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Multiplying square roots How do I simplify the following types of question:
$ \sqrt{x^2+5} \times \sqrt{x^2+20}$
Do I need to get both answer out of their roots first or not? This is how I would do it:
$ \sqrt{x^2+5} \times \sqrt{x^2+20} = x^2 + x\sqrt{20} + x\sqrt{5} + \sqrt{100}$ but I'm very certain this isn't the correct way.
|
For positive real $a,b,c,d$
$\sqrt{a+b}\cdot\sqrt{c+d}=\sqrt{(a+b)(c+d)}=\sqrt{ac+ad+bc+bd}$ which is $\ne \sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating the following limit: $\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $ I am trying to calculate this limit:
$$
\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}
$$
I've tried using conjugate of both denominator and numerator but I can't get the right result.
|
$$
\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\frac{x^2-x}{-x}\cdot\frac{1+\sqrt{x+1}}{\sqrt{x^2+1}+\sqrt{x+1}}\sim1\cdot\frac{2}{2}
$$
|
{
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"url": "https://math.stackexchange.com/questions/305497",
"timestamp": "2023-03-29T00:00:00",
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|
Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types
(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$
(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
|
(2) $$\int \frac{dx}{(x^4-1)^\frac14}=\int\frac{dx}{x(1-\frac1{x^4})^\frac14}=\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$
Let $1-\frac1{x^4}=y^4,4y^3dy=-4\frac{dx}{x^5}\implies \frac{dx}{x^5}=-y^3dy$ and $\frac1{x^4}=1-y^4\implies x^4=\frac1{1-y^4}$
$$\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$
$$=\int\frac{-y^3dy}{(1-y^4)y}=\int\frac{y^2dy}{y^4-1}=\frac12\left(\frac{dy}{y^2-1}+\frac{dy}{1+y^2}\right)$$
$$=\frac12\left(\frac12\ln\left|\frac{y-1}{y+1}\right|+\arctan y\right)+C$$ where $y^4=1-\frac1{x^4}$
(1) should be handled similarly by putting $1+\frac1{x^4}=y^4$
|
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|
Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality:
$$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
|
This is terse and skips some important steps, but I speculate that it can be turned nicely into a clean proof!
We show:
$$|\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{1}{\sqrt{2}}$$
We have
$$
\sqrt{\cos^2 2^n x } + \sqrt{\cos^2 2^{n+1} x}.
$$ And since $\cos^2 x = \frac{1}{2}(\cos(2x)+1)$ we get
$$
\frac{1}{\sqrt{2}}\sqrt{\cos (2^{n+1} x) + 1} + \sqrt{\cos^2 (2^{n+1} x)}.
$$ Substituting we get
$$
\frac{1}{\sqrt 2}\sqrt{z+1}+ |z|
$$ which has the minimum $\frac{1}{\sqrt 2}$ at $z=0$.
EDIT: Outline of complete proof
If $n$ is even, we are done since we have $n/2$ pairs.
For $n$ is odd there might be a simple proof, but I couldn't come up with anything better than this.
We show
$$|\cos 2^{n-1} x | + |\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{3}{2\sqrt{2}}$$
With tricks as above but also using $\cos(2x) = 2\cos^2(x) - 1$ on the last term we get
$$\frac{1}{2}\sqrt{\cos 2^{n} x +1} + |\cos 2^n x | + |2\cos^2 2^{n} x-1| \geq \frac{3}{2\sqrt{2}}$$
and thus
$$
\frac{1} {\sqrt 2}\sqrt{z+1}+ |z| + |2z^2-1|
$$
which has minimum
$$
\frac{\sqrt{2}+\sqrt{2-\sqrt{2}}}{2} > \frac{3}{2\sqrt{2}}
$$
(Since wolframalpha told me so (difference 0.02913, hard one)).
Now we can use this inequality to get an even number of pairs.
Anyone have a slicker proof for the odd case? This one sucks.
|
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|
Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$
Do you have an idea about this serie? If it converges what is the sum?
|
Use the equality
$$
\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}
$$
|
{
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|
Finding the derivative of $x$ to the power something that is a function of $x$ if $y = x^{(x+1)^\frac12}$
then how can I get the first derivative of $y$?
|
$$y=x^{(x+1)^{1/2}}=x^{\sqrt{x+1}}=e^{\sqrt{x+1}\log x}$$
and since
$$(\sqrt{x+1}\log x)'=\frac{\log x}{2\sqrt{x+1}}+\frac{\sqrt{x+1}}{x}=\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$
we get, applying as suggested the chaing rule, that
$$y'=y\frac{x\log x+2x+2}{2x\sqrt{x+1}}=x^{\sqrt{x+1}}\cdot\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$
|
{
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|
How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf
In problem (2), at the very end it says
$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$
I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?
Is one allowed to do the following?
$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$(0 + 0 + 0 + \dots)^{1/2} = 0$$
|
One can forget about the square root part for a while. Note that $\dfrac{1}{k^2}\lt \dfrac{1}{(k-1)k}$.
But $\dfrac{1}{(k-1)k}=\dfrac{1}{k-1}-\dfrac{1}{k}$. Thus your sum is less than
$$\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}+\cdots.$$
Note the wholesale cancellation: the above sum is $\dfrac{1}{n}$.
It follows that your original expression is less than $\dfrac{1}{\sqrt{n}}$.
Remark: Treating infinite "sums" as if they were long finite sums is a dangerous business that can all too easily give wrong answers. If one has experience with a particular series, such as the convergent series $\sum_1^\infty \frac{1}{n^2}$, then one can "see" that the tail must approach $0$. In fact, the issue is precisely the issue of the convergence of $\sum_1^\infty \frac{1}{n^2}$.
|
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|
Integrating a Rational Function I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it doesn't give a quick way either. If anybody has a way this integral can be solved quickly, as there will be a time crunch on my test.
$$\int\frac{x^3+x+2}{x^4+2x^2+1}dx$$
|
I think the integral can be solved in a very easy way.
$$
\begin{align}
\int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{x^4+2x^2+1}dx\\
&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{(x^2+1)^2}dx.
\end{align}
$$
In the RHS part, for the left integral uses substitution $u=x^4+2x^2+1\;\Rightarrow\;du=4(x^3+x)\,dx$ and for the right integral uses substitution $x=\tan\theta\;\Rightarrow\;dx=\sec^2\theta\;d\theta$. Therefore
$$
\begin{align}
\int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{1}{4u}du+\int\frac{2\sec^2\theta}{(\tan^2\theta+1)^2}d\theta\\
&=\frac{1}{4}\ln\,|u|+\text{C}+2\int\frac{\sec^2\theta}{\sec^4\theta}d\theta\\
&=\frac{1}{4}\ln\,\left|x^4+2x^2+1\right|+2\int\cos^2\theta\;d\theta+\text{C}.
\end{align}
$$
The rest should be easy to be solved. Just make sure you make appropriate limit for the integral when use substitution method.
$$\text{# }\mathbb{Q.E.D.}\text{ #}$$
|
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|
Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
|
Update: Upps, that is the same as that of @Vincent, sorry didn't see that first. However, it's a bit more explicte
A nicer way to use your results and proceed frm there is the following. You've got that
$$ a^2+b^2+c^2 \ge ab+bc+ca $$
Now use the fact that $(a+b+c)=S=3$ and multiply each side of your original inequality by S this is
$$ (a+b+c)S \ge S(ab+bc+ca)$$
or, (where we use now that $S=3$)
$$ (a+b+c)(a+b+c) \ge 3(ab+bc+ca)$$
Then
$$ a^2+b^2+c^2 + 2(ab+bc+ca) \ge 3(ab+bc+ca)$$
and
$$ a^2+b^2+c^2 \ge ab+bc+ca$$
which is exactly that, what you've already proven on your own.
|
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|
Cramer von Mises test statistic I am trying to derive the Cramer von Mises test statistic
$$nC_{n}=\frac{1}{12n}+\sum_{i=1}^{n}\left(U_{(i)}-\frac{2i-1}{2n}\right)^2$$
where $U_{(i)}=F_{0}(X_{(i)})$ the order statistics
from the original
$$C_{n}:=\int (\hat{F}_{n}(t)-F_{0}(t))^2dF_{0}(t)$$
Could anyone help me with this? I'm especially confused as to how to write the integral as a sum here.
|
Let $X_1,\dots,X_n$ be a random sample of size $n$ from $f(x)$.
The empirical CDF is
\begin{equation}
\hat{F}_n(x) = \frac{1}{n} \sum_{i=1}^n I(X_i \leq x)
= \begin{cases}
0, & x < X_{1:n},\\
i/n, & X_{i:n} \leq x < X_{i+1:n},\\
1, & X_{n:n} \leq x.
\end{cases}
\end{equation}
Let $U_i = F_0(X_{i:n})$, then
\begin{align*}
C_n &\triangleq \int \left[ \hat{F}_n (x) - F_0(x) \right]^2 \,\mathrm{d}{F_0(x)}
\\
&=\int_{-\infty}^{X_{1:n}} \left[ - F_0(x) \right]^2 \,\mathrm{d}{F_0(x)}
+ \sum_{i=1}^{n-1} \int_{X_{i:n}}^{X_{i+1:n}} \left[ \frac{i}{n} - F_0(x) \right]^2 \,\mathrm{d}{F_0(x)} + \int_{X_{n:n}}^{\infty} \left[ 1 - F_0(x) \right]^2 \,\mathrm{d}{F_0(x)}
\\
&= \frac{1}{3} U_1^3
+ \frac{1}{3} \sum_{i=1}^{n-1} \left[
\left( U_{i+1} - \frac{i}{n} \right)^3
- \left( U_{i} - \frac{i}{n} \right)^3
\right]
- \frac{1}{3} (U_n - 1)^3
\\
&= \frac{1}{3} U_1^3
+ \frac{1}{3} \sum_{i=1}^{n-1} \left[
U_{i+1}^3 - U_{i}^3
+ \frac{3 i^2}{n^2} \left(U_{i+1} - U_i\right)
- \frac{3 i}{n} \left(U_{i+1}^2 - U_i^2\right)
\right]
- \frac{1}{3} (U_n - 1)^3
\\
&= \frac{1}{3} U_1^3
+ \frac{1}{3} U_n^3 - \frac{1}{3} U_1^3
+ \left( U_n - \sum_{i=1}^n \frac{2 i - 1}{n^2} U_i \right)
- \left( U_n^2 - \sum_{i=1}^n \frac{1}{n} U_i^2 \right)
- \frac{1}{3} (U_n - 1)^3
\\
&= \frac{1}{3} + \frac{1}{n} \sum_{i=1}^n \left( U_i^2 - \frac{2 i - 1}{n} U_i \right)
\\
&= \frac{1}{3} + \frac{1}{n} \sum_{i=1}^n \left( U_i - \frac{2 i - 1}{2 n} \right)^2
- \frac{1}{n} \sum_{i=1}^n \left( \frac{2 i - 1}{2 n} \right)^2
\\
&= \frac{1}{12 n^2} + \frac{1}{n} \sum_{i=1}^n \left( U_i - \frac{2 i - 1}{2 n} \right)^2.
\qquad
\left[\text{by}~\sum_{i=1}^n (2 i - 1)^2 = \frac{1}{3} n (4 n^2 - 1) \right]
\end{align*}
|
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|
Problem involving permutation matrices from Michael Artin's book. Let $p$ be the permutation $(3 4 2 1)$ of the four indices. The permutation matrix associated with it is
$$ P =
\begin{bmatrix}
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$
This is the matrix that permutes the components of a column vector.
The problems asks you to decompose $p$ into transpositions and show that the associated matrix product equals the above matrix. However, I'm not getting that it does and I've run through it several times. Here are my calculations:
$p = (12)(14)(13)$
$$
P_{(12)} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
P_{(14)} =\begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix},
P_{(13)}=\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
However,
$$P_{(12)} (P_{(14)} P_{(13)}) =
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0
\end{bmatrix}
\neq P$$
I don't see where I've made a mistake.
|
The permutation is indeed $(1342)$, but this decomposes as $(13)(14)(12)$, so you should compute
$$
P_{(13)} P_{(14)} P_{(12)}
=
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
=
P.
$$
You should compose both permutations and matrices consistently left-to-right.
|
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|
Find $\lim\limits_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.
Find $\displaystyle \lim_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.
I tried to rationalize it, but doesn't help either. Please give me some hints. Thank you.
|
Hint:
$$
\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
|
$(a^5+b^5) = (a+b)(a^4+b^4) - ab(a^3+b^3)$.
So for any $n$, if $n \mid a^3+b^3$ and $n \mid a^4+b^4$ then $n \mid a^5+b^5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/319248",
"timestamp": "2023-03-29T00:00:00",
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|
Finding the area of a triangle using fractions?
To find the area of the triangle do you use Pythagorean theorem from what you have? Could this use similar triangles.
|
The area of $\triangle PST$ is the sum of the areas of $\triangle PSV$ and $\triangle VST$.
$$\begin{align}
\text{The area of }\triangle PSV &= \frac12 PV\cdot SV\\
&=\frac12 PV\cdot\frac23 QV\\
&=\frac13 PV\cdot QV,
\end{align}$$
$$\begin{align}
\text{and the area of }\triangle VST &= \frac12VT\cdot SV\\
&=\frac12 VT\cdot \frac23QV\\
&=\frac13VT\cdot QV.
\end{align}$$
\begin{align}
\text{So the area of }\triangle PST &= \frac13QV(PV+VT)\\
&=\frac13QV\cdot PT\\
&=\frac13QV\cdot\frac34PR\\
&=\frac14QV\cdot PR.
\end{align}
The area of $\triangle PQR = \frac12QV\cdot PR$.
\begin{align}
\text{So the ratio }\frac{\triangle PST}{\triangle PQR} &= \frac{\frac14 QV\cdot PR}{\frac12QV\cdot PR}\\
&=\frac{\frac14}{\frac12}\\
&=\frac12.
\end{align}
|
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|
Find the points on the given curve where the tangent line is horizontal or vertical $r=e^θ $ $($ Assume $0 ≤ θ ≤ 2π.)$
Apparently I keep getting this answer wrong. I dont know if i need to use $n $ in the answer or not...
|
As $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac yx$
Differentiating $$\sqrt{x^2+y^2}=e^{\arctan \frac yx}$$ wrt $x,$
$$\frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}$$
$$=e^{\arctan \frac yx}\cdot\frac1{1+\left(\frac yx\right)^2}\cdot\frac{\left(x\frac{dy}{dx}-y\right)}{x^2}$$
$$=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\right)}{x^2+y^2} \text { as } e^{\arctan \frac yx}=\sqrt{x^2+y^2}$$
$$\implies \frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\right)}{x^2+y^2}$$
$$\implies \frac{dy}{dx}=\frac{x+y}{x-y}$$
For horizontal tangent, $\frac{dy}{dx}=0\implies x+y=0\implies \theta=\arctan (-1)=n\pi-\frac{\pi}4$ where $n$ is any integer.
As $0\le \theta\le 2\pi,n=1,2$
Similarly for vertical tangent, $\frac{dx}{dy}=0\implies x-y=0\implies \theta=\arctan (1)=m\pi+\frac{\pi}4$ where $m$ is any integer.
As $0\le \theta\le 2\pi,m=0,1$
|
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|
Commuting Skew-symmetric Nilpotent 4x4 Matrices Suppose $A$ and $B$ are nonzero, commuting, skew-symmetric, nilpotent matrices in $M_4(k)$, $k$ a field (char $k\ne 2$). Must $A=\lambda B$ for some $\lambda\in k$? I have shown that this is true for $3\times 3$ matrices, and I believe it should also be true for $4\times 4$ matrices.
Thanks in advance to anyone willing to help me with this fairly dry question.
|
It's not true. Consider
$$ A = \left[ \begin {array}{cccc} 0&0&1&-i\\ 0&0&i&1\\ -1&-i&0&0\\ i&-1&0&0
\end {array} \right],\ B = \left[ \begin {array}{cccc} 0&1&-i&0\\ -1&0&0&i\\ i&0&0&1\\ 0&-i&-1&0\end {array}
\right]$$
where $i$ is a square root of $-1$.
EDIT: Or, a bit more generally, with the same $A$,
$$ B = \left[ \begin{array}{cccc} 0 & a & b & c \\ -a & 0 & c & -b \\
-b & -c & 0 & a \\ -c & b & -a & 0 \end{array}\right]$$
where $a^2 + b^2 + c^2 = 0$ and $a$, $b$, $c$ are not all $0$.
EDIT: Also, try
$$ A = \left[ \begin{array}{cccc} 0 & -a & b & c \\ a & 0 & -c & b \\
-b & c & 0 & a \\ -c & -b & -a & 0 \end{array}\right],\
B = \left[ \begin{array}{cccc} 0 & a & b & c \\ -a & 0 & c & -b \\
-b & -c & 0 & a \\ -c & b & -a & 0 \end{array}\right]$$
where again $a^2 + b^2 + c^2 = 0$ and $a$, $b$, $c$ are not all $0$.
In particular we get counterexamples over every field of nonzero characteristic
(and every field of Stufe $\le 2$).
Basically I found these examples by starting with a suitable $A$ and a general form for $B$ and (with Maple's help) solving the equations $A B - B A = 0$ and $B^2 = 0$.
|
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|
Further clarification needed on proof invovling generating functions and partitions (or alternative proof)
Show with generating functions that every positive integer can be written as a unique sum of distinct powers of $2$.
There are 2 parts to the proof that I don't understand. I will point them out as I outline the proof:
The generating function given is
$$
g^*(x) = (1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^k}) \cdots.
$$
Then I don't understand where the following comes from:
To show that every integer can be written as a unique sum of disticnt powers of $2$, we must show that the coefficient of every power of $x$ in $g^*(x) = 1 + x + x^2 + x^3 + \cdots = (1-x)^{-1} $ or equivalently, $(1-x)g^*(x)=1$.
Then they use the identity $(1-x^k)(1+x^k) = 1 - x^{2k}$ to manipulate $g^*(x)$:
$$\begin{align*}
(1-x)g^*(x) &= (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots\\
&= (1-x^2)(1+x^2) (1+x^4)(1+x^8)\cdots \\
&= (1-x^4)(1+x^4)(1+x^8)\cdots \\
&= \quad \quad \vdots\\
&= 1
\end{align*}$$
That is the the second part that I don't understand. If it's an infinite product, why can they "eventually eliminate all factors of $(1-x)g^*(x)$"?
