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Find the first three Laurent expansion terms of $\frac{1-z}{z^2} e^z$ I've this function $f(z)=\frac{1-z}{z^2} e^z$ and I've to find the first three Laurent expansion terms in $z=0$.
I've proceeded in this way:
First of all I've considered the expansion series of $e^z = \sum_{n=0}\frac{z^n}{n!}$ and I found the first three terms, so $$e^z=1+z+\frac{z^2}{2}+o(z^3)$$
At this point my function would be
$$f(z)=\frac{1}{z^2}(1-z)(1+z+\frac{z^2}{2}+o(z^3))$$
$$f(z)= \frac{1}{z^2}-\frac{1}{2}-\frac{z}{2}+o(z^3)$$
and it seems right, but using Wolfram Alpha, I've discovered that should be
$f(z)=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}-\frac{z^3}{30}+o(z^4)$.
As you can see, the third term is different. What I've to do?
|
First, you can either write
$$
e^z=1+z+\frac{z^2}{2}+o(z^2)
$$
with “little-oh,” or
$$
e^z=1+z+\frac{z^2}{2}+O(z^3)
$$
with “big-oh.” Let's assume that you meant the latter. Then your calculation gives
$$
(1-z)e^z = 1+z+\frac{z^2}{2}+O(z^3) - (z+z^2+\frac{z^3}{2}+O(z^4)) \\
= 1 - \frac{z^2}{2}- \frac{z^3}{2}+O(z^3)+O(z^4) \\
= 1 - \frac{z^2}{2} + O(z^3) \, .
$$
Note how the $-z^3/2$ is already covered by the $O(z^3)$. At that point, you calculated
$$
(1-z)e^z = 1 - \frac{z^2}{2} - \frac{z^3}{2}+O(z^4) \, ,
$$
which is wrong.
Finally,
$$
f(z)=\frac{1-z}{z^2} e^z = \frac{1}{z^2}(1 - \frac{z^2}{2} + O(z^3))
=\frac{1}{z^2} - \frac 12 + O(z)
$$
and that is consistent with
$$
f(z)=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}-\frac{z^3}{30}+O(z^4)
$$
from Wolfram Alpha.
In order to get also the $z$ term in $f(z)$ (which is actually the fourth term in the Laurent expansion) you have to start with four terms, i.e. with
$$
e^z=1+z+\frac{z^2}{2}+\frac{z^3}{6}+O(z^4)
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4479009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix} +\begin{pmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 1
\end{pmatrix}\right| =\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=n+1.$$
I can prove it by adding and subtracting rows and columns without changing the determinant:
*
*subtracting the first row from the others $
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*subtracting the other rows from the first $\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*adding the other columns to the first $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*and clearly $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right| =n+1$
but I feel there should be something more intuitive.
What would be a more intuitive or insightful way to prove this?
|
I'm going to elaborate on Exodd's comment. I'm inclined to agree that this is the simplest proof, provided you're comfortable with the basic constructions of linear algebra.
Lemma: If $A \in \mathbf{C}^{n \times n}$ has eigenvalues (counting multiplicity) $(\lambda_i)_{i=1}^n$, then $A + I$ has eigenvalues $(\lambda_i + 1)_{i=1}^n$.
Proof: Observe that $\det(xI - (A + I)) = \det((x-1)I - A)$. $\square$
Proposition: If $J \in \mathbf{C}^{n \times n}$ is the all-ones matrix, then $\det(I+J) = n+1$.
Proof: Since $J$ is rank 1, hence it has exactly one nonzero eigenvalue. Since $\text{trace}(J) = \sum_{i=1}^n \lambda_i$, it follows that the unique nonzero eigenvalue $\lambda_1 = n$. By the lemma, $\det(J+I) = \prod_{i=1}^n (\lambda_i + 1) = n+1$. $\square$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $n = 1$, the integrand $=x+x^3$
For $n = 2$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}$
For $n = 3$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}$
and so on.
For $n = 1000$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}-\dots+x^{7993}+x^{7995}$
Using MS-EXCEL with $n=1000$, I found that the value is approximately $0.56...$.
I do not know if $n \rightarrow \infty$ , will the required expression have a closed form or no.
Your help would be appreciated. THANKS!
|
We seek to calculate $$\begin{align}\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx&= \frac{1}{2}+\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{10}+\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\cdots \\&=\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots\right)\right)\end{align}$$ Now $$\arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots \implies \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\arctan{1}=\frac{\pi}{4}$$ and $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \implies \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)=\frac{\ln2}{2}$$ Hence our answer is $\frac{\pi}{8}+\frac{\ln2}{4}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
limit of function near a given point Find the limit of the given function as $x\to -2$ without using L’Hôpital’s Rule:
$$\displaystyle \lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}$$
I used the identity: $x^3 + 2^3 = (x+2)(x^2 -2x +4)$ at the denominator. Then i tried to use the same identity at the numerator and didnt succeed. So i tried to expand the numerator, but i got different answer by checking the limits from both of the sides near $x = -2$.
The answer is $\frac{1}{144}$ which is an approximation of the limit at the given point.
Any suggestions and support would be kindly appreciated.
Edit: Just watching it from the side and I can rewrite the 2 at the numerator as $\sqrt[3]2$. Yet, its not helping a lot.
|
As you have noticed that $x^3 + 2^3 = (x+2)(x^2 -2x +4)$, we first calculate the limit
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}.$$
Let $t=\sqrt[3]{x-6}$, then $t\to-2$ as $x\to-2$ and $x+2=t^3+8=(t+2)(t^2-2t+4)$, so
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}=\lim_{t\to-2}\frac{t + 2}{(t+2)(t^2-2t+4)}=\lim_{t\to-2}\frac1{t^2-2t+4}=\frac1{12}.$$
Therefore,
$$\lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}=\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}\cdot\lim_{x\to-2}\frac1{x^2-2x+4}=\frac1{12}\cdot\frac1{12}=\frac1{144}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt$ $$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$
Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$
$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2})}{t} dt=\int_0^1 \frac{\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(1-\frac{t}{z_2})}{t} dt$$
where $\ln(z_1)+\ln(z_2)=0$ and $\frac{1}{z_1}=z_2, \frac{1}{z_2}=z_1$
$$I=-\sum_{n=1}^\infty \frac{1}{n} \int_0^1 \frac{1}{t}\left(\frac{t}{z_1}\right)^n+\frac{1}{t}\left(\frac{t}{z_2}\right)^n dt = -\sum_{n=1}^\infty \frac{1}{n}\left( \frac{z_1^n+z_2^n}{n} \right)$$
where $z_1^n+z_2^n=2\cos(\frac{n\pi}{6})$
$$I=-2\sum_{n=1}^\infty \frac{\cos(\frac{n\pi}{6})}{n^2}$$
How to proceed next?
|
As @Frank W pointed out in comment...(although I saw the comment after I came up with the proof). Actually, I found the integral $I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$ in the book "Table of Integrals, Series, and Products"
by I.S. Gradshteyn and I.M. Ryzhik. Then I came up with the proof.
For $a \in (0, \pi/2]$, let
$$I(a) := \int_0^1 \frac{\ln(1 - 2t \cos a + t^2)}{t}\,\mathrm{d} t.$$
We have
\begin{align*}
I'(a) &= \int_0^1 \frac{2\sin a}{1 - 2t\cos a + t^2}\, \mathrm{d} t\\
&= 2\arctan \frac{t - \cos a}{\sin a}\Big\vert_0^1 \\
&= 2\arctan \frac{1 - \cos a}{\sin a}
+ 2\arctan\frac{\cos a}{\sin a}\\
&= \pi - a
\end{align*}
We have $I(a) = \pi a - \frac{a^2}{2} + C$
on $(0, \pi/2]$ for some constant $C$.
Since $I(a)$ is continuous on $[0, \pi/2]$, taking limit as $a \to 0^{+}$, we get
$$C = \int_0^1 \frac{2\ln(1 - t)}{t}\,\mathrm{d} t =
\int_0^1 (-2)\sum_{k=1}^\infty \frac{1}{k}t^{k-1}\,\mathrm{d} t = -2\sum_{k=1}^\infty \frac{1}{k^2} = - \pi^2/3.$$
Thus, we have
$I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$.
Thus, $I(\pi/6) = -\frac{13}{72}\pi^2$.
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Similar sums $\sum _{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ and $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi ^3}{32}$ Is it possible to find $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ if we know that $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$?
Any help is welcome.
Thanks in advance.
|
The short answer is no, but nevertheless they are some how related as we will shortly see below.
Lets start by the first series which is an instance of the Dirichlet eta function $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$, which is related to the Riemann Zeta function by the equation
\begin{align*}
\eta(s)=(1-2^{1-s})\zeta(s) \tag{1}
\end{align*}
In your case $s=3$, therefore, by $(1)$ we obtain
\begin{align*}
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}=\frac34\zeta(3)
\end{align*}
The second series is an instance of the Dirichlet Beta function which is defined by
\begin{align*}
\beta(s)=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^s}
\end{align*}
One relation between these two functions can be established by yet another function, the polylogarithm given by
\begin{align*}
\operatorname{Li}_s(x)=\sum_{n=1}^\infty \frac{x^n}{n^s} \tag{2}
\end{align*}
Setting $x=\sqrt{-1}=i$ in $(2)$ we obtain
\begin{align*}
\operatorname{Li}_s(i)&=\sum_{n=1}^\infty \frac{i^n}{n^s} \\
&=\frac{i}{1^s}-\frac{1}{2^s}-\frac{i}{3^s}+\frac{1}{4^s}+ \cdots\\
&=i\sum_{n=0}^\infty \frac{(-1)^s}{(2n+1)^s}+\sum_{n=1}^\infty \frac{(-1)^n}{(2n)^s}\\
&=i\beta(s)-\frac{1}{2^s}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}\\
&=i\beta(s)-2^{-s}\eta(s)\\
\end{align*}
For $s=3$ we obtain
\begin{align*}
\operatorname{Li}_3(i)&=i\beta(3)-\frac18\eta(3)\\
&=i \frac{\pi^3}{32}-\frac{3}{32}\zeta(3)
\end{align*}
One interesting fact to note is that the Riemann zeta function and consequently the Dirichlet eta function can be expressed in closed form for $n$ even, and the Dirichlet Beta function can be expressed in closed form for odd indices.
Hope this somehow helps you.
|
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|
If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align*}
b^{3} - b^{2} = a^{3} - a^{2} & \Longleftrightarrow b^{3} - a^{3} = b^{2} - a^{2}\\\\
& \Longleftrightarrow (b-a)(a^{2} + ab + b^{2}) = (b+a)(b-a)\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = b + a
\end{align*}
but what next?
|
We have:
\begin{align}
a^2+ab+b^2 = a+b &\implies 4(a+b)^2 - 4(a+b) = 4ab < (a+b)^2\\
&\implies 3(a+b)^2 - 4(a+b) < 0\\
&\implies (a+b)(3(a+b)-4) < 0\\
&\implies 3(a+b) - 4 < 0\\
&\implies (a+b) < \dfrac{4}{3}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
|
$$(x^3-3x+1)\sqrt{x^2-1}+x^4-3x^2+x+1=0$$
$$(x^3-3x+1)\sqrt{x^2-1}=-(x^4-3x^2+x+1)$$
$$(x^3-3x+1)^2(x^2-1)=(x^4-3x^2+x+1)^2$$
If you expand everything, and then subtract the two sides of the equation, you get
$$x^6-4x^4+2x^3+3x^2-4x+2 = 0$$
By the Rational Root Theorem, potential rational roots are $\pm 1$ and $\pm 2$, of which it turns out that $x = 1$ is the only one that works. Divide the polynomial by $x - 1$.
$$x^5+x^4-3x^3-x^2+2x-2=0$$
Unfortunately, there's no Quintic Formula. And no rational roots. But maybe we can find a quadratic factor somehow.
$$(x^3+ax^2+bx+c)(x^2+dx+e)=0$$
$$x^5+(a+d)x^4+(b+ad+e)x^3+(c+bd+ae)x^2+(cd+be)x+ce=0$$
It's a bit tedious, but we can match up the coefficients and solve the system of 5 equations for 5 variables, ultimately getting
$$(x^3+x^2-x+1)(x^2-2)=0$$
The quadratic factor obviously gives the roots $x = \pm \sqrt{2}$.
The cubic, AFAICT, has no simple solution, so I solved it numerically to get $x \approx -1.8392867552141612$. However, this turns out to be an extraneous solution that does not solve the original equation. The other two roots of the cubic are complex numbers, so I assume you're not interested in them.
Therefore, $x \in \lbrace 1, \sqrt{2}, -\sqrt{2} \rbrace$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$ The following question appeared in a JEE Mock exam, held two days ago.
Question:
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$
My Attempt:
I think if each term is $\frac1{\sqrt 2}$ then it would add upto $\sqrt 2$. So, $x=1$ is a solution.
How can we tell about other solutions, if any?
|
Let $x = 2 \cos^2 \theta, \theta \in \left[0, \dfrac{\pi}{2}\right]$
Then we have $\sin \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right) +\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right) = \sqrt 2 $
Now if $\sin \theta > \cos \theta $ we have $\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right)<\cos \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)$ and hence
LHS $< \sin \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)+\cos \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right)<\sqrt 2$ (by a well known result) which contradicts equality condition.
Similarly we can argue that $\sin \theta \not < \cos \theta$. Hence we have $\sin \theta = \cos \theta = \dfrac{\pi}{4}$ and at which LHS=RHS.
Thus $\boxed{x=1}$ is the unique solution
|
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|
Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $ Let $a \in (0,1)$, I would like to integrate
$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi t^*/2)^2 = a^2.$ Now $t^* <t$ such that the integral is well-defined.
Please let me know if you have any questions.
|
With $x\to 1-x$
\begin{align}
&\int_{t}^1 \frac{\sin^2\pi x}{(\sin^2\frac{\pi x}2-a^2)^2} \ dx\\
=&\ 4\int_0^{1-t}-1
+\frac{(1-2a^2)\cos^2\frac{\pi x}2+a^4}{(\cos^2\frac{\pi x}2-a^2)^2}\ dx\\
=& \ 4(t-1)+\frac2a\frac d{da}\int_0^{1-t} \frac{a^2-a^4}{\cos^2\frac{\pi x}2-a^2}\ dx\\
=& \ 4(t-1)+\frac2a\frac d{da}\bigg(
\frac{2 a\sqrt{1-a^2}}{\pi}\tanh^{-1}\frac{a\cot\frac{\pi t}2}{\sqrt{1-a^2}}\bigg)\\
=& \ 4\bigg(t-1
-\frac{\frac1\pi\cot\frac{\pi t}2}{a^2\csc^2 \frac{\pi t}2-1}+ \frac{1-2a^2}{\pi a\sqrt{1-a^2}}
\tanh^{-1}\frac{a\cot\frac{\pi t}2}{\sqrt{1-a^2}}\bigg)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain the 5,6,7 equation line? Thank you
|
$5th$ $step$
$1-\cos2y=\sin^2y$
$6th$ $step$
$expansion\ of\ \sin = y−\frac{y^3}{3!}+\frac{y^5}{5!}−⋯∞$
|
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|
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
|
Just for the fun (it is to long for a comment)
Using @Quanto's solution
$$I=\frac18\int_0^\pi\frac{x(\pi-x)}{\sin (x)}dx$$ and the $\large 1,400^+$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
$$I \simeq \frac{1}{128}\int_0^\pi \left(5 \pi ^2-4 (\pi -x) x\right)\,dx=\frac{13 \pi ^3}{384}$$ is in a relative error of $0.2$%
|
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"timestamp": "2023-03-29T00:00:00",
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|
Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as well but not the generalized form. Here, I conjecture a generalization -
$$ I = (-1)^{n+1} \binom{-1/2}{n-1} \cdot \frac{\pi}{2 \, a^{2n-1}} $$
This can be further simplified by
$$ \binom{-1/2}{n} = \left(\frac{-1}{4}\right)^n \binom{2n}{n} $$
Is this conjecture true?
|
Letting $x=a\tan \theta$ transform the integral into a Beta function
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}
&=\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta =\frac{1}{2 a^{2 n-1}} B\left(\frac{1}{2}, n-\frac{1}{2}\right)
\end{aligned}
$$
Using the property of Beta function, we have
$$
\begin{aligned}
B\left(\frac{1}{2}, n-\frac{1}{2}\right) &= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)} \\
&= \frac{1}{(n-1)!} \Gamma\left(\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\left(n-\frac{5}{2}\right) \cdots\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)\\&= \pi\frac{(2 n-3)(2 n-5) \cdots 1}{(n-1) ! 2^{n-1}} \times \frac{(2 n-2)(2 n-4) \cdots 2}{(2 n-2)(2 n-4) \cdots 2}\\&= \pi\frac{(2 n-2) !}{(n-1) ! 2^{2 n-2}(n-1) !}\\&= \frac{\pi}{2^{2 n-2}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
\end{aligned}
$$
Now we can conclude that
$$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}= \frac{\pi}{(2a)^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$
Let $a, b, c > 0$ with $a + b + c = 3$. Prove that
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$
This question was posted recently, closed and then deleted, due to missing of contexts etc.
By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.
My strategy is to split into many cases.
WLOG, assume that $a \ge b \ge c$.
If $a \ge 8/5$, true.
If $a \le 10/7$, let
$f(x) = x^{4/x} - 1 - 4(x - 1)$.
We have $f(x) \ge 0$ for all $x \in (0, 10/7)$.
If $10/7 < a < 8/5$ and $b \ge 4/5$, true.
(I stopped here since this approach is ugly. Actually, the proof of $x^{4/x} - 1 - 4(x - 1) \ge 0$ for all $x\in (0, 10/7)$ is complicated.)
I hope to see nice proofs.
|
I show the inequality (almost algebraically speaking) for $x\in[0,4/3]$ :
$$4\left(x-1\right)+1\leq x^{\frac{4}{x}}$$
In fact we have a stronger inequality for $x\in[3/4,4/3]$:
$$\left(x^{-\frac{1}{4}}\left(1+\left(\frac{1}{x}+\frac{1}{4}\right)\left(x-1\right)\right)\right)^{4}\leq x^{\frac{4}{x}}\tag{I}$$
To show it $(I)$ we use Bernoulli's inequality simply remarking :
$$x^{\frac{4}{x}}=\left(x^{\frac{1}{x}+0.25-0.25}\right)^4$$
Now see WolframAlpha for a factorization .
Except using derivative I cannot show this sixth degree polynomials is positive for $x\in[3/4,4/3]$
In fact we have clearly for $1\le x\le 4/3$:
$$x^{6}+30x^{5}-687x^{4}+400x^{3}+1632x^{2}-1280x+256>x^{4}+30x^{4}-687x^{4}+400x^{2}+1632x^{2}-1280x+256$$
We can use the second derivative or substitute by $y=x^2$ and factorize again a part.
For $3/4\le x\leq 1$ the sixth degree polynomials is increasing via derivative .
|
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|
Limit of $n^2$ and a recurrence relation with ceiling function For all positive integer $n$ we define a finite sequence in the following way:
$n_0 = n$, then $n_1\geq n_0$ and has the property that $n_1$ is a multiple of $n_0-1$ such that the difference $n_1 - n_0$ is minimal among all multiple of $n_0 -1 $ that are bigger than $n_0$. More generally $n_k \geq n_{k-1}$ and has the property that is a multiple of $n_0-k$ such that the difference $n_k - n_{k-1}$ is minimal among all multiple of $n_0 -k $ that are bigger than $n_{k-1}$. We stop the procedure when $k=n-1$ so that we define $f$ to be $f(n_0) = n_{n-1} $.
Question: What is the value of $\lim_{n \to \infty} \frac{n^2}{f(n)} $ ?
I'm able to prove that $$ \frac{8}{3} \leq \lim_{n \to \infty} \frac{n^2}{f(n)} \leq 4 $$
but I can't do better. Someone has an idea? I think that the limit is $ \pi $ but don't know how to prove that.
