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Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$
I have no idea where to go from there.
|
Another idea ,maybe
$$cos(x)+cos(2x)-cos(3x)=1\\
cos(x)-cos(3x)=1-\cos(2x)\\\cos
(2x-x)-\cos(2x+x)=1-\cos(2x)\\\cos(2x)\cos(x)+\sin(2x)\sin(x)-(\cos(2x)\cos(x)-\sin(2x)\sin(x))=1-\cos(2x)\\2\sin(2x)\sin(x)=1-\cos(2x)\\
2\sin(2x)\sin(x)=2\sin^2(\frac{2x}{2})\\
2\sin(2x)\sin(x)=2\sin^2(x)\\
\sin(2x)\sin(x)-\sin^2(x)=0\\\sin(x)(\sin(2x)-\sin(x))=0\\ \sin(x)=0 \to\\x=k\pi\\\sin(2x)=\sin(x) \to\\ 2x=x+2k\pi,2x=\pi-x+2k\pi$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
|
$$3\sin x+4\cos x=A$$
$$3\cos x-4\sin x=2$$
$$9\sin^2x+24\sin x\cos x+16\cos^2x=A^2$$
$$9\cos^2x-24\sin x\cos x+16\sin^2x=4$$
add both equations
$$9+16=A^2+4$$
$$A^2=21$$
$$A=\pm \sqrt{21}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2588061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
|
Would you not just introduce substitution $u=\sqrt{3x+1}$, $x=\frac{u^2-1}{3}$, $dx=\frac23 udu$ so:
$$\int (2x+3) \sqrt {3x+1} dx=\int \left(\frac23 (u^2-1)+3\right)u\cdot \frac23 udu$$
which is an integral of a polynomial in $u$ - easy to solve.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$ I have to solve this series transforming it into a telescopic sequence
$$\sum_{k=1}^\infty\frac{1}
{k(k+1)(k+2)(k+3)}$$
But I'm lost in the calculation!
|
You can also go the hard way: write
$$
\frac{1}{x(x+1)(x+2)(x+3)}=
\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3}
$$
so
$$
A(x+1)(x+2)(x+3)+Bx(x+2)(x+3)+Cx(x+1)(x+3)+Dx(x+1)(x+2)=1
$$
Now,
*
*for $x=0$: $6A=1$
*for $x=-1$: $-2B=1$
*for $x=-2$: $2C=1$
*for $x=-3$: $-6D=1$
Thus
$$
\frac{1}{k(k+1)(k+2)(k+3)}=
\frac{1}{6}\left(\frac{1}{k}-\frac{3}{k+1}+\frac{3}{k+2}-\frac{1}{k+3}\right)
$$
Hence your summation is $1/6$ of
\begin{align}
\sum_{k=1}^\infty \frac{1}{k}
-3\sum_{k=1}^\infty \frac{1}{k+1}
+3\sum_{k=1}^{\infty} \frac{1}{k+2}
-\sum_{k=1}^{\infty} \frac{1}{k+3}
&
=\sum_{k=1}^\infty \frac{1}{k}
-3\sum_{k=2}^\infty \frac{1}{k}
+3\sum_{k=3}^{\infty} \frac{1}{k}
-\sum_{k=4}^{\infty} \frac{1}{k}
\\[6px]
&=
\left(1+\frac{1}{2}+\frac{1}{3}\right)-3\left(\frac{1}{2}+\frac{1}{3}\right)
+3\frac{1}{3}
\\[6px]
&=\frac{1}{3}
\end{align}
Therefore the sum of your series is
$$
\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2589176",
"timestamp": "2023-03-29T00:00:00",
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|
using $x^9 - 1$ as a product of linear and quadratic factors with real coefficients to solve trig So here's where I'm at I know i can find the ninth roots of unity by using the nth roots of unity formula $$ \zeta_n = e^{\frac{i2\pi k}{n}} ,$$ when $k = 1,2,3,...,n-1$
so the roots of unity where $e^{i\frac{2\pi}{9}},e^{i\frac{4\pi}{9}},...,e^{i\frac{8\pi}{9}}$and the other 4 would be the same except the angles would just be negative.
The previous parts of the question was to show that $$(x-e^{i\theta})(x-e^{-i\theta})$$ could be expressed as a quadratic with real coefficients the answer I got was $$(x-e^{i\theta})(x-e^{-i\theta})=x^2-2x\cos(\theta)+1$$so I said well the $x^9 - 1= (x-e^{i\frac{2\pi}{9}})(x-e^{-i\frac{2\pi}{9}})...(x-e^{i\frac{8\pi}{9}})(x-e^{-i\frac{8\pi}{9}})(x-1)$ and then using the formula I proved said that, that was equal to $$(x^2-2x\cos(\frac{2\pi}{9})+1)(x^2-2x\cos(\frac{4\pi}{9})+1)...(x^2+2x\cos(\frac{8\pi}{9})+1)(x-1)$$
Somehow I need to use this to solve
$$\cos(\frac{2\pi}{9})+\cos(\frac{4\pi}{9})+\cos(\frac{6\pi}{9})+\cos(\frac{8\pi}{9})= -\frac{1}{2}$$
Any help ,much appreciated!!
|
You should have the four pairs of complex roots you have accounted for plus a factor $x-1$ as $x=1$ is a root of your polynomial. That gives you the linear term.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this the right solution for an unsolvable system? I have to find the values of $k$ such that this system $\left[
\begin{array}{ccc|cc}
1&0&-3&-3\\
2&k&-1&-2\\
1&2&k&1
\end{array}
\right]$ is:
a) unsolvable b) solvable with infinite solutions c) one solution.
After some steps we can see in this matrix $\left[
\begin{array}{ccc|cc}
1&0&-3&-3\\
0&k&5&4\\
0&2&k+3&4
\end{array}
\right]$ for $k=2$ the second and the thrid row are the same, so if the substract them on row becomes all zeros. (And the system is unsolvable)
The next step: $\left[
\begin{array}{ccc|cc}
1&0&-3&-3\\
0&k&5&4\\
0&0&k^2-7&4k-8
\end{array}
\right]$ now im not that sure but I think, the solutions to $k^2-7=0$ should be the values of $k$ such that there are infinite solutions.
Then every possbile value k besides 2 and the solutions for $k^2-7=0$ would return a solvable system with one solution?
|
from the first equation we get
$$x_1=-3+3x_3$$
plugging this in the second equation and solving for $x_3$ we have
$$x_3=\frac{4}{5}-\frac{k}{5}x_2$$
and for the third equation we get
$$x_2\left(2-\frac{k}{3}(3+k)\right)=4-\frac{4}{5}(3+k)$$
can you finish?
from the last equation we get
$$x_2(6-3k-k^2)=\frac {3}{5}(8-4k)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2597025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of convergence $R$ to be $2$, but I struggle to show that it converges on $x = \pm 2$. Can you give me a hint?
What I have so far: $R = \frac{1}{\limsup_{n \to \infty} (x_n)} = \frac{1}{\frac{1}{1+1}} = \frac{2}{1} = 2$, where
$$\begin{align} x_n := \sqrt[n]{|a_n|} &= \sqrt[n]{| \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2}|} = \frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{1} \\ &= \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &= \frac{n - 1}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{1 - \frac{1}{n}}{\sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}}} \end{align}$$
How do I determine if $\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} (\pm 2)^n$ converges? The root test doesn't work and the ratio test doesn't seem to be a smart move. What can I do?
|
Note that
$$\sqrt{n^2+n}=n\left(1+\frac1n\right)^{\frac12}\sim n+\frac12 \quad \sqrt{n^2+1}=n\left(1+\frac1{n^2}\right)^{\frac12}\sim n$$
$$\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n\sim\frac{1}{2^n}$$
thus
$$\frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n \sim \frac{x^n}{2^nn^2}$$
which diverges when $|x|>2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}$
Compute
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$
My Approach
Since $k^{3}+6k^{2}+11k+5= \left(k+1\right)\left(k+2\right)\left(k+3\right)-1$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}
= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{\left(k+3\right)!}\right)$$
But now I can't find this limit.
|
$\displaystyle
\begin{align}
\sum_{k=1}^\infty\left[\frac{1}{k!} - \frac{1}{(k+3)!}\right]
&=
\left\{\begin{array}{c}
\dfrac{1}{1!} &+\dfrac{1}{2!} &+\dfrac{1}{3!} &+\dfrac{1}{4!} &+\dfrac{1}{5!}
&+\dfrac{1}{6!} &+\dfrac{1}{7!} &+\dfrac{1}{8!} &+\dfrac{1}{9!} &+\cdots \\
& & &-\dfrac{1}{4!} &-\dfrac{1}{5!}
&-\dfrac{1}{6!} &-\dfrac{1}{7!} &-\dfrac{1}{8!} &-\dfrac{1}{9!} &-\cdots \\
\end{array} \right\}\\
&=\frac{1}{1!} +\frac{1}{2!} +\frac{1}{3!}
\end{align}$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive members but I couldn't manage to show something. It is obvious that $$\frac{n^n}{n!}$$ is increasing but that don't help us much (I think)
|
We have that $\frac{n}{(n!)^{\frac{1}{n}}}\leq\frac{n+1}{((n+1)!)^{\frac{1}{n+1}}}\Leftrightarrow\frac{n^{n(n+1)}}{(n!)^{n+1}}\leq\frac{(n+1)^{n(n+1)}}{((n+1)!)^{n}}\Leftrightarrow$ simplifying, $\frac{n^{n^2+n}}{n!}\leq(n+1)^{n^2}\Leftrightarrow\frac{n^{n^2+n}}{(n!)(n+1)^{n^2}}\leq1$;let us proceed by induction. For n=1 $\frac{1}{2}\leq1\Rightarrow$ it is true. If it is true for n, for n+1 we have $\frac{(n+1)^{(n+1)^2+n+1}}{((n+1)!)(n+1+1)^{(n+1)^2}}=\frac{(n+1)^{n^2+3n+1}}{(n!)(n+2)^{(n+1)^2}}=\frac{n^{n^2+n}}{(n!)(n+1)^{n^2}}\cdot\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}\leq\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}$, by the induction hypothesis. So, if we show that $\frac{(n+1)^{(n+1)^2+n^2+n}}{n^{n^2+n}(n+2)^{(n+1)^2}}\leq1$, then we have proved the induction step, and thus proved the thesis. Now, this last statement holds $\Leftrightarrow (n+1)^{(n+1)^2+n^2+n}=(n+1)^{(n+1)(2n+1)}\leq n^{n^2+n}(n+2)^{(n+1)^2}=n^{n(n+1)}(n+2)^{(n+1)^2}\Leftrightarrow(n+1)^{2n+1}\leq n^n(n+2)^{n+1}$ (because $n,n+1,n+2\geq1$), $\Leftrightarrow f(n)\leq0$, with
$f:(0,+\infty)\rightarrow\mathbb{R},x\mapsto (2x+1)\mathrm{ln}(x+1)-x\mathrm{ln}(x)-(x+1)\mathrm{ln}(x+2)$.
We have that $f(x)\to-\mathrm{ln}2<0$ as $x\to0^+$ (because $x\mathrm{ln}x\to0$) and $f(x)=x\mathrm{ln}\frac{x+1}{x}+(x+1)\mathrm{ln}\frac{x+1}{x+2}=x\mathrm{ln}(\frac{x+1}{x}\cdot\frac{x+1}{x+2})+\mathrm{ln}\frac{x+1}{x+2}$, where the second term tends to 0, and if $y=\frac{x+1}{x}\cdot\frac{x+1}{x+2}-1=\frac{1}{x(x+2)}>0\Leftrightarrow yx^2+2yx-1=0\Leftrightarrow x=\frac{\sqrt{y^2+y}-y}{y}$ (we have taken the + sign because we want x>0), $y\to0$ as $x\to+\infty$, and $\lim_{x\to+\infty}x\mathrm{ln}(\frac{x+1}{x}\cdot\frac{x+1}{x+2})=(\frac{\sqrt{y^2+y}-y}{y})\mathrm{ln}(1+y)=\frac{1}{\sqrt{y^2+y}+y}\mathrm{ln}(1+y)=\frac{1}{\sqrt{1+\frac{1}{y}}+1}\frac{\mathrm{ln}(1+y)}{y}\to0\cdot1=0$ as $x\to+\infty\Leftrightarrow y\to0^+$.
Next, we compute f' and f''.
$f'(x)=2\mathrm{ln}(x+1)+\frac{2x+1}{x+1}-\mathrm{ln}x-1-\mathrm{ln}(x+2)-\frac{x+1}{x+2}$, and $f''(x)=\frac{-4-3x}{x(x+1)^2(x+2)^2}<0$ for x>0, and so then f' is strictly decreasing for x>0. We have that $f'(x)\to+\infty$ as $x\to0^+$,$f'(x)\to0$ as $x\to+\infty$, so $f'(x)>0$ for x>0, and so $\forall x>0\,-\mathrm{ln}2<f(x)<0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $k$ is composite, then $2^{k} -1$ is composite for $k \geq 2$ I am trying to prove that if $k$ is composite, then $2^{k} -1$ is composite, $k \geq 2$.
I have already established the equality $$n^k - 1 = (n-1)(n^{k-1} + ... + 1) \tag{*}$$ If I let $n=2$ and $k=ab$, I don't really get anything useful. The hint says to let $n=2^a$ if $k=ab$, but this gives a $2^{a^2b}$ on the LHS of $*$...
Any help is appreciated
Update: I understand now...
|
We have in general
$$ 2^n-1 = \underset {n \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}}$$
If $n$ is composite, say $a \cdot b$ then
$$\begin{array}{lll} 2^n-1 &= \underset { a \cdot b \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}} \\ \phantom{X} \\
&=\underset {a \text{ times}} { \underbrace {
\underset {b \text{ times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
+ 2^b \underset {b \text{ times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
+ ...
+ 2^{b(a-1)}\underset {b \text{times}}{\underbrace{1+2^1+2^2+...+2^{b-1}}}
}} \\ \phantom{X} \\
&=(2^b-1) \cdot(1+2^b+2^{2b} + ... + 2^{(a-1)b})\\ \phantom{X} \\
&=(2^b-1) \cdot \left({ 2^{ab}-1 \over 2^b-1} \right) \end{array}$$
... is composite too (if $b>1$).
|
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|
Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up the following equation (for the purpose of this question I will refer to the leftmost matrix here as A).
$$
\begin{bmatrix}
1 - \lambda & 0 & 0 \\
3 & 3 - \lambda & -4\\
-2 & 1 & -\lambda -2 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
I) Substitute $\lambda_3 = 1$
$$
\begin{bmatrix}
0 & 0 & 0 \\
3 & 2 & -4\\
-2 & 1 & -3 \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2\\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
\end{bmatrix}
$$
II) Reduce the matrix with elementary row operations.
$R_2 \leftarrow R_2 - 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
-2 & 1 & -3 \\
\end{bmatrix}
$$
$R_3 \leftarrow 3R_2 + 2R_3$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
7 & 0 & 2\\
17 & 2 & 0 \\
\end{bmatrix}
$$
$R_2 \leftarrow \frac{1}{7} R_2$
$R_3 \leftarrow \frac{1}{17} R_3$$
$$
A =
\begin{bmatrix}
0 & 0 & 0 \\
1 & 0 & 2/7\\
1 & 2/17 & 0 \\
\end{bmatrix}
$$
III) multiply matrices to get a series of equations equal to 0 and rearrange them in terms of a common element.
$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$
$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$
IV) Substitute a value into the vector to get an eigenvector.
Let $\ x_1 = 1 \rightarrow X_3 = \left(\begin{smallmatrix}1\\-2/17\\-2/7\end{smallmatrix}\right)
$
Which at this point we can see is not a multiple of the expected $X_3$. Can anyone highlight my error for me?
Many thanks in advance.
|
Uh, that's just bad algebra- your fractions are upside down! You have $x_1= -\frac{2}{7}x_3$ and $x1= -\frac{2}{17}x_2$. Setting $x_1= 1$ gives $-\frac{2}{7}x_3= 1$ so $x_3= -\frac{7}{2}$ and $-\frac{2}{17}x_2= 1$ so $x_2= -\frac{17}{2}$, not what you have. that gives $X_3= \begin{pmatrix} 1 \\ -\frac{17}{2} \\ -\frac{7}{2} \end{pmatrix}$. That is a multiple of $\begin{pmatrix}2 \\- 17 \\ -7\end{pmatrix}$.
|
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|
Proof of (complicated?) summation equality This is a simplified case of something I'm trying to prove.
Suppose that $N,h$ are even. I want to show that
$$ \sum_{k=1}^{(N-h)/2} \frac{2^{2k}}{kN^{\underline{2k}}}\left(\frac{N}{2}\right)^{\underline{k}}\left(\frac{N-h}{2}\right)^{\underline{k}} = \sum_{j=1}^{(N-h)/2} \frac{2}{2j+h-1} $$
where $n^{\underline{k}} = n(n-1)...(n-k+1)$ is the falling factorial.
Now the LHS terms can be written as follows
\begin{align} \frac{2^{2k}}{kN^{\underline{2k}}}\left(\frac{N}{2}\right)^{\underline{k}}\left(\frac{N-h}{2}\right)^{\underline{k}} &= \frac{1}{kN^{\underline{2k}}} \prod_{j=0}^{k-1} (N-2j)(N-h-2j)
\\&= \frac{1}{k} \prod_{j=0}^{k-1} \frac{N-h-2j}{N-2j-1}
\\&= \frac{1}{k} \prod_{j=0}^{k-1} \left(1-\frac{h-1}{N-2j-1}\right).
\end{align}
Note the RHS can be written as $H_{\frac{N}{2}-\frac{1}{2}}-H_{\frac{h}{2}-\frac{1}{2}}$ where $H_n$ is the Harmonic number.
|
Consider the sum
$$
S^n_m=\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}};\quad n\ge m\ge1.\tag{1}
$$
We are going to prove:
$$
S^n_m=\frac{2m}{2n-2m+1}.\tag{2}
$$
It is easy to check that for $m=1$ and arbitrary $n\ge1$ the expression (2) is valid:
$$
S^{n}_1=4^1\frac{\binom{n}{1}\binom{1}{1}}{\binom{2n}{2}\binom{2}{1}}=\frac{2}{2n-1}.
$$
Assume that expression (2) is valid for some $S^{n-1}_{m-1}$ $(n\ge m\ge 2)$. It follows then that it is valid for $S^{n}_{m}$ as well:
$$\begin {align}
S^{n}_{m}&=\sum_{k=1}^{m}4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^{m}4^k\frac{\frac{n}{k}\binom{n-1}{k-1}\frac{m}{k}\binom{m-1}{k-1}}{\frac{2n(2n-1)}{2k(2k-1)}\binom{2n-2}{2k-2}\frac{2k(2k-1)}{k^2}\binom{2k-2}{k-1}}\\
&=\frac{2m}{2n-1}\sum_{k=1}^{m}4^{k-1}\frac{\binom{n-1}{k-1}\binom{m-1}{k-1}}{\binom{2n-2}{2k-2}\binom{2k-2}{k-1}}\\
&=\frac{2m}{2n-1}\sum_{k=0}^{m-1}4^{k}\frac{\binom{n-1}{k}\binom{m-1}{k}}{\binom{2n-2}{2k}\binom{2k}{k}}\\
& =\frac{2m}{2n-1}\left(S^{n-1}_{m-1}+1\right)\\
&\stackrel{I.H.}{=}\frac{2m}{2n-1}\left(\frac{2(m-1)}{2(n-1)-2(m-1)+1}+1\right)\\
& =\frac{2m}{2n-1}\cdot\frac{2n-1}{2n-2m+1}\\
&=\frac{2m}{2n-2m+1},\end {align}
$$
where $\stackrel{I.H.}{=}$ means substitution of the induction assumption. Thus, the equality (2) is proved.