Update: Alternative proofs are more than welcome.
|
For your second question, first note that
$$g^*(x)=\prod_{k\ge 0}\left(1+x^{2^k}\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)\dots$$
makes sense: if $n\le 2^m$, the coefficient of $x^n$ in $g^*(x)$ is its coefficient in the polynomial $$\prod_{k=0}^m\left(1+x^{2^k}\right)\;,$$ because the contribution of any factor $1+x^{2^k}$ with $k>m$ must be a factor of $1$, not of $x^{2^k}$. Similarly, the coefficient of $x^n$ in the product
$$(1-x)g^*(x)=(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\dots$$
is its coefficient in the polynomial
$$(1-x)\prod_{k=0}^m\left(1+x^{2^k}\right)\;,$$
for exactly the same reason. And by an easy induction
$$(1-x)\prod_{k=0}^m\left(1+x^{2^k}\right)=1-x^{2^{m+1}}\;,$$
so the coefficient of $x^0=1$ is $1$, and the coefficient of $x^n$ for $1\le n\le 2^m$ is $0$. Since this is true for every $m\in\Bbb Z^+$, for each $n>0$ the coefficient of $x^n$ in $(1-x)g^*(x)$ must be $0$, and since the constant term is $1$, $(1-x)g^*(x)=1$.
|
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|
Evaluation of definite integral using residue theorem $$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$$
I need to evaluate the above integral .
My idea is to consider the same integral but with the $x$'s as $z$'s, over the complex plane, have a closed contour integral over $\gamma$, and then use the residue theorem. i.e. consider:
$$ \int^{+\infty}_{-\infty} \frac{z-1}{z^3-1} dz$$
I'm stuck on how to formulate $\gamma$ though.
I know this has 3 poles: at
$z=1$, $z= \frac{-1}{2} + i\frac{\sqrt3}{2}$ and $z= \frac{-1}{2} - i\frac{\sqrt3}{2}$
How do I use this to divide up gamma over contours to which I can then use the residue theorem? And then do I have to either evaluate directly or apply the ML inequality to each individual contour?
|
Note that $\frac{x - 1}{x^3 - 1} = \frac{1}{x^2 + x + 1}$. So the integral is
$$
\int_{-\infty}^{+\infty} \frac{1}{x^2 + x + 1} \, dx.
$$
First take the integral from $-a$ to $a$ and then take the limit after. Just complete the square in the denominator and use substitution to get an integral in terms of $\arctan$.
$$
\int_{-a}^{a} \frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}} \, dx \\
=\frac{4}{3}\int_{-a}^{a} \frac{1}{\left(\frac{2}{\sqrt{3}}(x + \frac{1}{2})\right)^2 + 1} \, dx.
$$
Using the substitution $u = \frac{2}{\sqrt{3}}(x + \frac{1}{2})$, we have $dx = \frac{\sqrt{3}}{2}du$ and so the indefinite integral is
$$
\frac{4}{3}\frac{\sqrt{3}}{2}\int \frac{1}{u^2 + 1} \, dx \\
=\frac{2}{\sqrt{3}}\arctan(u) = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right).
$$
Thus, the definite integral is
$$
\left.\frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right)\right|_{-a}^a = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) - \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right).
$$
As $a \to \infty$, we have $\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) \to \frac{\pi}{2}$ and $\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right) \to -\frac{\pi}{2}$. Hence the integral is
$$
\int_{-\infty}^{+\infty} \frac{x - 1}{x^3 - 1} \, dx= \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} + \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} = \frac{2\pi}{\sqrt{3}}.
$$
|
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|
General term of $a_{n}= \sqrt{ a_{n-1}+6}$ What is the general term of $a_{n}= \sqrt{ a_{n-1}+6}$,$a_{1}=4$?
|
I doubt if such general closed form exists.
We have $\sqrt{a_{n-1}+6}$ to be $\sqrt{10}$ in the first step, and $\sqrt{6+\sqrt{10}}$ is hard to simplify as if one assume it has the form $a+\sqrt{b}$, then we must have $a^{2}+b=6,2a\sqrt{b}=\sqrt{10}$. This force $a^{2}b=5/2$, and one has to solve $x^{2}-6x+5/2$. This gives $a^{2}$ or $b$ equal to $\frac{6+\sqrt{36-10}}{2}=3+\frac{\sqrt{23}}{2}$, which is fairly ugly. Another guess might be $\sqrt{c}+\sqrt{d}$, but then we would have $6=c+d$ and $10=4cd$, which ends up to the same thing. Since $a_{2}$ is already such a mess it is hard to come up with general formula for $a_{i}$.
|
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|
"Fat" sets of integers and Fibonacci numbers Let us call a set of integers "fat" if each of its elements is at least as large as its cardinality. For example, the set $\{10,4,5\}$ is fat, $\{1,562,13,2\}$ is not.
Define $f(n)$ to count the number of fat sets of a set of integers $\{1...n\}$ where we count the empty set as a fat set.
eg: $f(4) = 8$ because $\{Ø, \{1\}, \{2\}, \{3\}, \{4\}, \{2,3\}, \{2,4\}, \{3,4\}\}$
Show that $f(n) = F_{n+2}$ where $F_n$ is the nth fibonacci number. (So for $n=4, f(4)=F_6=8$).
I was given a hint to first construct a recursive equation of $f(n)$ then use the initial condition to infer that the recursive equation has to be a fibonacci recurrence thus arriving at our goal. My main guess at the recursive equation was $f(n) = f(n-1)+f(n-2)$. As to how I arrived at this, I kind of cheated by assuming the identity to be true then split $F_{n+2} = F_{n+1} + F_{n}$ and applied the identity.
I want to show that this recurrence is true (which I guess is done by induction in some form or even better, a combinatorics argument as the structure looks rather familiar) then the rest I'm sure will follow given initial conditions.
Any advice in the right direction would be greatly appreciated.
|
This problem has a simple solution using ordinary generating functions.
Observe that the generating function of subsets of $\{k, n\}$ indexed
by number of elements ($u$) and total sum ($x$) is by inspection seen
to be
$$\prod_{m=k}^n (1+ux^m).$$
Therefore the generating function of the sets being considered is
$$\sum_{k=0}^n [u^k] \prod_{m=k}^n (1+ux^m)$$
where $[u^k]$ is the coefficient extraction operator.
For our purposes we don't need the sum parameter from this generating
function so we may set it to one, getting
$$\sum_{k=0}^n [u^k] \prod_{m=k}^n (1+u)
= \sum_{k=0}^n [u^k] (1+u)^{n-k+1}
= \sum_{k=0}^n {n-k+1\choose k}.$$
We will now compute the generating function $f(z)$ of this sum, getting
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n {n-k+1\choose k}
= \sum_{k\ge 0} \sum_{n\ge k} {n-k+1\choose k} z^n
= \sum_{k\ge 0} \sum_{n\ge 0} {n+1\choose k} z^{n+k}
\\= \frac{1}{1-z} +
\sum_{k\ge 1} \sum_{n\ge 0} {n+1\choose k} z^{n+k}
= \frac{1}{1-z} +
\sum_{k\ge 1} \sum_{n\ge k-1} {n+1\choose k} z^{n+k}
\\= \frac{1}{1-z} +
\sum_{k\ge 1} \sum_{n\ge 0} {n+k\choose k} z^{n+2k-1}
= \frac{1}{1-z} +
\sum_{k\ge 1} z^{2k-1} \sum_{n\ge 0} {n+k\choose k} z^n
\\= \frac{1}{1-z} +
\sum_{k\ge 1} z^{2k-1} \frac{1}{(1-z)^{k+1}}
= \frac{1}{1-z} + \frac{1}{1-z} \frac{1}{z}
\sum_{k\ge 1} z^{2k} \frac{1}{(1-z)^k}
\\ = \frac{1}{1-z} + \frac{1}{1-z} \frac{1}{z}
\frac{z^2/(1-z)}{1-z^2/(1-z)}
= \frac{1}{1-z} + \frac{1}{1-z} \frac{1}{z}
\frac{z^2}{1-z-z^2}.$$
This finally simplifies to
$$\frac{1}{1-z}
\left(1 + \frac{z}{1-z-z^2}\right)
= \frac{1}{1-z} \frac{1-z-z^2 + z}{1-z-z^2}
= \frac{1+z}{1-z-z^2}.$$
Now recall that the generating function of the Fibonacci numbers is
given by $$\frac{z}{1-z-z^2}$$ and therefore
$$[z^n] \frac{1+z}{1-z-z^2}
= F_{n+1} + F_n = F_{n+2}.$$
Addendum. An alternate evaluation of the OGF proceeds as follows:
$$f(z) =
\sum_{n\ge 0} z^n [w^{n+1}] (1+w)^{n+1}
\sum_{k\ge 0}
\frac{w^{2k}}{(1+w)^k}
\\ = \frac{1}{z}
\sum_{n\ge 0} z^{n+1} [w^{n+1}] (1+w)^{n+1}
\frac{1}{1-w^2/(1+w)}
\\ = \frac{1}{z}
\sum_{n\ge 0} z^{n+1} [w^{n+1}] (1+w)^{n+2}
\frac{1}{1+w-w^2}.$$
The contribution from $w$ is
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{n+2}} (1+w)^{n+2} \frac{1}{1+w-w^2}.$$
Now put $w/(1+w)=v$ so that $w = v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$ to
get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+2}} \frac{1}{1+v/(1-v)-v^2/(1-v)^2} \frac{1}{(1-v)^2}.$$
This gives for the OGF
$$\frac{1}{z} \sum_{n\ge 0} z^{n+1} [v^{n+1}]
\frac{1}{(1-v)^2+v(1-v)-v^2}
\\ = \frac{1}{z} \sum_{n\ge 0} z^{n+1} [v^{n+1}]
\frac{1}{1-v-v^2}
= \frac{1}{z} \left[ -1 + \frac{1}{1-z-z^2} \right]
\\ = \frac{1}{z} \frac{z+z^2}{1-z-z^2}
= \frac{1+z}{1-z-z^2},$$
the same as before.
|
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|
Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
|
Note that each term is of the form
$$\dfrac{df_k/dx}{f_k} = \dfrac{d(\log(f_k)))}{dx}$$
Hence,
$$\sum_{k=0}^n \dfrac{df_k/dx}{f} = \sum_{k=1}^n \dfrac{d(\log(f_k)))}{dx} = \dfrac{d(\sum_{k=0}^n \log(f_k)))}{dx} = \dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx}$$In your case, $f_0 f_1 f_2 \cdots f_n$ reduces to give a nice short expression, which can be easily differentiated.
Move the cursor over the gray area for a complete answer.
In your case, $f_0 = (x-y)$, $f_n = x^{2^{n-1}} + y^{2^{n-1}}$ for $n > 0$. Hence, $$f_0 f_1 f_2 \cdots f_n = (x-y)(x+y)(x^2+y^2)\cdots (x^{2^{n-1}} + y^{2^{n-1}}) = x^{2^n} - y^{2^n}$$ Hence, $$\dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx} = \dfrac{2^n x^{2^n-1}}{x^{2^n}-y^{2^n}}$$ In your case, $n=5$.
|
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|
Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$
Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$
Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$
$\log_4$ can be written as $\frac{1}{2}\log_2$
Please guide further..
|
Let $x=\sqrt{2+\sqrt 3}+\sqrt{2-\sqrt 3}$
$x^2=(2+\sqrt 3)+(2-\sqrt 3)+2\sqrt{(2+\sqrt 3)(2-\sqrt 3)}=6$
I think you get it from here.
|
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|
Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:
Monotonic:
The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$
$$a_1=1\leq 1+\frac{1}{2^2}=a_2$$
Need to show that $a_{n+1}\leq a_{n+2}$
$$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$
Thus the sequence is monotone and increasing.
Boundedness:
Since the sequence is increasing it is bounded below by $a_1=1$.
Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.
Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?
Thanks so much in advance!
|
Your work looks good so far. Here is a hint:
$$
\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}
$$
To elaborate, apply the hint to get:
$$
\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right)
$$
Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.
Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus:
$$
\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 - \frac{1}{n}
$$
Add $1$ to both sides to get:
$$
a_n \le 2 - \frac{1}{n} \le 2
$$
It follows that $a_n$ is bounded from above and hence convergent.
It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.
|
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|
Proof $ab-cd \ge 3$ $a$, $b$, $c$, and $d$ are real number
$a\ge b\ge c\ge d$
$a+b+c+d = 13$
$a^2+b^2+c^2+d^2 =43$
Proof that $ab-cd\ge 3$
|
Hint
$$ (a + b + c + d)^2 = 169 = \sum a^2 + 2 ( ab + bc + cd + da + ac + bd) $$
This gives: $ ab + bc + cd + da + ac + bd = 63 $. We already had $ ab + cd \leq \frac{43}{2} $ from $ (a - b)^2 + (c - d)^2 \geq 0 $.
Use the above results with $$ \left[ (a + b) - (c + d) \right]^2 \geq 0 $$
|
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|
How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number?
How to prove that $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$?
If $p$ is a prime, then the number of times $p$ divides $N!$ is $e_p(N)=\sum_{k=1}^\infty\left\lfloor\frac{N}{p^k}\right\rfloor$. So I need
$$e_p(5m)+e_p(5n)\geq e_p(m)+e_p(n)+e_p(3m+n)+e_p(3n+m).$$
What to do next? Thanks in advance.
|
I became aware of this question due to some recent edits. The shorter proofs seem to rely on
$$
\lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor
$$
My answer also uses this inequality. I have tried to provide a complete, yet shorter, proof of it.
Define
$$
f(x,y)=\lfloor5x\rfloor+\lfloor5y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor3x+y\rfloor-\lfloor3y+x\rfloor\tag1
$$
Note that $f(x+1,y)=f(x,y+1)=f(x,y)$. Thus, we only need worry about $x,y\in[0,1)$.
Let $x\in\left[\frac j5,\frac{j+1}5\right)$ and $y\in\left[\frac k5,\frac{k+1}5\right)$, where $j,k\in\{0,1,2,3,4\}$. Then
$$
\lfloor x\rfloor=\lfloor y\rfloor=0\tag{2a}
$$
$$
\lfloor5x\rfloor=j\qquad\text{and}\qquad\lfloor5y\rfloor=k\tag{2b}
$$
$$
\lfloor3x+y\rfloor\le\left\lfloor\frac{3j+k+3}5\right\rfloor
\qquad\text{and}\qquad
\lfloor x+3y\rfloor\le\left\lfloor\frac{j+3k+3}5\right\rfloor
\tag{2c}
$$
Apply $(2)$ to $(1)$:
$$
\begin{align}
f(x,y)
&=\lfloor5x\rfloor+\lfloor5y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor3x+y\rfloor-\lfloor3y+x\rfloor\\
&\ge j+k-\left\lfloor\frac{3j+k+3}5\right\rfloor-\left\lfloor\frac{j+3k+3}5\right\rfloor\\
&=g(j,k)\tag3
\end{align}
$$
Compute the values of $g(j,k)$ for $j,k\in\{0,1,2,3,4\}$:
$$
\begin{array}{c|cc}
g&0&1&2&3&4\\\hline
0&0&0&0&0&0\\
1&0&0&0&1&0\\
2&0&0&0&1&1\\
3&0&1&1&0&1\\
4&0&0&1&1&2
\end{array}\tag4
$$
Inequality $(3)$, table $(4)$, and the lattice periodicity of $f$, show that $f(x,y)\ge0$ for all $x,y$. That is,
$$
\lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor\tag5
$$
Inequality $(5)$, when combined with Legendre's Formula, yields
$$
m!\,n!\,(3m+n)!\,(m+3n)!\mid(5m)!\,(5n)!\tag6
$$
|
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|
$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$
For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge to $\infty$ as $n\rightarrow \infty$.
1 - How to find a bound for $f(n)$?
2- $f(n)\rightarrow\infty$, as $n\rightarrow\infty$?
Thank you for your time.
|
$$ \frac{x + n}{x + n^{-1}} = 1 + \frac{n-1}{nx + 1}$$
Using the expansion of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$
We see that the resulting expansion of the integrand is $$\frac{n - 1}{nx + 1} - \frac{(n - 1)^2}{2(n x+ 1)^2} + O\left (\left(\frac{n-1}{nx + 1} \right)\right)^3$$
Squaring this again, we get $(\frac{n - 1}{nx + 1})^2 + O((\frac{n-1}{n x + 1})^3)$
The integral of the left term is just $$\left.-\frac{(n - 1)^2}{n(1 + n x)}\right|^{\infty}_0 = \frac{(n-1)^2}{n}$$
|
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|
Find a polynomial of degree $4$ with the coefficient $x^2$ equal to $6$, and zeros $-3$, $ 2$, $ -1$, $-2$ I started off with:
$$f(x)= a(x-(-3)) (x-(2)) (x-(-1)) (x-(-2))$$
$$f(x)= a(x+3) (x-2) (x+1) (x+2)$$
|
You are right that $f(x)=a(x+3)(x+2)(x+1)(x-2)$
Expanding gives $$f(x)=a(x^{2}+3x+2x+6)(x^{2}+x-2x-2)=a(x^{2}+5x+6)(x^{2}-x-2)$$
Multiplying the quadratics, $$f(x)=a(x^{4}+5x^{3}+6x^{2}-x^{3}-5x^{2}-6x-2x^{2}-10x-12)$$
Collecting terms gives $f(x)=a(x^{4}+4x^{3}-x^{2}-16x-12)$
Hence, the coefficient on $x^{2}$ is $-a$
To make the coefficient on $x^{2}$ $6$, let $a=-6$.
|
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|
What may be the ratio of the perimeter of the trapezium to its midline? The diagonals of a symmetric trapezium are perpendicular to each other. What may be the ratio of the perimeter of the trapezium to its midline?
|
Let the perimeter $p = a+b+2c$, where $c$ are the sides and set $d = d_a + d_b$ to be the length of the diagonals with $d_a$ and $d_b$ being their parts. From the Pythagorean theorem we know that $d_a = \frac{a}{\sqrt{2}}$ and $d_b = \frac{b}{\sqrt{2}}$. Also, $c^2 = d_a^2+d_b^2$, hence $c = \frac{1}{\sqrt{2}}\sqrt{a^2+b^2}$. Finally $p = a+b+\sqrt{2}\sqrt{a^2+b^2}$ and $m = \frac{a+b}{2}$, thus, the ratio is
$$ 2\frac{a+b+\sqrt{2}\sqrt{a^2+b^2}}{a+b} =
2+2\frac{\sqrt{\frac{a^2+b^2}{2}}}{\frac{a+b}{2}}. $$
In other words, it is $2$ plus the two times ratio of quadratic and arithmetic means of the $a$ and $b$. This also follows from the geometric construction of quadratic mean, where the resulting symmetric trapezium looks as follows:
$\hspace{120pt}$
Edit: If you ask for possible values of the ratio (thanks to @joriki for noticing), then it follows from the properties of the means that it is in the interval $[4,2+2\sqrt{2})$ where $2+2\sqrt{2}$ is the degenerate case of $b = 0$, that is, the right and isosceles triangle.