My idea is to prove that
$$ f(n) = \frac{n^2}{\pi} + O(n) $$
the general term for $n_k$ is given by
$$ n_k = (n-k) \left \lceil \frac{n_{k-1}}{n-k} \right \rceil $$
So that
$$ n_k = (n-k) \left \lceil \frac{n-(k-1)}{n-k} \left \lceil \frac{n-(k-2)}{n-(k-1)} \left \lceil \ldots \left \lceil \frac{n-1}{n-2} \left \lceil \frac{n}{n-1} \right \rceil \right \rceil \right \rceil \right \rceil \right \rceil $$
|
Under the substitution
$$t_k = \frac kn, \quad a_k = \frac{n_k}{n(n - k)}, \quad Δt = \frac1n,$$
we have
$$t_0 = 0, \quad a_0 = Δt, \quad t_k = t_{k - 1} + Δt, \quad a_k = a_{k - 1} + \left\lceil\frac{a_{k - 1}}{1 - t_k}\right\rceil Δt,$$
which approximates* the differential equation
$$a(0) = 0, \quad a'(t) = \left\lfloor\frac{a(t)}{1 - t} + 1\right\rfloor.$$
The solution to this differential equation is piecewise linear; the piece with slope $i$ goes from $a(1 - b_{i - 1}) = (i - 1)b_{i - 1}$ to $a(1 - b_i) = ib_i$ where
$$b_0 = 1, \quad b_i = \frac{2i - 1}{2i}b_{i - 1}.$$
Observe that $\frac{1}{b_ia(1 - b_i)}$ is exactly the Wallis product that tends to $π$.
\begin{align*}
\frac{1}{b_ia(1 - b_i)} = \frac1{ib_i^2} &= \frac 21 ⋅ \frac21 ⋅ \frac43 ⋅ \frac43 ⋅ \frac65 ⋅ \frac65 ⋯ \frac{2i}{2i - 1} ⋅ \frac{2i}{2i - 1} ⋅ \frac1i \\
&= \frac21 ⋅ \frac23 ⋅ \frac43 ⋅ \frac45 ⋅ \frac65 ⋅ \frac57 ⋯ \frac{2i}{2i - 1} \cdot 2 → π.
\end{align*}
Therefore, at $b_i ≈ \frac1n$, we have
$$\frac{n^2}{n_{n - 1}} = \frac1{\frac1n a_{n - 1}} ≈ \frac{1}{\frac1n a{\left(1 - \frac1n\right)}} → π.$$
(* The approximation differs slightly from the forward Euler method in that the $1 - t_{k - 1}$ in the denominator is replaced by $1 - t_k$. With some rearrangement, in fact, we see that it’s extremely close to the implicit midpoint method:
\begin{align*}
a_k &= a_{k - 1} + \left\lceil\frac{a_{k - 1}}{1 - t_k}\right\rceil Δt = a_{k - 1} + \left\lceil\frac{a_{k - 1} + \frac{a_{k - 1}}{1 - t_k} ⋅ \frac{Δt}{2}}{1 - \frac{t_{k - 1} + t_k}{2}}\right\rceil Δt \\
&≈ a_{k - 1} + \left\lceil\frac{a_{k - 1} + \left\lceil\frac{a_{k - 1}}{1 - t_k}\right\rceil ⋅ \frac{Δt}{2}}{1 - \frac{t_{k - 1} + t_k}{2}}\right\rceil Δt = a_{k - 1} + \left\lceil\frac{\frac{a_{k - 1} + a_k}{2}}{1 - \frac{t_{k - 1} + t_k}{2}}\right\rceil Δt.
\end{align*}
The implicit midpoint method solves the related differential equation $a'(t) = \frac{a(t)}{1 - t} + c$ exactly(!), so we can expect some kind of $O(Δt^2)$ bound on the error, but there are some details to be checked here if we want it to be rigorous.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\varepsilon-\delta$ proof of this multivariable limit We've to prove that
$$
\lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0
$$
Kindly check if my proof below is correct.
Proof
We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that
$$
\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqrt{x^2+y^2}< \delta
$$
Start
$$
\left| \frac{x^3}{x^2+y^2} \right| <\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon
$$
Note
$$
\left| \frac{x^3}{x^2+y^2} \right|= \frac{x^2|x|}{x^2+y^2}>\frac{(x^2-y^2)|x|}{x^2+y^2}
$$
Therefore
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\varepsilon \tag{1}
$$
Note
$$
\frac{x^2-y^2}{x^2+y^2}|x|<|x|=\sqrt{x^2}<\sqrt{x^2+y^2}<\delta
$$
So
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\delta \tag{2}
$$
From $(1)$ and $(2)$, we can say
$$
\delta=\varepsilon
$$
|
That is not correct. You are supposed to find, for each $\varepsilon>0$, some $\delta>0$ such that$$\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^4}{x^2+y^2}\right|<\varepsilon,$$and that's not what you did.
Note that$$\left|\frac{x^3}{x^2+y^2}\right|=|x|\frac{x^2}{x^2+y^2}\leqslant|x|$$and that$$\left|\frac{y^4}{x^2+y^2}\right|=y^2\frac{y^2}{x^2+y^2}\leqslant y^2\leqslant|y|$$if $y\in[-1,1]$. So, take $\delta=\min\left\{1,\frac\varepsilon2\right\}$. Then, if $\sqrt{x^2+y^2}<\delta$, then $y\in[-1,1]$ and $|x|,|y|<\delta\leqslant\frac\varepsilon2$, and therefore$$\left|\frac{x^3+y^2}{x^2+y^2}\right|\leqslant|x|+|y|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$
|
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|
Is it possible for Riemann Sum and Standard Integration to have different answers? I have a specific question for this equation :
$$\frac{1}{x+1}
$$
Using Standard Integration,
$$\int_{0}^{1} \frac{1}{x+1}
$$
which is approximately 0.69
Using Riemann Sum (right end point), however, I get this
$$ \lim _{n \to \infty } \Sigma ^n _{i=1} \frac{1}{n+i}
$$
$$ = \lim _{n \to \infty } ( \frac {1} {n+1} + \frac {1} {n+2} + \frac {1} {n+3} + ... +\frac {1} {2n} )
$$
$$ = \lim _{n \to \infty } (\frac{1}{n})( \frac {1} {1+\frac{1}{n}} + \frac {1} {1+\frac{2}{n}} + \frac {1} {1+\frac{3}{n}} + ... +\frac {1} {2} )
$$
$$= (\frac{1}{\infty})( \frac {1} {1+0} + \frac {1} {1+0} + \frac {1} {1+0} + ... +\frac {1} {2} )
$$
$$= 0
$$
Is this right?
|
Yes, it's possible to show both expressions equal $\ln(2)$. Just like how others pointed out, we can't just "plug in $\infty$" all the time and get the right answer because $\infty$ is not a real number. To show our Riemann Sum converges to $\ln{(2)}$, we do
$$\eqalign{
\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n+i} &= \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n\left(1+\frac{i}{n}\right)} \cr
&= \lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{1}{1+\frac{i}{n}}\right)\left(\frac{1-0}{n}\right).
}$$
Recall the definition of Riemann Sums on $\left[a,b\right]$ using $n$ subintervals that $x_i = a+i\Delta x$ and $\Delta x = \frac{b-a}{n}$. Let $b=1$ and $a=0$ so that $\Delta x = \frac{1-0}{n}$ and $x_i = 0 + i\left(\frac{1-0}{n}\right) = \frac{i}{n}$. Then we get the sum to equal
$$\int_0^1 \frac{1}{1+x}dx,$$
which equals $\ln{(2)}$.
Does that make sense?
|
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|
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$
The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$
which is easy to prove by induction.
As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$
I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem?
Can anyone point me in the right direction for further study?
|
Starting the summations at $k=1$ and using Pochhammer symbols
$$\prod_{n=0}^i(k+n)=k (k+1)_i$$
$$\sum_{k=1}^N\frac 1{k (k+1)_i}=\frac{1}{i \,\Gamma (i+1)}-\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}$$ Now, using Stirling approximation for large values of $N$
$$\log\Bigg[\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\Bigg]=-(i \log (N)+\log (i))-\frac{i (i+1)}{2 N}+O\left(\frac{1}{N^2}\right)$$
$$\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\sim\frac{N^{-i}}{i}\,e^{-\frac{i (i+1)}{2 N}}$$
Edit
Going from $n$ to $2n$ as asked in comments is much more difficult since the result invoke the Gaussian hypergeometric functions.
$$\prod_{n=0}^i(k+2n)=2^i k \left(\frac{k}{2}+1\right)_i$$ and the summation up to $N$ gives
$$S_i=\frac{\sqrt{\pi } 2^{-i-1} \,
_2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma
\left(i+\frac{3}{2}\right)}+\frac{2^{-i-1} \, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}-$$
$$\frac{2^{-i-1} \Gamma \left(\frac{N}{2}+\frac{1}{2}\right) \,
_2F_1\left(1,\frac{N+1}{2};\frac{1}{2} (2 i+N+3);1\right)}{\Gamma
\left(i+\frac{N}{2}+\frac{3}{2}\right)}-$$ $$\frac{2^{-i-1} \Gamma
\left(\frac{N}{2}+1\right) \,
_2F_1\left(1,\frac{N+2}{2};i+\frac{N}{2}+2;1\right)}{\Gamma
\left(i+\frac{N}{2}+2\right)}$$ Once expanded, they write in a simple manner in terms of the gamma function.
$$S_1=\frac{1}{4} \left(3-\frac{(N+4) \Gamma \left(\frac{N}{2}+1\right)}{2 \Gamma
\left(\frac{N}{2}+3\right)}-\frac{(N+3) \Gamma \left(\frac{N+1}{2}\right)}{2
\Gamma \left(\frac{N}{2}+\frac{5}{2}\right)}\right)$$
$$S_2=\frac{1}{8} \left(\frac{11}{12}-\frac{(N+6) \Gamma \left(\frac{N}{2}+1\right)}{4 \Gamma
\left(\frac{N}{2}+4\right)}-\frac{(N+5) \Gamma \left(\frac{N+1}{2}\right)}{4
\Gamma \left(\frac{N}{2}+\frac{7}{2}\right)}\right)$$
$$S_3=\frac{1}{16} \left(\frac{7}{30}-\frac{(N+8) \Gamma \left(\frac{N}{2}+1\right)}{6 \Gamma
\left(\frac{N}{2}+5\right)}-\frac{(N+7) \Gamma \left(\frac{N+1}{2}\right)}{6
\Gamma \left(\frac{N}{2}+\frac{9}{2}\right)}\right)$$
$$S_4=\frac{1}{32} \left(\frac{163}{3360}-\frac{(N+10) \Gamma \left(\frac{N}{2}+1\right)}{8 \Gamma
\left(\frac{N}{2}+6\right)}-\frac{(N+9) \Gamma \left(\frac{N+1}{2}\right)}{8
\Gamma \left(\frac{N}{2}+\frac{11}{2}\right)}\right)$$ where you can see the simple patterns.
Expanding the gamme functions leads to the expressions you wrote.
When $N\to \infty$, the constant term is
$$T_i=2^{-i-1} \left(\frac{\sqrt{\pi } \,
_2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma
\left(i+\frac{3}{2}\right)}+\frac{\, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}\right)$$ that is to say
$$T_i=\frac 1{2^{i+1}\,i ^2} \left(\frac{i\sqrt{\pi } }{\Gamma \left(i+\frac{1}{2}\right)}+\frac{1}{\Gamma (i)}\right)=\frac 1{2^{i+1}}\frac{1}{i \,\Gamma (i+1)}\left(1+\frac{\sqrt{\pi }\, \Gamma (i+1)}{\Gamma \left(i+\frac{1}{2}\right)}\right)$$ and then
$$S_i\sim T_i-\frac {N^{-i}} i e^{-\frac{i (2 i+1)}{2 N} }$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Proving that a quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent. I would like a proof to a Theorem (via OnlineMathLearning.com) I found:
A quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent.
And more specifically, Wikipedia's "Isosceles trapezoid" entry says:
(An isosceles trapezoid is a) trapezoid in which both legs and both base angles are of equal measure.
If necessary, assume that the diagonals bisect the base angles.
I can bring more sources that simply state the properties of an isosceles trapezoid that corroborate the above claims, but no proofs.
I am really struggling finding any proof for the definition of an isosceles trapezoid (and especially for a proof that only starts with the assumption that the diagonals of the quadrilateral are congruent and/or that they bisect the base angles).
|
Label the trapezoid as the shown below. The base of the quadrilateral is $\overline{AD}$. The lengths of the sides $\overline{AB},\overline{BC},\overline{CD},\overline{AD}$ are respectively $b,c,d$, and $a$. The diagonals $\overline{AC}$ and $\overline{BD}$ both have length $r$. Angles $\angle BAC$ and $\angle CAD$ are both $\alpha$, while $\angle ADB$ and $\angle BDC$ are both $\beta$.
Now because of how we define the "inside" of the quadrilateral, it must be the case that $2\alpha<\pi$ and $2\beta<\pi$. Furthermore, since $\overline{AB}$ and $\overline{BC}$ must intersect, $2\alpha+\beta<\pi$. Likewise, $\alpha+2\beta<\pi$.
This valid region for $\alpha$ and $\beta$ is shown below, excluding the boundary.
Note that because of this range of allowable values of $\alpha$ and $\beta$, the sines and cosines of these angles are positive.
Now, by equating expressions for the positions of points $B$ and $C$, we get the following equations.
$$\begin{align*}
r\cos\alpha&=a-d\cos2\beta & &(1)\\
b\cos2\alpha&=a-r\cos\beta & &(2)\\
r\sin\alpha&=d\sin2\beta & &(3)\\
a\sin2\alpha&=r\sin\beta & &(4)\\
\end{align*}$$
Rearranging $(1)$
$$a-r\cos\alpha=d\cos2\beta$$
Dividing by $(3)$
$$\frac{a-r\cos\alpha}{r\sin\alpha}=\frac{d\cos2\beta}{d\sin2\beta}$$
Solving for $r$
$$a-r\cos\alpha=r\sin\alpha\cot2\beta$$
$$r=\frac{a}{\cos\alpha+\sin\alpha\cot2\beta}$$
Solving $(2)$ and $(4)$ for $r$
$$r=\frac{a}{\cos\beta+\sin\beta\cot2\alpha}$$
Combining these equations
$$\cos\alpha+\sin\alpha\cot2\beta=\cos\beta+\sin\beta\cot2\alpha$$
Multiplying by $\sin2\alpha\sin2\beta$
$$\sin2\alpha\sin2\beta\cos\alpha+\sin2\alpha\cos2\beta\sin\alpha\\=\sin2\alpha\sin2\beta\cos\beta+\cos2\alpha\sin2\beta\sin\beta$$
Applying double angle theorems
$$4\sin\alpha\cos^2\alpha\sin\beta\cos\beta+(1-2\sin^2\beta)2\sin^2\alpha\cos\alpha\\=4\sin\alpha\cos\alpha\sin\beta\cos^2\beta+(1-2\sin^2\alpha)2\sin^2\beta\cos\beta$$
Distributing and factoring
$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(1-\cos^2\alpha)\cos\alpha-2(1-\cos^2\beta)\cos\beta=0$$
$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(\cos\alpha-\cos\beta)-2(\cos^3\alpha-\cos^3\beta)=0$$
$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2\\-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))(\cos\alpha-\cos\beta)=0$$
Now
$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))<0$$
for all $\alpha$ and $\beta$ in the legal range†. (It is exactly zero at $\alpha=0,\beta=\frac{\pi}{2}$ and vice versa.) This means that
$$\cos\alpha-\cos\beta=0$$
$$\Rightarrow \alpha=\beta$$
Now clearly the base angles $2\alpha$ and $2\beta$ are equal. By SAS congruence, we get that $\triangle ABD$ and $\triangle DCA$ are congruent. Therefore the base side lengths $b$ and $d$ are equal.
†Verified graphically. A full proof would likely include function maximization to show that the expression does not exceed $0$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculate the sum $\sum_{n=0}^{\infty }\frac{1}{n!!}$ \begin{align*}
S =\sum_{n=0}^{\infty }\frac{1}{n!!}=\sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{1}{(2n+1)!!} \right )
&= \sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{(2n)!!}{(2n+1)!} \right )=\sum_{n=0}^{\infty }\left ( \frac{1}{2^nn!}+\frac{2^nn!}{(2n+1)!} \right )\\
& =e^{1/2}+\sum_{n=0}^{\infty }\frac{2^nn!}{(2n+1)!}; \; \sum_{n=0}^{\infty }\frac{n!x^{2n+1}}{(2n+1)!}=F(x)\\
&\Rightarrow \frac{\mathrm{d} F}{\mathrm{d} x}=\sum_{n=0}^{\infty }\frac{n!x^{2n}}{(2n)!}\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\left [ F(x)\exp \left \{ -\frac{x^2}{4} \right \} \right ]\\
&=\left ( \frac{\mathrm{d} F}{\mathrm{d} x}-\frac{1}{2}xF(x) \right )\exp \left \{ -\frac{x^2}{4} \right \}\\
& =\frac{1}{2}\left ( 2-x^2+\sum_{n=1}^{\infty }\left ( \frac{(n-1)!x^{2n}}{(2n-1)!}-\frac{n!x^{2n+2}}{(2n+1)!} \right ) \right )\exp \left \{ -\frac{x^2}{4} \right \}\\
& =\exp\left \{ -\frac{x^2}{4} \right \}\Rightarrow F(x)=\exp \left \{ \frac{x^2}{4} \right \}\int\limits_{0}^{x}\exp \left \{ -\frac{\xi ^2}{4} \right \}d\xi
\end{align*}
$$\left \{ \int\limits_{\mathbb{R}}e^{-x^2}dx \right \}^2=\iint\limits_{\mathbb{R}}e^{-x^2-y^2}dxdy=\int\limits_{0}^{2\pi}\int\limits_{0}^{\infty }e^{-r^2}rdrd\varphi =\frac{1}{2}\int\limits_{0}^{2\pi}\left ( 1-\lim_{r\rightarrow \infty } e^{-r^2}\right )d\varphi =\pi$$
\begin{align*}
\int\limits_{0}^{\infty }e^{-\xi ^2}d\xi =\frac{\sqrt{\pi}}{2}; \; \lim_{x\rightarrow \infty }\mathrm{erf}x=1
&\Rightarrow \int\limits_{0}^{x}\exp \left \{ -\frac{\xi ^2}{4} \right \}d\xi =\sqrt{\pi}\mathrm{erf}\frac{x}{2}\\
& \Rightarrow F(x)=\sqrt{\pi}\exp \left \{ \frac{x^2}{4} \right \}\mathrm{erf}\frac{x}{2}
\end{align*}
$$\sum_{n=0}^{\infty }\frac{1}{n!!}=\exp \left \{ \frac{1}{2} \right \}+\sum_{n=0}^{\infty }\frac{2^nn!}{(2n+1)!}=\sqrt{e}+\sqrt{\frac{\pi e}{2}}\mathrm{erf}\frac{1}{\sqrt{2}} $$
I showed my attempt at a solution above. This is the standard method of calculating the sum. The question is, is it possible to calculate this sum through a contour integral?
|
This is not an answer but it is too long for a comment.
Without integration, the partial sums can be computed (thanks to Mathematica).
$$S_m=\sum_{n=0}^{m }\frac{1}{n!!}$$
$$S_{2p}=\sqrt{\frac{e \pi }{2}} \,\frac{\Gamma
\left(p+\frac{1}{2},\frac{1}{2}\right)}{\Gamma
\left(p+\frac{1}{2}\right)}+\sqrt{e}\,\frac{ \Gamma
\left(p+1,\frac{1}{2}\right)}{\Gamma (p+1)}-\sqrt{\frac{e \pi }{2}}
\text{erfc}\left(\frac{1}{\sqrt{2}}\right)$$
$$S_{2p+1}=S_{2p}+\frac{\sqrt{\pi } }{2^{p+1}\,\Gamma \left(p+\frac{3}{2}\right)}$$ All of the above leads to
$$S_\infty=\sqrt e \left(1+\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{1}{\sqrt{2}}\right)\right)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4508620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\int_{\frac{\pi}{4}}^{0} \frac{\sin \left(\frac{\pi}{4}-y\right)+\cos \left(\frac{\pi}{4}-y\right)}{9+16 \sin 2\left[(\frac{\pi}{4}-y)\right]}(-d y) \\
\displaystyle &=\int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\sqrt{2}}(\cos y-\sin y)+\frac{1}{\sqrt{2}}(\cos y+\sin y)}{9+16 \cos 2 y} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\cos y}{9+16\left(1-2 \sin ^{2} y\right)} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{1}{\sqrt 2} } \frac{d z}{25-32 z^{2}} \text { , where } z=\sin y\\
\displaystyle &=\frac{\sqrt{2}}{10} \int_{0}^{\frac{1}{\sqrt 2} }\left(\frac{1}{5-4 \sqrt{2}z}+\frac{1}{5+4 \sqrt{2} z}\right) d z \\
\displaystyle &=\frac{\sqrt{2}}{10(4 \sqrt{2})}\left[\ln \left|\frac{5+4 \sqrt{2} z}{5-4 \sqrt{2} z}\right|\right]_{0}^{\frac{1}{\sqrt 2} } \\
\displaystyle &=\frac{1}{40}\ln 9 \end{aligned}$$
|
Same solution, made slightly more general.