The rest is rather straightforward. Let $l$ be an integer number $(1\le l\le n)$. Then:
$$\begin {align}
\sum_{m=1}^l\frac{2}{2n-2m+1}&\stackrel {(2)}=\sum_{m=1}^l\frac{1}{m}\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=
\sum_{m=1}^l\frac{1}{m}\sum_{k=1}^\infty4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^\infty4^k\frac{\binom{n}{k}}{\binom{2n}{2k}\binom{2k}{k}}\sum_{m=1}^l\frac{1}{m}\binom{m}{k}\\
&\stackrel{!}{=}\sum_{k=1}^\infty\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^l\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}.\tag{3}\end {align}
$$
The proof of identity:
$$
\sum_{m=1}^l\frac{1}{m}\binom{m}{k}=\frac{1}{k}\binom{l}{k};\quad k\ge1,
$$
used in $\stackrel{!}{=}$, is left as an exercise.
The equality (3) is identical to that one claimed in question. To see it redefine in the latter $N=2n$, $h=2(n-l)$ and reverse the summation order in RHS: $j\mapsto l+1-j$.
|
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|
Find the domain of convergence of the series
Find the domain of convergence of the series :
$(1)$ $$\sum\limits_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\Big(\frac{1-z}{z}\Big)^n$$
$(2)$ $$\frac{1}{2}z+\frac{1\cdot3}{2\cdot5}z^2+\frac{1\cdot3\cdot5}{2\cdot5\cdot8}z^3 +\dots$$
I was reading Rudin and I know the following theorems
$\star $ For any sequence $\{c_n\}$ of positive numbers, $$\lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n} \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ $$ \lim_{n\to\infty} \sup \sqrt[n]{c_n} \leq \lim_{n\to\infty} \sup \frac{c_{n+1}}{c_n}.$$
$\star $ Cauchy-Hadamard Theorem
My Attempt :
$(1)$ If $c_n = \frac{1\cdot3\cdot5\cdots(2n-1)}{n!},$ then $\lim_{n\to \infty}\frac{c_{n+1}}{c_n}=\lim_{n\to \infty}\frac{2n+1}{n+1}=2$
Hence the series converges if $|\frac{1-z}{z}|<1/2$ or if $|1-z|<|z|/2$ i.e., $|z|<3/2.$
$(2)$ Here similarly $\lim_{n\to \infty}\frac{c_{n+1}}{c_n}=\lim_{n\to \infty}\frac{2n+1}{3n-1}=2/3$
Hence series converges if $|z|<3/2.$
Is this correct? $
|
Let us first write
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n!}=\frac{1\cdot2\cdot3\cdot4\cdot\ldots\cdot(2n)}{2^nn!n!}=\frac{(2n)!}{2^n(n!)^2}$$
and now apply the quotient test:
$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(2n+2)!}{2^{n+1}\left((n+1)!\right)^2}\frac{(1-z)^{n+1}}{z^{n+1}}\cdot\frac{2^n(n!)^2}{(2n)!}\frac{z^n}{(1-z)^n}\right|=$$
$$=\frac{(2n+1)(2n+2)}{2(n+1)^2}\left|\frac{1-z}z\right|\xrightarrow[n\to\infty]{}2\left|\frac{1-z}z\right|\stackrel?<1\iff\left|\frac z{1-z}\right|>2$$
and putting $\;z=x+iy\;$ , the last inequality is the same as
$$x^2+y^2>4\left[(x-1)^2+y^2\right]\implies 3x^2-8x+4+3y^2<0\implies$$
$$3\left(x-\frac43\right)^2+3y^2<\frac43\iff \left|z-\frac43\right|<\frac23\;,\;\;z\neq1\; $$
which is almost an open disk of radius $\;\frac23\;$ around $\;\left(\frac43,0\right)\;$ on the complex plane.
|
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|
Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very easily using Muirhead. But my friend has no idea what Muirhead inequality is. So I want to know whether there is any other method to solve this problem except for Muirhead's inequality.
|
Using strictly familiar Angel form of the CS inequality and the traditional CS inequality itself we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{bc}= \displaystyle \sum\dfrac{(a^2)^2}{abc}\ge\dfrac{(a^2+b^2+c^2)^2}{abc+abc+abc}=\dfrac{(a^2+b^2+c^2)^2}{3abc}\ge\dfrac{(ab+bc+ca)^2}{3abc}= \dfrac{(ab)^2+(bc)^2+(ca)^2+2a^2bc+2ab^2c+2abc^2}{3abc}\ge\dfrac{(ab)(bc)+(bc)(ca)+(ca)(ab)+2a^2bc+2ab^2c+2abc^2}{3abc}= \dfrac{3a^2bc+3ab^2c+3abc^2}{3abc}=a+b+c$
|
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|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
Let $q(t)=(t-x)(t-y)(t-z)$ and $p_n=x^n+y^n+z^n$. Power sums give a base of the ring of symmetric polynomials in the variables $x,y,z$, and by Newton's identities
$$ e_1 = x+y+z = p_1=1, $$
$$ 2e_2 = 2xy+2yz+2xz = p_1^2-p_2 = 0, $$
$$3e_3 = 3xyz = e_2 p_1 - e_1 p_2 + p_3 = 0, $$
so $q(t) = t^3-t^2$ and $x,y,z$ are given by a permutation of $0,0,1$.
More generally, if we have $n$ variables $x_1,\ldots,x_n$ and the power sums from $p_1$ to $p_n$ all equal $1$, then $x_1,\ldots,x_n$ are given by a permutation of $0,0,\ldots,1$, since $t^n-t^{n-1}$ is the only monic polynomial associated to such power sums.
|
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|
How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?
How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$
I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
|
let $n = \frac 1x$
$\lim_\limits{n\to \infty} n((1+\frac 1n)^n - e)$
Binomial expansion:
$n(1 + 1 + \frac 12 (1-\frac 1n) + \frac 1{3!} (1-\frac 1n)(1-\frac 2{n})+\cdots +\frac {1}{n!} (1-\frac 1n)\cdots(1-\frac {n-1}{n}) - e)$
$e = 1 + 1 + \frac 1{2!} + \frac {1}{3!}\cdots$
$n( (-\frac 12 -\frac 1{3!} {3\choose 2} - \frac 1{4!} {4\choose 2} -\cdots - - \frac 1{n!} {n\choose 2})\frac 1{n} + o(\frac 1{n^2}))\\
-\frac 12(1+1+\frac 12 + \frac 1{3!}\cdots \frac {1}{n!}) + o(\frac 1n)\\
-\frac 12 e$
|
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|
Evaluate $\int_{1}^{2} \ \frac{\arctan (x)}{\arctan \left (\frac{1}{x^2-3x+3} \right )} dx$ In this exercise, we are given two integrals $I$ and $J$:
$$I = \int_{0}^{\pi/2}\frac{\sin^nx}{\sin^nx+\cos^nx} dx, \quad n\in \mathbb{N^*},$$
and
$$J = \int_{1}^{2} \frac{\arctan(x)}{\arctan \left (\frac{1}{x^2-3x+3} \right )} \, dx.$$
I know that $I = \pi/4$ but I do not understand the association between $I$ and $J$.
This exercise is found in a lesson on substitution so maybe there is no association between the two integrals, but $J$ should be $1.108205454429995$.
|
As you suspected, there may be one more typo in the textbook.
As I commented, I was wondering if approximation would be require. So, for the fun of it, I tried to.
$$I=\int_1^2\frac{\tan ^{-1}(x)}{\tan ^{-1}(2-x)-\tan ^{-1}(1-x)}\,dx=\int_0^1 \frac{\tan ^{-1}(y+1)}{\tan ^{-1}(y)-\tan ^{-1}(y-1)}\,dy$$
Using Taylor expansions built around $y=0$
$$\tan(y+a)=\tan ^{-1}(a)+\frac{y}{a^2+1}-\frac{a y^2}{\left(a^2+1\right)^2}+\frac{\left(3
a^2-1\right) y^3}{3 \left(a^2+1\right)^3}+\frac{\left(a-a^3\right)
y^4}{\left(a^2+1\right)^4}+O\left(y^5\right)$$ leads to
$$\frac{\tan ^{-1}(y+1)}{\tan ^{-1}(y)-\tan ^{-1}(y-1)}=\frac{\frac{\pi }{4}+\frac{y}{2}-\frac{y^2}{4}+\frac{y^3}{12}+O\left(y^5\right) }{\frac{\pi }{4}+\frac{y}{2}-\frac{y^2}{4}-\frac{5 y^3}{12}+O\left(y^5\right) }$$ Using long division, the integrand is
$$1+\frac{2 y^3}{\pi }-\frac{4 y^4}{\pi ^2}+O\left(y^5\right)$$ leading to
$$I_{(5)}=1+\frac{1}{2 \pi }-\frac{4}{5 \pi ^2}\approx 1.07810$$ which is not fantastic.
For sure, we could continue the expansions and get
$$I_{(6)}=1+\frac{1}{3 \pi }+\frac{31}{105 \pi ^2}+\frac{4}{21 \pi ^3}-\frac{16}{7 \pi ^4}\approx 1.11870$$
$$I_{(7)}=1+\frac{19}{48 \pi }+\frac{143}{840 \pi
^2}-\frac{145}{84 \pi ^3}+\frac{5}{7 \pi ^4}+\frac{4}{\pi ^5}\approx 1.10798$$
$$I_{(8)}=1+\frac{19}{48 \pi }-\frac{493}{1512 \pi ^2}-\frac{409}{756 \pi
^3}+\frac{33}{7 \pi ^4}-\frac{28}{9 \pi ^5}-\frac{64}{9 \pi ^6}\approx 1.10635$$
Now, I give up !
|
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|
Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
|
Let $ \epsilon >0$ be an arbitrary small number.
Choose a positive integer $n_0$ such that $$\frac {2}{n_0 +3} <\frac {\epsilon }{2}$$
Note that if $m \ge n_0$, and $n\ge n_0$ are integers, then $$ |f(n)-f(m)| = |\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|= $$ $$|\frac {2(m-n)}{(n+3)(m+3)} |\le $$ $$\frac {2(m+n)}{(n+3)(m+3)} <$$ $$\frac {2}{(m+3)}+\frac {2}{(n+3)} <\epsilon $$
Thus the sequence ($ \frac{3n+5}{2n+6}$) is a Cauchy sequence.
|
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|
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.
I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
|
$$
\begin{align}
\sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)}
&=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\frac12\left(\frac12-\frac1{2n+2}\right)\\
&=1-\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}-\frac14\\[3pt]
&=\frac34-\log(2)
\end{align}
$$
|
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|
Compute $\int\frac{x^2}{\tan{x}-x}dx$ for $x \in (0, \frac{\pi}{2})$
Compute the indefinite integral $\displaystyle{\int\frac{x^2}{\tan{x}-x}dx, x \in (0, \frac{\pi}{2})}$.
I have tried substituting $\tan{x} = t$ or write $\tan{x} = \displaystyle{\frac{\sin{x}}{\cos{x}}}$ and then try to solve it from there, but I didn't manage to do it. Also, $\displaystyle{\left(\frac{\sin{x}}{x}\right)' = \frac{\sin{x} - x\cos{x}}{x^2}}$, but in the integral I have the inverse of this...
|
$$\int\frac{x^2}{\tan{x}-x} =\int\frac{x^2 \cos x +x \sin x - x \sin x}{\sin{x}-x \cos x}=\int \left(\frac{x \sin x}{\sin{x}-x \cos x} - x \right) $$
Now notice that $$\sin x - x \cos x=t \implies x\sin x = dt$$
Hence $$\int \left(\frac{x \sin x}{\sin{x}-x \cos x} - x \right) = \ln |(\sin{x}-x \cos x) | - \frac{x^2}{2} +C $$
|
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Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$ whose slope is $$m_1=\frac{-4 \cot a}{5}$$
Also slope of $AP$ is given by
$$m_2=\frac{4 \sin a+5}{5 \cos a-3}$$
So both slopes are equal , with that we get
$$12 \cos a-25 \sin a=20 \tag{1}$$
Now distance $AP$ is given by
$$AP=\sqrt{(3-5 \cos a)^2+(5+4 \sin a)^2}=\sqrt{75-30 \cos a+40 \sin a} \tag{2}$$
Now using $(1)$ we have to find $\cos a$ and $\sin a$ and then substitute in $(2)$ which becomes lengthy.
Any better way?
|
let $$y=mx+n$$ then the equation of the line through $P(3;-5)$ is given by
$$y=m(x-3)-5$$ plug this in the equation of the Ellipse and solve this equation for $x$, since it should be a tangentline, the discriminante must be Zero
you will get $$16m^2-30m-9=0$$ solve this equation for $m$
|
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|
Why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$?
If $a,b\in\mathbb R_{>0}$ and $ b>a$, then why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$ ?
Usual change of variables doesnt bring anything, I think. Is there a special function involved here (because of $\pi$) ?
|
METHODOLOGY $1$:
Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\log\left(\frac{b^2+x^2}{a^2+x^2}\right)\,dx&=2\int_0^\infty \int_{a^2}^{b^2} \frac{1}{y+x^2}\,dy\,dx\\\\
&\overbrace{=}^{\text{Fubini}}2\int_{a^2}^{b^2}\int_0^\infty \frac{1}{y+x^2}\,dx\,dy\\\\
&=2\int_{a^2}^{b^2}\frac{\pi/2}{\sqrt y}\,dy\\\\
&=2\pi (b-a)
\end{align}$$
as expected!
METHODOLOGY $2$:
Alternatively, we can integrate $\int \log(a^2+x^2)\,dx$ by parts with $u=\log(a^2+x^2)$ and $v=x$ to find
$$\begin{align}
\int \log(a^2+x^2)\,dx&=x\log(a^2+x^2)-2\int \frac{x^2}{a^2+x^2}\,dx\\\\
&=x\log(a^2+x^2)-2x+2a\arctan(x/a)
\end{align}$$
And the rest is straightforward.
|
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|
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}-n)}{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}+\sqrt{n})} \right) \\
&= \lim_{n \to \infty}\left( \frac{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}-\frac{n}{\sqrt{n}})}{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}+ \frac{\sqrt{n}}{\sqrt{n}})} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{n}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{\sqrt{n}\sqrt{n}}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\sqrt{n}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{1}{2} - \frac{\sqrt{n}}{2} \right) \\
\end{align}$
Which is wrong.
Where could be my mistake?
|
$$\begin{align}
\sqrt{n+\sqrt{n}}-\sqrt{n} &=
\frac{(\color{red}{\sqrt{n+\sqrt{n}}}-\color{blue}{\sqrt{n}})(\color{red}{\sqrt{n+\sqrt{n}}}+\color{blue}{\sqrt{n}})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(\color{red}{\sqrt{n+\sqrt{n}}})^2-(\color{blue}{\sqrt{n}})^2}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(n+\sqrt{n})-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{\sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{1}{\sqrt{1+1/\sqrt{n}}+1} \\
\end{align}$$
and $1/\sqrt n \to 0$ hence
$$\lim_{n\to\infty}\sqrt{n+\sqrt{n}}-\sqrt{n} = \lim_{n\to\infty}\frac{1}{\sqrt{1+1/\sqrt{n}}+1} = \frac 1{\sqrt{1+0}+1} =\frac 12$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is this proof false? $\sqrt 2 + \sqrt 6 < \sqrt 15$ What is wrong with the given proof?
$$
\sqrt 2 + \sqrt 6 < \sqrt 15 \\
(\sqrt 2 + \sqrt 6)^2 < 15 \\
2 + 6 + 2\sqrt2\sqrt6 < 15 \\
2\sqrt2\sqrt6 < 7\\
2^2 \cdot 2 \cdot 6 < 49 \\
48 < 49
$$
|
As it stands, the proof is flawed, but not for any error of reasoning (really), but because it is argued in a slightly nonsensical manner. Typically, when you write a sequence statements, you want to connect them to each other. For example, "$A$ implies $B$" or "$A$ if and only if $B$". If no such connecting verbs are stated, the usual assumption is that early statements imply later statements. Thus, eliding all of the intermediate steps, one might reasonably read your proof as:
If $\sqrt{2} + \sqrt{6} < \sqrt{15}$, then $48 < 49$.
This is, in fact, the converse of what you want to show. There are two ways to make this more clear:
*
*You could use language or notation to show that one line is true if and only if the previous line is true. If you attacked it this way, you might write
\begin{align}
\sqrt{2} + \sqrt{6} < \sqrt{15}
&\iff \left( \sqrt{2} + \sqrt{6} \right)^2 < \sqrt{15}^2 && \text{(since both sides are positive)} \\
&\iff 2 + 2\sqrt{2}\sqrt{6} + 6 < 15 && \text{(distribution)} \\
&\iff 2\sqrt{2}\sqrt{6} < 7 && \text{(law of additive cancelation)} \\
&\iff 2^2\cdot 2 \cdot 6 < 7^2 && \text{(since both sides are positive)} \\
&\iff 48 < 49. && \text{(simplify)}
\end{align}
Since $48<49$, the proposition is proved.
*Alternatively, if you want to look clever, you write the proof as a series of implications: we know that $48 < 49$. Therefore
\begin{align}
48 < 49
&\implies \sqrt{48} < \sqrt{49} \\
&\implies 2 \sqrt{2}\sqrt{6} < 7 \\
&\implies 2 + 2\sqrt{2}\sqrt{6} + 6 < 15 \\
&\implies \left( \sqrt{2} + \sqrt{6} \right)^2 < 15 \\
&\implies \sqrt{2} + \sqrt{6} < \sqrt{15},
\end{align}
where $\sqrt{x}$ denotes the principle (or positive) square root.
Another rather pathological approach might be to prove by contradiction. Suppose that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$. Using the basic computations that were originally given, but replacing $<$ with $\ge$, we obtain
$$ 48 \ge 49, $$
which is obvious nonsense. Therefore it cannot be that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$, which proves the proposition.
|
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|
how to maximize $f(x,y)=\frac{x+y-2}{xy}$? How to maximize $f(x,y)=\frac{x+y-2}{xy}$ where $x,y \in \{1,2,\ldots,n\}$?
It seems that maximum will occur when $(x,y)=(1,n)$ or $(n,1).$
|
Note that $f(1,1)=0$ and $f(1,2)=f(2,1)=\frac {1}{2}$
Otherwise,
$$\begin{align}
f(x,y)&=\frac{x+y-2}{xy}\\
&=\frac {1}{y} +\frac {1}{x} - \frac {2}{xy}\\
&=\frac {1}{y} \left(1-\frac {1}{x}\right) +\frac {1}{x}\left(1-\frac {1}{y}\right)\\
&\le\frac {1}{2} \left(1-\frac {1}{x}\right) +\frac {1}{2}\left(1-\frac {1}{y}\right)\\
&\le1-\frac {1}{n}
\end{align}$$
Note that $$f(1,n)= 1-\frac {1}{n}$$
Thus the maximum is $ 1-\frac {1}{n}$ which is achieved at $(1,n)$.
|
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|
Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac{3}{4}\big)\implies\cos\big(\tan^{-1}x\big)=\cos\big(\tan^{-1}\frac{3}{4}\big)\\\implies\tan^{-1}x=2n\pi\pm\tan^{-1}\frac{3}{4}\\
\implies \tan^{-1}x=\tan^{-1}\frac{3}{4}\quad\text{ as }0\leq\tan^{-1}\frac{3}{4}\leq\frac{\pi}{2}\\
\implies x=\frac{3}{4}
$$
Is it correct or $\frac{-3}{4}$ also is a solutions ?
