \cdot 25$ thus getting $37^{m \cdot 82+1}= (37^k )^{25}$. By trial and error or using the GCD algorithm we get the smallest such positive $m=7$
$$7\cdot 82 + 1 = 575 = 576-1 = 24^2-1 = 23 \cdot 25$$
with the corresponding $k=23$.
Therefore
$$(37^{23})^{25}\equiv_{83} 37$$
so now we are left with calculating $37^{23}$ mod $83$. Instead of raising $37$ to the power $23$ and determining its remainder mod $83$ it is simpler to obtain $37^{23}$ mod $83$ using the properties of modular multiplication.
Note that
$$23 = 16 + 4 + 2 + 1 = 2^4 + 2^2 + 2^1 + 2^0$$
and so
$$37^{23} \equiv 37^{2^4} \cdot 37^{2^2}\cdot 37^2 \cdot 37$$
Note the rule $37^{2^n} = (37^{2^{n-1}} )^2$. Therefore:
\begin{eqnarray*}
37^1 &\equiv & 37\\
37^2 &\equiv & 41\\
37^4 &\equiv & 41^2 \equiv 21\\
37^8 &\equiv & 21^2\equiv 26\\
37^{16} &\equiv & 26^2 \equiv 12
\end{eqnarray*}
and so
$$37^{23} \equiv 12\cdot 21\cdot 41 \cdot 37 \equiv 69$$
We get
$$69^{25} \equiv_{83} 37$$
The solution $69$ is the unique one mod $83$ and thus $n=69$ is the smallest solution.
|
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|
Integrating $\int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx$ We need to evaluate $\displaystyle \int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx$
and some solution to this starts as,
$\displaystyle\int_0^{\pi/2} {\sin^2x \over 1 + \sin x\cos x}dx =
\int_0^{\pi/2} {\{\sin(\pi/2 -x)\}^2 \over 1 + \sin (\pi/2 -x)\cos (\pi/2 -x)}dx$.
We fail to understand how this step has been carried out. Even variable substitution
does not seem to work.
Do you think that you could give us a hint?
|
Using double-angle formula, we get
$$I=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2+\sin 2 x} d x$$
$$I\stackrel{x\mapsto\frac{\pi}{4}-x}{=} 2 \int_{0}^{\frac{\pi}{4}} \frac{1}{2+\cos 2 x} d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin 2 x}{2+\cos 2 x} d x$$
Since $\dfrac{1}{2+\cos 2 x}$ is even and $\dfrac{\sin 2 x}{2+\cos 2 x}$ is odd, therefore
$$ \displaystyle I=2 \int_{0}^{\frac{\pi}{4}} \frac{1}{1+2 \cos ^{2} x} d x$$
$$\displaystyle I=2 \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} x}{\sec ^{2} x+2} d x
\displaystyle \stackrel{t=\tan x}{=} 2 \int_{0}^{1} \frac{d t}{3+t^{2}}=\dfrac{\pi}{3 \sqrt{3}} $$
|
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|
Find $\cos(x+y)$ and $\sin(x+y)$ given that $\cos x + \cos y = a$ and $\sin x + \sin y = b$ If $\cos x + \cos y = a$ and $\sin x + \sin y = b$.
Find $\cos(x+y)$ and $\sin(x+y)$.
I only need some hints to start as I am not able to get any way to go forward to.
|
HINT:
Let $u = \cos x + i \sin x$, $v = \cos y + i \sin y$. Then
\begin{eqnarray}
u + v = a+ i b \\
\frac{1}{u} + \frac{1}{v} = a -i b
\end{eqnarray}
so
$$u v = (a+ib)/(a-ib)$$
But $uv = \cos(x+y) + i \sin(x+y)$. Now take real and imaginary parts.
|
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|
Proof irrationality $n\sqrt{11}$
Prove that $\sqrt{11}$ is irrational, subsequently prove that $n\sqrt{11}$ is also irrational for every $n \in \mathbb{N}$. You are allowed to use that if $p$ is prime, and $p | a^2$, then $p|a$.
Can't you also prove that $n\sqrt{11}$ is irrational simply by saying:
Assume $n\sqrt{11}$ is rational, then $n\sqrt{11} = \dfrac{p}{q}$, so $\sqrt{11} = \dfrac{p}{qn}$, which would mean that $\sqrt{11}$ is rational, which is false (assume we already proved that it's false).
So we know $n\sqrt{11}$ is irrational by contradiction.
|
I guess that your post also asked for a proof that $\sqrt{11}$ is irrational? Here it is.
Assume, for contradiction, that $\sqrt{11}$ is rational. Let $a$ and $b$ be relatively prime, positive integers. Then
\begin{align*}
\sqrt{11} &= \frac{a}{b} \\
11 &= \frac{a^2}{b^2}\\
11 b^2 &= a^2 .
\end{align*}
This means that $11 \vert a^2$ and so $11 \vert a$ as 11 is prime. So $a=11c$ and
\begin{align*}
11 b^2 &= (11c)^2 \\
b^2 &= 11c^2.
\end{align*}
This means that $11 \vert b^2$ and so $11 \vert b$ as 11 is prime. But that both $11 \vert a$ and $11 \vert b$ contradicts $a$ and $b$ being relatively prime, so $\sqrt{11}$ is not rational, which means that it is irrational. Q.E.D.
|
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|
Algebra problem solve for a,b,c and d? Can anyone find the values of these integers: a,b,c and d?
$$1+\sqrt{2}+\sqrt{3}+\sqrt{6} = \sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
a+b+c+d = ?
Thank you.
|
$$F=1+\sqrt2+\sqrt3+\sqrt6=(1+\sqrt2)(1+\sqrt3)$$
$$F^2=a+\sqrt{b+\sqrt{c+\sqrt d}}$$
Squaring we get $$F^2=(1+\sqrt2)^2(1+\sqrt3)^2=(3+2\sqrt2)(4+2\sqrt3)=12+8\sqrt2+6\sqrt3+4\sqrt6\implies a=12$$
$$F^4=a^2+b+\sqrt{c+\sqrt d}+2a\sqrt{b+\sqrt{c+\sqrt d}}$$
$$F^4=(3+2\sqrt2)^2(4+2\sqrt3)^2=(17+12\sqrt2)(28+16\sqrt3)$$
$\implies a^2+b=17\cdot28$ but $a=12$
Can you take it home from here?
|
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|
Integral ${\large\int}_0^1\left(-\frac{\operatorname{li} x}x\right)^adx$ Let $\operatorname{li} x$ denote the logarithmic integral
$$\operatorname{li} x=\int_0^x\frac{dt}{\ln t}.$$
Consider the following parameterized integral:
$$I(a)=\int_0^1\left(-\frac{\operatorname{li} x}x\right)^adx.$$
Can we find a closed form for this integral?
We can find some special values of this integral:
$$I(0)=1,\,\,I(1)=1,\,\,I(2)=\frac{\pi^2}6,\,\,I(3)\stackrel?=\frac{7\zeta(3)}2$$
The last value was suggested by numeric computations, and I do not yet have a proof for it.
Can we prove the conjectured value of $I(3)$?
One could expect that $I(4)$ might be a simple rational (or at least algebraic) multiple of $\pi^4$ but I could not find such a form.
Can we find closed forms for $I(4),I(5)$ and other small integer arguments?
|
In reference to Kirill's answer, I will show that indeed $$ I(3) = \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{xyz} \frac{dx \, dy \, dz}{x+y+z-2} = \frac{7}{2}\zeta(3).$$
I will make the same change of variables I made in my answer to your other more recent question.
$$ \begin{align} I(3) &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \int_{1}^{\infty} \frac{1}{y+z-2}\left( \frac{1}{x} - \frac{1}{x+y+z-2} \right) \, dx \, dy \, dz \\ &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{\log (y+z-1)}{y+z-2} \, dy \, dz \\ &= 2 \int_{1}^{\infty} \int_{1}^{z} \frac{1}{yz} \frac{\log (y+z-1)}{y+z-2} \, dy \, dz \\ &= 2 \int_{2}^{\infty} \int_{u-1}^{u^{2}/4} \frac{1}{v} \frac{\log(u-1)}{u-2} \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{1} \\ & = 4 \int_{2}^{\infty} \frac{\log(u-1)}{u-2}\int_{0}^{u-2} \frac{1}{u^{2}-t^{2}} \, dt \, du \tag{2} \\ &= 2 \int_{2}^{\infty} \frac{\log^{2}(u-1)}{u(u-2)} \, du \\ &= 2 \int_{1}^{\infty} \frac{\log^{2}(w)}{w^{2}-1} \, dw \\ &= 2 \int_{0}^{1} \frac{\log^{2}(s)}{1-s^{2}} \, ds \tag{3} \\ &= 2 \int_{0}^{1} \log^{2}(s) \sum_{n=0}^{\infty} s^{2n} \, ds \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{1} \log^{2}(s) s^{2n} \, ds \\ &= 4 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \\ &= 4 \left(\sum_{n=1}^{\infty} \frac{1}{n^{3}} - \sum_{n=1}^{\infty} \frac{1}{(2n)^{3}} \right) \\ &= 4 \left( \zeta(3) - \frac{\zeta(3)}{8}\right) \\ &= \frac{7}{2} \zeta(3) \end{align}$$
$(1)$ Make the change of variables $u=y+z, v=yz$.
$(2)$ Make the substitution $t^2 = u^{2}-4v$.
$(3)$ Make the substitution $s = \frac{1}{w}$.
|
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|
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.
How can I proceed?
|
We need to prove that
$$\prod_{cyc}\frac{a^2+b^2-c^2}{2ab}\leq\frac{1}{8}$$ or
$$a^2b^2c^2\geq\prod_{cyc}(a^2+b^2-c^2)$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)\geq0,$$
which is Schur.
Done!
|
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|
How find the matrix $X$ such $e^{X}$ is give it
Question:
let matrix $$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix} ,e^{X}=\begin{bmatrix}
-1&2\\
0&-1
\end{bmatrix}$$
and such $a+d=0$,
Find the matrix $X$
my idea
$$e^{Tr{(X)}}=det{(e^{X})}$$
so
$$1=1$$ is true
I think we can find the matrix function: $e^X$,but I have find this is ver ugly,so I think this problem have other methods
|
Here's a sketch. Since $d = -a$, you can write $X = \left(\begin{array}{rr} a & b \\ c & -a\end{array}\right)$, which gives
$$X^2 = \left(\begin{array}{cc} a^2 + bc & 0 \\ 0 & a^2 + bc\end{array}\right) = (a^2+bc)I.$$
This allows you to write a general expression for $X^n$: $X^{2n} = (a^2+bc)^n I$ and likewise $X^{2n+1} = (a^2+bc)^nX$.
With these two in hand, we can evaluate the exponential:
$$\exp X = \sum_{n=0}^{\infty} \frac{1}{n!} X^n = \sum_{n=0}^{\infty} \frac{1}{(2n)!} X^{2n} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}X^{2n+1}.$$
Making use of our above observation:
$$\exp X = \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n)!}\, I+ \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n+1)!}\,X.$$
Since $\exp X = \left(\begin{array}{rr} -1 & 2 \\ 0 & -1\end{array}\right)$, we see that $c= 0$, so that
$$\exp X = \sum_{n=0}^{\infty} \frac{a^{2n}}{(2n)!}\, I+\sum_{n=0}^{\infty} \frac{a^{2n}}{(2n+1)!}\,X.$$
From here, if you equate terms, you will find that $a = i\pi+2k\pi$ for some $k\in\Bbb Z$ and $b = -2$ as Nishant derived.
|
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|
How to evaluate the limit $\lim_{x\to 0} ((1+2x)^{1/3 }-1)/ x$ without using the l'Hospital's rule? $$\lim_{x\to 0} \frac{(1+2x)^{\frac{1}{3}}-1}{x}.$$
Please do not use the l'hospital's rule as I am trying to solve this limit without using that rule... to no avail...
|
Just in the same spirit as Paul, remember that, for small values of $y$, $$(1+y)^n=1+n y+\frac{1}{2} n(n-1) y^2+\frac{1}{6}n (n-1) (n-2) y^3+O\left(y^4\right)$$ So, replace $n$ by $\frac{1}{3}$ and $y$ by $2x$ and obtain $$(1+2x)^{\frac{1}{3}}=1+\frac{2 x}{3}-\frac{4 x^2}{9}+\frac{40 x^3}{81}+O\left(x^4\right)$$ $$\frac{(1+2x)^{\frac{1}{3}}-1}{x}=\frac{2}{3}-\frac{4 x}{9}+\frac{40 x^2}{81}+O\left(x^3\right)$$ which shows the limit as well as the manner it is approached
|
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|
Prove that $\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$ There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me :
If $a_1,a_2,a_3,...,a_n$ IS an arithmetic progression and $a_n$ is NOT equal to 0 then prove the following statement :
$\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$
Thanks in advance...
|
Here's sketch, not complete proof.
Since $A$ is an arithmetic progression, Let $a_{i+1}-a_i = d$
$$\begin{align}\dfrac{1}{a_ia_{i+1}} &= \dfrac{d}{d}\dfrac{1}{a_ia_{i+1}}
\\&= \dfrac{1}{d}\dfrac{a_{i+1}-a_i}{a_ia_{i+1}} \qquad \text{since } a_{i+1}-a_i = d
\\&= \dfrac{1}{d}\left(\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}}\right)\end{align}$$
Therefore, $$\begin{align} \dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots + \dfrac{1}{a_{n-1}a_n}&=
\sum\limits_{i = 1}^{n-1}\dfrac{1}{a_ia_{i+1}} \\&= \dfrac{1}{d}\sum\limits_{i = 1}^{n-1}\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}} \\&= \dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac{1}{a_2}+\dfrac{1}{a_2}-\cdots - \dfrac{1}{a_{n-1}} + \dfrac{1}{a_{n-1}}-\dfrac{1}{a_n} \right)\\&=\dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac1{a_n}\right)\end{align}$$
Now, take LCM and use the fact that $a_n = a_1 +(n-1)d$
|
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|
Prove that $(n!)^ 2 \gt n^n$ Prove the above by
*
*by mathematical induction
*By any other method.
I was just asked to prove this so I thought of using mathematical induction.
My effort :
I started first by verification and the inequality was true for n=3, 4...
Then the assumption step $(k!)^ 2 \gt k^k$
But after that while proving it for n=k+1 I was getting stucked inthe very next step.
|
Assuming that $(k!)^ 2 \gt k^k$ we write
$$((k+1)!)^ 2 = (k!)^2 \cdot (k+1)^2 \gt k^k \cdot (k+1)^2. \tag {*}$$
Now we compare the RHS (*) with $(k+1)^{k+1}\ $:
$$\dfrac{(k+1)^{k+1}}{ k^k \cdot (k+1)^2}=\dfrac{(k+1)^{k}}{ k^k \cdot (k+1) }=\dfrac{1}{k+1} \cdot {\left(1+\dfrac{1}{k}\right)^{k}}. $$
Now using the binomial formula, we have for $k \geqslant {3}$
$$
\left(1+\dfrac{1}{k}\right)^{k}=1+k\cdot \dfrac{1}{k}+\dfrac{k(k-1)}{2!\ k^2}+\ldots+ \dfrac{k(k-1)\cdots(k-(k-2)) }{(k-1)!\ k^{k-1}}+\dfrac{k(k-1)\cdots2\cdot{1}}{k!\ k^{k}}\\
< 1+1+\dfrac{1}{2!}+\ldots+\dfrac{1}{(k-1)!}+\dfrac{1}{k!} \\
< 1+1+\dfrac{1}{2}+\ldots+\dfrac{1}{2^{k-1}}+\dfrac{1}{2^k} < 3,
$$
thus for $k \geqslant {3}$
$$\dfrac{1}{k+1} \cdot {\left(1+\dfrac{1}{k}\right)^{k}} < 1.$$
Therefore,
$$ k^k \cdot (k+1)^2 > (k+1)^{k+1}.$$
Edit:
The multiplicative variant of Gauss's trick is very elegant. Writing squared factorial in the right and reverse order we have
$$(n!)^2 = [1\cdot 2 \cdots (n-1)\cdot n]\cdot[n\cdot (n-1) \cdots 2\cdot 1]= \prod_{k=1}^{n}{k(n+1-k)} . $$
In order to show that each term ${k(n+1-k)} \geqslant n$ we consider
$$k(n+1-k) - n = kn + k -k^2 - n = n(k-1) -k(k-1)=(k-1)(n-k) \geqslant 0$$
for $1 \leqslant k \leqslant n.$
Thus
$$\prod_{k=1}^{n}{k(n+1-k)} \geqslant \prod_{k=1}^{n}{n} = n^n.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integrating $\sin^2(x) \cos x $ become $\frac {1}{3}\sin^3(x) + C$? What mathematical process is being done? Why does the $\cos x$ disappear?
|
Notice that applying the chain rule to $\frac{1}{3}\sin^3(x)$ yields $\sin^2(x)\cos(x)$. Indeed you have $\int\sin^2(x)\cos(x)\,dx=\int\sin^2(x)\,d(\sin(x))=\frac{1}{3}\sin^3(x)+c$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this series.
|
Consider the series
\begin{align}
S = \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \ln\left(1 + \frac{1}{2k}\right) \right]
\end{align}
for which, by using the logarithm in series form, it becomes
\begin{align}
S &= \sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \frac{1}{2k} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n \, (2k)^{n}} \right] \\
&= \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \frac{(-1/2)^{n}}{n \, k^{n}} \\
&= \sum_{n=2}^{\infty} \frac{(-1/2)^{n}}{n} \, \zeta(n) \\
&= \sum_{n=1}^{\infty} \frac{(-1/2)^{n+1}}{n+1} \, \zeta(n+1).
\end{align}
Now the generating function for the zeta function is given by
\begin{align}
\sum_{n=1}^{\infty} \zeta(n+1) (-1)^{n+1} x^{n} = \gamma + \psi(x+1)
\end{align}
and by integration leads to
\begin{align}
\sum_{n=1}^{\infty} \frac{(-x)^{n+1}}{n+1} \, \zeta(n+1) = \gamma x + \ln\Gamma(x+1).