Let $a,b,c$ be real numbers such that
$$
c^2 = a^2 + b^2\ , \qquad a,c>0\ .
$$
Consider now the integral
$$
\begin{aligned}
J = J(a,b,c)
&:=
\int_0^{\pi/4}
\frac{\sin x+\cos x}{a^2 + b^2 \sin 2x}\; dx
=
\int_0^{\pi/4}
\frac{\sqrt 2\cos y}{a^2 + b^2 \cos 2y}\; dy
=
\int_0^{\pi/4}
\frac{\sqrt 2\cos y}{c^2 - 2b^2 \sin^2y}\; dy
\\
&=
\int_0^1
\frac1{c^2 - b^2 s^2}\; ds
=
\frac 1{2c}
\int_0^1
\left(
\frac1{c - bs} +
\frac1{c + bs}
\right)
\; ds
=\frac 1{2bc}\left[\log\frac{b+cs}{b-cs}\right]_0^1
\\
&=\frac 1{2bc}\log\frac{b+c}{b-c}
\ .
\end{aligned}
$$
The passage from $dx$ to $dy$ uses the substitution
$y=\pi/4-x$, the term in $\sin 2x$ from the denominator becomes $\cos 2y$, and in the numerator $\sin x+\cos x=\sqrt 2\cos(\pi/4-x)=\sqrt 2\cos y$.
Then we have substituted $\sqrt 2\sin y=s$.
In our case, $a,b,c$ are $3,4,5$. So the answer is
$$
\frac 1{2\cdot4\cdot5}\log\frac{5+4}{5-4}
=\frac 1{40}\log 9
=\frac 1{20}\log 3
\ .
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
prove that this expression equals the area of a specific triangle Prove that the area of a triangle with one of its angles $\frac{\pi}4$ rad, and the side opposite to this angle is $2$cm equals $\sin(2\theta)-\cos(2\theta)+1$, where $\theta$ is the angle adjacent to the $2$cm side.
and here is a picture to make things cleaner:
|
Let $a = 2, A = \frac {\pi} 4, B = \theta$
Then,
$$
\frac{b}{\sin \theta} = \frac{2}{\frac{\sqrt{2}}2}
\implies b = 2\sqrt 2 \sin \theta
$$
Then,
$$
\sin C = \sin(A+B) = \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)
$$
Then,
$$
S = \frac 12 ab\sin C =\frac 12 \times 2\times 2\sqrt 2\sin \theta\times \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)\\
= 2\sin \theta (\sin \theta + \cos \theta)=2\sin^2\theta+2\sin \theta \cos \theta = 1-\cos 2\theta + \sin 2\theta
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4511797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
|
In this answer I tried to generalize the answer given above.
Generalization:
Let,
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$
and
$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$
Then we have
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n\\
\begin{cases} m+n=a\\mn =\frac {b}{4}\end{cases}\\
t^2-at+\frac b4=0\\
4t^2-4at+b=0\\
m=\frac {a+\sqrt{a^2-b}}{2}\\
n=\frac {a-\sqrt{a^2-b}}{2}$$
This gives,
$$\sqrt {a\pm\sqrt b}=\sqrt{\frac {a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac {a-\sqrt{a^2-b}}{2}}$$
and
\begin{aligned}A&=2\sqrt m=\sqrt {4m}\\
&=\boxed{\sqrt {2\left(a+\sqrt {a^2-b}\right)}}.\end{aligned}
$\text {QED}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Graph Transformation Knowing dealing with graph transformations come handy MANY times. I searched on google to get a comprehensive graph transformation list but couldn't find one. Some good while back I learned them all with great enthusiasm but have now all forgotten them and like before, I'm not finding a great teacher neither a great search result. Can we have one here that will cover the following (at least):
$$\text {For }y=f(x):$$
$y=f(x+c)$
$y=f(x)+c$
$y=f(cx)$
$y=cf(x)$
$y=f(-x)$
$y=-f(x)$
$y=f(|x|)$
$y=|f(x)|$
$y+c=f(x)$
$cy=f(x)$
$-y=f(x)$
$|y|=f(x)$
Kindly feel to provide for more in case I have missed out any thing. Thank you :)
Note: Last few equations I know might be a repeat of the above cases but I have included them so as to clear any confusions.
|
\begin{array}{|c|c|c|c|}
\hline
\text{New function}(c>0,\lambda>1)& \text{Change of coordinate}& \text{Geometry of transformation}
\\ \hline
f(x) +c & (a, b) \mapsto(a,b+c)& \text{ shift up by} c\\ \hline
f(x) -c & (a, b) \mapsto(a,b-c)&\text{shift down by} c\\ \hline
\lambda f(x) & (a, b) \mapsto(a,\lambda b)&\text{vertical expansion by } \lambda\\ \hline \frac{1}{\lambda}f(x) & (a, b) \mapsto(a, \frac{1}{\lambda}b)&\text{vertical contraction by } \lambda \\\hline -f(x) &(a, b) \mapsto (a, -b) &\text{reflexion about $x$-axis}\\ \hline
\end{array}
Example:
$\boxed{\color{red}{f(x) =x^2}}$
$\boxed{\color{green}{f(x) =x^2+2}}(\text{ shifted upward $2$ units}) $
$\boxed{\color{black}{f(x)=x^2-2}}(\text{ shifted downward $2$ units}) $
$\boxed{\color{goldenrod}{f(x) =3x^2}}(\text{ vertically expanded $3$ units})
$
$\boxed{\color{blue}{f(x) =\frac{1}{3}x^2}} (\text{vertically contracted $\frac{1}{3}$ units}) $
$\boxed{\color{cyan}{f(x) =-x^2}} (\text{reflected about $x$- axis}) $
\begin{array}{|c|c|c|c|}
\hline
\text{New function}(c>0,\lambda>1)& \text{Change of coordinate}& \text{Geometry of transformation}
\\ \hline
f(x+c) & (a, b) \mapsto(a-c,b)& \text{ shift left by} c\\ \hline
f(x-c) & (a, b) \mapsto(a+c,b)&\text{shift right by} c\\ \hline
f(\lambda x) & (a, b) \mapsto(\frac{1}{\lambda}a,b)&\text{horizontal contraction by } \frac{1}{\lambda} \\ \hline f(\frac{1}{\lambda}x) & (a, b) \mapsto(\lambda a,b)&\text{horizontal expansion by } \lambda \\\hline f(-x) &(a, b) \mapsto (-a, b) &\text{reflexion about $y$-axis}\\\hline
\end{array}
Example:
$\boxed{\color{red}{f(x) =x^2}}$
$\boxed{\color{green}{f(x) =(x+2)^2}}(\text{ shifted left $2$ units}) $
$\boxed{\color{black}{f(x)=(x-2)^2}}(\text{ shifted right $2$ units}) $
$\boxed{\color{goldenrod}{f(x) =(3x)^2}}(\text{horizontally contracted $\frac{1}{3}$ units})$
$\boxed{\color{blue}{f(x) =(\frac{1}{3}x)^2}} (\text{horizontally expanded ${3}$ units}) $
$\boxed{\color{cyan}{f(x) =(-x) ^2}} (\text{reflected about $y$- axis}) $
$|f|=|\space |\circ f$
$|f(x)| =\begin{cases}f(x)& f(x) \ge 0\\-f(x)& f(x) <0\end{cases}$
Hence $|f|$ is same as $f$ on the upper half plane and the lower half plane $f$ is $-f$ .
In other words $|f|=\max\{f, -f\}$
To draw the graph of $|f|$ , first draw the graph of $f$ and the leave the portion of the graph as it is on the upper half plane and then reflect the portion of the graph of the lower half plane with respect to the $x$-axis.
$f(|x|) =(f\circ|\space |)(x) $
$f(|x|) =\begin{cases}f(x)& x\ge 0\\f(-x) & x <0\end{cases}$
Hence $f(|x|)$ is same as $f(x)$ on the right half plane and on the left half plane $f(|x|)$ is the reflection of $f(x)$ about $y$-axis.
To draw the graph of $f\circ |x|$ , first draw the graph of $f$ and the leave the portion of the graph as it is on the right half plane and then reflect the portion of the graph in the left half plane with respect to the $y$-axis.
Note: For a even function $f$, since $f(-x) =f(x) $ then $f(|x|) =f(x) $.Hence we won't see the difference between the graph of the function $f$ and $f\circ |\space |$.
Example:
$\boxed{\color{red}{f(x) =-x^3}}$
$\boxed{\color{blue}{|f(x)| =|-x^3|}}$
$\boxed{\color{green}{|f(x)| =-|x|^3}}$
$\boxed{\color{black}{|y|=f(x)}}$
$y =\begin{cases}f(x)& y\ge 0\\-f(x) & y <0\end{cases}$
Hence the graph of the function $|y|=f(x) $ is same as the graph of $y=f(x)$ on the upper half plane and on the lower half plane the graph of $|y|=f(x) $ is same as the graph of $y=-f(x) $.
To draw the graph of $|y|=f(x) $, first draw the graph of $y=f(x)$.Then left the graph on the upper half plane as it is and then reflect the graph of the upper half plane about $x$-axis.
Example:
$\boxed{\color{red}{f(x) =4-x^2}}$
The graph of $y=f(x)$
$\boxed{\color{green}{|y|=f(x)}}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4515138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find the plane through the points $(1,1,-1)$, $(2,0,2)$ and $(0,-2,1)$.
Answer:
The general form of a plane is:
$$ Ax + By + Cz = D$$
Sometimes the following constrain is added:
$$ A^2 + B^2 + C^2 = 1$$
By inspection, we can see this plane is not parallel to the x-axis, the y-axis or the z-axis. Hence,
we can assume that the plane is of the form:
$$ Ax + By + Cz = 1 $$
Now we setup the following system of linear equations.
\begin{align*}
A + B - C &= 1 \\
2A + 2C &= 1 \\
-2B + C &= 1 \\
\end{align*}
To solve this system of equations, we get rid of $A$ and $B$ in the first equation.
\begin{align*}
2A &= 1 - 2C \\
A &= \frac{ 1 - 2C }{2} \\
-2B &= 1 - C \\
B &= \frac{ C - 1 }{2} \\
\left( \frac{ 1 - 2C }{2} \right) + \left( \frac{ C - 1 }{2} \right) - C &= 1 \\
1 - 2C + C - 1 - 2C &= 2 \\
- 2C + C - 2C &= 2 \\
-3C &= 2 \\
C &= -\frac{2}{3} \\
B &= \frac{ -\frac{2}{3} - 1 }{2} = -\frac{2}{6} - \frac{1}{2} \\
B &= -\frac{5}{6} \\
A &= \frac{ 1 - 2\left( -\frac{2}{3} \right) }{2} = \dfrac{1 + \dfrac{4}{3} }{2} \\
A &= \dfrac{7}{6}
\end{align*}
Hence the equation is:
$$ \left( \dfrac{7}{6} \right) A + \left( -\frac{5}{6} \right) B + \left( -\frac{2}{3} \right) C = 1 $$
Clearing the fraction, we get the final answer of:
$$ 7A - 5B - 4C = 6 $$
As pointed out by Paul, the correct answer is:
$$ 7x - 5y - 4z = 6 $$
|
A plane can be defined by its normal $n$ and a point $p$ of the plane. The equation of the plane is, for any point $x$ in the plane, $x\cdot n=d$, where $d=n\cdot p$.
If we have three points in the plane, $a,b,c$ then $a.n=b.n=c.n=d$. So we can solve for $n$ by forming a matrix $$Q=\begin{pmatrix}a&b&c\end{pmatrix}^T$$
and the three equations $$Q n=\begin{pmatrix}d&d&d\end{pmatrix}^T$$ must all be true if $a,b,$ and $c$ are in the plane, then we can get $n$ by inverting the matrix, $n=A^{-1}\begin{pmatrix}d&d&d\end{pmatrix}^T$.
In the terminology of your answer, $n\equiv{1\over d}(A,B,C)^T$, so what you did was to find the normal to the plane by inverting the above matrix.
This method works except for the case that the plane goes through the origin, where $d=0$ and we cannot use the inverse.
|
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|
Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$ Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$
I tried adding and subtracting both equations, but didn't get anywhere. Would appreciated any ideas. Thanks!
|
Adding both equation gives:
$$a^4+b^4+8(a+b)=4(a^3+b^3)-8$$
i.e: $$(a^2-2a)^2-4(a^2-2a)+b^4-4b^3+8b+8=0$$
Consider this equation is a quadratic equation with variable $a^2-2a$
So the discriminant is:$(-2)^2-(b^4-4b^3+8b+8)=-b^4+4b^3-8b-4=-(b^2-2b-2)^2$
Then:$b^2-2b-2=0$, so $b=1-\sqrt{3}$ or $b=1+\sqrt{3}$ then you can easily find $a=1+\sqrt{3}$ and $b=1-\sqrt{3}$ is only solution which is satisfied. Then $a^4+b^4=56$
|
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|
How to find this indefinite integral? $\int\frac{1+x^4}{(1-x^4)\cdot \sqrt{1+x^4}}dx$ I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.
|
With parts of the calculation left as an exercise:
The surd-in-the-denominator expression I commented on is actually a subtle hint. Rewriting the numerator as$$1+x^4=\frac12(1-x^2)^2+\frac12(1+x^2)^2,$$we want to separately evaluate$$\frac12\int\frac{(1-x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1-x^2)}{(1+x^2)\sqrt{1+x^4}}dx$$and$$\frac12\int\frac{(1+x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx.$$Write the first part as$$\frac{1}{2\sqrt{2}}\int\frac{dy}{\sqrt{1-y^2}}=\frac{1}{2\sqrt{2}}\arcsin y+C_1$$with $y:=\frac{x\sqrt{2}}{1+x^2}$. Similarly, with $z:=\frac{x\sqrt{2}}{\sqrt{1+x^4}}$ the second part is$$\frac{1}{2\sqrt{2}}\int\frac{dz}{1-z^2}=\frac{1}{2\sqrt{2}}\operatorname{artanh}z+C_2.$$ Adding these, subsume the integration constants into one viz. $C=C_1+C_2$.
|
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|
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to definitively say that the statement is proved.
This is my "work" so far:
$a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$
$\frac {2a(b + 2a) + 2b(a + 2b) - 1(b + 2a)(a + 2b)}{2(a + 2b)(b+2a)} ≥ 0$
$\frac {4ab + 4a^2 + 4b^2 - 2b^2 - 5ab - 2a^2}{4b^2 + 10ab + 4a^2} ≥ 0$
$\frac {2a^2 - ab + 2b^2}{4b^2 + 10ab + 4a^2} ≥ 0$
|
You almost have it, just use
$$2a^2-ab+2b^2 = 2(a-b)^2+3ab > 0$$
(and the denominator is positive as well).
|
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|
Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive relationship of Bernoulli numbers and induction, but unsuccessful. Now I'm wondering if this is a viable proof strategy. Any comments are appreciated!
Edit: This is a conjecture, I evaluated it numerically for $n,m\leq 20$.
|
The instructive derivation of @RenéGy reduces OPs problem to show the nice symmetric Bernoulli number identity:
\begin{align*}
\color{blue}{(-1)^n \sum_{g=0}^m}&\color{blue}{ \binom{m}{g}\frac{B_{n+g+1}}{n+g+1}
+(-1)^m \sum_{g=0}^n \binom{n}{g}\frac{B_{m+g+1}}{m+g+1}}\\
&\color{blue}{= - \frac{1}{(n+m+1){\binom{n+m}{m}}}}\tag{1}
\end{align*}
Here we show the identity (1) follows from Theorem 2 in Bernoulli Numbers and a New Binomial Transform Identity by H.W. Gould.
Given a sequence $(a_n)_{n\geq 0}$ we consider the sequence of binomial transforms $(b_n)_{n\geq 0}$. We so have the binomial inverse pair
\begin{align*}
b_n=\sum_{k=0}^n\binom{n}{k}a_k\qquad\mathrm{and}\qquad a_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}b_k\tag{2}
\end{align*}
Setting $a_k=B_k$, the $k$-th Bernoulli number we have thanks to a well known recursion formula of the Bernoulli numbers the binomial inverse pair
\begin{align*}
a_k=B_k\quad\mathrm{and}\quad b_n=\sum_{k=0}^n\binom{n}{k}a_k=\sum_{k=0}^n\binom{n}{k}B_k=(-1)^nB_n\qquad\qquad n\geq 0
\end{align*}
Theorem 2 in the paper states the following identity for binomial inverse pairs:
\begin{align*}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n+k+s}{s}}a_{n+k+s}
&=\sum_{k=0}^n(-1)^{n-k}\frac{\binom{n}{k}}{\binom{m+k+s}{s}}b_{m+k+s}\\
&\quad+\sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}\binom{s-1-j}{i}\binom{s-1}{j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i)\binom{m+n+i}{n}}
\end{align*}
Setting $s=1$ the identity reduces to
\begin{align*}
\sum_{k=0}^m\binom{m}{k}\frac{a_{n+k+1}}{n+k+1}&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{b_{m+k+1}}{m+k+1}\\
&\qquad+\frac{(-1)^{n+1}a_0}{(m+n+1)\binom{m+n}{n}}\tag{3}
\end{align*}
Putting the Bernoulli numbers (2) in (3) we obtain with $a_0=B_0=1$:
\begin{align*}
\sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1}
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{(-1)^{m+k+1}B_{m+k+1}}{m+k+1}\\
&\qquad+\frac{(-1)^{n+1}}{(m+n+1)\binom{m+n}{n}}\\
\color{blue}{(-1)^n\sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1}}
&\color{blue}{=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{m+1}B_{m+k+1}}{m+k+1}}\\
&\qquad\color{blue}{-\frac{1}{(m+n+1)\binom{m+n}{n}}}
\end{align*}
and the claim (1) follows.
|
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|
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k$, provided that $b+c=a$?
$1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$
In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$,
$$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$
Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
|
$$g(x) = (b + c)x^2 + bx + c$$
$$g'(x) = 2(b + c)x + b$$
$$g''(x) = 2(b + c)$$
For $f$ to be continuous at $k$, we must have $g(k) = g'(k)$,
$$(b + c)k^2 + bk + c = 2(b + c)k + b$$
$$(b + c)k^2 + (- b - 2c)k + (c - b) = 0$$
$$k = \frac{b + 2c \pm \sqrt{(-b-2c)^2 - 4(b+c)(c-b)}}{2(b+c)}$$
$$k = \frac{b + 2c \pm \sqrt{5b^2+4bc}}{2(b+c)}$$
For $f$ to be differentiable at $k$, we must have $g'(k) = g''(k)$.
$$2(b + c)k + b = 2(b + c)$$
$$2(b + c)k = b + 2c$$
$$k = \frac{b + 2c}{2(b + c)}$$
Equating the two expressions for $k$ gives:
$$\frac{b + 2c \pm \sqrt{5b^2+4bc}}{2(b+c)} = \frac{b + 2c}{2(b + c)}$$
$$b + 2c \pm \sqrt{5b^2+4bc} = b + 2c$$
$$\pm \sqrt{5b^2+4bc} = 0$$
$$5b^2+4bc = 0$$
$$b(5b+4c) = 0$$
$$b = 0 \text{ or } b = -\frac{4c}{5}$$
If $b = 0$, then $k = \frac{2c}{2c} = 1$.
If $b = -\frac{4c}{5}$, then $k = \frac{-\frac{4c}{5} + 2c}{2(-\frac{4c}{5} + c)} = \frac{1.2c}{0.4c} = 3$.
So $k \in \{ 1, 3 \}$. The maximum value in this set is 3.
|
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials, I don't know why, let's just do it." I don't think that people just pointed a finger at the sky and came up with that formula.
So, how to prove $(1)$?
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As noticed by $a=-b$ we obtain $a^3+b^3=0$ therefore we can guess by homogeneity
$$a^3+b^3=(a+b)(Xa^2+Yab+Zb^2)$$
with $X$, $Y$ and $Z$ unknown, to obtain
$$(a+b)(Xa^2+Yab+Zb^2)=Xa^3+Ya^2b+Zab^2+Xa^2b+Yab^2+Zb^3=$$
$$=Xa^3+(Y+X)a^2b+(Z+Y)ab^2+Zb^3$$
which requires
*
*$X=1$
*$Z=1$
*$Y+X=0 \implies Y=-1$
*$Z+Y=0 \implies Y=-1$
|
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|
Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).