I hope this helps ;-)
|
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How prove this summation prove that:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\dfrac{n}{2}$$
I think can prove by the probability
my idea:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^2}{(n-k)!(n+k)!}$$
and
$$\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^2}{(n-k)!(n+k)!}=(n!)^2\displaystyle\sum_{k=1}^{n}\dfrac{1}{\Gamma{(n-k+1)}\Gamma{(n+k+1)}}$$
and use:http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Unendliche_Reihen:_Hypergeometrische_Reihen
and my student methods:
$$\Longleftrightarrow \displaystyle\sum_{k=1}^{n}\dfrac{kC_{n}^{k}}{C_{n+k}^{k}}=\dfrac{n}{2}$$
and we notice
$$\dfrac{kC_{n}^{k}}{C_{n+k}^{k}}=\dfrac{kC_{2n}^{n+k}}{C_{2n}^{n}}$$
$$\Longleftrightarrow \displaystyle\sum_{k=0}^{n}kC_{2n}^{n+k}=\dfrac{n}{2}C_{2n}^{n}$$
However,we have
$$\displaystyle\sum_{k=0}^{n}kC_{2n}^{n+k}=\displaystyle\sum_{k=0}^{n}C_{2n}^{k}-\displaystyle\sum_{k=0}^{n}kC_{2n}^{k}=\dfrac{n}{2}C_{2n}^{n}$$,done！
I wish to see other methods.
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Here is another method using the convolution of two generating functions.
Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the exponential
generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Re-write your sum as follows:
$$\sum_{k=0}^n \frac{k\times (n!)^2}{(n+k)! (n-k)!}
= n! \sum_{k=0}^n {n\choose k} \frac{k\times k!}{(n+k)!}.$$
We proceed to evaluate the sum term.
In the present case we have
$$A(z) = \sum_{k\ge 0} \frac{k\times k!}{(n+k)!} \frac{z^k}{k!}
= \sum_{k\ge 1} \frac{k}{(n+k)!} z^k$$
and $$B(z) = \sum_{k\ge 0} \frac{z^k}{k!} = \exp(z).$$
Simplifying $A(z)$ we see that it is given by
$$z \sum_{k\ge 1} \frac{k}{(n+k)!} z^{k-1}
= z \frac{d}{dz} \sum_{k\ge 1} \frac{z^k}{(n+k)!}
= z \frac{d}{dz} \frac{1}{z^n} \sum_{k\ge 1} \frac{z^{n+k}}{(n+k)!}
= z \frac{d}{dz}
\frac{1}{z^n} \left(\exp(z)-\sum_{q=0}^n \frac{z^q}{q!}\right)
\\= z \times
\left(-\frac{n}{z^{n+1}}
\left(\exp(z)-\sum_{q=0}^n \frac{z^q}{q!}\right)
+\frac{1}{z^n}
\left(\exp(z)-\sum_{q=0}^{n-1} \frac{z^q}{q!}\right)\right).$$
Proceeding to extract coefficients from $A(z) B(z)$ we get
$$[z^n] A(z) B(z)
\\= - n [z^{2n}]
\left(\exp(2z)-\exp(z)\sum_{q=0}^n \frac{z^q}{q!}\right)
+ [z^{2n-1}]
\left(\exp(2z)-\exp(z)\sum_{q=0}^{n-1} \frac{z^q}{q!}\right)
\\ =
-n\frac{2^{2n}}{(2n)!}
+n\sum_{q=0}^n \frac{1}{q!}\frac{1}{(2n-q)!}
+ \frac{2^{2n-1}}{(2n-1)!}
-\sum_{q=0}^{n-1} \frac{1}{q!}\frac{1}{(2n-1-q)!}
\\ =
n\sum_{q=0}^n \frac{1}{q!}\frac{1}{(2n-q)!}
-\sum_{q=0}^{n-1} \frac{1}{q!}\frac{1}{(2n-1-q)!}
\\ =
\frac{n}{(2n)!} \sum_{q=0}^n {2n\choose q}
- \frac{1}{(2n-1)!} \sum_{q=0}^{n-1} {2n-1\choose q}
\\ =
\frac{n}{(2n)!} \left(2^{2n-1}+\frac{1}{2}{2n\choose n}\right)
- \frac{1}{(2n-1)!} 2^{2n-2}
\\ =
\frac{2^{2n-2}}{(2n-1)!}
+ \frac{n}{2} \frac{1}{(2n)!} {2n\choose n}
- \frac{1}{(2n-1)!} 2^{2n-2}
= \frac{n}{2} \frac{1}{(n!)^2}.$$
The end result is that
$$n! \times n! \times [z^n] A(z) B(z)
= n! \times n! \times \frac{n}{2} \frac{1}{(n!)^2}
= \frac{n}{2}$$
as claimed. QED.
|
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When are both fractions integers? The sum of absolute values of all real numbers $x$, such that both of the fractions $\displaystyle \frac{x^2+4x−17}{x^2−6x−5}$ and $\displaystyle \frac{1−x}{1+x}$ are integers, can be written as $\displaystyle \frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?
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Note that
$$\frac{1+x}{1-x}=\frac{2}{1+x}-1$$
and
$$\frac{x^2+4x-17}{x^2-6x-5} = \frac{10x-12}{x^2-6x-5}+1.$$
We need both of these to be integers. So $\frac{2}{1+x}$ must be an integer, and happens when, for example, $x=-3,-2,0,1$ (you also need to check rational values). You can plug these into the second fraction to test its integrality. Then take the sum of the appropriate values.
|
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4.\binom{n}{r-2}+........+(-1)^r.(r+1).(r+2).\binom{n}{0}$
Now after that how can i calculate it
Thanks
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Felix Marin's approach is nice and short, but I thought I'd add another approach that uses standard Binomial tools: Vandermonde's Identity, the symmetry of Pascal's Triangle, and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$.
$$
\hspace{-12pt}\begin{align}
&\phantom{={}}\sum_{k=0}^r(-1)^k(k+1)(k+2)\binom{n}{r-k}\\
&=\sum_{k=0}^r(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}\tag{1a}\\
&=\sum_{k=0}^r(-1)^{r-k}[k(k-1)-2(r+1)k+(r+1)(r+2)]\binom{n}{k}\tag{1b}\\
&=\scriptsize n(n-1)\sum_{k=0}^r\binom{-1}{r-k}\binom{n-2}{k-2}-2(r+1)n\sum_{k=0}^r\binom{-1}{r-k}\binom{n-1}{k-1}+(r+1)(r+2)\sum_{k=0}^r\binom{-1}{r-k}\binom{n}{k}\tag{1c}\\
&=n(n-1)\binom{n-3}{r-2}-2(r+1)n\binom{n-2}{r-1}+(r+1)(r+2)\binom{n-1}{r}\tag{1d}\\
&=\left(n(n-1)\frac{r-1}{n-2}-2(r+1)n+(r+1)(r+2)\frac{n-1}r\right)\binom{n-2}{r-1}\tag{1e}\\
&=\frac{2(n-r-1)(n-r-2)}{r(n-2)}\binom{n-2}{n-r-1}\tag{1f}\\
&=\frac{2(n-3)}{r}\binom{n-4}{n-r-3}\tag{1g}\\
&=2\binom{n-3}{r}\tag{1h}\\
\end{align}
$$
Explanation:
$\text{(1a):}$ substitute $k\mapsto r-k$
$\text{(1b):}$ $(r-k+1)(r-k+2)=k(k-1)-2(r+1)k+(r+1)(r+2)$
$\text{(1c):}$ $\binom{n}{k}=\frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2}$ and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$
$\phantom{\text{(1c):}}$ $\binom{-1}{r-k}=(-1)^{r-k}[k\le r]$ (Iverson Brackets)
$\text{(1d):}$ Vandermonde's Identity
$\text{(1e):}$ $\binom{n-3}{r-2}=\frac{r-1}{n-2}\binom{n-2}{r-1}$ and $\binom{n-1}{r}=\frac{n-1}{r}\binom{n-2}{r-1}$
$\text{(1f):}$ simplify the rational function
$\phantom{\text{(1f):}}$ $\binom{n-2}{r-1}=\binom{n-2}{n-r-1}$ (symmetry of Pascal's Triangle)
$\text{(1g):}$ $\binom{n-2}{n-r-1}=\frac{(n-2)(n-3)}{(n-r-1)(n-r-2)}\binom{n-4}{n-r-3}$
$\text{(1h):}$ $\binom{n-4}{n-r-3}=\binom{n-4}{r-1}$ (symmetry of Pascal's Triangle)
$\phantom{\text{(1h):}}$ $\binom{n-3}{r}=\frac{n-3}r\binom{n-4}{r-1}$
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|
$x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$. How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.
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We have
$$x^4-4y^4=z^2\iff\left(2y^2\right)^2+z^2=\left(x^2\right)^2.$$
Therefore if a primitive solution exists then $2y^2=2nm, \ z=m^2-n^2, \ x^2=m^2+n^2.\;$ [Note: Primitive here means one such that $x^2, 2y^2$ and $z$ are coprime which has as a consequence that $n$ and $m$ are coprime.]
From $2y^2=2nm$ and because $n$ and $m$ are coprime we get that both $n$ and $m$ are perfect squares and therefore $n=k^2,m=\ell^2$ for some $k$ and $\ell$.
This means that $x^2=\ell^4+k^4$, which has no integer solutions (see this).
|
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|
proving or disproving that two tangent lines are parallel to a curve Im trying to prove or disprove that given the function, $f(x)=0.5\sqrt{1-x^{2}}$,
There are two different tangent lines to $f(x)$ so they are parallel.
I tried to derivative but with no success.
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$y=f(x)=0.5\sqrt{1-x^2}\implies 4y^2=1-x^2$ with $y\ge0$
Let us find the condition of $y=mx+c$ being a tangent to the curve
So, $4y^2=1-\left(\frac{y-c}m\right)^2\implies y^2(1-4m^2)+2cy+(m^2-c^2)=0$
This is a Quadratic Equation in $y$
For tangency, both the root must be same i..e, the discriminant must be $0$
So, $$(2c)^2=4(1-4m^2)(m^2-c^2)$$
$$\implies c^2=m^2-4m^4+4m^2c^2+c^2\implies 4c^2=1-4m^2\text{ if } m \text{ is finite and non-zero}$$
$c=\pm\sqrt{1-4m^2}$
So, $y=\frac{2c}{1-4m^2}=\frac1{2c}>0\implies c\ge 0$ so $c\not<0$
We won't get parallel tangents for finite and non-zero $m$
If $m=0, y=c \implies c\ge 0$
We won't get parallel tangents here.
If $\frac1m=0, y=0\implies 1-x^2=0\implies x=\pm1$
Evidently, the tangents passing through $(\pm1,0)$ are the only parallel tangents
|
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Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$?
$Approach$:
$N$=$11^2$.$13^4$.$17^6$
$N^2$=$11^4$.$13^8$.$17^{12}$
This means $N$ has $(2+1) * (4+1) * (6+1) = 105$ factors and $N^2$ has $ (4+1) * (8+1) * (12+1) = 585 $ factors.
Therefore, there are 480 numbers that are not a factor of N. They are any combination of :
$11^3$,$11^4$,$13^5$,$13^6$,$13^7$,$13^8$,$17^7$,$17^8$,$17^9$,$17^{10}$,$17^{11}$,$17^{12}$.
But how many of these combinations are less than N? Not really sure how to do that in a easy way.
Please guide me how to do so.
Any help will be appreciated. Thanks.
|
N= 11^2 * 13^4 * 17^6
No of Factors of N = 3*5*7 = 105
There will be one factor which is N, So remaining = 105-1 = 104
No of factors of N^2= 5*9*13 = 585, of these N will be one of the factors.
Out of the remaining 584 factors, half will be less than N and half will be grater than N , So no. of factors of N^2 which are less than N= 584/2= 292
Hence the number of factors of N^2 are less than N but not a factor of N = 292-104= 188
|
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Proof by induction for a recursive function $F(n) = F(n–1)+F(n–2)$ I'm having a lot of trouble with the following homework question:
Consider the following recursive function:
Base Case: $F(0) = 0,F(1) = 1$.
Recursive Step: $F(n)=F(n−1)+F(n−2)$ for all $n≥2$.
Prove that $F(n)\le \left ( \cfrac{1+ \sqrt5}{2} \right) ^{n-1}$
I have tried to do the induction step by rewriting this as $F(k+1) = F(k) + F(k-1)$ and then subbing in the inequalities for each of $F(k)$ and $F(k-1)$ but I feel that I am hopelessly barking up the wrong tree.
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Hint: $\displaystyle \frac{1+\sqrt{5}}{2}+1= \frac{3+\sqrt{5}}{2}=(\frac{1+\sqrt{5}}{2})^2$
The above implies that,
$$\displaystyle F(k+1)=F(k)+F(k-1)\le(\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{5}}{2})^{k-2} \text{ (By induction hypothesis)}$$
$$\displaystyle \Rightarrow (\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{5}}{2})^{k-2}\le (\frac{1+\sqrt{5}}{2})^{k-2}(\frac{1+\sqrt{5}}{2}+1)\le (\frac{1+\sqrt{5}}{2})^{k-2}(\frac{1+\sqrt{5}}{2})^{2}=(\frac{1+\sqrt{5}}{2})^{k} \text{ (Using Hint)}$$
|
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How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$
While solving one question, I am stuck, which looks obvious but without any feasible way to approach.
Few observations, not sure if it would help
$$
\begin{align}
\dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\
\dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7}
\end{align}
$$
|
Because $$\cos \frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{2\sin\frac{\pi}{7}\cos \frac{\pi}{7} - 2\sin\frac{\pi}{7}\cos\frac{2\pi}{7} + 2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{2\pi}{7}- \sin\frac{3\pi}{7}+\sin\frac{\pi}{7} + \sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{1}{2}=\cos\frac{\pi}{3}.$$
|
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Integer solutions for $x^3+2=y^2$? I've heard a famous result that $26$ is the only integer, such that $26-1=25$ is a square number and $26+1=27$ is a cubic number.In other words, $(x,y)=(5,3)$ is the only solution for $x^2+2=y^3$.
How if we make it like this $x^3+2=y^2$? Are there any integral solutions? If so, finite or infinite many?
I've checked first $100$ naturals and no solutions satisfy the equation. However, I have no ideas how to start the proof.
|
Edit: Misread the question. Will leave the answer for a while, since the writeup of the solution to the wrong problem may be interesting to MSE users.
We need to take a moderately lengthy detour through the arithmetic of $\mathbb{Z}[\sqrt{-2}]$, that is, the arithmetic of numbers of the form $a+b\sqrt{-2}$, where $a$ and $b$ are integers.
It turns out that the arithmetic is "nice," a version of the Unique Factorization Theorem holds. That crucial part of the work is not done in this answer.
The rest of the proof is quite natural. We do that "rest," in detail.
Suppose that our equation holds. Then $(x+\sqrt{-2})(x-\sqrt{-2})=y^3$. Any non-trivial common divisor of the two terms on the right must divide $2\sqrt{-2}$. But it is clear that $x$ is odd, for if $x$ is even then $x^2+2\equiv 2\pmod{4}$, so $x^2+2$ cannot be a perfect cube. Thus $x+\sqrt{-2}$
and its conjugate each have odd norm. But any non-trivial divisor of $2\sqrt{-2}$ has even norm. So $x+\sqrt{-2}$ and $x-\sqrt{-2}$ are relatively prime in $\mathbb{Z}[\sqrt{-2}]$.
By Unique Factorization, each of $x+\sqrt{-2}$ and its conjugate have the shape $\epsilon w^3$, where $\epsilon$ is a unit. The only units are $\pm 1$. If $\epsilon=-1$, it can be absorbed into $w$, so we may assume that
$x+\sqrt{-2}=w^3$.
Let $w=a+b\sqrt{-2}$. Expanding the cube, we find that
$$x+\sqrt{-2}=(a+b\sqrt{-2})^3= a^3-6ab^2+(3a^2b-2b^3)\sqrt{-2}.$$
This yields the equations $a^3-6ab^2=x$ and $3a^2b-2b^3=1$.
But $3a^2b-2b^3=1$ is a very restrictive condition. Since $b$ divides the left side, we must have $b=\pm 1$. If $b=1$ we get $a=\pm 1$, while if $b=-1$, there is no $a$ that works.
Thus $x=a^3-6ab^2$ with $b=1$ and $a=\pm 1$. That gives the solutions $x=\pm 5$, $y=3$.
|
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|
Proving $x\equiv 2^{11} \mod{97}$ In $\mathbb{N}$ we suppose the equation $(F): x^{35} \equiv 2 \mod{97}$, and $(97,x)=1$.
How could I prove that $x\equiv 2^{11} \mod{97}$ ?
|
Here we look for relations ('tricks') that will solve the problem without using Fermat's Little Theorem.
We interpret the OP's question as a 'how to calculate' ${2^{11}}^{35} \pmod{97}$.
Well $2^{11} \equiv 2^8 \times 2^3 \equiv 62 \times 8 \equiv 11 \pmod{97}$ so we've found our first relation,
$\tag 1 {2}^{11} {11}^{-1} \equiv 1 \pmod{97}$
We suspect that progress can be made by calculating ${11}^{-1} \pmod{97}$, i.e. solving the equation
$\quad 11x \equiv 1 \pmod{97}$
This can be done by hand using an algorithm discussed here. Proceeding,
Multiply by $ 9 $:
$\quad 99 x \equiv 9 \pmod{97}$
$\quad 2 x \equiv 9 \pmod{97}$
Multiply by $ 48 $:
$\quad 96 x \equiv 432 \pmod{97}$
$\quad -1 x \equiv 44 \pmod{97}$
$\quad 1 x \equiv 53 \pmod{97}$
So
$\quad 11^{-1} \equiv 53 \pmod{97}$
But then
$\quad 2^{1}11^{-1} \equiv 9 \pmod{97}$
$\quad 2^{2}11^{-2} \equiv 81 \equiv -16 \equiv - 2^4 \pmod{97}$
and we can write down another relation,
$\tag 2 2^{-2}11^{-2} \equiv -1 \pmod{97}$
The two (primitive) relations are all we need.