\end{align}
Letting $x = 1/2$ in this series leads to the value of the series for $S$, namely
\begin{align}
\sum_{k=1}^{\infty} \left[ \frac{1}{2k} - \ln\left(1 + \frac{1}{2k}\right) \right] = \frac{1}{2} \ln\left(\frac{\pi}{4}\right) + \frac{\gamma}{2}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$ I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
|
Starting with the Beta function
\begin{align}
B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt
\end{align}
differentiate with respect to $x$ and $y$ twice. This leads to
\begin{align}
I &= \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \ln^{2}(t) \, \ln^{2}(1-t) \, dt \\
&= \partial_{x}^{2} \partial_{y}^{2} B(x,y).
\end{align}
Now the integral in question is the case for $x=y=1$. For this, it is seen that
\begin{align}
\partial_{x}^{2} \left. B(x,y) \right|_{x=1} = \frac{1}{y} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right].
\end{align}
Differential with respect to $y$ yields
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=1} &= \frac{1}{y} \left[ - \psi'''(y+1) - 2 \psi''(y+1) \left( \psi(1) - \psi(y+1) \right) + 2 \left( \psi'(y+1) \right)^{2} \right] \\
& \hspace{5mm} - \frac{2}{y^{2}} \left[ - \psi''(y+1) - 2 \psi'(y+1) \left( \psi(1) - \psi(y+1) \right) \right] \\
& \hspace{10mm} + \frac{2}{y^{3}} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right].
\end{align}
Now
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} &= 4 - \psi'''(2) + 4 \psi''(2) - 4 \psi'(2) + 2 \left( \psi'(2) \right)^{2} .
\end{align}
Since
\begin{align}
\psi'(2) &= \frac{\pi^{2}}{6} -1 \\
\psi''(2) &= 2 - 2\zeta(3) \\
\psi'''(2) &= \frac{\pi^{4}}{45} - 6
\end{align}
then
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} = 24 - \frac{4 \pi^{2}}{3} - \frac{\pi^{4}}{90} - 8 \zeta(3).
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Remainder of a summation divided by $2^{12}$ For a positive integer $n$, let $f(n)$ be equal to $n$ if there is an integer $x$ such that $x^2-n$ is divisible by $2^{12}$, and let $f(n)$ be $0$ otherwise. Determine the remainder when $$\sum_{n=0}^{2^{12}-1} f(n)$$ is divided by $2^{12}$.
Source: Caltech Harvey Mudd Tournament
|
Basically, all we need is the fact that the squares in $(\mathbb{Z}/2^k\mathbb{Z})^\times$, for $k\geq 3$, are exactly those that are $1 \pmod{8}$. This follows from the standard fact that $(\mathbb{Z}/2^k\mathbb{Z})^\times \cong Z_2\times Z_{2^{k-2}}$, that all squares are $1 \pmod{8}$, and that there are exactly $2^{k-3}$ elements in the latter, and the same number of squares in the former.
The remaining squares in $\mathbb{Z}/2^n\mathbb{Z}$ are all divisible by $4$, so we can work inductively. If $F(k) = \sum_{n=0}^{2^k-1} f(n)$ ($k\geq 3$), then: $$F(k) - 4F(k-2) = \sum_{n\equiv 1\pmod{8}} n = \sum_{j=0}^{2^{k-3}-1} (8j+1) = 4\cdot 2^{k-3}(2^{k-3}-1) + 2^{k-3}$$ $$=2^{2k-4} - 2^{k-1} + 2^{k-3}$$
Since $F(2) = 1$, we have $F(12) = 4^5 + \sum_{j=2}^6 4^{6-j}(2^{4j-4} - 2^{2j-1} + 2^{2j-3}) = 1390080$.
Taken modulo $2^{12}$, this gives us $1536$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof of an identity involving binomial coefficients I have found numerically that the following identity holds:
\begin{equation}
\sum_{n=0}^{\frac{t-x}{2}} n 2^{t-2n-x}\frac{\binom{t}{n+x}\binom{t-n-x}{t-2n-x}}{\binom{2t}{t+x}} = \frac{x^2+t^2-t}{2t-1},
\end{equation}
where $n$, $t$, and $x$ are positive integers ($x \leq t$). To make it more visible, values of $n$ range from $0$ to $\frac{t-x}{2}$.
Any clue about how to prove it?
Thanks, Antonio
|
This is actually easier then I thought. All what we have to do in here is to use identities of the gamma function along with the Gauss' theorem for the hypergeometric function.
\begin{eqnarray}
&&\sum\limits_{n=0}^{\lfloor \frac{t-x}{2}\rfloor}
n 2^{t-x-2 n} \frac{\binom{t}{n+x} \binom{t-n-x}{t-2 n-x}}{\binom{2 t}{t+x}}=\\
&&\frac{t!(t+x)!}{(2 t)!x!} \sum\limits_{n=0}^{\lfloor \frac{t-x}{2}\rfloor}n 2^{t-x-2 n} \frac{(x-t)^{(2 n)}}{n! (1+x)^{(n)}}=\\
&&\frac{t!(t+x)!}{(2 t)!x!} \cdot \frac{\Gamma(t-x+1)}{\Gamma(\frac{t-x+1}{2})\Gamma(\frac{t-x+2}{2})} \sqrt{\pi} \sum\limits_{n=1}^{\lfloor \frac{t-x}{2}\rfloor} \frac{(\frac{x-t+1}{2})^{(n)} (\frac{x-t}{2})^{(n)}}{(n-1)! (1+x)^{(n)}}=\\
&& \frac{2^{-2+t-x}(x-t)(x-t+1)t!(t+x)!}{\Gamma(1+2 t) \Gamma(2+x)}\cdot F_{2,1}\left[\begin{array}{rr} \frac{x-t+3}{2} & \frac{x-t+2}{2} \\2+x \end{array};1\right]\\
&& \frac{2^{-2+t-x}(x-t)(x-t+1)t!(t+x)!}{\Gamma(1+2 t) \Gamma(2+x)}\cdot \frac{\Gamma(2+x) \Gamma(t-1/2)}{\Gamma(x/2+t/2+1/2)\Gamma(x/2+t/2+1)}=\\
&&\frac{1}{2} \frac{(t-x)(t-x-1)}{2 t-1}
\end{eqnarray}
In the second line from the above we replaced the binomial factors by Pochhammer symbols and simplified the expression. In the third line from the above we used the duplication formula for the Gamma function to reduce the term in sum to Pochhammers symbols with $n$ rather than $2 n$. In the fourth line we used the definition of the hypergeometric function and in the fifth line we used the Gauss' theorem for the hypergeometric function at unity. Finally in the last line we simplified the whole thing again making use of the duplication formula for the gamma function.
|
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|
The integral of $x^3/(x^2+4x+3)$ I'm stumped in solving this problem. Every time I integrate by first dividing the $x^3$ by $x^2+4x+3$ and then integrating $x- \frac{4x^2-3}{x+3)(x+1)}$ using partial fractions, I keep getting the wrong answer.
|
$$\frac{2x^3}{(x+1)(x+3)}=\frac{[(x+3)-(x+1)]x^3}{(x+1)(x+3)}=\frac{x^3}{x+1}-\frac{x^3}{x+3}$$
For $\displaystyle\int\dfrac{x^3}{x+a}dx,$ set $x+a=h$
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to solve $f''(e^{x} \cdot \sin x)$ How do I find the derivative $f''(e^{x} \cdot \sin x)$.
I start to find $f'(x)$ by using the product rule $f'(e^{x} \cdot \sin x) = e^{x} \cdot \sin x + e^{x} \cdot \cos x = e^{x}(\sin x + \cos x)$
Now when I have $f'(x)$ I use it to find $f''(x)$.
I split upp the problem by finding the derivative of $e^{x} \cdot \sin x$ first and then the derivative of $e^{x} \cdot \cos x$
$f'(e^{x} \cdot \sin x) = e^{x} \sin x + e^{x} \cos x$
$f'(e^{x} \cdot \cos x) = e^{x} - \sin x + e^{x} \cos x$
Then I plug this values back to $f''(x)$
$f''(x) = (e^{x} \sin x + e^{x} \cos x) + (e^{x} - \sin x + e^{x} \cos x)$
$f''(x) = 2 e^{x} (2 \cos x)$
I know this is wrong answer, the correct answer is $f''(x) = 2 e^{x} (\cos x)$
Where did it go wrong in my calculation?
Thanks!!
|
The last line is wrong :
$f''(x) = (e^{x} \sin x + e^{x} \cos x) + (e^{x} *(-\sin x) + e^{x} \cos x)$
$f''(x) = 2 e^{x} (\cos x)$
The $2 \cos x$ just came out of inattention !
|
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|
What is known about $x^m + y^m = z^n$ over $\mathbb{N}$ when $m,n \geq 2$ and $m \neq n$? So Fermat's Last Theorem resolves the question of positive integer solutions to $x^m + y^m = z^n$ when $m = n \geq 3$. But what about if $m \neq n$ and $m,n \geq 2$? Is anything general known about when there are positive integer solutions? Clearly if $gcd(m,n) \geq 3$ then there are no solutions by Fermat's Last Theorem. So let's assume $gcd(m,n) \leq 2$.
|
I attempt to solve the equation $x^3+y^3=z^n, x,y,z, n\in N \textrm{ and }n>3.$
Noting that $(1,1,2,1)$ is a solution and multiplying it by $2^{3k}$, we have
$$1^3+1^3=2^1 \Rightarrow \left(2^k\right)^{3 }+\left(2^k\right)^3=2^{3 k+1}$$
Therefore there are infinitely many solutions
$$(2^k,2^k, 2,3 k+1),$$
where $k\in N.$
Also $(1,2,3,2)$ is a solution and multiplying it by $3^{3k}$, we have
$$1^3+2^3=3^2 \Rightarrow (3^k)^{3 }+\left(2 \cdot 3^k\right)^3=3^{3 k+2}$$
Therefore there are infinitely many solutions
$$(3^k,2(3^k), 3,3 k+2) \textrm{ or } (2(3^k) ,3^k, 3,3 k+2), $$ where $k\in N.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplification of geometric series. Can someone please help simplify this series?
$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$
In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$
However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$.
I understand if the problem is $\sum_{k=1}^\infty\big(\frac 12\big)^k$. However, it is the $k$ term that I don't understand. The answer is suppose to be $2$.
Can someone help me? Thank you.
|
\begin{align*}
\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k &= \frac{1}{2} + 2 \frac{1}{2^2} + 3 \frac{1}{2^3} + 4 \frac{1}{2^4} \dots\\
&= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + 2 \frac{1}{2^3} + 3 \frac{1}{2^4} + \dots\right)\\
&= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \right) + \left(\frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^4} + \dots \right) + \dots\\
&= 1 + \frac{1}{2} + \frac{1}{2^2} + \dots\\
&= 2
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit of a Riemann Sum The function is $f(x) = 1-x^2$. I'm stuck as I can't factor the expression in the last line to find the limit.
|
The expression from your last line can be simplified:
\begin{align}
\frac{1}{n} \bigg[n-\frac{1}{n^2}\bigg(\frac{n(n+2)(2n+1)}{6}\bigg)\bigg]
& = \bigg[\frac{n}{n}-\frac{n(n+2)(2n+1)}{6n^3}\bigg] \\[0.1in]
& = 1-\frac{1}{6} \bigg( \frac{n+2}{n} \bigg)\bigg(\frac{2n+1}{n}\bigg) \\[0.1in]
& = 1-\frac{1}{6} \bigg( 1+\frac{2}{n} \bigg)\bigg( 2+\frac{1}{n} \bigg)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15}\right) = \pi$ How to prove the following identity
$$\sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15}\right) = \pi$$
I am totally clueless in this one. Would you help me, please? Any help would be appreciated. Thanks in advance.
|
Here is a summary of the proof given by Bailey, Borwein, Borwein, and Plouffe in The Quest for Pi in only a few lines of integration.
To begin they note the following definite integrals as summations, $n=1,\ldots,7$:
$$ \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^8} dx =
\int_0^{\frac{1}{\sqrt{2}}} \sum_{k=0}^\infty x^{n-1+8k} dx =
\frac{1}{2^{n/2}} \sum_{k=0}^\infty \frac{1}{16^k(8k+n)} $$
If the fractional factor of the summation in the Question is expanded by partial fractions:
$$ \frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15} =
\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6} $$
then the integrals above can be applied to give:
$$ \sum_{k=0}^\infty \frac{1}{16^k} \left( \frac{4}{8k+1} - \frac{2}{8k+4}
- \frac{1}{8k+5} - \frac{1}{8k+6} \right) $$
$$ = \int_0^{\frac{1}{\sqrt{2}}}\frac{4\sqrt{2} -8x^3 -4\sqrt{2}x^4 -8x^5}{1-x^8} dx
$$
At this point the authors claim a substitution of $y= x \sqrt{2}$, making:
$$ \int_0^{\frac{1}{\sqrt{2}}}\frac{4\sqrt{2} -8x^3 -4\sqrt{2}x^4 -8x^5}{1-x^8} dx =
\int_0^1 \frac{16y-16}{y^4-2y^3+4y-4} dy $$
Finally the last integral may be expanded by partial fractions to give:
$$ \int_0^1 \frac{4y}{y^2-2} dy - \int_0^1 \frac{4y-8}{y^2-2y+2} dy = \pi $$
By way of explanation the authors point out that this rigorous proof was sought only after the discovery of the apparent integer relations among summations and $\pi$ via the PSLQ algorithm.
|
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|
How to find the value of $m$ which is a power of a polynomial, when divided by a linear polinomial gives some remainder?
Q) The value of $m$ if $2x^m+x^3-3x^2-26$ leaves remainder of 226 when divided by $x-2$.
(1) 0
(2) 7
(3) 10
(4) all of these
How i solved it
let $p(x)=2x^m+x^3-3x^2-26$ and $g(x)=x-2=0$
therefore $x=2$
substituting the value of $x$ in $p(x)$
then:
$2(2)^m+(2)^3-3(2)^2-26=226$
$= ((2)^2)^m+8-12-26=226$
$= ((2)^2)^m = 226+30$
$= ((2)^2)^m =(16)^2 => ((2)^2)^m=((2)^4)^2 $
so on comparing both sides, since the base are same on both sides, taking the exponents
therefore, $ 2m=8 => m=4$!!
but there is not any option there....
so how to solve this?
thanks
|
It is $2\cdot 2^m=2^ {m+1}$ $(\ne (2^2)^m,$ which is the wrong point in your solution). So your equation should read
$$2^{m+1}=2^8$$ from where you get $m=7.$
|
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|
Continuous Functions such that $f(0) = 1$ and $f(3x) - f(x) = x$ Just had a midterm with the following problem:
Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.
I was just curious how this would end up.
During the exam I tried setting $f(1) = c$, and end up getting $f(3) = 2c, f(9) = 4c, \ldots$ and from that $c = 3/2$ with the $f(0) = 1$ condition, but I was unable to make more progress on it.
|
We have: $f(x) - f\left(\frac{x}{3}\right) = \dfrac{x}{3}$
$f\left(\frac{x}{3}\right) - f\left(\frac{x}{9}\right) = \dfrac{x}{9}$
.....
$f\left(\frac{x}{3^{n-1}}\right) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3^n}$.
Add the above equations:
$f(x) - f\left(\frac{x}{3^n}\right) = \dfrac{x}{3}\cdot \left(1 + \frac{1}{3} + ... + \frac{1}{3^{n-1}}\right)$.
Letting $n \to \infty$ in the above equation and using continuity of $f$ at $x = 0$ we have:
$f(x) - f(0) = \dfrac{x}{3}\cdot \dfrac{1}{1-\frac{1}{3}} = \dfrac{x}{2}$. Thus:
$f(x) - 1 = \dfrac{x}{2}$, and $f(x) = 1 + \dfrac{x}{2}$
|
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|
Integrals: Partial Fractions $$ \int \frac{x^2-x+12}{x^3+3x} $$
I factored the denominator to get $ x(x^2+3) $. I then seperated the x and the $x^2+3$ into the partials $\frac{A}{x}$ and $\frac{Bx+C}{x^2+3}$.
After combining the two, I came up with $$ \frac{A(x^2+3)+ x(Bx+C)}{x(x^2+3)}$$
This is where I'm stuck. I'm assuming that A would equal 1 because it is where $x^2$ is in the original problem. The same would go for b equaling -1 and $C = 12$. I'm not so sure though.
|
You are almots done. As you wrote $$ \frac{x^2-x+12}{x^3+3x}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$$ then $${x^2-x+12}=A(x^2+3)+(Bx+C)x$$Expand the rhs, so $${x^2-x+12}=(A+B)x^2+Cx+3A$$ Identify the coefficients for a given power of $x$; so $$A+B=1$$ $$C=-1$$ $$3A=12$$ So, $A=4$, $B=-3$,$C=-1$
|
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|
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
|
We have $a+ib=2w$ where $w^3=1$
Taking modulus on either sides $\sqrt{a^2+b^2}=2|w|$
Directly taking modulus on either sides on $(a+ib)^3=8,$
we get $(\sqrt{a^2+b^2})^3=8\implies (a^2+b^2)^3=8^2=(2^3)^2=2^6$
As $a^2+b^2>0, a^2+b^2=\sqrt[3]{2^6}=(2^6)^{\frac13}=?$
|
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|
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
|
If you want to use the power series way to do it(although I think it's overkill), by convex property of $f(x)=x^n$ we have $\frac{(a^n+b^n+c^n)}{3}\geq (\frac{a+b+c}{3})^n$ or equivalently $(a^n+b^n+c^n)\geq \frac{(a+b+c)^n}{3^{n-1}}$, where $n$ is positive integer.