The given answer is $\dfrac{79}{128}$.
|
Alternative approach:
Let $~\displaystyle x = \sin(\alpha), y = \cos(\alpha) = \implies
(x - y) = \frac{1}{2}, ~\left(x^5 - y^5\right) = \color{red}{what?}$
Tools
*
*$T_1 ~: ~(x - y)^2 = x^2 - 2xy + y^2.$
*$T_2 ~: ~(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.$
*$T_3 ~: ~(x - y)^5 = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5.$
Plan of Attack
Since $~x^2 + y^2 = 1~,$ there is sufficient information to compute $(xy)$. Using this you can compute $x^3 - y^3$, and using this you can compute $x^5 - y^5.$
From $T_1,$ you have that
$$\frac{1}{4} = 1 - 2xy \implies xy = \frac{3}{8}. \tag1 $$
Using (1), from $T_2,$ you have that
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3xy(x-y) \implies $$
$$\frac{1}{8} = \left(x^3 - y^3\right) - 3\left(\frac{3}{8}\right) \left(\frac{1}{2}\right) \implies (x^3 - y^3) = \frac{11}{16}. \tag2 $$
Using (1) and (2), from $T_3,$ you have that
$$\frac{1}{32} = \left(x^5 - y^5\right) -5(xy)(x^3 - y^3) + 10(xy)^2(x - y) \implies $$
$$\frac{1}{32} = \left(x^5 - y^5\right) - 5\left(\frac{3}{8}\right)\left(\frac{11}{16}\right) + 10\left(\frac{3}{8}\right)^2\left(\frac{1}{2}\right) \implies $$
$$\frac{4}{128} = \left(x^5 - y^5\right) - \frac{165}{128} + \frac{90}{128} \implies$$
$$\left(x^5 - y^5\right) = \frac{4 + 165 - 90}{128} = \frac{79}{128}.$$
|
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|
Real Solutions of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$ I recently stumbled across the following question:
Find all the real solutions $x$ of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$
It was given by one of my teachers (he was teaching us a few problems involving real and rational numbers).
Although I used a few typical algebraic manipulations (such as shifting terms from LHS to RHS and squaring both sides) all of them repeatedly led me towards wrong results. When I plugged the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1$ into Desmos, I came to see that all real numbers from 5 to 10 gave the value of zero when plugged into the expression. Could anyone show (through algebraic manipulations) why this is the case?
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$\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}=1$
Let $x=1+t^2$, then we re-write the eq. as
$$\sqrt{(t-2)^2}+\sqrt{(t-3)^2}=1$$
$$\implies |t-2|+|t-3|=1$$
When $2\le t \le 3$. then the equation becomes
$$ (t-2)+(3-t)=1\implies 1=1$$
So all roots are $2\le t\le 3$ \implies $5 \le x \le 10.$
Hence this equation has all numbers in $[5,10]$ as infinitely many roots.
Similarly one can check no roots in $(-\infty,5)$ and no root in $(10,\infty)$.
|
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
|
I suggest a solution different from OP; albeit simpler.
First, transform the condition $\forall x\in (0,2)$. Equivalent to this condition, introduce $x = \frac{2}{y}$ with $y >1$, so only one boundary has to be inspected and the term remains quadratic in $y$ and $a$:
The inequality $4x^2-2ax+a^2-5a+4>0$ becomes
$16-4ay+(a^2-5a+4)y^2>0$.
First note that only $a>0$ has to be inspected, since for $y > 1$ and $a \le 0$ we have
$16-4ay+(a^2-5a+4)y^2 > 16 - 9/4 + (-5/2 + a)^2 > 0$ so the condition is always obeyed for $a \le 0$.
Rewrite the question as
$$
16 - 4 a y + (a^2 - 5 a + 4) y^2 = (a y - 2 - \frac{5y}{2})^2 - \frac14 (9 y^2 + 40 y - 48) > 0 \tag{1}
$$
Since $9 y^2 + 40 y - 48 > 0$ always holds for $y >1$, we can discuss $g = (a y - 2 - \frac{5y}{2})^2$. For $a \to 0$, $g \to \frac14 (5y +4)^2 = \frac14 (16y^2 + 64) + \frac14 (9 y^2 + 40 y - 48)$ so the question's condition is always obeyed. Likewise, for very large $a$ it is clear that (1) will also be obeyed.
Conversely, for $a=4$ we have $16 - 4 a y + (a^2 - 5 a + 4) y^2 = 16(1-y)$ which will be negative for all $y >1$.
Hence the allowed regions for $a$ will be $a_1 > a $ and $a > a_2$ with $a_2 \ge 4 \ge a_1 > 0$.
What remains is to derive $a_1$ and $a_2$ from the roots of (1) and their dependence on $y$. This is standard curve discussion and we give the results.
We have for the smaller root:
$$\frac{1}{y}\Big(2 + \frac{5y}{2} - \frac12 \sqrt{9 y^2 + 40 y - 48}\Big)
\ge \frac{8}{5 + \sqrt{13}} = a_1 \simeq 0.93 $$
which holds since the minimum occurs at $ y = 5 + \sqrt{13} > 1$
We have for the larger root:
$$\frac{1}{y}\Big(2 + \frac{5y}{2} + \frac12 \sqrt{9 y^2 + 40 y - 48}\Big)
\le \frac23 (5 + \sqrt{13}) = a_2 \simeq 5.74 $$
which holds since the maximum occurs at $ y = 5 - \sqrt{13} > 1$. $\qquad \Box$
|
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|
Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$$
for all $n,m \in \mathbb{Z}$, $m>0$.
This is what I came up with. Is my proof correct?
Proof:
[see edit below]
Case 1: $m=1$ $$\left\lfloor \frac{n}{1} \right\rfloor = \lfloor n \rfloor = n$$ $$\left\lceil \frac{n-1+1}{1} \right\rceil = \lceil n \rceil = n$$
Case 2: $m>1$
If $\frac{n}{m}$ is an integer, then
$$\left\lceil \frac{n-m+1}{m} \right\rceil = \left\lceil \frac{n}{m} -1 + \frac{1}{m}\right\rceil = \frac{n}{m} -1 + \left\lceil \frac{1}{m} \right\rceil = \frac{n}{m} -1 + 1 = \frac{n}{m} = \left\lfloor \frac{n}{m} \right\rfloor$$
If $\frac{n}{m}$ is NOT an integer, then
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} -1 \right\rceil = \left\lceil \frac{n-m+1}{m} \right\rceil$$
$\blacksquare$
If it's correct but you know a simpler/better way to prove it, please include that in your answer. Thank you.
EDIT: As pointed out by user peterwhy, "Case 1: $m=1$" is simply a special case of "$\frac{n}{m}$ is an integer" and therefore is not needed; hence I have grayed it out, and we don't need to separate the $m=1$ and $m>1$ cases anymore.
|
As in your previous question on floor and ceiling, I am finding the interval of both sides with inequalities.
Both sides are defined and are integers. They are respectively within these intervals:
$$\begin{array}{rrcl}
\text{LHS:}&\dfrac nm-1 <&\left\lfloor\dfrac nm\right\rfloor&\le \dfrac nm\\\hline
\text{RHS:}&\dfrac{n-m+1}{m}\le& \left\lceil\dfrac{n-m+1}{m}\right\rceil &< \dfrac{n-m+1}{m}+1\\
\iff&\dfrac{n+1}{m}-1\le& \left\lceil\dfrac{n-m+1}{m}\right\rceil &< \dfrac{n+1}{m}\\
\end{array}$$
The $4$ endpoints are in this order:
$$\frac nm-1 < \frac{n+1}m-1 \le \frac nm < \frac{n+1}m$$
There are no possible integers strictly between $\dfrac nm-1 < \dfrac{n+1}m-1$, and none strictly between $\dfrac nm < \dfrac{n+1}m$. Then both LHS and RHS can only be the one integer within the overlapping interval:
$$\frac{n+1}m-1 \le (\text{Both LHS and RHS})\le \frac nm$$
(BTW in the question, if $\frac nm$ is not an integer, then $\left\lceil\frac nm\right\rceil = \left\lceil\frac nm + \frac 1m\right\rceil$. While intuitive, no integer exists strictly between $\frac nm < \frac{n+1}m$ is how I would explain it to myself)
|
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|
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Find value of $2x$
I worked it out as such:
$\sin^2x = \frac{2+ \sqrt{3}}{4} \implies \frac{1- \cos 2x}{2} = \frac{2+ \sqrt{3}}{4}$
$2 - 2 \cos 2x = 2 + \sqrt{3}$
$- 2 \cos 2x = + \sqrt{3}$
$ \cos 2x = \frac{- \sqrt{3}}{2}$
And $ 2x = \frac{5\pi}{6}$
I was told that the angle $2x$ has another value: $2x = 2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}$
Why is there another value and what is this value for or mean in this question context.
|
The reason why you have two solutions is because $\sin^2(x) = \sin^2(2\pi - x)$
therefore both solutions work.
|
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|
Let $p(x)$ and $q(x)$ be the predicates: $p(x):x^2-3x+2=0$ and $q(x):(x-1)^{3}(x-2)(x-3)x=0$. Show: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$. I would like to know if this is correct. Thanks in advance.
Let $p(x)$ and $q(x)$ be the predicates:
$p(x):x^2-3x+2=0$
$q(x):(x-1)^{3}(x-2)(x-3)x=0$
Show that: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
This is what I did:
$p(x):x^2-3x+2=0$
$$x^2-3x+2=0$$
$$(x-1)(x-2)=0$$
then,
$q(x):(x-1)^{3}(x-2)(x-3)x=0$
$$(x-1)^{3}(x-2)(x-3)x=0$$
$$(x-1)^{2}(x-1)(x-2)(x-3)x=0$$
$$(x-1)^{2}(0)(x-3)x=0$$
$$0=0$$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
|
I would rephrase like that:
then,
$(x-1)^{3}(x-2)(x-3)x= (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x=0$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
or
To show $q(x): (x-1)^{3}(x-2)(x-3)x=0$
$L.H.S.= (x-1)^{3}(x-2)(x-3)x = (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x = 0$
$R.H.S. = 0$
$L.H.S. = R.H.S.$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
|
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|
Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.
$I.$ Firstly, I used the simple expansion
$$\ln(x+1) = \sum_{n=1}^\infty (-1)^n\frac{x^n}{n}$$
which comes from the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I then noticed that
$$\ln(x) = \ln(1+(x-1))=\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}$$
$II$. $-f(x)=-\frac{1}{x^2}$ is the second derivative of $\ln(x)$, so that
$$\frac{d}{dx}\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}= \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2}$$
and
$$\frac{d}{dx} \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2} = \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}=-\frac{1}{x^2}$$
from what plainly follows that $$\frac{1}{x^2} = -\sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}$$
$III.$ Of course, this expansion is not in powers of $(x+2$). What I pressumed would be the logical thing to do is to repeat these steps but, instead of the using $\ln(1 + (x-1))$, using the equivalent
$$\ln(-2 + (x+2))=\ln((-2)(1+\frac{x+2}{(-2)})) = \ln(-2)+\ln(1+\frac{x+2}{(-2)})$$
I expected to be able to use this because the second term is of the form $\ln(1+ u)$ where $u$ is some function of $x$, and we know the expansion of such expression. However, $\ln(-2)$ is nonsense, since $\ln(x)$ is defined only for $\mathbb{R}^+$.
Is there an alternative, better way to do this or am I missing something? Thanks.
|
Let $u = x + 2$ so that $x = u - 2$. Then,
$$
\frac{1}{x^2} = \frac{1}{(u - 2)^2}
= \frac{1}{(2 - u)^2}
= \frac{1}{4 \bigl(1 - \tfrac12 u \bigr)^2}
= \frac{1}{4} \frac{1}{(1 - v)^2}
$$
where $v = \tfrac12 u$ or $u = 2v$.
We can derive a nice power series for $(1 - v)^{-2}$ centered at $v = 0$ by differentiating the geometric series in $v$. Try it yourself.
$$ \frac{1}{(1 - v)^2} = \frac{d}{dv} \frac{1}{1 - v} = \frac{d}{dv} \sum_{m=0}^\infty v^m = \sum_{m=0}^\infty \frac{d}{dv} v^m = \sum_{m=1}^\infty m\, v^{m-1} = \sum_{n = 0}^\infty (n+1)\, v^n,$$ where the sum is reindexed by $n = m - 1$ or $m = n+1$.
Now, back substitute $v \mapsto u \mapsto x$. Again, try to do this yourself before revealing the spoiler.
$$ \frac{1}{x^2} = \frac{1}{4} \frac{1}{(1 - v)^2} = \frac{1}{4} \sum_{n=0}^\infty (n+1)\, \bigl(\tfrac12(x + 2) \bigr )^n = \sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \, (x+2)^n$$
|
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|
Is there any natural number between $(N^2 + N)$ and $(N^2 + 1)$ that can divide $(N^2 + N)\times(N^2 + 1)$ My question is fairly simple and explained in title. I'm trying to prove that there are no natural number(s) between $(n^2+n)$ and $(n^2+1)$ that can divide $(n^2+1) \times (n^2 + n)$
[EDIT]
I was trying to solve the problem $n^4 + n^3 + n^2+n+1 = a^2$.
I said that $n^4 + n^3 + n^2+n = (a-1)\times(a+1)$. Refactoring left side, i got:
$(n^2+n)\times(n^2+1) = (a-1)+(a+1)$. It is obvious that $a-1$ and $a+1$ are the closest dividers of $a^2-1$ so i need to prove that, in natural numbers, $n^2+n$ and $n^2+1$ are the closest dividers of $a^2-1$ so i can assume that $(n^2+n) = a+1$ and $(n^2+1) = a-1$ and continue to solve the problem in proper way. Thanks for any help or hint in front.
|
A proof by contradiction. Assume there is a $n^{2}+1+k|(n^{2}+n)(n^{2}+1)$
where $1\leq\,k\,\leq\,n-1$. Then
$n^{2}+k+1|(n^{2}+n)(n^{2}+k-k+1)$ hence $n^{2}+k+1|(n^{2}+n)k$ and then
$n^{2}+k+1|(n^{2}+k+1-k-1+n)k$ hence $n^{2}+k+1|-(k+1-n)k$ and the latter
is positive ($k=n-1$ is not acceptable since it gives $n^{2}+k+1=n^{2}+n$)
which implies that
$n^{2}+1+k\,\leq\,-k^{2}-k+nk$ and hence
$(n-\dfrac{k}{2})^{2}+3\dfrac{k^{2}}{4}+2k+1\,\leq\,0$
which is clearly a contradiction! So there is no such $k$ and the result is proved!
|
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|
prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$
Let $S$ be the set of all $3\times 3$ matrices with entries in $\{0,1,-1\}$. Prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$.
Let $R = [-4,4]\cap \mathbb{Z}$, $T =\{\det(A) : A\in S\}$. To show $R\subseteq T,$ it suffices to show that $R\cap \mathbb{Z}^+\subseteq T,$ since clearly any matrix with an all-zero row or column (or a repeated row or column) has determinant zero and swapping any two distinct rows or columns of a matrix multiplies the determinant by $-1$ (the determinant is an alternating linear map). I know the determinant of an n by n matrix is also $n$-linear, meaning that it's linear with respect to any single row or column. Obviously the identity matrix has determinant 1. Let $A_2 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 0\\
0 & 0 & 1\end{pmatrix}, A_3 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & 0 & 1\end{pmatrix},A_4 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & -1 & 1\end{pmatrix}.$ Then using Laplace expansion along the first column, we can easily see that $\det A_i = i$ for $2\leq i\leq 4.$ Thus we just need to show $T\subseteq R$ to solve the question. Let $A = \begin{pmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\end{pmatrix}$. If $a_{i1}=0$ for $1\leq i\leq 3,$ then $\det(A) = 0\in R,$ so suppose otherwise. We just need to show that $|\det(A)|\leq 4$ since it is an integer (being a (multivariate) polynomial in the entries). WLOG, assume $a_{11} = 1$ (we may swap row $i\neq 1$ with row 1 and multiply the first row by $-1$ if necessary, which doesn't change the absolute value of the determinant). Perform the elementary row operations $R_i\mapsto R_i - R_1$ for $i=2,3$. Now note that if $a_{22} = a_{32} = 0,$ rows 2 and 3 are scalar multiples of each other or one of them equals the zero row, so the determinant is zero, which is in S. Hence, by swapping the two and multiplying row 2 by $-1$ if necessary, assume WLOG that $a_{22} > 0.$ Suppose first that $a_{22} = 1.$ Then $a_{12} = 0$ and $|a_{32}|\leq 1.$ Note that $|a_{23}|\leq 2,$ since $a_{13}$ was subtracted from it and has absolute value at most 1 and $a_{23}$ initially had absolute value at most 1. Also, we similarly have $|a_{33}|\leq 2.$ Subtracting row 2 from row 3 if necessary (i.e. if $a_{32}$ is nonzero), we get that $|a_{33}|\leq 4.$ But now the determinant equals $|a_{33}|\leq 4.$ Finally suppose $a_{22}= 2.$ Then $|a_{12}|=1$. Note that $|a_{23}| = 2$ only if $a_{23}' a_{13} = -1,$ where $a_{23}'$ is the original value of $a_{23}$ before subtracting from row 1. Similarly, $a_{22} = 2\Rightarrow a_{12}a_{22}' = -1.$ So $|a_{33}|=2$ and $|a_{23}|=2$ implies that $a_{33}' = a_{23}'$ and so $a_33 = a_{23}$. However, this approach seems very tedious and error-prone, so there's probably a better method.
|
Clearly $\det(A)$ is an integer. We first show that $|\det(A)|\leq4$. There are only two possibilities:
*
*Some row of $A$ is at most two nonzero elements. By Laplace expansion along that row, we see that $|\det(A)|$ is at most the sum of the absolute values of two minors in $A$.
*$A$ is entrywise nonzero. If $A$ is nonsingular, up to sign flips of rows and columns of $A$ and permutations of columns of $A$, we may assume that the first row of $A$ is $(1,1,1)$ and the second row is $(1,-1,-1)$. Therefore
$$|\det(A)|
=\left|\det\pmatrix{2&0&0\\ 1&-1&-1\\ a_{31}&a_{32}&a_{33}}\right|
=2\left|\det\pmatrix{-1&-1\\ a_{32}&a_{33}}\right|.$$
So, in both cases, $|\det(A)|$ is at most the sum of the absolute values of two minors in $A$. Since all elements of $A$ belong to $\{0,1,-1\}$, the absolute value of each such minor is at most $(1)(1)+(1)(1)=2$. Hence $|\det(A)|\leq4$.
Next, we show that all five values $0,1,2,3$ and $4$ can be realised by $|\det(A)|$. We have $\det(0)=0,\,\det(I_3)=1$,
$$\det\pmatrix{1&1&0\\ -1&1&0\\ 0&0&1}=2,
\ \det\pmatrix{1&1&0\\ -1&1&-1\\ 0&1&1}=3,
\ \det\pmatrix{1&1&0\\ -1&1&-1\\ -1&1&1}=4.$$
Therefore $\{\det(A):A\in S\}=\{d\in\mathbb Z:|d|\leq4\}$.
|
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|
Proving $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ I hope I'm allowed to ask this question here, but I have to prove that $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ using the following Fourier series:
$$
1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right]\cos(n{\pi}x) =
\begin{cases}
e^x & x\in[-1,0) \\
e^{-x} & x\in[0,1]
\end{cases}
$$
This is my progress so far:
Let $x=0$:
\begin{align*}
\therefore 1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= 1 \\
\sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= \frac{1}{e} \\
\sum_{n=0}^{\infty}\frac{2}{1+(2n+1)^2\pi^2}\left(1+\frac{1}{e}\right) + \sum_{n=1}^{\infty}\frac{2}{1+(2n)^2\pi^2}\left(1-\frac{1}{e}\right) &= \frac{1}{e}
\end{align*}
Does anyone know what I should do next?
|
Hint:
Try to see what happens for $x=1$.
Explanation:
Your progress so far has indicated that splitting the sum into even and odd sums may be useful. To that extend you could define
$$A=\sum_{n=0}^\infty \frac{2}{1+(2n+1)^2 \pi^2}$$ and
$$B=\sum_{n=1}^\infty \frac{2}{1+(2n)^2 \pi^2}.$$
Now, you already have one linear equation for $A,B$, which is (from your progress so far):
$$\left(1+\frac1e\right) \cdot A + \left(1-\frac1e\right)\cdot B = \frac1e$$
or, multiplying the whole equation by $e$ to avoid fractions, $$(e+1)A + (e-1)B = 1$$
Now, all you need is a second linear equation involving $A$ and $B$. You can get that by taking $x=1$, since the cosine factor expands nicely into $\cos(n\pi x) = \cos (n\pi) = (-1)^n$
Guidelines:
You should really try to solve the problem on your own from here on. But if you want to make sure you are still on the right track:
*
*the second equation you get should be something along the lines of $$(e+1)A + (e-1)B = 1$$
*The solution to the system of equations should then be $A=\frac{e-1}{2(e+1)}, B=\frac{3-e}{2(e-1)}$
*The final sum then comes out to $A+B=\frac{2}{e^2-1}$.
|
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|
Inverse of the function $f(x)=\sqrt{x-3}-\sqrt{4-x}+1$ I am trying to find the inverse of the function $$f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
First of all its domain is $[3,4]$
As far as my knowledge is concerned, since $f'(x)>0$, it is monotone increasing in $[3,4]$, so it is injective. Also the Range is $[0,2]$. So $$f:[3,4]\to [0,2],\:\:y=f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$ is Invertible.