What about the condition $0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}$, does this affect the solutions ?
|
Yes because $\cos$ is an even function.
It follows also from your way:
$$\arctan{x}=\pm\arctan\frac{3}{4},$$ which gives $x=\frac{3}{4}$ or $x=-\frac{3}{4}.$
|
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|
If $|a|\ge 2$,$|b|\ge 2$, $P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1$, prove that all roots of $P(x)$ are real
Given: $a,b\in \mathbb R$, $|a|\ge 2$,$|b|\ge 2$, $$P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1,$$
Prove or disprove: all roots of $P(x)$ are real.
From a math contest. The polynomial is reciprocal, but I can't see how to use this on the proof, if that is the case, and I tried to use Descartes' rule of signs, but without success. I'm a little lost on how to address the problem.
Hints or solutions are appreciated. Sorry if this is a duplicate.
|
Given the helpful hints and suggestions from David Quinn and Alex Francisco, I will try a full solution to my question, for sake of completeness.
First, by dividing $P(x)=x^4-(a+b)x^3+(ab+2)x^2-(a+b)x+1=0$ by $x^2$ we get
$$x^2-(a+b)x+(ab+2)-(a+b)\frac{1}{x}+\frac{1}{x^2}=0$$
$$\Leftrightarrow x^2+\frac{1}{x^2}-(a+b)(x+\frac{1}{x})+ab+2=0 $$
but, as $\displaystyle (x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2,$ the previous equation is equivalent to
$$(x+\frac{1}{x})^2-2-(a+b)(x+\frac{1}{x})+ab+2=0$$
$$\Leftrightarrow u^2-(a+b)u+ab=0~~\text{with the substitution}~~u=x+\frac{1}{x},$$
with obvious solution $u=a$ and $u=b$, real values by the assumption.
Now, as $u=x+\frac{1}{x}$, we can solve for $x$ noticing that
$$u=\frac{x^2+1}{x}\Leftrightarrow x^2-ux+1=0$$
Therefore, as $u=a$ and $u=b$ the original polynomial can be expressed by
$$P(x)=(x^2-ax+1)(x^2-bx+1)=0,$$
and the condition $|a|\ge 2$ and $|b|\ge 2$ ensures that the roots of both equations (that are the roots of $P(x)$) will be real numbers, as the discriminant for each second order degree polynomial will be non-negative with this condition.
|
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|
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by parts implies $$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{(x^2+a^2)-a^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{1}{(x^2+a^2)^{n}}-2na^2\int\frac{dx}{(x^2+a^2)^{n+1}}$$ but I don't think I'm doing this correctly since the power of the denominator $(x^2+a^2)$ is not decreasing.
|
Call your integral $I_n$ so that the result of your integration by parts is $$I_n=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$
Now rearrange so that $I_{n+1}$ is the subject of the formula.
Then replace all $n$ with $n-1$ and you have the formula you require.
|
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|
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$. $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.
|
Consider the sequence of consecutive integers beginning with one more than the $(n-1)$th triangular number and ending with $(n-1)$ more than the first term. This sequence has $n$ terms. If you take the terms of this sequence, cube them, and add them together, you get the $n$th term of the sequence that you want, which you can call $T_n$.
The $(n-1)$th triangular number is $\frac{n^2-n}2$. One more than that is $\frac{n^2-n+2}{2}$. And $(n-1)$ more than the latter is $\frac{n^2-n+2}{2} + (n-1) = \frac{n^2+n}{2}$.
Therefore $T_n$ can be computed as the difference between $C(\frac{n^2+n}{2})$ and $C(\frac{n^2-n}{2})$, where $C(m)$ represents the sum of consecutive cubes from $1$ to $m$, i.e.
$$T_n = \sum_{\frac{n^2-n}{2}}^{\frac{n^2+n}{2}}k^3 = C(\frac{n^2+n}{2}) - C(\frac{n^2-n}{2})$$
Now $C(m) = \frac 14m^2(m+1)^2$ (the sum of the cubes of the first $m$ positive integers is the square of the sum of the first $m$ positive integers, which is also the square of the $m$th triangular number - a standard result which is a particular case of the general Faulhaber's formula).
Hence,
$$\begin{align}T_n &= \frac 14 \left(\left(\frac{n^2+n}{2}\right)^2\left(\frac{n^2+n}{2}+1\right)^2 - \left(\frac{n^2-n}{2}\right)^2\left(\frac{n^2-n}{2}+1\right)^2 \right) \\&= \frac 18n^3(n^2+1)(n^2+3)\end{align}$$
The final simplification is tedious by hand, but an online computer algebra system makes it trivial. You can verify the formula works for the first few terms, e.g. $T_1 = 1, T_2 = 35, T_3 = 405$ and so forth.
|
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|
Limit using Taylor series expansion
Can someone help me with this limit. I know I have to expand in Taylor series
$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)$
$a^x=e^{x\ln a}=1+x\ln a+\frac{1}{2!}x^2(\ln a)^2+\frac{1}{3!}x^3(\ln a)^3+o(x^3)$.
But how to proceed next?
|
Note that from Taylor's expansion
$$\frac{ a^{\sqrt{x+1}}-a^{1+\frac{x^2}2-\frac{x^2}8} }{x^3}
=\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{x^3}
=\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{\frac{x^3}{16}+o(x^3)}\frac{\frac{x^3}{16}+o(x^3)}{x^3}$$
now use standard limits.
|
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|
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0$$
Now what do I do?
|
Since, $f(-x)=-f(x),$ we obtain:
$$|x-a|-|x-2|+b|x-5|+|x+a|-|x+2|+b|x+5|=0.$$
Now, for $x=2$ we obtain:
$$10b+|a-2|+|a+2|-4=0$$ and for $x=5$ we obtain:
$$10b+|a-5|+|a+5|-10=0,$$ which gives
$$|a-2|+|a+2|=|a-5|+|a+5|-6$$ or
$$-2\leq a\leq2,$$ which gives$$b=0.$$
Thus, $$f(x)=|x+a|-|x+2|$$ and
$$|x-a|-|x-2|+|x+a|-|x+2|=0,$$ which for $x=0$ gives
$$|a|=2$$ and we see that $a=2$ and $a=-2$ they are valid.
Id est, $a+b=2$ or $a+b=-2.$
|
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|
Unable to prove that $n^5-n$ is a multiple of 30. It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.
Basically, the issue is how to show that :
$(n-1)(n)(n+1)(n^2+1)$ for $n$ being odd or even, is a multiple of $30$.
Let us take two cases: (a)$n$ is odd, i.e. of the form $2k+1$, for a suitable integer $k$.
So, $(2k)(2k+1)(2k+2)(2k+1)$, as $n^2+1$ will also be an odd number.
This leads to a final expression of the form: $2k+2$.
But, not able to prove its being a multiple of $30$.
(b) Need not take even case, as the final even form is proved by one term being even; & still not able to prove it being a multiple of $30$.
So, try by smaller values, and that would not serve the cause until use induction (strong case might be needed, as need consider all lower value for the case of $k=1$).
Am I correct, & should proceed by strong induction?
|
Modulo $5$ we have $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)\equiv n(n^2-1)(n^2-4) =n(n-1)(n+1)(n-2)(n+2)$$ and the product of five successive integers is divisible by $5$, while $n(n^2-1)=n(n-1)(n+1)$ is divisible by $6$.
|
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|
Is $\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$ is divergent or not? I'm searching for a continuous function whose Fourier series diverges at x=0.
Now I've come up with an idea and don't know if it's right.
$$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$$
Is the series above divergent or not? I expect it be divergent.
I calculated its sum of first N terms (N varies from 1 to 1000) with help of Mathematica, there seems no sign that it converges.
|
$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$
Using Riemann sum for the integral:
$\int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\lim_{n -> \infty} {\frac{1}n}\sum_{k=0}^n$ $\frac{(-1)^k}{\log\log\frac{4}{\frac{k}{n}}}$
We get:
=$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}n}\sum_{k=0}^n$ $\frac{(-1)^k}{\log\log\frac{4n}{k}}$=
The sum can be decreased by using $log(log\frac{4n}{k})<\frac{4n}{k}$ where ${k}\le {n}.$
$>$ $\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}{4n^2}}\sum_{k=0}^n$ ${k}{(-1)^k}$= =$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}{4n^2}}\frac{n}{2}$=
$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\frac{1}{8}\sum_{n=1}^\infty {\frac{1}{n}} \longmapsto \infty$ if ${n \longmapsto \infty}$
Finally
$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$ $\gt \infty$
|
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|
Find the area of a quadrilateral in a parallelogram.
Assume quadrilateral $ABCD$ is a parallelogram and its area is $S$. And satisfy the following conditions: $AE=BE$, $BF=FC$, $AQ // PC$. Find the area of quadrilateral $APCQ$
|
Assume the origin of the coordinate system is $B$. so coordinate of point $B$ is $(0,0)$. And the $x$-axes is $BC$. Assume the coordinate of point $A$ is $(2a,2b)$. The coordinate of point $C$ is $(2c,0)$. Then we can calculate other points' coordinates. $E$ is $(a,b)$, $F$ is $(c,0)$ $D$ is $(2c+2a, 2b)$.
Then the equation of straight line $AF$ is $$y=\frac{-2b}{c-2a}(x-c)$$
the equation of straight line $ED$ is $$y-b=\frac{b}{2c+a}(x-a)$$
so the coordinate of point $P$ is $(\frac{2}{5}(3a+c),\frac{6}{5}b)$
then the equation of straight line $AQ$ is $$y-2b=\frac{3b}{3a-4c}(x-2a)$$
then the coordinate of point $Q$ is $(\frac{4c+7a}{5},\frac{7}{5}b)$
then $$|PQ|=\sqrt{\left(\frac{b}{5}\right)^2+\left(\frac{a}{5}+\frac{2c}{5}\right)^2}$$
and $$|ED|=\sqrt{b^2+(2c+a)^2}$$
then we know $|ED| = 5\times |PQ|$
then $S_{APCQ} = \frac{1}{5}S_{AECD}$ Assume the area of $ABCD$ is $S$. the area of $AECD$ is $\frac{3}{4}S$. So $S_{APCQ}= \frac{3}{4}S \times \frac{1}{5} = \frac{3}{20}S$
|
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|
Is $\int \frac{\sqrt{1-x^2}}{x}\;dx = \ln\lvert x\rvert + C$? I have the integral
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx
$$
I do a trig. substitution $x = \sin(\theta)$. Then I get $\sqrt{1-x^2} = \cos(\theta)$. So
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx = \int \frac{\cos(\theta)}{\sin(\theta)}\; d\theta = \ln\lvert \sin(\theta)\rvert + C= \ln\lvert x\rvert + C.
$$
But I am thinking that I made a mistake somewhere because the derivative of $\ln\lvert x\rvert$ is not $\frac{\sqrt{1-x^2}}{x}$.
Where is my mistake?
|
$$x=\sin\theta\implies dx=\cos(\theta)\,d\theta$$
$$\begin{align}
\int{ \sqrt{1-x^2} \over x}\,dx
&= \int{ \sqrt{1-\sin^2\theta} \over \sin\theta} \cos(\theta)\,d\theta \\
&= \int{ \sqrt{\cos^2\theta} \over \sin\theta } \cos(\theta)\,d\theta \\
&= \int{ \cos^2\theta \over \sin\theta}\,d\theta \\
&= \int{ -\sin^2(\theta) + 1 \over \sin\theta}\,d\theta \\
&= \int\bigl[ -\sin(\theta) + \csc(\theta)\bigr]\,d\theta
\end{align}$$
You got it from there.
|
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|
Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$ Prove that
$$\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$$
for $x \ge 0$, $y \ge 0$, $z \ge 0$ and $x+y+z \le 2$.
My work:
\begin{align*}
&\mathrel{\phantom{=}} \sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy}\\
&\le\sqrt3 \sqrt{x^2+yz+y^2+xz+z^2+xy}\\
&\le\sqrt3\sqrt{2x^2+2y^2+2z^2}=\sqrt6\sqrt{x^2+y^2+z^2}.
\end{align*}
|
It is enough to show the homogeneous inequality $\sum_{cyc} \sqrt{x^2+yz} \leqslant \frac32(x+y+z)$. WLOG, let $x \geqslant y \geqslant z$.
Note by AM-GM, we have $\sqrt{x^2+yz} \leqslant \sqrt{x^2+xz} \leqslant x + \frac12z$. This type of AM-GM is motivated by noting $(1, 1, 0)$ is a solution for equality.
Further by CS (or power means), we get
$$\sqrt{y^2+zx} + \sqrt{z^2+xy} \leqslant \sqrt{1+1} \sqrt{(y^2+zx)+(z^2+xy)}$$
Thus it is enough to show that
$$\sqrt2\sqrt{y^2+z^2+x(y+z)} \leqslant \frac12(x+3y+2z)$$
Simplifying, this is $(x-y-2z)^2+8z(y-z) \geqslant 0$ which is obvious. Equality is when any two variables are equal and the third zero.
|
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|
Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.
|
If two of $x,y,z$ are equal, wlog, $x=y$ then $x(6-x)=9 \to x=3$ and $z(6-3)=9 \to z=3$.
If all are $\neq$, wlog $0<x<y<z$, then $9=x(6-y)<z(6-x)=9$, abs.
Finally, the solution is $(3,3,3)$
|
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|
A limit question of 3-variable-functions. $\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$
$$\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$$
I checked that the limit does not exist but I cannot prove that.
I tried $y=mx$, $z=nx$ and also $y=x^m$, $z=x^n$ but they gave me nothing but the limit equals to zero.
Thanks a lot
|
Let consider
$$\left|\frac {xyz^2}{x^2+y^4+z^6}\right|=\frac {|x||y|z^2}{x^2+y^4+z^6}$$
and let
$$\begin{cases}
|x|=|X|\\\\
|y|=\sqrt{|Y|}\\\\
z=\sqrt[3] Z
\end{cases}$$
then
$$\frac {|x||y|z^2}{x^2+y^4+z^6}=\frac {|X|\sqrt{|Y|}\sqrt[3] {Z^2}}{X^2+Y^2+Z^2}=\frac{\rho^{1+\frac12+\frac23}f(\theta,\phi)}{\rho^2}=\frac{\rho^{\frac{13}6}f(\theta,\phi)}{\rho^2}=\rho^\frac16f(\theta,\phi)\to 0$$
therefore since
$$\frac {|x||y|z^2}{x^2+y^4+z^6}\to 0 \implies \frac {xyz^2}{x^2+y^4+z^6}\to 0$$
|
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|
Difficulty finding the integral of this rational function? I'm attempting to do the following integral
$$\int \frac{2x+1}{x^3 +2x^2 +1} \, dx$$
I wanted to try using partial fractions but I'm unsure how to factor the denominator. I've been unable to make any progress on this question because of this which is why I don't have any work for it. I was wondering whether there are any ways to take the integral of this function?
|
Hint
The denominator has three roots, only one neing real. So, write
$$x^3+2x^2+1=(x-a)(x^2+bx+c)$$ and use partial fractions
$$\frac{2x+1}{x^3 +2x^2 +1}=\frac 1 {a^+ab+c}\left(\frac{2 a+1}{x-a}-\frac{ (a+b-2 c)+(2 a+1) x}{x^2+bx+c} \right)$$
Now, work the problem of expressing $a,b,c$ which is not the most pleasant since
$$a=-\frac{2}{3}-\frac{4}{3} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{43}{16}\right)\right)$$
Edit
If this looks too difficult, name $a,b,c$ the roots of $x^3 +2x^2 +1=0$ and write
$$\frac{2x+1}{x^3 +2x^2 +1}=\frac{2x+1}{(x-a)(x-b)(x-c)}$$ Use partial fraction decomposition to get
$$\frac{2x+1}{x^3 +2x^2 +1}=\frac{2 a+1}{(a-b) (a-c) (x-a)}+\frac{2 b+1}{(b-a) (b-c) (x-b)}+\frac{2
c+1}{(c-a) (c-b) (x-c)}$$ which is simple, leading to a weighted sum of logarithms, two of them with complex arguments (you can recombine them later). The result is ugly.
|
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|
Any simple way for proving $\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$? How to prove
$$\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$$ with $\mathrm{erfc(x)} $ is the complementary error function, I have used integration by part but i don't succed
|
Recalling that $\text{erf} (x) = 1 - \text{erfc} (x)$, the integral can be rewritten as
$$\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx = \int_0^\infty \text{erfc}(x) \, dx - \int_0^\infty \text{erfc}^2 (x) \, dx.$$
As
$$\frac{d}{dx} \left (\text{erfc}(x) \right ) = -\frac{2}{\sqrt{\pi}} e^{-x^2},$$
integrating by parts gives
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= - \frac{2}{\sqrt{\pi}} \int_0^\infty x e^{-x^2} \, dx + \frac{4}{\sqrt{\pi}} \int_0^\infty x e^{-x^2} \text{erfc}(x) \, dx.
\end{align*}
And by parts again
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= \frac{1}{\sqrt{\pi}} - \frac{4}{\sqrt{\pi}} \left (\frac{1}{2} - \frac{1}{\sqrt{\pi}} \int_0^\infty e^{-2x^2} \, dx \right )\\
&= -\frac{1}{\sqrt{\pi}} + \frac{4}{\pi} \int_0^\infty e^{-2x^2} \, dx.
\end{align*}
In the last integral, enforcing a substitution of $x \mapsto x/\sqrt{2}$ leads to
\begin{align*}
\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx &= -\frac{1}{\sqrt{\pi}} + \frac{4}{\pi \sqrt{2}} \frac{\sqrt{\pi}}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-x^2} \, dx\\
&= -\frac{1}{\sqrt{\pi}} + \frac{2}{\sqrt{\pi} \sqrt{2}},
\end{align*}
or
$$\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx = \frac{\sqrt{2} - 1}{\sqrt{\pi}},$$
as expected.
|
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|
Using integration to solve a formula for the area of a ellipse Problem:
Set up a definite integral to find the area of an ellipse with axis lengths $a$ and $b$. Use a trigonometric substitution to find a formula for the area. What happens if $a=b$? Does this agree with a Geometry formula for a circle? Explain.
$$\frac {x^2}{a^2} + \frac{y^2}{b^{2}} = 1$$
where $a$ & $b$ are positive constants.
$B$ = Area of the first quadrant of the ellipse.