Therefore your left hand side(series expansion) is greater or equal to $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...=\frac{3}{2}$
|
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|
Solving the ODE $(x^2 - 1) y''- 2xy' + 2y = (x^2 - 1)^2$ I want to solve this ODE:
$$(x^2 - 1)y'' - 2xy' + 2y = (x^2 - 1)^2.$$
I found out that $y_1 = x$ and $y_2 = x^2+1$ are solutions of the associated homogeneous equation, $x$ by inspecting, and $x^2+1$ by multiplying $x$ by a function and solving. Now, I know that a particular solution of the problem is given by $y_p = k_1x + k_2(x^2+1)$, for some functions $k_1,k_2$, which must satisfy the following relations: $$\left\{ \begin{array}{l} k_1'x + k_2'(x²+1) = 0 \\k_1' + k_2'(2x) = (x^2-1)^2 \end{array}\right.$$
From the second equation, I have that $k_1' = (x^2-1)^2 - 2xk_2'$, and in the first equation we get that: $$ x(x^2-1)^2 - 2x^2k_2' + (x^2+1)k_2' = 0 \implies (-x^2+1)k_2' = -x(x^2-1)^2$$
and so $k_2' = x(x^2-1) = x³ - x$. Back in the second equation, we obtain: $$k_1' = (x^2-1)^2 - 2x^2(x^2-1) \implies k_1' = (x^2-1)(-x²-1) = -x⁴ + 1.$$
By simple integration, we get $k_1 = -\frac{x^5}{5}+x$ and $k_2 = \frac{x^4}{4} - \frac{x^2}{2}$. By this calculation, the particular solution would be: $$y_p =-\frac{x^6}{5} + x^2 + \frac{x^6}{4} -\frac{x^4}{2} + \frac{x^4}{4} - \frac{x^2}{2} \\ \implies y_p = \frac{x^6}{20} - \frac{x^4}{4} + \frac{x^2}{2}.$$
Differentiating, we get $y_p' = \frac{3x^5}{10} - x^3 + x$ and $y_p'' = \frac{3x^4}{2} - 3x^2 + 1$. Notice that the only constant term on the left side will be a lonely $-1$. But this can't be right, since it'll have to appear a $+1$ term on the right side, to match the $+1$ in $x^4-2x^2 + 1$ on the right side.
I don't have a single clue of why this isn't working. This is bothering me since yesterday, and I can't think of anything. Can someone help me?
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I have redone the calculations and solved the problem. I'll leave my solution here, in case it helps someone. This time, my mistake was not writing the equation as: $$y'' - \frac{2x}{x^2 - 1}y' + \frac{2}{x^2 - 1}y = (x^2 - 1)$$
before applying the method.Really, $y_1 = x$ and $y_2 = x^2+1$ are solutions for the homogeneous equation. Then, we go for:
$$\left\{ \begin{array}{l} k_1'x + k_2'(x²+1) = 0 \\k_1' + k_2'(2x) = (x^2-1) \end{array}\right.$$
Multiplying the second equation by $x$ and subtracting the first equation from it yields $(x^2-1)k_2' = x(x²-1)$, and hence $k_2' = x$. Back into the second equation, we have $k_1' = -x^2 - 1$. Integrating, we obtain $k_1 = -\frac{x^3}{3}-x$ and $k_2 = \frac{x^2}{2}$. The constants of integration are not needed. I have done the calculations with them, and all of the terms containing them kill each other. So, we have as a particular solution: $$y_p = -\frac{x^4}{3}-x^2 + (x^2+1)\frac{x^2}{2} \\ \implies y_p = -\frac{x^2}{2} + \frac{x^4}{6}.$$
Differentiating we get: $$\begin{align}y_p' &= -x + \frac{2x^3}{3} \\ y_p'' &= -1+2x^2 \end{align}$$
and it is easy to check now that it is really a solution. Hence, the general solution is given by: $$y = -\frac{x^2}{2} + \frac{x^4}{6} + c_1x + c_2(x^2+1), \qquad c_1,c_2 \in \Bbb R.$$
|
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Inverse Trigonometric Integrals How to calculate the value of the integrals
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx
$$
and
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$
|
The first integral is not that difficult to evaluate. Note that
$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_1^{\infty} dx \frac{\arctan^2{x}}{x^2}\\ &= \int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} - \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2\end{align}$$
The first integral may be evaluated by a simple substitution $x=\tan{u}$ to get
$$\int_0^{\infty} dx \frac{\arctan^2{x}}{x^2} = \int_0^{\pi/2} du \frac{u^2}{\sin^2{u}} $$
The latter integral is equal to $\pi \log{2}$; the derivation of this result may be found here.
The second integral is evaluated by expansion and repeated integration by parts, as follows:
$$\begin{align} \int_0^1 dx \left ( \frac{\pi}{2} - \arctan{x} \right )^2 &= \frac{\pi^2}{4} - \pi \int_0^1 dx \, \arctan{x} + \int_0^1 dx \, \arctan^2{x}\end{align} $$
The first integral on the RHS is
$$ \begin{align}\int_0^1 dx \, \arctan{x} &= \left [ x \arctan{x} \right ]_0^1 - \int_0^1 dx \frac{x}{1+x^2}\\ &= \frac{\pi}{4} - \frac12 \log{2} \end{align} $$
The second integral is a little more involved, but it is along similar lines:
$$\begin{align} \int_0^1 dx \, \arctan^2{x} &= \left [ x \arctan^2{x} \right ]_0^1 - 2 \int_0^1 dx \frac{x}{1+x^2} \arctan{x} \\ &= \frac{\pi^2}{16} - \left [\log{(1+x^2)} \arctan{x} \right ]_0^1 + \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} \end{align} $$
The latter integral on the RHS may be evaluated by recognizing that
$$\begin{align} \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} &= \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} + \int_1^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} \\ &= \int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} + \int_0^1 dx \frac{\log{(1+x^2)}- 2 \log{x}}{1+x^2}\end{align} $$
The latter integrals are obtained through a mapping $x \mapsto 1/x$ in the previous integral. Thus,
$$\begin{align}\int_0^1 dx \frac{\log{(1+x^2)}}{1+x^2} &= \frac12 \int_0^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} + \int_0^1 dx \frac{\log{x}}{1+x^2} \\ &= - \int_0^{\pi/2} du \, \log{\cos{u}} - G \\ &= \frac{\pi}{2} \log{2} - G \end{align} $$
where $G$ is Catalan's constant. The source of the first integral on the RHS is the same as that found in the above link (here).
Putting this all together, we get that
$$\begin{align}\int_0^1 dx \frac{\arctan^2{x}}{x^2} &= \pi \log{2} - \frac{\pi^2}{4} + \frac{\pi^2}{4} - \frac{\pi}{2} \log{2} - \frac{\pi^2}{16} + \frac{\pi}{4} \log{2} - \frac{\pi}{2} \log{2} + G \\ &= G + \frac{\pi}{4} \log{2} - \frac{\pi^2}{16} \end{align}$$
which matches up with other people's assertions.
|
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|
Average order of Eulers totient function squared I was wondering if one has a nice asymptotic formula for the sum
$$\sum_{n\le x} \phi(n)^2$$
and if so, how does one calculate it. I know that one has $\sum_{n\le x} \phi(n) = \frac{3}{\pi^2}x^2+O(x\log x)$, but I can't seem to find similar results for other powers of $\phi$.
|
We can get the first term of the asymptotics with the Wiener-Ikehara
theorem. For this purpose we need to evaluate
$$L(s) = \sum_{n\ge 1} \frac{\varphi(n)^2}{n^s}.$$
Recall that
$$ n = \prod_p p^v \quad\text{implies}\quad
\varphi(n) = \prod_{p|n} (p^v - p^{v-1})$$
which we may either quote or derive using the fact that $\varphi(n)$
is multiplicative and the fact that there are $p^{v-1}$ values not
coprime to $p^v$ in the interval $[1, p^v].$
This implies that $L(s)$ has the following Euler product:
$$\prod_p
\left(1 + \frac{(p-1)^2}{p^s} + \frac{(p^2-p)^2}{p^{2s}} +
\frac{(p^3-p^2)^2}{p^{3s}} + \frac{(p^4-p^3)^2}{p^{4s}} +
\cdots\right).$$
Expanding the squares we get
$$\prod_p
\left(1 + \sum_{q\ge 1} \frac{p^{2q}}{p^{qs}}
- 2 \sum_{q\ge 1} \frac{p^{2q-1}}{p^{qs}}
+ \sum_{q\ge 1} \frac{p^{2q-2}}{p^{qs}}\right).$$
This is
$$\prod_p
\left(1 +
\left(1 - \frac{2}{p} + \frac{1}{p^2} \right)
\sum_{q\ge 1} \frac{p^{2q}}{p^{qs}}
\right)
= \prod_p \left(1 +
\left(1 - \frac{2}{p} + \frac{1}{p^2} \right)
\frac{p^{2-s}}{1-p^{2-s}}
\right)
\\ = \prod_p \left(1 +
\left(1 - \frac{2}{p} + \frac{1}{p^2} \right)
\frac{p^{2-s}}{1-1/p^{s-2}}
\right)
\\ = \zeta(s-2)
\prod_p \left(1 - 1/p^{s-2} +
p^{2-s} - 2 p^{1-s} + p^{-s} \right)
\\ = \zeta(s-2)
\prod_p \left(1 - 2 \frac{1}{p^{s-1}} +
\frac{1}{p^s} \right).$$
There is a simple pole at $s=3$ (from the zeta function which contributes one to the residue and the product, which is readily seen to converge there) and hence by the
Wiener-Ikehara Theorem
we have
$$\sum_{n\le x} \varphi(n)^2
\sim \frac{x^3}{3}
\prod_p \left(1 - 2 \frac{1}{p^{2}} +
\frac{1}{p^3} \right).$$
Convergence of the product follows because we have
$$0 < \frac{2}{p^2} - \frac{1}{p^3} < 1 $$
and $$\sum_p \left( \frac{2}{p^{2}}
- \frac{1}{p^3} \right)$$
converges e.g. by comparison with $\zeta(2)$ and $\zeta(3).$
The numeric value of the constant is given by
$$\prod_p \left(1 - 2 \frac{1}{p^{2}} +
\frac{1}{p^3} \right)
\approx 0.428249.$$
This material is not original and can be found at
OEIS 127473.
|
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|
Limit calculation and discontinuity Having a function, which has a polynomial in the denominator like:
$$ \lim_{x \to 2}\,\dfrac{x+3}{x-2} $$
We see there is a discontinuity at x=2, because it sets the denominator to 0. But multiplying by its conjugate we get:
$$ \dfrac{x+3}{x-2}\cdot\dfrac{x-2}{x-2} = \dfrac{(x+3)(x-2)}{x^2-4} $$
And now there is no discontinuity in $x=2$ anymore, but should not that function be an equivalent of the original one? we are just multiplying by $1$.
UPDATE:
Apologies, I chose a very poor example.
Consider this one:
$$ \lim_{x \to -2}\,\dfrac{x+2}{x^3+8} $$
When doing f(-2) you get 0/0. By doing factor, cancel and plug in:
$$ \lim_{x \to -2}\, \dfrac{(x+2)}{(x+2)(x^2-2x+4)} $$
$$ \lim_{x \to -2}\, \dfrac{1}{x^2-2x+4} $$
Then f(-2)=1/12, it is not 0/0 anymore. So my question is that how come that two functions that should be equivalent, have different discontinuit?
|
The mistake you are making is that $\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to -2}\frac{x+2}{(x+2)(x^2-2x+4)} = \frac{0}{0}$ is not necessarily discontinuous. It is a so called indeterminant form. Typically, you have to do something algebraic to "get to" the actual limit, similar to what you did by factoring and cancelling. So indeed $$\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to2} \frac{1}{x^2-2x+4} = \frac{1}{12}. $$
|
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|
Find the coordinates of the expression $(\cos x + \sin x)^3$ in the basis {$1, \sin x, \cos x, \sin 2x, \cos 2x, \sin3x, \cos 3x$} I'm quite stumped here. I expanded out $(\cos x + \sin x)^3$ to $\sin^3x + \cos^3x + 3\sin x\cos^2x + 3\sin^2x\cos x$ And I've tried the trig identities for $\sin 3x = 3\sin x - 4\sin^3x$
I can't have a variable in the coordinates of the basis right? It has to be of some $\alpha_i$ that is a scalar to the linear combination; for example, I can't do something as $\alpha_1 = \cos x$?
|
When you expand, you get $\cos^3 x+3\cos^2 x\sin x+3\cos x\sin^2 x+\sin^3 x$.
Note that $3\cos^2 x\sin x=3(1-\sin^2 x)\sin x=3\sin x-3\sin^3 x$, and similarly
$3\cos x\sin^2 x=3\cos x-3\cos^3 x$.
So our expression is equal to $3\cos x+3\sin x-2\cos^3 x-2\sin^3 x$.
You know how to express $\sin^3 x$ in terms of $\sin 3x$ and $\sin x$. For from your expression we have $\sin^3 x=\frac{3\sin x-\sin 3x}{4}$.
Similarly, you can use $\cos 3x=4\cos^3 x-3\cos x$ to express $\cos^3 x$ as a linear combination of $\cos x$ and $\cos 3x$.
The functions $1$, $\cos 2x$, and $\sin 2x$ do not enter the game. Or if you want them to, they enter with coordinates $0$.
Remark: About your question, the coordinates have to be real numbers, not functions.
|
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|
A combinatorics and algebra problem I'm given two numbers $n$ and $m$. I have to make $n$ with $m$ numbers (only taking their sum). For example, if $n=6$, $m=3$, $6$ is formed with $3$ numbers in the following
way:
*
*$1+1+4=6$
*$2+2+2=6$
*$1+2+3=6$
For every element of the sets above, if I use $C(r,2)$, $0<r<n$ and take their summation I get like this.
*
*$C(1,2)+C(1,2)+C(4,2)=0+0+6=6$
*$C(2,2)+C(2,2)+C(2,2)=3$
*$C(1,2)+C(2,2)+C(3,2)=4$
Here,we have considered $C(1,2)=0$.
One of my friend said to me that
if we make $n$ with $m-1$ 1's plus $n-(m-1)$ then the sum of $C(r,2)$ will be maximum. In other words, the sum of $C(r,2)$ with $m-1$ 1's plus $n-(m-1)$ is always
greater than the rest sums of $C(r,2)$.
In the above exapmle, $6$ is greater than $3$ and $4$.
My question is that why such a occurence happens? What is the reasoning behind it?
|
It suffices to consider two summands $a,b$ and verify that $C(a,2)+C(b,2)<C(1,2)+C(a+b-1,2)$ whenever $a,b>1$. Indeed,
$$ C(a,2)+C(b,2) = \frac12(a^2-a+b^2-b)$$
$$ C(1,2)+C(a+b-1,2) = \frac12(a+b-1)(a+b-2)$$
and the difference is after simplification
$$C(1,2)+C(a+b-1,2) -(C(a,2)+C(b,2)) = \frac12(ab-a-b+1)=\frac12(a-1)(b-1)>0$$
|
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|
Closed form of $I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx$ Does the integral below have a closed-form:
$$I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx,$$
where $\tan^{-1} (\cdot)$ is inverse tangent function.
Neither Wolfram|Alpha nor Maple 13 can return possible closed-form of the integral. I cannot also find a similar integral in G&R 8th edition. Its numeric integral computed by Maple 13 is
$$I=-0.60581547102487915432009247784178206365553774419860 ...$$
This integral came up in discussion during symbolic computation seminar. Our professor asked us to help him to find the closed form of several integrals. This one comes from the study in topic special function: Inverse Tangent Integral.
As I said in my comment and I'll add some details, we have tried many substitutions, standard techniques such as: integration by parts, differentiation under integral sign, etc. We also tried method of countour integration, but none of them gave promising result so far. We have been evaluating this integral since 2 weeks ago but no success. So, I thought it's about time to ask you for help. Can you help me out to find its closed-form, please? Can someone help to prove the closed-form given by users Cleo and Anastasiya Romanova? Thanks.
|
Step 1: Introducing an extra parameter
Define
$$\phi(\alpha)=\int^\frac{\pi}{2}_0\arctan\left(\frac{\sin{x}}{\cos{x}-\frac{1}{\alpha}}\right)\cot{x}\ {\rm d}x$$
Differentiating yields
\begin{align}
\phi'(\alpha)
=&-\int^\frac{\pi}{2}_0\frac{\cos{x}}{1-2\alpha\cos{x}+\alpha^2}{\rm d}x\\
=&\frac{\pi}{4\alpha}-\frac{1+\alpha^2}{2\alpha(1-\alpha^2)}\int^\frac{\pi}{2}_0\frac{1-\alpha^2}{1-2\alpha\cos{x}+\alpha^2}{\rm d}x
\end{align}
Step 2: Evaluation of $\phi'(\alpha)$
For $|\alpha|<1$, the following identity holds.
$$\frac{1-\alpha^2}{1-2\alpha\cos{x}+\alpha^2}=1+2\sum^\infty_{n=1}\alpha^n\cos(nx)$$
Therefore,
\begin{align}
\phi'(\alpha)
=&\frac{\pi}{4\alpha}-\frac{1+\alpha^2}{2\alpha(1-\alpha^2)}\left(\frac{\pi}{2}+2\sum^\infty_{n=0}\frac{(-1)^n}{2n+1}\alpha^{2n+1}\right)\\
=&-\frac{\pi\alpha}{2(1-\alpha^2)}-\frac{\arctan{\alpha}}{\alpha}-\frac{2\alpha\arctan{\alpha}}{1-\alpha^2}\tag1
\end{align}
Step 3: The Closed Form
Integrating back, we get
\begin{align}
&\color{red}{\Large{\phi(\sqrt{2}-1)}}\\
=&\left(\frac{\pi}{4}+\arctan{\alpha}\right)\ln(1-\alpha^2)\Bigg{|}^{\sqrt{2}-1}_0-\int^{\sqrt{2}-1}_0\left[\color{#FF4F00}{\frac{\arctan{\alpha}}{\alpha}}+\color{#00A000}{\frac{\ln(1-\alpha^2)}{1+\alpha^2}}\right]{\rm d}\alpha\tag2\\
=&\frac{3\pi}{8}\ln(2\sqrt{2}-2)\color{blue}{\underbrace{\color{black}{-\int^\frac{\pi}{8}_0\color{#FF4F00}{2x\csc{2x}}\ {\rm d}x+\int^\frac{\pi}{8}_0\color{#00A000}{2\ln(\cos{x})}\ {\rm d}x}}}-\int^\frac{\pi}{8}_0\color{#00A000}{\ln(\cos{2x})}\ {\rm d}x\tag3\\
=&\frac{3\pi}{8}\ln(2\sqrt{2}-2)\color{blue}{-x\ln(\tan{x})\Bigg{|}^\frac{\pi}{8}_0+\int^\frac{\pi}{8}_0\ln\left(\frac{1}{2}\sin{2x}\right)\ {\rm d}x}-\int^\frac{\pi}{8}_0\ln(\cos{2x})\ {\rm d}x\tag4\\
=&\frac{\pi}{4}\ln(2\sqrt{2}-2)+\int^\frac{\pi}{8}_0\ln(\tan{2x})\ {\rm d}x\\
=&\frac{\pi}{4}\ln(2\sqrt{2}-2)-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{8}_0\cos\left((8n+4)x\right)\ {\rm d}x\tag5\\
=&\frac{\pi}{4}\ln(2\sqrt{2}-2)-2\sum^\infty_{n=0}\frac{\cos(n\pi)}{(2n+1)(8n+4)}\\
=&\frac{\pi}{4}\ln(2\sqrt{2}-2)-\frac{1}{2}\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tag6\\
=&\boxed{\color{red}{\Large{\displaystyle\frac{\pi}{4}\ln(2\sqrt{2}-2)-\frac{\mathbf{G}}{2}}}}\tag7
\end{align}
Explanation:
$(2):$ Integrated $(1)$ from $0$ to $\sqrt{2}-1$. Integrated $\dfrac{2\alpha\arctan{\alpha}}{1-\alpha^2}$ by parts.