Now to find $f^{-1}(x)$, we need to express $x$ in terms of $y$.
We have:
$$\begin{aligned}
&(y-1)^2=1-2 \sqrt{(x-3)(4-x)} \\
&\Rightarrow \quad 2\sqrt{(x-3)(4-x)}=1-(y-1)^2 \\
&\Rightarrow \quad 4(x-3)(4-x)=4 y^2+y^4-4 y^3 \\
&\Rightarrow \quad 4x^2-28 x+y^4-4 y^3+4 y^2+48=0
\end{aligned}$$
Which is a quadratic in $x$. But how to decide which root of $x$?
|
We want to solve
$$y=\sqrt{x-3}-\sqrt{4-x}+1$$
for $x$. Start by making the substitution $x=u^2+3$ (with $u\geq0$) so that
$$y=u-\sqrt{1-u^2}+1.$$
Rearranging and squaring we get
$$u^2-2u(y-1)+(y-1)^2=1-u^2.$$
Which can be rewritten as
$$u^2-u(y-1)+\frac{(y-1)^2-1}{2}=0.$$
Applying the appropriate quadratic formula we easily get that
$$u=\frac{y-1\pm\sqrt{2-(y-1)^2}}{2}.$$
We recall that $u=\sqrt{x-3}$, giving us that
$$\sqrt{x-3}=\frac{y-1\pm\sqrt{2-(y-1)^2}}{2}.$$
Since the left-hand side is non-negative, we must have the same for the right-hand side, and so we can replace the $\pm$ with $+$ on your given range for $y$ (check this for yourself). This it follows that
$$x=\left(\frac{y-1+\sqrt{2-(y-1)^2}}{2}\right)^2+3.$$
Thus, if
$$f(x)=\sqrt{x-3}-\sqrt{4-x}+1,$$
then
$$f^{-1}(x)=\left(\frac{x-1+\sqrt{2-(x-1)^2}}{2}\right)^2+3$$
(you can obviously try to simplify this if you want to).
|
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|
Integrating Question with U-Sub: $\int \tan^2\theta \sec^4\theta\ d\theta$ Question:
Find the value of the expression $$\int \tan^2\theta \sec^4\theta\ d\theta$$ by using the substitution $u=\tan\theta$.
My Working:
If $u=\tan\theta$, then
\begin{align}
\frac{du}{d\theta}&=\sec^2\theta\\
du&=\sec^2\theta\cdot d\theta\\
\int\tan^2\theta \sec^4\theta\ d\theta&=\int \tan^2\theta\cdot \sec^2\theta\cdot(\sec^2\theta\cdot d\theta)\\
&=^?\int u^2\cdot u^\prime\cdot du
\end{align}
After this, I do not know how to proceed as there is an extra remaining factor of $\sec^2\theta$, and I cannot use the product to sum formula as there is no $\sin x$ or $\cos x$. Could anyone please help? Thanks!
|
\begin{align}
\int\tan^2\theta \sec^4\theta\ d\theta&=\int \tan^2\theta\cdot \sec^2\theta\cdot(\sec^2\theta\cdot d\theta)\qquad(\text{say, }\tan\theta=u)\\
&=\int u^2\cdot (1+u^2)\cdot du\\
&=\int u^2 du +\int u^4 du\\
&=\frac{u^3}{3}+\frac{u^5}{5}+c\qquad(\text{some constant, }c)\\
&=\frac{\tan^3\theta}{3}+\frac{\tan^5\theta}{5}+c
\end{align}
|
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|
Solving Summation $(2 + 1 \sum_{n=1}^{m} \frac{2}{n}) - (2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1})$... I saw this in a textbook and I am not sure how to get from the LHS to the RHS:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum_{n=1}^{m-2} \frac{2}{n+2} + \frac{2}{m+1} + \frac{2}{m+2}\right) = 1- \frac{2}{m+1} + \frac{2}{m+2}$$
If somebody could show me the intermediate steps, that would be really helpful.
|
The series as written:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum_{n=1}^{m-2} \frac{2}{n+2} + \frac{2}{m+1} + \frac{2}{m+2}\right) $$
gives
$$ S_{m} = 1 - \frac{2}{m+1} + \frac{2}{m+2} + 2 \, f_{m} $$
where
\begin{align}
f_{m} &= \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=2}^{m-1} \frac{2}{n+1} + \sum_{n=1}^{m-2} \frac{1}{n+2} \\
&= \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=3}^{m} \frac{2}{n} + \sum_{n=3}^{m} \frac{1}{n} = \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=3}^{m} \frac{1}{n} \\
&= 1 + \frac{1}{2}
\end{align}
and leads to
$$ S_{m} = 4 - \frac{2}{m+1} + \frac{2}{m+2}. $$
This is not the result presented.
In order to obtain the claimed result it would seem to be more like:
$$ S_{m} = 2 \, \left(1 + \sum_{n=1}^{m} \frac{1}{n} \right) - 2 \, \left(1 + \frac{2}{m+1} + \sum_{n=1}^{m-1} \frac{2}{n+1} \right) + 2 \, \left(\frac{1}{m+1} + \frac{1}{m+2} + \sum_{n=1}^{m-2} \frac{1}{n+2} \right) $$
which gives
\begin{align}
S_{m} &= 2 \, \left[ - \frac{1}{m+1} + \frac{1}{m+2} + \sum_{n=1}^{m} \frac{1}{n} - \sum_{n=2}^{m} \frac{2}{n} + \sum_{n=3}^{m} \frac{1}{n} \right] \\
&= 2 \, \left[ - \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{1} + \frac{1}{2} - \frac{2}{2} \right] \\
&= 1 - \frac{2}{m+1} + \frac{2}{m+2}.
\end{align}
|
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|
The value of $(a^2+b^2)(c^2+1)$. The question is:
Given real numbers $a,b,c$ that satisfy
$$ab(c^2-1)+c(a^2-b^2)=12$$
$$(a+b)c+(a-b)=7$$
Find the value of $(a^2+b^2)(c^2+1)$
From what I've done, I got $7(3ac+3a+3bc-b)-2ab(c+1)(c-1)=(a^2+b^2)(c^2+1)$. I think I'm inching further from the actual solution. Can anyone give me a hint? Thanks in advance :).
|
We have
$$
(a+b)^2c^2+2(a^2-b^2)c + (a-b)^2=49
\\2ab(c^2-1) +2(a^2-b^2)c = 24
$$
Subtract them and we get
$$(a^2+b^2)c^2+2ab+(a-b)^2=25$$
Thus $$(a^2+b^2)(c^2+1) = 25.$$
|
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|
Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
|
That's a helpful start.
Notice $(t-2)^5+(t+2)^5$ is an odd function:
$$ ((-t)-2)^5+((-t)+2)^5 = -(t+2)^5-(t-2)^5 = -\left((t-2)^5+(t+2)^5\right) $$
Therefore $A = \frac{(t-2)^{5}+(t+2)^{5}}{2t}$ is an even function, so when it's multiplied out and collected to a basic polynomial form, that form must be $A = Bt^4+Ct^2+D$. This is encouragement that it won't be that bad, plus it means we'll be able to finish the problem by considering $A-121$ as a quadratic in $t^2$.
I get $$ A = t^4+40t^2+80 $$ so the equation is
$$ 2t(t^4+40t^2+80) = 242 t$$
$$ 2t(t^4+40t^2-41) = 0 $$
$$ t(t^2-1)(t^2+41) = 0 $$
The solutions (three real and one complex pair) are $t \in \{-1, 0, 1, i\sqrt{41}, -i\sqrt{41}\}$, $x \in \{-2,-1,0,-1+i\sqrt{41},-1-i\sqrt{41}\}$.
|
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|
Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fractions too. What happens when the power of $x$ is 4?
My thoughts are to do integration by parts (product rule)
$$\int u v' dx = uv - \int u' v dx$$
But I am unclear what do substitute $u$ and $v$ to
|
$$\int\frac3{2x}\cdot\frac{2x}{(x^2+5)^2}dx\\=\frac3{2x}\cdot\frac{-1}{(x^2+5)}-\int\frac{-3}{2x^2}\cdot\frac{-1}{x^2+5}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int\frac{x^2+5-x^2}{x^2(x^2+5)}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int(\frac1{x^2}-\frac1{x^2+5})dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10x}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{-3}{10x(x^2+5)}(5-x^2-5)+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c$$
|
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|
Number of strings of length $n$ with no consecutive $y$'s Suppose we have a set $S$ such that $\lvert S\rvert=k+1$. Fix an element $y$ in $S$. We want to find the recurrence relation on the number of $S$-strings of length $n$ that don't have two consecutive $y$'s, namely $yy$. Use $f(n)$ to denote the answer.
I already find some initial conditions: for $f(1)$, that is the number of $S$-strings of length $1$ without two consecutive $y$'s, which is obviously $f(1)=k+1$; For $f(2)$, there is only one string that has two consecutive $y$'s, which is $yy$, so $f(2)=(k+1)^2-1$; For $f(3)$, If we place the two consecutive $y$'s at the first and the second positions, there are $k+1$ strings and $yyy$ are among the $k+1$ strings; If we place $yy$ at the second and the third positions, there are another $k+1$ strings and $yyy$ are also among these $k+1$ strings. So we conclude $f(3)=(k+1)^3-(2k+1)$.
Yet I have no idea what the general case is like. Any help please?
|
This answer is based upon the Goulden-Jackson Cluster Method. We consider the set of words of length $n\geq 0$ built from an alphabet $S$ of $k+1$ letters with $y\in S$ and the set $B=\{yy\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $F(z)$ with the coefficient of $z^n$ being the number of wanted words of length $n$.
According to the paper (p.7) the generating function $F(z)$ is
\begin{align*}
F(z)=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{S}|=k+1$, the size of the alphabet and $\mathcal{C}$ is the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[yy])\tag{2}
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[yy])&=-z^2-z\cdot\text{weight}(\mathcal{C}[yy])\tag{3}\\
\text{weight}(\mathcal{C}[yy])&=-\frac{z^2}{1+z}\\
\end{align*}
so that
\begin{align*}
\text{weight}(\mathcal{C})=-\frac{z^2}{1+z}
\end{align*}
The additional term on the right-hand side of (3) takes account of the overlapping of $\color{blue}{y}y$ with $y\color{blue}{y}$.
We obtain according to (1) and (3)
\begin{align*}
\color{blue}{F(z)}&=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-(k+1)z+\frac{z^2}{1+z}}\\
&\; \color{blue}{=\frac{1+z}{1-kz(1+z)}}\tag{4}\\
\end{align*}
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We calculate from (4) $f(n)$ as
\begin{align*}
\color{blue}{f(n)}&=[z^n]F(z)=[z^n]\frac{1+z}{1-kz(1+z)}\\
&=[z^n](1+z)\sum_{q=0}^{\infty}k^qz^q(1+z)^q\tag{5.1}\\
&=\left([z^n]+[z^{n-1}]\right)\sum_{q=0}^{\infty}k^qz^q(1+z)^q\tag{5.2}\\
&=\sum_{q=0}^n[z^{n-q}]k^q(1+z)^q+\sum_{q=0}^{n-1}[z^{n-1-q}]k^q(1+z)^q\tag{5.3}\\
&=\sum_{q=0}^nk^q\binom{q}{n-q}+\sum_{q=0}^{n-1}k^q\binom{q}{n-1-q}\tag{5.4}\\
&\,\,\color{blue}{=k^n+\sum_{q=0}^{n-1}\left(\binom{q}{n-q}k+\binom{q}{n-1-q}\right)k^{n-1-q}}\tag{5.5}\\
\end{align*}
Comment:
*
*In (5.1) we use the geometric series expansion.
*In (5.2) we apply the rule $[z^{p-q}]F(z)=[z^p]z^qF(z)$.
*In (5.3) we apply the rule again and restrict the upper limits of the sum since other terms do not contribute.
*In (5.4) we select the coefficient of $[z^{n-q}]$ and $[z^{n-q-1}]$.
*In (5.5) we separate the term $k^n$ and merge the sums.
We verify (5.5) for small values $k=2,3$
\begin{align*}
\color{blue}{f(2)}&=k^2+\sum_{q=0}^1\left(\binom{q}{2-q}+\binom{q}{1-q}\right)k^q\\
&=k^2+\left(\binom{1}{1}+\binom{1}{0}\right)k\\
&\,\,\color{blue}{=k^2+2k}\\
\color{blue}{f(3)}&=k^3+\sum_{q=0}^2\left(\binom{q}{3-q}+\binom{q}{2-q}\right)k^q\\
&=k^3+\binom{1}{1}k+\left(\binom{2}{1}+\binom{2}{0}\right)k^2\\
&\,\,\color{blue}{=k^3+3k^2+k}
\end{align*}
in accordance with the result from other answers.
|
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|
In how many ways can points be assigned to five questions with one to four points per question so that the total is $14$? A professor must write an exam with $5$ questions. Question number i should give $p_i \in \mathbb{Z}$ points. The sum of the points must be $14$ and each question must give at least $1$ point and a maximum of $4$ points. In how many ways can he distribute the points on the questions?
I first turned $p_i= x_i - 1$ so I can get $0 \le x_i \le 3$ instead. Then I put the variable $x_i$ in the equation and got $x_1 + x_2 + x_3 + x_4 + x_5 = 9$. The total amount of combination is $C(tot) = C(5+9-1, 9)= C(13, 4)$. What I then did was find how many combinations there are when one $x$ can be greater than $3$, and then when $2$ can be greater than $3$. $C_1 = C(9,4)C(5,1)$ and $C_2 = C(5,4)C(5,2)$. So the answer is $C(13,4) - C(9,4)C(5,1) - C(5,4) C(5,2)$
What my answer sheet says is, $C = C(13,4) - C(9,4)C(5,1) + C(5,4) C(5,2) = 715-630+50= 135$ What I answered was $715 - 105 - 50 = 560$ How is $C(9,4)C(5,1)= 630$? Besides why did they do $+50$ instead of $-50$??
Can anyone help me ?
|
Our generating function is
$\begin {align*}
f(x)&=\left ( x+x^2+x^3+x^4 \right )^5\\&=x^5\left ( \frac{1-x^4}{1-x} \right )^5
\end{align*}$
So the answer is
$\begin {align*}
[x^{14}]f(x)&=[x^9]\left ( 1-5x^4+10x^8-h(x) \right )\sum_{k=0}^{\infty }\binom{-5}{k}(-x)^k\\&=\left ( [x^9]-5[x^5]+10[x] \right )\binom{k+4}{4}\\&=\binom{13}{4}-5\binom{9}{4}+10\binom{5}{4}\\&=\boldsymbol {135}
\end{align*}$
|
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|
How to solve $\int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $ I have problem to find the value of the integral:
$$ \int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $$
The first step is to split the absolute part.
I thank you in advance for help.
|
Using
\begin{eqnarray}
&&\int \frac{\log(b-x)}{x+a}\;dx\\
&=&\int \log(b-x)d\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\bigg(1-\frac{b-x}{a+b}\bigg)\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}+\text{Li}_2(\frac{b-x}{a+b})+C
\end{eqnarray}
one has
\begin{eqnarray}
&&\int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx\\
&=&\frac12\int_0^2\log(2-x)\bigg(\frac1{x+i}+\frac1{x-i}\bigg)\,dx\\
&=&\frac12\log2\log5-\frac12\bigg[\text{Li}_2\bigg(\frac{4}{5}-\frac{2i}{5}\bigg)+\text{Li}_2\bigg(\frac{4}{5}+\frac{2i}{5}\bigg)\bigg]
\end{eqnarray}
|
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|
Area enclosed between the roots of a quadratic Let $f(x)=ax^2+bx+c$
If $f(x)$ has roots $\alpha$ and $\beta$, what is the area enclosed by $f(x)$ and the $x$-axis between $x=\alpha$ and $x=\beta$ in terms of $a,b$ and $c$?
It is also given that $\alpha>\beta.$
If $a=1$, then I thought this might be easier since you get:
$$A=\int_\beta^\alpha{(x^2-(\alpha+\beta)x+({\alpha}{\beta})x) dx}$$
But even after evaluating this, I still wasn't even able to find an answer in terms of the coefficients.
I've been at this problem for a while now, and I would love some help. Any ideas?
|
To pursue your approach, not restricting the polynomial to $ \ a \ = \ 1 \ \ , $ we have
$$ A \ \ = \ \ a \int_\beta^\alpha \ \ x^2 \ - \ (\alpha+\beta)x \ + \ \alpha \beta \ \ dx $$
[you have an extra "x" on what should be the "constant term" : I presume that was a typo]
$$ = \ \ a·\left( \ \frac13x^3 \ - \ \frac{\alpha \ + \ \beta}{2} ·x^2 + \ \alpha \beta x \ \right) |_{\beta}^{\alpha} $$
$$ = \ \ a· \frac{2·(\alpha^3 \ - \ \beta^3) \ - \ 3·(\alpha \ + \ \beta)·(\alpha^2 \ - \ \beta^2) \ + \ 6·\alpha \beta·(\alpha \ - \ \beta)}{6} $$
$$ = \ \ \frac{a}{6}·(\alpha \ - \ \beta) · [ \ 2·(\alpha^2 \ + \ \alpha \beta \ + \ \beta^2) \ - \ 3·(\alpha \ + \ \beta)^2 \ + \ 6·\alpha \beta \ ] $$
[applying "differences of two cubes" and "two squares"]
$$ = \ \ \frac{a}{6}·(\alpha \ - \ \beta) · ( \ 2 \alpha^2 \ + \ 2\alpha \beta \ + \ 2\beta^2 \ - \ 3 \alpha^2 \ - \ 6 \alpha \beta \ - \ 3\beta^2 \ + \ 6 \alpha \beta \ ) $$
$$ = \ \ \frac{a·(\alpha \ - \ \beta)}{6} · ( \ - \alpha^2 \ + \ 2\alpha \beta \ - \ \beta^2 \ ) \ \ = \ \ - \frac{a·(\alpha \ - \ \beta)^3}{6} \ \ , $$
which is related to the result from Archimedes' quadrature that Doug M mentions. The result is negative since your integral evaluates an area "below" the $ \ x-$axis. So the area enclosed by the $ \ x-$axis and the parabola is the absolute-value of this, $ \ \frac{a·(\alpha \ - \ \beta)^3}{6} \ \ . \ $
As I mention in the comment to that post, the difference between the two real zeroes is $ \ \alpha - \beta \ = \ \frac{\sqrt{D}}{a} \ \ . \ $ With the vertex "below" the $ \ x-$axis, $ \ D \ = \ b^2 - 4ac \ > \ 0 \ \ , \ $ so we may at last write
$$ A \ \ = \ \ \frac{a· \left(\frac{\sqrt{D}}{a} \right)^3}{6} \ \ = \ \ \frac{( \ b^2 - 4ac \ )^{3/2}}{6·a^2} \ \ . $$
|
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|
Prove that $\frac{\cos2\theta}{\sin6\theta}+\frac{\cos6\theta}{\sin18\theta}+\frac{\cos18\theta}{\sin54\theta}=\frac12(\cot2\theta-\cot54\theta)$
Prove that $\frac{\cos2\theta}{\sin6\theta}+\frac{\cos6\theta}{\sin18\theta}+\frac{\cos18\theta}{\sin54\theta}=\frac12(\cot2\theta-\cot54\theta)$
I tried formulas for $\sin3\theta, \cos3\theta$ but couldn't conclude.
I tried taking LCM on LHS and using formula for $2\sin A\cos B$ but couldn't conclude.
I tried converting $\cot$ into $\sin, \cos$, taking LCM, using $2\sin A\cos B$ but couldn't conclude.
|
The triplication formulas for $\;\sin\;$ and $\;\cos\;$ are
$$ \sin(3x) \!=\! \sin(x)(3\cos(x)^2 \!-\! \sin(x)^2), \;
\cos(3x) \!=\! \cos(x)(\cos(x)^2 \!-\! 3\sin(x)^2). $$
Uisng these and $\;\cos(x)^2 + \sin(x)^2 = 1\;$ implies that
$$ 1 \!-\! \frac{\cot(3x)}{\cot(x)} \!=\!