$\quad \displaystyle {2^{11}}^{35} \equiv 11^{35} \equiv 11^{35} (-1) (11^{-34}) (2^{-34}) \equiv (-1) 11 \,{(2^{11})}^{-3}\, 2^{-1} \equiv (-1) \,({11}^{-2}) \, (2^{-1}) \equiv$
$\quad \quad (-1) \,\big [ ({11}^{-2}) \, (2^{-2}) \big ] \,2 \equiv (-1) \,(-1) \,2 \equiv 2 \pmod{97}$
|
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|
Proving $1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$ without factoring Is there a way to prove this identity without factoring?
$$1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$$
|
Motivated by the fact that $(a^2 + b^2 - ab)(a+b) =a^3 + b^3$, we see LHS = $1 - \frac{1}{2} \sin(2x) = \sin(x)^2 + \cos(x)^2 - \sin(x) \cos(x)$ and we are done.
|
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|
Finding roots of complex equation? Determine all roots of the equation $x^6+(3+i)x^3 + 3i = 0$ in $\mathbb{C}$ Express answers in standard form
|
Let $x^3 = z$. We then have that
$$z^2 + (3+i)z + 3i = 0 \implies (z+3)(z+i) = 0 \implies z = -3, -i \implies x^3 = -3,-i$$
Hence,
$$x = \sqrt[3]{-3},\sqrt[3]{-3} \omega, \sqrt[3]{-3} \omega^2, i, i\omega, i\omega^2$$where $\omega$ is the complex cube-root of $1$ and $\sqrt[3]{-3}$ is the real number $y$ such that $y^3 = -3$.
|
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|
Find Rational Roots and Factor $x^4 + x^3 - 4x^2 + x + 1$ Find all rational roots of the polynomial $x^4 + x^3 - 4x^2 + x + 1$. Factor this polynomial into a product of monic irreducible polynomials over the rationals.
I am confused about this question, but isn't the rational root theorem tell us the roots are only $\pm1$?
|
\begin{align}
x^4+x^3-4x^2+x+1 & = (x^4 - 2x^2 + 1) + (x^3 - 2x^2 + x) = (x^2-1)^2 + x(x-1)^2\\
& = (x-1)^2 \left( (x+1)^2 + x \right) = (x-1)^2(x^2 + 3x +1)\\
& = (x-1)^2 \left( \left(x+\dfrac32\right)^2 + 1 - \dfrac94\right)\\
& = (x-1)^2 \left( \left(x+\dfrac32\right)^2 - \dfrac54\right) = (x-1)^2 \left( \left(x+\dfrac32\right)^2 - \left(\dfrac{\sqrt5}2 \right)^2\right)\\
& = (x-1)^2 \left(x + \dfrac{3+\sqrt5}2\right) \left(x + \dfrac{3-\sqrt5}2\right)
\end{align}
|
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|
$2\int_0^\pi \sin x + \sin x \cos x\,dx$. Where am I going wrong?
$2\displaystyle\int_0^\pi \sin x + \sin x \cos x\,dx$
I let $u=\sin x \implies \dfrac{du}{dx}=\cos x \implies dx=\dfrac{du}{\cos x}$
$2\displaystyle\int_{x=0}^{x=\pi} 2u\,du$
When I try to change the limits I just get 0 and 0: Lower limit $=\sin 0 = 0$, upper limit $=\sin \pi=0$
$2\left[u^2\right]^0_0 = 0$. The answer should be 4, not 0.
Thanks,
|
As $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$
If $f(x)=\sin x+\sin x\cos x,$
$f(\pi+0-x)=\sin(\pi+0-x)+\sin(\pi+0-x)\cos(\pi+0-x)=\sin x-\sin x\cos x$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$
So, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi (\sin x - \sin x \cos x)dx=I\text{ say}$$
So, $$2I=\int_0^\pi (\sin x + \sin x \cos x)dx+\int_0^\pi (\sin x - \sin x \cos x)dx=2 \int_0^\pi\sin xdx$$
So, $$I=\int_0^\pi\sin xdx=(-\cos x)|_0^\pi=-\cos\pi-(-\cos 0)=2$$
Alternatively, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi \left(\sin x +\frac{\sin2x}2\right)dx=\left(-\cos x -\frac{\cos2x}4\right)_0^\pi$$
$$=\left(\cos x +\frac{\cos2x}4\right)_\pi^0=\cos0+\frac{\cos0}4-\left(\cos\pi+\frac{\cos\pi}4\right)=1+\frac14-\left(-1+\frac14\right)=2$$
In fact, $\int_0^\pi\sin2xdx=\int_0^{2\pi}\sin ydy=(-\cos y)_0^{2\pi}=1-1=0$
|
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|
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
My thought:
$365 = 5 \times 73$ where $5$ and $73$ are prime numbers.
So we can obtain $x^2 \equiv {-1} \pmod 5$ and $x^2 \equiv {-1} \pmod {73}$.
For $x^2 \equiv {-1} \pmod 5$,
we checked $5 \equiv 1 \pmod 4$, therefore it is solvable.
Using the Euler criterion, $(-1)^{\frac{5-1}{2}} \equiv 1 \pmod 5$.
Thus it has two solutions.
For $x^2 \equiv {-1} \pmod {73}$,
we checked $73 \equiv 1 \pmod 4$, therefore it is solvable.
Using the Euler criterion, $(-1)^{\frac{73-1}{2}} \equiv 1 \pmod{73}$.
Thus it has two solutions.
So, there are four solutions in total.
|
Because $5$ and $73$ are relatively prime, we have $x^2\equiv -1\pmod{5\cdot 73}$ if and only if $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{73}$.
Because $5$ is a prime of the form $4k+1$, the congruence $x^2\equiv -1 \pmod{5}$ has a solution, and hence two solutions. The same applies to $x^2\equiv -1\pmod{73}$.
Suppose that $(a,b)$ is an ordered pair such that $a^2\equiv -1\pmod{5}$ and $b^2\equiv -1\pmod{73}$. By the Chinese Remainder Theorem, there is a unique $c$ (modulo $5\cdot 73$) such that $c\equiv a\pmod{5}$ and $c\equiv b\pmod{73}$.
Then $c^2\equiv -1\pmod{5\cdot 73}$, so $c$ is a solution of $x^2\equiv -1\pmod{365}$. Conversely, any solution $x$ of $x^2\equiv -1\pmod{365}$ gives rise to such an ordered pair.
There are $4$ such ordered pairs, and hence $4$ solutions of the congruence modulo $365$.
Remark: The four solutions come in two $\pm$ pairs, so the task of finding them is only half as unpleasant as it looks. Actually, with the Chinese Remainder Theorem, once you have found $A$ and $B$ such that $73A\equiv 1 \pmod{5}$ and $5B\equiv 1\pmod{73}$, all four solutions can be written down mechanically. We do need to know that $\pm 2$ are the solutions of $x^2\equiv -1\pmod{5}$ and $\pm 27$ are the solutions of $x^2\equiv -1\pmod{73}$.
The idea generalizes: there are $8$ solutions modulo $5^3\cdot 73\cdot 97^2$.
|
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|
Determine the definite limit The following limit
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
evaluates to 1/12.
This is my progress so far:
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
$$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sqrt[3]{x^2}}{3(1 - x)}$$
$$\lim_{x\to 1}\frac{3(1 + \sqrt{x})- 2(1 + \sqrt[3]{x} + \sqrt[3]{x^2})}{6(1 - x)}$$
And that's as far as I go.
|
If we pose $x=1-h$ we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{2(1 - \sqrt{1-h})} - \frac{1}{3(1 - \sqrt[3]{1-h})}$$
and we have $$\sqrt{1-h}=1-\frac{1}{2}h-\frac{1}{8}h^2+o(h^2)$$ and $$(1-h)^{1/3}=1-\frac{1}{3}h-\frac{1}{9}h^2+o(h^2)$$
hence we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{h+\frac{1}{4}h^2+o(h^2)}-\frac{1}{h+\frac{1}{3}h^2+o(h^2)}\sim_0 \frac{\frac{1}{3}h^2-\frac{1}{4}h^2}{h^2}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$$
|
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|
Use Cauchy product to find a power series represenitation of $1 \over {(1-x)^3}$
Use Cauchy product to find a power series represenitation of
$$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$.
Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is there another way?
|
This is most easily done with the binomial theorem and the identity
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
which gives
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k
\end{align}
$$
To use the Cauchy products, we would use the $n=1$ version of the series above
$$
\frac1{1-x}=\sum_{k=0}^\infty x^k\tag{1}
$$
The Cauchy product method says that
$$
\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{k=0}^\infty b_k x^k\right)
=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)x^k\tag{2}
$$
$(2)$ applied to $(1)$ twice gives
$$
\begin{align}
\frac1{(1-x)^2}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k1\cdot1\right)x^k\\
&=\sum_{k=0}^\infty(k+1)x^k\tag{3}
\end{align}
$$
$(2)$ appplied to $(1)$ and $(3)$ yields
$$
\begin{align}
\frac1{(1-x)^3}
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k(j+1)\cdot1\right)x^k\\
&=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k\tag{4}
\end{align}
$$
|
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|
How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try?
$$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
|
Let $z=e^{ix}$ and then $\cos x=\frac{1}{2}(e^{ix}+e^{-ix})=\frac{1}{2}\left(z+\frac{1}{z}\right),dz=izdx$. We have
\begin{eqnarray*}
\int_0^\pi(\log(1-2a\cos x+a^2))^2dx&=&\frac{1}{2}\int_{-\pi}^{\pi}(\log(1-2a\cos x+a^2))^2dx\\
&=&\frac{1}{2}\int_{|z|=1}\left[\log(1-a\left(z+\frac{1}{z}\right)+a^2)\right]^2\frac{dx}{iz}
\end{eqnarray*}
Note that
\begin{eqnarray*}
\log(1-a\left(z+\frac{1}{z}\right)+a^2)=\log(1-a)-\log z+\log\left(1+\frac{az(1-az)}{1-a}\right)
\end{eqnarray*}
and $\log z$ is analytic in $|z|<1$ and
$$ \log\left(1+\frac{az(1-az)}{1-a}\right)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac{az(1-az)}{1-a}\right)^n. $$
Thus
\begin{eqnarray*}
\int_0^\pi(\log(1-2a\cos x+a^2))^2dx&=&\frac{1}{2i}2\pi i\text{Res}(\left(\log(1-a\left(z+\frac{1}{z}\right)+a^2)\right)^2,0)\\
&=&\pi\log^2(1-a).
\end{eqnarray*}
|
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|
How to derive a closed form of a simple recursion? Consider: $$T(n) = 2 T(n-1) + 1$$ with $T(1)$ a positive integer constant $a$.
I just stuck in finding a closed form for this simple recursion function. I would appreciate it, if someone gives me a hint.
|
A proof with generating functions: Let $$f(X) := \sum_{n=1}^{\infty} T(n) X^n$$ be a formal power series (we don't care about convergence); then you have $$f(X) = aX +\sum_{n=2}^{\infty} (2T(n-1) + 1)X^n$$ $$= aX + 2Xf(X) + \sum_{n=2}^{\infty} X^n$$ $$=2Xf(X) + \frac{X^2}{1-X} +aX.$$ Manipulate this to $$f(X) = \frac{X^2}{(1-X)(1-2X)} + \frac{aX}{1-2X}$$ $$= \frac{1}{2} + \frac{1/2}{1 - 2X} - \frac{1}{1-X} + \frac{aX}{1-2X}$$ $$=\frac{1}{2} + \sum_{n=0}^{\infty} 2^{n-1} X^n - \sum_{n=0}^{\infty} X^n + \sum_{n=1}^{\infty} 2^{n-1}a X^n$$ $$= 0 + \sum_{n=1}^{\infty} (2^{n-1} - 1 + 2^{n-1}a) X^n$$ and comparing coefficients gives $T(n) = 2^{n-1} - 1 + 2^{n-1}a.$
|
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|
Finding functional equation for generating function I'm given
$$
a_n = \sum_{i=2}^{n-2} a_ia_{n-i} \quad (n\geq 3), a_0 = a_1=a_2 = 1
$$
and I need to find the functional equation for the generating function satisfying the above equality. I obtained the correct answer, but I don't understand why some of the steps are allowed.
$$
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2} a_i a_{n-i} x^n\\
&=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2}a_ix^i a_{n-i}x^{n-i}\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{n-2} a_ix^i a_{n-i}x^{n-i} \text{ (since it doesn't make sense for $n$ to start at $0$})\tag1\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n-i}x^{n-i} \text{ (since terms past $n-2$ don't exist)} \tag2\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n}x^{n} \text{ (not sure)} \tag3
\end{align}
$$
Therefore,
$$
f(x) -1-x-x^2 = (f(x) -1 -x)^2.
$$
So the equations $(1),(2)$, and $(3)$, I'm not sure why those steps are allowed (or if they're even correct). Could someone please help?
|
For this problem, I would go in the other direction:
$\begin{align}
f^2(x)
&= (\sum_{n=0}^{\infty} a_n x^n) (\sum_{m=0}^{\infty} a_m x^m)\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n x^n a_m x^m\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n a_m x^{n+m}\\
&= \sum_{i=0}^{\infty} \sum_{j=0}^{i}a_j a_{i-j} x^{i}\\
&= \sum_{i=0}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +\sum_{j=2}^{i-2}a_j a_{i-j} )\\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +a_i )\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1\sum_{i=2}^{\infty} x^i a_{i-1}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1 x\sum_{i=2}^{\infty} x^{i-1} a_{i-1}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x\sum_{i=1}^{\infty} x^{i} a_{i}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x (f(x)-a_0)\\
&= a_0^2+ 2 a_0 a_1 x + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)- 2 a_0 a_1 x\\
&= a_0^2 + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)\\
&= 1 + f(x)(2 x+3)-3(x+1)\\
&= f(x)(2 x+3)-3x-2\\
\end{align}
$
This gives a quadratic equation for $f(x)$
(as expected, from the recursion for $a_n$):
$f^2 - (2x+3)f + 3x+2 = 0$.
Solving,
the discriminant is
$(2x+3)^2 - 4(3x+2)
= 4x^2+12x+9-12x-8
=4x^2+1
$,
so the solutions are
$f(x) = \frac{2x+3 \pm \sqrt{4x^2+1}}{2}
$.
Since $f(0) = 1$,
we have to take the minus sign,
so
$f(x) = \frac{2x+3 - \sqrt{4x^2+1}}{2}
$.
There is a significant chance that
there is some error here
(since I did it off the top of my head),
but the basic idea is that
this type of recurrence always leads to
a quadratic equation in the generating function.
Note that the coefficients of all the odd powers of $x$
(from $x^3$ on)
are zero, which agrees with Glenn O's comment
about $a_3$.
Also, I do not know whether or not it is a coincidence
that the discriminant does not have an $x$ term -
if it did, expanding the square root would be a lot harder.
|
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|
Find all values x, y and z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes.
It seems trivial that the only set of integers x, y and z that work are $(1^2 + 1)(2^2 + 1) = 3^2 + 1$, which is equivalent to $2 * 5 = 10$, but how would I go about proving this, or are there any other sets for which the following equation works out?
Thanks in advance.
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Odd primes: We first take care of the case $x^2+1$, $y^2+1$ both odd primes. If $x^2+1$ and $y^2+1$ are odd, then $x$, $y$, and $z$ are even.
Factor in the Gaussian integers. We get
$$(x-i)(x+i)(y-i)(y+i)=(z-i)(z+i).$$
Note that $x\pm i$ and $y\pm i$ are Gaussian primes. And since $z$ is even, the Gaussian integers $z-i$ and $z+i$ are relatively prime.
The Gaussian prime $x-i$ divides one of $z-i$ or $z+i$. The same remark applies to all the other Gaussian primes on the left. And we cannot have, for example, both $x-i$ and $x+i$ dividing $z-i$, else $x^2+1$ would, but it doesn't. Similarly, $y-i$ and $y+i$ cannot both divide $z-i$. Similar comments can be made about $z+i$.
It follows that $z-i$ is a unit times one of the products $(x-i)(y-i)$ or $(x-i)(y+i)$ or $(x+i)(y-i)$ or $(x+i)(y+i)$.
The arguments for all four cases are now essentially the same. Consider the first case, $(x-i)(y-i)$ equal to a unit times $z-i$. We have $(x-i)(y-i)=xy+1-i(x+y)$. Suppose that this is equal to a unit $\epsilon$ times $z-i$. Then
$$xy+1-i(x+y)=\epsilon(z-i).$$
If $\epsilon=\pm 1$, we are in trouble because $xy+1$ is odd and $\epsilon z$ is even. If $\epsilon=\pm i$, then $xy+1=\pm 1$, which is impossible.
One prime even: This is the case $x=1$. We get the Gaussian factorization $(1-i)(1+i)(y-i)(y+i)=(z-i)(z+i)$. The same argument as the one above shows that $z-i$ is a unit times $(1-i)(y\pm i)$. Multiplying out $(1-i)(y\pm i)$, we reach the conclusion that $y=\pm 2$. For example, if $z-i$ is a unit times $(1-i)(y+i)$, it is a unit times $1+y+i(1-y)$. If the unit is $\pm 1$, we get $1-y=\mp 1$. And if the unit is $\pm i$, then $1+y=\mp 1$.
.
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Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3.
In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So what I have done is said all primes $>3$ are odd $\implies n = 2k + 1$ for some $k \in \mathbb{N}$. Checking $2^1$ and $2^3$, we see that $2^{2k+1} \equiv 2 \mod 3$. From the lecture notes, I know that $2^{2k} \equiv 1 \mod 3$. I can then write
$$2^{2k + 1} = \underbrace{2^{2}\times 2^{2} \times \cdots \times 2^{2}}_\text{k times} \times 2 \equiv (1 \times 1 \times \cdots \times 1) \times 2 \equiv 2 \mod 3.$$
So we get that
$$2^{n} - 1 \equiv (2 \mod 3) - 1 \equiv 1 \mod 3$$
and so $3$ does not divide $2^n - 1$ when $n$ is odd. How does this show for it not dividing by $7$?
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Considering $7|2^n-1$.
$2^n-1= (2-1)(2^{n-1}+2^{n-2}+ \dots1)$
$7k=(2^{3m}-1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$
Let $7k=2^n-1=(2-1)(2^{n-1}+2^{n-2}+ \dots1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$
$(2^{n-1}+2^{n-2}+ \dots1)=(2^{3m-1}+2^{3m-2}+ \dots1) \implies n=3m$. But $n$ is a prime greater $3$, a contradiction.
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What is "prime factorisation" of polynomials? I have the following question:
Find the prime factorization in $\mathbb{Z}[x]$ of $x^3 - 1, x^4 - 1, x^6 - 1$ and $x^{12} - 1$. You will need to check the irreducibility in $\mathbb{Z}[x]$, of three quadratic polynomials and of one quartic. In the case of the quadratic, you will need to check that it has no integer zeros and does not factorize as a product of two quadratics with integer coefficients.