$(3):$ Applied the substitution $\alpha=\tan{x}$.
$(4):$ Integrated $-2x\csc{2x}$ by parts.
$(5):$ Used the Fourier series of $\ln(\tan{2x})$.
$(6):$ $\cos(n\pi)=(-1)^n$ for $n\in\mathbb{N}$.
$(7):$ Used the definition of Catalan's constant.
|
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|
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$
Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.
|
We have these identities. (Do you see why these are true? Hint: FOIL. Or, if you learned $``{(a+b)^2=a^2+2ab+b^2}
``$, use that.)
$$\left(a+\frac1a\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
$$\left(a^2+\frac1{a^2}\right)^2=\left(a^4+\frac1{a^4}\right)+2$$
Now, plug in $(a+1/a)=\sqrt3$.
$$\left(\sqrt3\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
So $3=(a^2+1/a^2)+2$, and $a^2+1/a^2=1$.
$$(1)^2=\left(a^4+\frac1{a^4}\right)+2$$
So $1=(a^4+1/a^4)+2$, or $a^4+1/a^4=-1$ and we have solved the problem.
|
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|
How can I show that $n! \leqslant \left(\frac{n+1}{2}\right)^n$?
Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$
I know that it can be done by induction but I always find line where I do not know what to do next.
|
Hint:
$$
(n!)^2 = 1\times n \times 2\times (n-1) \times \dots
= \prod_{k=1}^n k(n+1-k)
$$
then use
$$
k\times (n+1 -k) =
\left(\frac {n+1}2 + \frac {n+1}2 - k\right)\left(\frac {n+1}2 - \frac {n+1}2 + k\right)
\\= \left(\frac {n+1}2\right)^2 - \left( \frac {n+1}2 - k\right)^2
\le \left(\frac {n+1}2\right)^2
$$
to get
$$
(n!)^2 \le \left(\frac {n+1}2\right)^{2n}
$$
|
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|
Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions Have been working on this for years. Need a system which proves that there exists a number $C$ which has certain properties. I will give a specific example, but am looking for a system which could possibly be generalized.
Find (or prove there exists) $C$, such that the quadratic $x^2 - 47x - C = 0$ has integer roots, and furthermore, $C$ must have ALL OF $2$, $3$ and $5$ as its only prime factors (though each of these can be to any positive integer power).
|
Well if $ax^2+bx+c$ has integer roots, then $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ is an integer. Since in this case $a=1$ and $b=-47$ and we want to find $C$, $$\frac{47\pm \sqrt{(-47)^2+4C}}{2}=\frac{47\pm \sqrt{2209+4C}}{2}$$ is an integer. $47$ is odd. Since the fraction is an integer, the numerator must be divisible by $2$. Since $47\pm \text{odd_number}$ is even (and even means divisible by two), $\sqrt{2209+4C}$ must be an odd number. That in turn means $2209+4C$ must be an odd square.
We also know that $C=2^a\times 5^b\times 3^c$. So now we need to solve the equation $2209+4\times 2^a5^b3^c=x^2$ for integers. You can use a computer to solve it from here if you like.
$$2209+4\times 2^a5^b3^c=\text{a square bigger than 2209}\\ \Longrightarrow \text{a square bigger than 2209}-2209=4\times 2^a5^b3^c\\ \Longrightarrow (47+l)^2-47^2=4\times 2^a5^b3^c$$
For some integer $l$. See if you take it from here...
|
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|
Finding the range of equation. Any tricks? I m working on the following problem
For real numbers $a,b$, if $a+ab+b=3$, then find the range of $m=a-ab+b$. Is there any inequalities here to use?
|
You are given $a+ab+b=3$, which you can make $(a+1)(b+1)=4$. Let $x=a+1, y=b+1$, so $xy=4$ $m=a-ab+b=x-1-(x-1)(y-1)+y-1=-xy+2x+2y-3=2x+2y-7=2x+\frac 8x-7$
Can you find the range of that?
|
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|
Elementary proof of an inequality with $e^x$ when $|x|<1$. Assume $|x| <1 $ and we already know $0 \leq e^x - 1 - x$. Note that last inequality was god given, we know it is true, but we do not know how it was proved. Can we deduce from here that $e^x - 1 - x \leq x^2$ without using derivatives etc., just only with elementary algebraic manipulations and ideas of convexity?
|
If $f$ is convex and differentiable, we have $f(x)-f(y) \ge f'(y) (x-y)$ for all $x,y$.
The function $\exp$ is convex and differentiable, so we have $e^x-1 \ge x$ for all $x$.
Let $\phi(x) =e^x-1-x$, then $\phi(x) \ge 0$, $\phi(0)=\phi'(0) = 0$ and $\phi''(x) = e^x$.
The mean value theorem gives $\phi(x) = \phi(0) + \phi'(0) x + {1 \over 2} \phi''(c) x^2$ for some $c \in (0,x)$. Since $e^x \le 1$ for $x \le 0$, we have
$\phi(x) \le {1 \over 2} x^2$ for all $x \le 0$.
My estimate for $x \in (0,1]$ is messier:
Assuming $x>0$, we have $\phi(x) \le x^2$ iff ${ x^2 \over 2!} + { x^3 \over 3!} + \cdots \le x^2$ iff
${ x^3 \over 3!} + \cdots \le { x^2 \over 2!} $ iff
${ x \over 3!} + { x^2 \over 4!} +\cdots \le { 1 \over 2!} $.
We have the estimate
${ x \over 3!} + { x^2 \over 4!} +\cdots = { x \over 3!} (1 + { x \over 4} + {x^2 \over 4 \cdot 5 } + \cdots ) \le { x \over 3!} (1 + { x \over 4} + {x^2 \over 4^2 } + \cdots ) = { x \over 3!}{ 1 \over 1-{x \over 4}} \le {2 \over 9} x$, where we have used the fact that ${ 1 \over 1-{x \over 4}} \le {4 \over 3}$ when $x \in (0,1]$. Since ${2 \over 9} x \le {1 \over 2}$ for all
$x \in (0,1]$ we obtain $\phi(x) \le x^2$ for all $|x|<1$.
|
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|
Evaluate the sum. Evalute the following sum: $ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$.
I tried doing it but I keep getting the wrong answer. I've used known sums to solve it.
|
Reverse the sequence and add term-wise to the original:
$$ 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + . . . .+ (n-1) + n + n$$
$$ n + n + (n-1) + . . . . + 6 + 6 + 5 + 4 + 4 + 3 + 2 + 2 + 1$$
All triples will sum to $(n+1)+(n+2)+(n+1)=3n+4$, and each number is counted twice. But, beware, $n$ must be even and there are only $\frac n2$ triples !
$$S_n=\frac12\frac n2(3n+4)=\frac{3n^2+4n}4.$$
Check: $S_2=5$, $S_4=16$, $S_6=33$...
For odd $n$, we have to infer that the sum stops at the odd term, and
$$S_n=S_{n-1}+n=\frac{3(n-1)^2+4(n-1)}4+n=\frac{3n^2+2n-1}{4}.$$
Check: $S_1=1$, $S_3=8$, $S_5=21$...
|
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|
Limit proof for rational function $\frac{1}{x}$ A while ago I posted another one like this with a incorrect approach, please see this one!
Is this an accurate proof for limits for the function $\frac{1}{x}$
$\displaystyle \lim_{x\to1} \frac{1}{x} = \frac{1}{1} = 1$
Using $\epsilon-\delta$ So,
$\displaystyle \frac{|x-1|}{|x|} < \epsilon$ for some $\displaystyle |x-1| < \delta$
Lets assume $|x - 1| < \frac{1}{3}$
$\displaystyle \frac{2}{3} < |x| < \frac{4}{3}$
$= \frac{3}{2} > \frac{1}{|x|} > \frac{3}{4}$
So we have,
$|x - 1| < \delta$
$\frac{1}{x} < \frac{4}{3}$
Therefore we get: $\frac{|x-1|}{|x|} < \frac{3\delta}{4}$
It is the equation: $\frac{3\delta}{4} = \epsilon$, which is possible only if, $\delta_1 = \frac{4\epsilon}{3}$
Therefore, $\delta = \min(\frac{1}{3},\frac{4\epsilon}{3})$
|
I think you have the right idea, but it is a bit unorganized, separate the scratch work from the proof. This is how I would approach it.
Let $\epsilon>0$ be arbitrary, let $\delta=\min(1/2,\epsilon/2)$ such that $\vert x−1\vert<\delta$, then $$\left\vert\frac{1}{x}−1\right\vert= \frac{\vert x−1\vert}{\vert x\vert}< 2\vert x−1\vert <2\epsilon/2=\epsilon.$$
|
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|
how to find the solution to this system of equations Given the system of equations:
$
xy+xz=54+x^2 \\
yx+yz=64+y^2 \\
zx+zy=70+z^2 $
Need to find all of the solutions of $ x,y$ and $z$.
Tried to sum up all three equations but got stuck with nothing to factorize.
|
Adding, $x^{2}$ ,$y^{2}$, & $z^{2}$ to 1st,2nd & 3rd equations respectively we get,
$x(x+y+z)=2x^{2}+54$
$y(x+y+z)=2y^{2}+64$
$z(x+y+z)=2z^{2}+70$
So, $\dfrac{2x^{2}+54}{x}=\dfrac{2y^{2}+64}{y}=\dfrac{2z^{2}+70}{z}=x+y+z=k(say)$
From these, $2x^{2}-kx+54=0,2y^{2}-ky+64=0,2z^{2}-kz+70=0$.
So, $x=\dfrac{k\pm \sqrt{k^{2}-4.2.54}}{4}, y=\dfrac{k\pm \sqrt{k^{2}-4.2.64}}{4}, z=\dfrac{k\pm \sqrt{k^{2}-4.2.70}}{4}$.
Putting, $k=24$,& taking only positive sign we get, $x=9,y=8,z=7$.
Note that, $x+y+z=24$. So, if we take negative sign then $x=3,y=4,z=5$, in that case $x+y+z\ne 24$. So, it is not admissible.
Again, if we take, $k=-24$ then the solutions are $x=-9,y=-8,z=-7$ 7 in this case sum is also $-24$.
|
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|
Find the matrix $A$ with this condition.... If $\theta \in\mathbb{R}\setminus\{k\pi, k\in\mathbb{Z}\}$ and $A\in M_{2\times 2}(\mathbb{C})$ such that $$A^{-1} \begin{pmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta \\
\end{pmatrix} A= \begin{pmatrix}
e^{i\theta} & 0 \\
0 & e^{i\theta} \\
\end{pmatrix}.$$ Then Find $A?$
I just took $A=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$ , and subtituted in above. I try to find a,b,c and d. But i did't get correct ans.
|
This is not the 'right' way to answer the question, but if you use your idea, and note that you can scale $A$ so its determinant $ad-bc$ is 1 (giving a nice expression for the inverse), and multiply everything out, you get
\begin{align*}
\begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}&=
\begin{pmatrix}
e^{i\theta} & 0 \\
0 & e^{i\theta}
\end{pmatrix} \\
\begin{pmatrix}
d\cos\theta-b\sin\theta & -d\sin\theta-b\cos\theta \\
-c\cos\theta+a\sin\theta & c\sin\theta+a\cos\theta
\end{pmatrix}
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}&=
\begin{pmatrix}
e^{i\theta} & 0 \\
0 & e^{i\theta}
\end{pmatrix} \\
\begin{pmatrix}
(ad-bc)\cos\theta-(ab+cd)\sin\theta & -(b^2+d^2)\sin\theta \\
(a^2+c^2)\sin\theta & (ad-bc)\cos\theta+(ab+cd)\sin\theta
\end{pmatrix}&=
\begin{pmatrix}
e^{i\theta} & 0 \\
0 & e^{i\theta}
\end{pmatrix} \\
\begin{pmatrix}
\cos\theta-(ab+cd)\sin\theta & -(b^2+d^2)\sin\theta \\
(a^2+c^2)\sin\theta & \cos\theta+(ab+cd)\sin\theta
\end{pmatrix}&=
\begin{pmatrix}
\cos\theta+i\sin\theta & 0 \\
0 & \cos\theta-i\sin\theta
\end{pmatrix}
\end{align*}
We note an important fact: $u\sin\theta+v\cos\theta=s\sin\theta+t\cos\theta$ for all values of $\theta$ in an interval iff $u=s$ and $v=t$. This implies the following:
*
*$a^2+c^2=b^2+d^2=0\implies c=\pm ia,\,d=\pm ib$
*$ab+cd=-i\implies ab-(\pm a)(\pm b)=-i$
Since $ab-(\pm a)(\pm b)\neq0$, we must choose the signs for $c$ and $d$ in different ways. Then $ab-(\pm a)(\mp b)=2ab=i$.
In fact, if we choose any $a,b,c,d$ such that $c=\pm ai,\,d=\mp bi$, and $2ab=i$, then it can be checked that the two matrices are equal.
As an example, we could pick $a=d=\sqrt{2},\,b=c=\sqrt{2}i$.
|
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|
Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2 Prove that:
If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$
From Barnard & Child's "Higher Algebra".
I know that the highest power of a prime $p$ contained in $N$! is
$$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ...
$$
I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n > 2$.
|
It is an easy exercise to show that for all real numbers $x$ we have
$$
\lfloor 3x\rfloor=\lfloor x\rfloor+\lfloor x+\frac13\rfloor+\lfloor x+\frac23\rfloor.
$$
Thus for all $n$ and all prime powers $p^t\ge3$ we have
$$
\begin{aligned}
\lfloor \frac{3n}{p^t}\rfloor&=\lfloor \frac{n}{p^t}\rfloor+
\lfloor \frac{n}{p^t}+\frac13\rfloor+\lfloor \frac{n}{p^t}+\frac23\rfloor\\
&=\lfloor \frac{n}{p^t}\rfloor+
\lfloor \frac{n+\frac{p^t}3}{p^t}\rfloor+\lfloor \frac{n+\frac{2p^t}3}{p^t}\rfloor\\
&\ge \lfloor \frac{n}{p^t}\rfloor+
\lfloor \frac{n+1}{p^t}\rfloor+\lfloor \frac{n+2}{p^t}\rfloor.
\end{aligned}
$$
This leaves us to deal with the case $p^t=2$. But because $n>2$, we see that $3n$ exceeds one power of two higher than any of $n,n+1,n+2$. If $n>4$ (thanks, Petite Etincelle!) we have $3n>2(n+2)$, and in the cases $n=3,4$ we have $3n>8>n+2$. This gives us a necessary extra term compensating for the deficiency at $p^t=2$.
More precisely, if $n=2k+1$ is an odd integer, then $\lfloor \dfrac{3n}2\rfloor=3k+1$ in comparison to
$$
\lfloor\frac n2\rfloor+\lfloor\frac{n+1}2\rfloor+\lfloor\frac{n+2}2\rfloor=k+(k+1)+(k+1)=3k+2.
$$
On the other hand, if $n=2k$ is even, then $\lfloor \dfrac{3n}2\rfloor=3k$ and
$$
\lfloor\frac n2\rfloor+\lfloor\frac{n+1}2\rfloor+\lfloor\frac{n+2}2\rfloor=k+k+(k+1)=3k+1.
$$
In either case we are missing a single factor two, so having that single extra term suffices.
Summing the above inequalities for $p^t\ge3$ and coupling the terms corresponding to $p^t=2$ and $p^t=2^\ell$, where $\ell$ is the largest integer such that $2^\ell\le 3n$ shows that for all primes $p$ we have
$$
\sum_{t>0}\lfloor\frac{3n}{p^t}\rfloor\ge
\sum_{t>0}\lfloor\frac{n}{p^t}\rfloor+\sum_{t>0}\lfloor\frac{n+1}{p^t}\rfloor+\sum_{t>0}\lfloor\frac{n+2}{p^t}\rfloor.
$$
The claim follows from this.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving the following formula of $\ln(2)$ Proving the following formula of $\ln2$
$$\ln2=\frac{1}{2}\left(\frac{3}{2}\right)-\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{6}\left(\frac{9}{8}\right)-\frac{1}{8}\left(\frac{15}{16}\right)+\frac{1}{10}\left(\frac{33}{32}\right)-\frac{1}{12}\left(\frac{63}{64}\right)\cdots$$
|
Your formula is :
\begin{align}
S&:=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}\frac{2^n-(-1)^n}{2^n}\\
&=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}+\sum_{n=1}^\infty \frac{(-1)^{2n}}{2\,n}\frac 1{2^n}\\
&=-\frac 12\sum_{n=1}^\infty \frac{(-1)^n}{n}+\frac 12\sum_{n=1}^\infty \frac 1{n} \left(\frac 12\right)^n\\
&=\frac 12\log(1-(-1))-\frac 12\log(1-1/2)\\
&=\log 2\\
\end{align}
( using only $\,\displaystyle -\log(1-x)=\sum_{n=1}^\infty \frac{x^n}n\;$)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Question on modulo Find the last two digits of $3^{2002}$.