1 \!-\! \frac{ \cos(x)^2 \!-\! 3\sin(x)^2}
{3\cos(x)^2 \!-\! \sin(x)^2} \!=\!
\frac{2(\cos(x)^2 + \sin(x)^2)} {3\cos(x)^2 \!-\! \sin(x)^2} \!=\! \frac{2\sin(x)}{\sin(3x)}. $$
Now this implies that
$$ \frac{\cos(x)}{\sin(3x)} = \cot(x)\frac{\sin(x)}{\sin(3x)} =
\frac{\cot(x)}2 \left(1 \!-\! \frac{\cot(3x)}{\cot(x)}\right) =
\frac{\cot(x)\!-\!\cot(3x)}2. $$
Define their common value as a function
$$ f(x) := \frac{\cos(x)}{\sin(3x)} =
\frac{\cot(x)\!-\!\cot(3x)}2. $$
Your telescoping sum is now
$$ f(2x) + f(6x) + f(18x) = \\
\frac{\cot(2x)-\cot(6x)}2 +
\frac{\cot(6x)-\cot(18x)}2 + \frac{\cot(18x)-\cot(54x)}2 =\\
\frac{\cot(2x) - \cot(54x)}2. $$
|
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|
Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\
\vdots\\
&=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\
&=-\frac{2}{3}
\end{align}
$$
|
$\cot^2x-\frac1{x^2}=\csc^2x-1-\frac1{x^2}=\frac1{\sin^2x}-\frac1{x^2}-1=\frac{x^2-\sin^2x}{x^2\sin^2x}-1=\frac{(x-\sin x)(x+\sin x)}{x^4\frac{\sin^2x}{x^2}}-1$
In the numerator, using $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$, thus,
$\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4\frac{\sin^2x}{x^2}}-1$
Using $\lim_{x\to0}\frac{\sin x}{x}=1$, we get,
$$\lim_{x\to0}\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4}-1\\=\lim_{x\to0}\frac{(\frac1{3!}-\frac{x^2}{5!}+..)(2-\frac{x^2}{2!}+...)}1-1\\=\frac26-1=-\frac23$$
|
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|
Show there's no integer between $x_n$ and $y_n$ For positive integers $n$, let $x_n, y_n$ be defined as:
$x_n = \sqrt{n} + \sqrt{n+1}$ and $y_n = \sqrt{4n + 2}$
Show that there is no whole number between $x_n$ and $y_n$.
Observation:
Obviously $x_n \not\in \mathbb{N}$ since $n$ and $n+1$ cannot both be perfect squares and $y_n \not\in \mathbb{N}$ since $4n +2 \equiv 2 \textrm{ (mod 4)}$ and any square must be $\equiv 0 \textrm{ or } 1 \textrm{ (mod 4)}$.
Question from 2009 German Math Olympiad, 2nd round
|
A very straightforward approach can be as below;
Assume $m^2\lt n \le (m+1)^2$. Hence;
$$2m \lt \sqrt n+ \sqrt {n+1}\lt 2m+3,$$
and;
$$2m\lt \sqrt {4n+2}\lt 2m+3.$$
Therefore $k$ (the integer between $x_n$ and $y_n$) is either $2m+1$ or $2m+2$. But It is impossible for $k$ to be $2m+2$ because, in this manner, $n$ has to be $(m+1)^2$ while;
$$(m+1)+(m+1)\lt \sqrt n+ \sqrt {n+1}\lt 2m+2,$$
which is a contradiction.
Now, if $k=2m+1$, we must have:
$$\sqrt {4n+2}\gt 2m+1\implies n\gt m^2+m-\frac{1}{4} \implies n\ge m^2+m. $$
But, in this case, we must also have:
$$\sqrt{m^2+m}+\sqrt {m^2+m+1}\le \sqrt n+ \sqrt {n+1}\lt 2m+1, $$
which is impossible; because for every integer $m$, we have:
$$\sqrt{m^2+m}+\sqrt {m^2+m+1}\gt 2m+1.$$
|
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Finding all the rational points of the curve $x^2 -2y^2 =1$ This question was left as an exercise in my class of number theory and I want to verify my solution.
Question: Let $C$ be a curve given by $x^2 -2 y^2=1$. Find all the rational points on $C$.
Attempt: $( 1,0 )$ is a solution and let m be the slope of the line passing through $(1,0)$ . Now the equation of the line is $y-0 = t(x-1)$ (say $L$). So, $x= 1+y/t$.
As, the curve is an hyperbola, so this line say L will intersect the curve at least $1$ more time.
Putting $x= y/t+1$ in $x^2 -2y^2 =1 $, I get y = $\frac {2t } { 2 t^2 -1} $ and $x= + \frac{ 2t^2 +1} {2t^2-1} , - \frac{ 2t^2 +1} {2t^2-1}$. So, these are the rational points of the curve along with $(1,0)$.
Is my solution fine?
|
What you did is basically OK. However, there are several places that could be improved.
A detailed solution.
If $x=1$, then $y=0$.
Suppose $(x,y)$ is a rational solution $(x,y)$ with $x\not=1$. Let the slope of the line through $(1,0)$ and $(x,y)$ be $t=\frac{y-0}{x-1}=\frac y{x-1}$, which must be a rational number.
$t=\frac y{x-1}$ implies $y=t(x-1)$. Putting $y=t(x-1)$ in $x^2-2y^2 =1$, we get $$(x-1)((2t^2-1)x-(2t^2+1))=0.$$ Since $x\not=1$, we get $x=\frac{ 2t^2 +1} {2t^2-1}$ and then y = $\frac {2t } { 2 t^2 -1} $.
Note that when $t$ is a rational number, $2t^2-1\not=0$ and both $\frac{2t^2+1}{2t^2-1}$ and $\frac{2t}{2t^2-1}$ are rational numbers.
So, all rational points are $(1,0)$ and $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$.
Wait, don't we also have $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$?
Yes, $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$ are also rational points on $C$. However, they do not provide any point that is not $(1,0)$ or $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for some rational $t$. This fact is clear since the meticulous solution process above must have found all rational points on $C$.
It turns out that there are many parametrized solutions.
Besides the one given by $(1,0)$ and $(\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$, all rational points on $C$ can also be $(-1,0)$ and $(-\frac{2t^2+1}{2t^2-1}, \frac{2t}{2t^2-1})$ for all rational $t$, where $t$ can be understood as the negative of the slope of the line through $(-1,0)$ and $(x,y)$.
For anther example, consider $(3,2)$ on $C$. Let $t$ be the slope of the line through $(3,2)$ and $(x,y)$. We can find that all rational points on $C$ are $(3,\pm2)$ and $(\frac{6t^2-8t+3}{2t^2-1}, \frac{-4t^2+6t-2}{2t^2-1})$ for all rational $t$.
More generally, suppose $(a,b)$ is a rational point on $C$. Then all rational points on $C$ are $(a, \pm b)$ and $(\frac{a(2t^2+1)-4bt}{2t^2-1}, \frac{-b(2t^2+1)+2at}{2t^2-1})$ for all rational $t$.
|
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How do we evaluate $\sum_{k=1}^{\infty} \frac{1}{3k-1}-\frac{1}{3k}$ We have to evaulate $$\sum_{k=1}^{\infty} \frac{1}{3k-1}-\frac{1}{3k}$$
I have tried to expand it to see some kind of pattern, but there seems to be no obvious one! Then I tried to just mindlessly integrate and after solving the integral $\int_1^{\infty} \frac{1}{3k-1}-\frac{1}{3k} dk$ the answer came to be $\frac{1}{3}\log(\frac{3}{2})$ which contrasts the answer from wolfram alpha by a significant margin which is $\frac{1}{18}(9\log3-\sqrt{3}\pi)$.
I have found problems of similar kind on this site but I am not really able to understand them, specifically, how did this particular sum got converted into a gamma function (they don't seem to be related on the surface).
I have some idea about gamma and beta function but please try to be as detailed as possible (at least for the conversion of the sum into integral part) with your solutions.
|
I think @ThomasAndrews' comment is worthy of becoming an answer in its own right.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\infty}}\color{blue}{\left(\frac{1}{3k-1}-\frac{1}{3k}\right)}&=\sum_{k=1}^{\infty}\int_{0}^1\left(x^{3k-2}-x^{3k-1}\right)dx\\
&=\int_0^1\sum_{k=0}^{\infty}\left(x^{3k+1}-x^{3k+2}\right)dx\tag{1.1}\\
&=\int_{0}^1\left(\frac{x}{1-x^3}-\frac{x^2}{1-x^3}\right)dx\tag{1.2}\\
&\,\,\color{blue}{=\int_{0}^1\frac{x}{1+x+x^2}dx}\tag{1.3}
\end{align*}
Comment:
*
*In (1.1) we exchange sum and integral and shift the index of the sum to start with $k=0$.
*In (1.2) we apply the geometric series expansion.
*In (1.3) we use $1-x^3=(1-x)\left(1+x+x^2\right)$.
We derive from (1.3) by elementary transformations
\begin{align*}
\color{blue}{I:=\int_{0}^1}&\color{blue}{\frac{x}{1+x+x^2}dx}\\
&=\int_{0}^1\frac{2x+1}{1+x+x^2}dx-\int_{0}^1\frac{1+x}{1+x+x^2}dx\\
&=\ln\left(1+x+x^2\right)\Big|_0^1-I-\int_0^1\frac{dx}{1+x+x^2}\\
&=\ln (3)-I-\int_{0}^1\frac{dx}{\frac{3}{4}+\left(x+\frac{1}{2}\right)^2}\\
&=\ln (3)-I-\frac{4}{3}\int_{0}^1\frac{dx}{1+\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)^2}\\
&=\ln (3)-I-\frac{2\sqrt{3}}{3}\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{du}{1+u^2}\tag{2.1}\\
&=\ln (3)-I-\frac{2\sqrt{3}}{3}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(\frac{1}{\sqrt{3}}\right)\right)\\
&=\ln (3)-I-\frac{2\sqrt{3}}{3}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\
&\,\,\color{blue}{=\ln (3)-I-\frac{\pi}{9}\sqrt{3}}\tag{2.2}
\end{align*}
Comment:
*
*In (2.1) we substitute
$
u=\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right),
du=\frac{2}{\sqrt{3}}dx
$
and set the limits of the integral accordingly.
Putting (1.3) and (2.2) together we finally get
\begin{align*}
\color{blue}{\sum_{k=1}^{\infty}}&\color{blue}{\left(\frac{1}{3k-1}-\frac{1}{3k}\right)
=\frac{1}{2}\ln (3)-\frac{\pi}{18}\sqrt{3}}
\end{align*}
|
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|
Prove that $\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$ for $x > 0$
Given $x>0$ , prove that $$\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$$
I have tried to construct $F(x)=\frac{\ln x}{x}+\frac{1}{e^x}$ and find the derivative function of $F(x)$ to find the maximun value, but I can't solve the transcendental equation.
So I tried another way.I tried to use the inequality $e^{-x}\le \frac{1}{x+1}$ (when $x>0$ ) to prove the inequality $\frac{\ln x}{x}<\frac{1}{2}-\frac{1}{e^x}$ , but I can't connect these two inequalities well, and I did't solve the problem in the end.
How can I do?
|
Update: I found a simpler proof.
It suffices to prove that
$$\frac{x}{2} - \ln x - x\mathrm{e}^{-x} > 0. $$
Let $p := \frac{23}{37} - \frac{14}{23}\mathrm{e}^{-37/23}$.
Since $p < \frac12$, it suffices to prove that
$$f(x) := px - \ln x - x\mathrm{e}^{-x} > 0. \tag{1}$$
We have
$$f'(x) = p - \frac{1}{x} + \mathrm{e}^{-x}(x-1)$$
and
$$f''(x) = \frac{1}{x^2} - (x - 2)\mathrm{e}^{-x}
= \frac{\mathrm{e}^{-x}}{x^2}[\mathrm{e}^x - (x-2)x^2]> 0. \tag{2}$$
(Note: We have $\mathrm{e}^x - (x-2)x^2
\ge 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}
+ \frac{x^5}{120} - (x-2)x^2 > 0$.)
Thus, $f(x)$ is convex on $x > 0$. Also, we have
$$f'\left(\frac{37}{23}\right) = 0.$$
Thus, the (global) minimum of $f(x)$ on $x > 0$ is attained at $x = \frac{37}{23}$.
Thus, we have
$$f(x) \ge f\left(\frac{37}{23}\right)
= 1 - \frac{1369}{529}\mathrm{e}^{-37/23} - \ln \frac{37}{23} > 0.$$
We are done.
|
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|
How can we describe this factorization process? We know that, there exist infinite number of algebraic expressions, so that we can not factorise these with real coefficients. For example,
*
*$a^2+b^2$
*$a^2+b^2+c^2$
The simplest example, $a^2+b^2$ can not be factored over $\Bbb R$.
But, I want to consider the factorization like the one below.
$$a^2+b^2=(a+b)^2-2ab=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$
But, what is the mathematical name of this type of factorisation? Is this a really factoring operation?
Edit:
Ah, I see that the, people misunderstood my question, maybe I was not clear.
I know that,
$$a^2+b^2=(a+b)^2-2ab=(a+b)^2-(\sqrt {2ab})^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$
I am aware $$a^2-b^2=(a-b)(a+b)$$
But, my question is different here.
I wonder how this factoring operation is positioned in mathematics.
I mean, in which mathematical sense, this can be considered as factoring?
$$a^2+b^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$
How can we describe/define this factorization process?
$$a^2+b^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$ over $?$
|
Note that if
$a=c^2$ and $b=2d^2$
then
$a^2+b^2=(a+b)^2-2ab=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$
becomes
$c^4+4d^4
=(c^2+2d^2)^2-4c^2d^2
=(c^2+2d^2-2cd)(c^2+2d^2+2cd)
$
which is an
unexpected factorization
with integer coefficients.
|
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|
Sum of two odd cubes plus four times a cube is zero Let $x,y$ be odd integers and let $z$ be an integer. The question is to find all solutions to the equation,
$$x^3+y^3+4z^3=0$$
Of course we have the trivial solution $(x,-x,0)$. Are there any others?
By considering the equation modulo $4$ we see that wlog $x=4k+1$ and $y=4m+3$. Numerical experimentation shows that there are no solutions with $|k|, |m| < 10000$. In fact since the cubic residues modulo $9$ are $0,\pm 1$ then we see that $z$ is a multiple of $9$ and moreover, dividing by $9^3$ if necessary, we see that $x^3$ and $y^3$ must have opposite residue modulo $9$. The same holds by considering the equation modulo $7$. I'm also aware of Broughan's theorem, but it's not clear to me whether it helps.
|
$x^3+y^3+4z^3=0$
Let $X=x/z, Y=y/z$ then we get $X^3+Y^3=-4$.
In general, $X^3+Y^3=n$ can be transformed to elliptic curve $v^2=u^3-432n^2$.
Hence we get $v^2=u^3-6912$.
According to LMFDB, this elliptic curve has rank $0$ and has no integer solution.
Hence $x^3+y^3+4z^3=0$ has no nontrivial solution.
|
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|
Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
=
\frac{(4^{\arcsin(0)} - 1)(\sqrt[10]{1 - \arctan(0)} - 1)}{(1-\cos\tan0)\ln(1-\sqrt{\sin 0})}
=
\frac{(1 - 1)(\sqrt[10]{1 - 0} - 1)}{(1-1)\ln(1-\sqrt{0})}
=
\frac{0*0}{0*0}
=
\frac{0}{0}
$$
But as you can see at $0$ there is limit Indeterminacy ($0/0$). How to play around and solve it?
|
As the comment suggests, you shall use Taylor expansion up to $O(3)$, that is:
$$4^{\arcsin(x^2)} \approx 1+x^2 \log (4)+O\left(x^4\right)$$
$$\sqrt[10]{1- \arctan(3x^2)} \approx 1-\frac{3 x^2}{10}+O\left(x^4\right)$$
$$\cos(\tan(6x)) \approx 1-18 x^2+O\left(x^4\right)$$
$$\sqrt{\sin(x^2)} \approx x + O(x^4)$$
Plug the values inside and you will get
$$\lim_{x\to 0} \dfrac{x^2 \log (4) \cdot \left(-\frac{3 x^2}{10}\right)}{-18 x^2 \cdot x} = \lim_{x\to 0} \dfrac{3\log(4)}{180}x = 0$$
|
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|
In $\triangle ABC$, $D$ lies on $BC$, $\angle ADB=60$, $\angle ACB=45$ and $BD=2CD$. Find $\angle ABC$ The question is as the title states. In the following figure, with some given angles and two sides, the goal is to find the measure of $\angle ABC$. This problem is inspired by one which appeared in a local Math contest in Turkey. I'll post my approach to this below as an answer! Please show your own methods of solving this!
|
Let $\angle ABC = x $, then applying the law of sines on $\triangle ADC $:
$\dfrac{AD}{\sin 45^\circ} = \dfrac{a}{\sin 15^\circ} $
Applying the law of sines on $\triangle ADB$:
$\dfrac{AD}{\sin x} = \dfrac{2 a}{\sin(120^\circ - x) } $
Dividing the two equation to eliminate $AD$ and $a$
$ \dfrac{ \sin x }{ \sin 45^\circ } = \dfrac{1}{2} \dfrac{ \sin(120^\circ - x) }{ \sin 15^\circ } $
By cross-multiplying,
$ 2 \sin x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ \cos x - \cos 120^\circ \sin x \bigg) $
Dividing through by $\cos x $
$ 2 \tan x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ - \cos 120^\circ \tan x \bigg)$
Hence,
$ \tan x = \dfrac{ \sin 45^\circ \sin 120^\circ }{ 2 \sin 15^\circ + \sin 45^\circ \cos 120^\circ } $
It follows that
$ x = 75^\circ $
|
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|
Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Step
Let n = 1
⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$
I then proved that the remainder is 0 using polynomial long division.
$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0
∴ s(1) is true
Assumption Step
Assume that s(m) is true
⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial
Inductive Step
To prove that s(m+1) is true
⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial
⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)
⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)
However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!
|
you could use the relationship between S(m) and S(m+1) such that:
$ x^{6m+2} − x^{6m+1} + 1 $ and $ x^6 * (x^{6m+2} − x^{6m+1} )+1 $ if you use the famous trick of +1 -1 in () we get :
S(m+1) = $ x^6 * (x^{6m+2} − x^{6m+1} +1 )/(x^2 −x +1) +(1-x^6) / (x^2 −x +1) $
= $ (x^6 )* S(m) + (1-x^6) / (x^2 −x +1) $
=$ (x^6 )* Q(x) + (-x^4 -x^3 +x +1 ) $ for some Q(x) polynomial
= K(x) polynomial as sum of 2 polynomials
so we get if S(m) is true so is S(m+1) , from the induction it always holds (you already calculated the case S(1) ).
|
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|
The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = 1$ and $k = 1$, so the new formula of the shifted parabola would be: $y=(x-1)^2+1$
What I don't understand is the answer of my exercise sheet, which says that the new equation would be $y= -x^2+6x-4$
Can someone help me understand this?
Thanks
|
"Shifted" here means translated.
In particular, after being shifted by $p$ units in the $y$ direction and $q$ units in the $x$ direction, the shifted parabola satisfies the equation
$$
(y-p)
= (x - q)^2 + 4(x - q) + 8.
$$
Now, the problem is to solve the above for $p$ and $q$ such that $(x, y) =(1, 1)$ and $(3, 5)$ satisfy the equation.
Substituting these points, we find that
\begin{align*}
1 - p
&= -(1-q)^2 - 4(1 - q) + 8, \\
5 - p
&= -(3-q)^2 - 4(3 - q) + 8.
\end{align*}
Expanding these yeilds
\begin{align*}
p &= q^2 + 2q - 10, \\
p &= q^2 - 2q - 6,
\end{align*}
which we can solve to find $p = -7, q = 1$, which corresponds to the equation
\begin{align*}
(y + 7)
&= (x - 1)^2 + 4(x - 1) + 8,\\
y
&= -x^2 + 6x - 4.
