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Determining if a Telescoping Series is convergent or not How do you determine if a telescoping series is convergent or not? If it converges, what value does it converge to?
It seems like you need to do partial fraction decomposition and then evaluate each term individually?
For example:
$$ \sum_{n=2}^\infty \frac{1}{n^3-n} $$
|
Partial fraction decomposition in your example yields
$$
\frac{1}{n^3 - n} = \frac{1}{2} \left[ \frac{1}{n - 1} - \frac{2}{n} + \frac{1}{n + 1} \right]
$$
To see the telescoping of the sum, it's nice to arrange terms like this:
$$
\begin{array}{*{13}{c}}
2 \sum_{n = 2}^\infty \frac{1}{n^3 - n} &=& \frac{1}{1} &-& \frac{2}{2} &+& \frac{1}{3} && && && \\
&& &+& \frac{1}{2} &-& \frac{2}{3} &+& \frac{1}{4} && && \\
&& && &+& \frac{1}{3} &-& \frac{2}{4} &+& \frac{1}{5} && \\
&& && && &+& \frac{1}{4} &-& \frac{2}{5} &+& \frac{1}{6} \\
&& && && && &+& \cdots &&
\end{array}
$$
Now it's clear that the $N$th partial sum $(N \ge 3)$ is
$$
s_N = \sum_{n = 2}^N \frac{1}{n^3 - n} = \frac{1}{2} \left[ \frac{1}{1} - \frac{1}{2} - \frac{1}{N} + \frac{1}{N + 1} \right].
$$
The infinite series is the limit of these partial sums:
$$
\sum_{n = 2}^\infty \frac{1}{n^3 - n} = \lim_{N \to \infty} s_N = \frac{1}{2} \biggl( \frac{1}{1} - \frac{1}{2} \biggr) = \frac{1}{4}.
$$
|
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|
Please tell me if i did this inverse laplace correctly. Thanks The question is : find the inverse laplace transformation of
$$\frac{13s^2+3s+6}{(s-2)(s^2+9)}.$$
Please tell me if i did this correctly
Here is my work:
Using partial factions:
\begin{align}
Y(s) &= \frac{13s^2 + 3s + 6}{(s - 2)(s^2 + 9)}
\\
&= \frac{64}{13 (-2 + s)} + \frac{3 (83 + 35 s)}{13 (9 + s^2)}
\end{align}
And: $$\frac{3 (83 + 35 s)}{13 (9 + s^2)} = \frac{83}{9 + s^2} + \frac{35s}{9 + s^2}$$
Finally, identity in partial fraction is:
\begin{equation}
Y(s) = \frac{64}{13 (-2 + s)} + \frac{83}{9 + s²} + \frac{35 s}{9 + s²}
\end{equation}
And then:
$Y(s) ↔ y(t)$, using the Laplace Transformation table.
The answer is:
$$y(t) = (64/13) e^{2t} u(t) + (83/3) \sin(3t) u(t) + 35\cos(3t) u(t)$$
|
The partial fraction expansion yields:
$\displaystyle
\frac{3 (35 s+83)}{13 (s^2+9)} + \frac{64}{13 (s-2)} = \frac{3(35 s)}{13 (s^2+9)} + \frac{3(83)}{13 (s^2+9)}+ \frac{64}{13 (s-2)}$
Now, we put that result into the desired forms:
$\displaystyle \frac{3(35 s)}{13 (s^2+3^2)} + \frac{3(83)}{13 (s^2+3^2)}+ \frac{64}{13 (s-2)}$
From this, we can see the forms we need.
This yields:
$$\displaystyle y(t) = \frac{1}{13}\left(105 \cos 3t + 83 \sin 3t + 64 e^{2t}\right)$$
|
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|
Proof that the sum of the cubes of any three consecutive positive integers is divisible by three. So this question has less to do about the proof itself and more to do about whether my chosen method of proof is evidence enough. It can actually be shown by the Principle of Mathematical Induction that the sum of the cubes of any three consecutive positive integers is divisible by 9, but this is not what I intend to show and not what the author is asking. I believe that the PMI is not the authors intended path for the reader, hence why they asked to prove divisibility by 3. So I did a proof without using the PMI. But is it enough?
It's from Beachy-Blair's Abstract Algebra Section 1.1 Problem 21. This is not for homework, I took Abstract Algebra as an undergraduate. I was just going through some problems that I have yet to solve from the textbook for pleasure.
Question: Prove that the sum of the cubes of any three consecutive positive integers is divisible by 3.
So here's my proof:
Let a $\in$ $\mathbb{Z}^+$
Define \begin{equation} S(x) = x^3 + (x+1)^3 + (x+2)^3 \end{equation}
So,
\begin{equation}S(a) = a^3 + (a+1)^3 + (a+2)^3\end{equation}
\begin{equation}S(a) = a^3 + (a^3 + 3a^2 + 3a + 1) + (a^3 +6a^2 + 12a +8) \end{equation}
\begin{equation}S(a) = 3a^3 + 9a^2 + 15a + 9 \end{equation}
\begin{equation}S(a) = 3(a^3 + 3a^2 + 5a + 3) \end{equation}
Hence, $3 \mid S(a)$.
QED
|
Your approach and deduction is absolutely fine
Just for the sake of completeness, here is an alternative prove
Known
$$(a+b+c)^3 = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 a^{2} c + 3 a b^{2} + 3 a c^{2} + 3 b^{2} c + 3 b c^{2} + 6 a b c $$
$$ = a^{3} + b^{3} + c^{3} + 3f(a,b,c)$$
Let $a=k-1,b=k,c=k+1$,then, $a,b,c$ represents consecutive numbers
So we have
$$(a+b+c)^3=(k-1+k+k+1)^3=(3k)^3=27k^3$$
Thus
$$a^{3} + b^{3} + c^{3} = (a+b+c)^3 - 3\cdot f(a,b,c) = 27k^3-3\cdot f(a,b,c)$$
$$\Rightarrow a^{3} + b^{3} + c^{3}=3\cdot(9k^3-f(a,b,c))$$
|
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|
Equation of the line passing through the origin and parallel to the planes $x+y+z=-1$ and $x-y+z=1$
Find a vector equation of the line that passes through the origin and is parallel to the planes $x+y+z=-1, x-y+z=1$
Is the answer $2x-2z=0$? I took the normals of the two planes which are $(1, 1, 1)$ and $(1, -1, 1)$ and used the cross-product to get the normal of the new plane, which is $(2, 0, -2)$. Since the line passes the origin, I would get $2x-2z=0$. Is this the right approach?
|
Your analysis is correct.
Here is a different way to describe the line parallel to the intersection of the two planes. Just row reduce the corresponding homogeneous system of equations. (These planes are parallel to your given ones but passing through the origin.)
$$
\left\{
\begin{align}
x + y + z &= 0 \\
x - y + z &= 0
\end{align}
\right.
$$
As matrices,
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & -1 & 1
\end{bmatrix}
\leadsto
\begin{bmatrix}
1 & 1 & 1 \\
0 & -2 & 0
\end{bmatrix}
\leadsto
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix}.
$$
We have
$$
\left\{
\begin{align}
x + z &= 0 \\
y &= 0
\end{align}
\right.,
$$
which is (by renaming the free variable $z = t$) equivalent to
$$
\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -t \\ 0 \\ t \end{bmatrix} = t \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}.
$$
|
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|
Proving $\binom{2n}{n}\le 4^n$ for all $n$ by smallest counterexample
Prove $$\binom{2n}{n}\le 4^n$$ for all natural numbers $n$ by smallest (minimal) counterexample.
My attempt:
First, $$\binom{2n}n = \frac{(2n)!}{(n!)^2} \le 4^n\;.$$ We know that $x\ne 0$ because $\frac{(2\cdot 0)!}{(0!)^2} = 1$ which is true. So $x\ge 2$. Now consider $x-1\in \Bbb N$. Also note that $x-1 <x$ and is the smallest counterexample. So, $n=x-1$.
$$\frac{(2(x-1))!}{((x-1)!)^2} \le 4^{x-1}$$
$$\frac{(2x-2)!}{((x-1)!)^2} \le 4^{x-1}$$
So this is where I'm stuck. Do I keep on expanding? If so, how?
|
HINT: In order to complete your proof, you need to show that the inequality $$\frac{(2x-2)!}{((x-1)!)^2} \le 4^{x-1}\tag{1}$$ implies that $$\frac{(2x)!}{x!^2}\le 4^x\;.$$
Now $$\frac{(2x)!}{x!^2}=\frac{2x(2x-1)(2x-2)!}{x^2(x-1)!^2}=\frac{2x(2x-1)}{x^2}\cdot\frac{(2x-2)!}{(x-1)!^2}\le\frac{2x(2x-1)}{x^2}\cdot4^{x-1}$$ by virtue of the hypothesis $(1)$. You’ll be done if you can show that
$$\frac{2x(2x-1)}{x^2}\le 4\;.$$
Can you finish it now.
|
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|
Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables
$x=x $
$y=y$
$ z= 2 \cdot(4-(x^{2} + y^{2}))^{1/2}$
$Det(u,v)= \begin{bmatrix}
i & j& k \\
1 & 0 & \frac {-x}{(4-(x^{2} + y^{2}))^{1/2}} \\
0 & 1 & \frac {-y}{(4-(x^{2} + y^{2}))^{1/2}} \\
\end{bmatrix}=(\frac {x}{(4-(x^{2} + y^{2}))^{1/2}})i + (\frac {y}{(4-(x^{2} + y^{2}))^{1/2}})j + k$
$dA=(\frac {x^{2}+y^{2}}{(4-(x^{2} + y^{2}))} +1)^{1/2}$
$\int \int_{S} (\frac {(x^{2}+y^{2})^{3}}{(4-(x^{2} + y^{2}))}+(x^{2}+y^{2})^{2})^{1/2}$
Projecting when z=1 and z=2 we have $x^{2} + y^{2}= 4-1$ $\to r= 0,(3)^{1/2}$
going to polar we have:
$(\frac {r^{6}}{(4-r^{2})}+r^{4})^{1/2}rdrd\theta=(\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$
my problem is $2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} (\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ there is no nice way i can think of to integrate this. it can also be written as:
$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$
|
Although the sine substitution probably is the easiest method, here's another one:
If the integrand is a fraction with a square root in the denominator, see if you can write the integrand as the derivative of the square root times another function to pave the way for an integration by parts:
$I = \int \frac{r^3}{\sqrt{4-r^2}}\, \mathrm{d}r = \int -r^2 \frac{-r}{\sqrt{4-r^2}}\, \mathrm{d}r $
$\: = -r^2 \sqrt{4-r^2} + 2 \int r \sqrt{4-r^2}\, \mathrm{d}r$
$\: = -r^2 \sqrt{4-r^2} - \frac{2}{3} (4-x^2)^{\frac{3}{2}}$
|
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|
Is $2^{218!} +1$ prime?
Prove that $2^{218!} +1$ is not a prime number.
I can prove that the last digit of this number is $7$, and that's all.
Thank you.
|
Hint $\ $ If $\rm\: k\:$ is odd then $\rm\:a^n\!+\!1\mid a^{nk}\!+\!1\ $ by $\rm\ mod\ a^n\!+\!1\!:\ a^n\!\equiv -1\:\Rightarrow\:a^{nk}\!\equiv (a^n)^k\!\equiv (-1)^k\equiv -1.\:$
Or $ $ Factor Theorem $\rm\:\Rightarrow\: x\!-\!c\mid x^k\!-\!c^k\: $ in $\rm\:\Bbb Z[x],\:$ so $\rm\:c=-1\:\Rightarrow\: x\!+\!1\mid x^k\!+\!1,\:$ hence evaluating at $\rm\:x = a^n\:$ yields $\rm\:a^n\!+\!1\mid a^{nk}\!+\!1\:$ in $\,\Bbb Z.\:$ Thus this integer divisibility results is a special case of a polynomial divisibility result. Factors of this form are sometimes called algebraic factors.
Similar to the example above, often number identities are more perceptively viewed as special cases of function or polynomial identities. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x)\, =\, C_n(x)^2\! - n\, x\, D_n(x)^2\;$. These play a role in factoring integers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations.
$$\begin{eqnarray}
\rm x^4 + 2^2 &=\,&\rm (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\rm \frac{x^6 + 3^3}{x^2 + 3} &=\,&\rm (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\rm \frac{x^{10} - 5^5}{x^2 - 5} &=\,&\rm (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\rm \frac{x^{12} + 6^6}{x^4 + 36} &=\,&\rm (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{eqnarray}$$
For more on this and related topics see Sam Wagstaff's introduction to the Cunningham Project.
|
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|
Finding a point in an ellipsoid I know the semi-principal axes $(x,y,z)$ of the ellipsoid $E$ (centered at the origin). Given the normalized direction vector $\vec{v}=(a,b,c)$ pointing from the origin to the surface, how can I find the factor $f$ so that $f\vec{v} \in E$?
|
The ellipsoid $E$ is the locus of points $(x, y, z) \in \mathbb{R}^3$ satisfying
$$
\frac{x^2}{r^2} + \frac{y^2}{s^2} + \frac{z^2}{t^2} = 1
$$
If $\vec{v} = (a, b, c)$ and $f\vec{v} \in E$ with $f \in \mathbb{R}$, then we can find the scalar $f$ by putting the coordinates of $f\vec{v}$ into the equation.
$$
f\vec{v} = (fa, fb, fc)
$$
Now,
$$
\begin{align}
1 &= \frac{(fa)^2}{r^2} + \frac{(fb)^2}{s^2} + \frac{(fc)^2}{t^2} \\
&= \left( \frac{a^2}{r^2} + \frac{b^2}{s^2} + \frac{c^2}{t^2} \right)f^2.
\end{align}
$$
Therefore,
$$
f = \pm \sqrt{\left( \frac{a^2}{r^2} + \frac{b^2}{s^2} + \frac{c^2}{t^2} \right)^{-1}}.
$$
|
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|
Minimum Value of expression Given that $x$, $y$ and $z$ are positive real numbers satisfying $xyz=32$, find the minimum value of:
$$x^2+4xy+4y^2+2z^2$$
Perhaps AM-GM and manipulation but I'm not quite sure how?
Source BMO.
|
Yes AM-GM is the right approach.
By AM-GM,
$$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6} \geq (16 \cdot x^4 \cdot y^4 \cdot z^4)^{\frac{1}{6}}=$$
$$=(2^{24})^{\frac{1}{6}}=16$$ with equality iff $x=z=2y$.
So the minimum value of $x^2+4xy+4y^2+2z^2$ is $16 \cdot 6=96$. QED
|
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|
Finding the solution to this specific recurrence relation What would be the solution to $a_n = 7a_{n−2} + 6a_{n−3}$ with $a_0 = 9$,
$a_1 = 10$, and $a_2 = 32$
I can find it for a specific value of (n), but not for just a general solution. Thanks!
|
Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write the recurrence as:
$$
a_{n + 3} = 7 a_{n + 1} + 6 a_n \quad a_0 = 9, a_1 = 10, a_2 = 32
$$
By properties of ordinary generating funtions:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3} = \frac{A(z) - a_0}{z} + 6 A(z)
$$
Writing as partial fractions:
$$
A(z) = 4 \cdot \frac{1}{1 - 3 z}
- 3 \cdot \frac{1}{1 + 2 z}
+ 8 \cdot \frac{1}{1 + z}
$$
Expanding the various geometric series:
$$
a_n = 4 \cdot 3^n - 3 \cdot (-2)^n + 8 \cdot (-1)^n
$$
|
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Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$ Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$
I am able to find $b = \pm \sqrt{2}$ using L'Hopitals Rule, but unable to do anything for $a$.
|
Another way to do it is with Taylor expansion of the numerator,
$$-1 + ax + (1 + bx + \frac{b^2}{2}x^2 + \mbox{higher order terms}) = (a+b)x + \frac{b^2}{2} + \mbox{ h.o.t.}).$$
Now dividing by $x^2$ and sending $x \to 0$ it is clear that the higher order terms vanish, and $a = -b$ to make sure the limit exists. We are left with
$$\lim_{x\to 0} \frac{b^2}{2} \frac{x^2}{x^2} = \frac{b^2}{2}.$$
And as such $b^2 = 2$.
|
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|
Evaluating this integral $ \small\int \frac {x^2 dx} {(x\sin x+\cos x)^2} $ The question:
Compute$$
\int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2}
$$
Tried integration by parts. That didn't work.
How do I proceed?
|
$$\text{Observe that, }\frac{d(x\sin x+\cos x)}{dx}=x\cos x$$
$$ \int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2} =\int \frac x{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x)^2}dx$$
So, if $z=x\sin x+\cos x, dz=x\cos xdx$
So, $\int \frac{x\cos x}{(x\sin x+\sin x)^2}dx=\int \frac{dz}{z^2}=-\frac1z=-\frac1{x\sin x+\cos x}$
So, $$I=\frac x{\cos x}\int \frac{x\cos x}{(x\sin x+\cos x)^2}dx-\int \left(\frac{d(\frac x{\cos x})}{dx}\int \frac{x\cos x}{(x\sin x+\cos x)^2}dx\right)dx$$
$$=-\frac x{\cos x(x\sin x+\cos x)}+\int \left(\frac{x\sin x+\cos x}{\cos^2x}\right)\left(\frac1{x\sin x+\cos x} \right)dx$$
$$=-\frac x{\cos x(x\sin x+\cos x)}+\int\sec^2xdx$$
$$=-\frac x{\cos x(x\sin x+\cos x)}+\tan x+C$$ where $C$ is an arbitrary constant of indefinite integral
$$\text{Another form will be } \frac{\sin x-x\cos x}{x\sin x+\cos x}+C$$
|
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|
Evaluating Sums Algebraically or Combinatorially Consider
(1) $$\sum_{k=0}^{n}\binom{n}{k}2^{k-n}$$
(2) $$\sum_{k=0}^{n}\binom{n}{k}\frac{k!}{(n+k+1)!}$$
These sums appear too difficult (in my mind) to evaluate combinatorially. What are some good methods to attack these problems algebraically?
|
A combinatorial argument for (1) isn’t too hard. First rewrite the sum:
$$\sum_k\binom{n}k2^{k-n}=2^{-n}\sum_k\binom{n}k2^k\;.$$
Now $\binom{n}k2^k$ is the number of ways of choosing a $k$-subset $S$ of $[n]$ and then a subset $T$ of $S$, so $\sum_k\binom{n}k2^k$ is the number of ways of splitting $[n]$ into three subsets. This is clearly $3^n$, so the original expression is $\left(\frac32\right)^n$.
I don’t at the moment see a useful combinatorial interpretation of the second summation, but I can evaluate it. If
$$f(n)=\sum_{k=0}^n\binom{n}k\frac{k!}{(n+k+1)!}\;,$$
then we have the following data:
$$\begin{array}{c|r}
n&f(n)\\ \hline
0&1=\frac{2^0}{1!!}\\
1&\frac12+\frac16=\frac23=\frac{2^1}{3!!}\\
2&\frac16+\frac2{24}+\frac2{120}=\frac4{15}=\frac{2^2}{5!!}\\
3&\frac1{24}+\frac1{40}+\frac1{120}+\frac1{840}=\frac8{105}=\frac{2^3}{7!!}\\
4&\frac1{120}+\frac1{180}+\frac1{420}+\frac1{1680}+\frac1{15120}=\frac{16}{945}=\frac{2^4}{9!!}
\end{array}$$
Apparently $f(n)=\dfrac{2^n}{(2n+1)!!}=\dfrac{4^nn!}{(2n+1)!}$. Now
$$\begin{align*}
\frac{(2n+1)!}{n!}f(n)&=\frac{(2n+1)!}{n!}\sum_{k=0}^n\binom{n}k\frac{k!}{(n+k+1)!}\\\\
&=\frac1{n!}\sum_{k=0}^n\binom{n}kk!(2n+1)^{\underline{n-k}}\\\\
&=\frac1{n!}\sum_{k=0}^nn^\underline k(2n+1)^{\underline{n-k}}\\\\
&=\sum_{k=0}^n\frac{(2n+1)^{\underline{n-k}}}{(n-k)!}\\\\
&=\sum_{k=0}^n\frac{(2n+1)!}{(n-k)!(n+k+1)!}\\\\
&=\sum_{k=0}^n\binom{2n+1}{n-k}\\\\
&=\sum_{k=0}^n\binom{2n+1}k\;.
\end{align*}$$
(Here $x^{\underline k}$ is the falling factorial.) And
$$\sum_{k=0}^n\binom{2n+1}k=\sum_{k=0}^n\binom{2n+1}{2n+1-k}=\sum_{k=n+1}^{2n+1}\binom{2n+1}k\;,$$
so
$$\frac{(2n+1)!}{n!}f(n)=\frac12\sum_{k=0}^{2n+1}\binom{2n+1}k=2^{2n}=4^n\;,$$
and the conjecture that $f(n)=\dfrac{2^n}{(2n+1)!!}=\dfrac{4^nn!}{(2n+1)!}$ is established.
|
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|
Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
|
$$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}}$$
Mutliply top and bottom by $\frac{1}{x^2}$
$$lim_{x \to \infty} \frac{\sqrt{x^4 + 1}}{\sqrt[3]{x^6+1}} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = lim_{x \to \infty} \frac{\sqrt{\frac{x^4}{x^4} + \frac{1}{x^4}}}{\sqrt[3]{\frac{x^6}{x^6}+\frac{1}{x^6}}}$$
$$= lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x^4}}}{\sqrt[3]{1+\frac{1}{x^6}}} = \frac{\sqrt{1}}{\sqrt[3]{1}} = 1$$
|
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|
Why does the harmonic series diverge but the p-harmonic series converge I am struggling understanding intuitively why the harmonic series diverges but the p-harmonic series converges. I know there are methods and applications to prove convergence, but I am only having trouble understanding intuitively why it is. I know I must never trust my intuition, but this is hard for me to grasp. In both cases, the terms of the series are getting smaller, hence are approaching zero, but they both result in different answers.
$$\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \text{diverges}$$ $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\text{converges}$$
|
We produce two series that are close in spirit to the series you mentioned. Perhaps the divergence of the first, and the convergence of the second, will be clearer.