How should I approach this question using modulo? I obtained 09 as my answer however the given answer was 43.
My method was as follows:
$2002\:=\:8\cdot 250+2$
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
The last two digits is the remainder when divided by 100. Thus we need to compute the sum mod 100.
$3^8\equiv 61\:\left(mod\:100\right)$
Thus,
$3^{2002}=\:3^{8\cdot 250+2}\:=\:3^{8\cdot 250}\cdot 3^2$
$\equiv 61^{250}\cdot 9\:\equiv 61^{5\cdot 50}\cdot 9\equiv \:9\:\left(mod\:100\right)$
Therefore my answer would be 09. Am I doing it the right way?
Thanks in advanced
|
Observe that $\text{gcd(3,100)} = 1$, by Euler totient formula: $3^{\phi(100)} \cong 1(\mod 100)$. But $\phi (100) = 100\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{5}\right) = 40$. Thus: $3^{40} \cong 1(\mod 100)$. So $3^{2002} = (3^{40})^{50}\cdot 3^2 \cong 1\cdot 9(\mod 100) \cong 9(\mod 100)$. Thus the remainder is $9$ as claimed.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
computation on hyper surface $z=x^2+y^2$ I have problem with following exercise
Consider the hypersurface $M$ parametrized by $z=x^2+y^2$.
Endow this with the Riemannian metric induced from the $\mathbb{R}^3$.
Compute the sectional curvature.
|
This method was using Monge surface parametrization, which was introduced by @THW. in my former question.
Endowment of $\mathbb{R}^3$, we can do Gauss approach as follows.
\begin{align}
\mathbf{x}(r,\theta)=(f(r),r\cos(\theta), r\sin(\theta))
\end{align}
where $f(r)=r^2=z$.
\begin{align}
&\mathbf{x}(r,\theta)=\left(f(r),r\cos(\theta), r\sin(\theta) \right) \quad \mathbf{x}_r = \left(f'(r), \cos(\theta), \sin(\theta) \right) \\
& \mathbf{x}_{rr} = \left( f''(r),0,0 \right) \quad \mathbf{x}_\theta = \left( 0,-r\sin(\theta), r\cos(\theta) \right) \\
&\mathbf{x}_{\theta\theta} = \left( 0,-r\cos(\theta), -r\sin(\theta) \right) \quad \mathbf{x}_{r\theta} = \left(0, -\sin(\theta), \cos(\theta)\right)
\end{align}
Here $'$ denote the $r$-derivative.
Then
\begin{align}
I=ds^2 =(\mathbf{x}_r du + \mathbf{x}_\theta dv) \cdot(\mathbf{x}_r dr + \mathbf{x}_\theta d\theta) = Edr^2 + 2F drd\theta+ G d\theta^2
\end{align}
with $E = 1 +(f')^2$, $F=0$, $G=r^2$. Note
\begin{align}
&\mathbf{x}_r \times \mathbf{x}_\theta = \left(r, -f'(r) r \cos(\theta), -f'(r) r \sin(\theta) \right) \\
&||\mathbf{x}_r \times \mathbf{x}_\theta|| = \sqrt{EG-F^2} = r\sqrt{1 + (f')^2} \\
& U=\frac{\mathbf{x}_r \times \mathbf{x}_\theta}{||\mathbf{x}_r \times \mathbf{x}_\theta||}= \frac{\left(1, -f' \cos(\theta), -f' \sin(\theta) \right)}{\sqrt{1 + (f')^2}}
\end{align}
and
\begin{align}
&e= \mathbf{x}_{rr} \cdot U=\frac{f''}{\sqrt{1 + (f')^2}}, \qquad f=\mathbf{x}_{r\theta}\cdot U=0 , \qquad g=\mathbf{x}_{\theta\theta} \cdot U=\frac{f'r}{\sqrt{1 + (f')^2}} \\
&\kappa_\mu = \frac{e}{E} =\frac{f''}{\sqrt{\left(1 +(f')^2 \right)^3}}, \quad \kappa_{\pi} = \frac{g}{G}=\frac{f'}{r\sqrt{1+(f')^2}}
\end{align}
Thus we obtain Gauss Curvature
\begin{align}
K = \kappa_\mu \kappa_\pi =\frac{f'f''}{r\left( 1 +(f')^2 \right)^2}\stackrel{f =r^2 }=\frac{4}{\left( 1 +4r^2 \right)^2} >0
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
|
$$3\left(2x-1\right)^{2}=147\iff\left(2x-1\right)^{2}-49=0\iff$$$$\left[\left(2x-1\right)-7\right]\left[\left(2x-1\right)+7\right]=0\iff\left(2x-8\right)\left(2x+6\right)=0$$
A product $p\times q$ equals $0$ if and only if $p=0$ or $q=0$.
Applying that here gives $2x-8=0$ or $2x+6=0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Express this sum of radicals as an integer? I have read somewhere that the radical $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ and I don't understand it. How do you solve this(when the RHS is unknown)?
|
Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Write $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$. Then
$$
x^3=
a+b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2}
=
a+b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})
=4-3x
$$
Now, the derivative of $x^3+3x-4$ is always positive, which means there is only one real root. By inspection, $x=1$ is that root.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Discriminant of quadratic formula I noticed something that i find interesting about the quadratic formula. I hope someone can explain it to me.
$f(x) = Ax^2 + Bx + c$
$f'(x) = 2Ax + B$
If I say that $f'(x) = 0$ I get that $x = -\frac{B}{2A}$. From this I get that $f(-\frac{B}{2A})$ is a max/min point.
The quadratic formula is $x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} = -\frac{B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}$. Because $f(-\frac{B}{2A})$ is a max/min point this means that the distance between the max/min point to the points where $f(x)$ is $\frac{\sqrt{B^2-4AC}}{2A}$.
I tried to put$x = -\frac{B}{2A}$ into f(x) and got the following result:$f(-\frac{B}{2A})=A (-\frac{B}{2A})^{2} + B(-\frac{B}{2A}) + C = \frac{AB^2}{4A^2} -
\frac{B^2}{2A} + C = \frac{B^2 -2B^2+4AC}{4A} = \frac{-B^2+4AC}{4A}$
If I take the root of it I get $\frac{\sqrt{4AC-B^2}}{2\sqrt{A}}$. It reminds me of the 2nd part of the quadratic formula$\frac{\sqrt{B^2-4AC}}{2A}$. Are they related? Why?
|
I follow you up to
$$f\left(-\frac{B}{2A}\right) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} + C = \frac{4AC - B^2}{4A}.$$
But the square root of this gives you a $2\sqrt{A}$ in the denominator, so it's not quite what you have.
Here's the connection I see. In order to get real roots, $B^2 - 4AC$ needs to be non-negative.
We can rewrite the equation above as
$$f\left(-\frac{B}{2A}\right) = \frac{B^2 - 4AC}{-4A}.$$
We know $A \neq 0$. If $B^2 - 4AC = 0$, then $f(-B/2A) = 0$, which makes sense: a double real root.
If $B^2 - 4AC > 0$, then the sign of $A$ tells us the sign of $f(-B/2A).$ If $A$ is negative, then $f(-B/2A)$ is positive, and the parabola opens down to intersect the $x$-axis at the two roots. Likewise, if $A$ is positive, then $f(-B/2A)$ is negative, and the parabola opens up to intersect the $x$-axis at the two roots.
That might not be everything, but that's what I see.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find this sum $S$ using Real-analysis methods only $$S = \sum_{k=1}^{\infty}\frac{2H_k}{(k+1)(k+2)^3}$$
I have tried a lot and failed, any help is appreciated. $H_k$ is the harmonic number.
Thanks (real method only please)
|
We have:
$$\frac{1}{2}\log^2(1-x)=\sum_{k=1}^{+\infty}\frac{H_k}{k+1}x^{k+1}$$
but since $\int_{0}^{1} x^n\log^2 x\,dx = \frac{2}{(n+1)^3}$ it follows that:
$$\begin{eqnarray*} S &=& \frac{1}{2}\int_{0}^{1} \log^2(1-x)\log^2 x\,dx=\frac{1}{2}\left.\frac{\partial^4}{\partial^2 a\,\partial^2 b}B(a,b)\,\right|_{a,b=1}\end{eqnarray*}.$$
Since $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ and $\Gamma'(z)=\psi(z)\Gamma(z)$ by differentiating multiple times we get a long expression depending only on the values of $\psi,\psi',\psi''$ and $\psi'''$ in $1$ and $2$:
$$ S = 12 - 4\,\zeta(2) -4\,\zeta(3)-\frac{1}{2}\,\zeta(4).$$
Addendum: An interesting approach is given by writing $\log(x)\log(1-x)$ as a linear combination of shifted Legendre polynomials. From the Rodrigues formula, exploiting integration by parts, we get:
$$\log x\log(1-x)=2-\frac{\pi^2}{6}-\sum_{j=1}^{+\infty}\frac{4j+1}{2j^2(2j+1)^2}P_{2j}(2x-1)$$
hence the Parseval's identity gives:
$$ S = \frac{1}{2}\left(\left(2-\frac{\pi^2}{6}\right)^2+\sum_{j=1}^{+\infty}\frac{(4j+1)}{4j^4(2j+1)^4}\right) $$
and the problem boils down to showing that:
$$\sum_{j=1}^{+\infty}\frac{(4j+1)}{4j^4(2j+1)^4}=20-4\zeta(2)-8\zeta(3)-\frac{7}{2}\zeta(4).\tag{1}$$
This problem seems to be feasible to creative telescoping techniques:
$$\frac{1}{(2n)^4}-\frac{1}{(2n+1)^4}=\frac{1+4n}{16n^4(2n+1)^4}+\frac{1+4n}{4n^3(2n+1)^3},$$
$$\frac{1}{(2n)^3}-\frac{1}{(2n+1)^3}=\frac{1}{8n^3(2n+1)^3}+\frac{3}{4n^2(2n+1)^2},$$
$$\frac{1}{(2n)^2}-\frac{1}{(2n+1)^2}=\frac{4n+1}{4n^2(2n+1)^2}=\frac{1}{4n^2}-\frac{1}{(2n+1)^2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A limit coming from high school I'm sure you guys can do this in many ways, using integrals, Taylor series, but I need a
high school way, like the use of a simple squeeze theorem. Can we get such a way?
$$\lim_{n\to\infty}\left(1+\log\left(1+\frac{1}{n^2}\right)\right)\left(1+\log\left(1+\frac{2}{n^2}\right)\right)\cdots\left(1+\log\left(1+\frac{n}{n^2}\right)\right)$$
EDIT: The proof I need I plan to present to some kids, so everything should be clear for them.
|
Approach 1
We can use the inequality
$$
x-\frac12x^2\le\log(1+x)\le x\tag{1}
$$
for $x\ge0$, to good result.
First, note that
$$
\begin{align}
\color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}
&=\frac{n^2+n}{2n^2}\\
&=\frac12+\frac1{2n}\tag{2}
\end{align}
$$
Next, since $\frac{k}{n^2}\le\frac1n$, we get
$$
\begin{align}
\color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}-\frac1{2n}
&=\sum_{k=1}^n\left[\frac{k}{n^2}-\frac1{2n^2}\right]\\
&\le\color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}\\
&\le\color{#0000FF}{\sum_{k=1}^n\frac{k}{n^2}}\tag{3}
\end{align}
$$
Finally, each of the terms $\log\left(1+\frac{k}{n^2}\right)$ is less that $\frac1n$, so we get
$$
\begin{align}
\color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}-\frac1{2n}
&=\sum_{k=1}^n\left[\log\left(1+\frac{k}{n^2}\right)-\frac1{2n^2}\right]\\
&\le\color{#C00000}{\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)}\\
&\le\color{#00A000}{\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)}\tag{4}
\end{align}
$$
Putting together $(2)$, $(3)$, and $(4)$ implies
$$
\frac12-\frac1{2n}\le\color{#C00000}{\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)}\le\frac12+\frac1{2n}\tag{5}
$$
An application of the Squeeze Theorem yields
$$
\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=\frac12\tag{6}
$$
and therefore,
$$
\lim_{n\to\infty}\prod_{k=1}^n\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=e^{1/2}\tag{7}
$$
Approach 2
Here we only assume $\log(1+x)\lt x$ for all $x\gt-1$. Then, because $1-\frac{x}{1+x}=\frac1{1+x}$,
$$
\frac{x}{1+x}\le-\log\left(1-\frac{x}{1+x}\right)=\log(1+x)\le x\tag{8}
$$
Then, for $0\le x\le\frac1n$, $(8)$ implies
$$
\frac{n}{n+1}x\le\frac{x}{1+x}\le\log(1+x)\le x\tag{9}
$$
Applying $(9)$ twice shows that for $0\le x\le\frac1n$,
$$
\left(\frac{n}{n+1}\right)^2x\le\log(1+\log(1+x))\le x\tag{10}
$$
Summing $(10)$, we get
$$
\left(\frac{n}{n+1}\right)^2\sum_{k=1}^n\frac{k}{n^2}\le\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)\le\sum_{k=1}^n\frac{k}{n^2}\tag{11}
$$
Applying $(2)$ gives
$$
\frac12\frac{n}{n+1}\le\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)\le\frac12\frac{n+1}{n}\tag{12}
$$
An application of the Squeeze Theorem yields
$$
\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=\frac12\tag{13}
$$
and therefore,
$$
\lim_{n\to\infty}\prod_{k=1}^n\left(1+\log\left(1+\frac{k}{n^2}\right)\right)=e^{1/2}\tag{14}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this limits $\lim_{N\to\infty}\sum_{n=1}^{N}\frac{1}{(n+1)}\sum_{i=1}^{n}\frac{1}{i(n+1-i)}$ Find this limits
$$\lim_{N\to\infty}\sum_{n=1}^{N}\left(\dfrac{1}{(n+1)}\left(\dfrac{1}{1\cdot n}+\dfrac{1}{2\cdot(n-1)}+\cdots+\dfrac{1}{(n-1)\cdot 2}+\dfrac{1}{n\cdot 1}\right)\right)$$
I know this
$$\dfrac{1}{1\cdot n}+\dfrac{1}{2\cdot(n-1)}+\cdots+\dfrac{1}{(n-1)\cdot 2}+\dfrac{1}{n\cdot 1}=\sum_{i=i}^{n}\dfrac{1}{i(n+1-i)}=\dfrac{1}{n+1}\sum_{i=1}^{n}\left(\dfrac{1}{i}+\dfrac{1}{n+1-i}\right)=\dfrac{2H_{n}}{n+1}$$
so this limits is
$$\sum_{n=1}^{\infty}\dfrac{2H_{n}}{(n+1)^2}$$
then I can't,Thank you
|
$\displaystyle \begin{align} \sum_{n=1}^{\infty}\dfrac{H_n}{(n+1)^2} &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n+1}}{(n+1)^2} - \frac{1}{(n+1)^3}\right) \\ &= \sum_{n=2}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= 2\zeta(3) - \zeta(3)\end{align}$
The last sum $\displaystyle \sum_{n=1}^{\infty}\dfrac{H_{n}}{n^2} = 2\zeta(3)$ is proved here for the general case $\displaystyle \sum_{n=1}^{\infty}\dfrac{H_{n}}{n^q}$.
|
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|
Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
|
It's quite easy to make that polynomial a square. Let,
$$P = (4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2$$
Then,
$$P = (a d)^2,\quad \text{if}\, a = 2b\tag{1}$$
and all other variables are free. Similarly,
$$P = (b c)^2,\quad \text{if}\, c = 2d\tag{2}$$
$$P = (2b d)^2,\quad \text{if}\, a = 2b,\, \text{or}\, c = 2d\tag{3}$$
where the last one is equivalent to restrest's solution. (Note that $a = 6$ and $b = 3$ is $a=2b$.)
Are there any other conditions on $P$?
|
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|
Number of length $8$ binary strings with no consecutive $0$'s How many $8$ bit strings are there with no consecutive $0$'s?
I just sat an exam, and the only question I think I got wrong was the above(The decider for a high-distinction or a distinction :SSS)
I took Number with consecutive $0$'s
1 zero $0$
2 zeros $\frac{7!}{6!}$
3 zeros above + $\frac{6!}{5!}$
4 zeroes above + $\frac{5!}{4!}$
5 zeros above + $\frac{4!}{3!}$
6 zeroes above + $\frac{3!}{2!}$
7 zeroes above + $\frac{2!}{1!}$
8 zeroes above + $1$
$(7*7)+(6*6)+(5*5)+(4*4)+(3*3)+(2*2)+1=140$
and we have $2^8 - 140=116$
Is this correct?
|
A bit string with no consecutive zeros is composed of 1 and 10 units, but may also begin with one 0 unit. (That is, every zero is either preceeded by a one or is at the start of the string).
Let $N_{x,y}$ count the permutations of a bit string composed of $x$ 1 and $y$ 10 units.
$$N_{x,y} = \frac{(x+y)!}{x!\, y!}$$
So $N_{6,1}$ counts permutations of an eight bit string of $6$ 1 and $1$ 10.
$N_{5,1}$ counts permutations of an eight bit string beginning with 0, and the remainder composed of $5$ 1 and $1$ 10. (The starting 0 is fixed and cannot move.)
And so forth ...