Total area = $4B$
\begin{align}
& y^{2} = \left(\frac{b^{2}}{a^{2}}\right)({a^{2}-x^{2}}) \\
\implies & y = \frac{b}{a}\sqrt{a^{2}-x^{2}}
\end{align}
Hence,
\begin{align}
B & = \int_0^a{\frac{b}{a}\sqrt{a^{2}-x^{2}}}\;dx = \frac{b}{a}\int_0^a{\sqrt{a^{2}-x^{2}}}\;dx
\end{align}
To remove $\sqrt{\quad}$, make a trig sub.
$$1 - \sin^{2} \Theta = \cos^{2}\Theta$$
$$x = a\,\sin\Theta$$
\begin{align}
B & = \frac{b}{a}\int_0^a{\sqrt{a^{2}-a^{2}\sin^{2}\Theta}\;dx} = \frac{b}{a}\int_0^a{\sqrt{a^{2}(\cos^{2}\Theta)}}\;dx\\
& = \frac{b}{a}\int_0^a{\sqrt{a^{2}}\,\sqrt{\cos^{2}\Theta}}\;dx = \frac{b}{a}\int_0^a{a\,\cos\Theta}\;dx
\end{align}
\begin{align}
\frac{dx}{dΘ} = (a \, \sin\Theta)' \implies dx = a \, \cos \Theta \, d\Theta
\end{align}
Therefore,
\begin{align}
B & = \frac{b}{a}\int_0^a{a \, \cos\Theta(a \, \cos \Theta \, d \Theta)} = \frac{a^{2} \, b}{a}\int_0^a{\cos^{2}\Theta}\;d\Theta = (a \, b)\int_0^a{\cos^{2}\Theta}\;d\Theta
\end{align}
I am a little lost up until this point and the formula doesn't seem to be going in the direction it needs to so that it will become the area of an ellipse. I feel like I made a mistake somewhere along the way. Any help is much appreciated!
|
The integral is pretty trivial. Spoiler:
$$\int \sqrt{a^2-t^2}\ dt = \frac{1}{2} \left(t \sqrt{a^2-t^2}+a^2 \tan ^{-1}\left(\frac{t}{\sqrt{a^2-t^2}}\right)\right)$$
This can be obtained by a simple substitution $t = a\sin(z)$, $dt = a\cos(z)\ dz$
Which turns the integral into
$$a^2\int \cos^2(z)\ dz$$
Which is trivial and straightforward.
Remember that $z = \arcsin\left(\frac{t}{a}\right)$
The final result will be thence:
$$\int_0^a \sqrt{a^2-x^2}\ dx = \frac{1}{4} \pi a^2$$
Now, adjust with the constants.
|
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|
Unable to evaluate limit correctly I want to find the limit of
$$\frac{e^x + \frac1{e^x} - 2\cos x}{x\tan x}$$
as $x$ tends to $0$.
My attempt:
The limit of $\frac{e^x + 1/e^x - 2cosx}{xtanx}$ should be the same as the limit of $\frac{2 - 2\cos x}{x\tan x}$, which can evaluated using the standard limits of $x/\sin x$ and $((1-\cos x)/x^2)$. But this gives me the answer as $1$, while the correct answer is $2$. I suspect the error is in writing it as $\frac{2 - 2cosx}{xtanx}$, but am unable to see why. Please help.
To give context - I can use only the basic limit laws for algebraic combinations of limits and some standard limits.
|
You cannot apply partial limits in addition and subtraction. Though, it can be used in products.
\begin{align} \lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x \tan x} &=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \frac{x}{\tan x }\\
&=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \lim_{x \to 0} \frac{x}{\tan x }\\
&=\lim_{x \to 0} \frac{e^x \color{red}{-1 -x } + 1/e^x \color{red}{-1 +x +2} - 2 \cos x}{x^2}\\
&=\lim_{x \to 0} \frac{ \color{blue}{e^x -1 -x} + \color{red}{e^{-x} -1 +x} +2 \sin ^2 (x/2)}{x^2}\\
&=\lim_{x \to 0} \frac{ \color{blue}{e^x -1 -x}}{x^2} + \lim_{x \to 0} \frac{\color{red}{e^{-x} -1 +x}}{x^2}+ \lim_{x \to 0} \frac{2 \sin ^2 (x/2)}{x^2}\\
\end{align}
Can you proceed using standard limits now?
|
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|
What is the value of $\delta $ if $\epsilon=0.01$?
Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2}, & \text{if
$(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$
Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0)
\vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-
1.$\delta <0.01$
2.$\delta <0.001$
3.$\delta <0.02$
4.no such $\delta $ exists
solution:since $f(0,0)=0$,then consider $\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{xy(2x+3y)}{x^2+y^2}\right\vert\le \frac{(2x+3y)}{2}$ as $xy\le \frac{x^2+y^2}{2}$
From here, how to proceed further...
|
To get a larger possible value of $\delta$, use the polar coordinate transformation $x = r \cos t, y = r \sin t$ to rewrite the equality in the question.
$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{r^3 \cos t \sin t (2\cos t+3\sin t)}{r^2}\right\vert = r |\cos t| |\sin t| |2 \cos t + 3 \sin t|
$$
In another answer, choosing $|x|,|y| \le \sqrt{x^2+y^2}$ is similar to bounding $|\sin t|,|\cos t|$ by $1$. IMHO, it's too brutal as all information about the variable $t$ is lost. $2 \cos t + 3\sin t = \sqrt{2^2+3^2} \sin(t + \alpha) = \sqrt{13} \sin(t+\alpha)$, where $\tan \alpha = \dfrac23$.
From this
$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}\right\vert
= \frac{\sqrt{13}}{2} r |\sin 2t \sin(t+\alpha)| \le \frac{\sqrt{13}}{2} \delta < \epsilon$$
for any $t \in [0,2\pi]$ and $r < \delta$.
The choice $\delta < \dfrac{2}{\sqrt{13}} \epsilon$ guarantees that $|f(x,y)-f(0,0)| < \epsilon$. When $\epsilon = 0.01$, it becomes
$$\delta < \frac{2}{\sqrt{13}} \cdot \frac{1}{100} = \frac{1}{50\sqrt{13}} \approx 0.005547. \quad \text{(cor. to 4 d.p.)}$$
This improves the bound given in another answer.
The graph of $g(t) = \cos t \sin t (2 \cos t + 3 \sin t)$
Observe that the amplitude is between $1.5$ and $2$. This suggest that choosing $\dfrac52 \delta < \epsilon$ is too restrictive. We can allow more points to enter this $\delta$-ball in the domain of $f$ by choosing a larger $\delta$. For a fixed $\epsilon$, this means choosing a smaller coefficient of $\delta$. The choice $\dfrac{\sqrt{13}}{2} \approx 1.8028$ is nearer to the amplitude than $2.5$.
Another answer shows that (2) "$\delta < 0.001$" is true and (4) "no such $\delta$ exists" is false. To complete this question, it remains to discuss the possibility of (1) and (3). To illustrate the sharpness of the bound $\delta < \dfrac{1}{50 \sqrt{13}}$ given above, choose $(r,t) = \left(0.0057, \dfrac{13}{50} \pi\right)$. This gives $(x,y) \approx (0.0039, 0.0042)$, and $f(x,y) \approx 0.010114 > 0.01 = \epsilon$.
Hence, (1) and (3) do not hold.
|
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|
Conjecture: $N=x_1^2+x_2^2-y^3$ has integer solutions for all $N$, with $x_1>0$,$x_2>0$ and $y>0$ Background.
I constructed some code for Algorithm for diophantine equation and decided to reuse it to investigate $N=x_1^2+x_2^2+z^3$ with $z$ integer. Negative values for $z$ seemed to produce plentiful small $N$ values, so I elected to concentrate my efforts here, and changed $z$ to $-y$.
Question.
Can anyone prove or disprove this conjecture, or help me find a method to do this, please?
I would also appreciate any useful background information.
My efforts.
Under nine minutes brute-force to find a solution for each $N=-10^6$ to $10^6$.
I’ve searched around on the net to find similar solutions. Perhaps a method based on this link http://www.dms.umontreal.ca/~mlalin/Lagrange.pdf would do?
Examples.
$$0=2^2+11^2-5^3$$
$$11=6^2+10^2-5^3$$
$$-3=5^2+6^2-4^3$$
$$999999=40^2+1718^2-125^3$$
$$-999999=8^2+1365^2-142^3$$
|
This was a problem posed by Noam Elkies and Irving Kaplansky in the January 1995 M.A.A. Monthly. They did not ask that your $y$ be positive, but that is how it comes out anyway, with a very few exceptions for the variable $x$ below being, for example $0,1,2.$ These can then be fixed by a quick search.
Later, Kap and I published a little paper where we pointed out that $x^2 + y^2 - z^9$ fails to represent infinitely many numbers
The closest looking thing to your original that fails to represent infinitely many numbers would be $x^2 + 27 y^2 - 7 z^3.$
Here is a solution by Andrew Adler, in the June-July 1997 issue:
$$ 2x+1 = \left( x^3 - 3 x^2 + x \right)^2 + \left(x^2 - x - 1 \right)^2 -\left( x^2 - 2 x\right)^3 $$
$$ 4x+2 = \left(2 x^3 - 2 x^2 - x \right)^2 + \left(2 x^3 - 4 x^2 - x + 1 \right)^2 -\left( 2 x^2 - 2 x - 1 \right)^3 $$
$$ 8x+4 = \left(x^3 + x + 2 \right)^2 + \left(x^2 - 2x - 1 \right)^2 -\left(x^2 + 1 \right)^3 $$
$$ 16x+8 = \left( 2 x^3 - 8 x^2 + 4 x + 2 \right)^2 + \left(2 x^3 - 4 x^2 - 2 \right)^2 -\left(2 x^2 - 4 x \right)^3 $$
$$ 16x = \left(x^3 + 7 x - 2 \right)^2 + \left(x^2 + 2 x + 11 \right)^2 -\left( x^2 + 5 \right)^3 $$
|
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What is the probability that some people are chosen in different groups? Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What is the probability that there will be at least a malicious person in every group?
|
Not sure if this is the correct answer, but maybe my reasoning can help you nonetheless.
The total number of arrangements is: ${n \choose 6} \cdot { 6 \choose 3}$.
${n \choose 6}$ represents the total ways of choosing 6 people out of the n people
${ 6 \choose 3}$ represents the number of ways of splitting the people into two groups
Now I will (attempt) to find the number of event in which one of the groups has no malicious persons...which is the complement of what you require.
This requires 4 separate cases:
Case 1: None of the chosen 6 people are malicious
${n-m \choose 6}\cdot {6 \choose 3}$
Case 2: one person is malicious
$2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3}$
Case 3: Two persons are malicious and placed in the same group
$2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3}$
Case 4: Three persons are malicious and placed in the same group
$2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}$
So the final answer is: $$1-\dfrac{{n-m \choose 6}\cdot {6 \choose 3}}{{n-m \choose 6}\cdot {6 \choose 3} + 2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3} + 2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3} + 2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}}$$
|
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|
On the sides $BC$, $CA$, $AB$ of $\Delta ABC$, points $D$, $E$, $F$ are taken in such a way that $\frac{BD}{DC} = \frac{CE}{EA} = $ How to find the area ?
The Diagram and the question.
|
Let $G\in EC$ such that $DG||BE$ and $GC=2x$.
Thus, by Thales $$\frac{EG}{2x}=\frac{BD}{DC}=2,$$ which gives $EG=4x$, $EC=6x$ and $AE=3.$
Thus, by Thales again we obtain:
$$\frac{AP}{PD}=\frac{AE}{EG}=\frac{3x}{4x}=\frac{3}{4}.$$
Now, let $I\in DC$ such that $EI||AD$ and $DI=y$.
Thus, by Thales $IC=2y$ and from here $DC=3y$ and $BD=6y$ and by Thales again we obtain:
$$\frac{BP}{PE}=\frac{BD}{DI}=\frac{6y}{y}=6.$$
Id est, $$AP:PR:RD=BQ:QP:PE=CR:RQ:QF=3:3:1.$$
Hence, $$S_{\Delta PQR}=\frac{1}{2}S_{\Delta QPC}=\frac{1}{2}\cdot\frac{3}{4}S_{\Delta QEC}=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{7}S_{\Delta BEC}=$$
$$=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{7}\cdot\frac{2}{3}S_{\Delta ABC}=\frac{1}{7}S_{\Delta ABC}=\frac{1}{7}\cdot106.61=15.23.$$
|
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|
Area of a triangle using a matrix Find the area of a triangle whose vertices are (1,0), (2,2), and (4,3)
$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}1&0&1\\2&2&1\\4&3&1\end{vmatrix}$$
$$=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$
$$=1(-1)+0+1(-2)$$
$$=-3$$
Then you multiply by $-\frac{1}{2}$ to get a positive area of $\frac{3}{2}$.
My question is where does $-1$ come from in the second line of the equation and why is it being squared first, then cubed, then raised to a power of 4?
|
It comes from
$$\det(A) = \sum_{j=1}^n (-1)^{i+j} a_{i,j} M_{i,j}$$
with $n=3$ and minor $M_{i,j}$, where $i=1$. So in this case
$$\det(A) = \sum_{j=1}^3 a_{1,j}(-1)^{1+j} M_{1,j}=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$
|
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|
For $x, y \in \mathbb{R}$, prove that $\max(x, y) = \frac{x + y + |x - y|}{2},$ and $\min(x, y) = \frac{x + y - |x - y|}{2}$. Prove that for all real numbers $x$ and $y$,
$$\max(x, y) = \dfrac{x + y + |x - y|}{2},$$
and
$$\min(x, y) = \dfrac{x + y - |x - y|}{2}.$$
For any real number $x$, the absolute value of $x$, denoted $|x|$ is defined as follows:
\begin{equation}
|x| = \begin{cases} x; & \text{ if } x \geq 0 \\
-x; & \text{ if } x< 0
\end{cases}
\end{equation}
What I understand from this is that $|x| = x$ if $x \geq 0$ or $|x| = −x$ if $x<0$. Other than that I don't really know how to start this.
|
$|x-y|$ means the distance between $x$ and $y$ in the number line.
On the other hand, $\frac{x+y}{2}$ means the midpoint between $x$ and $y$.
(you can check those two statements if you didn't know them!)
Therefore, $\frac{x+y+|x-y|}{2}=\frac{x+y}{2}+\frac{|x-y|}{2}$, which means adding half the distance between $x$ and $y$ to the midpoint of $x$ and $y$, is $\max(x, y)$.
Similarly, $\frac{x+y-|x-y|}{2}=\frac{x+y}{2}-\frac{|x-y|}{2}$, which means subtracting half the distance between $x$ and $y$ from the midpoint of $x$ and $y$, is $\min(x, y)$.
|
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|
Elementary Matrix and Row Operations Find an elementary matrix E such that EA = B
$$A = \begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}$$
$$B = \begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$
$$\begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}*E=\begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$
So the Elementary Matrix is an Identity Matrix that has one elementary row operation performed on it. Multiplying an Elementary Matrix by A should result in a matrix that is equivalent to having that elementary row operation performed onto A.
I can see that only the first row of A is modified to obtain B and I can tell that the first row of A is scaled by a value of -2.
Therefore the Elementary Matrix should be the Identity Matrix with the first row scaled by -2.$$\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$$
However, $$\begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}*\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}≠\begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$
What am I doing incorrectly?
|
Since the operation is $B=EA$ we have
\begin{equation}
\begin{pmatrix}
-2 & 0\\0 & 1
\end{pmatrix}\begin{pmatrix}
2 & -1\\5 & 3
\end{pmatrix} = \begin{pmatrix}
-4 & 2\\5 & 3
\end{pmatrix}.
\end{equation}
Your elementary matrix is correct but you meant to multiply it to $A$ on the left not on the right.
|
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|
Difference Calculus of the Factorial Function for Negative Integer Powers of $x$ Under the difference calculus, we have that
$$\Delta f(x)=f(x+1)-f(x)$$.
We also have factorial polynomials (sometimes referred to the rising and or falling factorials). The text I am reading defines the factorial function as
$$x^{(n)}=x(x-1)(x-2)...(x-(n-1))$$
with $x^{(0)}=1$ by definition. Applying Delta to the factorial function gives
$$\Delta x^{(n)}=(x+1)^{(n)}-x^{(n)}$$
$$=(x+1)[x(x-1)...(x-(n-2))-[x(x-1)...(x-(n-2))](x-(n-1))$$
$$=[(x+1)-(x-n+1)][x(x-1)...(x-(n-2))]$$
$$=nx^{(n-1)}$$
and this holds for positive $n$. It is then noted that
$$x^{(m+n)}=x(x-1)...(x-(m-1))\cdot(x-m)(x-m-1)...(x-m-(n-1))$$
and so
$$x^{(m+n)}=x^{(m)}\cdot (x-m)^{(n)}$$
Now by the above, if we let $m+n=0$, then $m=-n$. So we have
$$1=x^{(0)}=x^{(-n+n)}=x^{(-n)}(x-(-n))^{(n)}=x^{(-n)}(x+n)^{(n)}$$
So since $1=x^{(-n)}(x+n)^{(n)}$, we have that
$$x^{(-n)}=\frac{1}{(x+n)^{(n)}}$$
Now I am in the game to show that
$$\Delta x^{(n)}=nx^{(n-1)}$$
for negative integers. Which I interpret as trying to find
$$\Delta x^{(-n)}$$ So when I attempt the calculation, I get
$$\Delta x^{(-n)}=(x+1)^{(-n)}-x^{(n)}$$
$$=\frac{1}{(x+1+n)^{(n)}}-\frac{1}{(x+n)^{(n)}}$$
$$=\frac{1}{(x+1+n)(x+n)...(x+2)}-\frac{1}{(x+n)(x+(n-1))...(x+1)}$$
$$=\frac{1}{(x+1+n)[(x+n)...(x+2)]}-\frac{1}{[(x+n)...(x+2)](x+1)}$$
$$=\frac{1}{x+1+n}\frac{1}{(x+n)...(x+2)}-\frac{1}{x+1}\frac{1}{(x+n)...(x+2)}$$
$$=\left[\frac{1}{x+1+n}-\frac{1}{x+1}\right]\frac{1}{(x+n)...(x+2)}$$
$$=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\frac{1}{(x+n)^{(n-1)}}$$
And this is where i get stuck...
|
Everything looks fine: We obtain
\begin{align*}
\color{blue}{\Delta x^{(-n)}}&=(x+1)^{(-n)}-x^{(n)}\\
&=\cdots\\
&=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\cdot\frac{1}{(x+n)^{(n-1)}}\\
&=-n\cdot\frac{1}{(x+n+1)\cdot (x+n)^{(n-1)}\cdot (x+1)}\\
&=-n\cdot\frac{1}{(x+n+1)^{(n+1)}}\\
&=\color{blue}{-nx^{(-n-1)}}
\end{align*}
and we conclude the rule is valid for negative $n$ as well.
Hint: You might be interested in the calculation following (2.52) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
|
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|
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake?
Thank you.
|
If $n + 1$ is even, then $n$ must be odd. Let $n = 2k + 1$ since it is in the form of an odd number. Now substitute into the quadratic expression: $$\begin{align} n^2 - 2n + 7 &= (2k+1)^2 - 2(2k+1) + 7 \\ &= (2k)^2 + 1 + 4k - 4k - 2 + 7 \\ &= 4k^2 + 6 \\ &= 2(2k^2 + 3).\end{align}$$ Since the quadratic is an even number, this also completes the proof. For more sensibility, since $n + 1$ is even, we can let $n + 1 = 2r$ and then let $n = 2r - 1$, but the same is implied anyway.
|
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|
How to compute $\int_{0}^{1}x\sin (bx) J_0\!\!\left(a\sqrt{1-x^2}\right)\!\mathrm dx$? I need a help with integral below,
$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx$$
where $a \geq 0$ and $b$ are constants and $J_0(x)$ is the zeroth-order of Bessel function of the first kind.
This integral has a closed form solution ?
I found some integrals similar to the integral above, but I don't have any idea on how to apply it.