Consider the series
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\cdots.$$
So there is $1$ term equal to $\frac{1}{2}$, then a block of $2$ each equal to $\frac{1}{4}$, then a block of $4$ each equal to $\frac{1}{8}$, then a block of $8$ each equal to $\frac{1}{16}$, and so on forever. Each block has sum $\frac{1}{2}$, so if you add enough terms, your sum will be very big. But it will take an awful lot of terms to add up to $1000$, many more more terms than there are atoms in the universe. Note that each term is less than the corresponding term in the harmonic series, so if you add together enough terms of the harmonic series, the sum will be very big.
Now consider the series
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\cdots.$$
Each term is $\ge$ the corresponding term in the series $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\cdots$.
Again, we find the sums of the blocks. The first block has sum $1$. The second has sum $\frac{1}{2}$. The third has sum $\frac{1}{4}$. The fourth has sum $\frac{1}{8}$, and so on forever. So if we add up "all" the terms, we get sum $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$, a familiar series with sum $2$.
|
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|
How to find $\cos A \cos B - \sin A \sin B$? Given that:
$\tan A=1$
and
$\tan B = \sqrt{3}$
How would you find $\cos A \cos B - \sin A \sin B$?
EDIT: This is what I've tried after reading bhattacharjee's answer:
$$ \tan(A+B) = \tan A+\tan B−\tan A\tan B$$
so,
$\tan(A+B)= {1+\sqrt{3} \over 1-\sqrt{3}}$
from this I get $1 \over \cos(A+B)^2 $ $=1+ \left ( {1+\sqrt{3} \over 1-\sqrt{3}}\right )^2$
=> $ 1 \over \cos^2(A+B) $ $=$ $ 8 \over 4-2 \sqrt{3}$
Is this right, because it seems like a dead end to me? How am I supposed to proceed from here?
|
since $tan A =1$, then $A = 45^0$ and since $tan B = \sqrt{3}$, then $ B = 60^0$.
By drawing triangles you can find that $sin 45^0 = 1/\sqrt{2}$ and $cos 45^0 = 1/\sqrt{2}$ and $sin 60^0 = \sqrt{3}/2$ and $cos 60^0 = 1/2$.
|
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|
Finding the complex trig integral using the method of residues for $\int_{-\pi}^{\pi}\frac{d\theta}{1 + (\sin^{2}\theta)} = {\pi}{\sqrt{2}}$ $$\int_{-\pi}^{\pi}\frac{d\theta}{1 + (\sin^{2}\theta)} = {\pi}{\sqrt{2}}$$
I can't seem to factor this question into the solution the textbook got
so $(\sin^{2}\theta)$ = ${((1/2i)(z - 1/z))^2}$
After factoring I got
$(4/i)(z)/(4z^2 - (z^2 - 2z + 1)^2)$ which doesn't come close to the final solution.
|
$$\dfrac1{1+\sin^2(\theta)} = \sum_{k=0}^{\infty}(-1)^k \sin^{2k}(\theta)$$
Now recall that
$$\int_{-\pi}^{\pi} \sin^{2k}(\theta) d \theta = \dfrac{(2k)!}{(k!)^2 \cdot 4^k} 2 \pi$$
Hence,
$$I = \int_{-\pi}^{\pi} \dfrac{d \theta}{1+\sin^2(\theta)} = \underbrace{\int_{-\pi}^{\pi}\sum_{k=0}^{\infty}(-1)^k \sin^{2k}(\theta) d \theta = \sum_{k=0}^{\infty}(-1)^k \int_{-\pi}^{\pi}\sin^{2k}(\theta) d \theta }_{(\text{Why?})}$$
Hence, we get that
$$I = 2 \pi \left(\sum_{k=0}^{\infty} \dbinom{2k}k \left(-\dfrac14\right)^k \right) \tag{$\star$}$$
Recall that
$$\sum_{k=0}^{\infty} \dbinom{2k}k (-x)^k = \dfrac1{\sqrt{1+4x}}$$
Hence, $(\star)$ becomes,
$$I = 2 \pi \times \dfrac1{\sqrt{1+4 \times \dfrac14}} = \pi \sqrt2$$
You can proceed by your method as well. Let $z=e^{i\theta}$. Note that $$1+ \sin^2(\theta) = 1 + \left(\dfrac{z-1/z}{2i} \right)^2 = 1 - \left(\dfrac{z^2-1}{2z} \right)^2$$
Hence,
$$\dfrac1{1+\sin^2(\theta)} = \dfrac{4z^2}{(2z)^2-(z^2-1)^2} = \dfrac{4z^2}{(2z+z^2-1)(2z-z^2+1)} = \dfrac{4z^2}{((z+1)^2-2)(2-(z-1)^2)}$$ Also, we have $dz = i e^{i \theta} d \theta \implies d \theta = \dfrac{dz}{iz}$. Hence, we get the integral as
$$I = \int_{-\pi}^{\pi} \dfrac{d\theta}{1+\sin^2(\theta)} = \dfrac4i \oint_{C} \dfrac{z}{((z+1)^2-2)(2-(z-1)^2)}dz$$
$$\oint_{C} \dfrac{z}{((z+1)^2-2)(2-(z-1)^2)}dz = \dfrac14 \oint_{C} \dfrac{dz}{z^2+2z-1} - \dfrac14 \oint_{C} \dfrac{dz}{z^2-2z-1}$$
$$\oint_{C} \dfrac{dz}{z^2+2z-1} = \oint_{C} \dfrac{dz}{(z+1)^2-2} = \oint_{C} \dfrac{dz}{(z+1+\sqrt2)(z+1-\sqrt2)} = \dfrac{2\pi i}{2\sqrt2} = \dfrac{\pi i }{\sqrt2}$$
$$\oint_{C} \dfrac{dz}{z^2-2z-1} = \oint_{C} \dfrac{dz}{(z-1)^2-2} = \oint_{C} \dfrac{dz}{(z-1+\sqrt2)(z-1-\sqrt2)} = \dfrac{2\pi i}{-2\sqrt2} = -\dfrac{\pi i }{\sqrt2}$$
Hence, we get that
$$I = \dfrac4i \times \dfrac14 \times \left(\dfrac{\pi i }{\sqrt2} + \dfrac{\pi i }{\sqrt2}\right) = \pi \sqrt2$$
|
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|
Derivation of the general forms of partial fractions I'm learning about partial fractions, and I've been told of 3 types or "forms" that they can take
(1) If the denominator of the fraction has linear factors:
$${5 \over {(x - 2)(x + 3)}} \equiv {A \over {x - 2}} + {B \over {x + 3}}$$
(2) If the denominator of the fraction has quadratic factors that don't factorise:
$${{2x + 3} \over {(x - 1)({x^2} + 4)}} \equiv {A \over {x - 1}} + {{Bx + C} \over {{x^2} + 4}}$$
(3) If the denominator has a factor that repeats:
$${{5x + 3} \over {(x - 2){{(x + 3)}^2}}} \equiv {A \over {x - 2}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}$$
I'd appreciate it if someone could explain to me how these equivalence relationships are derived, essentially how do they take the forms that they do? I sort of understand the first one, but the other two I don't.
Thank you
|
To look at your third problem, consider the following example: $${{x^2 + 4} \over {(x)(x + 2){{(3x-2)}}}} = {A \over {x}} + {B \over {x + 2}} + {C \over {{{(3x-2)}}}}$$
Next, set the numerators equal. $$x^2+4 = A(x+2)(3x-2) + Bx(3x-2) + Cx(x+2)$$ Now pick few x values and find the constants. For $x=0,$ $4=A(2)(-2).$ So $A=-1$. Following the same procedure for $x=-2$ and $x=2/3,$ we find that $B = 1/2$ and $C = 5/2.$ Substituting our values give us $${{x^2 + 4} \over {(x)(x + 2){{(3x-2)}}}} = {1 \over {x}} + {1/2 \over {x + 2}} + {5/2 \over {{{(3x-2)}}}}$$
This is not an answer to your question but to show you how its usually done.
|
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|
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:
\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}
And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.
|
Hint: Another method to approach the problem (more by brute force than Andre's clever technique) is $$f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)(x^2+1)}.$$ Now use partial fraction decomposition and geometric series.
|
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|
What's the Maclaurin Series of $f(x)=\frac{1}{(1-x)^2}$? This function seemed to be pretty much straight forward, but my solution is incorrect.
I have two questions:
1. Where did I make a mistake?
2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?
$$f(x)=\frac{1}{(1-x)^2} \qquad f(0)=1$$
$$f'(x)=\frac{2}{(1-x)^3} \qquad f'(0)=2$$
$$f''(x)=\frac{6}{(1-x)^4} \qquad f''(0)=6$$
$$f'''(x)=\frac{24}{(1-x)^5} \qquad f'''(0)=24$$
$$f''''(x)=\frac{120}{(1-x)^6} \qquad f''''(0)=120$$
The series is then
$$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$
and when simplified is
$$1+x+x^2+x^3+x^4$$
But the correct answer is
$$1+2x+3x^2+4x^3+5x^4 $$
|
Recall that
$$f(x) = f(0) + \dfrac{f'(0)}{1!} x + \dfrac{f''(0)}{2!}x^2 + \cdots$$
Your derivative computation is correct, i.e., $f^{(n)}(0) = (n+1)!$. But you have them matched to the wrong denominators. Hence, you will get
$$\dfrac1{(1-x)^2} = 1 + \dfrac{2!}{1!}x + \dfrac{3!}{2!}x^2 + \dfrac{4!}{3!}x^3 + \cdots = 1 + 2x + 3x^2 + 4x^3 + \cdots$$
|
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|
Finite difference implicit schema for wave equation 1D not unconditionally stable? The wave equation 1D with constant density is defined as:
\begin{equation}
\frac{\partial^2 U}{\partial t ^2} = V^2 \frac{\partial^2 U}{\partial x ^2}
\label{eqa}
\end{equation}
And the implicit difference schema:
\begin{equation}
U_j^{n+1} = \left( \frac{\Delta t V_j^n}{\Delta s} \right) ^2 \left( U_{j+1}^{n+1} - 2 U_j^{n+1} + U_{j-1}^{n+1} \right) + 2 U_j^n - U_j^{n-1}
\label{eq:b}
\end{equation}
Where $\Delta x = \Delta s $.
Von Newman stability analysis.
Since $ u(x,t) = e^{i(wt+kx)} $ is a solution for the discrete schema above can be written as:
\begin{eqnarray}
u(x_j,t_n) &=& e^{i(wt_n+kx_j)} \nonumber \\
&=& e^{iwn\Delta t} e^{ikj\Delta s} \nonumber \\
&=& \epsilon^n e^{ikj\Delta s}
\label{eq:c}
\end{eqnarray}
Where $\epsilon = e^{iw \Delta t} $ and $ i = \sqrt{-1} $ .
To maintain stability we should make sure that $ \epsilon \leq 1 $ not growing exponentially with increasing time steps. So applying the previous equation in the implicit differences schema we can analyse the growth.
\begin{multline}
U_j^{n+1} = \left( \frac{\Delta t V_j^n}{\Delta s} \right) ^2 \left( U_{j+1}^{n+1} - 2 U_j^{n+1} + U_{j-1}^{n+1} \right) + 2 U_j^n - U_j^{n-1} \\
\epsilon^{n+1} e^{ikj\Delta s} = r^2 \left( \epsilon^{n+1} e^{ik(j+1)\Delta s} - 2 \epsilon^{n+1} e^{ikj\Delta s} + \epsilon^{n+1} e^{ik(j-1)\Delta s} \right) \\
+ 2 \epsilon^n e^{ikj\Delta s} - \epsilon^{n-1} e^{ikj\Delta s} \\
1 = r^2 (e^{ik\Delta s} -2 + e^{-ik\Delta s}) + 2 \epsilon^{-1} - \epsilon^{-2} \\
\epsilon^{-2} - 2 \epsilon^{-1} - r^2 (e^{ik\Delta s} -2 + e^{-ik\Delta s}) + 1 = 0 \\
\epsilon^{-2} - 2 \epsilon^{-1} - 4 r^2 \sin^2 \left( k \Delta s / 2 \right) + 1 = 0
\label{eq:d}
\end{multline}
Where $ r = \left( \frac{\Delta t V_j^n}{\Delta s} \right) $
At the previous equation we can substitute $ \phi = \epsilon^{-1} $ turning it to a second degree bellow:
\begin{eqnarray}
\phi^2 - 2 \phi -4 r^2 \sin^2 \left(k\Delta s/2 \right) +1 = 0 \nonumber \\
\phi^2 - 2 \phi + c = 0 \nonumber
\end{eqnarray}
With:
$$ c = 1 -4 r^2 \sin^2 \left(k\Delta s/2 \right) $$
Has roots $ \phi^{'} $ and $ \phi^{''} $ as bellow. Replacing $c$ also.
\begin{eqnarray}
&=& \frac{2 \pm \sqrt{4 - 4c}}{2} = 1 \pm \sqrt{1-c} \nonumber \\
&=& 1 \pm \sqrt{4 r^2 \sin^2 \left(k\Delta s/2 \right) } \nonumber \\
&=& 1 \pm 2 r \sin \left(k\Delta s/2 \right) \nonumber \\
\phi^{'} &=& 1 + 2 r \sin \left(k\Delta s/2 \right) \\
\phi^{''} &=& 1 - 2 r \sin \left(k\Delta s/2 \right)
\end{eqnarray}
Going back to $\epsilon$ we get :
\begin{eqnarray}
\epsilon^{'} &=& \frac{1}{1 + 2 r \sin \left(k\Delta s/2 \right)} \\
\epsilon^{''} &=& \frac{1}{1 - 2 r \sin \left(k\Delta s/2 \right) }
\end{eqnarray}
But with $r \sin \left(k\Delta s/2 \right) \to 0$ we get that $\epsilon^{''} \to \infty $
Should it not be unconditionally stable? What is wrong in here?
Update: After awesome answer by Jitse, fixing the signal we get:
\begin{eqnarray}
\phi^{'} &=& 1 + i 2 r \sin \left(k\Delta s/2 \right) \\
\phi^{''} &=& 1 - i 2 r \sin \left(k\Delta s/2 \right)
\end{eqnarray}
Analyzing the modulus, since $ \phi^{'} $ and $ \phi^{''} $ are conjugate pairs of the same complex number the modulus is the same for both.
We get:
$$ \| \phi^{'} \| = \| \phi^{''} \| = \sqrt{1+ 4 r^2 \sin ^2 \left(k\Delta s/2 \right)}$$
So since $ \| \phi \| \geq 1 \ \ \forall \ r, \ k,\ \Delta s $ than
$$ \| \epsilon \| = \frac{1}{ \sqrt{1 + 4 r ^2 \sin ^2 \left(k\Delta s/2 \right)}} $$
$$ \| \epsilon \| \leq 1 \ \ \forall \ r, \ k,\ \Delta s$$
Awesome, unconditionally stable!!
|
Your method for Neumann stability analysis is correct, and indeed I would expect the scheme to be unconditionally stable.
I do not understand the step where you introduce the sine. I think that
$$ e^{ix} - 2 + e^{-ix} = -4\sin^2(\tfrac12x). $$
The minus sign seems to have been lost in your computation.
|
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|
How to derive the Golden mean by using properties of Gamma function? The Golden mean known as $\frac{1+\sqrt{5}}{2}$.
How could one show the Golden mean can be expressed as
$$
\frac{2\cdot 3\cdot 7\cdot 8\cdot 12\cdot 13\cdots}{1\cdot 4\cdot 6\cdot 9\cdot 11\cdot 14\cdots}
$$
This is the limiting case of the Rogers-Ramanujan continued fraction. But how would you prove this by using properties of Gamma function?
Any help would be appreciated!
|
Consider the combination
\begin{align}f_N=\frac{\Gamma\left(N+\frac{2}{5}\right)\Gamma\left(N+\frac{3}{5}\right)}{\Gamma\left(N+\frac{1}{5}\right)\Gamma\left(N+\frac{4}{5}\right)}=
\frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}f_{N-1}=\ldots=\\=
\frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}\times\ldots\times \frac{8\cdot 7}{9\cdot6}\times\frac{3\cdot 2}{4\cdot1}f_0.
\end{align}
But $$\displaystyle \frac{1}{f_0}=\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}=\frac{\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}}=\frac{1+\sqrt5}{2}$$
and $f_{N\rightarrow\infty}=1$ since
$$\lim_{x\rightarrow +\infty}\frac{\Gamma(x-a)\Gamma(x+a)}{\Gamma(x-b)\Gamma(x+b)}=1.$$
The infinite fraction representation follows. $\blacksquare$
|
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|
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
|
This kind of simplification occurs if and only if $p \pm q\sqrt d$ has a cube root of the form $x \pm y\sqrt d$ with rational $x,y$. So, to get all instances of this, start by choosing $x+y\sqrt d$, and cube it to get the values for $p$ and $q$.
Setting up the system $(x + y\sqrt d)^3 = p + q\sqrt d$, we get $x^3+3dxy^2 = p$ and $3x^2y+dy^3 = q$. This system has $9$ solutions in $\Bbb C^2$.
To get a particular solution, pick a cube root $a$ of $p + q\sqrt d$, a cube root $b$ of $p - \sqrt d$, and build $x = (a+b)/2, y = (a-b)/2\sqrt d$ :
Then, $8(x^3+3dxy^2) = (a+b)^3+3(a+b)(a-b)^2 = 4a^3 + 4b^3 = 8p$, and $8\sqrt d(3x^2y+dy^3) = 3(a+b)^2(a-b)+(a-b)^3 = 4a^3-4b^3 = 8q\sqrt d$, which proves that those $9$ couples are solution.
To show that it has only $9$ solution, we can show $x$ is a root of a degree $9$ polynomial :
squaring the second equation we get $q^2 = 9x^4y^2 + 6dx^2y^4 + d^2y^6$. Multiply by $27dx^3$ and use the first equation to get $27dx^3q^2 = 81x^6(p-x^3) + 18x^3(p-x^3)^2 + (p-x^3)^3$, which reduces to $$64x^9-48px^6+(27dq^2-15p^2)x^3-p^3 = 0$$
Finally, given a value for $x$, we can solve for $y$ : The first equation gives $y^2 = \frac{p-x^3}{3dx}$, and plugging this into the second we get $y = \frac{3qx}{8x^3+p}$. Hence the $9$ solutions I gave earlier are the only solutions to the system.
More importantly, if we can find a pair of cube roots such that $x$ is rational, then the corresponding $y$ is also rational, and what we really have found is that $p+q\sqrt d$ has a cube root in $\Bbb Q(\sqrt d)$.
Now, this makes all problems of the form "show that $\sqrt[3]{p+q\sqrt d} + \sqrt[3]{p-q\sqrt d} = 2x$" solvable immediately by computing $y= \frac{3qx}{8x^3+p}$ and then checking that $(x+y\sqrt d)^3 = p + q\sqrt d$. If this doesn't work, then the problem was wrong to begin with.
I should also point out that the degree $9$ polynomial in $x$ factors over $\Bbb Q(\sqrt{-3},\sqrt[3]{p^2-dq^2})$ into a product of $3$ cubics :
$$(4x^3 - 3\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j^2\sqrt[3]{p^2-dq^2}x - p) = 0$$ (where $j$ is a primitive cube root of $1$).
Attempting to use Cardan's formula on those will give you back the original expression $2x = \sqrt[3]{p+q\sqrt d}+\sqrt[3]{p-q\sqrt d}$.
Attempting to use Cardan's formula on the cubic for $x^3$ and then taking a cube root seems to be an even worse idea, and the moral of the story is : unlike square roots, you can't algebraically find the cube roots of $p + q\sqrt d$.
|
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|
Generating functions of partition numbers I don't understand at all why:
\begin{equation}
\sum\limits_{n=0}^\infty p_n x^n = \prod\limits_{k=1}^\infty (1-x^k)^{-1}
\end{equation}
Where $p_n$ is the number of partitions of $n$. Specifically how can the sum of positive numbers ever be factored into a fraction or factors to the $-1$ power? Is this something purely notational or am I missing something big?
Thanks for all the help!
|
We have $\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}+\cdots$. These are combined formally in a process known as generating functions.
More details, as requested. We multiply out $(1+x+x^2+x^3+\cdots)(1+x^2+x^4+x^6+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^4+x^8+\cdots)\cdots$, and gather the powers of $x$, as if they were polynomials. For example, $x^3$ has a coefficient of $3$ -- one term from $x^3\cdot 1\cdot 1\cdots$, one term from $x^1\cdot x^2\cdot 1\cdots$, and one term from $1\cdot 1\cdot x^3\cdots$. The first corresponds to $3=1+1+1$ (three 1's), the second corresponds to $3=2+1$ (one 2 and one 1), and the third corresponds to $3=3$ (one 3).
|
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|
Evaluating $\int_0^{\infty} \text{sinc}^m(x) dx$ How do I evaluate $$I_m = \displaystyle \int_0^{\infty} \text{sinc}^m(x) dx,$$ where $m \in \mathbb{Z}^+$?
For $m=1$ and $m=2$, we have the well-known result that this equals $\dfrac{\pi}2$. In general, WolframAlpha suggests that is seems to be a rational multiple of $\pi$.
\begin{array}{c|c|c|c|c|c|c|c}
m & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
\hline
I_m & \dfrac{\pi}2 & \dfrac{\pi}2 & \dfrac{3\pi}8 & \dfrac{\pi}3 & \dfrac{115\pi}{384} & \dfrac{11\pi}{40} & \dfrac{5887 \pi}{23040}\\
\end{array}
$(1)$. Can we prove that $I_m$ is a rational multiple of $\pi$ always?
$(2)$. If so, is there a nice formula, i.e., if $I_m = \dfrac{p(m)}{q(m)} \pi$, where $p(m),q(m) \in \mathbb{Z}^+$, are there nice expressions for $p(m)$ and $q(m)$?