You want to calculate: $N_{8,0} + N_{7,0} + N_{6,1} + N_{5,1} + N_{4,2}+ N_{3,2} + N_{2,3} + N_{1,3} + N_{0,4}
\\ = \frac{8!}{8!\, 0!}+ \frac{7!}{7!\, 0!}+ \frac{7!}{6!\, 1!}+ \frac{6!}{5!\, 1!}+\frac{6!}{4!\, 2!}+\frac{5!}{3!\, 2!}+\frac{5!}{2!\, 3!}+\frac{4!}{1!\, 3!}+\frac{4!}{0!\, 4!}
\\ = 1 + 1 + 7 + 6 + 15 + 10 + 10 + 4 + 1
\\ = 55
$
|
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|
convergence of this series $a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$?? The sequence with nth term is given as
$a_{n}=\frac{1}{n}+\frac{1}{n+1}+....+\frac{1}{2n}$
this sequence will converge to??
what I did is:
for this we can use sandwich theorem i.e $f(x)\leq g(x)\leq h(x)$ then $lim h(x)=lim g(x)=lim f(x)$
here we can write...
$0\leq m\leq n$
$0+n\leq m+n\leq n+n$
$\frac{1}{n}\geq \frac{1}{m+n}\geq \frac{1}{2n}$
adding this n times we get
$1\geq\frac{1}{m+n}\geq \frac{n}{2n}$
applying lim we get 1/2..
..but its wrong.. how will it converge to log2??
|
Simply note that,
$\displaystyle \begin{align} a_n = \sum\limits_{k=n}^{2n}\frac{1}{k} = \sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n-1}\frac{1}{k} &= \sum\limits_{k=1}^{2n}\frac{1}{k} - 2\sum\limits_{k=1}^{n-1}\frac{1}{2k} \\ &= \left(\sum\limits_{k=1}^{n}\frac{1}{2k-1} + \sum\limits_{k=1}^{n}\frac{1}{2k} - 2\sum\limits_{k=1}^{n}\frac{1}{2k}\right) + \frac{1}{n} \\ &= \left(\sum\limits_{k=1}^{n}\frac{1}{2k-1} - \sum\limits_{k=1}^{n}\frac{1}{2k}\right) +\frac{1}{n} \\&= \frac{1}{n} + \sum\limits_{k=1}^{2n}\frac{(-1)^{k-1}}{k} =\frac{1}{n} + \left(1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2n}\right)\end{align}$
The later is the alternating harmonic series which converges to $\ln 2$.
As, $\displaystyle \sum\limits_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} = \sum\limits_{k=1}^{\infty} (-1)^{k-1}\int_0^1 x^{k-1}\,dx = \int_0^1 \sum\limits_{k=1}^{\infty}(-1)^{k-1}x^{k-1}\,dx = \int_0^1 \frac{1}{1+x}\,dx = \ln 2$.
|
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|
Finding the maximum area of polygon? Firstly , I divided the polygon to three triangles and I used the Heron's formula to find the area of triangles which formed the polygon $$A=\sqrt{p(p-a)(p-b)(p-c)}$$.I couldn't find easily the relationship between $d$ and area for the three triangles. Now I need the distance $d$ which give me a maximum area of polygon.
|
$$ h1=\sqrt{5^2-(\frac{d}{2})^2} \\$$upper triangle area is $$ s1=\frac{1}{2}d\sqrt{5^2-(\frac{d}{2})^2}$$
$$h2=\sqrt{4^2-(\frac{d-5}{2})^2}\\$$so 2 lower triangle area =$$s2=2*\frac{1}{2}\frac{d-5}{2}\sqrt{4^2-(\frac{d-5}{2})^2} $$ rectangle area is =$$s3=5*\sqrt{4^2-(\frac{d-5}{2})^2} $$the problem reduce to max s=s1+s2+s3
$$s_{total}=s1+s2+s3=s1=\frac{1}{2}d\sqrt{5^2-(\frac{d}{2})^2}+2*\frac{1}{2}\frac{d-5}{2}\sqrt{4^2-(\frac{d-5}{2})^2}+ 5*\sqrt{4^2-(\frac{d-5}{2})^2}$$
|
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|
Disproving existence of real root in some interval for a quintic equation Disprove the statement: There is a real root of equation $\frac{1}{5}x^5+\frac{2}{3}x^3+2x=0$ on the interval (1,2).
I am not sure whether to prove by counter-example or by assuming the statement is true and then proving by contradiction.
|
It's sufficient to show that:
*
*f(1) > 0
*For all n > 0: f(1 + n) > f(1)
Step #2 means that the function is increasing at all points to the right of x=1. From this you can say that f(x) doesn't equal zero (has a real root) in the range (1,2) because f(x) is always above 1 in that range.
Working out the algebra:
f(1 + n) = (1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5
Then to show that f(1 + n) > f(1), you have to show:
(1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > (1/5 + 2/3 + 2)
Canceling the (1/5 + 2/3 + 2) gives:
4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > 0
It should be easy to see that as n increases, this inequality remains true, as long as n > 1.
Edit: here's one way to work this out:
Divide: by 4: n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0
We know that n>1 so we can add a second inequality which removes the initial n +:
n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0
Subtract out the middle pieces to make the middle zero:
n > 0 > -n^2 - (3/4)n^3 - (1/4)n^4 - (1/20)n^5
|
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|
Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3. Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3.
How do I start with this question? Rationalising the denominator? But how would I expand it?
|
$$\frac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}=\frac{(x-3)^2\sqrt{(x-3)^2+4}+2(x-3)^2}{(x-3)^2}$$ $$\frac{(x-3)^2\sqrt{(x-3)^2+4}+2(x-3)^2}{(x-3)^2}=\sqrt{(x-3)^2+4}+2$$
|
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|
Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers) Question is to Find Rectangle of Constant Perimeter whose diagonal is maximum (My attempt with Lagrange Multipliers) .
I took rectangle with sides $x$ and $y$ .
Since Perimeter is constant so i took $2(x+y) = 2k$ , where k is any constant
And diagonal is $\sqrt{x^{2}+y^{2}}$
So i took $f(x,y)$ = $x^{2}+y^{2}$
And $g(x,y)$ which is constraint as $2(x+y-k)=0$
On solving By routine Lagrange Muliplier Method i get $x = k/2 = y$ ...So rectangle is a square ..Are conditions i have taken for to apply Lagrange Multiplier is correct here ?? (i men f , g )...Thanks
|
A simpler approach:
If the perimeter is
$2a$,
the sides are
$x$ and $a-x$.
If $d$ is the diagonal,
$\begin{array}\\
d^2
&=x^2+(a-x)^2\\
&=x^2+a^2-2ax+x^2\\
&=2x^2-2ax+a^2\\
&=2(x^2-ax+a^2/2)\\
&=2(x^2-ax+a^2/4+a^2/4)
\text{ (completing the square)}\\
&=2((x-a/2)^2+a^2/4)\\
\end{array}
$
This is maximized when
$(x-a/2)^2$ is maximized,
i.e., when
$x=0$ or $x=a$
(since $0 \le x \le a$)
and has a value of
$a^2$,
so
$d = a$.
If you are not allowed
to have a rectangle
with a side of length zero,
then you can get as cloase as you want,
but can never reach this value.
This is minimized when
$x=a/2$
and has a value of
$a^2/2$,
so the result is a square
with diagonal
$a/\sqrt{2}$.
|
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|
How to integrate $\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx$ How do I integrate
\begin{equation}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx,
\end{equation}
which has arisen from a problem I'm working on? I've noticed I can do the following:
\begin{align}
\int_0^1\sqrt{-x^6+x^4-x^2+1}\:dx & =\int_0^1\sqrt{\left(-x^4\right)\left(x^2-1\right)-\left(x^2-1\right)}\:dx \\[3ex]
& = \int_0^1\sqrt{\left(-x^4-1\right)\left(x^2-1\right)}\:dx \\[3ex]
& = i\int_0^1 \sqrt{\left(x^4+1\right)\left(x+1\right)\left(x-1\right)}\:dx
\end{align}
But where do I go from here? Also, I'm a bit unsure about my last step above, i.e. not sure if it would be the right route to take.
Thanks in advance!
|
$\int_0^1\sqrt{-x^6+x^4-x^2+1}~dx$
$=\int_0^1\sqrt{x^4(1-x^2)+1-x^2}~dx$
$=\int_0^1\sqrt{(1-x^2)(x^4+1)}~dx$
$=\int_0^\frac{\pi}{2}\sqrt{(1-\sin^2x)(\sin^4x+1)}~d(\sin x)$
$=\int_0^\frac{\pi}{2}(\cos^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}(1-\sin^2x)\sqrt{\sin^4x+1}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{4n}x}{4^n(n!)^2(1-2n)}dx-\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{4n+2}x}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(4n)!\pi}{2^{6n+1}(n!)^2(2n)!(1-2n)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(4n+2)!\pi}{2^{6n+3}(n!)^2(2n+1)!(1-4n^2)}$ (according to http://en.wikipedia.org/wiki/Wallis%27_integrals#Recurrence_relation.2C_evaluating_the_Wallis.27_integrals)
|
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|
If $|\alpha|\leq 1$ and $|\beta|\leq 1$, prove that $|\alpha+\beta|\leq |1+\overline{\alpha}\beta|$ Note $\alpha$ and $\beta$ are complex numbers and $\overline{\alpha}$ is the conjugate of $\alpha$. I've tried using variations of the triangle inequality and I couldn't find anything to work.
|
$|\alpha + \beta|^2 = |(x_1+x_2) + (y_1+y_2)i|^2 = (x_1+x_2)^2+(y_1+y_2)^2$
$|1+\bar{\alpha}\beta|^2 = |1+(x_1-y_1i)(x_2+y_2i)|^2 = |1+x_1x_2+y_1y_2 + (x_1y_2-x_2y_1)i|^2 = (1+x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2$. Thus we prove:
$x_1^2+x_2^2+2x_1x_2 + y_1^2+y_2^2+2y_1y_2 \leq 1+x_1^2x_2^2+y_1^2y_2^2 + 2x_1x_2 + 2y_1y_2 + 2x_1y_1x_2y_2 + x_1^2y_2^2 - 2x_1y_1x_2y_2 + x_2^2y_1^2$ or:
$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + x_1^2x_2^2 + y_1^2y_2^2 + x_1^2y_2^2 + x_2^2y_1^2$
or:
$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + (x_1^2+x_2^2)(y_1^2+y_2^2)$
or:
$(x_1^2+x_2^2 - 1)(y_1^2+y_2^2 - 1) \geq 0$
which is true since $|\alpha|, |\beta| \leq 1$.
|
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|
Proving binomial coefficients identity: $\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}$
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that:
$$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.$$
Tried proving it by induction but got stuck. Any help with proving it by induction or any other proof technique is appreciated.
|
Let $E = \{1,2, \dots , n+1\}$. The number $\binom{n+1}{r+1}$ is the number of subsets $A$ of $E$ with $r + 1$ elements.
Classify these subsets $A$ according to their largest element $b$, which can be any number among $r + 1$, $r + 2$, ..., $n + 1$. The number of $(r+1)$-element subsets of $E$ with largest element $b$ is the same as the number of $r$-element subsets of $\{1, 2, \dots, b-1\}$, which is $\binom{b-1}{r}$.
Now $b$ can be any number $r + 1, r + 2, \ldots, n + 1$, and there are $\binom{r}{r}$, $\binom{r + 1}{r}, \ldots, \binom{n}{r}$ possible subsets $A$ in each case. This proves the desired equality.
|
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|
Solve $x^4 - 2x^3 + x = y^4 + 3y^2 + y \wedge (x,y) \in \mathbb{Z}^2$ I want to solve equation $x^4 - 2x^3 + x = y^4 + 3y^2 + y$ in integers. The task comes from the LXVI Polish Mathematical Olympiad. Series with this task ended twenty days ago, so it is legal to talk about it.
It is of course equal to $x(x-1)(x^2-x-1) = y(y^3 + 3y + 1) \wedge (x,y) \in \mathbb{Z}^{2}$.
It is obvious that $\left(\nexists x \in \mathbb{Z}\right)\left((x^2-x-1) = 0\right)$ and $\left(\nexists y \in \mathbb{Z}\right)\left( (y^3 + 3y + 1) = 0\right)$.
$$x(x-1)(x^2-x-1) = 0
\Longrightarrow x \in \lbrace 0;~1 \rbrace
\Longrightarrow y = 0$$
Therefore pairs $(0,0), (1,0)$ satisfy equation.
We can see
$|y| \neq 1 \Rightarrow y \nmid (y^3+3y+1)$
or
$\left(|x| \neq 1 \Rightarrow x \nmid (x-1) \wedge x \nmid (x^2 - x -1)\right)$, and lot similar.
Should note too, that any multiplier have to divide some multiplier from second side.
$$\frac{x(x-1)(x^2-x-1)}{y} = y^3+3y+1 \in \mathbb{Z} \Longrightarrow y \mid x(x-1)(x^2-x-1)\\
$$
And so on. With lot of transformations and special cases we can obtain, that there is no more solutions (if I didn't make mistake). But it looks badly, is so long and it is easy to confuse. Have anyone simple idea? I strongly believe here some exist.
|
Hint: $x^4-2x^3+x=y^4+3y^2+y\let\leq\leqslant\let\geq\geqslant$ is equivalent to $(2x^2-2x-1)^2-(2y^2+3)^2=4y-8$.
We have $2x^2-2x-1+2y^2+3\mid4y-8$, hence $|2x^2-2x-1+2y^2+3|\leq|4y-8|$. As $x^2\geq x$ for $x\in\mathbb Z$ this gives $y^2+1\leq|2y-4|$. For $y\geq2$ this is $(y-1)^2\leq-5$; for $y\leq2$ this is $(y+1)^2\leq3$. It remains to check $y\in[-2,0]$.
$y=-2$: $x^2-x+y^2+1\mid 2y-4$ gives $x^2-x=3$, impossible.
$y=-1$: $x^2-x+2\mid6$ means $x^2-x=4$, impossible, or $x=0,1$. None of them works.
$y=0$: $x^2-x+1\mid4$ gives $x^2-x=0$, hence $x=0,1$. Both are solutions.
|
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|
Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
|
Divide the numerator & the denominator by $\cos^4x$ to find
$$I=\int\frac{\sec^4x}{\tan^4x+1+\tan^2x}dx$$
Putting $\tan x=y,$
$$I=\int\frac{1+y^2}{1+y^2+y^4}dy=\int\frac{1+\dfrac1{y^2}}{\dfrac1{y^2}+1+y^2}dy$$
$$=\int\frac{1+\dfrac1{y^2}}{\left(y-\dfrac1y\right)^2+3}dy $$
Finally set, $u=y-\dfrac1y=\dfrac{y^2-1}y=-2\cdot\dfrac1{\dfrac{2\tan x}{1-\tan^2x}}=-2\cot2x$
|
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|
Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
|
It's Quite Simple..
You know, $a^2 - b^2 = (a-b)(a+b)$
Now, $(x-5)^2 - 4 = (x-5)^2-2^2$
Let $a=x-5 $ and $b=2$
So, $a^2-b^2 = (x-5)^2-2^2 = (a-b)(a+b)=(x-5-2)(x-5+2)$
|
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|
Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$ This induction problem is giving me a pretty hard time:
$$\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
I am struggling because my math teacher explained us that in this case ($n^2$) when we prove that $n+1$ satisfies the property we have to write it like this: $$ \frac{1}1+\frac{1}4+\frac{1}9+\cdots+ \frac{1}{n^2} +\frac{1}{(n+1)^2} < \frac{4(n+1)}{(2(k+1)+1)}$$
and I probably got lost in the process.
|
Hint: What is $\frac{4(n+1)}{2(n+1)+1}-\frac{4n}{2n+1}$? (What is the corresponding difference on the other side of the inequality when you move from $n$ to $n+1$?)
|
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|
Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
|
Do you know that $(\cos^{-1}t)'=-\frac{1}{1+t^2}$?
So $$f'(x)=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot (\frac{1-x}{1+x})'
=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot \frac{-(1+x)-(1-x)}{(1+x)^2} $$
Is it okay?
|
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|
Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ How can we prove that:
$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$
|
Different approach:
Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$
By using the identity $4ab=(a+b)^2-(a-b)^2$ and setting $a=\ln(1-x)$ and $b=\ln(1+x)$ we have
$$4I=\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx-\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx$$
The first integral:
$$\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx=\frac18\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx=\frac14\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1 x^n\ln^2x\ dx\\=\frac12\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac12\zeta(5)=\boxed{\zeta(5)-\frac12\zeta(2)\zeta(3)}$$
The second integral:
We proved here that
$$\ln^2\left(\frac{1-x}{1+x}\right)=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$$
Then we can write
$$\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx=16\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}\\
=8\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}2\sum_{n=1}^\infty\frac{H_n}{n^4}=\boxed{\frac{31}{4}\zeta(5)-\frac72\zeta(2)\zeta(3)}$$
and the answer follows by substituting these boxed results.
|
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|
Limit of a difference equation Given $y_{k+1} = 1 + \sqrt{y_k}$ for $k \geq 0$ and $y_0 = 0$, we have a limit of the form $L = \lim_{k \rightarrow \infty} y_k = 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ... \sqrt{1 + \sqrt{2}}}}}$.
Apparently this can be rewritten as $L = \frac{3 + \sqrt{5}}{2}$, but I am at a loss for how to approach this. Tips would be greatly appreciated.
I see that $L = \frac{3 + \sqrt{5}}{2} = 1 + \frac{1}{2} + \frac{\sqrt{5}}{2}$, but this doesn't really take me very far.
|
To show that the limit actually exists, we can proceed as follows:
Let
$$
\textstyle a_n=\overbrace{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots+\sqrt{1+\sqrt2}}}}}^{n\ 1\text{s}}\tag{1}
$$
so that
$$
(a_n-1)^2=a_{n-1}\tag{2}
$$
Subtracting $(2)$ substituting $n\mapsto n-1$ from $(2)$ and dividing by $(a_n-1)+(a_{n-1}-1)$, gives
$$
a_n-a_{n-1}=\frac{a_{n-1}-a_{n-2}}{(a_n-1)+(a_{n-1}-1)}\tag{3}
$$
Since $a_1=1+\sqrt2\gt2=a_0$, $(3)$ says that $a_n$ is monotonically increasing
Furthermore, we have that for $n\ge1$, $(a_n-1)+(a_{n-1}-1)\ge1+\sqrt2$. Therefore,
$$
0\le a_n-a_{n-1}\le\frac{a_{n-1}-a_{n-2}}{1+\sqrt2}\tag{4}
$$
$(4)$ implies that $a_n$ is Cauchy, and therefore, converges.