For starters, what does prime factorization mean? If we look at $x^3 - 1$ as an example, I can see that this can't be factorized any further, so it is irreducible in $\mathbb{Z}[x]$. Does that mean there is no prime factors for $x^3 - 1$?
If we then look at $x^4 -1 = (x^2 - 1)(x^2 + 1)$. Here, we see that it is irreducible and so would it simply be that the prime factorization of $(x^4 -1 ) = (x^2 + 1)(x - 1)(x + 1)$?
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For quadratics and cubics, irreducibility is easy to test, for if there are no rational roots, the polynomial must be irreducible. And the polynomials $x^2+x+1$ and $x^2-x+1$ don't even have real roots. In more complicated cases, the Rational Roots Theorem could be helpful.
Let's deal with $x^{12}-1$. This immediately factors as $(x^6-1)(x^6+1)$. We leave further decomposition of $x^6-1$ to you. For $x^6+1$, note that it is equal to $(x^2+1)(x^4-x^2+1)$.
We would like to show that $x^4-x^2+1$ is irreducible. It certainly cannot be written as a linear polynomial with coefficients in $\mathbb{Z}$ times a cubic, since $x^4-x^2+1=0$ has no rational roots. Can we express $x^4-x^2+1$ as a product of quadratics with integer coefficients?
If we can, without loss of generality we would have that the decomposition has shape $(x^2+ax+b)(x^2+cx+d)$. Expand. Since $x^4-x^2+1$ has no $x^3$ term, we must have $a+c=0$, so $c=-a$.
Also, we must have $b=1$, $d=1$ and $b=-1$, $d=-1$. This gives the two possibilities (i) $(x^2+ax+1)(x^2-ax+1)$ and (ii) $(x^2+ax-1)(x^2-ax-1)$.
We check there is no factorization of type (i). For the coefficient of $x^2$ in the product $(x^2+ax+1)(x^2-ax+1)$ is $2-a^2$. This cannot be $-1$ for integer $a$.
Similarly, one can show (ii) can't work.
Remark: We want to express each polynomial as a product of irreducibles. The factorization $x^4-1=(x-1)(x+1)(x^2+1)$ is such a factorization. It is not quite unique. For example we also have $x^4-1=(-x+1)(x+1)(-x^2-1)$, and two other obvious variants. But the decomposition is essentially unique, so giving one decomposition is enough.
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Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
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You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are $A,B,C$ opposite to the sides $a,b,c$, respectively. Then
$$ c^2=a^2+b^2-2ab\cos C, S=\frac{1}{2}ab\sin C $$
and hence
\begin{eqnarray*}
a^2+b^2+c^2-4S\sqrt{3}&=&2[a^2+b^2-ab(\cos C+\sqrt{3}\sin C)]\\
&=&2[a^2+b^2-2ab\sin(\frac{\pi}{6}+C)]\\
&\ge&2(a^2+b^2-2ab)\\
&=&2(a-b)^2\\
&\ge&0
\end{eqnarray*}
since $\sin(\frac{\pi}{6}+C)\le 1$. The equal sign holds iff $a=b$ and $\sin(\frac{\pi}{6}+C)=1$, which implies that $a=b, C=\frac{\pi}{3}$ or $a=b=c$.
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|
Summation of sequence $a_n - a_{n-1} = 2n$ $(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to?
$a_1$ = 2
$a_2$ = 2 + 2x2
$a_3$ = 6 + 2x3
$a_4$ = 12 + 2x4
$a_5$ = 20 + 2x5
$a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn
$b_3$-$b_2$=4
$b_4$-$b_3$=6
$b_5$-$b_4$=8
$b_6$-$b_5$=10
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I have 2 methods of doing this:
Method one:
$a_2-a_1=4$, i.e. $a_2=6$
$a_3-a_2=6$, i.e. $a_3=12$
$a_4-a_3=8$, i.e. $a_4=20$
You will notice that $a_k = 2+4+6+8+ \cdots +2k = \frac{k}{2}(2k+2)=k^2 + k$ (using summation formula)
$a_1+a_2+ \cdots + a_{20} = \displaystyle\sum_{k=1}^{20} (k^2+k)=\frac{(20)(20+1)(2\times20+1)}{6}+20\times\frac{1+20}{2}=3080$
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Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...
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The idea of discriminant works nicely, as @Jyrki, @Andre and @Mike answered. Yet, I think of another (perhaps more "number theory"ish :) way.
First, the equation is symmetric in $x$ and $y$, hence without loss of generality, we assume $x\ge y$.
Since $xy+1$ divides $2^z$ and is not equal to $1$, then it should be even, meaning that $x$ and $y$ should both be odd. Let $x=2a+1$ and $y=2b+1$. A replacement yields
$$
4(a+b+1)(2ab+a+b+1)=2^z.
$$
Since $a+b+1$ is a power of $2$, $2ab+a+b+1$ will too be another power of $2$ only if $2ab=a+b+1$ or $2ab=0$ (a simple proof follows from assuming contrary). The latter case yields $b=0$ (or $a=0$) which means $y=1$ and $(x+1)^2=2^z$. Therefore, a set of solutions is found among $(x,y,z)=(2^u-1,1,2u)$ (and obviously $(x,y,z)=(1,2^u-1,2u)$). The former case yields
$$
{2ab=a+b+1\implies
\\(2a-1)(2b-1)=3\implies
\\(x-2)(y-2)=3\implies
\\x=5,y=3\implies z=7
\\\text{ or }
\\x=3,y=5\implies z=7
}.
$$
In conclusion, all the solutions are
$$
(x,y,z){\in
\left\{(2^u-1,1,2u):u\in\Bbb N\right\}
\\\cup
\left\{(1,2^u-1,2u):u\in\Bbb N\right\}
\\\cup
\left\{(3,5,7),(5,3,7)\right\}
}
$$
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|
In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
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Following answer is for general case.
Denote by $c_m(1,n)=\binom{n-1}{m-1}$ the number of composition of $n$ into $m$ positive parts, and by $c_m(2,n)$ number of compositions of $n$ into $m$ parts greater than $1$.
Each composition of $n$ into $m$ parts greater than $1$, if all parts decrease by 1, becomes a composition of $n-m$ into $m$ positive parts. That means $$c_m(2,n)=c_m(1,n-m)=\binom{n-m-1}{m-1}$$
Number of all compositions of $n$ into parts greater than $1$ is $$c(2,n)=\sum_{m=1}^{n}c_m(2,n)=\sum_{m=1}^{n}\binom{n-m-1}{m-1}$$
In our case $n=12$
$$c(2,12)=\sum_{m=1}^{12}\binom{11-m}{m-1}=\sum_{m=1}^{6}\binom{11-m}{m-1}=$$
$$=\binom{10}{0}+\binom{9}{1}+\binom{8}{2}+\binom{7}{3}+\binom{6}{4}+\binom{5}{5}=$$
$$1+9+28+35+15+1=89$$
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Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ The question is basically in the title: Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$
I get how to do it from $7\mid $ and $7\mid y$ to $7\mid x^2+y^2$, but not the other way around.
Help is appreciated! Thanks.
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$x^2+y^2 \equiv \mod 7 \implies x^2 \equiv k \mod 7$ and $y^2 \equiv 7-k \mod 7$
And any $a^2 \equiv 0,1,4,2\mod 7$(Why?) $\implies k=0$
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Determine the values of $a,b,c$ for which the function is continuous at $x=0$
Determine the values of $a,b,c$ for which the function is continuous at $x=0$
$$
f(x) =
\begin{cases}
\frac{\sin(a+1)x+sinx}{x} \qquad \text{if} \ x<0 \ ; \\ \\
c \qquad\quad\qquad \text{if} \ x=0 \ ; \\ \\
\frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}} \ \ \ \text{if} \ x>0 \ .
\end{cases}
$$
I tried to solve the problem like this:
$$\lim_{x\to 0}\frac{\sin(a+1)x+sinx}{x}$$
$$=\lim_{x\to 0}\frac{acosx}{x}+cosax +1$$
$$=2+a$$
and
$$\lim_{x\to 0}\frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}}$$
$$\lim_{x\to 0}\frac{\sqrt{1+bx}-1}{bx}$$
$$=?$$
Can you help me solve this problem?
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Hint: $$\frac{\sqrt{1+bx}-1}{bx}=\frac{\sqrt{1+bx}-1}{bx}\cdot\frac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1}=\frac{1}{\sqrt{1+bx}+1}$$
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Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. Maybe distribute the $i$? Wondering if I can get some help.
|
Take a Weierstrass substitution $u = \tan \frac{\theta}{2}$ after reducing the integration range using $\cos(2\pi - \theta) = \cos(\theta)$ in the second integral below:
$$
\int_0^{2\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} = \int_0^{\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} + \int_{\pi}^{2\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta}
$$
giving
$$
\int_0^{2\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} = 2 \int_0^{\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta}
$$
Now apply the substitution, using $\cos\theta = \frac{1-u^2}{1+u^2}$, and $\mathrm{d}\theta = \frac{2\mathrm{d}u}{1+u^2}$:
$$
2 \int_0^{\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} = 4 \int_0^\infty \frac{\mathrm{d}u}{a^2-a+1 + (a^2+a+1) u^2} = 4 \cdot \frac{\pi}{2 \sqrt{(a^2-a+1)(a^2+a+1)}} = \frac{2\pi}{\sqrt{a^4+a^2+1}}
$$
where we used the assumption $a^2-a+1>0$ and $a^2+a+1>0$, valid for all real $a$.
|
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For positive numbers a,b,c ,$abc=1$ prove$ \frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$ For positive numbers $a$,$b$,$c$ , with $abc=1$ prove $$\frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$$
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The problem as stated is false. Take $a=b=0.1$, $c=100$. $\frac{1}{a+b}=5$.
However, if $\min(a,b,c)\ge 1$ (for example, if they are integers), then it's true, by the other posted solution.
|
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|
Finding the locus of the midpoint of chord that subtends a right angle at $(\alpha,\beta)$ There is a circle $x^2+y^2=a^2$. On any line that cuts the circle in two distinct points(it is a secant), the points of intersection with circle are taken and at those two points I draw the tangents that intersect at some point, say $(\alpha,\beta)$. It's given that the tangents intersect at right angles at the point $(\alpha,\beta)$. I need to find the locus of the midpoint of the chord.
THE BOOK'S WAY:
Let the midpoint be $(h,k)$.
Using $T=S_1$ for the equation of chord, the chord is $hx+ky=h^2+k^2$
If the chord intersects the circle at $(x_1,y_1)$ and $(x_2,y_2)$, then by condition of perpendicularity
$$m_1m_2=-1$$
$$({y_1-\beta \over x_1-\alpha})({y_2-\beta \over x_2-\alpha})=-1$$
$$(x_1x_2+y_1y_2)+(\alpha^2+\beta^2)=\alpha(x_1+x_2)+\beta(y_1+y_2)$$
Then using the equation of chord and separately eliminating $x,y$ we obtain quadratics
$$\lambda x^2-2\lambda hx+\lambda^2-a^2k^2=0$$
$$\lambda y^2-2\lambda ky+\lambda^2-a^2h^2=0$$
where $\lambda=h^2+k^2$. Using the values of product of roots and sum of roots, the locus is
$$x^2+y^2-\alpha x-\beta y+{1 \over 2}(\alpha ^2+\beta ^2-a^2)=0$$
MY WAY:
I translate the origin to $(\alpha,\beta)$ so that $x=X+\alpha$, $y=Y+\beta$
Circle becomes $X^2+Y^2+2\alpha X+2\beta Y+\alpha^2+\beta^2-a^2=0$
To get the equation of pair of lines that join the circle and chord intersection points to origin(translated), I homogenise the equation of circle, then retranslate the axes to previous origin by using $X=x-\alpha$, $Y=y-\beta$ and then put the sum of coefficent of $x^2$ and coefficient of $y^2$ equal to zero. Then I get an equation that doesn't match the book's answer. What's the flaw?
Thanks in advance[Sorry if that's a long question]
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Parameterize circle as $(a \cos \alpha, a \sin \alpha)$ with $\alpha \in \left[ 0 , 2 \pi \right]$, now the tangent vector is given as $(-a \sin \alpha , a \cos \alpha)$. Let the two points correspond to end of chord be at angle $\alpha$ and at angle $\beta$, then:
$$ (a \cos \alpha, a \sin \alpha) \cdot ( a \cos \beta , a \sin \beta)=0$$
Since the tangents intersect at ninty degrees, then:
$$ \cos (\alpha - \beta)=0$$
WLOG, let $\alpha>\beta$ then $\alpha - \beta= \frac{\pi}{2}$ or $ \alpha - \beta = \frac{3 \pi}{2}$ the position vector for mid point of the chord is given as:
$$ (X,Y)=( a \frac{ \cos \alpha + \cos \beta}{2} , a \frac{ \sin \alpha + \sin \beta} {2})=(a \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha-\beta}{2},a \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2})$$
Taking the modulus:
$$ X^2 +Y^2 = a^2 \cos^2 \frac{\alpha - \beta}{2}$$
But, $\alpha - \beta= \frac{\pi}{2}$ or $ \alpha - \beta = \frac{3\pi}{2}$, hence: $ \cos^2 \alpha - \beta = \frac{1}{2}$, simplfying our equation to:
$$ X^2 +Y^2 = \frac{a^2}{2}$$
QED
More generally locus of midpoint of chord, between point making angle $\alpha$ and $\beta$ is given as:
$$ X^2 +Y^2 = a^2 \cos^2 ( \frac{\alpha - \beta}{2})$$
|
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|
Finding a coefficient using generating functions I need to find a coefficient of $x^{21}$ inside the following expression:
$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}$$
I think the only way (using generating functions) is to express the parentheses content with a generating function.
The generating function for the $(1, x, x^{2}, x^{3}, ...)$ sequence is equal to $\frac{1}{1-x}$.
However, well...the expression at the top is not infinite, so I can't really express it as a generating function.
How can I do this?
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$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}=(x^2(1+x+x^2+x^3+x^4))^8=$$
$$=x^{16}\left(\frac{1-x^5}{1-x}\right)^8=x^{16}(1-x^5)^8(1-x)^{-8}=$$
$$=x^{16}\sum_{k=0}^{8}\binom{8}{k}(-x)^{5k}\sum_{l=0}^{\infty}\binom{8+l-1}{l}x^l=$$
$$=\sum_{l=0}^{\infty}\sum_{k=0}^{8}(-1)^{5k}\binom{8+l-1}{l}\binom{8}{k}x^{16+5k+l}$$
$$16+5k+l=21\Rightarrow k=1,l=0 \text{and}k=0,l=5$$
coefficient next $x^{21}$ is
$$(-1)^0\binom{8+5-1}{5}\binom{8}{0}+(-1)^5\binom{8+0-1}{0}\binom{8}{1}=\binom{12}{5}-8$$
|
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
|
Let $p = a+b+c$, then $ab+bc+ca = \frac{1}{2}(p^2 - p)$ and $abc=1$. So $a,b,c$ are solutions to the equation
$$x^3 - px^2 + \frac{1}{2} (p^2-p)x - 1 = 0$$
The discriminant is $(\frac{1}{2}(p^2-p))^2 - 4(\frac{1}{2}(p^2-p))^3 - 4p^3 - 27 + 18 p (\frac{1}{2}(p^2-p)) \ge 0$, since the roots are real.
This is equivalent to
$$p^6 - 4p^5 + 5p^4 - 22p^3 + 36p^2 + 108 \leq 0$$
But by calculus or other means, one can check that the left hand side has minimum 0 only at $p = 3$. Thus $p = 3$, and $ab+bc+ca = 3$, $abc = 1$. So $a,b,c$ are roots of $x^3-3x^2+3x-1 = (x-1)^3$, i.e. $a,b,c = 1$, i.e. $x=y=z$.
|
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|
The number $ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}$ For a real number $a$ and a positive integer $k$, denote by $(a)^{(k)}$ the number $a(a+1)\cdots (a+k-1)$ and $(a)_k$ the number
$a(a-1)\cdots (a-k+1)$. Let $m$ be a positive integer $\ge k$. Can anyone show me, or point me to a reference, why the number
$$ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}= \frac{2^{2k}(m)^{(k)}(m)_k}{(2k)!}$$ is always an integer?
|
Using Knuth's notation:
$$
x^{\underline{k}} = x (x - 1) \ldots (x - k + 1) = (x)_{(k)} \qquad
x^{\overline{k}} = x (x + 1) \ldots (x + k - 1) = (x)^{(k)}
$$
What you are looking for is:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
$$
As if $k > m$ then $m^{\underline{k}} = 0$, while all other factors are positive, the range of interest is $0 \le k \le m$. We also have:
$$
\frac{m^{\underline{k}}}{k!} = \binom{m}{k}
$$
which is an integer, we still need:
$$
m^{\overline{k}} = (m + k - 1)^{\underline{k}}
$$
$$
\begin{align*}
(1/2)^{\overline{k}} &= (1/2) \cdot (3 / 2) \cdot \ldots \cdot (1/2 + k - 1) \\
&= \frac{1 \cdot 3 \cdot \ldots \cdot (2 k - 1)}{2^k} \\
&= \frac{1 \cdot 2 \cdot \ldots \cdot 2k}
{2^k \cdot 2^k \cdot k!} \\
&= \frac{(2 k)!}{2^{2 k} k!}
\end{align*}
$$
Thus:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
= \frac{m (m + k - 1)^{\underline{2 k-1}} 2^{2 k} }{(2 k)!}
$$
|
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|
How to solve a polynomial with power fractions like $a-ax+x^{0.8}-x^{0.2}=0$ I have something like
$a-ax+x^{0.8}-x^{0.2}=0$
with parameter a>0 and variable x>0.
I know by trial and error that the equation has three real roots for parameter a greater than certain value, otherwise it has only one root. x=1 is one obvious root. I want to find the value of parameter a that yields multiple roots. Is there a pen and paper solution to that, not using any computational algorithms like Newton's method etc.?
|
Let $y=x^{0.2}>0$, then the problem reduces to finding all $a>0$ such that the polynomial equation $a-ay^5+y^4-y=0$ has only 1 positive real root $y=1$.
For $y=1$ to be a double root, we must have $-5a(1)^4+4(1)^3-(1)=0$, so $a=\frac{3}{5}$. Putting this into the equation gives $$0=\frac{3}{5}-\frac{3}{5}y^5+y^4-y=-\frac{(y-1)^2}{5}(3y^3+y^2-y-3)=-\frac{(y-1)^3}{5}(3y^2+4y+3)$$
The quadratic $3y^2+4y+3$ has no real roots, so $a=\frac{3}{5}$ gives only 1 real root $y=1$ with multiplicity $3$.