P.S: This integral came up when I was trying my method to answer this question, by writing $\dfrac{\sin(x)}{x+\sin(x)}$ as $$\dfrac{\sin(x)}{x+\sin(x)} = \text{sinc}(x) \cdot \dfrac1{1+\text{sinc}(x)} = \sum_{k=0}^{\infty} (-1)^k \text{sinc}^{k+1}(x)$$
|
Notice $\lim_{x\to 0} \frac{\sin x}{x}$ is bounded at $x = 0$,
$$\begin{align}\int_0^{\infty} \left(\frac{\sin x}{x}\right)^m dx
&= \frac12 \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^m dx\tag{*1}\\
&= \lim_{\epsilon\to 0} \frac12 \left(\frac{1}{2i}\right)^m \oint_{C_{\epsilon}} \left(\frac{e^{ix} - e^{-ix}}{x}\right)^m dx\tag{*2}
\end{align}$$
We can evaluate the integral $(*1)$ as a limit of a integral over a deformed
contour $C_{\epsilon}$ which has a little half-circle of radius $\epsilon$ at origin:
$$C_{\epsilon} = (-\infty,-\epsilon) \cup \left\{ \epsilon e^{i\theta} : \theta \in [\pi,2\pi] \right\} \cup ( +\epsilon, +\infty)$$
We then split the integrand in $(*2)$ in two pieces, those contains exponential factors $e^{ikx}$ for $k \ge 0$ and those for $k < 0$.
$$(*2) = \lim_{\epsilon\to 0} \frac12 \left(\frac{1}{2i}\right)^m \oint_{C_{\epsilon}} \left( \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} + \sum_{k=\lfloor\frac{m}{2}\rfloor+1} ^{m} \right) \binom{m}{k} \frac{(-1)^k e^{i(m-2k)x}}{x^m} dx$$
To evaluate the $1^{st}$ piece, we need to complete the contour in upper half-plane. Since the completed contour contains the pole at $0$, we get:
$$\begin{align}
\sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \text{ in }(*2)
&= \frac12 \left(\frac{1}{2i}\right)^m (2\pi i)\sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \binom{m}{k} \frac{(-1)^k i^{m-1}(m-2k)^{m-1}}{(m-1)!}\\ &= \frac{\pi m}{2^m} \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \frac{(-1)^k (m-2k)^{m-1}}{k!(m-k)!}\tag{*3}\end{align}$$
To evaluate the $2^{nd}$ piece, we need to complete the contour in lower half-plane instead. Since the completed contour no longer contains any pole, it contributes nothing and hence $I_m$ is just equal to R.H.S of $(*3)$.
Update
About the question whether $I_m$ is decreasing. Aside from the exception $I_1 = I_2$, it is strictly decreasing.
For $m \ge 1$, it is clear $I_{2m} > I_{2m+1}$ because the difference of corresponding integrands is non-negative and not identically zero. For the remaining cases, we have:
$$\begin{align}&I_{2m+1}-I_{2m+2}\\
= & \int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^{2m+1}\left(1 - \frac{\sin x}{x}\right) dx\\
= & \left(\sum_{n=0}^{\infty} \int_{2n\pi}^{(2n+1)\pi}\right) \left(\frac{\sin x}{x}\right)^{2m+1}\left[1 - \frac{\sin x}{x} - \left(\frac{x}{x+\pi}\right)^{2m+1}\left(1 + \frac{\sin x}{x + \pi}\right)\right] dx
\end{align}$$
Over the range $\cup_{n=0}^{\infty} (2n\pi,(2n+1)\pi)$, the factor $\left(\frac{\sin x}{x}\right)^{2m+1}$ is positive. The other factor $\Big[\cdots\Big]$ in above integral is bounded below by:
$$
\begin{cases}
1 - \frac{\sin x}{x} - \left(\frac{x}{x+\pi}\right)^3\left(1 + \frac{\sin x}{x + \pi}\right), & \text{ for } x \in (0,\pi)\\
1 - \frac{1}{x} - \frac{x}{x+\pi}\left(1 + \frac{1}{x}\right)
= \frac{(\pi - 2)x - \pi}{x(x+\pi)} & \text{ for } x \in \cup_{n=1}^{\infty}(2n\pi,(2n+1)\pi)
\end{cases}
$$
A simple plot will convince you both bounds are positive in corresponding range. This
implies the integrand in above integral is positive and hence $I_{2m+1} > I_{2m+2}$.
|
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|
Evaluate $ \lim\limits_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) } dt$
$$ \lim_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) dt}$$
This is what I've tried:
Let $F(x) = \displaystyle\int_0^x{\sin(t^2) dt}$, and let $f(x) = {\sin(t^2)}$.
Then $F'(x) = f(x) \Rightarrow F'(0) = f(0) = \sin 0 = 0$.
Where do I go from here?
Thanks for help.
|
Using L'Hospital's rule once, we have
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x}.
$$
We examine the limit
$$
\lim_{x\rightarrow 0} \frac{\sin x^2}{\sin^2x} = \lim_{x\rightarrow 0} \frac{2x\cos x^2}{2\sin x \cos x} =\lim_{x\rightarrow 0} \frac{x\cos x}{\sin x} = 1
$$
using the fact that $\lim_{x\rightarrow 0} \frac{x}{\sin x} = 1$. Thus,
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x} = \lim_{x\rightarrow 0} \frac{1}{3\cos x} = \frac{1}{3}.
$$
|
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|
Calculus Reduction Formula
For any integer $k > 0$, show the reduction formula
$$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
for some constant $C_{k}$.
(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?
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I like to find the reduction formula for the integral by rationalization followed by integration by parts.
$$
\begin{aligned}
I_{k} &=\int_{-2}^{2} x^{2 k} \sqrt{4-x^{2}} d x \\
&=\int_{-2}^{2} \frac{x^{2 k}\left(4-x^{2}\right)}{\sqrt{4-x^{2}}} d x \\
&=-\int_{-2}^{2} x^{2 k-1}\left(4-x^{2}\right) d\left(\sqrt{4-x^{2}}\right) \\
&=-\left[x^{2 k-1}\left(4-x^{2}\right)^{\frac{3}{2}}\right]_{-2}^{2}+\int_{-2}^{2}\left[4(2 k-1) x^{2 k-2}-(2 k+1) x^{2 k} \right] \sqrt{4-x^2}dx\\
&=4(2 k-1) I_{k-1}-(2 k+1) I_{k} \\\\
I_{k} &=\frac{2(2 k-1)}{k+1} I_{k-1}
\end{aligned} \tag*{}
\\$$
Hence $C_k= \dfrac{2(2 k-1)}{k+1} .$
|
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|
How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
|
*
*$RHS=2^x+2^{-x}=2^x+\frac{1}{2^x}\geq 2$ by $AM\geq GM$, since $2^x>0$ for all $x\in\mathbb{R}$
*$LHS\leq 2$ for all $x\in\mathbb{R}$. This is obvious.
*One can verify $x=0$ is a solution.
*$f(x)=2^x+2^{-x}$ implies $f'(x)>0$ for $x>0$.
These should suffice to conclude the unique solution at $x=0$.
|
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|
Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square. Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.
What I have done:
This has to either be done with contradiction or contraposition, I was thinking contradiction more likely.
|
$$a,b,c=\text{odd}\quad\iff\quad a=2A+1\quad;\quad b=2B+1\quad;\quad c=2C+1$$ $$\Delta=b^2-4ac=n^2\quad\iff\quad(2B+1)^2-4\,(2A+1)(2C+1)=n^2$$ $$(4B^2+4B+1)-4\,(4AC+2A+2C+1)=n^2$$ $$4\,\Big[(B^2+B)-(4AC+2A+2C+1)\Big]+1=n^2$$ $$\iff n=2k+1\iff n^2=4k^2+4k+1\iff$$ $$\underbrace{\underbrace{B(B+1)}_{even}-\underbrace{2\,(2AC+A+C)}_{even}-1}_{odd}=\underbrace{k(k+1)}_{even}$$ Contradiction !
|
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|
Inverse of a symmetric tridiagonal filter matrix How to get the inverse of this matrix:
$\left(\begin{array}{ccccccc}
2&-1\\-1&2&-1\\&-1&2&-1\\&&&\ddots\\&&&&\ddots\\&&&&-1&2&-1\\&&&&&-1&2
\end{array}\right)$
where the blank elements are all zeros. Thank you.
|
Lets do an example using your matrix with dimension $4x4$, symmetric, tridiagonal.
We have the matrix:
$$\displaystyle A = \begin{bmatrix}2&-1&0&0\\-1&2&-1&0\\0&-1&2&-1\\0&0&-1&2\end{bmatrix}.$$
The inverse of this matrix:
$$\displaystyle A^{-1} = \frac{1}{5}\begin{bmatrix}4 & 3 & 2 & 1\\3 & 6 & 4 & 2\\2 & 4 & 6 & 3\\1 & 2 & 3 &4\\\end{bmatrix}.$$
Look at the structure of the "first and fourth" and "second and third" rows.
If you are looking for the eigenvalues and eigenvectors, we have:
$$\displaystyle \lambda_1 = \frac{1}{2} (5+\sqrt{5}), ~v_1 = (-1, \frac{1}{2} (1+\sqrt{5}), \frac{1}{2} (-1-\sqrt{5}), 1)$$
$$\displaystyle \lambda_2 = \frac{1}{2} (3+\sqrt{5}), ~v_2 = (1, \frac{1}{2} (1-\sqrt{5}), \frac{1}{2} (1-\sqrt{5}), 1)$$
$$\displaystyle \lambda_3 = \frac{1}{2} (5-\sqrt{5}), ~v_3 = (-1, \frac{1}{2} (1-\sqrt{5}), \frac{1}{2} (-1+\sqrt{5}), 1)$$
$$\displaystyle \lambda_4 = \frac{1}{2} (3-\sqrt{5}), ~v_4 = (1, \frac{1}{2} (1+\sqrt{5}) \frac{1}{2} (1+\sqrt{5}), 1)$$
That should help you follow the Wiki algorithm and @joriki provided the algorithm for using the normalized eigenvector and eigenvalue as another approach.
|
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|
Integral Of $\int\sqrt{\frac{x}{x+1}}dx$ I want to solve this integral
$$\int\sqrt{\frac{x}{x+1}}dx$$
And think about:
1) $t=\frac{x}{x+1}$
2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$
Now I need your advice! Thanks!
|
Let $$\sqrt{\dfrac{x}{x+1}} = t \implies \dfrac{x}{x+1} = t^2 \implies x+1 = \dfrac1{1-t^2} \implies dx = \dfrac{2tdt}{(t^2-1)^2}$$
Hence,
$$\int \sqrt{\dfrac{x}{x+1}} dx = \int \dfrac{2t^2}{(t^2-1)^2}dt$$
I trust you can take it from here using partial fractions.
$$\dfrac{2t^2}{(t^2-1)^2} = \dfrac12 \left(\dfrac1{(t-1)^2} + \dfrac1{(t-1)} + \dfrac1{(t+1)^2} - \dfrac1{(t+1)}\right)$$
|
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|
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists? Here is my candidate for
the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists.
I would be interested in seeing others.
$***$ Added after some comments:
I prove here by very elementary means that the limit exists.
Calling the limit "$e$" names it.
$***$
It only needs Bernoulli's inequality (BI)
in the form
$(1+x)^n \ge 1+nx$
for $x > -1$ and
$n$ a positive integer,
with equality only if
$x = 0$ or $n = 1$.
This is easily proved by induction:
It is true for $n=1$,
and $(1+x)^{n+1}
= (1+x)(1+x)^n
\ge (1+x)(1+nx)
= 1+(n+1)x+nx^2
\ge 1+(n+1)x
$.
(If $-1 < x < 0$,
if $1+nx \ge 0$,
the above proof goes through, and if
$1+nx < 0$,
$1+mx < 0$ for all $m \ge n$
so certainly $(1+x)^m > 1+mx$.)
This proof originally appeared in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563
and uses the
arithmetic-geometric mean inequality (AGMI)
in the form
$\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$
(all $v_i$ positive) with equality if and only if all the $v_i$ are equal.
Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.
We will prove that $a_n$ is an increasing sequence
and $b_n$ is an decreasing sequence.
Since $a_n < b_n$, this implies,
for any positive integers $n$ and $m$ with $m < n$
that $a_m < a_n < b_n < b_m$.
For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.
Since their sum is $n+2$
and their product is $(1+1/n)^n$,
by the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n$,
or $(1+1/(n+1))^{n+1} > (1+1/n)^n$,
or $a_{n+1} > a_n$.
For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.
Since their sum is $n$
and their product is $(1-1/n)^n$,
by the AGMI,
$(n/(n+1))^{n+1} > (1-1/n)^n$
or $(1+1/n)^{n+1} < (1+1/(n-1))^n$,
or $b_n > b_{n+1}$.
Since $b_n-a_n
= (1+1/n)^{n+1} - (1+1/n)^n
= (1+1/n)^n(1/n)
=a_n/n
$
and every $a_n$ is less than any $b_n$
and $b_3 = (1+1/3)^4 = 256/81 < 4$,
$b_n-a_n < 4/n$,
so $b_n$ and $a_n$ converge to a common limit.
These proofs do not seem to be really elementary,
since they use the AGMI.
However, they use a special form of the AGMI,
where all but one of the values are the same,
and this will now be shown to be implied by BI,
and thus be truly elementary.
Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.
The AGMI for these values is
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
with equality if and only if $u = v$.
We will now show that this is implied by BI:
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
is the same as
$(u+(v-u)/n)^n \ge u^n(v/u)$.
Dividing by $u^n$,
this is equivalent to
$(1+(v/u-1)/n)^n \ge v/u$.
By BI,
since $(v/u-1)/n > -1/n > -1$,
$(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$
with equality only if $n=1$ or $v/u-1 = 0$.
Thus BI implies this version of the AGMI.
|
We can prove directly from BI at the expense of some algebra.
$(1+(v/u-1)/(n+1))^{n+1} > v/u$ with $u=1+1/n$ and $v=1$ is
$$\frac{1}{1+\frac{1}{n}} < \left(1+\frac{\frac{1}{1+\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right)^{n+1}$$
which gives $a_{n+1} >a_n$ in Marty's notation. With $u=1-1/n$ and $v=1$
$$\frac{1}{1-\frac{1}{n}} < \left(1+\frac{\frac{1}{1-\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n-1}}{1+\frac{1}{n}} \right)^{n+1}$$
Since $(1-\frac{1}{n})(1+\frac{1}{n-1}) = 1$
this gives $b_n >b_{n+1}$.
|
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|
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$
(I'm not sure if that's right)
and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)
I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$
but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)
|
Let's write your recurrence relation for $n$ and $n+1$:
$a_{n}-6a_{n-1}+5a_{n-2}-1=0$
$a_{n+1}-6a_{n }+5a_{n-1}-1=0$
Now we subtract one from another:
$a_{n+1}-7a_{n }+11a_{n-1}-5a_{n-2}=0$ (relation 2)
Then, from Theorem on wiki we build a characteristic polynomial
$x^3-7x^2+11x-5 $ with roots $1,1,5$. Hence, by that theorem, the solution of our recurrence relation (2) is $c_1 1^n+ c_2n1^n+c_3 5^n$. All you have to do now is to find those constants $c_i$ from initial conditions and from the fact that it should satisfy your initial recurrence relation.
|
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|
Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$ I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots
$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$
|
Using middle term factor, $$(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0$$
$$\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0$$
$$\implies (x^2-4)(x^2-1)=0$$
$\implies x^2-1=0$ or $x^2-4=0$
Alternatively, using quadratic equation formula for $x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9$ or $6$
|
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|
Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} 1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{pmatrix};
A_5 = \begin{pmatrix} 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2\\
1& 2& 3& 3& 3\\
1& 2& 3& 4& 4\\
1& 2& 3& 4& 5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix} 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4\\
1& 2& 3& 4& 5& 5\\
1& 2& 3& 4& 5& 6
\end{pmatrix};
A_7 = \begin{pmatrix} 1& 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4& 4\\
1& 2& 3& 4& 5& 5& 5\\
1& 2& 3& 4& 5& 6& 6\\
1& 2& 3& 4& 5& 6& 7
\end{pmatrix}
$$
|
$$A_2 = \begin{bmatrix}1 & 1\\ 1 & 2 \end{bmatrix} = \begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$$
$$A_3 = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\1 & 2 & 3 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}$$
$$A_4 = \begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{bmatrix}$$
Can you see the pattern here? Prove this is the case in general.
Use this, along with the fact that $\det(XY) = \det(X) \cdot \det(Y)$ and determinant of a triangular matrix is the product of the diagonal entries to conclude the answer.
EDIT
Note that $$A_{ij} = \min(i,j) = \sum_{k=1}^{\min(i,j)} 1 = \sum_{k=1}^{n} \mathbb{I}_{k \leq i}\mathbb{I}_{k \leq j} = \sum_{k=1}^n B_{ik} C_{kj}$$
where $B_{ik} = \begin{cases} 1 & k \leq i\\ 0 & \text{else}\end{cases}$ and $C_{kj} = \begin{cases} 1 & k \leq j\\ 0 & \text{else}\end{cases}$.
This should enable you to get the decomposition.
|
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|
Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$ I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.
Provided: $$a+b+c=0$$
Find the value of: $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$
Like I'm not sure where to start, and the provided clue doesn't even make sense. There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.
Thanks!
|
$$ −a^2b−b^2c−c^2a+a^2c+b^2a+c^2b
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\\
\overset{\tiny{+abc-abc}}{=}
\Big[ abc+bc^2-b^2c-ac^2 \Big]
+
\Big[ -a^2b-abc+ab^2+a^2c \Big]
\\
=c(ab+bc-b^2-ac)-a(ab+bc-b^2-ac)
\\
=(c-a)(ab+bc-b^2-ac)
\\
=(c-a)(a-b)(b-c)
. $$
|
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|
Details about a Recurrence Relation problem. I am trying to understand Recurrence Relations through the Towers of Hanoi example, and I am having trouble understanding the last step:
If $H_n$ is the number of moves it takes for n rings to be moved from the first peg to the second peg, then:
$H_n = 2H_{n−1} + 1$
$H_n = 2(2H_{n−2} + 1) + 1 = 2^2H_{n−2} + 2 + 1$
$H_n = 2^2(2H_{n−3} + 1) + 2 + 1 = 2^3H_{n−3} + 2^2 + 2 + 1 $
$\vdots$
$H_n = 2^{n−1}H_1 + 2^{n−2} + 2^{n−3} +· · ·+2 + 1$
$H_n = 2^{n−1} + 2^{n−2} +· · ·+2 + 1$
$H_n = 2^n − 1.$
I am having trouble understanding the last two steps. How does one go from this:
$H_n = 2^{n−1} + 2^{n−2} +· · ·+2 + 1$
To this:
$H_n = 2^n − 1$
?
|
The last few lines contain errors. The line
$$H_n=2^{n-1}H_1+2^{n-2}+2^{n-3}+\ldots+2+1$$
is correct, but the next line is not: since $H_1=1$, it should be
$$H_n=2^{n-1}+2^{n-2}+2^{n-3}+\ldots+2+1\;.$$
Apparently the exponents got dropped down onto the main line of text. This is now the sum of a geometric series:
$$H_n=\sum_{k=0}^{n-1}2^k=\frac{2^n-1}{2-1}=2^n-1\;,$$
and again it appears that an exponent was accidentally dropped down to the main line.
|
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|
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$ I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$
Is it OK how I want to do?
$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$
I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?
I thinks is not because if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.
Thanks :)
|
Here's another approach, using the OP's trick of getting the square roots into the denominator, but with a different trig identity, in this case $\sin(a+b)=\sin a\cos b+\cos a\sin b$.
$$\begin{align}
|\sin(\sqrt{x+1})-\sin\sqrt x|
&=|\sin(\sqrt x+\sqrt{x+1}-\sqrt x)-\sin\sqrt x|\\
&=|\sin\sqrt x ( \cos(\sqrt{x+1}-\sqrt x)-1)+\cos\sqrt x\sin(\sqrt{x-1}-\sqrt x)|\\
&\le|\cos(\sqrt{x+1}-\sqrt x)-1|+|\sin(\sqrt{x-1}-\sqrt x)|\\
&=\left|\cos\left(1\over\sqrt{x+1}+\sqrt x \right)-1\right|+\left|\sin\left(1\over\sqrt{x+1}+\sqrt x \right) \right|\\
&\to|\cos0-1|+|\sin0|=|1-1|+|0|=0
\end{align}$$
|
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|
quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$
official answer at the back of the book has only one real solution of $-4$
where did I go wrong?
|
Why do we have to reject the solution $x=2$?
Hint: What happens when we put $x=2$ in the original equation?
Review your equations. Make sure that you didn't multiply by 0.
|
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|
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
|
In polar form,
$$x = 35+18 i \sqrt{3} = \sqrt{13}^3 \cdot \exp\left( i \cdot \rm{atan} \frac{18\sqrt{3}}{35}\right)$$
so that
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \; \cos\left(\frac{1}{3} \rm{atan} \frac{18\sqrt{3}}{35} \right)$$
which magically evaluates to $7$ in a calculator.
(I know, I know: It's not appropriate to call any one value the cube root of another. Be that as it may ...)
To verify that the calculation is exact, we'll seek $T = \tan\theta$ such that $\tan 3\theta = 18\sqrt{3}/35$.
Invoking the triple-angle tangent formula, we have
$$\tan3\theta = \frac{T ( 3 - T^2 )}{1 - 3 T^2} = \frac{18\sqrt{3}}{35}$$
If I were convinced that $T$ is a relatively nice number, I'd be willing to believe that the $T$ in the left-hand numerator accounts for the $\sqrt{3}$ in the right-hand numerator. Write $T := S \sqrt{3}$, so that:
$$\frac{S \; ( 1 - S^2 )}{1 - 9 S^2 } = \frac{6}{35}$$
Now, note that (negative) $35$ is just one away from (negative) $36$, a perfect square (and multiple of $9$, to boot), so that comparing denominators suggests $S = \pm 2$. By happy coincidence, $S=2$ gives precisely the numerator we'd need.
Therefore, $\tan\theta = T = S\sqrt{3} = 2 \sqrt{3}$, so that $\cos\theta = \frac{1}{\sqrt{1 + T^2}} = \frac{1}{\sqrt{13}}$. This gives
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 2$$
and we're don--- hey, waydaminnit ... $2\neq 7$! What happened?