Letting
$$
a=\lim_{n\to\infty}a_n\tag{5}
$$
if we take the limit of $(2)$, we get
$$
(a-1)^2=a\tag{6}
$$
and since $a_n\ge2$, we must have
$$
a=\frac{3+\sqrt{5}}2\tag{7}
$$
|
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|
How to calculate $\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$? Question :
$$\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$$
Here I tried $\tan x$ and $\sin x$ expansion in numerator and denominator which are as follows :
$$\tan x =x +\frac{x^3}{3}+\frac{2x^5}{15} \cdots$$
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}\cdots$$
but this method is not working and also other alternative of using L'Hospital's rule is not working one this ... please help how to tackle this limit prblem thanks.
|
It looks like a bad guy. I used Wolfram Mathematica to compute
$$
\tan^3 x -x^3 = x^5+\frac{11}{15}x^7 + O(x^8)
$$
and
$$
e^{\tan^3 x -x^3}=1+x^5+\frac{11}{15}x^7 + O(x^8).
$$
Therefore
$$
e^{\tan^3 x}-e^{x^3}=x^5+\frac{11}{15}x^7 + O(x^8).
$$
Since
$$
\log(1+x^3 \sin^2 x) = x^5 +O(x^6),
$$
the limit exists and is equal to $1/2$.
But it can be very boring to perform these computations by hand: all the terms up to power $5$ cancel out, so you should take care of really many terms in each expansion, otherwise you'll get nothing useful.
|
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|
Evaluting $\tan15°$ using difference formula
Evalute $\tan15°$ using difference formula
Steps I took:
$$\begin{align}
\tan(45-30)&=\frac { \tan(45)-\tan(30) }{ 1+\tan(45)\tan(30) }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \cdot \frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1-\frac { \sqrt3 }{ 3 } }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } } \\
\end{align}$$
Where do I go from here to arrive at the correct answer? And what mistakes did I make, if any?
|
$$\frac{1-\dfrac1{\sqrt3}}{1+\dfrac1{\sqrt3}}=\frac{\dfrac{\sqrt3-1}{\sqrt3}}{\dfrac{\sqrt3+1}{\sqrt3}}$$
$$=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}=?$$
|
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|
How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})
\\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$
|
Hint: Divide both sides of $8bc=a[4b^2 + (b^2-c^2)^2]$ by $bc$. You will end up with this:
$$8 = a[4\frac{b}{c}+bc(\frac{b}{c}-\frac{c}{b})^2]$$
Take the RHS and you can prove that it is equal to 8.
Process: Calculate $\frac{b}{c}$ from the given equations.
$$b = \frac{\sin (\theta + \phi)}{\cos \theta. \cos \phi}$$
$$c = \frac{\cos \theta + \cos \phi}{\cos \theta. \cos \phi}$$
$$\frac{b}{c} = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})}{2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}$$
$$\frac{b}{c} = \frac{\sin(\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})}$$
You will end up with:
$$\frac{b}{c}-\frac{c}{b}=-\frac{\cos \theta. \cos \phi}{\cos(\frac{\theta - \phi}{2}).sin(\frac{\theta + \phi}{2})}$$
$$bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{\sin (\theta + \phi)(\cos \theta + \cos \phi)}{\cos^2 \theta. \cos^2 \phi}\frac{\cos^2 \theta. \cos^2 \phi}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$
$$ = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$
$$ = \frac{4\cos^2 (\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})\sin (\frac{\theta + \phi}{2})}$$
$$4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{4}{\cos (\frac{\theta - \phi}{2})}(\sin(\frac{\theta + \phi}{2})+\frac{\cos^2 (\frac{\theta + \phi}{2})}{\sin(\frac{\theta + \phi}{2})})$$
$$= \frac{4}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$
$$a[4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2] = 4\frac{\sin \theta + \sin \phi}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$
$$ = 4\frac{2\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$
$$= 8$$
|
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|
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.
Therefore the nth term of given series is :
$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)
By using partial fraction :
and solving for A,B and C we get A = 1/2, B = -1, C =1/2
Putting the values of A,B and C in (1) we get :
$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$
But by putting $r =1,2,3, \cdots$ I am not getting the answer. Please guide how to solve this problem . Thanks.
|
HINT:
If the $r$th term
$$T_r=\frac1{2r}-\frac1{r+1}+\frac1{2(r+2)}$$
$$2T_r=\frac1r-\frac2{r+1}+\frac1{(r+2)}=\left(\underbrace{\frac1r-\frac1{r+1}}\right)-\left(\underbrace{\frac1{r+1}-\frac1{r+2}}\right)$$
Recognize the two Telescoping series
|
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|
Number of solutions to equation, range restrictions per variable Find the number of solutions of the equation $x_1+x_2+x_3+x_4=15$ where variables are constrained as follows:
(a) Each $x_i \geq 2.$
(b) $1 \leq x_1 \leq 3$ , $0 \leq x_2$ , $3 \leq x_3 \leq 5$, $2 \leq x_4 \leq 6$
I believe I understand part A. I can fix the values at greater than two by adding two to each term, as in:
$(x_1 + 2) + (x_2 + 2) + (x_3+2) +(x_4+2) = 15$
$x_1+x_2+x_3+x_4=7$
So, $C(7+4-1,7) = {10! \over 7!3!} = 120$
For part B, i really have no idea where to start. Thanks for any insight!
|
By similar reasoning to (a) we can simplify the problem to:
\begin{eqnarray*}
x_1+x_2+x_3+x_4 &=& 9 \qquad\qquad\text{(*)} \\
\text{with } && 0 \leq x_1 \leq 2 \\
&& 0 \leq x_2 \\
&& 0 \leq x_3 \leq 2 \\
&& 0 \leq x_4 \leq 4.
\end{eqnarray*}
Define sets
\begin{eqnarray*}
S\;\, &=& \{\text{all solutions to (*) without upper bounds on the $x_i$}\} \\
S_1 &=& \{\text{all solutions in $S$ where $x_1 \geq 3$}\} \\
S_2 &=& \{\text{all solutions in $S$ where $x_3 \geq 3$}\} \\
S_3 &=& \{\text{all solutions in $S$ where $x_4 \geq 5$}\} \\
\end{eqnarray*}
By the same method used in (a) we calculate:
\begin{eqnarray*}
|S| &=& \binom{9+4-1}{9} = \binom{12}{9} \\
|S_1| = |S_2| &=& \binom{6+4-1}{6} = \binom{9}{6} \\
|S_3| &=& \binom{4+4-1}{4} = \binom{7}{4} \\
|S_1 \cap S_2| &=& \binom{3+4-1}{3} = \binom{6}{3} \\
|S_1 \cap S_3| = |S_2 \cap S_3| &=& \binom{1+4-1}{1} = \binom{4}{1} \\
|S_1 \cap S_2 \cap S_3| &=& 0.
\end{eqnarray*}
Then we require, where set complement is with respect to $S$,
\begin{eqnarray*}
\text{Ans.} &=& |S_1^c \cap S_2^c \cap S_3^c| \\
&=& |S| - |S_1 \cup S_2 \cup S_3| \qquad\qquad\text{by de Morgan's Law} \\
&=& |S| - (|S_1| + |S_2| + |S_3|) + (|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|) - |S_1 \cap S_2 \cap S_3| \\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{by the Inclusion-Exclusion Principle} \\
&=& \binom{12}{9} - 2\binom{9}{6} - \binom{7}{4} + \binom{6}{3} + 2\binom{4}{1} - 0 \\
&=& 220 - 168 - 35 + 20 + 8 \\
&=& 45.
\end{eqnarray*}
|
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|
Prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent How do I prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent for every x in real numbers except for $x=0$?
I tried using the ratio test, but it doesn't seem to be conclusive.
|
You could use the limit comparison test.
Use $\displaystyle b_n = \frac{1}{n^2}$
Let $\displaystyle a_n = \frac{1}{n^2x^2 + 1}$
$\displaystyle \frac{a_n}{b_n} = \frac{n^2}{n^2x^2 + 1}$
$x$ is a constant $x \ne 0$
$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{1}{x^2}$
Provided $x \ne 0, x^2 > 0, \frac{1}{x^2} > 0$ we find that since $\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges to $\displaystyle \frac{\pi^2}{6}$ we find that $\sum a_n$ converges as well. (By Limit Comparison Test)
|
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|
Finding $\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$ The WolfarmAlpha couldn't give me the sum of $$\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$$ therefore I thought that this problem is difficult so I used my calculator to get $(1/24)$
Is this value right or not?
If this right, why the WolfarmAlph couldn't find it?
|
Consider the sum:
$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} $$
This sum is a bit easier than it looks, if you know the residue theorem. The main observation is that the numerator either takes the value $\sqrt{3}/2$, $-\sqrt{3}/2$, or $0$. Then you may rewrite the sum as
$$\frac{\sqrt{3}}{2} \sum_{n=1}^{\infty} \left [\frac1{(3 n-2)^5} - \frac1{(3 n-1)^5}\right ] = \frac{\sqrt{3}}{4 \cdot 3^5} \sum_{n=-\infty}^{\infty} \left [\frac1{(n-2/3)^5} - \frac1{(n-1/3)^5}\right ]$$
We may apply the residue theorem to this sum by considering the following result:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [ \cot{\pi z} \, f(z) ]$$
where $z_k$ are the non-integer poles of $f$. All we need to do is compute the residues at the poles, which in this case are at $z=1/3$ and $z=2/3$. For example,
$$\begin{align}\operatorname*{Res}_{z=2/3} \left [ \cot{\pi z} \, \left ((z-2/3)^{-5}-(z-1/3)^{-5} \right ) \right] &= \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z} \right ]_{z=2/3} - \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z} \right ]_{z=1/3} \\ &= -\frac{8\pi^4}{3 \sqrt{3}}\end{align}$$
The sum is then
$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} = \frac{2 \pi^5}{729}$$
The result follows.
|
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|
How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. I then set $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$. However, I afterwards realized that I had squared an inequality to get this. The following is what my proof looked like before I realized this mistake:
Let $\varepsilon > 0$. Choose $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$ and let $n > N$. Then $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. Thus $2n\varepsilon > 1 - \varepsilon^2$, $\varepsilon^2 + 2n\varepsilon > 1$, $\varepsilon^2 + 2n\varepsilon + n^2 = (\varepsilon + n)^2 > 1 + n^2$, $\varepsilon + n > \sqrt{1 + n^2}$, and $\varepsilon > \sqrt{n^2 + 1} - n$. Hence $| \sqrt{n^2 + 1} - n - 0 | = \sqrt{n^2 + 1} - n <
\varepsilon.$ Therefore, $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$.
How should I find $N$? Also, are there any other errors in this proof?
|
Another way using Taylor series $$A=\sqrt{n^2 + 1} - n=n\Big(\sqrt{1+\frac{1}{n^2}}-1\Big)$$ Remembert that, for samm values of $x$, $$\sqrt{1+x} =1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Replace $x$ by $\frac{1}{n^2}$ so $$\sqrt{1+\frac{1}{n^2}}-1=1+\frac{1}{2 n^2}-\frac{1}{8}
\left(\frac{1}{n^2}\right)^2+\cdots-1=\frac{1}{2 n^2}-\frac{1}{8}
\left(\frac{1}{n^2}\right)^2+\cdots$$ Finally, $$A=\frac{1}{2 n}-\frac{1}{8n^3}+\cdots$$ which shows the limit and how it is approched.
|
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|
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
|
let f(n)=3n-2
let s(n)=[f(1)+f(2)+f(3)+...+f(n-1)+f(n)]
s(n) =(3(1)-2)+(3(2)-2)+(3(3)-2)+...+(3(n-1)-2)+(3(n)-2)
s(n) =(3(1) + 3(2) + 3(3) +...+ 3(n-1) + 3(n) )-2n
s(n) = 3(1 + 2 + 3 +...+ (n-1) + n )-2n
s(n) =3(sum of all #s from 1 to n) -2n
s(n) =3(sum of all #s from 1 to n-1)+3(n)-2n
s(n) =3(sum of all #s from 1 to n-1)+n
Now we reverse the order:
s(n) =3((n-1)+(n-2)+...3+2+1+(n-n))+n
s(n) =3(n^2)-3(sum of all #s from 1 to n)+n
s(n) =3(n^2)-3(sum of all #s from 1 to n)+n+n-n
s(n) =3(n^2)-3(sum of all #s from 1 to n)+2n-n
s(n) =3(n^2)-3(sum of all #s from 1 to n)-(-2n)-n
s(n) =3(n^2)-[3(sum of all #s from 1 to n)-2n]-n
We've got another value for s(n) on the right side now.
s(n) =3(n^2)-[s(n)]-n
2[s(n)] =3(n^2)-n
2[s(n)] =n(3n)-n(1)
2[s(n)] =n(3n-1)
s(n) =n(3n-1)/2
|
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|
Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Find the number of elements of order $1$, order $2$, order $3$, and order $5$. But this scenario can't happen. Why not?
Let $G$ be a group of order $30=2 \cdot 3 \cdot 5$.
The number of Sylow $2$-subgroup $n_2$ divides $15$ and has the form $n_2=2k+1$ by the Sylow theorems. Therefore $n_2=1,3,5,15$.
The number of Sylow $3$-subgroup $n_3$ divides $10$ and has the form $n_2=3k+1$ by the Sylow theorems. Therefore $n_3=1,10$.
The number of Sylow $5$-subgroup $n_5$ divides $6$ and has the form $n_5=5k+1$ by the Sylow theorems. Therefore $n_5=1,6$. However, since Sylow $5$-subgroup isn't normal $n_5 \neq 1$.
\begin{array}{|c|c|c|c|}
\hline
n_2 & n_3 & n_5 & number \,of \,elements & Possible \\ \hline
1 & 1 & 6 & 1+1\cdot2+6\cdot4=26 & Yes \\ \hline
1 & 10 & 6 & 1+10\cdot2+6\cdot4=45 & No \\ \hline
3 & 1 & 6 & 3+1\cdot2+6\cdot4=29 & Yes \\ \hline
3 & 10 & 6 & 3+10\cdot2+6\cdot4=47 & No \\ \hline
5 & 1 & 6 & 5+1\cdot2+6\cdot4=31 & No \\ \hline
5 & 10 & 6 & 5+10\cdot2+6\cdot4=49 & No \\ \hline
15 & 1 & 6 & 15+1\cdot2+6\cdot4=41 & No \\ \hline
15 & 10 & 6 & 15+10\cdot2+6\cdot4=59 & No \\ \hline
\end{array}
Where do I go from here?
|
My way: you deduced that for such a $G$ it must be $n_5=6$. These six subgroups are cyclic of order $5$ (every group of prime order is necessarely cyclic). Let $K_1,\dots,K_6$ be these $5$-Sylow. It must be $K_i\cap K_j=1$ for $i\neq j$ and every element different from the identity must has order 5. Hence $G$ contains exactly $6\cdot(5-1)=24$ elements of order $5$.
Again by your Sylow computation $n_3=1,10$. If was $n_3=10$ the by the same argument as above we'd have $10\cdot(3-1)=20$ elements of order $3$. So $G$ would contain more than $30$ elements. Hence $n_3=1$. So $G$ contains exactly $2$ elements of order $3$.
So till now we have $|G|=2\cdot3\cdot5$ hence it could contain only elements of order $2,3$ or $5$. And by Cauchy theorem it must contain at least one element of order $2,3$ and $5$.
Moreover we counted (togheter with the identity element which is the only element of order $1$) $24+2+1=27$ elements in $G$. The three remaining elements must be of order $2$, there is no other way (and from this we deduce $n_2=3$).
Summing up, we have that if $G$ is a group of order $30$ with a $5$-Sylow not normal, it must be:
$\bullet$ $n_2=3,\;\;n_3=1,\;\;n_5=6$;
$\bullet$ $G$ has exactly $1$ element of order $1$, $3$ elements of order $2$, $2$ elements of order $3$ and $24$ elements of order $5$.
|
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|
Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
|
Hint
Let $$A=\left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}$$ and take logarithms of both sides $$\log(A)=9x^2 \log\Big(\frac{x^2+2}{x^2-4}\Big)=9x^2\log\Big(\frac{1+\frac{2}{x^2}}{1-\frac{4}{x^2}}\Big)$$ $$\log(A)=9x^2 \Big[\log \left(1+\frac{2}{x^2}\right)-\log \left(1-\frac{4}{x^2}\right) \Big]$$ Now, use the fact that, for mall $y$, $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ Apply that to each of the logarithms separately, replace $y$ by its expression as a function of $x$ (that is to say $\frac{2}{x^2}$ for the first and $\frac{-4}{x^2}$ for the second) and simplify.