Consider $y \not =1$, then we can write $$a=\frac{y^4-y}{y^5-1}=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$$
We know that $y=1$ is a triple root at $a=\frac{3}{5}$, so this motivates us to consider
\begin{align}
& \frac{y^3+y^2+y}{y^4+y^3+y^2+y+1} \leq \frac{3}{5} \\
\Leftrightarrow & 3y^4+3y^3+3y^2+3y+3 \geq 5y^3+5y^2+5y \\
\Leftrightarrow & (y-1)^2(3y^2+4y+3) \geq 0
\end{align}
The function $f(y)=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$ achieves the values $\frac{3}{5}$ and $0$ at $y=1, 0$ respectively, so $f(y)$ can take all values from $0$ to $\frac{3}{5}$ exclusive (remember $y \not =0, 1$) since $f$ is continuous.
Thus $\forall a \in (0, \frac{3}{5})$, there exists $y>0, y \not =1$ s.t. $a=\frac{y^3+y^2+y}{y^4+y^3+y^2+y+1}$, and $\forall a \in [\frac{3}{5}, \infty)$, $y=1$ is the only positive real root.
|
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|
Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({{x + a\ln a} \over a}\right) \cr
{1 \over y}{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \times x{a^x} \cr
{{dy} \over {dx}} &= {{x{a^{x + 1}} + {x^3}{a^x} + {a^{x + 1}}{x^2}\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {a^x} + {x^2}{a^{x - 1}} + {a^x}x\ln a \cr
{{dy} \over {dx}} &= {a^x}\left(1 + {x^2}{a^{ - 1}} + x\ln a\right) \cr
{{dy} \over {dx}} &= {a^x}\left({{a + {x^2} + x\ln a} \over a}\right) \cr} $$
However the answer in the back of the book is:
$${dy \over dx} = a^x (1 + x\ln a)$$
What have I done wrong?
|
When going from line
$$\ln y = \ln x + x \ln a$$
to
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + (x \times \frac{1}{a} + \ln a \times 1)$$
you made a mistake: when you tried to differentiate $x \ln a$ with respect to x via the product rule, you differentiated $\ln a$ with respect to a, turning it into $\frac{1}{a}$, when you should have done so with respect to x, since you were differentiating the whole product with respect to x. With respect to x, $\ln a$ is a constant. Differentiating the constant (with respect to x) $\ln a$ yields $0$, and thus you'd have $(x \times 0 + \ln a \times 1) = \ln a$. So the above line should be
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \ln a$$.
Now multiply by $y = xa^x$ and you should have your answer.
Note, of course, that you don't actually need the product rule: since $\ln a$ is constant, you can just use the rule $\frac{d}{dx} k f(x) = k \frac{df}{dx}$. And this gives $\frac{d}{dx} x \ln a = \frac{d}{dx} (\ln a)x = \ln a \frac{d}{dx} x = \ln a$.
Edit: There's a second mistake in going from the second-to-last line to the last one. $a^x (1 + x^2 a^{-1} + x \ln a) = a^x (\frac{a + x^2 + ax \ln a}{a})$ and not $a^x (\frac{a + x^2 + x \ln a}{a})$ -- but the first mistake makes this irrelevant.
|
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|
Write the expression in terms of $\log{a}$ and $\log{b}$ Write in terms of $\log{a}$ and $\log{b}$ : $\log{\frac{a^3}{b^4}}$?
using log laws $log(\frac{x}{y})$ = $\log{x} - \log{y}$
$\log{\frac{a^3}{b^4}} = \log{a^3} - \log{b^4}$
and using log law that $\log{x^y} = y\log{x}$
$\log{a^3} = 3 \log{a}$ & $\log{b^4} = 4\log{b}$
therefore $\log{\frac{a^3}{b^4}} = 3\log{a} - 4\log{b}$
does that seem correct? please show your solution.
|
It seems correct, you used well the logarithm laws.
|
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|
finding a primitive root. It says for part A to Find a primitive root r of 38? Im not sure if I did it right.
I first calculated $\phi(38)=\phi(19*2)=18$. So there are 18 numbers that are relatively prime to 38. Listing them out we get 1,3,5,7,9,11,13,15,17,21,... so on. So I decided to test out 1.
But $ord_1 38$ does not equal $\phi(38)$. But $ord_3 38=\phi(38)$.
So I calculated it until i found a power of 3 that was congruent to $1 \pmod {38}$
So I get $$3^1≡ 3 \pmod {38}$$ $$3^2 ≡ 9 \pmod {38}$$ $$3^3 ≡ 27 \pmod {38}$$ $$3^4 ≡ 5 \pmod {38}$$ $$ 3^5 ≡ 15 \pmod {38}$$ $$3^6 ≡ 7\pmod {38}$$ $$3^7 ≡ 21 \pmod {38}$$
$$3^8 ≡ 25 \pmod {38}$$ $$3^9 ≡ -1 \pmod {38}$$ $$3^{16} ≡ 17 \pmod {38}$$ $$3^{17} ≡ 13 \pmod {38}$$ $$3^{18} ≡ 1 \pmod {38}$$ So $ord_3 38=18=\phi(38)$.
|
As $\phi(19)=18$ and $ord_{19}2$ must divide $\phi(19),$ we only need to test for the powers $1,2,3,6,9,18$
$3^2=9;3^3=27\equiv-11\pmod{38};3^6\equiv(-11)^2\equiv121\equiv7\pmod{38}$
So, $3^9=3^3\cdot3^6\equiv(-11)(7)\equiv-77\equiv-1\pmod{38}$
$\implies 3 $ is a primitive root $\pmod{38}$
Alternatively, let's start with finding a primitive root of $19$
Starting with $2$, the smallest positive integer $>1$
$2^2=4,2^3=8,2^6=64\equiv7\pmod{19},$
$2^9=2^3\cdot2^6\equiv 8\cdot7\pmod{19}\equiv-1$
$\implies 2 $ is a primitive root $\pmod{19}$
Now, as $(2,38)=2>1,2$ can not be a primitive root $\pmod {38}$
But, $ord_{19}(2+19k)=ord_{19}2=18$
Now, $ord_2(2+19k)$ will be $1$ if $(2+19k)$ is odd i..e, if $k$ is odd
So, $ord_{(2\cdot19)}(2+19k)=$lcm$(1,18)=18=\phi(38)$ if $k$ is odd
Putting $k=2r+1$ where $r$ is any integer, $2+19k=2+19(2r+1)\equiv21\pmod{38}$
|
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Proving/disproving $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}$ Let $a\in\mathbb{R}\setminus\mathbb{Z}$. Prove or disprove that there do not exist three distinct $k_1, k_2, k_3\in\mathbb{N}$ such that $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}\neq 0$, where $\{x\}=x-\lfloor x \rfloor$.
|
This is not a complete answer, but I thought the approach might help someone find a solution. We can ask when the following condition holds for positive integers $a>b>c$ and $a>d\ge c$:
$$
x^a - x^d - m = P(x)(x^b - x^c - n),
$$
where $P(x)$ is some polynomial over $x$, and $m,n\in\mathbb{Z}$. If it does, then $\{x^a\}=\{x^d\}$ and $\{x^b\}=\{x^c\}$ whenever $x$ is a root of $x^b-x^c-n$. If, moreover, $d=b$ and the right-hand trinomial has a root in $\mathbb{R}$ for which $x^a\not\in\mathbb{Z}$, then OP's question has a positive answer.
Clearly $P(x)$ must start with $x^{a-b}$; so
$$
x^a-x^d-m=(x^b - x^c - n)(x^{a-b}+Q(x))=x^a-x^{a-b+c}-nx^{a-b}+Q(x)(x^b-x^c-n),
$$
or
$$
Q(x)(x^b-x^c-n)=x^{a-b+c}+nx^{a-b}-x^d-m.
$$
The constant term on the left and right must be $-m$, so $Q(x)=m/n+xR(x)$, giving
$$
(xR(x)+m/n)(x^b-x^c-n)=xR(x)(x^{b}-x^{c}-n)+(m/n)x^b-(m/n)x^c-m \\ =x^{a-b+c}+nx^{a-b}-x^d-m,
$$
or
$$
R(x)(x^b-x^c-n)=x^{a-b+c-1}+nx^{a-b-1}-x^{d-1}-(m/n)x^{b-1}+(m/n)x^{c-1}.
$$
We must have $c=1$ at this point, so
$$
R(x)(x^b-x-n)=x^{a-b}+nx^{a-b-1}-x^{d-1}-(m/n)x^{b-1}+(m/n).
$$
Now, for the highest-order terms to match, we need either $a-b \ge b$ or $d=b+1$. The latter case does not help with OP's question, but setting $d=b+1$ gives $R(x)=-1$ and
$$
-x-n=-x^{a-b}-nx^{a-b-1}+(m/n)x^{b-1}-(m/n),
$$
so $m=n=1$ and $(a,b,c,d)=(5,3,1,4)$. This gives the known but nontrivial result that $x^5-x^4-1=P(x)(x^3-x-1)$; and $x^3-x-1$ has as a real root the plastic constant $P=1.3247…$. We conclude that $\{P^5\}=\{P^4\}$ and $\{P^3\}=\{P\}$, which is interesting but not exactly what we want.
If we instead set $a=2b+k$ for $k\ge 0$, then
$$
R(x)(x^b-x-n)=x^{b+k}+nx^{b+k-1}-x^{d-1}-(m/n)x^{b-1}+(m/n).
$$
Restricting ourselves to $d=b$,
$$
R(x)(x^b-x-n)=x^{b+k}+nx^{b+k-1}-\left(\frac{m}{n}+1\right)x^{b-1}+\left(\frac{m}{n}\right).
$$
It's then possible to try various values for $k$. Interestingly, if $k=1$, then the solution $(a,b,c,d)=(5,2,1,2)$, $m=18$, $n=-3$ arises. This corresponds to
$$
x^5-x^2-18 = P(x)(x^2-x+3),
$$
which is very close to a positive result. Unfortunately $x^2-x+3$ has only complex roots: $r=1/2 + i\sqrt{11}/2$ and its complex conjugate. But indeed
$$
r^5 = 31/2 + i\sqrt{11}/2, \\
r^2 = -5/2 + i\sqrt{11}/2, \\
r = 1/2 + i\sqrt{11}/2,
$$
so this can be seen as a complex solution to $\{x^5\}=\{x^2\}=\{x\}\neq 0$.
|
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|
Prove by mathematical induction that $1 + 1/4 +\ldots + 1/4^n \to 4/3$ Please help. I haven't found any text on how to prove by induction this sort of problem:
$$
\lim_{n\to +\infty}1 + \frac{1}{4} + \frac{1}{4^2} + \cdots+ \frac{1}{4^n} = \frac{4}{3}
$$
I can't quite get how one can prove such. I can prove basic divisible "inductions" but not this. Thanks.
|
i have a solution, but not seem mathematical induction
$s = 1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \cdots$
$\dfrac1{4}s = \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^5} + \cdots$
$\dfrac1{4}s = - 1 + (1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^5} + \cdots)$
$\dfrac1{4}s = - 1 + s$
$\dfrac3{4}s =1$
$s = \dfrac4{3}$
|
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Distance between point and a line - problems simplifying the minimised distance equation Someone asked how to prove the distance between a point $(x_1,y_1)$ and a line $Ax + By + C = 0$ is:$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$ The currently accepted answer shows that a point on the line can be expressed as $(x, {-Ax-C\over B})$, then explains an approach using calculus to find the value of $x$ when the square distance is minimised, and then plug it into the distance formula: $$\tag{1}D=\sqrt{ (x_1-x)^2 + {\Bigl(y_1- { \textstyle{-Ax-C\over B}}\Bigr)^2 }}$$ and simplify.
I understand the approach. My problem is with the simplification part. I found $x$ to be minimised when $x = \dfrac{B^2x_1 - A(By_1 + C)}{A^2 + B^2}$. How would you plug this into $(1)$ and simplify effectively?
My first approach was to plug $x$ into the two places it appears in the $(1)$ but expanding $(x_1 - \frac{B^2x_1 - A(By_1 + C)}{A^2 + B^2})^2$ (let alone the rest of it) made me give up as it just seemed to get even more complex. I tried some substitutions:
but this route didn't seem to be getting me closer to the formula to be proven.
I understand there are other far simpler ways to arrive at the formula (see other answers to the linked question), but it irked me that my algebraic manipulation skills are too weak to follow through this one. What I'm wondering is how an expert would approach this. How would you break the problem down and proceed?
I was deliberating over whether to post this question because part of me is telling me I'm lazy for not just trying to plough on through it.
|
Since you wanted to see how this is done algebraically, we'll need to look at a particular property of quadratic functions. As I've said in a comment, if $r > 0$, then $ f(x) = rx^2 + sx + t \ $ represents an "upward-opening" parabola. By completing the square, we can write this as
$$f(x) \ = \ r \cdot (x^2 \ + \ \frac{s}{r} \cdot x \ + \ \frac{s^2}{4r^2}) \ + \ (t \ - \ r \cdot \frac{s^2}{4r^2}) \ = \ r \cdot ( \ x + \frac{s}{2r} \ )^2 \ + \ ( \ t - \frac{s^2}{4r} \ ) \ . $$
The vertex of this parabola lies at $ \ x = -\frac{s}{2r} \ $ , with the y-coordinate of the vertex being the minimum value of the function $ \ f(-\frac{s}{2r}) \ = \ t - \frac{s^2}{4r} .$ We will exploit this to reduce the amount of writing we will need to do.
We will label the closest point on the line to $ \ (x_1 , y_1) \ $ as $ \ (X,Y) \ $ . The distance-squared function is
$$D^2 \ = \ ( X - x_1 )^2 \ + \ (Y - y_1)^2 \ = \ ( X - x_1 )^2 \ + \ ( \ [-\frac{AX + C}{B} \ ] - y_1)^2 \ $$
$$= \ ( X - x_1 )^2 \ + \ ( \frac{A}{B} \cdot X \ + \ \frac{C}{B} + \ y_1)^2 \ = \ ( X - x_1 )^2 \ + \ ( \frac{A}{B} \cdot X \ + \ K \ )^2 \ , $$
where we will make a temporary substitution, $ \ K = \frac{C+By_1}{B} \ $ , to further save some writing. (This is often a good idea when working out complicated algebraic expressions, as it reduces the risk of embarrassing, time-wasting errors.) We will see that we can hold out for some time before needing to replace $K$ .
From here, we have
$$D^2 \ = \ ( \ X^2 \ - \ 2x_1X \ + \ x_1^2 \ ) \ + \ ( \ \frac{A^2}{B^2} \cdot X^2 \ + \ 2\frac{A}{B} K \cdot X \ + \ K^2 \ ) \ $$
$$= \ ( \ 1 + \frac{A^2}{B^2} \ ) \cdot X^2 \ + \ 2 ( \ \frac{A}{B} K - x_1 \ ) \cdot X \ + \ ( \ K^2 \ + x_1^2 \ ) , $$
which now has the "general form" for a quadratic function. Since we already know how to convert this to "standard form" by completing the square, we can simply use the results we discussed back at the start. The vertex of the parabola described by this equation lies at
$$X \ = \ -\frac{s}{2r} \ = \ -\frac{2 ( \ \frac{A}{B} K - x_1 \ )}{2( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2} \ = \ \frac{ B^2 x_1 \ - \ AB K}{A^2 \ + \ B^2 \ } $$
$$= \ \frac{ B^2 x_1 \ - \ AB \cdot (\frac{C+By_1}{B})}{A^2 \ + \ B^2 \ } \ = \ = \ \frac{ B^2 x_1 \ - \ A \cdot (C \ + \ By_1)}{A^2 \ + \ B^2 \ } \ . $$
So here is the x-coordinate of the point on the line that is closest to $(x_1,y_1)$, confirming the stated result without the use of calculus.
We do not bother substituting this into either the quadratic function or the equation for the line, since we know from the standard form of the quadratic function that its minimal value is
$$D^2_{min} \ = \ t - \frac{s^2}{4r} \ = \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{[ \ 2 ( \ \frac{A}{B} K - x_1 \ ) \ ]^2}{4( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2} $$
$$= \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{ ( \ A K \ - \ Bx_1 \ )^2}{ A^2 + B^2} \ = \ \frac{( \ K^2 \ + \ x_1^2 \ )(A^2 + B^2) \ - \ ( \ A K - Bx_1 \ )^2}{ A^2 + B^2} $$
$$= \ \frac{( \ A^2K^2 \ + \ B^2K^2 \ + \ A^2x_1^2 \ + \ B^2x_1^2 \ ) \ - \ ( \ A^2K^2 \ - \ 2ABKx_1 \ + \ B^2x_1^2 \ )}{ A^2 + B^2} $$
$$= \ \frac{ A^2x_1^2 \ + \ 2ABKx_1 \ + \ B^2K^2 \ }{ A^2 + B^2} \ = \ \frac{ ( \ Ax_1 \ + \ BK \ )^2 }{ A^2 + B^2}$$
$$= \ \frac{ ( \ Ax_1 \ + \ B \cdot [\frac{C+By_1}{B}] \ )^2 }{ A^2 + B^2} \ = \frac{ ( \ Ax_1 \ + \ By_1 \ + C \ )^2 }{ A^2 + B^2} \ .$$
Since $ \ \sqrt{x^2} = |x| \ , $ we at last have the relation for the minimum distance,
$$D_{min} \ = \ \frac{ | \ Ax_1 \ + \ By_1 \ + C \ | }{ \sqrt{A^2 + B^2}} \ .$$
It took me a few goes to find a reasonably clean way to develop this by algebra alone. The straight answer to your question about how "an expert" would set this up and proceed is that they wouldn't, but instead would use a more efficient method (such as the ones I linked to in my other answer). That may also explain why one doesn't have much luck finding an approach like this by looking around the internet. In any event, I hope this is what you were inquiring about.
|
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|
Find integral $\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$ (most likely substitution) $$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$
I tried letting $x^2=\tan \theta$ but it didn't work. What should I do?
Please don't give full solution, just a hint and I will continue.
|
Integrate as follows
\begin{align}
&\int_0^1\frac{\ln(1+x^2)}{1+x^2} dx\\
=&\int_0^1\int_0^1 \frac{2x^2y}{(1+x^2)(1+x^2y^2)}dy\ dx=\int_0^1\frac{\frac\pi2 y-2\tan^{-1}y}{y^2-1}
\ \overset{y\to \frac{1-y}{1+y}}{dy}\\
= &\ \frac\pi2\int_0^1 \frac{1}{1+y}dy
-\int_0^1\frac{\tan^{-1}y}{y}dy=\frac\pi2\ln2-G
\end{align}
|
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|
To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way:
The problem is as follows:
Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.
My solution:
We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.