Well, notice that we actually solved
$$\frac{S(1-S^2)}{1-9S^2} = \frac{-6}{-35} = \tan(3\theta+\pi)$$
("Wait, what's the difference?" "Just keep reading.") so that we actually got
$$T = \tan(\theta+\pi/3) = 2\sqrt{3}$$
That is, what we took for $\theta$ was in fact $\theta+\pi/3$. We can get at the $\theta$ we wanted thusly:
$$\tan\theta = \tan\left((\theta+\pi/3)-\pi/3\right) = \frac{T-\tan\frac{\pi}{3}}{1+T\tan\frac{\pi}{3}} = \frac{2\sqrt{3}-\sqrt{3}}{1+2\sqrt{3}\cdot\sqrt{3}} = \frac{\sqrt{3}}{7}$$
so that $\cos\theta = \frac{7}{2\sqrt{13}}$ and therefore $\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 7$, as expected.
So, why the $2$ before?
When we solved using $\frac{-6}{-35}$ instead of $\frac{6}{35}$, we implicitly replaced $x$ ---a complex number having positive real and imaginary components--- with $-x$. (Recall that $\tan 3\theta$ is the ratio of these components. It can't distinguish $x$ from $-x$.) Consequently, the $T$ we first obtained corresponds to the relation:
$$\sqrt[3]{-x} = \sqrt{13} \left( \frac{1}{\sqrt{13}} + i \frac{2\sqrt{3}} {\sqrt{13}} \right) = 1 + 2 i \sqrt{3}$$
which is to say: $(1+2i\sqrt{3})^3 = - \left( 35 + 18 i \sqrt{3}\right)$.
Then, since the arguments of $x$ and $-x$ differ by $\pi$, the arguments of the calculated cube roots must differ by $\pi/3$, and our second computation gives $\theta$ corresponding to this result:
$$\sqrt[3]{x} = \sqrt{13} \left( \frac{7}{2\sqrt{13}} + i \frac{\sqrt{3}}{2\sqrt{13}}\right) = \frac{1}{2}\left( 7 + i \sqrt{3} \right)$$
|
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|
Finding the range of a vector valued function For a single valued function, I can infer if the function is monotone from its derivative.
For a vector valued function, is it possible to infer monotonicity from the directional derivative?
For example, define
$$
D=[1,2]\times[1,2],
$$
and
$$
f(x,y)=\left( \frac{2}{1/x+1/y},\sqrt{xy} \right).
$$
Is it possible to show that $f$ maps $D$ to $D$ from its gradient $\nabla f$?
The gradient is
$$
\nabla f = \begin{pmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{pmatrix},
$$
whence, for $(x,y)$ in $D$, the directional derivative
$$
\left(\nabla f(x,y)\right)\begin{pmatrix} x \\ y \end{pmatrix},
$$
is always positive and I would like to conclude that, on $D$,
$$
\text{$f$ is minimal at $(1,1)$},\\
\text{$f$ is maximal at $(2,2)$}.
$$
Is it the right way to proceed?
The graph of each component of $f$ looks like this
|
Note that the Jacobian determinant is
$$\det J_f(x,\, y)=\det\begin{bmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{bmatrix}=\\
=\dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} x} - \dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} y}= \\
=\dfrac{y-x }{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} (xy)^{\frac{3}{2}}},$$ therefore $\det J_f(x,\, y)=0$ on the line $y=x.$
Addition:
Denote $$G_1=\{(x, \ y)\colon \;\; {1} < {x} < {2},\;\; x < {y} < {2} \}, \\
G_2=\{(x, \ y)\colon \;\; {1} < {y} < {2},\;\; y < {x} < {2} \}$$ Then $J_f (x,\ y)\ne{0}, \;\;\; \forall(x, \ y)\in{G_1\cup G_2},$
thus $f(x, \ y)$ is a diffeomorphism on each $G_1$ and $G_2$ and
$$f(\partial{G_1})=\partial{f(G_1 )}, \\
f(\partial{G_2})=\partial{f(G_2 )}.$$
As noted by Ted Shifrin, $$\gamma_1=\left\{ f(1,\ y),\;\; 1 < y < 2\right\}=\left\{\left(\dfrac{2y}{y+1},\sqrt y\right)\colon \;\; 1 < y < 2\right\}, \\
\gamma_2=\left\{ f(x,\ 2),\;\; 1 < x < 2\right\} = \left\{\left(\frac{4x}{x+2},\sqrt{2x}\right)\colon \;\; 1 < x < 2\right\}, \\
\gamma_3=\left\{ f(x,\ x),\;\; 1 < x < 2\right\} = \left\{\left(x,x\right)\colon \;\; 1 < x < 2\right\}.$$
Therefore the interior of $\partial{G_1}$ is mapped by $f$ into the interior of $f(\partial{G_1}) = \gamma_1\cup\gamma_2\cup \gamma_3.$
Due to the symmetry of $f$ with respect to $x$ and $y$ the interior of $\partial{G_2}$ is mapped by $f$ also into the interior of $\gamma_1\cup\gamma_2\cup \gamma_3.$
Different signs of the Jacobian determinant on both sides of the diagonal indicate the opposite orientation of parts of the image of $D.$
|
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|
Value of series, Partialsum? given is the following series
$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$
And I need to find its value.
How can I start finding it?
Thanks for all
does the Telescop-Summing work here as well?:
$\sum_{n=1}^\infty \frac{1}{4n^2-1} $
now:
$\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$
Now I have to "add the sum":
$\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} -
\sum_{n=1}^\infty \frac{1}{2n+1}] = \frac{1}{2} - \frac{1}{4n+2} $ And than for $n \to \infty$ it is $\frac{1}{2}$ ??
|
split it up the numerator to 2n + 2 - 1 then factor to 2(n+1) -1.
By spliting the fraction you can cancel terms.
This should allow you to see this is a finite result.
|
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|
How to resolve this algebra equation? $$f = X^3 - 12X + 8$$ $a $- complex number, $a$ is a root for $f$
$b = a^2/2 - 4 $.
Show that $f(b) = 0$
This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .
|
\begin{align*}
f(b) &= f(\frac{a^2}{2} - 4) \\
&= (\frac{a^2}{2} - 4)^3 - 12(\frac{a^2}{2} - 4) + 8 \\
&= \frac{a^6}{8} - 3a^4 + 24 a^2 - 64 - 6a^2 + 48 + 8 \\
&= \frac{a^6}{8} -3a^4 + 18a^2 -8
\end{align*}
On the other hand,
$$
f(a) = a^3 - 12a + 8 = 0
$$
and therefore
\begin{align*}
0 = 0^2 &= (a^3 - 12a + 8)^2 \\
&= a^6 - 24a^4 + 16a^3 + 144 a^2 - 192a + 64 \\
&= (a^6 - 24a^4 + 144 a^2 - 64) + (16a^3 - 192a + 128) \\
&= 8(\frac{a^6}{8} -3a^4 + 18a^2 -8) + 16(a^3 - 12a + 8)\\
&= 8f(b) + 16f(a) \\
&= 8f(b)
\end{align*}
Hence,
$$
f(b)=0.
$$
|
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|
Linear equation of 4 variables I'm stuck on this Math problem :
How many solutions does the equation
$x_{1} + x_{2} + 3x_{3} + x_{4} = k$
have, where $k$ and the $x_{i}$ are non-negative integers such that $x_{1} \geq 1$, $x_{2} \leq 2$, $x_{3} \leq 1$
and $x_{4}$ is a multiple of 6.
I tried to write the possible cases for $x_{2}$ and $x_{3}$ since they are bounded.
Case $x_{2} = 2$ :
*
*Case $x_{3} = 1$ : $x_{1} + 5 + x_{4} = k$
*Case $x_{3} = 0$ : $x_{1} + 2 + x_{4} = k$
Case $x_{2} = 1$ :
*
*Case $x_{3} = 1$ : $x_{1} + 4 + x_{4} = k$
*Case $x_{3} = 0$ : $x_{1} + 1 + x_{4} = k$
Case $x_{2} = 0$ :
*
*Case $x_{3} = 1$ : $x_{1} + 3 + x_{4} = k$
*Case $x_{3} = 0$ : $x_{1} + x_{4} = k$
But now, I'm stuck. Should I try to resolve all this equations ? Am I in the right direction ? Any help would be grealty appreciated.
EDIT :
I found that the number of solutions will be given by the coefficient $a_{k}$ of $x_{k}$, i.e :
$(x^0+x^1+x^2)(x^0+x^1)(\displaystyle\sum\limits_{k=1}^\infty x^k)(\displaystyle\sum\limits_{k=0}^\infty x^{6k})$
$=(x^0+x^1+x^2)(x^0+x^1)(\frac{1}{(1-x^3)(1+x^3)})(\frac{x}{1-x})$
$=\frac{x+2x^2+2x^3+x^4}{(x^3-1)(x^3+1)(x-1)}$
Now I don't know how to find the $a_{k}$ coefficient.
|
This will give you a start: it's the coefficient of $x^k$ in $$(x+x^2+x^3+\dots)(1+x+x^2)(1+x^3)(1+x^6+x^{12}+\dots)$$
|
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|
Proving the remainder is $1$ if the square of a prime is divided by $12$ Given, $p$ is a prime number and $p>3$. How do we prove that the remainder $r$ is always $1$ if $p^2$ is divided by $12$?
|
If $(a,12)=1, (a,3)=1$ and $ (a,2)=1$
$(a,3)=1\implies a\equiv\pm1\pmod 3\implies a^2\equiv1\pmod 3$
$(a,2)=1\implies a$ is odd $=2b+1$(say) where $b$ is some integer
$(2b+1)^2=4b^2+4b+1=8\frac{b(b+1)}2+1\equiv1\pmod 8$
$\implies a^2\equiv1\pmod { \text{lcm}(3,8)}$
Now, lcm $(3,8)=24$
|
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|
Proving $\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$ Let $p_n$ denote the $n$th prime number.
How could one prove that:
$$\prod \limits_ {k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_j}\!\!,\;\;1\le j\!<\!n$$
Examples:
$n=3812,\;j=81\qquad\implies\quad\large{\prod \limits _{k=81}^{3812} \,\frac{p_{k+1}}{p_k}= \frac{p_{3813}}{p_{81}} = \frac{35897}{419}}$
$n=20019,\;j=1\qquad\implies\quad\large{\prod \limits _{k=1}^{20019} \frac{p_{k+1}}{p_k} = \frac{p_{20020}}{p_{1}} = \frac{224993}{2}}$
$n=129181,\;j=35\quad\implies\quad\large{\prod \limits _{k=35}^{129181} \!\!\frac{p_{k+1}}{p_k}= \frac{p_{129182}}{p_{35}} = \frac{1715059}{149}}$
|
$$\prod _{k=j}^n \frac{p_{k+1}}{p_k} = \frac{p_{n+1}}{p_n} \times \frac{p_{n}}{p_{n-1}} \dots \frac{p_{j+2}}{p_{j+1}}\times \frac{p_{j+1}}{p_j}=\frac{p_{n+1}}{p_j}$$
|
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|
Proof of Astroid? How can I prove that an astroid is an envelope of all line segments of length 1 from the x-axis to the y-axis?
I read one proof of this online at the link Link
but I don't understand how this proof works.
Therefore, it would be much appreciated if someone could show me another proof or make the online proof more understandable. For example, for the first step, why is it x**/t^2**+y/(a^2-t^2)? I mean, I get that this is because of the Pythagorean theorem, but why are these the denominators of x and y?
|
For a reverse proof, compute the tangent of the astroid, find the intersection with the axis and show they distance is 1:
Start with the parametric equation $(x,y)=(\cos^3\theta, \sin^3\theta)$, whose derivative relative to $\theta$ is $(x',y')=(3 \cos^2\theta \sin\theta, -3 \sin^3\theta \cos\theta)$. The tangent at $\theta$ has equation $\frac {x - \cos^3\theta} {3 \cos^2\theta \sin\theta} = \frac {x - \sin^3\theta} {-3 \cos\theta \sin^2\theta}$, which simplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$. The tangent cuts the axis in $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$. As $PQ=1$, this proves that the tangent of the astroid at any point is a segment of length 1 from the x-axis to the y-axis.
For the constructive proof, take the segment computes its equation $y=F_\theta(x)$, and, accordingly to the books, solve the system $y=\frac {\partial F_\theta} {\partial \theta} (x)$, which express that you choose the point on the segment in such y way that the segment is tangent to the envelope.
Take $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$, the segment equation is $\frac x {\cos\theta} + \frac y {\sin\theta}=1$, which symplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$ (as above). Derivating (relative to $\theta$!) leads to $-x \sin\theta + y \cos\theta= \cos^2\theta - \sin^2\theta$. So, the equation of the envelope is the solution of the system:
$\begin{cases} +x \cos\theta + y \sin\theta &= \cos\theta \sin\theta\\ -x \sin\theta + y \cos\theta &= \cos^2\theta - \sin^2\theta \end{cases}$
To solve the system multiply the first line by $\sin\theta$, the second by $\cos\theta$ and add to eliminate $x$, which leaves $y=\sin^3\theta$. Then multiply the first line by $\cos\theta$, the second by $\sin\theta$ and add to eliminate $y$, which leaves $x=\cos^3\theta$. So the envelope parametric equation is $(x,y)=(\cos^3\theta, \sin^3\theta)$. You can eliminate the parameter by replacing $(\cos\theta, \sin\theta)=(x^\frac 1 3, y^\frac 1 3)$ in $\cos^2 \theta+\sin^2\theta=1$ to get the familiar $x^\frac 2 3 + y^\frac 2 3 =1$.
I may seem quick with mental calculations, but a lot of simplifications will appear a long the way, as long as you keep your formulas as symmetric as possible.
|
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|
Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$ Problem:
If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then
(a) $S$ is an odd integer
(b) $S + 1$ is not divisible by $9$
(c) The integer next above $(5+2\sqrt{6})^{2n+1}$ is divisible by $10$
(d) $S-1 = \frac{t}{t-1}$
Please guide me how to proceed with this question...
|
Hint: note that $(5+2\sqrt 6)^{2n+1}+(5-2\sqrt 6)^{2n+1}$ is an integer, and $(5-2 \sqrt 6) \lt 1$, so a power of it will be small and positive.
|
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|
Not so easy optimization of variables? What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for a much more efficient way than that.
This question came from a math competition from the Math Honor Society, Mu Alpha Theta.
|
Since this question popped up to the top of the stack recently, I thought I'd add another answer, since there is a way to solve the problem without needing to determine the intersection points themselves (although they can be obtained with a small amount of additional calculation).
The hyperbola is $ \ 2x^2 + 5xy + 3y^2 = 2 \ $ , the ellipse is $ \ 6x^2 + 8xy + 4y^2 = 3 \ $ .
The conic sections described by the given equations, although rotated with respect to the coordinate axes, are centered on the origin, and thus retain their symmetry about the origin ["if the point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x , -y) $ " ] . The function $ \ x^2 + y^2 \ $ , the "distance-from-the-origin-squared" function, is also symmetric about the origin. This means that its extrema on such curves will lie at points so symmetrically placed, and thus will lie on lines passing through the origin, $ \ y = mx \ $ (or, in certain situations, on the $ \ y-$ axis). I suspect this is the key observation that the poser of the problem may have intended, since this greatly reduces the amount of calculation, as we shall see. (It is understandable that this might be missed, as the effect of symmetries in optimization seems to be rather under-emphasized in teaching the topic; I found this through my own experience with such problems.)
Applying this additional condition, the function to be optimized becomes $ \ x^2 + (mx)^2 \ = \ (m^2 + 1) \ x^2 \ . $ While it still appears that we will need to solve for two variables, this condition permits us to eliminate one of them. The equations of the conic sections become
$$ 2x^2 \ + \ 5mx^2 \ + \ 3m^2x^2 \ = \ 2 \ \ \text{and} \ \ 6x^2 \ + \ 8mx^2 \ + \ 4m^2x^2 \ = \ 3 \ \ . $$
Dividing both equations through by $ \ x^2 \ , $ multiplying each equation by the appropriate constant, and equating the two, we obtain
$$ 6 \ + \ 15m \ + \ 9m^2 \ = \ \frac{6}{x^2} \ = \ \ 12 \ + \ 16m \ + \ 8m^2 $$
$$ \Rightarrow \ \ m^2 \ - \ m \ - \ 6 \ = \ ( m + 2 ) \ ( m - 3 ) \ = \ 0 \ \ \Rightarrow \ \ m \ = -2 \ , \ 3 \ \ , $$
thereby confirming the factorization and the lines through the origin shown by Bill Kleinhans (and André Nicolas in the other posting of this problem). We next return to either of the conic section equations to write
$$ 3m^2 \ + \ 5m \ + \ 2 \ = \ \frac{2}{x^2} \ \ \Rightarrow \ \ x^2 \ = \ \frac{2}{3m^2 + 5m + 2} \ \ \text{or} $$
$$ 4m^2 \ + \ 8m \ + \ 6 \ = \ \frac{3}{x^2} \ \ \Rightarrow \ \ x^2 \ = \ \frac{3}{4m^2 + 8m + 6} , $$
allowing us to write our function to be extremized as
$$ \frac{2 \ (m^2 + 1) }{3m^2 + 5m + 2} \ \ \text{or} \ \ \frac{3 \ (m^2 + 1) }{4m^2 + 8m + 6} \ \ . $$
Inserting our two slope values into either of these expressions yields
$$ m \ = \ -2 \ \ \rightarrow \ \ \frac{5}{2} \ \ , \ \ m \ = \ 3 \ \ \rightarrow \ \ \frac{5}{11} \ \ . $$
Hence, the minimum value for $ \ x^2 + y^2 \ , \ \frac{5}{11} \ , $ occurs at the intersection points of the two conic sections lying on the line $ \ y = 3x \ , $ and the maximum value, $ \ \frac{5}{2} \ , $ occurs for those intersections on the line $ \ y = -2x \ . $ The graph below illustrates the situation.
Although they are not needed to answer the question, we can determine the intersection points themselves from the foregoing:
$$ m \ = \ -2 \ \ \rightarrow \ \ x^2 \ = \ \frac{1}{2} \ \ \Rightarrow \ \ ( \ \pm \frac{\sqrt{2}}{2} \ , \ \mp \sqrt{2} \ ) \ \ ; $$
$$ m \ = \ 3 \ \ \rightarrow \ \ x^2 \ = \ \frac{1}{22} \ \ \Rightarrow \ \ ( \ \pm \frac{1}{\sqrt{22}} \ , \ \pm \frac{3}{\sqrt{22}} \ ) \ \ . $$
These do in fact agree with the extremal values for the "distance-squared" function.
$$ \ \ $$
The approach described by response will also work (it was the first method I looked at, as well), although it still needs to be taken together with the rest of the equations. The intersection points do lie on the "vertical" hyperbola $ \ 4y^2 - 14x^2 = 1 \ $ (shown in magenta in the graph above), and the symmetrical arrangement of the intersections can again be used to find the extrema of our function and locate the points:
$$ 4m^2x^2 \ - \ 14x^2 \ = \ 1 $$
$$ \Rightarrow \ \ 3m^2 \ + \ 5m \ + \ 2 \ = \ \frac{2}{x^2} \ = \ 8m^2 \ - \ 28 \ \ \Rightarrow \ \ 5m^2 \ - \ 5m \ - \ 30 \ = \ 0 \ \ \text{or} $$
$$ \Rightarrow \ \ 4m^2 \ + \ 8m \ + \ 6 \ = \ \frac{3}{x^2} \ = \ 12m^2 \ - \ 42 \ \ \Rightarrow \ \ 8m^2 \ - \ 8m \ - \ 48 \ = \ 0 \ \ , $$
leading us once more to the slope values we obtained above, to be used in the function $ \ \frac{m^2 + 1}{2 \ (2m^2 - 7 )} \ . $
|
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|
How to find the following indefinite integral? $$ \int {dx \over {\sin^3 x+\cos^3 x}}$$
Can this integral be found by substitution or any other method such as complex number?
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$$I=\int \frac{dx}{\sin^3x+\cos^3x}=\int \frac{dx} {\left(\sin(x)+\cos(x)\right)\left(1-\sin(x)\cos(x)\right)}$$ Write $$\sin(x)\cos(x)= \frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}$$ So
$$ I=\int \frac{dx} {\left(\sin(x)+\cos(x)\right) \left(1-\frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}\right)}$$ Now Use Substitution $$\sin(x)-\cos(x)=t$$
$$ \left(\cos(x)+\sin(x)\right)dx=dt \implies dx=\frac{dt}{\left(\sin(x)+\cos(x)\right)}$$ So
$$ \frac{dx}{\left(\cos(x)+\sin(x)\right)}=\frac{dt}{\left(\sin(x)+\cos(x)\right)^2} =\frac{dt}{1+2\sin(x)\cos(x)}=\frac{dt}{1+2\frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}}=\frac{dt}{2-t^2}$$
So $$I=\int \frac{2dt}{\left(t^2+1\right)\left(2-t^2\right)}=\frac{2}{3}\left( \int \frac{dt}{t^2+1}-\int \frac{dt}{t^2-2}\right)$$ So
$$ I=\frac{2}{3} \tan^{-1}(t)-\frac{1}{3\sqrt{2}}\ln\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|$$
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|
Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is grateful. Thanks.
|
I will compute the limits
of the numerator and denominator separately.
To make this more rigorous, imagine that
the two limits are being done
at the same time,
so the final division is justified.
$\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}
=\sqrt{x+1}(1+\sqrt{x-1}-\sqrt{x^2-x+1})
$
so,
putting $x = 1+y$
and using $\sqrt{1+z} \approx 1+z/2$ for small $z$,
$\begin{align}
\lim_{x\to 1} \sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}
&= \lim_{x\to 1}\sqrt{x+1}(1+\sqrt{x-1}-\sqrt{x^2-x+1})\\
&= \lim_{y\to 0}\sqrt{y+2}(1+\sqrt{y}-\sqrt{1+2y+y^2-y-1+1})\\
&= \lim_{y\to 0}\sqrt{2}(1+\sqrt{y}-\sqrt{1+y+y^2})\\
&\approx \sqrt{2}(1+\sqrt{y}-(1+y/2))\\
&\approx \sqrt{2y}
\end{align}
$
Similarly,
$\begin{align}
\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}
&=\sqrt{y}+\sqrt{1+1+2y+y^2}-\sqrt{1+1+4y+6y^2+4y^2+y^4}\\
&=\sqrt{y}+\sqrt{2}(\sqrt{1+y+y^2/2}-\sqrt{1+2y+3y^2+2y^3+y^4/2})\\
&\approx \sqrt{y}+\sqrt{2}((1+y/2)-(1+y))\\
&= \sqrt{y}+\sqrt{2}(-y/2)\\
\end{align}
$
so
$\lim_{x\to 1} \sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}
= \lim_{y\to 0} \sqrt{y}+\sqrt{2}(-y/2)
= \lim_{y\to 0} \sqrt{y}
$.