I found this integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:
$$\int_{0}^{a} \cos (cx)J_0\left(b\sqrt{a^2-x^2}\right)\,\mathrm dx = \frac{\sin (a\sqrt{b^2+c^2})}{\sqrt{b^2+c^2}} \quad [b\geq 0]$$
I try use the definition of zeroth-order of Bessel function of the first kind to solve this integral:
$$J_0(z) = \sum_{k=0}^\infty (-1)^k\frac{(\frac{1}{4}z^2)^k}{(k!)^2}$$
then i found:
$$J_0(a) \sum_{k=0}^\infty \int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx $$
On Wolfram Alpha i found that:
$$\int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx = \frac{1}{4}\sqrt{\pi}a\Gamma (k+1)\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big)$$
where $\tilde{F}_1$ is the Regularized Hypergeometric Function, where:
$$\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big) = \frac{J_{k+\frac{3}{2}}(a) }{(\frac{a}{2})^{k+\frac{3}{2}}}$$
therefore:
$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx = J_0(a)\frac{1}{4}\sqrt{\pi}a \sum_{k=0}^\infty \Gamma (k+1) J_{k+\frac{3}{2}}(a)\Big(\frac{a}{2}\Big)^{-(k+\frac{3}{2})} $$
The solution found above is right ?
Thanks in advance.
|
Suggestion by user tired:
Integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:
$$\int_0^1 \cos (b x) J_0\left(a \sqrt{1-x^2}\right) \, dx=\frac{\sin
\left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}}$$
Differentiate for variable b:
$$\int_0^1 \frac{\partial \left(\cos (b x) J_0\left(a \sqrt{1-x^2}\right)\right)}{\partial b} \,
dx=\frac{\partial }{\partial b}\frac{\sin \left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}}$$
$$\color{blue}{\int_0^1 x \sin (b x) J_0\left(a \sqrt{1-x^2}\right) \, dx=-\frac{b \cos
\left(\sqrt{a^2+b^2}\right)}{a^2+b^2}+\frac{b \sin \left(\sqrt{a^2+b^2}\right)}{\left(a^2+b^2\right)^{3/2}}}$$
|
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|
Prove that $x^4 + y^4 - 3xy = 2$ is compact The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.
In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case?
My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?
Many thanks,
|
Note that
$$\eqalign{x^4+y^4-2xy&={1\over2}(x^2+y^2)^2+{1\over2}(x^2-y^2)^2-2xy\geq{1\over2}(x^2+y^2)^2-(x^2+y^2)\cr
&=(x^2+y^2)^2\left({1\over2}-{1\over x^2+y^2}\right)\geq{1\over4}(x^2+y^2)^2\geq4\ ,\cr}$$
as soon as $x^2+y^2\geq4$. It follows that the constraint defines a closed and bounded, hence compact, set in the plane.
|
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$
$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$
$y\geq 0.5$
Could some help me where I have wrong, thanks
|
MrYouMath has already provided a good answer.
This answer uses your method.
You already have a quadratic equation on $t$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0\tag1$$
where $y\not=\frac 12$.
Note here that we want to find $y$ such that $(1)$ has at least one real solution $t$ satisfying $-1\le t\le 1$.
It seems that you've missed the condition $-1\le t\le 1$.
Let $f(t)$ be the LHS of $(1)$.
Then, since the vertex of the parabola $Y=f(t)$ is on $t=-2$, we have
$$f(-1)f(1)\le 0,$$
i.e.
$$\frac 59\le y\le 1$$
|
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|
Distance between two parallel lines by having linear equations I wonder where does this formula is coming from?
It is for finding the distance between two parallel lines when we have their linear equation:
First line is:$ax+by+c=0$
Second line is:$ax+by+c_1=0$
Their distance :$$\frac{|c-c_1|}{\sqrt{a^2+b^2}}$$
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Alternatively: the perpendicular line that passes through origin is: $bx-ay=0$. It crosses the two parallel lines at:
$$\left(-\frac{ac}{a^2+b^2},-\frac{bc}{a^2+b^2}\right) \ \ \text{and} \ \ \left(-\frac{ac_1}{a^2+b^2},-\frac{bc_1}{a^2+b^2}\right).$$
The distance between these points is:
$$d=\sqrt{\left(\frac{a(c-c_1)}{a^2+b^2}\right)^2+\left(\frac{b(c-c_1)}{a^2+b^2}\right)^2}=\frac{|c-c_1|}{\sqrt{a^2+b^2}}.$$
|
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|
Can a parallelogram have whole-number lengths for all four sides and both diagonals?
Is it possible for a parallelogram to have whole-number lengths for all four sides and both diagonals?
One idea I had was to arrange four identical right triangles such that the right angles are adjacent. For example if we take four triangles that are all 3-4-5 right triangles, and arrange them so that their legs form a cross with two arms that are 3 units and two arms that are 4 units, their hypotenuses will form a parallelogram in which all sides are 5 units – that is, a rhombus.
A second idea was to take two pairs of isosceles triangles with all legs the same length, where one vertex angle is supplementary to another, and arrange them so that their vertex angles are adjacent. For example we could take two triangles with two sides of 6 units that form a 25° angle, and another two triangles with two sides of 6 units that form a 155° angle, and arrange them to form a quadrilateral with 12-unit diagonals – that is, a rectangle.
That's all I can imagine. My hunch is that a parallelogram can have whole-number lengths for all four sides and both diagonals only if it is either a rhombus or a rectangle.
Is that right? If so, can this limitation be explained elegantly?
|
Your hunch is wrong. A non-rectangular non-rhomboid parallelogram with integer side and diagonal lengths exists:
Suppose that the parallelogram is $ABCD$ with $AB=a$, $AD=b$ and $BD=c$. By the law of cosines:
$$\cos\angle DAB=\frac{a^2+b^2-c^2}{2ab}$$
Since $\angle CDA=\pi-\angle DAB$, $\cos\angle CDA=-\cos\angle DAB$. Then
$$AC^2=d^2=a^2+b^2-2ab\cos\angle CDA=a^2+b^2+2ab\cdot\frac{a^2+b^2-c^2}{2ab}=2a^2+2b^2-c^2$$
$$\color{red}{c^2+d^2=2(a^2+b^2)}$$
If $c=d$ we have a rectangle; if $a=b$ we have a rhombus. Thus, we look for a number that is a sum of two unequal squares $a$ and $b$ whose double is also a sum of two unequal squares $c$ and $d$, with triangle inequalities satisfied to ensure the parallelogram is non-degenerate: $|a-b|<c,d<a+b$.
Fixing $a$ and $b$, a trivial choice is $c=a-b$ and $d=a+b$ because $2(a^2+b^2)=(a-b)^2+(a+b)^2$. However, these assignments do not satisfy the triangle inequalities, making numbers that have multiple representations as sums of two squares valuable for this problem. I used OEIS A025426 to find them; the first number I saw was $145=9^2+8^2=12^2+1^2$, whose double is $290=13^2+11^2=17^2+1^2$. The first representations listed here allowed me to quickly construct the parallelogram above, although it is not the smallest: there is a parallelogram with sides 4 and 7, diagonals 7 and 9.
The numbers being squared in the above equations relate to the above parallelogram as follows. The lengths of its sides are 9 and 8, and of its diagonals 13 and 11. If its left side is extended upwards along the same straight line by 8 (so that it is now 16), the resulting quadrilateral has sides 16, 9, 8 and 11, and its diagonals are 17 and 13.
Here is a way to efficiently generate infinitely many such integer parallelograms. Let $r$ and $s$ be two coprime integers with $r>s>0$ and $(r,s)\ne(3,1)$. The product $(2^2+1^2)(r^2+s^2)$ can be written as a sum of two squares in two different ways (the Brahmagupta–Fibonacci identity):
$$(2^2+1^2)(r^2+s^2)=(2r+s)^2+(2s-r)^2=\color{blue}{(2s+r)^2+(2r-s)^2}$$
From these two representations, we can write twice the product as two different sums of two squares too:
$$2(2^2+1^2)(r^2+s^2)=\color{blue}{(2(r+s)-(r-s))^2+(2(r-s)+(r+s))^2}
=(2(r-s)-(r+s))^2+(2(r+s)^2+(r-s))^2$$
To satisfy the triangle inequalities we choose
$$\color{blue}{a=2r-s\qquad b=2s+r\qquad c=2(r+s)-(r-s)\qquad d=2(r-s)+(r+s)}$$
These are guaranteed to form a non-rectangular non-rhomboid integer parallelogram with the given restrictions on $r$ and $s$. The one pictured at the top of this answer corresponds to $(r,s)=(5,2)$ and the smallest instance (the one with side lengths 4 and 7) corresponds to $(r,s)=(3,2)$.
|
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|
Strange Pattern in Decimal Expansion I noticed something weird when I was fooling around with my calculator.
I calculated several powers of $30$ of the form $30^{\left(\frac{10^n-1}{10^n}\right)}$ and I noticed a pattern in the fractional part:
$$30^{\left(\frac{9999}{10000}\right)}=29.9897981429$$
$$30^{\left(\frac{99999}{100000}\right)}=29.9989796581$$
$$30^{\left(\frac{999999}{1000000}\right)}=29.9998979643$$
The rest of the powers just keep sticking a $9$ in the tenths place and shifting all the other digits down.
Why is the decimal expansion for $30^{\left(\frac{10^n-1}{10^n}\right)}$ when $n\ge 5$ $29.9\cdots 989796$?
|
Short answer: it's because of the approximation
$$
(1 + x)^{1/10} \approx 1 + \frac{x}{10}
$$
valid when $x$ is very close to $0$. (This approximation is the linear truncation of the Taylor series $(1 + x)^{1/10} = 1 + \binom{1/10}{1}x + \binom{1/10}{2}x^2 + \binom{1/10}{3}x^3 + \dotsb$.)
A different way of writing this approximation is as
$$
y^{1/10} \approx \frac{9+y}{10}
$$
valid when $y \approx 1$, by the simple substitution $y = 1+x$.
The operation that takes $30^{\frac{99}{100}}$ to $30^{\frac{999}{1000}}$ to $30^{\frac{9999}{10000}}$ is the map $t \mapsto t^{1/10} \cdot 30^{9/10}$.
When $t$ is very close to $30$, then $\frac{t}{30}$ is very close to $1$, so we have the approximation
$$
\left(\frac{t}{30}\right)^{1/10} \approx \frac{9}{10} + \frac{t/30}{10}
$$
which becomes
$$
t^{1/10} \cdot 30^{9/10} \approx \frac{9}{10} \cdot 30 + \frac{1}{10} \cdot t
$$
when we multiply both sides by $30$. On the right, we have the next term of the sequence of powers: if $t = 30^{99/100}$, then $t^{1/10} \cdot 30^{9/10} = 30^{999/1000}$. On the left, the expression
$$
\frac{9}{10} \cdot 30 + \frac{1}{10} \cdot t
$$
is a weighted average that says "take a number ten times as close to $30$ as $t$ is". Or to put it another way, the difference $30 - (\frac{9}{10} \cdot 30 + \frac{1}{10} \cdot t)$ simplifies to $\frac1{10}(30 - t)$: a tenth of the distance $t$ was from $30$. When you cut the distance from $30$ in ten, you do this by adding an extra $9$ in the decimal expansion, so the error goes, for example, from
$$
30 - 30^{0.99999} \approx 0.0010203...
$$
to
$$
30 - 30^{0.999999} \approx 0.00010203...
$$
|
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Numbers up to 1000 divisible by 2 or 3 and no other prime My task requires to find all numbers from $1-1000$ such that they are divisible by $2$ or $3$ and no other primes. I know that $2$ divides even numbers and I can use the formula $\left \lfloor{\frac{1000}{2}}\right \rfloor $ and the numbers divisible by three $\left \lfloor{\frac{1000}{3}}\right \rfloor $. Also we rule out the numbers that are divisible by both, so by $6$ $\left \lfloor{\frac{1000}{6}}\right \rfloor $. In total we get that there are $500+333-166=667$ numbers divisible by $2$ or $3$. However I also need to make sure that I $\textit{only}$ count numbers divisible by either of these $2$. Is there a quick way to do it?
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Such a number should be in the form of $2^m3^n$ where $m$ and $n$ are non-negative integers such that $m$ and $n$ are not both zero.
If $n=0$, $m\ge1$ and $2^m\le1000$. So $1\le m\le 9$.
If $n=1$, $2^m\le\frac{1000}{3}$. So $0\le m\le 8$.
If $n=2$, $2^m\le\frac{1000}{9}$. So $0\le m\le 6$.
If $n=3$, $2^m\le\frac{1000}{27}$. So $0\le m\le 5$.
If $n=4$, $2^m\le\frac{1000}{81}$. So $0\le m\le 3$.
If $n=5$, $2^m\le\frac{1000}{243}$. So $0\le m\le 2$.
If $n=6$, $2^m\le\frac{1000}{729}$. So $m= 0$.
If $n=7$, $2^m\le\frac{1000}{2187}$, which is impossible.
Number of possibilities is $9+9+7+6+4+3+1=39$.
|
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|
Conditional Probability: Problem with 6 persons
There are $6$ persons, let's name them $ A, B, C, D, E$ and$ F $. They are ordered according to their popularity, but without letting them know their order, so they guess all ways they can be ordered are equally possible.
Taking as a fact that those persons learn that $A$ is more popular than $B$, what do they think the probability for $ A$ to be more popular than $C$ is?
This question confuses me quite a lot, especially the way it's stated. All I know is that it's about conditional probability. Even a small hint would be a great help.
|
I am going to make use of Bayes' Theorem:
$$P({A > C | A > B})=\frac{P(A>C \cap A>B)}{P(A>B)}$$
Consider lining up the letters as follows:
_ _ _ _ _ _
where popularity is rated from left to right.
By symmetry $$P(A>B)=\frac{1}{2}$$
We must solve for $$P(A>C \cap A>B)$$
which is the probability that $A$ is bigger than both $B$ and $C$.
If $A$ is in the leading position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $1$
If $A$ is in the second position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{4}{5}\cdot\frac{3}{4}$
If $A$ is in the third position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{3}{5}\cdot\frac{2}{4}$
If $A$ is in the fourth position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{2}{5}\cdot\frac{1}{4}$
If $A$ is in the fifth of final position then it is impossible for it to be greater than both $B$ and $C$.
Thus we have
$$\begin{align*}
P({A > C | A > B})
&=\frac{P(A>C \cap A>B)}{P(A>B)}\\\\
&=\frac{\left(\frac{1}{6}\cdot1\right)+\left(\frac{1}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\right)+\left(\frac{1}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)+\left(\frac{1}{6}\cdot\frac{2}{5}\cdot\frac{1}{4}\right)}{\frac{1}{2}}\\\\
&=\frac{2}{3}
\end{align*}$$
|
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|
proving the sum of a few determinant values as a constant Let
$$ \Delta_a = \begin{vmatrix}
a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n
\end{vmatrix} $$
My task is to show that
$$ \sum_{a = 1}^n \Delta_a = c
$$
where c is some constant.
I tried taking a-1 and n common from $C_1$ and $C_2$ respectively and tried to simplify the resulting determinant in the normal method. However, I ended up with a very large result which clearly didn't seem to satisfy the given condition.Please Help.
|
Recall that the determinant is linear in the columns of the matrix. Since $\Delta_a$ differ only in the first column, we have:
$$\sum_{a=1}^n \Delta_a = \sum_{a=1}^n\begin{vmatrix}
a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n
\end{vmatrix} = \begin{vmatrix}
\sum_{a=1}^n (a-1) & n & 6 \\ \sum_{a=1}^n (a-1)^2 & 2n^2 & 4n - 2 \\ \sum_{a=1}^n (a-1)^3 & 3n^2 & 3n^2 - 3n
\end{vmatrix}$$
Recognizing the familiar sums $$\sum_{a=1}^n (a-1)= \frac{n(n-1)}2$$$$\sum_{a=1}^n (a-1)^2= \frac{n(n-1)(2n-1)}6$$$$\sum_{a=1}^n (a-1)^3= \left(\frac{n(n-1)}2\right)^2$$
we obtain
$$\begin{vmatrix}
\frac{n(n-1)}2 & n & 6 \\ \frac{n(n-1)(2n-1)}6 & 2n^2 & 4n - 2 \\ \left(\frac{n(n-1)}2\right)^2 & 3n^2 & 3n^2 - 3n
\end{vmatrix}$$
Now notice that
$$\pmatrix{6 \\ 4n-2 \\ 3n^2-3n} -\frac{12}{n(n-1)} \pmatrix{\frac{n(n-1)}2 \\ \frac{n(n-1)(2n-1)}6 \\ \left(\frac{n(n-1)}2\right)^2} = 0$$
so the columns are linearly dependent. We conclude that the determinant is $0$, which is certainly a constant.
|
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|
Characterizing elements with square norms in quadratic integer rings Given the ring $\mathbb{Z}\left[\sqrt{D}\right]$ (where $D$ is a positive square-free integer) can we characterize all elements $\alpha$ with positive norms for which $N(\alpha)$ is a perfect square in $\mathbb{Z}$?
Obviously this is true for all integers, but there are also plenty of non-integers for which this could be true. For example, if we set $D=15$, then $N(8+2\sqrt{15})=4=2^2$.
|
Seeing Rene's answer: I am talking about arranging $x^2 - Dy^2 = w^2$ with $\gcd(x,y) = 1.$
$$ 8^2 - 15 \cdot 1^2 = 7^2 $$
$$ 16^2 - 15 \cdot 3^2 = 11^2 $$
$$ 23^2 - 15 \cdot 4^2 = 17^2 $$
$$ $$
$$ 83^2 - 15 \cdot 8^2 = 77^2 $$
$$ $$
$$ \left( 5 u^2 + 3 v^2 \right)^2 - 15 \left( 2uv \right)^2 = \left( 5 u^2 - 3 v^2 \right)^2 $$
$$ \left( u^2 + 15 v^2 \right)^2 - 15 \left( 2uv \right)^2 = \left( u^2 - 15 v^2 \right)^2 $$
You can accomplish this for any prime $p$ that does not divide $2D$ and for which Legendre symbol $(D|p) = 1.$ Also any product of such primes. This is nothing more than Gauss composition for binary quadratic forms of the same discriminant. All primes such as I describe are (primitively) represented by a primitive form of discriminant $4D,$ the way you are writing this. The form composed with its opposite is the principal form $x^2 - D y^2.$
|
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|
Counting Principle - Dependent Events There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.
I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{210}{1320}$
b) $1 - \left(\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} + \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10}\right) = \frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?
|
$(a)$
$$\frac{7 \choose 3}{12 \choose 3}$$
You can show that this equals what you did.
$$\begin{align*}
\frac{7 \choose 3}{12 \choose 3}
&=\frac{\frac{7!}{3!4!}}{\frac{12!}{3!9!}}\\\\
&=\frac{7\cdot6\cdot5}{\color{red}{3\cdot2\cdot1}}\frac{\color{red}{3\cdot2\cdot1}}{12\cdot11\cdot10}\\\\
&=\frac{7}{12}\cdot\frac{6}{11}\cdot\frac{5}{10}
\end{align*}$$
$(b)$
$$1-\frac{7 \choose 3}{12 \choose 3}-\frac{5 \choose 3}{12 \choose 3}$$
|
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|
Convolution of i.i.d. with uniform distribution I am trying to solve another exercise about convolution of continuous variables. Here is what is asked:
Let $X_1,\cdots,X_n$ be i.i.d. such that $X_1 \sim U(0,1)$. Verify that$$
P(X_1+\cdots+X_n\leq x)=\frac{1}{n!}\sum_{k=0}^{n-1}(-1)^k {n\choose k}(x-k)_+^n, \quad \forall 0\leq x \leq n$$
where$$
(x-k)_+=\begin{cases}
0; & \text{if } x <k \\
x-k; & \text{otherwise}
\end{cases}
$$
My first thought would be to find $P(S\leq s)$, where $S=\sum\limits_{i=1}^{n}X_i$. But I am wondering how I can solve such a huge convolution? My integral would look like:
$$\int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty}f_{X_1}(x_1)\cdots f_{X_n}(x_n)\,\mathrm{d}x_1\cdots\mathrm{d}x_n.$$
At first I had just $X_1$ and $X_2$ to add and I could find the CDF of $X_1 +X_2$ through convolution like this: $$\int_0^z f_{X_1}(z-v)f_{X_2}(v)\,\mathrm{d}v.$$
My question would be how can I solve it? Just even a hint in order to get started. My teacher say it is a simple generalization but many people suggest me to use conditional expectation, which I did not study yet.