\end{align*}
|
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|
Calculate $\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$ for $t > 0, n \ge 1$ The problem is as follows:
Show that for all $t > 0, n \ge 1$,
$$
\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx = {2n-2 \choose n -1}\frac{\pi}{(2t)^{2n -1}}
$$
What I have so far:
Let $f(t) = \int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$
$$
\frac{df}{dt} = \int_{0}^{\infty}\frac{\partial}{\partial t}\frac{1}{(x^2+t^2)^n}dx
$$
After simplification
$$
\frac{df}{dt} = \int_{0}^{\infty}-2nt(x^2+t^2)^{-n-1}dx
$$
However, I am not sure how to integrate this and WolframAlpha outputs a complicated solution, so any help would be greatly appreciated.
|
Let $$ I_{n} = \int_{0}^{\infty} \frac{dx}{(x^2 + t^2)^n} $$ then by using integration by parts this becomes
\begin{align}
I_{n} &= \left[\frac{x}{(x^2 + t^2)^{n+1}}\right]_{0}^{\infty} + 2 \, n \, \int_{0}^{\infty} \frac{x^2 \, dx}{(x^2 + t^2)^{n+1}} \\
&= 2 \, n \, \int_{0}^{\infty} \frac{(x^2 + t^2) - t^2}{(x^2 + t^2)^{n+1}} \, dx \\
I_{n} &= 2 \, n \, \left(I_{n} - t^2 \, I_{n+1}\right).
\end{align}
This gives
$$I_{n+1} = \frac{2 n -1}{2 \, n \, t^2} \, I_{n}. $$
When $n=1$ the integral $I_{1}$ gives the result
$$ I_{1} = \int_{0}^{\infty} \frac{dx}{x^2 + t^2} = \frac{1}{t} \, \left[ \tan^{-1}\left(\frac{x}{t}\right) \right]_{0}^{\infty} = \frac{\pi}{2 \, t}.$$
From here the values of $I_{n}$ can be obtained.
A method using a generating function goes as follows.
\begin{align}
\sum_{n=1}^{\infty} I_{n} \, y^n &= \sum_{n=1}^{\infty} \, \int_{0}^{\infty} \left(\frac{y}{x^2 + t^2}\right)^n \, dx \\
&= \int_{0}^{\infty} \frac{y \, dx}{x^2 + t^2 - y} \\
&= \frac{\pi}{2} \, \frac{y}{\sqrt{t^2 - y}} \\
&= \frac{\pi}{2} \, \sum_{k=0}^{\infty} \binom{-1/2}{k} \, \frac{(-1)^{k}}{t^{2k+1}} \, y^{k+1} = \frac{\pi}{2} \, \sum_{n=1}^{\infty} \binom{-1/2}{n-1} \, \frac{(-1)^{n-1}}{t^{2n-1}} \, y^n.
\end{align}
This leads to
$$ I_{n} = \frac{\pi}{2} \, \binom{-1/2}{n-1} \, \frac{(-1)^{n-1}}{t^{2n-1}} = \frac{\pi \, n \, C_{n-1}}{(2 t)^{2n-1}}, $$
where $C_{n}$ is the $n^{th}$ Catalan number.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2\theta}}d\theta
$$
I used to identity $1 - \sin^2\theta = \cos^2\theta$ and simplified the denominator as follows
$$
\int\frac{\cos\theta}{\sin\theta-\cos\theta}d\theta
$$
I then rewrote $\cos\theta$ as $\frac{\sin\theta + \cos\theta}{2} - \frac{\sin\theta - \cos\theta}{2}$ and rewrote the integrand as follows
$$
\int\frac{1}{2}\frac{\sin\theta+\cos\theta}{\sin\theta - \cos\theta} - \frac{1}{2}d\theta
$$
I then split the integral as follows
$$
\frac{1}{2}\int\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}d\theta - \frac{1}{2}\int1d\theta
$$
For the first integral, I substituted $\phi = \sin\theta-\cos\theta$, with $d\theta = \frac{1}{\sin\theta+\cos\theta}d\phi$
We can then rewrite our integral as
$$
\frac{1}{2}\int\frac{1}{\phi}d\phi
$$
This is trivial and after undoing the substitutions we have a result of
$$
\frac{\ln({x - \cos(\arcsin(x))})}{2}
$$
The second integral is also trivial and just evaluates to $\frac{x}{2}$
Combining everything together gives us a final simplified answer of
$$
I = \frac{\ln({x - \sqrt{1-x^2}})-x}{2} + C
$$
However, both IntegralCalculator and WolframAlpha give very different answers, so if someone could tell me where I made a mistake or another approach entirely that would be greatly appreciated.
|
I approached this integral by trigonometric substitution and partial fraction.
Assume the followings:
$x=\sin\theta\ \ldots(1)\\
\text{Hence, }dx=\cos\theta\ d\theta\ \ldots(2)\\
\text{Also, }\cos\theta=\sqrt{1-x^2}\ \ldots(3)\\$
Then,
$\operatorname{\Large\int}\dfrac{1}{x-\sqrt{1-x^2}}dx=\operatorname{\Large\int}\dfrac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2\theta}}d\theta=\operatorname{\Large\int}\dfrac{\cos\theta}{\sin\theta-\cos\theta}d\theta$
$=\operatorname{\Large\int}\dfrac{1}{\tan\theta-1}d\theta=\operatorname{\Large\int}\dfrac{1}{\sec^2\theta\ (\tan\theta-1)}d\tan\theta=\operatorname{\Large\int}\dfrac{1}{(\tan^2\theta+1)(\tan\theta-1)}d\tan\theta$
$\stackrel{\text{partial fraction}}{=}\dfrac{1}{2}\operatorname{\Large\int}(\dfrac{1}{\tan\theta-1}-\dfrac{1+\tan\theta}{1+\tan^2\theta})\ d\tan\theta$
$=\dfrac{1}{2}(\operatorname{\Large\int}\dfrac{1}{\tan\theta-1}d\tan\theta-\operatorname{\Large\int}\dfrac{\sec^2\theta\ (1+\tan\theta)}{1+\tan^2\theta}\ d\theta)$
$=\dfrac{1}{2}(\operatorname{\Large\int}\dfrac{1}{\tan\theta-1}d\tan\theta-\operatorname{\Large\int}(1+\tan\theta)\ d\theta)$
$=\dfrac{1}{2}[\ln|\tan\theta-1|-(\theta-\ln|\cos\theta|)]+C$
$=\dfrac{1}{2}(\ln|\sin\theta-\cos\theta|-\theta)+C$
$=\underline{\underline{\dfrac{1}{2}(\ln|x-\sqrt{1-x^2}|-\arcsin x)+C}}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finding the set of values of k for which a modulus equation has exactly 4 roots In my assignment, I have the following question:
Find the set of values of k for which the equation $|x^2-1|+x=k$ has
exactly four roots.
What I've tried:
Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-1)=0$. Calculated their discriminants such that $D_{1}$ and $D_{2}$ (the discriminants of the first and second equations, respectively) are both positive, I got $k∈(-5/4,5/4)$. After that, since in the first equation $x∈(-∞, -1]∪[1,∞)$, I used the quadratic formula and got $±√D_{1}∈(-∞,-1]∪[3,∞)$ and proceeding similarly for the second equation, I got $±√D_{2}∈(-1,3)$. I am not able to proceed any further, please help.
|
Find the set of values of k for which the equation $|x^2-1|+x=k$ has exactly four roots.
What I've tried:
Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-1)=0$.
You have two cases
*
*$(x^2 - 1) \leq 0.$
*$(x^2 - 1) \geq 0.$
Note the overlap in the two cases, where $~x^2 - 1 = 0.~$ This overlap will be discussed following the discussion of the two individual cases.
$\underline{\text{Case 1} ~: x^2 - 1 \leq 0}$
Any roots of $x$ must be in the region $-1 \leq x \leq 1$
and the pertinent equation is
$1 - x^2 + x = k \implies x^2 - x + (k-1) = 0 \implies $
$$x = \frac{1}{2} \left[1 \pm \sqrt{1 - 4(k-1)}~\right]
~=~ \frac{1}{2} \left[1 \pm \sqrt{5 - 4k}~\right]. \tag1 $$
To complete Case 1, you have to find all values of $k$ such that:
*
*$5 - 4k > 0$
*The corresponding roots represented in (1) above are both in the range $-1 \leq x \leq 1.$
For the moment, let $A$ denote $\sqrt{5 - 4k} \implies A > 0$.
Then, the two roots will be given by
$$\frac{1 + A}{2}, ~~\text{and}~~ \frac{1 - A}{2}. \tag2 $$
So, it is clear that both of the roots represented by (2) above will be in the range $~-1 \leq x \leq 1,~$ if and only if $A \leq 1.$
Therefore, you have that
$$0 < \sqrt{5 - 4k} \leq 1 \implies $$
$$0 < (5 - 4k) \leq 1 \implies 1 \leq k < \frac{5}{4}.$$
$\underline{\text{Case 2} ~: x^2 - 1 \geq 0}$
Any roots of $x$ must be in the region
$x \leq -1~$ or $~x \geq 1$
and the pertinent equation is
$x^2 - 1 + x = k \implies x^2 + x - (k+1) = 0 \implies $
$$x = \frac{1}{2} \left[-1 \pm \sqrt{1 + 4(k+1)}~\right]
~=~ \frac{1}{2} \left[-1 \pm \sqrt{5 + 4k}~\right]. \tag3 $$
To complete Case 2, you have to find all values of $k$ such that:
*
*$5 + 4k > 0$
*The corresponding roots represented in (3) above are both in the range $~x \leq -1~$ or $~x \geq 1.$
For the moment, let $A$ denote $\sqrt{5 + 4k} \implies A > 0$.
Then, the two roots will be given by
$$\frac{-1 + A}{2}, ~~\text{and}~~ \frac{-1 - A}{2}. \tag4 $$
So, it is clear that both of the roots represented by (4) above will be in the range $~x \leq -1~$ or $~x \geq 1 ~$ if and only if $A \geq 3.$
Therefore, you have that
$$\sqrt{5 + 4k} \geq 3 \implies 5 + 4k \geq 9 \implies k \geq 1.$$
$\underline{\text{Final Analysis}}$
This portion of my response is intentionally going to be very slow and pedantic, because the target audience is Math students new to the topic. For a more sophisticated audience, this response would have been significantly shorter.
The results so far are:
*
*The combined assumption that $1 \leq k < \dfrac{5}{4}$
and $(x^2 - 1) \leq 0$, will yield two distinct roots that both satisfy $~(x^2 - 1) \leq 0.$
*The combined assumption that $1 \leq k $
and $(x^2 - 1) \geq 0$, will yield two distinct roots that both satisfy $~(x^2 - 1) \geq 0.$
So, superficially, you could wrongly infer that the correct answer is $1 \leq k < \frac{5}{4}.$ In order to get the correct answer, you have to consider the original question more carefully.
It is required to identify all values of $k$ that result in four distinct roots. Case 1 identified a range for $k$ that would provide two distinct roots that satisfied $(x^2 - 1) \leq 0.$ Case 2 identified a range for $k$ that would provide two distinct roots that satisfied $(x^2 - 1) \geq 0.$
In order to complete this (very tricky) problem, you have to consider which values of $k$ will result in four distinct roots from Case 1 and Case 2 combined.
This will be done by:
*
*Changing the Case 1 specification to force $~(x^2 - 1) < 0,~$ and changing the Case 2 specification to force $~(x^2 - 1) > 0.$
*These changes will guarantee four distinct roots.
*Then, consider Case 1 and Case 2 separately. In each case, manually identify what happens if a value of $k$ is permitted that results in at least one of the roots specifically satisfying $x^2 - 1 = 0.$
In Case 1, if the requirement had (instead) been that $(x^2 - 1) < 0$, then the resulting range would have been
$$1 < k < \frac{5}{4}.$$
In Case 2, if the requirement had (instead) been that $(x^2 - 1) > 0$, then the resulting range would have been
$$1 < k.$$
Therefore, before deciding on a final answer, it remains to consider exactly what happens when $k = 1.$
Consider the equation
$$|x^2 - 1| + x = 1.$$
If the Case 1 assumption is made that $x^2 - 1 \leq 0$, then the two roots are $0$ and $1$.
If the Case 2 assumption is made that $x^2 - 1 \geq 0$, then the two roots are $-2$ and $1$.
So, the specific value of $k = 1$ generates only three distinct roots, rather than four.
Therefore, the final answer is
$$1 < k < \frac{5}{4}.$$
|
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|
How to show that every non-trivial orbit of ODE is a periodic circle I am working with this system of first-order ODEs.
\begin{align}
\dot{x} & = y^2 - x^2 \ ,\\
\dot{y} & = - 2 x y\ .
\end{align}
We were told to convert the system using $z=x+iy$ where I found that $z'=-z^2$. From there we were asked to show that every other orbit is a circle. I am not sure where to begin. I also looked at the rest point which occurs at the origin and believe it to by hyperbolic.
|
I will ask you to criticize my decision, because I just thought of it, there may be inaccuracies.
I assumed that
$$ t, x(t),y(t) \in \mathbb{R}$$
$$ z(t) = x(t) + y(t)i , $$
$$ \frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt}i$$
$$ -z^{2} = (y^2 - x^2) - 2x\cdot y \cdot i $$
$$ \frac{dz}{dt} = -z^2$$
$$ \frac{1}{z} = t + C$$
where $C$ is an arbitrary constant
$$ \frac{x-yi}{x^2+y^2} = \frac{x}{x^2+y^2} + \left[\frac{-y}{x^2+y^2}\right ] i = t + C $$
we understand that $ t + C = t + (A + Bi) \in \mathbb{C}
\Rightarrow$
\begin{equation*}
\begin{cases}
\frac{x(t)}{x^{2}(t)+y^2(t)} = t + A,
\\
\frac{-y(t)}{x^2(t)+y^2(t)} = B,
\end{cases}
\end{equation*}
from the second equation we obtain the equation of a circle with a center at a point $(0 ; - \frac{1}{2B})$ and a radius $\frac{1}{2|B|}$
$$ x^2 + y^2 = \frac{-y}{B}$$
$$ x^2 + y^2 + \frac{2y}{2B} + \left(\frac{1}{2B}\right)^2 = \left(\frac{1}{2B}\right)^2 $$
$$ x^2 + \left(y + \frac{1}{2B} \right)^2 = \left(\frac{1}{2B}\right)^2 $$
If you decide to count this decision as correct, then I want to warn you that Lutz Lehmann helped me
|
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|
A closed form expression of $\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n$
I am working on the following exercise: Use the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ and elementary operations on power series (addition, multiplication, integration, differentiation) to find closed form expressions of the following power series:
*
*$\sum_{n \ge 0} n^2z^n$
*$\sum_{n \ge 0} \frac{n}{n+1}z^n$
*$\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n.$
I have solved the first two points in the following way:
*
*Using differentiation on the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ we obtain
$$\frac{1}{(1-z)^2} = \sum_{n \ge 0} nz^{n-1}$$
and multiplying the above equality with $z$ yields $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ and again using differentiation and multplication with $z$ on this term yields $\frac{z^2+z}{(1-z)^3} = \sum_{n \ge 0} n^2z^n$ as desired.
*Integrating the term $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ from before yields
$$\ln(\vert 1-z \rvert)+\frac{1}{1-z} = \sum_{n \ge 0} \frac{n}{n+1}z^{n}$$
I realise that the third part is probably somehow related to the exercises before, but I do not see how. Could you please help me?
|
Where these series converge,
\begin{align}
&\Bigl( \frac11 \Bigr) \, z
+ \Bigl( \frac11 + \frac12 \Bigr) \, z^2
+ \Bigl( \frac11 + \frac12 + \frac13 \Bigr) \, z^3
+ \cdots \\
&\qquad = \frac11 \Bigl( z + z^2 + z^3 + \cdots \Bigr)
+ \frac12 \Bigl( z^2 + z^3 + \cdots \Bigr)
+ \frac13 \Bigl( z^3 + \cdots \Bigr) \\
&\qquad = \frac11 z \Bigl(1 + z + z^2 + \cdots \Bigr)
+ \frac12 z^2 \Bigl(1 + z + z^2 + \cdots \Bigr)
+ \frac13 z^3 \Bigl(1 + z + z^2 + \cdots \Bigr) \\
&\qquad = \Bigl( \frac11 z + \frac12 z^2 + \frac13 z^3 + \cdots \Bigr)
\Bigl(1 + z + z^2 + \cdots \Bigr) \\
&\qquad = -\frac{\ln(1 - z)}{1-z}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$
I don't know how can I proceed from here.
Is there a way so that not using the complicated expansion of polynomials?
I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.
|
Firstly $x>0$
Let $\theta=\sin^{-1}{x} \Rightarrow x=\sin \theta$
We get
$$\sin \theta(1+\cos \theta)=\sqrt{1+\sin ^2 \theta} $$
Squaring both sides:
$$\begin{aligned}
& \left(1-\cos ^2 \theta\right)(1+\cos \theta)^2=2-\cos ^2 \theta \\
\Rightarrow \quad & \cos ^4 \theta+2 \cos ^3 \theta-\cos ^2 \theta-2 \cos \theta+1=0 \\
\Rightarrow \quad & \left(\cos ^2 \theta+\frac{1}{\cos ^2 \theta}\right)+2\left(\cos \theta-\frac{1}{\cos \theta}\right)-1=0
\end{aligned}$$
$$\begin{aligned}
& \cos \theta-\frac{1}{\cos \theta}=u \\
\Rightarrow & u^2+2+2 u-1=0 \\
\Rightarrow \quad & u=-1 \\
\Rightarrow \quad & \cos ^2 \theta+\cos \theta-1=0 \\
\Rightarrow \quad & \cos \theta=\frac{-1+\sqrt{5}}{2}
\end{aligned}$$
$$\Rightarrow \quad x=\sin \theta=\sqrt{\frac{\sqrt{5}-1}{2}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I do to plot this function by myself not by WolframAlpha. Thank you in advance.
|
$$
\begin{align}
& (x^{\frac{1}{2}}+y^{\frac{1}{2}})^{2} = 5\\
\implies & x^{\frac{1}{2}}+y^{\frac{1}{2}} = \pm \sqrt{5}\\
\implies & \sqrt{x}+\sqrt{y} = \pm \sqrt{5}
\end{align}
$$
The inequality $0\le x,y\le 5$ holds for real $x$ and $y$, as mentioned in the previous answers.
Method 1: Plot $f(x,y)=|\sqrt{x} + \sqrt{y} - \sqrt{5}|$
We take real numbers $x$ and $y$ from $0$ to $5$ and plot $f(x,y)=|\sqrt{x} + \sqrt{y} - \sqrt{5}| < \epsilon$ for very small $\epsilon$ (here $1e-9$ on a $100\times 100$ grid):
Method 2: Plot individual points
$$
x = 4y\\
\implies y = \dfrac{5}{9}\\
\implies x = \dfrac{20}{9}
$$
Similarly, $x=9y\implies (x,y)=(45/9,5/9)$, $x=16y\implies (x,y)=(80/25,5/25)$, and in general $x=n^{2}y\implies (x,y)=(\dfrac{5n^2}{(n+1)^2},\dfrac{5}{(n+1)^2})$.
Since the equation is symmetric for in $x$ and $y$, we plot both $(x,y)$ (shown in blue) and $(y,x)$ (shown in red).
Note: the point at the center of the curve is both red and blue (overlapping). Different choices of $x(y)$ give different distribution of points.
|
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|
Finding integer roots of an equation
Find all negative integer solutions of x and y such that
$$
x^3 + 3y^2 = xy^2 + 27
$$
To begin, I isolated $x$ and $y$ to see if there is anything I could do and got
$$
y = \pm \sqrt{x^2+3x+9} \\
x = \dfrac{1}{2} \left(-3 \pm \sqrt{y^2-27}\right)
$$
Since $x$ and $y$ are negative here I thought I could remove the positive branch and got
$$
y = - \sqrt{x^2+3x+9} \\
x = - \dfrac{1}{2} \left(3 + \sqrt{y^2-27}\right)
$$
And I really don't know what to do at this point. I don't have much knowledge on finding integer solutions, any help is appreciated.
|
The equation $x^3 + 3y^2 = xy^2 + 27$ can be written as
$$x^3-27+3y^2-xy^2=0$$
$$(x-3)(x^2+3x+9)+y^2(3-x)=0$$
$$(x-3)(x^2+3x+9-y^2)=0$$
Now, $x^2+3x+9-y^2=0$ can be written as
$$\bigg(x+\frac 32\bigg)^2+\frac{27}{4}-y^2=0$$
Multiplying the both sides by $4$ gives
$$(2x+3)^2+27-(2y)^2=0$$
and so
$$(2y-2x-3)(2y+2x+3)=27$$
I think that you can continue from here.
|
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|
Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.
|
Using $ac - b^2 > 0$ and $2ab - (a + b)c > 0$, we have
\begin{align*}
&bc - (b^2 + c^2 - a^2)\\
>\,&bc - (b^2 + c^2 - a^2) - (ac - b^2) - [2ab - (a + b)c]\\
=\,& a^2 - 2ab + 2bc - c^2\\
=\,&(a - c)(a + c - 2b)\\
>\,& 0
\end{align*}
where we have used $a > c$ and $a + c > 2b$.