The ratio of these two is thus
$ \lim_{y\to 0} \frac{ \sqrt{2y}}{\sqrt{y}}
= \sqrt{2}$
|
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There are no integers $x,y$ such that $x^2-6y^2=7$ How to show that there is no here are no integers $x,y$ such that $x^2-6y^2=7$?
Help me. I'm clueless.
|
We work modulo $8$. Clearly $x$ cannot be even. So $x^2\equiv 1\pmod{8}$.
If $y$ is even, then $6y^2\equiv 0\pmod{8}$, so $x^2-6y^2\equiv 1\pmod{8}$. But $7\not\equiv 1\pmod{8}$.
If $y$ is odd, then $y^2\equiv 1\pmod{8}$, so $6y^2\equiv 6\pmod{8}$. It follows that $x^2-6y^2\equiv -5\pmod{8}$. But $7\not\equiv -5\pmod{8}$.
Remark: The $7$ in the problem, combined with the $6$, shout "work modulo $7$," so I would consider that the natural approach. This is particularly the case if the question is in the quadratic congruences chapter, since the key fact is that $-1$ is not a quadraic residue of $7$.
|
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|
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
|
$$\frac{1 - \frac{1-x}{1 - 2x}}{1 - 2\frac{1-x}{1-2x}}$$
Multiplying through by $1-2x$ gives
$$\frac{1 - 2x - (1 - x)}{1 - 2x - 2(1 - x)}.$$
Expanding the brackets and then simplifying gives
$$\frac{1 - 2x - 1 + x}{1 - 2x - 2 + 2x}$$
$$= \frac{-x}{-1} = x.$$
NOTE: This method works well because you have the same "denominator" of $1 - 2x$ every where. If you had different ones, I would recommend the finding the lowest common denominator method that is your accepted answer.
|
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How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
|
Let $x=\cos A+i\sin A$
Then we have,
$\displaystyle x-\frac{1}{x}=2i\sin A$
$\displaystyle (x-\frac{1}{x})^{6}=-2^6\sin^6 A$
Then we have by expanding,
$x^6+\frac{1}{x^6}-6(x^4+\frac{1}{x^4})+15(x^2+\frac{1}{x^2})-20=2\cos6 A-6\cos{4}A+15\cos2 A-20$
So we have,
$\sin^6 A=\frac{-1}{2^6}(2\cos 6 A-12\cos{4}A+30\cos2A-20)$
I have used the fact that $x^k=\cos kA+i\sin kA\Rightarrow x^k+\frac{1}{x^k}=2\cos kA$.(for $k\in Z$)
From here on I think it will be easy.
|
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|
Why does the tangent of numbers very close to $\frac{\pi}{2}$ resemble the number of degrees in a radian? Testing with my calculator in degree mode, I have found the following to be true:
$$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$
Why is this? What is the proof or explanation?
|
First, note that by standard trigonometric identities,
$$\tan\left(90^\circ-\frac{1^\circ}{10^n}\right)= \frac{\sin\left(90^\circ-\frac{1^\circ}{10^n}\right)}{\cos\left(90^\circ-\frac{1^\circ}{10^n}\right)}=\frac{\cos\left(\frac{1^\circ}{10^n}\right)}{\sin\left(\frac{1^\circ}{10^n}\right)}=\frac{\cos\left(\frac{1}{10^n}\times\frac{\pi}{180}\right)}{\sin\left(\frac{1}{10^n}\times\frac{\pi}{180}\right)}.$$
Also, the power series for $\sin$ and $\cos$ are
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$
so that, for small $x$, we have
$$\sin(x)\approx x\hphantom{,}$$
$$\cos(x)\approx 1,$$
and therefore
$$\tan\left(90^\circ-\frac{1^\circ}{10^n}\right)\approx\frac{1}{\frac{1}{10^n}\times\frac{\pi}{180}}=\frac{180}{\pi}\times 10^n.$$
|
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|
$\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$. I'm starting Spivak's Calculus and finally decided to learn how to write epsilon-delta proofs.
I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that
$$\lim\limits_{x \to 1} \frac{1}{x} = 1.$$
Here's how I started my proof,
$$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| < \epsilon \implies \left| x-1 \right| < \epsilon |x|$$
I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?
|
$\tag 1 |\frac{1}{x} - 1| < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; -\varepsilon < \frac{1}{x} - 1 < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
$ \text{ iff } \; \frac{-\varepsilon}{1 + \varepsilon} < x - 1 < \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
We are trying to find our $\delta \gt 0$ 'setup', $|x -1| \lt \delta$ contrained by
$\tag 2 \frac{-\varepsilon}{1 + \varepsilon} \le -\delta \lt x - 1 \lt +\delta \le \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$
so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\text{iff }$-logic-chain" in reverse).
Setting
$\quad \delta = \frac{\varepsilon}{1 + \varepsilon}$
sets up the left side of $\text{(2)}$, but also takes care of the rights side since
$\quad \delta = \frac{\varepsilon}{1 + \varepsilon} \le \frac{\varepsilon}{1 - \varepsilon}$
Note: Here we don't use the 'min delta' approach. Using the same mechanical technique found here, we simply 'turn-the-crank'. As we work things out we realize that we want to constrain $\varepsilon$.
Instructional Example 1: You are challenged with an $\varepsilon$ that is greater than or equal to $1/2$.
Then set $\delta = 1/3$.
Note that if $\varepsilon = 1/500$, we take $\delta = 1/501$, a 'looser' number than the $\varepsilon/2 = 1/1000$ found in other (excellent) answers in this thread.
|
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|
Confirm the meaning of Prime and Primitive in a Galois(2) polynomial. Here it discusses primality (or more accurately irreducibility) and primitivity of polynomials in $G(2)$. More specifically it states that $x^6 + x + 1$ is irreducible and primitive.
But here I can divide $x^7 + 1$ by $x^6 + x + 1$ and get $x$ remainder $x^2+x+1$. Surely this means the the order of $x^6 + x + 1$ is at most $7$ and therefore nowhere near the required $2^6-1 = 63$ as required for primitivity.
What mistake am I making?
I guess it is my interpretation of
The order of a polynomial $f(x)$ for which $f(0)$ is not $0$ is the smallest integer $e$ for which $f(x)$ divides $x^e+1$. A polynomial over GF(2) is primitive if it has order $2^n-1$.
|
According to the definition in the first link you provided, to see that the order of
$x^6+x+1$ is $2^6-1$, you only need to check that $(x^6+x+1)\nmid (x^{n}+1)$ for all $n<2^6-1$, and that $x^6+x+1$ does divide $x^{2^6-1}+1$. A computer check easily confirm that this is the case.
I don't see any reason why the division of $x^7+1$ by $x^6+x+1$, giving quotient x and remainder $x^2+x+1$, would imply that the order is at most $7$, as you claimed.
|
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|
Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem :
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried :
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed
|
What you did is a very good first step.
For the second integral, make the substitution $u=1-x^2$. We end up with the easy $\int -\frac{1}{2}u^{-3/2}\,du $.
For the first integral, note that the bottom simplifies to $(1-x^2)\sqrt{1-x^2}$, since the square root is only defined when $|x|\le 1$. There is nice cancellation, and we are integrating $\frac{1}{\sqrt{1-x^2}}$, easy, we get an $\arcsin$.
|
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Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$
Mathematica is unable to give a closed form for the indefinite integral.
How can we prove this result? Please help me.
EDIT
Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$
Please take a look here.
|
This isn't an answer to the question, but I thought I should post some of my work here.
Consider the function
$$
F(a) \;=\; \int_0^1 \frac{\log(1+x^a)}{1+x}dx.
$$
The question asks us to prove that $F(2+\sqrt{3}) = \dfrac{\pi^2}{12}(1+\sqrt{3}) + \log(2)\log(1+\sqrt{3})$.
*
*Mathematica isn't able to compute a closed form for $F(2+\sqrt{3})$, but it can compute $F(a)$ for certain other values of $a$:
*
*$F(0) \;=\; (\log 2)^2.$ (Very easy.)
*$F(1) \;=\; \dfrac{1}{2}(\log 2)^2.$ (Also easy to do by hand.)
*$F(2) \;=\; -\dfrac{\pi^2}{48} \,+\, \dfrac{3}{4}(\log 2)^2$
*$F(3) \;=\; \dfrac{\pi^2}{9}+\dfrac{1}{2}(\log 2)^2 \,+\, \dfrac{1}{2}\mathrm{Li}_2\left(-\dfrac13\right) - \mathrm{Li}_2\left(\dfrac{1+i\sqrt{3}}6\right) - \mathrm{Li}_2\left(\dfrac{1-i\sqrt{3}}6\right)$
It can also compute $F(1/3)$, $F(1/2)$, $F(-2)$, $F(-1)$, $F(-1/2)$, and $F(-1/3)$, which follow from the symmetry properties below, as well as very long expressions for $F(4)$, $F(-4)$, $F(1/4)$, and $F(-1/4)$.
In the formula for $F(3)$, the function $\mathrm{Li}_2$ is the dilogarithm, which is defined by the integral
$$
\mathrm{Li}_2(a) \;=\; -\int_0^1 \frac{\ln(1-ax)}{x}dx.
$$
Interestingly, the only values of $a$ for which $\mathrm{Li}_2(a)$ is known to have a closed form are $-1$, $0$, $1/2$, $1$, $2$, and various expressions involving $\sqrt{5}$. (See here.)
*The function $F(a)$ defined above has a few symmetry properties. It is easy to show that
$$
F(-a) \;=\; F(a) - a\int_0^1 \frac{\log x}{1+x}dx.
$$
which gives
$$
F(-a) \;=\; F(a) - \frac{\pi^2 a}{12}.
$$
In addition, integration by parts followed by a substitution can be used to show that
$$
F(a) \,+\, F(1/a) \;=\; (\log 2)^2
$$
for any positive value of $a$. Since $(2+\sqrt{3})^{-1} = 2-\sqrt{3}$, it follows that the given question is equivalent to the equation
$$
F(2-\sqrt{3}) \;=\; \frac{\pi^2}{12}(\sqrt{3}-1)\,+\,\log(2)\log(\sqrt{3}-1).
$$
*It's not too hard to find a series for $F(\alpha)$. We have
$$
\begin{align*}
\frac{\log(1+x^\alpha)}{1+x} \;&=\; \left(\frac{1}{1+x}\right)\log(1+x^\alpha) \\[6pt]
&=\; \left(\sum_{k=1}^\infty (-1)^{k+1}x^{k-1}\right)\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{\alpha n}}{n} \\[6pt]
&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{x^{\alpha n+k-1}}{n}
\end{align*}
$$
We can now (by Abel's Theorem) integrate this series term-by-term to get
$$
F(\alpha) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n(\alpha n+k)}
$$
In particular
$$
F(2+\sqrt{3}) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n\bigl((2+\sqrt{3})n+k\bigr)}
$$
The $2+\sqrt{3}$ in the denominator of this series is unfortunate. However, we already know the value of the sum $F(2+\sqrt3)+F(2-\sqrt3)$, so all we really want is the value of the difference $F(2+\sqrt3)-F(2-\sqrt3)$. In general:
$$
\begin{align*}
F(\alpha)-F(\alpha^{-1}) \;&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\left(\frac{1}{n(\alpha n+k)}-\frac{1}{n(\alpha^{-1} n+k)}\right) \\[6pt]
&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{(\alpha^{-1} n+k)-(\alpha n+k)}{n(\alpha n+k)(\alpha^{-1} n+k)} \\[6pt]
&=\; \bigl(\alpha^{-1}-\alpha\bigr)\sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + (\alpha+\alpha^{-1})nk + k^2}
\end{align*}
$$
So:
$$
F(2+\sqrt{3})-F(2-\sqrt{3}) \;=\; -2\sqrt{3} \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + 4nk + k^2}
$$
This series is more pleasant, but I still have no idea how to evaluate it.
|
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Show that $e^x \geq (3/2) x^2$ for all non-negative $x$ I am attempting to solve a two-part problem, posed in Buck's Advanced Calculus on page 153. It asks "Show that $e^x \geq \frac{3}{2}x^2$ $\forall x\geq 0$. Can $3/2$ be replaced by a larger constant?"
This is after the section regarding Taylor polynomials, so I have been attempting to leverage the Taylor expansion for $e^x$ at $0$. $e^x \geq 1+x+\frac{x^2}{2}$, and by quadratic formula, we have $1+x+\frac{x^2}{2}\geq \frac{3}{2}x^2$ for $x\in [0, \frac{1+\sqrt{5}}{2}]$.
Now also $e^x\geq 1+x+\frac{x^2}{2}+\frac{x^3}{6}$. We know there exists a $c\in \mathbb{R}^{\geq 0}$ such that for all $x\geq c$, we have $1+x+\frac{x^2}{2}+\frac{x^3}{6}\geq \frac{3}{2}x^2$. I want to find this point without messing with the cubic formula, etc. I think I am missing a simpler way. Any ideas?
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Consider a quadratic $y=a x^2$ that is tangent to $y=e^x$. That is, we match function and derivative at a point $x=x_0$. Then
$$e^{x_0}=a x_0^2$$
$$e^{x_0}=2 a x_0$$
Then $a x_0^2=2 a x_0$ so that $x_0=2$. Plugging this back into one of the above equations, we get that
$$a=\frac{e^2}{4} \approx 1.84726$$
Consider, as @julien suggests, the function $g(x)=a x^2 e^{-x}$ which has derivative $a (2-x) x e^{-x}$ and second derivative $a (x^2-4 x+2) e^{-x}$. This function represents the ratio of the quadratic to the exponential and is a maximum at $x=2$ for all $a>0$. As $a x^2$ monotonically increases in $a$, we have
$$e^x \ge \frac{e^2}{4} x^2 \gt \frac{3}{2} x^2$$
|
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|
Finding Big-O with Fractions I'd want to know how I can find the lowest integer n such that f(x) is big-O($x^n$) for
a) $f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}$
I've fooled around with this a bit and tried going from
$\frac {x^4 + x^2 + 1}{x^3 + 1} \le \frac {x^4 + x^2 + x}{x^3 + 1} \le \frac {x^4 + x^2 + x}{x^3}$
but I'm honestly just taking wild stabs at the problem. The single fractional problem in my textbook talks about making the fraction larger by making the numerator bigger and/or the denominator smaller. That's the only basis I have for what is written above and I don't know how to progress from there.
b) $f(x) = \frac {x^4 + 5 \log x}{x^4 + 1}$
I've got no clue where to start with this one. I tried looking at it like as a combination of functions ($x^4 +5 \log x)$ but I have no idea what to do with 5 log x.
When writing your reply please try to keep it as simple as possible.
Edit (June 24):
Because the question asks the answer to use the lowest exponents possible, the book lists $2x$ where x > 1 as a possible solution for a) and $2x^0$ where x > 1 as a possible solution to b).
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Intuitively, the idea is that the "order" of a function is that of its highest-order term. In $x^4 + x^2 + 1$, you can usually ignore everything except the $x^4$, and say that it is $O(x^4)$ and not $O(x^d)$ for any smaller $d < 4$. Similarly $x^3 + 1$ is of the order of $x^3$, and in general a polynomial of degree $d$ is of the order of $x^d$.
So in your first function $f(x) = f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}$, the numerator is of order $x^4$ and the denominator is of order $x^3$, so the fraction $f(x)$ is of order $\frac{x^4}{x^3} = x$.
That's the idea of what's really going on, but to prove it formally, you can do the division: write $f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1} = x + \frac{x^2 - x + 1}{x^3 + 1}$, and observe that the remainder term $\frac{x^2 - x + 1}{x^3 + 1} = \frac{1}{x+1} \to 0$ as $x \to \infty$, so $f(x) = O(x)$ and not $O(x^{1 - \epsilon})$ for any $\epsilon > 0$.
You don't even have to do the division completely: just get rid of the $x^3$ terms from both numerator and denominator. Observe that
$$f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}
= \frac{(x^4 + x^2 +1)/x^3}{(x^3 + 1)/x^3}
= \frac{x + \frac{1}{x} + \frac{1}{x^3}}{1 + \frac{1}{x^3}}$$
and you can prove directly from this that $f(x) = O(x)$.
Similarly, for the second function, $f(x) = \frac {x^4 + 5 \log x}{x^4 + 1}$, both numerator and denominator are of the same order $x^4$, so you should expect the fraction to be $O(1)$. You can prove it by writing
$$f(x)
= \frac {x^4 + 5 \log x}{x^4 + 1}
= \frac {(x^4 + 5 \log x)/x^4}{(x^4 + 1)/x^4}
= \frac{1 + \frac{5 \log x}{x^4}}{1 + \frac{1}{x^4}}
$$
and proving from there that $f(x) = O(1) = O(x^0)$.
(E.g. you can show that, for $c_1 = \frac12$ and $c_2 = 2$ (say) and sufficiently large $x$, we have $c_1 \le f(x) \le c_2$, since $f(x) \ge \frac{1}{1 +\frac1{x^4}} \ge \frac{1}{2}$ when $x \ge 1$, and similarly $f(x) \le \frac{1 + \frac{5\log x}{x^4}}{1} \le 2$ when $\frac{5 \log x}{x^4} \le 1$, which again is true for sufficiently "large" $x$. Actually in this case any constants $c_1 < 1 < c_2$ will do, since $\lim_{x\to\infty} \frac{f(x)}{x} = 1$, but we don't need all this just to prove that $f(x) = O(1)$.)
|
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Showing that $\lim\limits_{n\to\infty}x_n$ exists, where $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$ Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$
a) Show that $x_{n} < x_{n+1}$
b) Show that $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$
Hint : Square $x_{n+1}$ and factor a 2 out of the square root
c) Hence Show that $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.
Any help? I don't know where to start.
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10 days old question, but .
a) Is already clear, that $ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n+1}}}}$ , because $\sqrt{n} <\sqrt{n} + \sqrt{n+1}$ which is trivial.
My point here is to give some opinion about b) and c), for me it's better to do the c) first. We know that : $$ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{p+\sqrt{p+\sqrt{p+ ... }}} $$
But it is only true for $q\leq p<\infty $ for $q \in \mathbb{Z}^{+}$. Because it is trivial that
$$ \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} > \sqrt{1+\sqrt{1+\sqrt{1+ ... }}} $$
Let $x=\sqrt{2+\sqrt{2+\sqrt{2+ ... }}}$, then $x^2=2+ \sqrt{2+\sqrt{2+\sqrt{2+ ... }}} \rightarrow x^2-x-2=0 $, thus $x=2$, because $x>0$.
Now let's probe this equation :
$$\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} \leq \sqrt{2+\sqrt{2+\sqrt{2+ ... }}}=2 \tag{1}$$
2 is bigger than 1 , with their difference is 1. so for $x_{n}$ to be bigger than 2, it is required for $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \geq 3$
but if square both sides of (1) and substract, we get that $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \leq 3$.
for (b) , first square both sides, the '1' is gone , square again until the '2' is gone, and we arrive to this equation :
$$\sqrt{3+\sqrt{4+...\sqrt{n}}} \leq 2.\sqrt{2+\sqrt{3+...\sqrt{n}}}$$
which is true, because from (1) we know that $\sqrt{3 +\sqrt{4 ...+\sqrt{n}}} \leq 2$ and $ \sqrt{2+\sqrt{3+...\sqrt{n}}} >0 $
In fact, if you can prove (b) then (c) is trivial and vice versa.
|
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Repeated Eigenvalues Two problems from Differential Equations; Dynamical Systems, and an Introduction to Chaos (Morris W. Hirsch,Stephen Smale.Robert L. Devaney), examples page 112,113:
If $$A= \begin{pmatrix}
2 & 0 & -1 \\
0 & 2 & 1 \\
-1 & -1 & 2 \\
\end{pmatrix}$$
… $\lambda =2 , m_{\lambda}=3$ $rank(A-2I)=2 , n_{\lambda}=1 ,k= m_{\lambda}- n_{\lambda}+1=3$ so
$$p= \begin{pmatrix}
1& 0 & 1 \\
-1 & 0 & 0 \\
0 & -1 & 0 \\
\end{pmatrix}$$ by $v_2=(A-2I)v_3, v_1=(A-2I)v_2, v_3 : \begin{cases}
(A-2I)^3v_3= 0\\
(A-2I)^2v_3 \not = 0\\
\end{cases}
$But $p^{-1}Ap \not =J$ $$p= \begin{pmatrix}
0& 0 & 0 \\
0& 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} \not = \begin{pmatrix}
2& 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2 \\
\end{pmatrix}$$ what’s wrong?!
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I will assume you understand how they derived $P$. (If not, please respond).
We have:
$$P= \begin{pmatrix}1& 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0 \end{pmatrix}$$
This give us:
$$P^{-1}= \begin{pmatrix} 0 & -1 & 0 \\0 & 0 & -1 \\1 & 1 & 0 \end{pmatrix}$$
From this we get:
$$P^{-1}AP = \begin{pmatrix} 0 & -1 & 0 \\0 & 0 & -1 \\1 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix}2 & 0 & -1 \\0 & 2 & 1 \\-1 & -1 & 2 \end{pmatrix} \cdot \begin{pmatrix}1& 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0 \end{pmatrix} = \begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\0 & 0 & 2\end{pmatrix}$$
This validates the authors' solution.
Did you calculate $P^{-1}$ or the product $P^{-1}AP$ incorrectly?
|
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Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
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Suppose $$\frac{2(k+1)-1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$
$$ \frac{2k+1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$
$$\frac{(2k+1)^2}{4(k+1)^2 (3k+1)}\geq \frac{1}{3(k+1)+1}$$
$$ 3(k+1)(2k+1)^2+(2k+1)^2 \geq 4(k^2+2k+1)(3k+1)$$
$$ (3k+3)(4k^2+4k+1)+4k^2+4k+1 \geq (k^2+2k+1)(12k+4)$$
$$12k^3+12k^2+12k^2+12k+3k+3+4k^2+4k+1 \geq 12k^3+24k^2+12k+4k^2+8k+4$$
$$19k \geq 20k$$
$$ 1\geq k$$
But .....
|
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If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:
If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.