Thank you for your help.
|
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$The base case of induction is, in fact, $n = 1$.
Denote $S_n = \sum\limits_{k = 1}^n X_k$ for any $n$. For $n = 1$,$$
P(S_1 \leqslant x) = P(X_1 \leqslant x) = x_+. \quad \forall 0 \leqslant x \leqslant 1
$$
Suppose the proposition holds for $n$. For any $0 \leqslant x \leqslant n + 1$, because $S_n$ and $X_{n + 1}$ are independent, then\begin{align*}
&\peq P(S_{n + 1} \leqslant x) = P(S_n + X_{n + 1} \leqslant x)\\
&= \iint\limits_{y + z \leqslant x} f_{X_{n + 1}}(y) f_{S_n}(z) \,\d y\d z = \int_0^1 f_{X_{n + 1}}(y) \,\d y \int_0^{(x - y)_+} f_{S_n}(z) \,\d z\\
&= \int_0^1 P(S_n \leqslant (x - y)_+) f_{X_{n + 1}}(y) \,\d y = \int_0^1 P(S_n \leqslant (x - y)_+) \,\d y\\
&= \int_0^1 \frac{1}{n!} \sum_{k = 0}^{n - 1} \binom{n}{k} (-1)^k ((x - y)_+ - k)_+^n \,\d y. \tag{1}
\end{align*}
If $0 \leqslant x < 1$, then\begin{align*}
(1) &= \frac{1}{n!} \int_0^1 (x - y)_+^n \,\d y = \frac{1}{n!} \int_0^x (x - y)^n \,\d y\\
&= \frac{1}{(n + 1)!} x^{n + 1} = \frac{1}{(n + 1)!} \sum_{k = 0}^n \binom{n + 1}{k} (-1)^k (x - k)_+^{n + 1}.
\end{align*}
Otherwise, suppose $m = [x] \geqslant 1$, then\begin{align*}
(1) &= \frac{1}{n!} \sum_{k = 0}^{n - 1} \binom{n}{k} (-1)^k \int_0^1 ((x - y)_+ - k)_+^n \,\d y\\
&= \frac{1}{n!} \sum_{k = 0}^{m - 1} \binom{n}{k} (-1)^k \int_0^1 (x - y - k)^n \,\d y + \frac{1}{n!} \binom{n}{m} (-1)^m \int_0^1 (x - y - m)_+^n \,\d y.
\end{align*}
Because\begin{align*}
&\peq \frac{1}{n!} \sum_{k = 0}^{m - 1} \binom{n}{k} (-1)^k \int_0^1 (x - y - k)^n \,\d y\\
&= \frac{1}{n!} \sum_{k = 0}^{m - 1} \binom{n}{k} (-1)^k · \frac{1}{n + 1} ((x - k)^{n + 1} - (x - k - 1)^{n + 1})\\
&= \frac{1}{(n + 1)!} \left( \sum_{k = 0}^{m - 1} \binom{n}{k} (-1)^k (x - k)^{n + 1} - \sum_{k = 1}^m \binom{n}{k - 1} (-1)^{k - 1} (x - k)^{n + 1} \right)\\
&= \frac{1}{(n + 1)!} x^{n + 1} + \frac{1}{(n + 1)!} \sum_{k = 1}^{m - 1} \left( \binom{n}{k} + \binom{n}{k - 1} \right) (-1)^k (x - k)^{n + 1}\\
&\peq - \frac{1}{(n + 1)!} \binom{n}{m - 1} (-1)^{m - 1} (x - m)^{n + 1}\\
&= \frac{1}{(n + 1)!} \sum_{k = 0}^{m - 1} \binom{n + 1}{k} (-1)^k (x - k)^{n + 1} + \frac{1}{(n + 1)!} \binom{n}{m - 1} (-1)^m (x - m)^{n + 1},
\end{align*}
and\begin{align*}
&\peq \frac{1}{n!} \binom{n}{m} (-1)^m \int_0^1 (x - y - m)_+^n \,\d y = \frac{1}{n!} \binom{n}{m} (-1)^m \int_0^{x - m} (x - y - m)^n \,\d y\\
&= \frac{1}{n!} \binom{n}{m} (-1)^m · \frac{1}{n + 1} (x - m)^{n + 1} = \frac{1}{(n + 1)!} \binom{n}{m} (-1)^m (x - m)^{n + 1},
\end{align*}
then\begin{align*}
(1) &= \frac{1}{(n + 1)!} \left( \sum_{k = 0}^{m - 1} \binom{n + 1}{k} (-1)^k (x - k)^{n + 1} + \left( \binom{n}{m} + \binom{n}{m - 1} \right) (-1)^m (x - m)^{n + 1} \right)\\
&= \frac{1}{(n + 1)!} \sum_{k = 0}^m \binom{n + 1}{k} (-1)^k (x - k)^{n + 1} = \frac{1}{(n + 1)!} \sum_{k = 0}^{n + 1} \binom{n + 1}{k} (-1)^k (x - k)_+^{n + 1}\\
&= \frac{1}{(n + 1)!} \sum_{k = 0}^n \binom{n + 1}{k} (-1)^k (x - k)_+^{n + 1}.
\end{align*}
End of induction.
|
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|
$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}=-ie^{
\frac{i2 \pi k}{11}}$$
so,$$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=-i \sum_{k=1}^{10}e^{
\frac{i2 \pi k}{11}}=-i \times e^{
\frac{i2 \pi }{11}} \times \frac{{\{e^{
\frac{i2 \pi }{11}}\}}^{10} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
$$=-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
now how can i proceed and simplify the result?
Answer $=i$
|
\begin{align}-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}&=-i\frac{\exp(\frac{i22\pi}{11})-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1} \\
&=-i\frac{\exp(i2\pi)-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1}
\\
&=-i\frac{1-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1}\\
&=-i(-1)\\
&=i
\end{align}
|
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|
How to find the eigen values of the given matrix
Given the matrix
\begin{bmatrix}
5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3
\end{bmatrix}
find its eigen values(preferably by elementary row/column operations).
Since I don't know any other method other than elementary operations to find eigen values so I tried writing the characteristic polynomial of the matrix which is follows:
\begin{bmatrix}
x-5&-1&-1&-1&-1&-1\\-1&x-5&-1&-1&-1&-1\\-1&-1&x-5&-1&-1&-1\\-1&-1&-1&x-4&-1&0\\-1&-1&-1&-1&x-4&0\\-1&-1&-1&0&0&x-3
\end{bmatrix}
Using $R1=R1-(R2+R3+R4+R5+R6)$
\begin{bmatrix}
x&-x+8&-x+8&-x+6&-x+6&-x+4\\-1&x-5&-1&-1&-1&-1\\-1&-1&x-5&-1&-1&-1\\-1&-1&-1&x-4&-1&0\\-1&-1&-1&-1&x-4&0\\-1&-1&-1&0&0&x-3
\end{bmatrix}
|
The characteristic polynomial $\chi_t$ of this matrix can be factored nicely, i.e.,
$$
\chi_t(A)=(t - 2)(t - 3)(t - 4)^3(t - 9).
$$
So the six eigenvalues are $2,3,4,4,4,9$.
|
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|
Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?
So this track seems to difficult, another approach. Can I square the area first, find the derivative of that to solve for x?
So the Area = $4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$Area^2 = 16x^2 \cdot (4 - \frac{4x^2}{9}$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} Area^2 = 128x - \frac{256x^3}{9}$$
$$128x(1-\frac{2x^2}{9}$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals 0 when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of x into y we get that $y = \sqrt{2}$ so the Area is 3.
Is this valid? If so why? Does squaring not cause any problems?
|
I think you are in right track but your expression for the first derivative of $A$ need a little touch. I put it in more clear way:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}=4x \left(4-\frac{4x^2}{9}\right)^{1/2}$$ Thus,
$$A' = 4 \left(4-\frac{4x^2}{9}\right)^{1/2} + 4x \left(4-\frac{4x^2}{9}\right)^{-1/2}\left(\frac{-8x}{9} \right)=4 \left(4-\frac{4x^2}{9}\right)^{1/2} - \frac{32x^2}{9} \left(4-\frac{4x^2}{9}\right)^{-1/2}=\frac{36 \left(4-\frac{4x^2}{9}\right) - 32x^2} {9\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{4 \left(36-4x^2\right) - 288x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{144 - 304x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}$$
To find the maximum for $A$, set $A'=0$ since $-3\le x\le +3$, and hence $\left(4-\frac{4x^2}{9}\right)^{1/2}\ne0$. Therefore, $144 - 304x^2=0$ when $A'-0$ and $x=\pm\sqrt\frac{19}{9}$.
$$A_{max}=4\left(\sqrt\frac{19}{9}\right) \left( \sqrt{4 - \frac{4\left(\sqrt\frac{19}{9}\right)^2}{9}}\right)=8\left(\sqrt\frac{19}{9}\right) \left( \frac{\sqrt{62}}{9}\right)=\frac{8\sqrt{1178}}{27}$$
Note: The $\pm x$ values for rectangle are symmetrical points on major axis of ellipse where vertical legs of rectangle meet.
|
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|
Is this sequence a integer? For $n \in \mathbb{N}$ and $n>(q+1)/4$. Where $q>1$ and integer. With $p\in \mathbb{Z}$
$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \big(\frac{1}{4n-3}-
\frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \big)$
I suspect that it's not integer. $1-1/3+1/5-...$ converges to $\pi/4$. And the rest of the sums are integers.
|
Let $n=1, q=2, p=3$. Then $4n>q+1$ but $$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \left(\frac{1}{4n-3}-
\frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \right)$$ becomes $$9-2-24\left(1-\frac13+\frac\pi4\right)=7-24+8-6\pi=-9-6\pi\not\in\mathbb{Z}$$ so the result does not hold.
|
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|
Inverse of a modular matrix I have the matrix$$
\begin{pmatrix}
1 & 5\\
3 & 4
\end{pmatrix} \pmod{26}
$$
and I need to find its inverse. I do it according to this website.
I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). Then I have $$\frac{1}{19}×\begin{pmatrix}4 & -5\\-3 & 1\end{pmatrix}.$$ I calculate that$$-5 \equiv 21 \pmod{26},\ -3 \equiv 23 \pmod{26}.$$
No matter what I do I am not able to get the solution they have on the website, which is$$\begin{pmatrix}2 & 17\\5 & 7\end{pmatrix}.$$
Can someone help me with this? What am I doing wrong?
|
Doing aritmetic modulo $\;26\;$ all the time, we get that
$$\det A=\begin{vmatrix}1&5\\3&4\end{vmatrix}=4-15=-11=15\implies$$
$$\implies A^{-1}=\frac1{15}\begin{pmatrix}4&-5\\-3&1\end{pmatrix}=7\begin{pmatrix}4&21\\23&1\end{pmatrix}=\begin{pmatrix}2&17\\5&7\end{pmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
\end{align*}
and I'm stuck, should I divide $63$ by $9$ to get $7$? but it does not comply to the given equation as when $x=7$, it would become $16 \pmod{47}$.
|
We wish to solve the congruence $9x \equiv 3 \pmod{47}$. You correctly found that $21$ is the multiplicative inverse of $9$ modulo $47$. However, you made an error in this step:
$$9 \cdot 21 \equiv \color{red}{3} \pmod{47}$$
If we multiply one side of the congruence by $21$, we must multiply the other side of the congruence by $21$. You should have obtained
\begin{align*}
21 \cdot 9x & \equiv 21 \cdot 3 \pmod{47}\\
1x & \equiv 63 \pmod{47}\\
x & \equiv 16 \pmod{47}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x .$
For fixed $c>0$, show
$$\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x ,$$
where $p$ is a prime.
I am not sure how to show this result. We know that $$\sum_{p \leq x} \log \left( 1-\frac{1}{p} \right) = -\log \log x - B + O\left (\frac{1}{\log x} \right)$$ for some constant $B>0$, but I am not sure if this helps much as the RHS does not seem to give a hint of the RHS of what we want to prove.
|
Note that
$$\frac{t}{1+t} < \log (1+t) < t$$
for $t> -1$ and $t \neq 0$. Then
\begin{align*}
c\log \left( 1 - \frac{1}{p} \right) & > \frac{-c/p}{1 - 1/p} \\ & =
\frac{p}{p-1} \left(- \frac{c}{p} \right) \\ & >
- \frac{c}{p} \\ &
> \log \left(1 - \frac{c}{p} \right)
\end{align*}
for $p >c$. Exponentiate both sides to obtain
$$\left( 1 - \frac{c}{p} \right) < \left( 1 - \frac{1}{p} \right) ^c.$$
So,
\begin{align*}
\prod_{c<p<x} \left(1 - \frac{c}{p} \right) & <
\prod_{c<p<x} \left(1 - \frac{1}{p} \right) ^c \\ & \ll
\prod_{p<x} \\ & \ll
\log ^{-c} x
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big(\frac{\frac{2A}2}2\Big)\cos\frac02 =
2\sin\frac{A}2
$$
where $\cos\frac02 = \cos0 = 1$
|
Hint:
Your first step is wrong because
$\sin(x)$ is not a linear function so: $$\sin x \ne \sin(\frac{x}{2})+\sin(\frac{x}{2})$$
|
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|
$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$ $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$
I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2(1+\frac{2}{n}+\frac{1}{n^2})})=n^{\alpha+\frac{2}{5}} (\sqrt[5]{1+\frac{1}{n}}-\sqrt[5]{1+\frac{2}{n}+\frac{1}{n^2}})$
Then
$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}+ \infty & \alpha>-\frac{2}{5}\\
2 & \alpha=-\frac{2}{5}\\0 & \alpha<-\frac{2}{5}\end{cases}$
Is it right?
|
By binomial expansion
*
*$\sqrt[5]{n^2+n}=\sqrt[5]{n^2}\sqrt[5]{1+1/n}=\sqrt[5]{n^2}(1+\frac1{5n}+o(1/n))$
*$\sqrt[5]{n^2+2n+1}=\sqrt[5]{n^2}\sqrt[5]{1+2/n+1/n^2}=\sqrt[5]{n^2}(1+\frac2{5n}+o(1/n))$
then
$$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=-\frac{n^\alpha\sqrt[5]{n^2}}{5n}+o\left(\frac{n^\alpha\sqrt[5]{n^2}}{n}\right)=-\frac15 n^{\alpha-\frac35}+o\left(n^{\alpha-\frac35}\right)$$
therefore
$$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}
- \infty & \alpha>\frac{3}{5}\\
-\frac15 & \alpha=\frac{3}{5}\\
0 & \alpha<\frac{3}{5}\end{cases}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find coefficient of multiple binomial expansions quickly?
Find the coefficient of $(x^{-2})$ in the expansion of $(x-1)^3(\frac1{x}+2x)^6$
How can this be done quickly without expanding the two brackets?
I tried writing the general formula in terms of $r$ for both brackets and equated $-2$ to the sum of the powers of $x$ in terms of $r$.
$${3 \choose r} (x^{3-r}) (-1)^r {6 \choose r} \left(\frac1{x}^6-r\right)(2x)^r$$
I understand that there are multiple ways of getting $x^{-2}$, but shouldn't this be accounted for in the general formula method? (multiple $x^{-2}$ terms can be simplified to one number which should obey the general term formula).
I'd really appreciate your help;)
|
Here is a variation which goes consecutively through the factors expanding one by one up to terms which are needed ignoring other terms. We also start with the factors with greatest exponents which keep the number of terms which are to expand small.
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{-2}]}&\color{blue}{(x-1)^3\left(\frac{1}{x}+2x\right)^6}\\
&=[x^4](1+2x^2)^6(x-1)^3\tag{1}\\
&=[x^4]\left(1+\binom{6}{1}2x^2+\binom{6}{2}4x^4\right)(x-1)^3\tag{2}\\
&=\left([x^4]+12[x^2]+60[x^0]\right)(x-1)^3\tag{3}\\
&=12\binom{3}{2}(-1)+60\binom{3}{0}(-1)^3\tag{4}\\
&\,\,\color{blue}{=-96}
\end{align*}
Comment:
*
*In (1) we factor out $\frac{1}{x^6}$ and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (2) we expand the $(1+2x^2)^6$ up to terms which contribute to $[x^4]$. Other terms can be ignored.
*In (3) use the linearity of the coefficient of operator.
*In (4) we select the coefficients of $(x-1)^3$ accordingly.
|
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|
Proving that a set has empty interior For $n=0,1,2...$ define $$A_n = \bigcup^{3^n}_{k=1}\left(\frac{k-\frac23}{3^n},\frac{k-\frac13}{3^n}\right)$$ Let $A=\bigcup_{n=0}^{\infty}A_n$, and let $B=[0,1]\cap \overline A$. I wish to show that $B$ has an empty interior. First, I suppose it has an interior. So I let $x\in B$ and $0\leq x \leq \frac{k-\frac23}{3^n}$ and assume the existence of $\delta>0$ such that $(x-\delta,x+\delta)\subset B$. Then I tried to choose $n$ such that $\frac{\frac 16}{3^n}<\delta$ then add $\frac{\frac 16}{3^n}$ to $x$ to yield $\frac{\frac 16}{3^n}<x+\frac{\frac 16}{3^n}<\frac{k-\frac12}{3^n}$ which is in $A_n$ for k=1. Is this the right direction of the proof? Thank you in advance.
|
The direction of the proof is great but it needs some tiny adjustments. In short, we should choose $n$ after $\delta$.
Suppose $x\in B$ and assume the existence of $\delta>0$ such that $(x-\delta,x+\delta)\subset B$. Then we can choose $n$ such that $\frac{\frac 13}{3^n}<\delta$. Since $x\in B$, there exists $k$ such that $\frac{k-\frac 13}{3^n}\leq x \leq \frac{k +1-\frac 23}{3^n}=\frac{k +\frac 13}{3^n}$.
Now there are two cases.
If $\frac{k-\frac 13}{3^n}\leq x \leq \frac{k}{3^n}$, then we minus $\frac{\frac 13}{3^n}$ from $x$ and get $\frac{k-\frac 23}{3^n}<x-\frac{\frac 13}{3^n}<\frac{k-\frac 13}{3^n}$. Thus $x-\frac{\frac 13}{3^n}\in A_n$.
If $\frac{k}{3^n}\leq x \leq \frac{k + \frac 13}{3^n}$, then we add $\frac{\frac 13}{3^n}$ to $x$ and get $\frac{k+\frac 13}{3^n}<x+\frac{\frac 13}{3^n}<\frac{k+\frac 23}{3^n}$. Thus $x+\frac{\frac 13}{3^n}\in A_n$ (for $k+1$).