Now we solve the equation $ax+by+c=0$ for $y$:
$ax + by + c=0$
$ax + by = -c$
$by = -ax -c$
$y = \frac {-ax-c}{b}$
$\because x = 0$ at y-intercept,
$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$
$y = -\frac cb$.
The point at which the line crosses the y-axis is $(0,-\frac cb)$
Now we solve the equation $ax+by+c=0$ for $x$:
$ax+by+c=0$
$ax+by=-c$
$ax = -by-c$
$x = \frac {-by-c}{a}$
$\because y = 0$ at x-intercept
$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$
$x = -\frac ca$
The point at which the line crosses the x-axis is $(-\frac ca,0)$
|
easier way is this equation of line intercept form
$$ \frac xa+\frac yb=1$$
so write the given eqn in this form
$$ax+by+c=0$$
$$ax+by=-c$$
$$\frac {ax}{-c}+\frac {by}{-c}=1$$
$$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$
so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)
|
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|
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
|
Note that $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization...
|
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|
Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help
|
I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.
|
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|
Class Group of $\mathbb Q(\sqrt{-35})$ As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.
We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.
$-35\equiv 5$ mod $8$, so $2$ is inert. Also, $-35\equiv 1$ mod $3$, so $3$ splits, i.e. $(3)=Q\overline Q$ with $Q=(3,1+\sqrt{-35})$, $\overline Q=(3,1-\sqrt{-35})$. The ideals $Q,\overline Q$ are not principal, because there are no solutions to $x^2+35y^2=12$, i.e. no elements of norm 3.
Now we know that there are at most $3$ elements (or do we?), namely $(1),Q,\overline Q$. Mathematica tells me that the class number is $2$, so $Q$ and $\overline Q$ must be in the same equivalence class and $Q^2$ has to be a principal ideal. But how can I show this?
|
Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$.
If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$
(because
$\dfrac{1-\sqrt{-35}}{2}$ and $2$ are coprime, and so generate the ideal $1$).
I find the computation a bit easier by phrasing the factorization of $(3)$
in the following alternative way:
$(3) = (3, (1 + \sqrt{-35})/2) (3,(1- \sqrt{-35})/2)$, and
$$(3,(1+\sqrt{-35})/2)^2 = (9, 3(1+\sqrt{-35})/2,(-17+\sqrt{-35})/2) = ( (1+\sqrt{-35})/2 )$$ is principal.
As a consistency check, note that $ (9) = (3) (3) = Q \overline{Q} Q \overline{Q}
= Q^2 \overline{Q}^2,$ but also $9 = ( (1+\sqrt{-35})/2) ( ( 1 - \sqrt{-35})/2),$ so we must have $Q^2$ equal to one of $( (1 \pm \sqrt{-35})/2).$
|
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|
How to prove the existence of odd numbers $a$ and $b$ in $2m\equiv a^{20}+b^{11}\pmod{2^n}$
Show that for any natural numbers $m,n$, there exist odd numbers $a,b$ such that
$$2m\equiv a^{20}+b^{11}\pmod{2^n}$$
Thank you everyone.
|
We proceed by induction on $n$.
When $n=1$, we have $1^{20}+1^{11} \equiv 0 \equiv 2m \pmod{2^1}$.
Suppose that the statement holds for $n=k$, so $\exists a, b \in \mathbb{Z}$, $a, b$ odd, such that $2m \equiv a^{20}+b^{11} \pmod{2^k}$. Thus $a^{20}+b^{11} \equiv 2m, 2m+2^k \pmod{2^{k+1}}$. If $a^{20}+b^{11} \equiv 2m \pmod{2^{k+1}}$, we are done. Otherwise we have $a^{20}+b^{11} \equiv 2m+2^k \pmod{2^{k+1}}$. Then $a^{20}+(b+2^k)^{11} \equiv a^{20}+b^{11}+11b^{10}2^k \equiv (a^{20}+b^{11})+2^k \equiv 2m \pmod{2^{k+1}}$. We are thus done by induction.
Therefore for any $m, n \in \mathbb{Z}^+$, there exist odd $a, b \in \mathbb{Z}$ such that $2m \equiv a^{20}+b^{11} \pmod{2^n}$.
|
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|
finding nth term Let
3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
|
Note: I just added a derivation of an explicit formula
for the terms as the OP requested.
From 3,8,17,32,57,
if the term is n,
the next term is 2n+k
where k = 2, 1, -2, -7.
The differences of k are
-1, -3, -5.
If we assume that the next difference is -7,
the next k is -7-7=-14
and the next term is 2*57-14 = 100.
To get a formula from this,
since the sum of the odd numbers are the squares,
the values of k are $2-m^2$
starting with $m = 0$.
Letting $s(0) = 3$,
$s(n+1) = 2s(n)+2-n^2$.
This is made explicit below.
Check:
$s(1) = 2*3+2-0 = 8$,
$s(2) = 2*8+2-1 = 17$,
$s(3) = 2*17+2-4 = 32$,
$s(4) = 2*32+2-9 = 57$.
To get an explicit form for
$s(0) = 3$,
$s(n+1) = 2s(n)+2-n^2$
let $s(n) = 2^n t(n)$.
Then $2^{n+1}t(n+1) = 2^{n+1}t(n)+2-n^2$,
or
$t(n+1) = t(n) + 1/2^{n}-n^2/2^n$,
so
$t(n+1) - t(n) = 1/2^{n}-n^2/2^n$.
Summing this,
$t(m)-t(0) = \sum_{n=0}^{m-1} (t(n+1) - t(n))
= \sum_{n=0}^{m-1} (1/2^{n}-n^2/2^n)
= 1-1/2^m - \sum_{n=0}^{m-1}n^2/2^n
$.
To evaluate the last sum,
let $f(x) = \sum_{n=0}^{m-1} x^n
= (1-x^m)/(1-x)
$.
Then $x f'(x) = \sum_{n=0}^{m-1} nx^n$
and $x (x f'(x))' = \sum_{n=0}^{m-1} n^2 x^n$.
We want $x (x f'(x))'$ at $x = 1/2$.
Note that $x (x f'(x))'
= x (f'(x) + x f''(x))
= x f'(x) + x^2 f''(x)
$.
I'm feeling lazy, so I won't bother getting
$f'$ and $f''$.
|
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|
Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$ If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of $\sqrt n$ in denominator
|
You could as OP suggests, add 2, and take the square root.
For example, $\dfrac{1}{(7+4\sqrt3)}$ is $(7-4 \sqrt3)$, so the sum of these numbers is $14$. You add $2$ to it to get $16$, and take the square root.
The new number is $4 = x + \dfrac {1}x$, which leads to $x = \dfrac{\sqrt{4+2} + \sqrt{4-2}}{2}$ = 1.93185165259.
|
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|
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$
I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.
Any help is appreciated! Thanks.
Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$
Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$
|
A general technique is taught by Wilf's "generatingfunctionology". Define $S(z) = \sum_{n \ge 0} s_n z^n$ and write ($s_0$ you get from the recurrence "backwards", mostly for not having to mess around with indices):
$$
s_{n + 1} = 3 s_n + 2^{n - 1} - 1 \qquad s_0 = \frac{1}{6}
$$
Multiply the recurrence by $z^n$, add over $n \ge 0$ to get:
$$
\begin{align*}
\frac{S(z) - s_0}{z}
&= 3 S(z) + \frac{1}{2} \sum_{n \ge 0} 2^n z^n - \sum_{n \ge 0} z^n \\
&= 3 S(z) + \frac{1}{2} \frac{1}{1 - 2 z} - \frac{1}{1 - z}
\end{align*}
$$
Solving for $S(z)$ and expanding in partial fractions:
$$
S(z)
= \frac{1}{6} \frac{1}{1 - 3 z}
- \frac{1}{2} \frac{1}{1 - 2 z}
+ \frac{1}{2} \frac{1}{1 - z}
$$
Everything in sight is geometric series:
$$
s_n = \frac{1}{6} \cdot 3^n - \frac{1}{2} \cdot 2^n + \frac{1}{2}
= \frac{1}{2} \left( 3^{n - 1} - 2^{n - 1} + 1 \right)
$$
|
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|
Formulas for calculating pythagorean triples I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:
$$\begin{align}
&a = m^2-n^2 \\
&b = 2mn\\
&c = m^2+n^2
\end{align}$$
With numerous parameters.
So are there other formulas/methods?
|
Here is the method I use:
*
*Take a number $c >= 1$
*Find all pairs $a, b$ such that $ab = 2c^2$
*For every pair of $a, b$ you get an unique solution $x = a + 2c, y = b + 2c, z = a + b + 2c$. It can be easily proven that $x^2 + y^2 = z^2$ is reduced to $4c^2 = 2ab$ which is satisfied at point 2.
The advantage of this method is that you can get all possible triplets. For example with $c = 3, a = 3, b = 6$ a triplet $9, 12, 15$ is found, which cannot be found with euclid's formula.
It can be proven that this method give all possible triplets. If there is a pythagorean triplet $x, y, z$ than it can be represented in a form $x = a + 2c, y = b + 2c, z = a + b + 2c$ such that $ab = 2c^2$:
*
*We have $x^2 + y^2 = z^2$
*$a = z - y, b = z - x, c = (x + y - z) / 2$
*Now we have to prove that $ab = 2c^2$
*$ab = (z - y)(z - x) ? 2c^2 = (x + y - z)^2 / 2$
*$z^2 - yz - xz + xy ? (x^2 + y^2 + z^2 + 2xy - 2xz - 2yz) / 2$
*$2z^2 - 2yz - 2xz + 2xy ? x^2 + y^2 + z^2 + 2xy - 2xz - 2yz$
*After removing all repeating parts we get $z^2 ? x^2 + y^2$ which are equal. Done.
|
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|
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
|
In this answer, the substitution $x=\frac{1-y}{1+y}$ is used to get
$$
\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x=\frac\pi8\log(2)\tag{1}
$$
We can use the substitution $x\mapsto1/x$ to get
$$
\begin{align}
\int_1^\infty\frac{\log(1+x)}{1+x^2}\mathrm{d}x
=\int_0^1\frac{\log(1+x)-\log(x)}{1+x^2}\mathrm{d}x\tag{2}
\end{align}
$$
which implies
$$
\int_0^\infty\frac{\log(1+x)}{1+x^2}\mathrm{d}x
=2\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x-\int_0^1\frac{\log(x)}{1+x^2}\mathrm{d}x\tag{3}
$$
Therefore, we can use
$$
\begin{align}
\int_0^1x^k\log(x)\,\mathrm{d}x
&=\frac1{k+1}\int_0^1\log(x)\,\mathrm{d}x^{k+1}\\
&=-\frac1{k+1}\int_0^1x^{k+1}\,\mathrm{d}\log(x)\\
&=-\frac1{k+1}\int_0^1x^k\,\mathrm{d}x\\
&=-\frac1{(k+1)^2}\tag{4}
\end{align}
$$
to get
$$
\begin{align}
\int_0^\infty\frac{\log(1+x)}{1+x^2}\mathrm{d}x
&=2\int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x-\int_0^1\frac{\log(x)}{1+x^2}\mathrm{d}x\\
&=\frac\pi4\log(2)-\int_0^1\sum_{k=0}^\infty(-1)^kx^{2k}\log(x)\,\mathrm{d}x\\
&=\frac\pi4\log(2)+\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\\[4pt]
&=\frac\pi4\log(2)+\mathrm{G}\tag{5}
\end{align}
$$
|
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|
linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}
Is this right so far? If not, what am I doing wrong?
|
Note that you're hiding the core of the proof, which is simply that if $p,q$ are polynomials, we define their sum, which is also a polynomial, and $(p+q)(x)=\sum_{i=1}^n (a_i+b_i)x^i=p(x)+q(x)$.
Thus, if $T$ is your transformation, $$T((p+q)(x))=(p+q)(x+1)=p(x+1)+q(x+1)=T(p(x))+T(q(x))$$ and $$T(\lambda p(x))=(\lambda p)(x+1)=\lambda p(x+1)=\lambda T(p(x))$$
Maybe it is easier to see why the transformation is linear by seeing how it acts on the coefficients of $P$. If $P$ is $(a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ then $$T(P)=(a_0+a_1+a_2,a_1+2a_2,a_2)$$ because $$\begin{align}a_0+a_1(x+1)+a_2(x+1)^2&=a_0+a_1x+a_1+a_2x^2+2a_2x+a_2\\&=(a_0+a_1+a_2)+(a_1+2a_2)+a_2x^2\end{align}$$
|
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|
Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
|
3 √x+13 = x + 9
Squaring both sides, we get
9 (x + 13) = x^2 + 81 + 18x
9x + 117 = x^2 + 81 + 18x
x^2 + 18x - 9x + 81 - 117 = 0
x^2 + 9x - 36 = 0
x^2 + 12x - 3x - 36 = 0
x(x + 12) - 3(x + 12) = 0
(x + 12)(x - 3) = 0
Therefore, x = -12 or 3
However, when the problem equation does not get satisfied when x is substituted by -12, but it gets satisfied when x is substituted by 3.
Therefore, 3 is the required solution or root of the equation.
Hope that you understood the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/398051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$ Let $\Gamma(z)$ be the gamma function.
In a paper I'm reading, the author states that $$ |\Gamma(z)| = |\Gamma(a+ib)| \sim \sqrt{2 \pi} |b|^{a-\frac{1}{2}} e^{-|b|\frac{\pi}{2}}$$ as $|b| \to \infty$.
Can this asymptotic behavior be derived from Stirling's formula?
EDIT:
I think I have something.
Assume that $a,b >0$ and that $b$ is very large.
Then it would seem that
$$ \begin{align}|\Gamma(a+ib)| &\sim \left|\sqrt{\frac{2 \pi}{a+ib}} \Big(\frac{a+ib}{e}\Big)^{a+ib} \right| \\ &= \sqrt{2 \pi} \left|(a+ib)^{a-\frac{1}{2}} \right| \left| (a+ib)^{ib} \right| \left|e^{-a-ib} \right| \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} \ \left|\left(\sqrt{a^{2}+b^{2}} e^{i \arg \left(\frac{b}{a}\right)} \right)^{ib} \right| e^{-a} \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} e^{-b \arg \left(\frac{b}{a} \right)} e^{-a} \\ &\sim \sqrt{2 \pi} \, b^{a - \frac{1}{2}} e^{-b \frac{\pi}{2}} {\color{red}{e^{-a}}}. \end{align}$$
But apparently this is not quite correct.
|
Assume $|y|\gg|x|$. Note that
$$
\begin{align}
y\arctan\!2(y,x)
&=|y|\arctan\!2(|y|,x)\\[6pt]
&=\frac\pi2|y|-|y|\arctan\!2(x,|y|)\\
&\sim\frac\pi2|y|-x
\end{align}
$$
and
$$
x^2+y^2\sim|y|^2
$$
Therefore,
$$
\begin{align}
\left|\sqrt{2\pi}\,\frac{\color{#C00}{z^{z-1/2}}}{\color{#090}{e^z}}\right|
&=\sqrt{2\pi}\color{#C00}{\left(x^2+y^2\right)^{x/2-1/4}}e^{\color{#090}{-x}\color{#C00}{-y\arctan\!2(y,x)}}\\
&\sim\sqrt{2\pi}\,|y|^{x-1/2}e^{-\frac\pi2|y|}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/398260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime The number $30$ has a curious property:
All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!
I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.
Are there any such numbers $\gt 30$?
|
Let $n>30$ be such a number. Observe that $p^2<n$ implies $p|n$.
Thus $2|n$, $3|n$, $5|n$ follow directly from $n>25>9>4$.
Thus $n$ is a multiple of $30$ and $n>30$, which implies $n\ge 60>49$ and hence $7|n$.
Up to now, $n$ is divisible by the four smallest primes.
Lemma. The product of four consecutive primes is greater than the square of the next prime:
$$\tag1p_kp_{k+1}p_{k+2}p_{k+3}> p_{p+4}^2.$$
Proof:
Using Bertrand's postulate, we have $p_{k+2}>\frac{p_{k+3}}2$, and $p_{k+4}\le 2p_{k+2}$, hence $(1)$ follows from
$$p_kp_{k+1}p_{k+2}p_{k+3}>3\cdot 5\cdot \frac{p_{k+3}}{2}p_{k+3}> 4p_{k+3}^2>p_{k+4}^2$$
at least if $p_k\ge 3$. The case $p_k=2$ is verified directly: $2\cdot3\cdot 5\cdot 7>11^2$. $_\square$
Now if $n$ is divisible by four consecutive primes $p_k,\ldots,p_{k+3}$, we conclude from $(1)$ that $n$ is also divisible by the next prime, because otherwise $p_{k+4}^2<n$ is relatively prime to $n$. Hence $n$ is divisible by the four consecutive primes $p_{k+1},\ldots,p_{k+4}$.
By induction, we conclude that $n$ is divisible by all primes, but that is of course absurd.
Hence there is no such number beyond $30$.
|
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"url": "https://math.stackexchange.com/questions/398399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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|
how to solve this multivariate quadratic equation? Any hope to acquire an analytic solution to such equations:
Solve:
$$\sum_{j=1}^n a_{ij} x_i x_j = b_i$$
for $i=1,\ldots,n$, where $a_{ij}$'s and $b_i$'s are known constants and $x_i$'s are unknowns to be solved.
Thanks a lot!
P.S. Thanks for alex.jordan's interesting comment! Let's consider this problem in a positive setting, i.e. let's require all coefficients($a_{ij}$'s and $b_i$'s) to be positive so that the solution seems to exist. Also $a$ is symmetric, i.e. $a_{ij}=a_{ji}$. If there could be any fast numerical solution it's also useful.
P.S. Just for the information of all readers: the original form of this problem is formulated as:
$$ x_i\cdot\left(\sum_{j=1}^n a_{ij}x_j\right) = \sum_{j=1}^n a_{ij}q_{ij} $$ where $q_{ij}\in(0, 1)$. I simplified the RHS into $b_{ij}$ but now it seems better not to do so(sorry for that!) to at least give the existence of real solutions a better chance.
|
Let $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}$, $C=A^{-1}B$.
There is a possible numerical solution when $A$ is invertible and $b_i\neq 0$ ($i$ = 1, 2, $\cdots$, $n$ ).