Working :
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$
$$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$
but it seems this is the wrong way of doing this...please suggest. Thanks!
|
Using
$2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$$A=\frac{\sin 6^\circ\cdot \sin 42^\circ}{\cos 6^\circ\cdot \cos 42^\circ}=\frac{\cos36^\circ-\cos48^\circ}{\cos36^\circ+\cos48^\circ}$$
Applying Componendo and dividendo,
$$\frac{1+A}{1-A}=\frac{\cos36^\circ}{\cos48^\circ}$$
Similarly, Using
$2\sin A\cos B=\sin(A+B)+\sin(A-B)$ and $2\cos A\sin B=\sin(A+B)-\sin(A-B),$
$$B=\frac{\cos66^\circ \cos72^\circ}{\sin66^\circ \sin72^\circ}=\frac{\cos66^\circ \sin18^\circ}{\sin66^\circ \cos18^\circ}=\frac{\sin84^\circ-\sin48^\circ}{\sin84^\circ+\sin48^\circ}$$
Applying Componendo and dividendo,
$$\frac{1+B}{1-B}=\frac{\sin84^\circ}{\sin48^\circ}$$
$$\implies \frac{1+A}{1-A}\cdot\frac{1+B}{1-B}=\frac{\sin84^\circ\cdot \cos36^\circ}{\sin48^\circ\cdot \cos48^\circ}=\frac{2\sin84^\circ\cdot \cos36^\circ}{\sin(2\cdot48)^\circ}=2\cos36^\circ$$
as $\sin96^\circ=\sin(180-96)^\circ=\sin84^\circ$
Now $\cos36^\circ$ can be found here
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/432322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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|
A function inequality Is it true that the following function
$$\frac{\pi ^2 \left(t^2-4 (-1+t) \text{cos}\left[\frac{\pi }{m}\right]^2\right) \text{csc}\left[\frac{\pi (-2+t)}{m}\right]^2}{m^2}, t\in[0,1]$$ attains its maximum in 0 and 1. Here $m>3$.
|
Let us denote $$f(t) := \left(t^2-4(-1+t)\cos^2\left(\frac \pi m \right)\right)\csc^2 \left(\frac {\pi(-2+t)} m \right) .$$ We start from the verification $$f(0)=f(1)= \csc^2\left( \frac \pi m \right) . $$ Next, in order to prove the statement under consideration, it is enough to prove the convexity of $f(t)$ on $(0,1)$. Because $f(t) \in C^2(0,1)$, the convexity is equivalent to $f''(t) \ge 0$ there. After calculating
$$f''(t)=-2\, \left( \csc \left( {\frac {\pi \, \left( t-2 \right) }{m}}
\right) \right) ^{2}$$ $$ \left(12\pi^{2} \left( \cos \left(
\frac {\pi} {m} \right) \right)^{2} \left( \cot \left( \frac {\pi
\left( t-2 \right) }{m} \right) \right) ^{2}t-\right.$$ $$12\,{\pi }^{2}
\left( \cos \left( {\frac {\pi }{m}} \right) \right) ^{2} \left(
\cot \left( {\frac {\pi \, \left( t-2 \right) }{m}} \right) \right) ^
{2}-$$ $$3\,{\pi }^{2} \left( \cot \left( {\frac {\pi \, \left( t-2
\right) }{m}} \right) \right) ^{2}{t}^{2}+4\,{\pi }^{2} \left( \cos
\left( {\frac {\pi }{m}} \right) \right) ^{2}t-$$ $$8\,\pi \, \left( \cos
\left( {\frac {\pi }{m}} \right) \right) ^{2}\cot \left( {\frac {
\pi \, \left( t-2 \right) }{m}} \right) m-4\,{\pi }^{2} \left( \cos
\left( {\frac {\pi }{m}} \right) \right) ^{2}-$$ $$\left.{\pi }^{2}{t}^{2}+4\,
\pi \,\cot \left( {\frac {\pi \, \left( t-2 \right) }{m}} \right) mt-{
m}^{2} \right) {m}^{-2}
$$
we find its asymtotics in $m$ with help of Maple:
$$
f''(t)=\frac 2 3 {\frac {{\pi }^{2} \left( {t}^{4}-8\,{t}^{3}+24\,{t}^{2}-20\,t+28
\right) }{ \left( t-2 \right) ^{4}}}
+O\left(\frac 1 {m^2}\right), m \to \infty.
$$
This implies the convexity of $f(x)$ on $(0,1)$ for big values of $m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/432454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Finding limit function $\lim_{n \rightarrow \infty} n ((x^2 +x + 1)^{1/n} -1)$ \begin{align}
f(x) &= \lim_{n \rightarrow \infty} n ((x^2 +x + 1)^{1/n} -1) \\&= \lim_{n \rightarrow \infty} n ((\infty)^{1/n} -1) \\&= \lim_{n \rightarrow \infty} n (1 -1)\\& =
\lim_{n \rightarrow \infty} n \cdot 0 \\&= 0
\end{align}
Did I solve it correctly??
|
Note that $$\lim\limits_{n \rightarrow \infty} n ((x^2 +x + 1)^{\tfrac{1}{n}} -1) =\lim\limits_{n\rightarrow \infty} \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}. $$
Denoting $h(x)=x^2 +x + 1,$ we have
$$\lim\limits_{n\rightarrow \infty} \dfrac{(x^2 +x + 1)^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty} \dfrac{(h(x))^{\tfrac{1}{n}} -1}{\dfrac{1}{n}}=\ln{h(x)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/433101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve $\frac{dx}{dt} = x^3 + x$ for $x$ This is a seemingly simple first order separable differential equation that I'm getting stuck on. This is what I have so far:
$$\frac{dx}{dt} = x^3+x$$
goes to
$$\frac{dx}{x(1+x^2)} = dt$$
Now using partial fractions to integrate the left-hand side:
$$\frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx+C}{1+x^2}$$
Solving for A, B, C:
$1 = A(1+x^2) + (Bx+C)x$, and using coefficient matching, I get $A=1, B=-1, C=0$.
So the integral yields:
$$\int\frac{1}{x} - \frac{x}{1+x^2}dx = \int dt$$
This yields: $$\ln x -\frac{1}{2}\ln(1+x^2) =t + C$$
So I tried using log rules and such to solve for $x$. I think this is where my source of error is.
My attempt:
$$\ln x - \ln(1+x^2)^\frac{1}{2} = t+C$$
$$\ln\frac{x}{(1+x^2)^{\frac{1}{2}}} = t + C$$
$$\frac{x}{(1+x^2)^{\frac{1}{2}}}= Ce^t$$
But this seems wrong. I apologize in advance if it's a silly mistake that I didn't see.
|
Assume that $x\ge 0$.
$$\frac{x}{(1+x^2)^{\frac{1}{2}}}= \frac{\sqrt{x^2}}{(1+x^2)^{\frac{1}{2}}}=\sqrt{\frac{x^2}{1+x^2}}$$
and you can probably take it from there.
If $x<0$, then $x=-\sqrt{x^2}$ and proceed similarly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/433966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Plotting graphs using numerical/mathematica method From the author's equation 13, 14 We can write by inserting V''(A)=0,
Solving for R we get,
$$R= \frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}$$
Now inserting the V into the article equation (11)$$E= \left(\frac{\pi }{2}\right)^{D/2} R^D V,$$ we get,
$$E= \left(\frac{\pi }{2}\right)^{D/2} \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+\frac{D}{2 R^2}\right)\right) R^D$$
Now inserting the value of R, we get,
$$E= \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+2^{-1-\frac{D}{2}} 3^{-D/2} \left(-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2\right)\right)\right) \left(\frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}\right)^D \left(\frac{\pi }{2}\right)^{D/2}$$
For $D= 3$
we finally get,
$$E= \frac{27 6^{3/4} \left(-\frac{2}{3} \sqrt{\frac{2}{3}} A^3+\frac{A^4}{8 \sqrt{2}}+A^2 \left(1+\frac{-12 \sqrt{6}+48 A-9 \sqrt{3} A^2}{12 \sqrt{6}}\right)\right) \pi ^{3/2}}{\left(-12 \sqrt{6}+48 A-9 \sqrt{3} A^2\right)^{3/2}} \tag{1}$$
the graph for equation (1) must satisfy the article graph (FIG 2)
My graph:
Plot[(27 6^(3/4) (-(2/3) Sqrt[2/3] A^3 + A^4/(8 Sqrt[2]) +A^2 (1 + (-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)/( 12 Sqrt[6]))) \[Pi]^(3/2))/(-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)^(3/2), {A, 0.5, 2.5}]
But the author got,
Output :
Am I doing wrong in simulation?
Then The author got like this in Fig 3
`
|
Look at your function in a much more simple way:
$$R(A) = \frac{\sqrt{12}}{\sqrt{24 A - 9 A^2-12}}$$
for $D=2$. The function should have a horizontal tangent when the derivative of the radicand is zero:
$$\frac{d}{dA} (24 A - 9 A^2-12) = 0 \implies A = \frac{4}{3}$$
which seems to agree with your result. Also, $R$ blows up when the denominator goes to zero, or when $A=2$ and $A=2/3$, which also agrees with your plot. So far, I'd say your plot looks good.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/434768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Partial fractions to integrate$\int \frac{4x^2 -20}{(2x+5)^3}dx$ $$\int \frac{4x^2 -20}{(2x+5)^3}dx$$
I can't use the coverup method that I learned since making anything zero in this makes everything zero. I would probably just use random test points because I don't have any other tricks memorized. Is there some specific trick I need to use on this so it is possible to compute on a timed test?
|
$$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac A{(2x + 5)} + \dfrac B{(2x + 5)^2} + \dfrac C{(2x + 5)^3}\, dx$$
Partial fractions can be thought of as the reverse of "finding the common denominator".
That is, we can equate the numerator of the original integrand with the expansion of each of the numerators of the desired fractions, times the factor we'd need to find the common denominator, if we were adding the fractions in the desired integrand.
Doing this gives us $$A(2x + 5)^2 + B(2x + 5) + C = 4x^2 - 20\tag{1}$$
Solve for C first: to do this we can put $x = -\frac 52$ to zero out all but $$C = 4(-5/2)^2 - 20 = 4\cdot \frac {25}{4} - 20 = 25 - 20 = 5$$
Then expand the left-hand side, substitute $C = 5$, and match up coefficients, (you can easily find $A$ that way, since it will be the only term on the left hand side that is a coefficient of a term with $x^2$. Substitute, solve for $B$ knowing $A$. In other situations, you could use two arbitrary values for $x$ to obtain two equations in two unknowns.
Expanding $(1)$: $$A(4x^2 + 20 x + 25) + B(2x + 5) + 5 = 4x^2 - 20 \tag{2}$$
$$\implies 4Ax^2 = 4 \implies A = 1$$
Now, we have $A = 1, C = 5$, and can substitute this into our equation $(2)$:
$$4Ax^2 + 20 Ax + 25A + 2Bx + 5B + 5 = 4x^2 - 20$$ $$ \iff (20 A + 2B)x = 0\cdot x $$ $$\iff 20 A + 2B = 0 $$ $$\iff 20\cdot 1 + 2B = 0 \iff B = -10 $$
$$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac 1{(2x + 5)} + \dfrac {-10}{(2x + 5)^2} + \dfrac 5{(2x + 5)^3}\, dx$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/437153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?
|
Clearly, $x\ne0$
$$\sqrt{x-1/x} + \sqrt{1 - 1/x} = x$$
$$\implies \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt x$$
$$\implies \sqrt{x-1}(\sqrt{x+1}+1)=x\sqrt x$$
$$\implies \sqrt{x-1}\frac{x+1-x}{\sqrt{x+1}-1}=x\sqrt x\text{ (rationalizing the numerator) }$$
$$\implies \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}-\sqrt x$$
$$\implies \sqrt x+ \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}$$
Squaring we get, $$x+x-1+2\sqrt{x^2-x}=x^2+x\iff 2\sqrt{x^2-x}=x^2-x+1$$
Putting $x^2-x=a^2,a^2=2a-1\implies (a-1)^2=0\implies a=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438452",
"timestamp": "2023-03-29T00:00:00",
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|
Using complete induction, prove that if $a_1=2$, $a_2=4$, and $a_{n+2}=5a_{n+1}-6a_n$, then $a_n=2^n$ Could anyone please explain to me how to do this problem by using the principle of complete induction? Thanks. :)
Let $a_1=2$, $a_2=4$, and $a_{n+2}=5a_{n+1}-6a_n$ for all $n\geq 1$. Prove that $a_n=2^n$ for all natural numbers $n$.
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For induction we wish to show the following:
*
*If $a_{n+2} = 2^{n+2}$ and $a_{n+1} = 2^{n+1}$ for some particular positive integer $n$ then: $a_{n+3} = 2^{n+3}$ must be true for that same value $n$
*$a_1 = 2^1$ and $a_2 = 2^2$
Since if 1 is true and 2 is true then we can repeatedly use 1 from the truth of 2 to establish that
*
*$a_n = 2^n$ for all $n >= 1$
Proving point number 2 is trivial. It is already given to us. We now must prove point number 1.
*
*Assume that
$$a_{n+2} = 2^{n+2}$$
$$a_{n+1} = 2^{n+1}$$
For some arbitrary positive integer n.
*
*Notice that using the original definition:
$$a_{n+3} = 5*a_{n+2} - 6*a_{n+1} = 5*2^{n+2} - 6*2^{n+1}$$
Now we will simplify this expression:
$$5*2^{n+2} - 6*2^{n+1} =2*(5*2^{n+1})- (5*2^{n+1}) - 2^{n+1}$$
We now factor and reorganize:
$$2*(5*2^{n+1})- (5*2^{n+1}) - 2^{n+1} =(5*2^{n+1})(2 - 1) - 2^{n+1} = (5*2^{n+1}) - 2^{n+1} $$
And now simplify:
$$ (5*2^{n+1}) - 2^{n+1} = 4*2^{n+1} = 2*2^{n+2} = 2^{n+3} $$
Thus we have shown: that if
$$a_{n+2} = 2^{n+2}$$
$$a_{n+1} = 2^{n+1}$$
and
$$a_{n+3} = 5*a_{n+2} - 6*a_{n+1}$$
Then:
$$a_{n+3} = 2^{n+3}$$
Thus we are now done with the induction.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a linear system with complex eigenvalues I have the system:
\begin{equation}
x' = \begin{pmatrix}5&10\\-1&-1\end{pmatrix}x
\end{equation}
The corresponding characteristic equation is:
\begin{equation}
\lambda^2-4\lambda+5 \\
\implies \lambda_1 = 2+i \land \lambda_2 = 2-i
\end{equation}
I am having trouble solving for the eigenvector:
\begin{align}
(A-\lambda I)x_1 =& 0\\
\begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}x_1 =&0
\end{align}
I really don't know what to do with the complex numbers. I thought about multiplying by the conjugate in the rows, but then I will have complex numbers in opposite columns again...
*
*How do I solve for the eigenvectors in the case of complex numbers?
|
If we set $x_1 = (a, b)^T$, we get
$$
\begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}x_1 = \begin{pmatrix}3-i&10\\-1&-3-i\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\\\\
= \begin{pmatrix}(3-i)a + 10b\\-a - (3 + i)b\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}
$$
which is just a set of two equations of two (complex) unknowns. Multiplying the lower equation by $-(3-i)$ (multiplying by conjugates is a good trick against both square roots and complex numbers, you're right bout that) shows that these equations are linearly dependant (which is good), and we get that $b = \left(-\frac{3}{10} + \frac{i}{10}\right)a$ solves them both:
$$
(3-i)a +10b = 0\\\\
(3-i)a = -10b\\\\
\frac{3-i}{-10}a = b \\\\
\left(-\frac{3}{10} + \frac{i}{10}\right)a = b
$$
And there you have your eigenspace for the first eigenvalue.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by 47? Can any one please tell the approach or solve the question
what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by $47$?
I can solve remainder of $45!$ divided by $47$ using Wilson's theorem but I don't know what must be the approach for this model problems, as $47$ is a prime number I cannot convert it into another factorial and divide.
If any one of you viewing have any idea regarding the approach, please post your approach here.
Thanks in advance.
Regards,
Pavan Kumar
|
Just to compose table:
\begin{array}{|c|r|}
\hline
n! & \equiv \ldots (\bmod \:47) \\
\hline \\
1! & 1 \\
2! & 2\cdot 1 = 2 \\
3! & 3 \cdot 2 = 6 \\
4! & 4 \cdot 6 = 24 \\
5! & 5 \cdot 24 = 120 \equiv 26 \\
6! & 6 \cdot 26 = 156 \equiv 15 \\
7! & 7 \cdot 15 = 105 \equiv 11 \\
\cdots \\
44! & 44 \cdot 8 = 352 \equiv 23 \\
45! & 45 \cdot 23 = 1035 \equiv 1 \\
\hline
\end{array}
$45$ steps/rows in total.
Then to find sum:
$S = 1+2+6+24+26+15+11+\ldots+23+1 = \color{#E0E0E0}{1052 \equiv 18 (\bmod \: 47)}$.
Here we use idea:
if $\qquad$ $k! \equiv s (\bmod \: p)$,
then $\;$ $(k+1)! \equiv (k+1)\cdot s (\bmod \: p)$,
and apply it step-by-step.
|
{
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|
Saddle Point Integral I want to calculate ,
$$I = \int_0^\infty dx \,x^{2n}e^{-ax^2 -\frac{b}{2}x^4} $$
for real positive a, b and positive integer n. n is the large parameter. Using Saddle Point Integration
I find saddle points by setting the derivative P'(x) = 0 where
$$ P(x) = n\log(x^2) -ax^2 -\frac{b}{2}x^4$$
In order to do this I never know which saddle point to use ! I see there are two imaginary ones and two real ones. I think I want the one that is positive and real but I have no idea why. (My professor hinted at this one).
By the way the reason there are two real solutions and two imaginary ones is actually not completely obvious to me but I believe that is the case by inspecting the function you get
$$ 0 = n -ax^2 -bx^4$$
this function has two real roots so the other two must be imaginary.
My professor said: "Just plot the integrand at positive
psi and you will see what saddle point to use"
I looked at the plot using coefficient n=a=b= 1 but i didn't get how that tells me which saddle point to use.
Any help would be appreciated !thanks!
|
$\int_0^\infty x^{2n}e^{-ax^2-\frac{b}{2}x^4}~dx$
$=\int_0^\infty x^{2n}e^{-x^2\left(a+\frac{b}{2}x^2\right)}~dx$
$=\int_0^\infty\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^{2n}e^{-\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\left(a+\frac{b}{2}\left(\frac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)^2\right)}~d\left(\dfrac{\sqrt{2a}\sinh x}{\sqrt{b}}\right)$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a\sinh^2x(a+a\sinh^2x)}{b}}\sinh^{2n}x\cosh x~dx$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{2a^2\sinh^2x\cosh^2x}{b}}\left(\dfrac{e^x-e^{-x}}{2}\right)^{2n}\dfrac{e^x+e^{-x}}{2}dx$
$=\dfrac{2^{n+\frac{1}{2}}a^{n+\frac{1}{2}}}{b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\sinh^22x}{2b}}\dfrac{e^x+e^{-x}}{2}\left(\dfrac{(-1)^nC_n^{2n}}{4^n}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}C_{n-k}^{2n}(e^{2kx}+e^{-2kx})}{4^n}\right)dx$
$=\dfrac{\sqrt2a^{n+\frac{1}{2}}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2}{2b}\frac{\cosh4x-1}{2}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{2kx}+e^{-2kx})(e^x+e^{-x})}{2(n+k)!(n-k)!}\right)dx$
$=\dfrac{\sqrt2a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^nb^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!(e^x+e^{-x})}{2(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(e^{(2k+1)x}+e^{-(2k-1)x}+e^{(2k-1)x}+e^{-(2k+1)x})}{2(n+k)!(n-k)!}\right)dx$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh4x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh x}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!(\cosh((2k+1)x)+\cosh((2k-1)x))}{(n+k)!(n-k)!}\right)d(4x)$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\int_0^\infty e^{-\frac{a^2\cosh x}{4b}}\left(\dfrac{(-1)^n(2n)!\cosh\dfrac{x}{4}}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(\cosh\dfrac{(2k+1)x}{4}+\cosh\dfrac{(2k-1)x}{4}\right)}{(n+k)!(n-k)!}\right)dx$
$=\dfrac{a^{n+\frac{1}{2}}e^\frac{a^2}{4b}}{2^{n+\frac{3}{2}}b^{n+\frac{1}{2}}}\left(\dfrac{(-1)^n(2n)!K_\frac{1}{4}\left(\dfrac{a^2}{4b}\right)}{(n!)^2}+\sum\limits_{k=1}^n\dfrac{(-1)^{n+k}(2n)!\left(K_\frac{2k+1}{4}\left(\dfrac{a^2}{4b}\right)+K_\frac{2k-1}{4}\left(\dfrac{a^2}{4b}\right)\right)}{(n+k)!(n-k)!}\right)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$? If $x,y,z>0$ are not equal and positive and if $x+y+z=1$ the expression
$$\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$$
is greater than what quantity?
|
By AM-GM we obtain:
$$\frac{\prod\limits_{cyc}(1+x)}{\prod\limits_{cyc}(1-x)}=\frac{\prod\limits_{cyc}(2x+y+z)}{\prod\limits_{cyc}(x+y)}=\frac{\prod\limits_{cyc}(x+y+x+z)}{\prod\limits_{cyc}(x+y)}\geq\frac{8\prod\limits_{cyc}\sqrt{(x+y)(x+z)}}{\prod\limits_{cyc}(x+y)}=8.$$
The equality occurs for $x=y=z=\frac{1}{3}$, which says that we got the answer: $8$.
|
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|
If $a,b,c > 0$ satisfy $a^2+b^2+c^2=3$ then $\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}$
Given $a,b,c > 0$ satisfying the condition $$a^2+b^2+c^2=3,$$
prove that
$$\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}.$$
Thank you all
|
This can be done automagically using Lagrange multipliers (like most inequalities of this kind...)