But we already assumed that $(x-\delta,x+\delta)\subset B$ and $\frac{\frac 13}{3^n}<\delta$. Such contradiction proves the claim.
|
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|
What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? I'd like to find the four independent units in (the ring of integers of ) $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{R}$ We also have that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$, as a field extension.
I just want to find the Norm, $\mathfrak{N}(x)$ for $x = a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The conjugates are like this:
$$ \big(a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6}\big)
\big(a - b \sqrt{2} + c \sqrt{3} - d\sqrt{6}\big)
\big(a + b \sqrt{2} - c \sqrt{3} - d\sqrt{6}\big)
\big(a - b \sqrt{2} - c \sqrt{3} + d\sqrt{6}\big)$$
If we multiply all four of these things together, we obtain a mess. I used sympy:
$$
a^4 - 4\,a^2b^2 - 6\,a^2c^2 - 12\,a^2d^2 + 48\,abcd + 4\,b^4 - 12\,b^2c^2 - 24\,b^2d^2 + 9\,c^4 - 36\,c^2d^2 + 36\,d^4
$$
Instead we can rearrange the terms it looks almost manageable:
$$ (a^4 + 4\,b^4 + 9\,c^4 + 36\,d^4)- (4\,a^2 b^2 + 6 \, a^2 c^2 + 12\,a^2 d^2 + 12\,b^2c^2 + 24\,b^2 d^2 + 36\, c^2 d^2 ) + (48\, abcd)$$
and Dirichlet's Unit theorem says we can find integers $a,b,c,d \in \mathbb{Z}$ such that this thing $=1$.
Fortunately, I can find two subfields off the bat: $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{2})] = 2$ and $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{3})] = 2$ and we get that :
$$ 1, 3 + 2\sqrt{2}, 2 + \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3}) $$
are still units in this quartic field (by Pell Eq). There's one left. Which is it?
Related:
*
*Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
|
The fundamental unit of $\mathbb{Q}(\sqrt{6})$ is $$5+2\sqrt{6}$$ Does this help ?
|
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|
Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3)^2}}$ , but still didn't get an answer. I got -$\infty$, and everytime I am getting $\infty -\infty$ .
|
For $n\gt2$ we have
$$\left(x-{1\over x^2}\right)^3=x^3-3+{3\over x^3}-{1\over x^6}\lt x^3-3+{3\over4}\lt x^3-1$$
Hence, since $x^3-1\lt x^3$ is clear, we have
$$0\lt x-\sqrt[3]{x^3-1}\lt x-\left(x-{1\over x^2}\right)={1\over x^2}\to0\quad\text{as }x\to\infty$$
So by the Squeeze Theorem,
$$\lim_{x\to\infty}\left(x+\sqrt[3]{1-x^3}\right)=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Binomial sum involving power of $2$
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$
Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\binom{n}{n}\frac{x^{n+1}}{n+1}$$
Could some help me to solve it , Thanks
|
$$f(x)=(1+x)^n =\displaystyle \sum^{n}_{k=0}x^{k}\binom{n}{k} $$
Thus
$$g(x)=\int f(x) dx = \frac{(1+x)^{n+1}-1}{n+1}= \displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)}\cdot x^{k+1}\binom{n}{k}$$
Where the constant of integration was chosen so that $g(0)=0$. Also,
$$\int g(x) dx= \frac{(1+x)^{n+2}-1}{(n+1)(n+2)}-\frac{x}{n+1}=\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot x^{k+2}\binom{n}{k}$$
So that $$\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k} = \frac{(1+2)^{n+2}-1}{(n+1)(n+2)}-\frac{2}{n+1}= \frac{3^{n+2}-2n-5}{(n+1)(n+2)}$$
|
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|
Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a similar pattern for 5 x 5, 6 x 6, ..., n x n matricies.
How would we prove that we would achieve this pattern for all inverses of n x n matricies?
|
Hmm, you can also prove it from basic understanding of inverses of triangular matrices, and a proof by induction.
$$ \left( \begin{array}{c | c}
A_{TL} & A_{TR} \\ \hline
0 & A_{BR}
\end{array} \right)^{-1} =
\left( \begin{array}{c | c}
A_{TL}^{-1} & -A_{TL}^{-1} A_{TR} A_{BR}^{-1}\\ \hline
0 & A_{BR}^{-1}
\end{array} \right)
$$
Now, let's use $ A_{n \times n} $ to equal a matrix of the given form, of size $ n \times n $,
and $ x_n = \left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
$, of size $ n $.
Notice that
$$ A_{n \times n} =
\left( \begin{array}{c | c}
A_{k \times k} & (-1)^k x_k x_{(n-k)}^T \\ \hline
0 & A_{(n-k) \times (n-k)}
\end{array} \right)
$$
where
$$
x_k x_{n-k}^T = \left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
\left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)^T
=
\left( \begin{array}{c}
1 \\
-1 \\
1 \\
\vdots
\end{array} \right)
\left( \begin{array}{c c c c}
1 & -1 & 1 & \cdots
\end{array}
\right)
=
\left( \begin{array}{c c c c}
1 & -1 & 1 & \cdots \\
-1 & 1 & -1 & \cdots \\
1 & -1 & 1 & \cdots \\
\vdots & \vdots & \vdots &
\end{array} \right)
$$
So...
$$ A_{n \times n}^{-1} =
\left( \begin{array}{c | c}
A_{k \times k}^{-1} & - (-1)^k A_{k \times k}^{-1} x_k x_{(n-k)}^T A_{(n-k) \times (n-k)}^{-1} \\ \hline
0 & A_{(n-k) \times (n-k)}^{-1}
\end{array} \right)
$$
Now,
$$
A_{k \times k}^{-1} x_k =
-(-1)^{k} e_L
$$
where $ e_L $ is a vector of all zeroes except the last entry, which equals one.
Similarly,
$$
x_{(n-k)}^T A_{k \times k}^{-1} =
e_F
$$
where $ e_F $ is a vector of all zeroes except the first entry, which equals one.
Now, I am notorious for loosing negative signs and minor typos, but you get the idea.
|
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|
Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\left ( b+ c \right )^{2}\geq 4bc$$
$$\left ( c+ a \right )^{2}\geq 4ca$$
So I want to prove $abc\geq 1$. I need to the help. Thanks!
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$(9uv^2-w^3)^2\geq64u^3w^3$$ or $f(w^3)\geq0$, where
$$f(w^3)=(9uv^2-w^3)^2-64u^3w^3.$$
But, $$f'(w^3)=-2(9uv^2-w^3)-64u^3<0,$$ which says that $f$ decreases.
Thus, it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Since $f(w^3)\geq0$ is homogeneous already, it's enough to assume $b=c=1$ and we need to prove that
$$27(a+1)^4\cdot4\geq64a(a+2)^3,$$ which is
$$(a-1)^2(11a^2+34a+27)\geq0.$$
Done!
|
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|
Prove this inequality $\sum_{cyc}\sqrt[4]{\frac{a}{b+c}}\ge \sqrt[4]{16+\frac{196abc}{(a+b)(b+c)(c+a)}}$ Let $a;b;c\ge0$. Prove the inequality $$\sqrt[4]{\dfrac{a}{b+c}}+\sqrt[4]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\sqrt[4]{16+\dfrac{196abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}$$
I think we should let $\frac{a}{b+c};\frac{b}{c+a};\frac{c}{a+b}\rightarrow x;y;z$ and use Schur's ineq but i don't know how to annul the radicals and the inequality isn't homogeneous so i can't solve it.
|
We need to prove that
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4\geq16\prod_{cyc}(a+b)+196abc.$$
Now, by Holder, C-S,AM-GM and Schur we obtain:
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4=\sum_{cyc}a(a+b)(a+c)+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)+abc)^3(b^2(a+b+c)+abc)}+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)+abc)^3(c^2(a+b+c)+abc)}+$$
$$+6\sum_{cyc}\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}+$$
$$+12\sum_{cyc}\sqrt[4]{a^2bc(a+b)^3(a+c)^3(b+c)^2}\geq$$
$$\geq\sum_{cyc}(a^3+a^2b+a^2c+abc)+4\sum_{cyc}\left(\left(\sqrt{a^3b}+\sqrt{a^3c}\right)(a+b+c)+2abc\right)+$$
$$+6\sum_{cyc}(ab(a+b+c)+abc)+144abc\geq$$
$$\geq\sum_{cyc}\left(a^3+7a^2b+7a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+77abc\right)\geq$$
$$\geq\sum_{cyc}\left(8a^2b+8a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+76abc\right)$$ and since
$$16\prod_{cyc}(a+b)+196abc=\sum_{cyc}(16^2b+16a^2c+76abc)\geq0,$$ it's enough to prove that
$$\sum_{cyc}\left(4\sqrt{a^5b}+4\sqrt{a^5c}-8a^2b-8a^2c+8\sqrt{a^3b^3}\right)\geq0$$ or
$$\sum_{cyc}\sqrt{ab}\left(a^2-2\sqrt{a^3b}+2ab-2\sqrt{ab^3}+b^2\right)\geq0$$ or
$$\sum_{cyc}\sqrt{ab}(a+b)(\sqrt{a}-\sqrt{b})^2\geq0.$$
Done!
|
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Converting a generating function from fraction form to a power series Given a fraction $\frac{(1 + ax^{n})^{m}}{(1 + bx^{p})^{q}}$, how does one convert it to a series of the form $a_{0}x^{0} + a_{1}x^{1} + a_{2}x^{2} + a_{3}x^{3} . . .$ ?
I was unable to find instructions for this, and I could use the information to to solve this problem:
Given the equation $x_{1} + x_{2} + x_{3} + x_{4} + y_{1} + y_{2} = 6$,
where $0 ≤ x_{i} ≤ 2$ and where $y_{i}$ is divisible by $3$,
find the number of possible solutions in natural numbers by calculating the coefficient of $x^{6}$ in the generating function:
$f(x) = (1 + x + x^{2})^{4} (1 + x^{3} + x^{6} + . . .)^{2}$.
Hint:
$1 + x^{3} + x^{6} + . . . = \frac{1}{1 - x^{3}};$
$1 + x + x^{2} = \frac{1 - x^{3}}{1 - x}$.
|
Looking at the hints, we have:
$$1 + x^{3} + x^{6} + \cdots = \frac{1}{1 - x^{3}}$$
and
$$1 + x + x^{2} = \frac{1 - x^{3}}{1 - x}$$
so
$$\begin{align}f(x) &= (1 + x + x^{2})^{4} (1 + x^{3} + x^{6} + \cdots)^{2}\\[1ex]&=\frac{(1 - x^{3})^4}{(1 - x)^4(1-x^3)^2}\\[1ex]&=\frac{(1 - x^{3})^2}{(1 - x)^4}\\[1ex]&=\frac{(1 - 2x^{3}+x^6)}{(1 - x)^4}\\[1ex]&=(1 - 2x^{3}+x^6)\sum_{k\ge 0}\binom{k+3}{3}x^k\\[1ex]&=\sum_{k\ge 0}\left(\binom{k+3}{3}-2\binom{k}{3}+\binom{k-3}{3}\right)x^k\tag{1}\end{align}$$
The penultimate step uses the negative binomial expansion:
$$(1-x)^{-4}=\sum_{k\ge 0}\binom{k+3}{3}x^k$$
So the $x^6$ coefficient of $(1)$ is:
$$\binom{9}{3}-2\binom{6}{3}+\binom{3}{3}=45\tag{Answer}$$
This method can be easily generalised by using the general binomial and negative binomial expansions:
$$(1+x)^n=\sum_{k= 0}^{n}\binom{n}{k}x^k$$
and
$$(1-x)^{-m}=\sum_{k\ge 0}\binom{k+m-1}{m-1}x^k$$
respectively.
|
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Proving the trig identity $\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ without cross-multiplying I need to prove the following identity.
$$\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$$
I want to prove it by deduction rather than cross multiplying.
|
$$\frac{1-\cos\theta}{\sin\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)}=\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}$$
$$\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}=\frac{\sin\theta}{1+\cos\theta}$$
Let $\displaystyle t=\tan\frac{\theta}{2}$
$$\frac{1-\cos\theta}{\sin\theta}=\frac{1-\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}}=t$$
$$\frac{\sin\theta}{1+\cos\theta}=\frac{\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}=t$$
|
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Interesting 3 Variable Inequality for Real Numbers I recently saw an olympiad style inequality that seemed very difficult. I tried to use elementary inequalities such as AM-GM or Cauchy-Schwarz, but neither have helped in making significant progress. Could anyone provide a rigorous proof, preferably using more elementary inequalities?
Problem: "Prove the inequality $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)}\geq\frac{27}{2(x+y+z)^2}$ if $x,y,z$ are positive reals."
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The inequality is homogenous in $(x, y, z)$, therefore we can assume
that $xyz=1$. Then
$$
\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \ge \dfrac{3}{2}
$$
as for example demonstrated in
If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ or A inequality proposed at Zhautykov Olympiad 2008.
Also
$$
1 = \sqrt[3]{xyz} \le \frac{x+y+z}3 \Longrightarrow
(x+y+z)^2 \ge 9 \, .
$$
from the inequality between the geometric and the arithmetic mean.
Combining these estimates gives the desired inequality.
|
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If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$b=2R \sin B$$
$$c=2R \sin C$$
Then,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
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Since
$$4 = 2\sin^2 A + 2\sin^2 B + 2\sin^2 C$$
we have
$$(1-2\sin^2 A ) + (1-2\sin^2B) + 2 - 2\sin^2C=0.$$
or
$$2\cos^2 C + (\cos(2A) + \cos(2B)) = 0$$
Since $$\cos(2A)+\cos(2B) = 2\cos(A+B) \cos(A-B) = -2\cos C\cos(A-B),$$
we have
$$2\cos C(\cos C - \cos(A-B)) = 0$$
Replace $\cos C = -\cos(A+B)$, we get
$$ \cos C(\cos(A+B) + \cos(A-B)) = 0$$
equivalently
$$\cos A \cos B \cos C = 0.$$
The conclusion follows.
|
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Integration of $x^2$ using double substitution my professor has asked me to calculate: $$\int x^2dx = \frac{1}{3}x^3 + C$$ using the substitution of $$u=x$$ $$v=x$$
meaning that $$du=dx$$ $$dv = dx$$
The next step is where I think I'm wrong but I'm unsure why $$dx = \frac{du + dv}{2}$$
$$ \int uv(\frac{du+dv}{2}) = \frac{1}{2}\int uvdu + \frac{1}{2}\int uvdv = \frac{1}{4}u^2v + \frac{1}{2}uv^2 + C $$
Substituting back results in
$$ \frac{1}{4}x^2x + \frac{1}{4}xx^2 + C = \frac{1}{2}x^3 + C$$
However, $$ \frac{1}{2}x^3 + C \neq \frac{1}{3}x^3 + C $$
So I am unsure where I went wrong. Any help would be appreciated thanks.
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You can't use two substitutions in the same integral for the same variable. It dosen't make sense..
Note that
$$\int uv(\frac{du+dv}{2}) = \frac{1}{2}\int uvdu + \frac{1}{2}\int uvdv = \frac{1}{4}u^2v + \frac{1}{2}uv^2 + C$$
You have uv as variable but du...and also $u=v$ so normally
$$\int uvdu=\int u^2du=\frac {u^3}3+K$$
|
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Evaluating $\lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$ $$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$$
I understand that we need to divide by the highest power, but first we want to factor out the $x^6$ from the radical. This gives
$$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{x^6}\sqrt{4+\frac{1}{x^6}}}.$$
My confusion is when I plug this limit into a program such as symbolab is that it says this fraction should simplify to $$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3-x^3\sqrt{4+\frac{1}{x^6}}}$$ because as $x\to-\infty\Rightarrow\sqrt{x^6}=-x^3$. Why is it that we apply this computation here, and not anywhere else in this fraction? I'm mostly confused as to why we need to make this computation in the first place.
|
In general, for any real $x$, $\sqrt{x^6}=|x|^3$. As $x\to-\infty$, we may assume that $x<0$ and therefore $\sqrt{x^6}=|x|^3=-x^3$. This simplification allows us to easily divide the numerator and the denominator by $x^3$, and find that
$$\lim_{x\to-\infty}\frac{x^3+1}{x^3-x^3\sqrt{4+\frac{1}{x^6}}}=\lim_{x\to-\infty}\frac{1+\frac{1}{x^3}}{1-\sqrt{4+\frac{1}{x^6}}}.$$
Can you take it from here?
P.S. On the other hand, as $x\to+\infty$, we may assume that $x>0$, $\sqrt{x^6}=|x|^3=x^3$ and
$$\lim_{x\to+\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}=\lim_{x\to+\infty}\frac{x^3+1}{x^3+x^3\sqrt{4+\frac{1}{x^6}}}=\lim_{x\to+\infty}\frac{1+\frac{1}{x^3}}{1+\sqrt{4+\frac{1}{x^6}}}.$$
|
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Why does one of the units, namely $50 - 7\sqrt{51}$, in $\mathbb{Z}[\sqrt{51}]$ have a decimal expansion that is the Catalan numbers? $50 - 7\sqrt{51}$ is one of the units in $\mathbb{Z}[\sqrt{51}]$. In fact, corresponding to $n = 51$, $x = 50$ and $y = 7$ is the smallest integer solution to $x^2 - ny^2 = 1$.
The Catalan numbers have a great deal of combinatorial significance, and they appear in the decimal expansion of $50 - 7\sqrt{51} \approx 0.010001000200050014004201320429.$
Why is this the case?
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wikipedia names $$ c(x) = \sum_0^\infty \; C(n) \; x^n $$
and says that it satisfies
$$ c(x) = 1 + x \, c^2(x) $$
Define $$ A = c \left( \frac{1}{10000} \right) $$
Then we have
$$ A = 1 + \frac{1}{10000} \; A^2 $$
so
$$ A^2 - 10000A + 10000 = 0. $$ So
$$ A = \frac{10000 \pm \sqrt{100000000 - 40000}}{2} = \frac{10000 \pm 100 \sqrt{10000 - 4}}{2}$$
$$ A = \frac{10000 \pm 200 \sqrt{2500 - 1}}{2}= 5000 \pm 100 \sqrt {2499} $$
Note $$ 2499 = 51 \cdot 49 $$
Actually, later they say
$$ c(x) = \frac{1 - \sqrt{1 - 4 x}}{2x} = \frac{2}{1 + \sqrt{1 - 4x}} $$
which confirms that we should take
$$ A = 5000 - 100 \sqrt {2499} = 100 \left( 50 - 7 \sqrt{51} \right)$$
|
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Writing integers as a product of as few elements of $\{\frac21, \frac32, \frac43, \frac54, \ldots\}$ as possible This question is inspired by question 2 of the 2018 European Girls' Mathematical Olympiad. Also posted on mathoverflow.
Any integer $x \ge 2$ can be written as a product of (not necessarily distinct) elements of the set $A = \{\frac21, \frac32, \frac43, \frac54, \ldots\}$, as can be seen from a simple telescoping argument. Let $f(x)$ be the minimum number of elements of $A$ required.
For example, $f(11)=5$ because $11 = \frac{33}{32} \cdot \frac{4}{3} \cdot \frac{2}{1} \cdot \frac{2}{1} \cdot \frac{2}{1}$ (or, alternatively, $11 = \frac{11}{10} \cdot \frac{5}{4} \cdot \frac21 \cdot \frac21 \cdot \frac21$), but $11$ cannot be written as the product of $4$ or less elements of $A$. In general, it seems difficult to compute $f(x)$ directly.