From the conditions, it will result
$$
\begin{pmatrix}
x_1 a_{11} & \cdots & x_1 a_{1n} \\
\vdots & \ddots & \vdots \\
x_n a_{n1} & \cdots & x_n a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix}.
$$
or
$$ \begin{pmatrix}
x_1 \\
& \ddots \\
& & x_n
\end{pmatrix}
\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix}.
$$
As $\sum\limits _{j=1} ^n a_{ij} x_i x_j=b_i$, i.e., $x_i \sum\limits _{j=1} ^n a_{ij} x_j=b_i$, since $b_i \neq 0$, then $x_i \neq 0$.
so
$$
\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
\begin{pmatrix}
x_1^{-1} \\
& \ddots \\
& & x_n^{-1}
\end{pmatrix}
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix}.
$$
As $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}$, $C=A^{-1}B$, then
$$ A
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
B
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix},
$$
so
$$
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
A^{-1} B
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix},
$$
that is to say,
$$
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix} =
C
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix}.
$$
Let $
\begin{pmatrix}
x_1^{(0)} \\
\vdots \\
x_n^{(0)}
\end{pmatrix} =
\begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix}$,
by the recurrence
$$
\begin{pmatrix}
x_1^{(k+1)} \\
\vdots \\
x_n^{(k+1)}
\end{pmatrix} =
C
\begin{pmatrix}
{\left(x_1^{(k)}\right)} ^{-1} \\
\vdots \\
{\left(x_n^{(k)}\right)}^{-1},
\end{pmatrix},
$$
we can get a numerical solution.
If the result does not converge, we can use another one
$$
\begin{pmatrix}
{\left(x_1^{(k+1)}\right)} ^{-1} \\
\vdots \\
{\left(x_n^{(k+1)}\right)}^{-1},
\end{pmatrix} =
C^{-1}
\begin{pmatrix}
x_1^{(k)} \\
\vdots \\
x_n^{(k)}
\end{pmatrix}.
$$
If you want the analytic solution, you may have a try by using the Groebner-Shirshov Bases. (The software Maple or Mathematica can do it.) But I am not sure the analytic solution can be find easily.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)
|
You want to integrate
$$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$
where $c = \dfrac{b^2-a^2}{a^2}$.
We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor series, we have
$$\dfrac1{(1+cy^2)^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k y^{2k}$$
Hence (swapping integral and infinite summation), we get that
$$\int_0^{\pi/2}\dfrac{dx}{(1+c\sin^2(x))^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \int_0^{\pi/2}\sin^{2k}(x)dx$$
From here, we have
$$\int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$$
Hence,
$$I = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \dbinom{2k}k \dfrac{\pi}{2^{2k+1}} = \dfrac{\pi}2 \sum_{k=0}^{\infty} (k+1) \dbinom{2k}k \left(-\dfrac{c}4\right)^k$$
Now from Taylor series, we have
$$\sum_{k=0}^{\infty} (k+1) \dbinom{2k}k x^k = \dfrac{1+4x\sqrt{1-4x}-2x-\sqrt{1-4x}}{(1-4x)^{3/2}}$$
Hence,
$$J = \dfrac{\pi}2 \left(\dfrac{1-c\sqrt{1+c}+c/2-\sqrt{1+c}}{a^4(1+c)^{3/2}} \right)$$
Now plug in the value of $c$ and get the value of the original integral $J$.
|
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"url": "https://math.stackexchange.com/questions/402223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving that sequence $a_n = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}} = x^{1-2^{-n}}$ Let $x>0$.
For sequence $a_n$, such that $n$ denotes the $n$th term:
$$\begin{align} a_1&= \sqrt{x}\\
a_2&= \sqrt{x \sqrt{x}}\\
a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\
a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\
&\vdots\\
a_{n-1}&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x}}}}\\
a_n&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x \sqrt{x}}}}}\end{align}$$
How could one prove that:
$${a_n = x^{1-2^{-n}}}?$$
|
Hint:
\begin{align}
a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}}
\end{align}
Can you use induction in these footsteps?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$
$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$
$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$
$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$
$$=\dfrac{6x^3-2x^2}{9x^3-x}$$
This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$
Edit, missed a difference of squares:
$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$
Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$
|
$$\dfrac{(6x^3-2x^2)}{x(9x^3-x)} = \dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} $$
Cancel common factor of $x$ in numerator and denominator gives us:
$$\dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} = \frac{2x(3x-1)}{\left[9x^2 - 1\right]}$$
Now we have a difference of squares in the denominator, and can factor it:
$$\frac{2x(3x-1)}{\color{blue}{\bf \left[9x^2 - 1\right]}}= \frac{2x(3x-1)}{\color{blue}{\bf \left[(3x)^2 - 1\right]}} = \frac{2x(3x - 1)}{(3x - 1)(3x + 1)}$$
Now, cancel like terms: Note that $\color{blue}{(3x - 1)}$ is a factor in the numerator and in the denominator, so we proceed to cancel:
$$ \frac{2x\color{blue}{\bf (3x - 1)}}{\color{blue}{\bf(3x - 1)}(3x + 1)} = \frac{ 2x}{3x + 1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subtracting roots of unity. Specifically $\omega^3 - \omega^2$ This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these?
The answer they give is $2cos(\dfrac{3\pi}{5})$
Please, some help will be greatly appreciated. I cant seem to find help over the internet either.
|
$$\omega=\cos(\pi/5)+i\sin(\pi/5)$$
$$\omega^3-\omega^2=\cos(3\pi/5)+i\sin(3\pi/5)-(\cos(2\pi/5)+i\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=-2\sin(\pi/2)\cdot\sin(\pi/10)-2i\cos(\pi/2)\cdot\sin(\pi/10)$$
$$\omega^3-\omega^2=-2\sin(\pi/10)$$
$$\omega^3-\omega^2=-2\cdot\dfrac{\sqrt{5}-1}{4}$$
$$\omega^3-\omega^2=\dfrac{1-\sqrt{5}}{2}$$
use some trig. formula:
$\cos C-\cos D=2\cdot \sin\dfrac{C+D}{2}\cdot \sin \dfrac{D-C}{2}$
and $\sin C-\sin D=2\cdot \cos\dfrac{C+D}{2}\cdot \sin \dfrac{C-D}{2}$
and $\sin (\pi/10)=\dfrac{\sqrt5-1}{4}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up getting
$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$
Dont know what to do...
EDIT: Thanks to Zev, what I did was:
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$
|
Having rewritten the recurrence, we get an equation:
$$10R_{n-2}-7R_{n-1}+R_n=0$$
that can easily be solved.
The roots of a characteristical polymomial $10-7k+k^2=0$ are $k_1=2,k_2=5$.
So general solution is $R_n=c_12^{n}+c_25^{n}$, where $c_1$ and $c_2$ are arbitrary constants.
|
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|
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
|
If you type $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$ into most pocket calculators you will get the answer $1$. This is because the calculator applies the next operation to the result of the previous one, effectively computing$$(1+1+1+1+1+1+1+1+1+1+1+1)⋅0+1=0+1=1$$
When using a calculator you have to respect the logic of the machine. When writing mathematics you need to be aware of the standard conventions for brackets, which give the answer $12$ in this case. I suspect that the confusion arises because the two are not the same. Understanding why will help you to make better use of a calculator and to write correct mathematical equations.
|
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|
Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test?
If I remove the first term I can see $\dfrac{u_n}{u_{n+1}}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}
\\=\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^2}
\\={\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^{-2}}\\=\left(1+\dfrac{5}{2n}+\dfrac{3}{2n^2}\right)\left(1-\dfrac{1}{n}\ldots\right)\\=1+\dfrac{3}{2n}+O\left(\dfrac{1}{n^2}\right)$
So the series is convergent.
Is it a correcct attempt?
|
Ratio test is not enough.
Example 1:
$$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$
$$\dfrac{u_n}{u_{n+1}}=\dfrac{\left(\dfrac{1}{n}\right)}{\left(\dfrac{1}{n+1}\right)}=\frac{n+1}{n}=1+\frac{1}{n}$$
Ratio will be $1$ if $n--->\infty$
But we know that S series is divergent.
So the ratio test says us If the ratio is 1. maybe the series will be convergent. We need more test to determine the result.
I will give you a method how you can approach such series.
$$f(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$
$$f'(x)=1+\dfrac{1}{2}x^2+\dfrac{1\cdot 3}{2\cdot 4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\cdots$$
$$f''(x)=x+\dfrac{1\cdot 3}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$
$$\int_0^x \frac{f''(x)}{x} dx=x+\dfrac{1}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$
$$\int_0^x \frac{1}{x}\int_0^x \frac{f''(t)}{t} dt dx=f(x)$$
$$\int_0^x \frac{f''(x)}{x}dx=xf'(x)$$
$$ \frac{f''(x)}{x}=xf''(x)+f'(x)$$
$$ \int_0^x \frac{f''(x)}{f'(x)} dx=\int_0^x\frac{x}{1-x^2} dx$$
$$ \ln f'(x)=-\frac{1}{2}\ln{(1-x^2)} $$
$$ f'(x)=\frac{1}{\sqrt{1-x^2}} $$
$$ f(x)=\arcsin(x) $$
$$\arcsin(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$
$x=1$
$$\arcsin(1)=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$
$$\dfrac{\pi}{2}=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$
You can use the same method and to show the series $$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$ is divergent.
$$f(x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$
$$f'(x)=1+x+x^2+x^3+x^4\cdots=\frac{1}{1-x}$$
$$\int_0^x f'(x) dx=f(x)=-\ln{(1-x)}$$
$x=1$
$$f(1)=-\ln{0}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$
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|
How to simplify $a^n - b^n$? How to simplify $a^n - b^n$?
If it would be $(a+b)^n$, then I could use the binomial theorem, but it's a bit different, and I have no idea how to solve it.
Thanks in advance.
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$$
a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} - b^n = (a-b)b^{n-1} + {n \choose 1}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n = (a-b)(b^{n-1} + {n \choose 1}(a-b)b^{n-2} + {n \choose 2}(a-b)^2b^{n-3} + {n \choose 3}(a-b)^3b^{n-4} + ... + {n \choose n-1} (a-b)^{n-2}b + (a-b)^{n-1} )
$$
too complicated to simplify if you use bionomial expansion.
$$
a^n -b^n = a^n -a^{n-1}b + a^{n-1}b -b^n = a^{n-1}(a-b) + b(a^{n-1}-b^{n-1}) = a^{n-1}(a-b) + b(a^{n-1} -a^{n-2}b + a^{n-2}b -b^{n-1}) = a^{n-1}(a-b) + b(a^{n-2}(a-b) + b(a^{n-2}-b^{n-2})) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2(a^{n-2}-b^{n-2}) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2 a^{n-3}(a-b) + b^3(a^{n-3}-b^{n-3})
= ...(repeat ) = a^{n-1}(a-b) + b a^{n-2}(a-b) + b^2 a^{n-3}(a-b) + b^3 a^{n-4}(a-b) + ... + b^{n-2}a(a-b) + b^{n-1}(a-b) = (a-b)(a^{n-1} + b a^{n-2} + b^2 a^{n-3} + ... + b^{n-2}a + b^{n-1})$$
the power of $a$ and $b$ add up to $n-1$ in the second factor.
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|
What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
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without L-hospital law
$$\lim_{x\to\dfrac \pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\frac{1-\sqrt{3}\dfrac {\sin x}{\cos x}}{6\left(\dfrac\pi 6-x\right)}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac{1\cdot\cos x-\sqrt{3} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac{\dfrac12\cdot\cos x-\dfrac {\sqrt{3}}{2} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac {\sin \dfrac\pi6\cdot\cos x-\cos \dfrac \pi6 \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {\sin\left(\dfrac \pi6-x\right)}{3\left(\dfrac\pi 6-x\right)\cos x}$$
since $$\lim_{x\to a} \dfrac {\sin a}{a}=1$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {1}{3\cos \dfrac \pi6}$$
$$\dfrac{2}{3\sqrt3}\implies \dfrac {2\sqrt3}{9}$$
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|
Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he divided $x^3 - 3x^2 + 2x - 1$ by $x^2 - 4x + 5$.
The quotient was $x + 1$ and the remainder $x - 6$. Now since $\rm dividend = quotient\cdot divisor + remainder$, he concluded that $x^3 - 3x^2 + 2x - 1 = x-6$ since the divisor is $0$.
Plugging $2 + i$ into $x - 6$, we get $-4 + i$.
But my question is, how was he able to divide by zero in the first place? Why did dividing by zero work in this case?
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The division your teacher did is polynomial division. He did not divide by zero; he divided by $x^2 - 4x + 5$.
The long division he did is just an algorithm that allows you to get the following identity:
$$x^3 - 3x^2 + 2x - 1 = (x^2 - 4x + 5)(x+1) + x-6$$
This is an identity involving multiplication, not division. Now, when you plug in $x = 2 + i$, you're not dividing by zero; you're multiplying by zero, which you should agree is allowed.
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|
How find $\int(x^7/8+x^5/4+x^3/2+x)\big((1-x^2/2)^2-x^2\big)^{-\frac{3}{2}}dx$ How can I compute the following integral:
$$\int \dfrac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{\left(\left(1-\frac{x^2}{2}\right)^2-x^2\right)^{\frac{3}{2}}}dx$$
According to Wolfram Alpha, the answer is
$$\frac{x^4 - 32x^2 + 20}{2\sqrt{x^4 -8x^2 + 4}} + 7\log\bigl(-x^2 - \sqrt{x^4 - 8x^2 + 4} + 4\bigr) + C$$
but how do I get it by hand?
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Perform the substitutions $y=\frac{x^2}{2}, y=2+\sqrt{3}\sec{\theta}$ to get:
\begin{align}
&\int{\frac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{((1-\frac{x^2}{2})^2-x^2)^{\frac{3}{2}}} dx} \\
& =\int{\frac{y^3+y^2+y+1}{(y^2-4y+1)^{\frac{3}{2}}} dy} \\
&=\int{\left(\frac{y+5}{\sqrt{(y-2)^2-3}}+\frac{20y-4}{\sqrt{((y-2)^2-3)^3}}\right) dy} \\
&=\int{\left[\frac{7+\sqrt{3}\sec{\theta}}{\sqrt{3}\tan{\theta}}+\frac{36+20\sqrt{3}\sec{\theta}}{3\sqrt{3}\tan^3{\theta}}\right](\sqrt{3}\sec{\theta}\tan{\theta}) d\theta} \\
&=\int{\left(7\sec{\theta}+\sqrt{3}\sec^2{\theta}+12\csc{\theta}\cot{\theta}+\frac{20}{\sqrt{3}}\csc^2{\theta}\right) d\theta} \\
&=7\ln{|\tan{\theta}+\sec{\theta}|}+\sqrt{3}\tan{\theta}-12\csc{\theta}-\frac{20}{\sqrt{3}}\cot{\theta}+c \\
&=7\ln{|\frac{\sqrt{y^2-4y+1}+y-2}{\sqrt{3}}|}+\sqrt{y^2-4y+1}-\frac{12(y-2)}{\sqrt{y^2-4y+1}}-\frac{20}{\sqrt{y^2-4y+1}}+c \\
&=7\ln{|\frac{\sqrt{y^2-4y+1}+y-2}{\sqrt{3}}|}+\frac{y^2-16y+5}{\sqrt{y^2-4y+1}}+c \\
&=7\ln{|\sqrt{x^4-8x^2+4}+x^2-4|}+\frac{x^4-32x^2+20}{2\sqrt{x^4-8x^2+4}}+c'
\end{align}
where $c,c'$ are arbitrary constants of integration.
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|
If $\frac{\cos^4\theta}{\cos^2\phi}+\frac{\sin^4\theta}{\sin^2\phi}=1$, show $\frac{\cos^4\phi}{\cos^2\theta} +\frac{\sin^4\phi}{\sin^2\theta}=1$
If $\dfrac{\cos^4\theta}{\cos^2\phi}+\dfrac{\sin^4\theta}{\sin^2\phi}=1$, prove that $\dfrac{\cos^4\phi}{\cos^2\theta} +\dfrac{\sin^4\phi}{\sin^2\theta}=1$.
Unable to move further ...request you to please suggest how to proceed ..Thanks..
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Hint: Let $ x = \cos \theta$, $y = \cos \phi$.
Show by expansion (and clearing denominators) that both equations are equivalent to $x^4 - 2x^2 y^2 + y^4 =0$, hence these statements are equivalent to each other.
Note: This shows that the condition is satisfied iff $x = \pm y$. This is not required, but very strongly hinted at in the question.
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|
Integrate: $\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$ How to evaluate
$$\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$$
I tried trigonometric substitution $x + a = b \sec \theta$ and I encountered
$$\int \frac{\tan \theta}{ (b - a \cos \theta) \sqrt{\tan^2 \theta}}d\theta$$
how to handle this term $\displaystyle \frac{\tan \theta}{|\tan \theta|}$?
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I substituted $x=1/y$ and got
$$-\int \frac{dy}{\sqrt{1+2 a y+(a^2-b^2) y^2}}$$
I then completed the square in the square root to get
$$-\frac{1}{\sqrt{a^2-b^2}} \int \frac{dy}{\displaystyle\sqrt{\left(y+\frac{a}{a^2-b^2}\right)^2-\left(\frac{a}{a^2-b^2}\right)^2}}$$
Now let
$$y=\frac{a}{a^2-b^2} (\cosh{u}-1)$$
Then the integral becomes
$$-\frac{1}{\sqrt{a^2-b^2}} \int du \frac{\sinh{u}}{\sinh{u}} = -\frac{1}{\sqrt{a^2-b^2}} u+C$$
where $C$ is a constant of integration. Back substituting, we get
$$\int \frac{dx}{x \sqrt{(x+a)^2-b^2}} = -\frac{1}{\sqrt{a^2-b^2}}\text{arccosh}{\left(\frac{a^2-b^2}{a x}+1 \right)} + C $$
Use $\text{arccosh}{z} = \log{(z+\sqrt{z^2-1})}$ to express the result in terms of logs.
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|
What are the odds of rolling 3 pairs with six dice? Given a roll of six standard six-side dice, how do you calculate the odds of rolling 3 pairs? This should included non-unique pairs like 2,2,2,2,3,3 or even 5,5,5,5,5,5.
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Let's split into $3$ cases:
Case 1: All dice show the same number. There are $6$ ways this can happen.
Case 2: $4$ dice show the same number, and $2$ dice show a different number. There are $\binom{6}{2}=15$ different ways to choose the $2$ dice that show a different number, and $\binom{6}{2}=15$ ways to choose which two numbers appear. We then multiply by $2$ to choose which number appears $4$ times, and which number appears $2$ times. Thus there are $15\cdot 15\cdot 2=450$ possible outcomes for Case 2.
Case 3: There are three distinct pairs. Here, there $\binom{6}{3}=20$ ways to choose which three numbers appear, and $6!/2^3=90$ ways to arrange them. So here we have $1800$ possibilities.
Altogether, there are $6^6$ different outcomes, which gives us a probability of:
$$\frac{6+450+1800}{6^6}=\frac{47}{972}\approx.04835$$
It looks like this will happen about once every $21$ throws.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.