Take $f=\frac{a}{b+c+3}+\frac{c}{a+b+3}+\frac{b}{a+c+3}$, $g=a^2+b^2+c^2$, and find the critical points of $f$ subject to the condition that $g=3$. In Mathematica, this can be done as follows:
In[15]:= f = a/(b + c + 3) + b/(a + c + 3) + c/(a + b + 3);
In[16]:= g = a^2 + b^2 + c^2;
In[17]:= Solve[Flatten[{g == 3,
Table[D[f + q g, v] == 0, {v, {a, b, c}}]}],
{a, b, c, q}, Reals]
Out[17]= {{a -> -1, b -> -1, c -> -1, q -> 3/2}, {a -> 1, b -> 1, c -> 1, q -> -(3/50)}}
This shows that there is one critical point (with positive coordinates) at $(a,b,c)=(1,1,1)$.
All we have left is to check that at this point we have a minimum (or a maximum... :P )
|
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|
An identity which applies to all of the natural numbers Prove that any natural number n can be written as $$n=a^2+b^2-c^2$$ where $a,b,c$ are also natural.
|
Consider $n\ge 6$.
If $n$ is odd, $n=2m+1$, then
$$
n = 2m+1 = 2^2 + (m-1)^2 - (m-2)^2;
$$
If $n$ is even, $n=2m$, then
$$
n = 2m = 1^2 + m^2 - (m-1)^2;
$$
Small $n$:
$1=1^2+1^2-1^2$,
$2=3^2+3^2-4^2$,
$3=4^2+6^2-7^2$,
$4=2^2+1^2-1^2$,
$5=4^2+5^2-6^2$.
|
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|
Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
|
We have,
$$\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$$
and
$$\cos(3x) = 4\cos^{3}(x) - 3 \cos(x).$$
Thus, your expression reduces to
$$3 - 4\sin^{2}(x) + 4\cos^{2}(x) - 3$$
which is equal to
$$4 \cos(2x),$$
as $\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$.
|
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|
How to find the limit $x_n=\frac{(a)(a+1)(a+2)..(a+n)}{(b)(b+1)(b+2)..(b+n)}$ Let a,b be positive number then how to find the limit of $x_n=\frac{(a)(a+1)(a+2)..(a+n)}{(b)(b+1)(b+2)..(b+n)}$ when $n\rightarrow \infty $ when a=b obviously the limit is 1 but what about $a<b$ ?
this problem reminded me of the limit $0<A_n=\frac{1*3*5*..*(2n-1)}{2*4*6*..*(2n)} \rightarrow 0 $ I can show this by denote $B_n=\frac{2*4*6*..*(2n)}{3*5*7*..*(2n+1)}$ then $A_n<B_n$ and $A_nB_n=\frac{1}{(2n+1)}$ so $A_n<\sqrt{\frac{1}{2n+1}}\rightarrow0$ when $n\rightarrow \infty$ but the same method seem to be useless in the above problem.I now also consider using delta-epsilon method or even using Gamma function to assure that the limit should be zero.What method should I use?
|
(Since $x_n>x_{n+1}>0$ $\lim_{n\to\infty} x_n$ does exist and $\geq0$.) Applying the inequality $1+x\leq e^x$ (which is valid for any real $x$) we obtain
$$
\frac{a+k}{b+k}=\frac{b+k+a-b}{b+k}=1+\frac{a-b}{b+k}\leq e^{\frac{a-b}{b+k}}.
$$
Using this inequality we obtain
$$
0\leq x_n=\frac{(a)(a+1)(a+2)..(a+n)}{(b)(b+1)(b+2)..(b+n)}\leq\exp\left( (a-b)\sum_{k=0}^{n}\frac{1}{b+k} \right).
$$
Here $a-b<0$ and $\sum_{k=0}^{n}\frac{1}{b+k}$ tends to $\infty$ as $n\to\infty$. So the limit is $0$.
|
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|
Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$ Let $a, b, c, d$ be real numbers. I need to prove this innocent inequality
$$
(a-d)^2+(b-c)^2\geq 1.6
$$
if $a^2+4b^2=4$ and $cd=4$.
I was told that there exist nice and sweet elementary solutions.
|
Using $a^2 + 4b^2 = 4$ and $cd = 4$, we have
\begin{align*}
&(a - d)^2 + (b - c)^2 + \frac12(a^2 + 4b^2 - 4) - \frac{2\sqrt 2}{3}(cd - 4)\\[6pt]
=\,& \frac16(3a - 2d)^2 + \frac13(3b - c)^2 + \frac13(d - c\sqrt 2)^2 + \frac{8\sqrt 2}{3} - 2 \\[6pt]
\ge\, &\frac{8\sqrt 2}{3} - 2\\
>\,& \frac85.
\end{align*}
We are done.
|
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|
Maclaurin series of $f(x) = e^x \sin x$ $$f(x) = e^x \sin x$$
I tried applying the given formula in my book but it didn't work.
The maclaurin for $e^x$ is given as $\displaystyle \sum \frac{x^n}{n!}$ and $\sin x$ $\displaystyle \sum \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
I attempted to multiply them together, failed teh books answer. I tried inputting values for them and then multiplying and that fails as well. Why?
$\displaystyle \frac{x^n}{n!} \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
Fails
|
It’s just like multiplying polynomials.
Just as
$$(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf$$
is the sum of all possible products of one term from the first factor and one term from the second factor, so also is the product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6+\ldots\right)\left(x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\right)\;.$$
Thus, it must be
$$\begin{align*}
x&-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\\
&+x^2-\frac{x^4}6+\frac{x^6}{120}-\frac{x^8}{5040}+\ldots\\
&+\frac{x^3}2-\frac{x^5}{12}+\frac{x^7}{240}-\frac{x^9}{10080}+\ldots\\
&+\frac{x^4}6-\frac{x^6}{36}+\frac{x^8}{720}-\frac{x^{10}}{30240}+\ldots\\
&+\ldots\;.
\end{align*}$$
Of course some of these terms can be combined, since they involve the same power of $x$, but we can already see that the first four powers of $x$ that will appear in this product are $x,x^2,x^3$, and $x^4$: the constant term is $0$. What products of one term of
$$1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+\ldots$$
and one term of $$x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots$$ are of the form $ax^k$ with $k=1,2,3$, or $4$? Obviously the second factor can’t be an $x^5$ term or higher. The second factor is always at least $x$, so the first factor can’t be an $x^4$ term or higher. Thus, the only partial products that contribute to the first four terms of the product are ones found in the polynomial product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6\right)\left(x-\frac{x^3}6\right)\;,$$
and not all of those will be needed. Specifically, we need only
$$1\cdot x-1\cdot\frac{x^3}6+x\cdot x-x\cdot\frac{x^3}6+\frac{x^2}2\cdot x+\frac{x^3}6\cdot x\;;$$
every other partial product yields a power of $x$ higher than the fourth power and therefore does not contribute to the first four terms of the desired series.
|
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|
Triple integral (check solution) The function given is
$f(x,y,z) = \displaystyle\frac{1}{(x+y+z+1)^2}$
$D = \{(x,y,z) \in \mathbb{R}^3 : x \geq 0, y \geq 0, z \geq 0, x+y+z \leq 1 \}$
It seems that the domain would be the under a plane that contains the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$; then if i consider the plane $XZ$ i have the set $D_{XZ}= \{(x,z) : z=1-x, 0 \leq x \leq 1 \}$ also $D_{XY}= \{(x,z) : y=1-x, 0 \leq x \leq 1\}$ and $D_{YZ}= \{(x,z) : z=1-y, 0 \leq y \leq 1\}$.
Then i could set $0\leq x\leq z-1$, $0\leq y\leq z-1$ and $0\leq z\leq 1$, and the integral would be given by:
$$\begin{align*} \displaystyle\int\displaystyle\int\displaystyle\int_D f(x,y,z)\,dx\,dy\,dz
&= \displaystyle\int\displaystyle\int\displaystyle\int_D \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz
\\&= \displaystyle\int_0^1\displaystyle\int_0^{1-z}\displaystyle\int_0^{1-z} \displaystyle\frac{1}{(x+y+z+1)^2}\,dx\,dy\,dz
\\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{(1-z)+y+z+1)}- \displaystyle\frac{1}{(0)+y+z+1)}\bigg)\,dy\,dz
\\ &= -\displaystyle\int_0^1\displaystyle\int_0^{1-z} \bigg(\displaystyle\frac{1}{y+2}- \displaystyle\frac{1}{y+z+1}\bigg)\,dy\,dz
\\ &= -\displaystyle\int_0^1 (\log(y+2)-\log(y+z+1))|_{y=0}^{y=1-z}\,dz
\\ &= -\displaystyle\int_0^1 \{[(\log(3-z)-\log(2)] - [\log(2)-\log(z+1)]\}\,dz
\\ &= -\displaystyle\int_0^1 [(\log(3-z)+\log(z+1)-2\log(2)]\,dz
\\ &= 2\log(2)+\{[(3-z)\log(3-z) \\ & \;\;\;- (3-z)]|_{z=0}^{z=1} - [(z+1)\log(z+1)-(z+1)]|_{z=0}^{z=1}\}
\\ &= 2\log(2) - 3\log(3) +4
\end{align*} $$
Did i solved it correctly?
|
I don't think so. I think the integral looks like
$$\int_0^1 dx \, \int_0^{1-x} dy \, \int_0^{1-x-y} dz \frac{1}{(1+x+y+z)^2}$$
which is equal to
$$-\int_0^1 dx \, \int_0^{1-x} dy \, \left [\frac{1}{2} - \frac{1}{1+x+y} \right ]$$
which is equal to
$$-\int_0^1 dx \, \left [\frac12 (1-x) - \log{2} + \log{(1+x)} \right ] $$
which in turn is equal to
$$-\frac14 + \log{2} - \left [ (1+x) \log{(1+x)} - (1+x)\right]_0^1 = \frac{3}{4}-\log{2}$$
This result was verified in Mathematica.
|
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|
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\cos2\alpha\sin\alpha$ can be expressed in three ways:
$$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha - \sin\alpha$$
$$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha - 2\sin^3\alpha$$
I tried adding these, but nothing came close to the required answer.
So then I tried calculating $\sin4\alpha$ (from the required answer):
$$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$
$$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$
Still looking at the answer, I calculated $\cos4\alpha$
$$\cos4\alpha = 1- \sin^2(2\alpha)$$
If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$
then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$
and $$\cos4\alpha = 1 - 4\sin^2\alpha\cos^2\alpha$$
and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$
I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.
|
use $$\sin(A+B) = \sin A\cos B + \cos A\sin B $$on LHS and
$$\sin(A-B) =\sin A\cos B - \cos A\sin B$$ on RHS
so $$\sin(3\alpha) = \sin(3\alpha)$$
|
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|
Conversion of sum of series into product form: $\sum_{n=1}^\infty \frac1{n(n+1)} = \frac12 \prod_{n=2}^\infty \left( 1+\frac{1}{n^2-1} \right)$ Show that the following series and product are equivalent:
$$
\sum_{n=1}^\infty \left[ \dfrac{1}{n(n+1)} \right] = \dfrac{1}{2} \prod_{n=2}^\infty \left[ 1+\dfrac{1}{n^2-1} \right]
$$
Thought of solving this through induction as the relation is true for 1....but don't know how to proceed with the n part...Please help!!
|
$\frac 1{n(n+1)}= \frac1n-\frac1{n+1}$ so this is a telescoping sum, almost all terms will cancel out:
$$ \sum_{n=1}^N \frac1{n(n+1)}= 1 - \frac12 + \frac12 - \frac13 + \dots + \frac1{N-1}-\frac1N + \frac1N-\frac1{N+1}=1-\frac1{N+1}.$$
Similarly, you can see that in the product of the terms of the form $\frac{n^2}{n^2-1}=\frac{n^2}{(n+1)(n-1)}$ a lot of terms cancel out, you only need to check which of them.
$$\prod_{n=2}^N \frac{n^2}{(n+1)(n-1)} = \frac{2^2}{3\cdot 1} \cdot \frac{3^2}{2\cdot 4} \dots \frac{(N-1)^2}{(N-2)N} \cdot \frac{N^2}{(N+1)(N-1)}=
2 \cdot \frac{N}{N+1}.$$
Now it only remains to look what happens for $N\to\infty$ or to notice that already partial sums are the same as partial products, as pointed out in achille hui's comment.
|
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|
Gauge fractions with exponents - No Calculator How does one (without the use of Calculator) determine that $5/6$ is less than $(35/36)^6$? How is this done mentally?
|
If you look at your second number, it can be written as
$$( 1 - \frac{1}{36} )^6 \ = \ 1 \ - \ 6 \cdot 1^5 \cdot \frac{1}{36} \ + \ \binom 62 \cdot 1^4 \cdot (\frac{1}{36})^2 \ - \ ... $$
with the remaining (unwritten) terms being very small (the first two terms equal $ \ \frac{5}{6} \ $ ) . So $ \ \frac{5}{6} \ $ ends up being slightly smaller than $ ( 1 - \frac{1}{36} )^6 \ $ (by a little less than $ \ 15 \cdot \frac{1}{36^2} \ = \ \frac{5}{432} \ \approx \ 0.0116 \ $ ) .
[This is what Daniel Fischer is briefly describing.]
|
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|
Difficult Gaussian Integral Involving Two Trig Functions in the Exponent: Any Help? Here's the integral:
$$\int_d^e \exp\left(-a\left((b+c)\cos(x)-\sqrt{b^2 - (b+c)^2 \sin^2(x)}\right)^2 \right) \, dx$$
I've tried using Mathematica: it fails.
Can anyone help evaluate it?
Perhaps some kind of series expansion of the exponential might help...
Have any of you dealt with these kinds of Gaussians before?
Offering Respect,
Alex
|
Assume $b\neq0$ , $c\neq0$ and $b+c\neq0$ to maintain the key meaning of the question.
$\int_d^e\exp\left(-a\left((b+c)\cos x-\sqrt{b^2-(b+c)^2\sin^2x}\right)^2 \right)~dx$
$=\int_d^e\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n\left((b+c)\cos x-\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2n}}{n!}dx$
$=\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^nC_{2m}^{2n}a^n(b+c)^{2n-2m}\cos^{2n-2m}x\left(\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2m}}{n!}dx-\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^nC_{2m-1}^{2n}a^n(b+c)^{2n-2m+1}\cos^{2n-2m+1}x\left(\sqrt{b^2-(b+c)^2\sin^2x}\right)^{2m-1}}{n!}dx$
$=\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{(-1)^n(2n)!a^n(b+c)^{2n-2m}\cos^{2n-2m}x\left(b^2-(b+c)^2+(b+c)^2\cos^2x\right)^m}{n!(2m)!(2n-2m)!}dx-\int_d^e\sum\limits_{n=0}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^n(2n)!a^n(b+c)^{2n-2m+1}\cos^{2n-2m+1}x\left(b^2-(b+c)^2\sin^2x\right)^{m-\frac{1}{2}}}{n!(2m-1)!(2n-2m+1)!}dx$
Note that both $\int\cos^{2n-2m}x\left(b^2-(b+c)^2+(b+c)^2\cos^2x\right)^m~dx$ and $\int\cos^{2n-2m+1}x\left(b^2-(b+c)^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ , where $m$ and $n$ are any non-negative integers have close-form.
|
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|
A further question on the irrationality of $x^2+y^2=3$ (Apologies for a further question on the same problem)
On page 79 of Julian Harvil's book "The Irrationals" he sets out to prove (by contradiction) that all the points on the circle described by $x^2+y^2=3$ are irrational.
To paraphrase his proof:
*
*Let $\left(\frac{p}{q},\frac{r}{s}\right)$ be a point on the circle, where $p,q,r,s$ are all integers
*Hence $(ps)^2 + (qr)^2 = 3(qs)^2$
*We can restate this as $a^2 + b^2 = 3c^2$ and we know that one of $a,b$ must be odd and the other even
*So $a^2 + b^2 = 4(m^2 + n^2 + n) + 1 = 4N + 1 = 3c^2$
*We can say that $c$ could be of the form $4M, 4M + 1, 4M + 2$ or $4M + 3$
And this is where I get lost (this time):
*This means $c^2$ is of the form $4N, 4N + 1, 4N, 4N + 1$ so $3c^2$ must be of the form $4N, 4N + 3, 4N, 4N + 3$
Could someone explain the reasoning for this last line?
|
Remember that the square $c^2$ of an integer $c$ is either $\equiv 0\pmod 4$ (if $c$ is even) or $\equiv 1\pmod 8$ (if $c$ is odd); in fact $1\pmod 4$ instead of $1\pmod 8$ is good enough here, as we obtain $3c^2\equiv 0\pmod 4$ or $3c^3\equiv 3\pmod 4$, but definitelys not $3c^3\equiv 1\pmod 4$.
|
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|
What is the probability of getting a sum of 7 or at least one 5 when you roll two die Please tell me how to approach this problem.
(Sum of 7) = {4+3, 3+4, 6+1, 1+6, 5+2, 2+5} = 6
(At Least one 5) = {1+5, 2+5, 3+5, 4+5, 6+5, 5+1, 5+3, 5+4, 5+5, 5+6} = 10
so the answer will be 16/36 = 4/9 ?
|
You double counted some possibilities (the two events are not mutually exclusive!). By taking the union of the sets of desirable outcomes, we obtain:
$$ \left\{\begin{align*}
&(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),\\
&(5,1),(5,2),(5,3),(5,4),~~~~~~~~~~~(5,6),\\
&(1,6),(6,1),(3,4),(4,3)~~~~~~~~~~~~~~~~~~~~~~~
\end{align*}\right\}$$
so the probability is:
$$
\dfrac{15}{36}=\dfrac{5}{12}
$$
|
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Initial value problem differential equation $y' = (x-1)(y-2)$ $$y' = (x-1)(y-2)$$
$y(2)= 4$
$$\frac{1}{y-2}dy = (x-1)dx$$
$$\int \frac{1}{y-2}dy =\int (x-1)dx$$
$$\ln(y-2) = \frac{x^2}{2} - x + c$$
$$y - 2 = e^{\frac{x^2}{2} - x + c} $$
$$y = e^{\frac{x^2}{2} - x + c} + 2$$
plug in the inital value
$$y = e^{\frac{2^2}{2} - 2 + c} + 2$$
$$y = e^c+ 2$$
I feel like this is wrong anyways, so where did I go wrong?
|
Continuing from your last step:
$$
y = e^c + 2 = 4\\
\Rightarrow c = \ln2
$$
finally giving
$$
y = e^{\frac{1}{2}x^2-x+\ln2} +2
$$
or equivalently,
$$
y = 2e^{\frac{1}{2}x^2-x} +2
$$
|
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|
Convergence of sum related to $\sum_k 1/p_k$ Motivation (skip if desired): We can show that $$S =\frac{1}{2}+ \frac{1}{2\cdot 3}+ \frac{(2)}{2\cdot (3)\cdot5}+\frac{(2\cdot 4)}{2\cdot(3\cdot5)\cdot 7}+\frac{(2\cdot4\cdot6)}{2\cdot(3\cdot5\cdot7)\cdot11}+...$$etc. converges.
We know
$$\frac{1}{2}\sum_{k=1}\frac{1}{p_k} = \infty$$ and
$$ \sum_{n=1}^{\infty} \frac{1}{n} = \infty$$
and that the sum of factors (in parentheses above)
$$ \frac{2}{3} + \frac{2\cdot 4}{3\cdot 5} + \frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}+\frac{2\cdot 4\cdot6\cdot 10}{3\cdot 5\cdot 7\cdot 11}+... = \infty.$$
Question: Let $a(3) = \frac{2}{3}, a(4) = \frac{2\cdot 4}{3\cdot 5},$ etc. Does $$T = \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{2}{2\cdot3\cdot 5}+\frac{2\cdot 4}{2\cdot3\cdot5\cdot6}+\frac{2\cdot 4\cdot6}{2\cdot3\cdot5\cdot7\cdot7}+~ ...~\text{etc.}$$
$$= \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{1}{2}\sum_{k=3}\frac{a(k)}{k+2} $$
converge?
Note: $a(n) = \prod_{k=3}^n \frac{p_{k-1}-1}{p_{k-1}}.$
|
As the commenters have suggested: Mertens proved the asymptotic formula
$$
\prod_{p\le x} \bigg( 1-\frac1p \bigg) \sim \frac{e^{-\gamma}}{\log x}
$$
(where $\gamma$ is Euler's constant). Therefore
$$
\prod_{k=3}^{n} \frac{p_{k-1}-1}{p_{k-1}} = 2 \prod_{p\le p_{n-1}} \bigg( 1-\frac1p \bigg) \sim \frac{2e^{-\gamma}}{\log p_{n-1}} \sim \frac{2e^{-\gamma}}{\log n}.
$$
Therefore
$$
\frac{a(k)}{k+2} \sim \frac{2e^{-\gamma}}{k \log k},
$$
and so $\sum_k \frac{a(k)}{k+2}$ diverges by the limit comparison test.
|
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|
Proving that a discrete stochastic variable is binomial distributed. Given a discrete stochastic variables, with the probability function;
$$
p_{X}\left(x\right)=\left\{
\begin{array}{cc}
\frac{1}{4} & \text{if } x = -1 \\
\frac{1}{4} & \text{if } x = 0 \\
\frac{1}{2} & \text{if } x = 1 \\
0 & \text{otherwise}
\end{array}
\right.
$$
How do I prove that;
$$|X| \sim b\left(1, \frac{3}{4}\right)$$
What I've done so far: Found $P(|X| = 1) = \frac{3}{4}$, and then I've unsuccessfully tried to derive the probability function (above) from the binomial probability function (below);
$$
p\left(x\right) = \left(
\begin{array}{c}
n \\ x
\end{array}\right) \pi^{x} \left(1 - \pi\right)^{n-x}
$$
However without any luck, as $|X| \neq X$.