Clearly we have $f(xy) \le f(x) + f(y)$ for any $x,y \ge 2$. The EGMO question asks to show that this inequality is strict infinitely often (an example is $x=5$, $y=13$). Here we ask:
For integral $x \ge 2$, let $f(x)$ be the smallest $k$ so that $x$ can be written as product of $k$ elements of $\{\frac21, \frac32, \frac43, \ldots\}$.
Is it true that $f(xy) =f(x) + f(y) - O(1)$?
In other words, is the difference between $f(x)+f(y)$ and $f(xy)$ bounded?
Some observations:
*
*$f(x) \ge \log_2(x)$ as $A$ has no elements larger than $2$;
*if it were true that $f(x) = \log_2(x) + O(1)$, this would imply that the question above has a positive answer.
|
We can compute an upper bound to $f(x)$ by adding $\lfloor \log_2(x)\rfloor$ to the number of $1$s in the binary expansion of $x$ and subtracting $1$. This gives $f(5)=2+2-1=3,5=\frac 54 \frac 21 \frac 21, f(13)=3+3-1=5, 13=\frac {13}{12} \frac 32(\frac21)^3,f(65)=6+2-1=7,65=\frac {65}{64}(\frac 21)^6$
To prove that we can find an expansion like this, write $x=x_0$ in binary. We will make a descending chain of $x_i$ ending with $1$ where $\frac {x_i}{x_{i+1}}\in A.$ If $x_i$ is odd, $x_{i+1}=x_i-1$. If $x_i$ is even, $x_{i+1}=\frac {x_i}2$. This gets an expression for $x_0$ with the stated number of steps. We have one $-1$ step for every $1$ in the expansion except the leading one and one divide by $2$ step for every power of $2$ in the leading bit. This does not prove that there is not a shorter expression. user133281 gives the example of $43$ which I represent as $\frac {43}{42}2\frac {21}{20}2^2\frac 542^2$ for $8$ terms but can also be $\frac {129}{128}\frac 432^5$ for seven.
Given this expression for $f(x)$ we want to show that we can find $x$ and $y$ such that $xy$ has arbitrarily many fewer $1$ bits in binary than the sum of the number of bits in $x$ and $y$. We use the fact that if $b$ is odd $2^{ab}+1=(2^a+1)(2^{a(b-1)}-2^{a(b-2)}-2^{a(b-3)}-\ldots+1)$ so let $x=2^a+1$ with two bits on, $y=2^{a(b-1)}-2^{a(b-2)}-2^{a(b-3)}-\ldots+1$ with $a(b-1)/2$ bits on and $xy$ has only two bits on, so $f(x)+f(y)-f(xy)=a(b-1)/2$ We still need to justify that there are $y$s that don't have significantly shorter representations.
|
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Binomial coefficients $\binom{n}{b} \pmod{n}$, $n > b$, $n$ odd, $n,b$ positive integers, AKS test I am looking at the values of Binomial coefficients $\binom{n}{b} \pmod{n}$, where $n > b$, $n$ odd, and $n,b$ positive integers. I am planning to use the results in an implementation in python of the AKS primality test.
Examples of binimial coeffs mod $n$:
*
*$b=3, n=5$, $$\binom{5}{3} \pmod{5} \cong 0$$
*$b=10, n=15$, $$\binom{15}{10} \pmod{15} \cong 3$$
Since it is mentioned in comments, Lucas's theorem is for $n$ prime, but I am considering odd $n$.
I have found that there are several patterns, that I think may depend on the factorisation of $b$
Examples of patterns found (using assumptions stated):
*
*For $b=6$:
$\binom{n}{6} \pmod{n} \cong \frac{n}{3}$ when $n \pmod{3} \cong 0$ and $n \pmod{9} \cong 0$
$\binom{n}{6} \pmod{n} \cong \frac{2n}{3}$ when $n \pmod{3} \cong 0$ and $n \pmod{9} \cong 6$
$\binom{n}{6} \pmod{n} \cong 0$ when $n \pmod{3}$ is not cong to $0$
*For $b=7$:
$\binom{n}{7} \pmod{n} = \frac{n}{7}$ when $n \pmod{7} \cong 0$
$\binom{n}{7} \pmod{n} = 0$ when $n \pmod{7} $ not cong to $0$
*For $b=8$:
$\binom{n}{8} \pmod{n} \cong 0$
Questions:
*
*What is the relationship between $\binom{n}{b} \pmod{n}$ and $b$ for odd $n$?
*Seeking reference requests.
|
As a consequence of theorem 4 in
*
*Andrew D. Loveless: A Congruence for Products of Binomial Coefficients modulo a Composite, in: INTEGERS 7 (A44 10 Jan 2007)
we have $\binom{n}{b}\equiv 0\pmod{n}$ if (but not only if) $\gcd(n,b)=1$. The other theorems and corollaries in the paper may be of interest to you as well.
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Evaluate $\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}$ Find value of following integral $$\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}\text{dx}$$
the numerator is $\text{d[sec(x)]}$ but that isnt work due to $x$ in denominator. First we can simplify as $$\int\frac{x\sin(x)}{(\sin(x)-x\cos(x))^2} \text{dx} = \int \frac{x}{(\sin(x)-x\cos(x))} \text{dx}+\int \frac{x^2 \cos(x)}{(\sin(x)-x\cos(x))^2} \text{dx}$$
but again its not manipulative. Suggest a useful substitution or method.
Thanks a lot!
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Consider the function $f(x) = \frac{1}{(\sin x - x\cos x)}$. Once differentiation gives:
$$f'(x) = \frac{-(\cos x +x\sin x-\cos x)}{(\sin x - x\cos x)^2} = \frac{-x \sin x}{(\sin x - x\cos x)^2}$$
So your integral is:
$$\int \frac{x \sin x}{(\sin x - x\cos x)^2} dx = \frac{-1}{(\sin x - x \cos x)}$$
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How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
I have omitted the Base Case due to lack of specific relevance. An excerpt from the book's page:
Assume the assertion has been proved for a specific value of $n$, say $n = k$. That is, assume we have proved
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 \lt \frac{k^3}{3}$$
for a fixed $k \ge 1$. Now using this, we shall deduce the corresponding result for $k + 1:$
$$A(k + 1): 1^2 + 2^2 + \cdots + k^2 \lt \frac{(k + 1)^3}{3}.$$
Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
$$1^2 + 2^2 + \cdots + k^2 \lt \frac{k^3}{3} + k^2.$$
To obtain $A(k + 1)$ as a consequence of this, it suffices to show that
$$\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}.$$
But this follows at once from the equation
$$\frac{(k + 1)^3}{3} = \frac{k^2 + 3k^2 + 3k + 1}{3} = \frac{k^3}{3} + k^2 + k + \frac13.$$
Therefore, we have shown that $A(k + 1)$ directly follows from $A(k)$.
I can not understand the last two steps. Why do the expressions seem to flip on the inequality in step 4 such that $\frac{k^3}{3} + k^2$ is now on the other side of the less than symbol, and how does the final equation prove the assertion?
Thank you.
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$A(k)$ states
$$1^2 + \cdots + (k-1)^2 < \frac{k^3}{3}.$$
Adding $k^2$ to both sides yields
$$1^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
If we prove $\frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$
then we can tack this onto the end of the above inequality to get
$$1^2 + \cdots + k^2 < \frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$$
which is the desired inequality $A(k+1)$.
It remains to verify the unjustified claim $\frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$.
The final equation $\frac{(k+1)^3}{3} = \frac{k^3}{3} + k^2 + k + \frac{1}{3}$ in your post is just obtained by expanding $(k+1)^3$. The right-hand side is greater than $\frac{k^3}{3} + k^2$ since $k+1$ is positive.
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Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-2)\gt0}$
${3-x^2-2x\gt0}$
${-x^2-2x+3\gt0}$
${-1(x^2+2x-3)\gt0}$
$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$
${(x-1)(x+3)\lt0}$
At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.
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Why do you make things more complex than they are?
The function $x(x-2)$ is increasing and positive on $(2,+\infty)$, hence
$\;\dfrac2{x(x-2)}\;$ is decreasing and positive on this interval.
Furthermore $f(3)=1$…
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?
Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?
My attempts :
$\sin(x+y) = \sin x\cos y + \cos x \sin y$
now $\sin(n +\frac{1}{n}) = \sin n \cos \frac{1}{n} + \cos n\sin\frac{1}{n}$
now $\sum_{n=2}^{\infty} \frac {\sin n \cos 1/n }{\log\log n} + \sum_{n=2}^{\infty}{\frac {\cos n \sin 1/n}{\log\log n}}$
After that I can not able to proceed further.
|
\begin{align}
\sum_{n=2}^{\infty} \frac {\sin( n + 1/n) }{\log\log n}&=\sum_{n=2}^{\infty} \frac {\sin n \cos 1/n }{\log\log n} + \sum_{n=2}^{\infty}{\frac {\cos n \sin 1/n}{\log\log n}}\\
&=\sum_{n=2}^{\infty} \sin n\,\frac {1 }{\log\log n} +\sum_{n=2}^{\infty} \frac {(\sin n)(\cos 1/n-1) }{\log\log n}+ \sum_{n=2}^{\infty}{\cos n\,\frac { \sin 1/n}{\log\log n}}.
\end{align}
By Dirichlet's test, the first and last series converge. The second one is absolutely convergent, since $\cos 1/n-1\sim -1/(2n^2)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{2\pi}{11},2\cos\frac{8\pi}{11},2\cos\frac{14\pi}{11}$$
I tried to eliminate options using the sum and product of roots but I can't figure out if $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$$
or
$$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$
|
Put $x=2t$
Hence $$8t^3-6t+1=0$$
Now put $t=\cos \theta$
Hence $$2\cos (3\theta)=-1$$
And the rest is simple trigonometric equation
|
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"timestamp": "2023-03-29T00:00:00",
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|
For rotation matrix $A$, find $B = A^4- A^3 + A^2 - A$
Find the value of $B = A^4- A^3 + A^2 - A$ where $A$ is the matrix given below $$ A=
\left [ \begin{matrix}
\cos\alpha & \sin \alpha \\
-\sin\alpha & \cos\alpha
\end{matrix} \right ]
$$
It's actually quite simple when you look at it. Find $A$, $A^2$, $A^3$, $A^4$ by doing some matrix multiplication and then add and subtract following the question.
I did all of the tedious work mentioned above.
But the matrix $B$ I obtain is filled with garbage. I assuming either we must use some trigno mumbo jumbo identities, or there's something beautiful that I missed from the question.
Edit: Could the sum of a geometric series help?
|
By induction,
$$ A^n = A^{n-1}A = \begin{pmatrix} \cos{(n-1)\theta} & \sin{(n-1)\theta} \\
-\sin{(n-1)\theta} & \cos{(n-1)\theta} \end{pmatrix} \begin{pmatrix} \cos{\theta} & \sin{\theta} \\
-\sin{\theta} & \cos{\theta} \end{pmatrix} = \begin{pmatrix} \cos{(n-1)\theta}\cos{\theta} - \sin{\theta}\sin{(n-1)\theta} & \sin{(n-1)\theta}\cos{\theta} + \cos{(n-1)\theta}\sin{\theta} \\
-\sin{(n-1)\theta} & \cos{(n-1)\theta}\cos{\theta} - \sin{\theta}\sin{(n-1)\theta} \end{pmatrix} \\
= \begin{pmatrix} \cos{n\theta} & \sin{n\theta} \\
-\sin{n\theta} & \cos{n\theta} \end{pmatrix}. $$
The prosthaphaeresis formulae are
$$ \sin{A}+\sin{B} = 2\sin{\tfrac{1}{2}(A+B)}\cos{\tfrac{1}{2}(A-B)} \\
\sin{A}-\sin{B} = 2\cos{\tfrac{1}{2}(A+B)}\sin{\tfrac{1}{2}(A-B)}\\
\cos{A}+\cos{B} = 2\cos{\tfrac{1}{2}(A+B)}\cos{\tfrac{1}{2}(A-B)}\\
\cos{A}-\cos{B} = -2\sin{\tfrac{1}{2}(A+B)}\sin{\tfrac{1}{2}(A-B)}. $$
Thus
$$ \cos{4\theta}-\cos{3\theta} = -2\sin{\tfrac{7}{2}\theta}\sin{\tfrac{1}{2}\theta}, \\
\cos{2\theta}-\cos{\theta} = -2\sin{\tfrac{3}{2}\theta}\sin{\tfrac{1}{2}\theta}, $$
and adding gives
$$ \cos{4\theta}-\cos{3\theta} + \cos{2\theta}-\cos{\theta} = -2\sin{\tfrac{1}{2}\theta} ( \sin{\tfrac{7}{2}\theta} + \sin{\tfrac{3}{2}\theta} ) \\
= -4\sin{\tfrac{1}{2}\theta}\sin{\tfrac{5}{2}\theta}\cos{\theta}. $$
Similarly,
$$ \sin{4\theta}-\sin{3\theta} + \sin{2\theta}-\sin{\theta} = 4\sin{\tfrac{1}{2}\theta}\cos{\tfrac{5}{2}\theta}\cos{\theta}, $$
so
$$ B = 4\sin{\tfrac{1}{2}\theta}\cos{\theta} \begin{pmatrix} -\sin{\tfrac{5}{2}\theta} & -\cos{\tfrac{5}{2}\theta} \\ \cos{\tfrac{5}{2}\theta} & -\sin{\tfrac{5}{2}\theta} \end{pmatrix}. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result
Is this an ok proof?
|
You have:
$$d\mid n \Rightarrow n=ad$$
Then:
$$x^n-1=x^{ad}-1=(x^d)^a-1$$
Setting $x^d=y$ (just for simplifying the process) we have:
$$y^a-1=(y-1)(y^{a-1}+\dots+y+1)=(x^d-1)((x^d)^{a-1}+\dots + x^d+1)$$
In other words we showed that:
$$x^n-1= (x^d-1)((x^d)^{a-1}+\dots + x^d+1) $$
Which obviously implies that:
$$x^d-1 \mid x^n-1$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.
In addition, this identity is supposed to be true:
$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$
|
If $x=0$, the equality is trivial. So, it is sufficient to consider $x\ne 0$.
Let $y=ax, z=abx$. Then:
$$x+ax+abx=0 \Rightarrow ab=-1-a.$$
Then:
$$\frac{x^5+a^5x^5+(ab)^5x^5}{5}=\frac{x^2+a^2x^2+(ab)^2x^2}{2}\cdot \frac{x^3+a^3x^3+(ab)^3x^3}{3} \Rightarrow$$
$$\frac{1+a^5+(-1-a)^5}{5}=\frac{1+a^2+(-1-a)^2}{2}\cdot \frac{1+a^3x^3+(-1-a)^3}{3} \Rightarrow$$
$$-a-2a^2-2a^3-a^4=(1+a+a^2)(-a-a^2).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\displaystyle m\left( \bigcup_{x \in C} \left( x - \frac{1}{20}, x+\frac{1}{20} \right) \right)$ where $C$ is Cantor set.
Question: Let $C$ be the Cantor set, obtained from the unit interval $[0,1].$
For each $x \in C$, let $I_x$ be the interval $\displaystyle \left(x-\frac{1}{20}, x + \frac{1}{20}\right).$
Note that $\displaystyle \bigcup_{x \in C}I_x$ is open.
Evaluate $\displaystyle m\left( \bigcup_{x \in C}I_x \right).$
My attempt:
It seems to me that $\bigcup_{x\in C}I_x$ is an open cover for $C.$
Since $0\in C$ and $1\in C,$ therefore
$$m\left( \bigcup_{x\in C}I_x \right) = 1+\frac{1}{20} + \frac{1}{20} = 1.1$$
But I am not sure whether this makes sense.
EDITED (based on hint provided by Ross Millikan): Observe that intervals missing from $\bigcup_{x\in C} I_x$ are
$$\bigg[\frac{1}{3} + \frac{1}{20} , \frac{2}{3} - \frac{1}{20}\bigg], \bigg[ \frac{2}{9} + \frac{1}{20}, \frac{3}{9} - \frac{1}{20} \bigg] \text{ and } \bigg[ \frac{7}{9} + \frac{1}{20} ,\frac{8}{9} - \frac{1}{20} \bigg] .$$
Therefore,
$$m\left( \bigcup_{x\in C} I_x\right) = \frac{11}{10} - \bigg( \frac{1}{3} - \frac{1}{10} \bigg) - 2\times \bigg( \frac{1}{9} - \frac{1}{10} \bigg) = \frac{76}{90}.$$
|
Since the whole interval $(\frac 13,\frac 23)$ is excluded from the Cantor set the interval $(\frac 13+\frac 1{20},\frac 23-\frac 1{20})$ is excluded from your union. The measure is then less than $1.1$. You need to find all the intervals that are missing from your set, add them up, and subtract from $1.1$. There aren't very many of them.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2748040",
"timestamp": "2023-03-29T00:00:00",
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|
Verify that $x^2 + cy^2 = 1$ is an implicit solution to $\frac{dy}{dx} = \frac{xy}{x^2 - 1}.$ When I differentiate $x^2 + cy^2 = 1$ implicitly and solve for $\dfrac{dy}{dx}$, I get $\dfrac{dy}{dx} = \dfrac{-x}{cy}.$ I thought I had to multiply by a fraction in order to make it similar to $\dfrac{dy}{dx} = \dfrac{xy}{x^2 - 1}$\$ but nothing worked.
I then checked the answer and someone put up these steps:
\begin{align*}
y^2 &= -(x^2 - 1) * \dfrac{1}{c} \\
y^2 &= c(x^2 - 1) \\
y &= \sqrt{c(x^2 - 1)} \\
y &= c\sqrt{x^2 - 1)}
\end{align*}
I don't get how $\dfrac{1}{c}$ changes to $c$ and it's square root disappears.
|
Notice that:
$$
x^2 + cy^2 = 1 \iff cy = \frac{1 - x^2}{y}
$$
So:
$$
\frac{dy}{dx} = \frac{-x}{cy} = \frac{-x}{\left(\frac{1 - x^2}{y} \right)} = \frac{-xy}{1 - x^2} = \frac{xy}{x^2 - 1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2748212",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that in a field $F, \exists a,b,c \in F$ such that $x^3+x^2+1$ is a divisor of $x^{2018} + ax^3 + bx + c$ I have no idea how to even start this! Maybe using the Euclidean algorithm and showing that the extra term is 0?
|
Apply Euclidean algorithm to $x^{2018}$ and $x^3+x^2+1$, we can find a polynomial $q(x)$ and quadratic polynomial $\alpha x^2 + \beta x + \gamma$ such that
$$x^{2018} = q(x)(x^3+x^2+1) + \alpha x^2 + \beta x + \gamma\\
\Downarrow\\
x^{2018} + ax^3 + bx + c = q(x)(x^3+x^2+1) + \underbrace{ax^3 + \alpha x^2 + (\beta+b)x + (\gamma + c)}_{\mathcal{M}}$$
If we take $a = \alpha, b = -\beta, c =\alpha - \gamma$, we have
$$\mathcal{M} = \alpha (x^3 + x^2 + 1)
\quad\implies\quad
x^{2018} + ax^3 + bx + c = (q(x)+\alpha)(x^3+x^2+1)$$
For such a choice of $a,b$ and $c$, $x^3+x^2+1$ is a factor of $x^{2018} + ax^3 + bx + c$.
|
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"timestamp": "2023-03-29T00:00:00",
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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.