I am sure that you can take from here.
|
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|
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
|
The hard way, by Taylor:
First establish the derivatives,
$$\begin{align}
f(x)(x-1)&=x^6-1\\
f'(x)(x-1)+f(x)&=6x^5\\
f''(x)(x-1)+2f'(x)&=30x^4\\
f'''(x)(x-1)+3f''(x)&=120x^3\\
f''''(x)(x-1)+4f'''(x)&=360x^2\\
f'''''(x)(x-1)+5f''''(x)&=720x\\
f''''''(x)(x-1)+6f'''''(x)&=720\\
f'''''''(x)(x-1)+7f''''''(x)&=0\\
\dots\\
f^{(n+1)}(x)(x-1)+(n+1)f^{(n)}(x)&=0\\
\end{align}$$
Then solve for $x=0$:
$$\begin{align}
f(0)&=1,\\
f'(0)=f(0)&=1,\\
f''(0)=2f'(0)&=2,\\
f'''(0)=3f''(0)&=3!\\
f''''(0)=4f'''(0)&=4!\\
f'''''(0)=5f''''(0)&=5!,\\
f''''''(0)=6f'''''(0)-720&=0\\
\dots\\f^{(n)}&=0\end{align}$$
Conclusion,
$$f(x)=1+x+x^2+x^3+x^4+x^5.$$
|
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|
Weak principle of induction for $5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$
Show that
$$5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$$
Proving the base case $n(1)$:
$5(1)= \frac{5(1)(1+1)}{2}$
$5 = \frac{5(2)}{2}$
$5 = 5$
Induction hypothesis:
$n = k$
$5+10+15+\ldots+5k = \frac{5k(k+1)}{2}$
Induction step (adding $k+1$):
$5+10+15+\ldots+5k+5k+1 = \frac{5k+1(k+1+1)}{2}$
Substituting $\frac{5k(k+1)}{2}$ for $5k$:
$\frac{5k(k+1)}{2}+5k+1 = \frac{(5k+1)(k+1+1)}{2}$
Simplifying:
$\frac{5k(k+1)+2(5k+1)}{2} = \frac{(5k+1)(k+2)}{2}$
$\frac{5k^2+5k+10k+2}{2} = \frac{5k^2+10k+k+2}{2}$
These aren't equal, so what did I do wrong here?
|
When you plug in $n=k+1$ you have that
$$5+10+\cdots+5k+5(k+1)\not=5+10+\cdots+5k+5k+1$$
|
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|
What is the smallest possible value of their sum? The product of two positive numbers is 36. What is the smallest possible value of their sum?
so far I got
$$xy=36$$
$$y=\frac{36}{x}$$
|
First note that
$$f(x)=x+\frac{36}{x}=x+36x^{-1}$$
$$\frac{d}{dx}[f(x)]=1-36x^{-2}=1-\frac{36}{x^2}$$
So now let's find the critical points
$$\frac{d}{dx}[f(x)]=0 $$
$$ 1-\frac{36}{x^2}=0 $$
$$ 1=\frac{36}{x^2} $$
$$ x^2=36= (\pm 6)^2 $$
$$ x=\sqrt{(\pm 6)^2}=\pm 6 $$
Since we're only considering positive values, we'll omit the negative value. Next we must check if $x=6$ is a local minima of $f(x)$. So now
$$\frac{d^2}{dx^2}[f(x)]=72x^{-3}=\frac{72}{x^3}$$
$$\frac{d^2}{dx^2}[f(6)]=\frac{72}{6^3}\gt 0 $$
Therefore, $f(x)=x+\frac{36}{x}$ has a local minima at $x=6$. Which implies that the smallest sum of positive values is
$$f(6)=6+6=12$$
|
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|
Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3^2=5\cdot 2^1$$
$$2^2+4^2=5\cdot 2^2$$
$$2^2+6^2=5\cdot 2^3$$
|
Hint:-
$5\cdot2^n=x^2+y^2 \implies 5\cdot 2^{n+2}=(2x)^2+(2y)^2$
|
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|
Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} = \frac{2\tan A}{\sec^2 A} = 2 \tan A\cos^2 A = 2 \sin A \cos A = \sin 2A $$
Since $x=A/2$, $\sin 2A = \sin x$
Let $t=\tan \frac{x}{2}$
$$ \int \frac{1}{\sin x} dx = \int \frac{2t}{1+t^2} dt $$
Let $u= 1+t^2$, $ du = 2t\,dt$
$$\int \frac{2t}{1+t^2} dt = \int \frac{2t}{u}\cdot\frac{1}{2t} du = \int \frac{1}{u} \,du = \ln u + C \\ = \ln(1+t^2) + C = \ln\left(1+\tan^2 \frac{x}{2} \right) + C $$
Then what do I do?
How do I show this is equal to $-\cos x + C$ ?
|
Since $t=\tan \frac{x}{2}, \tan^{-1}t=\frac{x}{2}, x=2\tan^{-1}t$, and $dx=\frac{2}{1+t^2}dt$.
Then $\displaystyle\int\frac{1}{\sin x} dx=\int\frac{1+t^2}{2t}\cdot\frac{2}{1+t^2}dt=\int\frac{1}{t}dt=\ln|t|+C=\ln\left|\tan\frac{x}{2}\right|+C$
|
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|
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
|
Hint $\ $ Times $2$ it is equivalent to $\,{\rm mod}\ 11\!:\ 8^n - 3^n\equiv 0,\,$ i.e.$\ 3^n\equiv (-3)^n\equiv \color{#c00}{(-1)^n} 3^n.\,$ Thus it suffices to prove $\ n$ even $\,\Rightarrow\, \color{#c00}{(-1)^n}\equiv 1,\,$ which is straightforward (by induction or not).
|
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|
If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.$
So $\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.$
Now $\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi$
Now How can I calculate $\sin 2\phi.$
Help me, Thanks
|
Like Sameer Kailasa,
$\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \dfrac{5}{4}\ \ \ \ (1)$
and let $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) =x \ \ \ \ (2)$
$(1)-(2)\implies2(\sin \phi+\cos\phi)=\dfrac{5}{4}-x$
$\implies \sin \phi+\cos\phi=\dfrac{5-4x}8 $
Squaring we get $1+2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2\iff2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2-1$
$(1)+(2)\implies2(1+\sin \phi\cos\phi)=x+\dfrac{5}{4}\implies 2\sin \phi\cos\phi=\dfrac{4x-3}4$
Equate the values of $2\sin \phi\cos\phi$
|
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|
Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of
$$
\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}
$$
My idea was to use ratio test:
$$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$
$$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$
Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all
$\displaystyle n$ , but
$$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$
so I am not quite sure if I can apply ratio test.
|
How about computing the partial sums? For any $N>2$ we have:
$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$
hence:
$$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$
ensures convergence (towards $\frac{3}{2}$).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify a quick sum of sines Simplify $\sin 2+\sin 4+\sin 6+\cdots+\sin 88$
I tried using the sum-to-product formulae, but it was messy, and I didn't know what else to do. Could I get a bit of help? Thanks.
|
$$\sum_{n=1}^{N} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(N+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$
$$\sum_{n=1}^{44} {\sin(nx)} = \frac{1}{2} \cot{\frac{x}{2}} - \frac{\cos{(44+\frac{1}{2})x}}{2\sin{\frac{x}{2}}}$$
$$\sum_{n=1}^{44} {\sin(2 n)} = \frac{1}{2} \cot{\frac{2}{2}} - \frac{\cos{[(44+\frac{1}{2})2]}}{2\sin{\frac{2}{2}}}$$
$$\sum_{n=1}^{44} {\sin(2 n)} = \frac{1}{2} \cot{1} - \frac{1}{2}\frac{\cos{99}}{\sin{1}}$$
$$\sum_{n=1}^{44} {\sin(2 n)} \approx 0.2974$$
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities
|
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|
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
|
Consider the contour integral
$$\oint_C dz \frac{(z^2-1) \log^2{z}}{1+z^4} $$
where $C$ is a keyhole contour about the positive real axis having an outer radius $R$ and an inner radius $\epsilon$. As $R \to \infty$ and $\epsilon \to 0$, the integral may be shown to be equal to
$$-i 4 \pi \int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} dx \frac{x^2-1}{1+x^4} $$
The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, which are at $e^{i (2 k+1) \pi/4}$ for $k=0,1,2,3$, or
$$i \frac{\pi}{2} \left [\frac{(i-1) (-\pi^2/16)}{e^{i 3 \pi/4}} - \frac{(i+1) (-9\pi^2/16)}{e^{i \pi/4}} + \frac{(i-1) (-25 \pi^2/16)}{e^{-i \pi/4}} - \frac{(i+1) (-49\pi^2/16)}{e^{-i 3 \pi/4}} \right ]$$
which simplifies to $-i (\pi^3/32) 16 \sqrt{2} = -i \pi^3/\sqrt{2}$. Equating real and imaginary parts, we find that
$$\int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} = \frac{\pi^2}{4 \sqrt{2}} $$
|
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|
Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.
|
Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides.
$$ \begin{align} \int_0^\infty\log\left(\frac{x^2+2kx\cos(b)+k^2}{x^2+2kx\cos(a)+k^2}\right)\frac{dx}{x}
&=\int_0^\infty\log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \tag{1} \\ &= 2 \int_{0}^{1} \tag{2} \log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \\ &= 4 \int_{0}^{1} \frac{1}{u} \sum_{k=1}^{\infty} (-u)^{k} \frac{\cos(ka) - \cos(kb)}{k} \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) -\cos(kb)}{k} \int_{0}^{1} u^{k-1} \ du \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) - \cos(kb)}{k^{2}} \\ &= 4 \left(\frac{a^{2}}{4} - \frac{b^{2}}{4} \right) \tag{3} \\ &= a^{2}- b^{2} \end{align} $$
$ $
$(1)$ Let $ \displaystyle u= \frac{x}{k}.$
$(2)$ Separate the integral into two integrals, namely one over the interval $(0,1)$ and one over the interval $(1, \infty)$, and replace $u$ with $\frac{1}{u}$ in the second integral.
$(3)$ By integrating the Fourier series $ \displaystyle \sum_{k=1}^{\infty} (-1)^{k} \frac{\sin(ka)}{k} = - \frac{a}{2} \ , \ |a| < \pi$, we get $$ \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka)}{k^{2}} = \frac{a^{2}}{4} - \frac{\pi^{2}}{12} \ , \ |a| < \pi .$$
|
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|
Understanding Generating Function I have been looking at This Problem and Answer about generating functions. The problem asked for the generating function of: $$a_n=4a_{n-1}-4a_{n-2}+{n\choose 2}2^n+1$$ I understand how Ron Gordon simplified it to: $$\sum_{n=2}^{\infty} (a_n-4 a_{n-1}+4 a_{n-2}) x^n = \sum_{n=2}^{\infty} n(n-1)2^{n-1} x^n + \sum_{n=2}^{\infty} x^n$$ and where he defined the generating function as: $$g(x) = \sum_{n=0}^{\infty} a_n x^n$$but then he broke the summation into $3$ seperate summations and this is where I start to get messed up. I do understand in the first summation how we can get: $$\sum_{n=2}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^n-(a_0+a_1x)$$I don't understand the next two summations though.
Question:
1) In the second and third summation where did the $x$ and $x^2$ come from?
2) Do you always multiply by $x^n$ in generating functions? or does it change?
3) I believe I remember my professor saying that every generating function would always be $f(x)=\sum_{n=0}^na_nx^n$ but why is that the case?
4) I guess my overall trouble is understanding what the goal of a generating function is.
Any help would be much appreciated!
(I have looked at other questions, but I can't find a clear answer. So if there is one out there that I missed that answers this I'd be happy to delete my question. Thanks.)
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Easiest is to define:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
then shift indices so there aren't subtractions there:
$$
a_{n + 2} = 4 a_{n + 1} - 4a_n + \binom{n + 2}{2} 2^{n + 2} + 1
$$
Multiply by $z^n$ and sum over $n \ge 0$:
$$
\sum_{n \ge 0} a_{n + 2} z^n
= 4 \sum_{n \ge 0} a_{n + 1} z^n - 4 \sum_{n \ge 0} a_n z^n
+ \sum_{n \ge 0} \binom{n + 2}{2} 2^{n + 2} + \sum_{n \ge 0} z^n
$$
We recognize:
$\begin{align}
\sum_{n \ge 0} a_{n + k} z^n
&= \frac{A(z) - a_0 - a_1 z - \dotsb - a_{k - 1} z^{k - 1}}{z^k} \\
\sum_{n \ge 0} \binom{n + 2}{2} 2^{n + 2}
&= \sum_{n \ge 0} \binom{n}{2} 2^n z^n \\
&= \frac{(2 z)^2}{(1 - 2 z)^3} \\
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z}
\end{align}$
Replace in the sum above:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z}
- 4 A(z)
+ \frac{(2 z)^2}{(1 - 2 z)^3} - 1 - 4 z
+ \frac{1}{1 - z}
$$
Solve for $A$:
$$
A(z)
= \frac{a_0
+ (a_1 + 11 a_0) z
- (7 a_1 - 46 a_0 - 1) z^2
+ (18 a_1 - 92 a_0 - 6) z^3
- (20 a_1 - 88 a_0 - 16) z^4
+ (8 a_1 - 32 a_0 - 12) z^5
}
{1 - 11 z + 50 z^2 - 120 z^3 + 160 z^4 - 112 z^5 + 32^6}
$$
Expand this as partial fractions, and read off the coefficients.
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables I have come as far as deducing that the denominator can be written as a geometric series. Hence, for $m=2$,
\begin{align*}
\int_{-\infty}^\infty \frac{1-x}{1-x^5} dx &= 2 \pi i ( B_1 + B_2 ) - \int_{C_R} \frac{1-z}{1-z^5} dz \\
&= 2 \pi i \left( \frac{1 - \exp(i \frac{2 \pi}{5})}{-5 \exp(i\frac{8 \pi}{5})} + \frac{1 - \exp(i \frac{4 \pi}{5})}{-5 \exp(i\frac{16 \pi}{5})} \right) - 0\\
&\overset{*}{=} \frac{\pi}{5} \cot\left( \frac{\pi}{10} \right) \sec \left( \frac{\pi}{5} \right).
\end{align*}
where $B_1$ and $B_2$ denote the residues in the upper-half plane.
However, I can't work out why the equality $\overset{*}{=}$ should hold. I'm sure it's the result of some clever manipulation but I've been staring at it for a while without much progress. Is my solution wrong?
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There seems to be a tiny error in the last step, otherwise your solution is perfect.
You have that
\begin{align}
2\pi i \left(\frac{1-e^{i2\pi/5}}{-5e^{i8\pi/5}}+ \frac{1-e^{i4\pi/5}}{-5e^{i16\pi/5}} \right)
& = -\frac{2\pi i}{5}\left(e^{-i8\pi/5} - e^{-i6\pi/5} + e^{-i16\pi/5} - e^{-i12\pi/5}\right)\\
& = -\frac{2\pi i}{5}\left(e^{i2\pi/5} - e^{i4\pi/5} + e^{i4\pi/5} - e^{-i2\pi/5}\right)\\
& = \frac{4\pi}{5}\frac{e^{i2\pi/5} - e^{-i2\pi/5}}{2i}\\
& = \frac{4\pi}{5} \sin\left(\frac{2\pi}5\right).
\end{align}
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"url": "https://math.stackexchange.com/questions/1070106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Is there a solution to $a^4+(a+d)^4+(a+2d)^4+(a+3d)^4+\dots = z^4$? Is there a,
$$a^4+(a+d)^4+(a+2d)^4+(a+3d)^4+\dots = z^4\tag1$$
in non-zero integers? One can be familiar with,
$$31^3+33^3+35^3+37^3+39^3+41^3 = 66^3\tag2$$
I found that,
$$29^4+31^4+33^4+35^4+\dots+155^4 = 96104^2\tag3$$
which has $m=64$ addends. The equation,
$$a^4+(a+b)^4+(a+2b)^4+\dots+(a+63b)^4 = y^n\tag4$$
or,
$$64 a^4 + 8064 a^3 b + 512064 a^2b^2 + 16257024 a b^3 + 206447136 b^4 = y^n\tag5$$
for $n=2$ can be reduced to an elliptic curve, so there is an infinite number of primitive integer solutions. However, for $n=4$, it is now a superelliptic curve, so is harder to solve.
Update: Courtesy of Antony's answer, a little experimentation showed that,
$$(\text{Excluding}\; 19^4):\quad 5^4+6^4+7^4+\dots+38^4 = 64^4$$
So close! Perhaps there is a non-zero solution to $(1)$ yet.
Questions:
*
*What is the general formula for $m$ addends of,$$F(k)=a^k+(a+b)^k+(a+2b)^k+(a+3b)^k+\dots$$ for $k=4?$ (The case $k=2,3$ can be found here, and the special case $a=b=1$ is given by Faulhaber's formula.)
*For some $m$ addends, does $F(4)=y^4$ have a solution in non-zero integers $a,b,y$?
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Let,
$$S(n)=a^4+(a+d)^4+\dots+(a+nd)^4\tag1$$
Then we can write
$$S(n)
=a^4+(a^4+4a^3d+6a^2d^2+4ad^3+d^4)+\dots+(a^4+4a^3dn+6a^2d^2n^2+4ad^3n^3+d^4n^4)\\
=(n+1)a^4+4a^3d(1+2+\dots+n)+6a^2d^2(1^2+2^2+\dots+n^2)+4ad^3(1^3+2^3+\dots+n^3)+d^4(1^4+2^4+\dots+n^4)$$
Using Faulhaber's formula for $F(x) = 1^x+2^x+\dots+n^x$ we get
$$S(n)=(n+1)a^4+4a^3d \cdot F(1)+6a^2d^2 \cdot F(2)+4ad^3\cdot F(3)+d^4 \cdot F(4)$$
Explicitly
$$S(n)=(n+1)\big(a^4+2a^3dn+a^2d^2n(2n+1)+ad^3n^2(n+1)+\tfrac{1}{30}d^4n(2n+1)(3n^2+3n-1)\big)\tag2$$
In general, let
$$s(n)=a^x+(a+d)^x+\dots+(a+nd)^x\\
=a^x+(a^x+C^1_xa^{x-1}d+\dots+d^x)+\dots+\big(a^x+C_x^1a^{x-1}dn+\dots+(dn)^x\big)\\
=(n+1)a^x+C_x^1a^{x-1}d(1+2+\dots+n)+C_x^2a^{x-2}d^2(1^2+2^2+\dots+n^2)+\dots+d^x(1^x+2^x+\dots+n^x)\\
=(n+1)a^x+C_x^1a^{x-1}d\cdot F(1)+C_x^2a^{x-2}d^2\cdot F(2)+\dots+d^x\cdot F(x)$$
So
$$s(n)=(n+1)a^x+\sum_{i=1}^x C_x^ia^{x-i}d^iF(i)$$
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"url": "https://math.stackexchange.com/questions/1072318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
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