So my question is somewhat simple; How would I go about proving that a discrete stochastic variable is binomial distributed?
|
You already did all of the work. You showed that
$$
p_{\left|\mathbf{X}\right|}\left(x\right)
=
\left(\begin{array}{c}
1\\
x
\end{array}\right)\frac{3}{4}^{x}\frac{1}{4}^{1-x}
=
\begin{cases}
\frac{3}{4} & \text{if }x=1\\
\frac{1}{4} & \text{if }x=0
\end{cases}
$$
|
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|
Find $a$ such that $P(X\le a)=\frac 1 2$
Given probability density function $f(x)=\begin{cases} 1.5(1-x^2),&0<x<1\\0& \mbox{otherwise}\end{cases}$, calculate for which '$a$', $P(X\le a)=\frac 1 2$ (the solution is $2\cos\left(\frac{4\pi}{9}\right))$.
I calculated the integral and got $P(X\le a)=\displaystyle\int_0^af_X(t)dt=1.5a\left(1-\frac {a^2}{3}\right) \stackrel?= \frac 1 2$. I substituted $a=\cos\theta$ and then got $$\frac 1 2=1.5\cos\theta-0.5\cos^3\theta$$$$1=3\cos\theta -\cos\theta(\cos^2\theta)=3\cos\theta-\cos\theta(\cos^2\theta)=3\cos\theta-\cos\theta\left(\frac {1-\cos 2\theta} 2\right)$$ and then using some identites I got $$-2.75\cos\theta-0.25\cos3\theta+1=0$$ or after multiplying by (-4 )$$11\cos\theta+\cos3\theta-4=0$$ How can I solve this equation?
|
All you should do is just finding roots of the following polynomial which lie within $[0,1]$
$$
1.5a\left(1-\frac {a^2}{3}\right)-\frac 1 2 \\
\Rightarrow a^3-3a+1
$$
Now you can solve it using methods available for solving cubic polynomials or you can write it in this way :
$$
\begin{align}
a^3-3a+1 &= (a-x_1)(a-x_2)(a-x_3)\\ &= a^3 -(x_1 + x_2+x_3)a^2+(x_1x_2+x_1x_3+x_2x_3)a-x_1x_2x_3
\end{align}
$$
By equating coefficients of $a^n, \quad n=0,1,2,3$ on two sides , You will find that :
$$
\begin{cases}
x_1 \approx -1.8794\\
x_2 \approx 0.3473\\
x_3 \approx 1.5321
\end{cases}
$$
And obviously the second one is desired one (why?).
Also following the method you've used in question: set $ a = 2\cos \theta\\$
$$
\begin{align}
(2\cos \theta)^3-3(2\cos \theta) + 1
& = 2 (4\cos^3 \theta-3\cos \theta) + 1\\
&= 2\cos 3\theta + 1\\
& = 0
\end{align}\\
\Rightarrow \cos 3\theta = -\frac12 \Rightarrow \theta
= \frac{2\pi}9 (3n \pm 1) , \qquad n \in \mathbb Z
$$
Note that taking $a=2\cos \theta$ restrict the solutions to the interval $[-2,2]$ and since $a \in [0,1]$ are desired (otherwise the first equality doesn't hold) this substitution would not miss any desired solutions.
|
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|
Some identities with the Riemann zeta function Can someone either help derive or give a reference to the identities in Appendix B, page 27 of this, http://arxiv.org/pdf/1111.6290v2.pdf
Here is a reproduction of Appendix B from Klebanov, Pufu, Sachdev and Safdi's $2012$ preprint (v2) 'Renyi Entropies for Free Field Theories' (from the source at arxiv.org and hoping there is no problem citing it here...(RM)).
B Useful mathematical formulae
In this section we present some useful mathematical formulae. We begin with zeta function identities.
For $0 < a \leq 1$ we have the identity
$$\tag{B.1} \zeta(z, a) = \frac{2 \Gamma(1 - z)}{(2 \pi)^{1-z}} \left[\sin \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^{1-z}}
+ \cos \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\sin 2 \pi a n}{n^{1-z}} \right] \,$$
Taking derivatives at $z=0, -1, -2$ gives
\begin{align}
\zeta'(-2, a) &= - \frac{1}{4 \pi^2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^3}
- \frac{1}{4 \pi^3} \sum_{n=1}^\infty \frac{(2 \log (2 \pi n) + 2 \gamma - 3) \sin 2 \pi a n}{n^3} \,, \\
\tag{B.2}\zeta'(-1, a) &= \frac{1}{4 \pi} \sum_{n=1}^\infty \frac{\sin 2 \pi q n}{n^2}
- \frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma - 1) \cos 2 \pi a n}{n^2} \,, \\
\zeta'(0, a) &= \frac{1}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n}
+ \frac{1}{\pi} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma) \sin 2 \pi a n}{n} \,.
\end{align}
Two other useful identities are the regularized sums
\begin{align}
\tag{B.3}\sum_{n \in \mathbb{Z}} \log \left( \frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \sinh (\pi q |a|) \right] \,, \\
\sum_{n \in \mathbb{Z} + \frac 12} \log \left(\frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \cosh (\pi q |a|) \right] \,.
\end{align}
These sums follow from the more general formula
$$\tag{B.4}\sum_{n \in \mathbb{Z}} \log \left( \frac{(n + \alpha)^2}{q^2} + a^2 \right)
= \log \left[2 \cosh (2 \pi q |a|) - 2 \cos (2 \pi \alpha) \right] \,.$$
This relation in turn follows from the Poisson summation formula
$$\tag{B.5} \frac{1}{ 2 \pi q} \sum_{n \in \mathbb{Z}} \hat f \left( \frac{n + \alpha}{q} \right) =\sum_{k \in \mathbb{Z}} e^{-i 2 \pi k \alpha} f(2 \pi q k) \,$$
applied to
$$\tag{B.6}\hat f(\omega) = \log \left( \omega^2 + a^2 \right) \,.$$
For $t \neq 0$ one can simply calculate the inverse Fourier transform of $\hat f$:
$$\tag{B.7}f(t) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} e^{-i \omega t} \log \left(\omega^2 +a^2 \right) = - \frac{e^{-|a|\;|t|}}{|t|} \,.$$
The case $t=0$ requires special care because the expression for $f(0)$ is divergent and requires regularization:
$$\tag{B.8}f(0) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} \log \left(\omega^2 +a^2 \right)
= -\frac{d}{ds} \int \frac{d\omega}{2 \pi} \frac{1}{\left(\omega^2 +a^2 \right)^s} \Biggr\rvert_{s=0}
= |a| \,.$$
Using $(B.6)-(B.8)$ one can show that $(B.5)$ reduces to $(B.4)$.
|
$(B.1)$ is the functional equation of the Hurwitz zeta function and Knopp and Robins' proof is available here.
$$\tag{B.1}\zeta(z,a)=\frac{2\,\Gamma(1-z)}{(2\pi)^{1-z}}\left[\sin\frac {z\pi}2\sum_{n=1}^\infty\frac{\cos2\pi an}{n^{1-z}}+\cos\frac {z\pi}2\sum_{n=1}^\infty\frac{\sin2\pi an}{n^{1-z}}\right]$$
The author of your paper proposed a derivation of $(B.3)$ in reverse order since starting from $B.7)$. But let's try another derivation using the logarithm of the infinite products with $x:=\pi\,q\,|a|$ :
$$\tag{1}\sinh(x)=x\prod_{k=1}^\infty \left(1+\frac {x^2}{\pi^2k^2}\right)$$
$$\tag{2}\cosh(x)=\prod_{k=1}^\infty \left(1+\frac {4\,x^2}{\pi^2(2k-1)^2}\right)$$
which may be found for example here or in the online references.
The derivation will not be direct since these products are convergent while $(B.3)$ is clearly divergent and needs some regularization.
\begin{align}
\frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\frac 12\log\bigl(a^2\bigr)+\sum_{n=1}^\infty\log\left(\frac {n^2}{q^2}+a^2\right)\\
&=\log|a|+\sum_{n=1}^\infty\log\frac {n^2}{q^2}+\log\left(1+\frac{q^2a^2}{n^2}\right)\\
\tag{3}&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+2\sum_{n=1}^\infty\log n-\log q\\
\end{align}
The last sum at the right (as often in QFT) is heavily divergent so let's use zeta regularization to rewrite it in a finite form :
$$f(z):=\sum_{n=1}^\infty\frac{\log n-\log q}{n^z}=-\zeta'(z)-\zeta(z)\log q,\quad \text{for}\ \Re(z)>1$$
(since $\;\displaystyle\frac d{dz}n^{-z}=\frac d{dz}e^{-z\log n}=-\frac{\log n}{n^z}$)
From this we deduce the 'zeta regularized sum' (using analytic extension of $f(z)$ down to $0$) :
$$\sum_{n=1}^\infty\log n-\log q=\lim_{z\to 0^+}f(z)=-\zeta'(0)-\zeta(0)\log q=\frac{\log 2\pi}2+\frac 12\log q=\frac{\log 2\pi q}2$$
and get :
\begin{align}
\frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+\log 2\pi +\log q\\
&=\log\left[2\,\pi q|a|\prod_{n=1}^\infty \left(1+\frac{q^2a^2}{n^2}\right)\right]\\
\\
&=\log[2\,\sinh(\pi\;q\,|a|)],\quad\text{using}\ (1)\ \text{for}\;\;q\,|a|=\frac x{\pi}\\
\end{align}
Which is the first part of $(B.3)$ :
$$\tag{B.3}\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)=2\;\log[2\,\sinh(\pi\;q\,|a|)]$$
I'll let you get the corresponding equation for $\cosh$.
Hoping this clarified things.
|
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|
Prove the Inequality: $\sum\frac{x^3}{2x^2+y^2}\ge\frac{x+y+z}{3}$ Let $x, y, z>0$. Prove that:
$$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
|
I have a direction which may be fruitful but didn't work out all cases. Rewrite the inequality as:
$$
\frac{x^3}{2x^2+y^2}-\frac{x}{3}+\frac{y^3}{2y^2+z^2}-\frac{y}{3}+\frac{z^3}{2z^2+x^2}-\frac{z}{3}\ge0
$$
Let $f(t)=\frac{1-t^2}{2+t^2}$, then the inequality is equivalent to: $xf(a)+yf(b)+zf(c)\ge0$, where $a=y/x, b=z/y $ and $c=x/z$.
Case 1. Assume that $a,b,c \in (0,2]$. Note that $f(t)\ge \frac{1-t}{2}$ for each $t \in (0,2]$. Then we have $$xf(a)+yf(b)+zf(c) \ge \frac{x}{2}(1-\frac{y}{x})+\frac{y}{2}(1-\frac{z}{y})+\frac{z}{2}(1-\frac{x}{z})=0$$
Case 1 is settled.
Case 2 $x\ge y\ge z$ and $x\ge 2z$ remains to be settled.
Case 3 $z\ge y\ge x$ and $z\ge 2y$ OR $y\ge 2x$ remains to be settled.
|
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|
A first order non-linear ordinary differential equation How do we solve the first order non-linear ordinary differential equation ,
$$y^3=xy^2 \frac{dy}{dx} + x^4 \Big(\frac{dy}{dx}\Big)^2$$
|
Firstly we note that: $$y(x)=0 \tag{Solution 1}$$ is a solution, hereafter we take $y(x)\ne0$.
Substitute: $$y(x)=\dfrac{1}{h(x)}$$ into:$$ y \left( x \right) ^{3}=x y \left( x \right)
^{2}{\frac {d}{dx}}y \left( x \right) +{x}^{4} \left( {\frac
{d}{dx}}y \left( x \right) \right) ^{2}\tag{1}$$
to get:
$$\begin{align}
h \left( x \right) &=-x{\frac {d}{dx}}h \left( x \right) +{x}^{4}
\left( {\frac {d}{dx}}h \left( x \right) \right) ^{2} \\
h \left( x \right) +\dfrac{1}{4x^2}&={x}^{4} \left( {\frac {d}{dx}}h
\left( x \right) -\dfrac{1}{2x^3} \right) ^{2}\\
&={x}^{4} \left( {\frac {d}{dx}}\left(h
\left( x \right) +\dfrac{1}{4x^2}\right) \right) ^{2}\tag{2}
\end{align}$$
and we see that:
\begin{align} h(x)&=-\dfrac{1}{4x^2}\\
y(x)&=-4x^2 \tag{Solution 2}\end{align}
is a solution (as found by @Daniel Littlewood and @user64494). Note: in $(2)$ we divided by $1/h(x)^4$ under the assumption it was not zero everywhere, if it were we would recover $\text{Solution 1}$. Then substitute:
$$h(x)=g \left( x \right) -\dfrac{1}{4x^2}$$
into $(2)$ to get:
$$g \left( x \right) ={x}^{4} \left( {\frac {d}{dx}}g \left( x \right)
\right) ^{2}\tag{3}$$
and change the variable to $x=\dfrac{1}{\xi}$ in $(3)$ and write $g(x)=f(x)^2$ to get:
\begin{align}g \left( \xi \right) &=\left( {\frac {d}{d\xi}}g \left( \xi \right)
\right) ^{2}\\
f\left( \xi \right)^2 &=4f\left( \xi \right)^2\left( {\frac {d}{d\xi}}f\left( \xi \right)
\right) ^{2}\tag{4}\end{align}
we then divide by $f(\xi)^2$ under the assumption it is not zero everywhere, if it were we would recover $\text{Solution 2}$, and we have:
\begin{align} {\frac {d}{d\xi}}f \left( \xi \right)&=\pm\dfrac{1}{2}\\
f \left( \xi \right)&=C\pm\dfrac{\xi}{2} \tag{5}\end{align}
but by noting that $g=f^2$ we can encompass either sign possibility in the arbitrary constant $C$ and write:
$$\begin{align}
g(\xi)&=\left(C+\dfrac{\xi}{2}\right)^2 \\
g(x)&=\left(C+\dfrac{1}{2x}\right)^2\tag{6}
\end{align}$$
Working backwards we then have:
$$y \left( x \right) ={\frac {1}{\left(\dfrac{C}{2}\right)^2 +\left(\dfrac{C}{2}\right)\dfrac{1}{x}}}. \tag{7}$$
To fix the constant in $(7)$ we find that the starting value is useless because, irrespective of $C$:
$$\lim_{x\rightarrow0}y(x)=0$$
but we can use the starting value of the derivative (Neumann boundary condition) to find that:
$$C=\dfrac{2}{y'(0)}$$
$$y \left( x \right) ={\frac {y'(0)^2}{1 +y'(0)\dfrac{1}{x}}}={\frac {xy'(0)^2}{x +y'(0)}} \tag{Solution 3}$$
so this solution is bilinear in $x$. We also note that $\text{Solution 1}$ is encompassed in $\text{Solution 3}$ but $\text{Solution 2}$ is distinct.
|
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|
Inequality related to the continued fraction expansion of $\sqrt 3$ I am working on a problem related to the continued fraction expansion of $\sqrt 3$. If $p_k$ and $q_k$ denote the numerator and denominator, respectively, of the $k$th convergent, I should show that
$$\left|\sqrt{3}-\frac{p_{2n+1}}{q_{2n+1}}\right| \lt \frac{1}{2\sqrt{3}q_{2n+1}^2}\;.$$
I have determined that the continued fraction expansion of $\sqrt 3$ is $[1;\overline{1,2}]$ and I am able to show the equality apart from the factor $2\sqrt3$ in the denominator.
Any suggestions?
Thanks in advance!
|
From the period length of $2$, you obtain that for all $n$, you have
$$p_{2n+1}^2 - 3q_{2n+1}^2 = 1,$$
and from that you can deduce
$$\begin{align}
\frac{p_{2n+1}}{q_{2n+1}} - \sqrt{3} &= \frac{p_{2n+1} - \sqrt{3}q_{2n+1}}{q_{2n+1}}\\
&= \frac{(p_{2n+1} - \sqrt{3}q_{2n+1})(p_{2n+1} + \sqrt{3}q_{2n+1})}{q_{2n+1}(p_{2n+1} + \sqrt{3}q_{2n+1})}\\
&= \frac{1}{\left(\frac{p_{2n+1}}{q_{2n+1}}+\sqrt{3}\right)q_{2n+1}^2}\\
&< \frac{1}{2\sqrt{3}q_{2n+1}^2}
\end{align}$$
since $\frac{p_{2n+1}}{q_{2n+1}}>\sqrt{3}$ (which you can either deduce from the general fact that odd convergents are larger than [or equal to] the value of the continued fraction, or from the fact mentioned above).
|
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|
A $4$ variable inequality If $a,b,c,d$ are positive numbers such that $c^2+d^2=(a^2+b^2)^3$, prove that
$$\frac{a^3}{c} + \frac{b^3}{d} \ge 1,$$
with equality if and only if $ad=bc$.
Source: Don Sokolowsky, Crux Mathematicorum, Vol. 6, No. 8, October 1980, p.259.
|
Because by Holder
$$\left(\frac{a^3}{c}+\frac{b^3}{d}\right)^2(c^2+d^2)\geq(a^2+b^2)^3$$
and we are done!
|
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|
Probability of first and second drawn balls of the same color, without replacement I have an urn with 10 balls: 4 red and 6 white. What is the probability that the first two drawn balls have the same color?
Approach No. 1:
$|U| = 10! = 3628800$
White balls on first and second draw: on the first and second draw, I can pick any of the 6 white balls on the first draw, and 5 on the second draw. After that, I have 8 balls left, of which I don't care what order they are in. This will become $6\times5\times8\times7\times6\times5\times4\times3\times2\times1=1209600$.
$\left(\frac{1209600}{3628800} \right)=\frac{1}{3}$
If we have red balls on the first and second draw, then: I can pick any of the 4 red balls on the first draw, and on the second draw any of the left over 3. This will be$4\times3\times8\times7\times6\times5\times4\times3\times2\times1=483840$.
$\left(\frac{483840}{3628800} \right)=\frac{2}{15}$
The sum of these two will be: $\frac{1}{3}+\frac{2}{15}=\frac{7}{15}$
Approach number two:
Chance of a red ball on first draw: $\frac{4}{10}$. Chances of a red ball on second draw: $\frac{3}{9}.$
$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$.
Chance of a white ball on first draw: $\frac{6}{10}$. Chances of a white ball on second draw: $\frac{5}{9}$.
$\frac{6}{10}\times\frac{5}{9}=\frac{5}{15}$.
$\frac{5}{15}+\frac{2}{15}=\frac{7}{15}$
What I want to know, is this answer correct? Somehow I can't get it straight in my head that this is the right answer, since the $8!$ seems a bit weird to me. I mean don't the chances of the other red balls in the case of red balls on first and second draw count?
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S = {Set of all possible combination of 2 balls from the available 4 Red and 6 White Balls}
n(S) = $^{10} P_{2}$ ways = $10 \times 9 = 90$ (Since, 2 balls can be selected from $10$ in $^{10} P_{2}$ ways)
Now,
E = Both are of same colour
n(E) = $2$ Red balls can be selected from $4$ in $^4 P_{2}$ ways AND $2$ White balls can be selected from $6$ in $^6 P_{2}$ ways) Total = $^4 P_{2} + ^6 P_{2} =(12 + 30) =42$ ways.
P(E) = $\frac{n(E)}{n(S)} = \frac{42}{90} = \frac{7}{15}$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/472819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^2 + ax + 1 = 0$$
has at least one real solution.
Thanks in advance.
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[The following solution relies on the ideas in the previous two answers.]
Let the roots of $x^4+ax^3+bx^2+ax+1=0$ be given by $c, \frac{1}{c}, d, \frac{1}{d}$ where $c\in\mathbb{R}$, and
let $t=c+\frac{1}{c}$ and $l=d+\frac{1}{d}$.
Then $|t|\ge2$ since $c+\frac{1}{c}\ge2$ if $c>0$ and $c+\frac{1}{c}\le2$ if $c<0$.
Since $x^4+ax^3+bx^2+ax+1=(x-c)(x-\frac{1}{c})(x-d)(x-\frac{1}{d})=(x^2-tx+1)(x^2-lx+1)$,
we have that $a=-(t+l)$ and $b=2+lt$; so
$a^2+b^2=(t+l)^2+(2+lt)^2=t^2+2lt+l^2+4+4lt+l^2t^2=(l^2+1)t^2+6lt+l^2+4$.
We consider two cases:
$\mathbf{A)}$ If $d\in\mathbb{R}$, then $|l|\ge2$ (for the same reason that $|t|\ge2$); so
$\;\;\;\;a^2+b^2=(l^2+1)t^2+6lt+l^2+4\ge20+6lt+8\ge4$ $\;$(since $|lt|\le4\implies lt\ge-4$).
$\mathbf{B)}$ If $d\not\in\mathbb{R}$, then $\frac{1}{d}=\bar{d}\implies|d|=1$, so $d=e^{i\theta}=\cos\theta+i\sin\theta$ for some $\theta\in\mathbb{R}$. Then
$\;\;\;\;l=d+\frac{1}{d}=d+\bar{d}=2\cos\theta$, so $|l|\le2$.
With $l$ fixed, let $f(t)=a^2+b^2=(l^2+1)t^2+6lt+l^2+4$.
Since the vertex of the graph of $f$ is at $t=\frac{-3l}{l^2+1}$,
and $-2<\frac{-3l}{l^2+1}<2\;\;\;\;$ (since $2l^2+3l+2>0$ and $2l^2-3l+2>0$),
the minimum value of f for $|t|\ge2$ will occur when $t=2$ or $t=-2$:
1) If $t=2$, $a^2+b^2=4(l^2+1)+12l+l^2+4=5l^2+12l+8$, and
$\;\;\;g(l)=5l^2+12l+8$ has its minimum when $l=-\frac{6}{5}$, with $g(-\frac{6}{5})=\frac{4}{5}$.
$\;\;\;\;$(Notice that $|-\frac{6}{5}|\le2$).
2) If $t=-2$, $a^2+b^2=4(l^2+1)-12l+l^2+4=5l^2-12l+8$, and
$\;\;\;g(l)=5l^2-12l+8$ has its minimum when $l=\frac{6}{5}$, with $g(\frac{6}{5})=\frac{4}{5}$.
$\;\;\;\;$(Notice that $|\frac{6}{5}|\le2$).
From parts A and B, we can conclude that the minimum value of $a^2+b^2$ is $\;\;4/5$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/474507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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