(Explanations:
By AM-GM, we have
$2b < 2\sqrt{ac} \le a+c$.
Also, by AM-GM, we have
$$c < \frac{2ab}{a + b} \le \frac{2ab}{2\sqrt{ab}} = \sqrt{ab} < \sqrt{a \sqrt{ac}}$$
which results in $c < a$.)
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
|
For the antiderivative, a little bit faster could be $${\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}=\frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}$$
$$\theta=\tan ^{-1}(t)\quad \implies I= \int \frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}\,d\theta=\int \frac{1-t^2}{t^4+1}\,dt$$
$$\frac{1-t^2}{t^4+1}=\frac{1-t^2}{(t^2+i)(t^2-i)}=-\frac{\frac{1-i}{2}}{t^2+i}-\frac{\frac{1+i}{2}}{t^2-i}$$ making
$$I=\frac{1}{2 \sqrt{2}}\log \left(\left| -\frac{t^2+\sqrt{2} t+1}{t^2-\sqrt{2} t+1}\right| \right)$$
|
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|
Finding the angles of a right triangle if $\frac{\text{area of triangle}}{\text{area of incircle}}=\frac{2\sqrt{3}+3}{\pi}$
Let $ABC$ be a right triangle with $\measuredangle ACB=90^\circ$. If $k(O;r)$ is the incircle of the triangle and
$$\dfrac{S_{ABC}}{S_k}=\dfrac{2\sqrt3+3}{\pi}$$ find the angles of the triangle.
($S$ denotes the areas of the triangle and the circle.)
The area of the triangle is $S=pr$, where $p$ is the semiperemeter, the area of the circle is $S_k=\pi r^2,$ so $$\dfrac{pr}{\pi r^2}=\dfrac{2\sqrt3+3}{\pi}\\\dfrac{p}{r}=2\sqrt3+3\\\dfrac{\frac{a+b+c}{2}}{\frac{a+b-c}{2}}=2\sqrt3+3\\\dfrac{\frac{a}{c}+\frac{b}{c}+1}{\frac{a}{c}+\frac{b}{c}-1}=2\sqrt3+3\\\ \dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}=2\sqrt3+3$$ I don't know if my approach is reasonable and don't see what can be done from here.
|
Your approach is reasonable. You have correctly expressed the ratio of the areas of the triangle and the incircle in terms of the side lengths of the triangle and the radius of the incircle.
To proceed, and find find the angles you can follow these steps:
Use the formula for the sine of the angle $\alpha$ in terms of the sides of the triangle:
$$\sin \alpha = \frac{b}{c}$$
Substitute this into the equation:
$$\frac{\frac{a}{c} + \frac{b}{c} + 1}{\frac{a}{c} + \frac{b}{c} - 1} = 2\sqrt{3} + 3$$
to obtain:
$$\frac{\frac{a}{c} + \sin \alpha + 1}{\frac{a}{c} + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Use the formula for the cosine of the angle $\alpha$ in terms of the sides of the triangle:
$$\cos \alpha = \frac{a}{c}$$
Substitute this into the equation above to obtain:
$$\frac{\cos \alpha + \sin \alpha + 1}{\cos \alpha + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Use the identity $\cos^2 \alpha + \sin^2 \alpha = 1$ to express $\cos \alpha$ in terms of $\sin \alpha$:
$$\cos \alpha = \pm \sqrt{1 - \sin^2 \alpha} = \pm \sqrt{1 - \frac{b^2}{c^2}}$$
Substitute this into the equation above to obtain:
$$\frac{\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha + 1}{\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha - 1} = 2\sqrt{3} + 3$$
Simplify this equation to obtain:
$$(\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha - 2\sqrt{3})(\pm \sqrt{1 - \frac{b^2}{c^2}} + \sin \alpha + 2\sqrt{3}) = 1$$
Solve this quadratic equation for $\sin \alpha$ using the quadratic formula.
Once you have found $\sin \alpha$, use it to find $\cos \alpha$.
Use these values to find the angles of the triangle using the formulas:
$$\alpha = \sin^{-1} \left( \frac{b}{c} \right)$$
$$\beta = \cos^{-1} \left( \frac{a}{c} \right)$$
$$\gamma = 90^\circ - \alpha - \beta$$
Note that there are two solutions for $\sin \alpha$, corresponding to the two possible values of $\cos \alpha$. You should check both solutions to ensure that you have found all possible solutions for the angles of the triangle
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|
Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$. Prove or disprove that the inequality
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$
is valid if $x,y,z$ are positive numbers and $$xyz=1.$$
My solution is:
Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\dfrac{c}{a}.$$
So we have
$$\dfrac{1}{\sqrt{1+\dfrac{a}{b}}}+\dfrac{1}{\sqrt{1+\dfrac{b}{c}}}+\dfrac{1}{\sqrt{1+\dfrac{c}{a}}}=\dfrac{1}{\sqrt{\dfrac{a+b}{b}}}+\dfrac{1}{\sqrt{\dfrac{b+c}{c}}}+\dfrac{1}{\sqrt{\dfrac{c+a}{a}}}=\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}=\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}}.$$
Then we can use this:
$$\dfrac{1}{\sqrt{xy}}\geq\dfrac{2}{x+y}.$$
So we have
$$\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}} \geq \dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c}.$$
Then we can use this:
$$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \geq 3 \sqrt[3]{\dfrac{abc}{xyz}}.$$
So we have
$$\dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c} \geq 3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}}.$$
The question is what should I do next? Is it already obvious that
$$3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}} \geq 1?$$
Any hint would help a lot! Thanks!
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Let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and
$$\sum_{cyc}\frac{1}{\sqrt{1+x}}=\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a+b}}\geq\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$
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|
Solve $\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$ using hypergeometric functions How can we solve this integral?
$$\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$$
I used geometric series
$$=\int_0^1(x+1)^b\sum_{n=0}^\infty\left(\frac{x}{\alpha}\right)^{n+1}dx=\sum_{n=0}^\infty\int_0^1\left(\frac{x}{\alpha}\right)^{n+1}(x+1)^bdx$$
Using
$$\int_0^1t^b(1-t)^c(1-zt)^adt=\Gamma(b+1)\Gamma(c+1){_2}F_1(-a,b+1;c+b+2;z)$$
I got $$\sum_{n=0}^\infty a^{-n-1}\Gamma(n+2){_2}F_1(-b,n+2;n+3;-1)$$
The problem is I don't know how to simplify this any further. How can ${_2}F_1(-b,n+2;n+3;-1)$ be simplified?
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Answer for $\alpha> 1$ and $b\notin\Bbb Z$
For $\alpha>1$, we can write the integral as
$$
I=\int_0^1\frac{x(x+1)^b}{\alpha-x}dx=\int_0^1\frac{x(x+1)^b}{\alpha+1-(x+1)}dx
=
\frac{1}{\alpha+1}\int_0^1\frac{x(x+1)^b}{1-\frac{x+1}{\alpha+1}}dx,
$$
where $0<\frac{1}{\alpha+1}\le\frac{x+1}{\alpha+1}\le \frac{2}{\alpha+1}<1$. Hence, the term $\frac{1}{1-\frac{x+1}{\alpha+1}}$ can be expanded by its Taylor's series. We can then write the integral as
$$
I{=\frac{1}{\alpha+1}\int_0^1x(x+1)^b\sum_{n=0}^\infty(\frac{x+1}{\alpha+1})^ndx,
\\=
\frac{1}{\alpha+1}\int_0^1x\sum_{n=0}^\infty\frac{(x+1)^{n+b}}{(\alpha+1)^n}dx,
\\=
\frac{1}{\alpha+1}\int_0^1(x+1-1)\sum_{n=0}^\infty\frac{(x+1)^{n+b}}{(\alpha+1)^n}dx,
\\=
\frac{1}{\alpha+1}\int_0^1\sum_{n=0}^\infty\frac{1}{(\alpha+1)^n}[(x+1)^{n+b+1}-(x+1)^{n+b}]dx
\\=
\frac{1}{\alpha+1}\sum_{n=0}^\infty\frac{1}{(\alpha+1)^n}\left[\frac{2^{n+b+2}-1}{n+b+2}-\frac{2^{n+b+1}-1}{n+b+1}\right]
.
}
$$
The value of the integral is $\infty$ for $\alpha=1$ and is not explicitly defined for $0<\alpha<1$, unless through the limit of Cauchy integral. The case for $\alpha<0$ should also follow similarly.
|
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Find the greatest integer in the expression below as a function of the given conditions Find the greatest integer less than $\sqrt{2^{100}+10^{10}}.$
Answer: $2^{50}$
I tried, but I can't finish:
$\sqrt{2^{100}+(2\cdot5)^{10}}=\sqrt{2^{10}(2^{90}+5^{10})}=2^5\sqrt{2^{90}+5^{10}}$
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It results that
$2^{50}=\sqrt{2^{100}}<\sqrt{2^{100}+10^{10}}<\sqrt{2^{100}+16^{10}}=$
$=\sqrt{2^{100}+2^{40}}<\sqrt{2^{100}+2^{51}+1}=\sqrt{\left(2^{50}+1\right)^2}=2^{50}+1$
Hence ,
$2^{50}<\sqrt{2^{100}+{10^{10}}}<2^{50}+1\;.$
Consequently,
$\left\lfloor\sqrt{2^{100}+10^{10}}\right\rfloor=2^{50}\;.$
|
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|
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
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$(\sqrt{2})^9 \quad vs.\quad 9^\sqrt{2}\qquad$// square both side
$2^9=512 \quad vs.\quad
3^{4\sqrt{2}} \;≈\; 3^{5.657} \;<\; 3^5\,3^\frac{2}{3}$
$\displaystyle 3^5\,3^\frac{2}{3} = 243×2×\sqrt[3]{1+\frac{1}{8}} \;<\;486×\left(1+\frac{1}{3×8}\right) = 506.25$
$→ (\sqrt{2})^9 \;>\; 9^\sqrt{2}\qquad$
|
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|
Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\
=& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1
\end{align}
All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get
\begin{align}
g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\
&=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\
&= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt
\end{align}
for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$.
$$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$
Let $p(x):=2x^3+3x^2-2x-7$. It can be shown that $p(x)\le p(1)=-4, \forall x\in[0,1]$. We can take $a=\frac13$ since we can show, with a bit of work, $f''(\frac13)>0$.
(to be continued)
|
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
$$\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\
=& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1
\end{align}$$
All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get
$$\begin{align}
g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\
&=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\
&= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt
\end{align}$$
for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$.
$$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$
Let $p(x):=2x^3+3x^2-2x-7$. It can be shown that $p(x)\le p(1)=-4, \forall x\in[0,1]$. We can take $a=\frac13$ since we can show, with a bit of work, $f''(\frac13)>0$.
(to be continued)
|
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|
find $\int^6_0f(x)dx$ by definition , $f(x)=\left\lfloor\frac{x}{3}\right\rfloor$
Let $f(x)=\lfloor\frac{x}{3}\rfloor$,
Is $f(x)$ integrable at $[0,6]$?
If so, find $\int^6_0f(x)dx$ by definition.
The function is integratable because it is bounded and it has a finite number of points which are not continuous in $[0,6]$.
So first we need to divide the interval into even parts, I did that by $\frac{6-0}{n}=\frac{6}{n}=\Delta x$.
And then choose the integration points: let $x_k^* \in [0,6]$ and the points are $x_k^*=1+\frac{5n-1}{n}$.
Then calculate the summation:
$$S_n=\sum^n_{k=1}f(x_k^*)\cdot \Delta x = \sum^n_{k=1} \left\lfloor\frac{x_k^* }{3}\right\rfloor \cdot \frac{6}{n} = \sum^n_{k=1} \left\lfloor\frac{6-\frac{1}{n}}{3}\right\rfloor \cdot \frac{6}{n} = \sum^n_{k=1} \left\lfloor\frac{6n-1 }{3n}\right\rfloor \cdot \frac{6}{n} $$
Now if I calculate the limit of $S_N$ as $n$ approaches infinity I get that the answer is $0$ which is wrong according to the book it is supposed to be $3$ but it only provides the final answer.
Can anyone give me any tips and hints on that? I find this topic complicated.
Hopefully my translations are understandable. Thank you.
EDIT - my answer is also wrong even if I use $\Delta x=\frac{b-a}{n}$ and $x^*=a+ \Delta x \cdot k$
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by your definition $\Delta x = \frac{b-a}{n} = \frac{6}{n}$, so $x_k = \frac{k \Delta x}{n}$. $k$ starts from $0$ to $n$. so the sum would be:
$$
S_n = \sum_{k=0}^n f(x_k)\Delta x = \sum_{k=0}^n \left\lfloor \frac{6k}{3n} \right\rfloor \frac{6}{n} = \sum_{k=0}^n \left\lfloor \frac{2k}{n} \right\rfloor \frac{6}{n}
$$
$2k/n$ is less than $1$ if $k<\lceil n/2 \rceil$, so $\lfloor \frac{2k}{n} \rfloor$ is zero. only nonzero elements are for $ \lceil n/2 \rceil \leq k \leq n$ and $\left\lfloor \frac{2k}{n} \right\rfloor$ is just 1. so the summation becomes :
$$
S_n = \sum_{k=\lceil n/2 \rceil}^n \frac{6}{n} = \frac{6}{n} \sum_{k=\lceil n/2 \rceil}^n1 = \frac{6}{n} ( n - \left\lceil \frac{n}{2} \right\rceil).
$$
if you take limit for $n \rightarrow \infty$ , you get:
$$
\lim_{n \rightarrow \infty}S_n = \frac{6}{n} \left(\frac{n}{2}\right) = 3.
$$
you can find this value easily by linearity of integral:
$$
\int_0^6 \left\lfloor \frac{x}{3} \right\rfloor dx = \int_0^3 0 dx + \int_3^6 1 dx =\int_3^6 dx = 6-3 = 3.
$$
you didn't use correct formula for $x_k$ because there is no $k$ in your formula and hence in your sum!
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|
Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve for it but it'll be very lengthy. Is there any other short and nicer method.
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Let $\displaystyle \sin(x)+\cos(x)=z$
Then $[\sin(x)+\cos(x)]^2=z^2\Longrightarrow \sin(2x)=z^2-1$
So $\displaystyle \sin(x)+\cos(x)+\frac{2(\sin^2x+\cos^2x)}{\sin(2x)}+\frac{2(\sin x+\cos(x))}{\sin(2x)}=7$
$\displaystyle z+\frac{2}{z^2-1}+\frac{2z}{z^2-1}=7$
$\displaystyle z+\frac{2}{z-1}=7\Longrightarrow z^2-8z+9=0$
$\displaystyle z=4-\sqrt{7}\ $ because $-\sqrt{2}\leq z\leq \sqrt{2}$
So $\sin(2x)=z^2-1=(4-\sqrt{7})^2-1$
So $\displaystyle \sin(2x)=22-8\sqrt{7}=a-b\sqrt{7}$
$\displaystyle \Longrightarrow \frac{a+b}{10}=\frac{22+8}{10}=3$
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|
Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$. Prove or disprove that the inequality $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2.$
What do I do?
First I use this $$c^2+b^2 \geq 2bc.$$ If $$c^2+b^2 \geq 2bc,$$ then $$\dfrac{1}{c^2+b^2} \leq \dfrac{1}{2bc}.$$
So we have $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^{2}}+\dfrac{u^2}{\left(u^2+1\right)^{2}} \leq \dfrac{x^2}{(2x)^{2}}+\dfrac{y^2}{(2y)^2}+\dfrac{z^2}{(2z)^2}+\dfrac{u^2}{(2u)^2}=\dfrac{x^2}{4x^2}+\dfrac{y^2}{4y^2}+\dfrac{z^2}{4z^2}+\dfrac{u^2}{4u^2}=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=1 \geq \dfrac{16}{25}.$$
But as I understand $1$ is just a maximum, so the initial inequality can still be less than $\dfrac{16}{25}$.
Any hint would help a lot! Thanks in advance!
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Partial answer :
As user @alet show the inequality as the variable are not in $[0.5,2/3]$ I complete it :
Let $a,c,d\in[0.5,1/\sqrt{3}]$ and $b\in[0,0.5]$ such that $a+b+c+d=2$ and $a\geq d\ge c\ge b$ then we have :
$$af\left(a\right)+b\left(b\right)+cf\left(c\right)+df\left(d\right)\leq \left(2-b\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{2-b}\right)+bf\left(b\right)-\frac{16}{25}\leq 0\leq 16/25$$
where :
$$f\left(x\right)=\frac{x}{\left(x^{2}+1\right)^{2}}$$
The LHS is just weighted Jensen's inequality because for $x\in[0,1]$ :
$$f''(x)=\frac{12x\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{4}}\leq 0$$
For the LHS we have $b=x$:
$$g(x)=\left(2-x\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{2-x}\right)+xf\left(x\right)-\frac{16}{25}\leq \left(2-x\right)f\left(\frac{1.25-x+\left(0.5-x\right)^{2}}{2-x}\right)+xf\left(x\right)-\frac{16}{25}$$
Or :
$g(1/2-1/2*1/(x+1))=\frac{-8(14175x^{10}+151830x^{9}+725973x^{8}+2037080x^{7}+3707116x^{6}+4559360x^{5}+3823368x^{4}+2146160x^{3}+764600x^{2}+153600x+12800)}{(25(-5x^{2}-14x-10)^{2}(-5x^{2}-8x-4)^{2}(9x^{2}+18x+10)^{2})}< 0$
|
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|
How many integer solutions of $2^x+3^y-7n=0,~n \in \mathbb Z$? Consider the equation $2^x+3^y=7n,~n \in \mathbb Z$.
How many integer (positive) solutions are there for this equation? Can it have infinitely many positive integer solutions?
What is an example of a positive rational solution?
To reduce the number of variables, we fix $x$ as follows:
Fix $x=3$, then the equation becomes $$2^3+3^y=7n \Rightarrow 3^y=7n-8 \Rightarrow 3^y=27\Rightarrow y=3,$$ taking $n=5$. So $(3,3)$ is an integer solution. Is there other integer solutions for other $n$ ?
So when $x=3$, we can draw $3^y=7z-8$ and see the graph is asymptotic in nature. I doubt there is other positive integer solution.
If we fix $x=5$, then the equation becomes $$2^5+3^y=7n \Rightarrow 3^y=7n-32 \Rightarrow 3^y=3\Rightarrow y=3,$$ taking $n=1$. So $(5,1)$ is an integer solution. Again, if we draw the curve $3^y=7z-32$ and see it is asymptotic in nature. I doubt there is other positive integer solution except $(5,1)$.
My intuition says, if we fix $x$ to be positive integer, then $y$ can take only one positive integer. There might be positive rational solution, but I don't know off them.
Therefore the equation $2^x+3^y=7n,~n \in \mathbb Z$ and $x,y \in \mathbb N$ has "countably" many solutions, for each $x \in \mathbb N$ there is only one $y \in \mathbb N$.
Are there any examples of positive rational solutions that are not integer solutions?
I appreciate your answer.
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Rewriting the equation in modulo 7, we have
$$3^y\equiv -2^x \quad \pmod 7 $$
Noting that $2^3\equiv 1 \quad \pmod 7$, we get $3$ cases for $x$,
$\quad $ A. $x=3h$,$ \quad $ B. $x=3k+1 \quad $ and $\quad $ C. $3k+2$.
A. When $x=3h, $
$$
3^y \equiv-(2^{3}) ^h\equiv-1 \equiv 6 \quad \pmod 7
$$
By the Fermat Little Theorem, $3^6\equiv 1 \quad \pmod 7 $, we have
$$3^{6 m+3} \equiv \left(3^6\right)^m \cdot 3^3 \equiv 1^m \cdot 3^3
\equiv 6 \quad \pmod 7
$$
Therefore $y=6m+3$.
B. When $x=3k+1, $
$$
3^y \equiv-2(2^{3}) ^k \equiv-2 \equiv 5 \quad \pmod 7
$$
By the Fermat Little Theorem, $3^6\equiv 1 \quad \pmod 7 $, we have
$$3^{6 m+5} \equiv \left(3^6\right)^m \cdot 3^5 \equiv 1^m \cdot 3^5
\equiv 5 \quad \pmod 7
$$
Therefore $y=6m+5$.
C. When $x=3k+2, $
$$
3^y \equiv-4(2^{3}) ^k \equiv-4 \equiv 3 \quad \pmod 7
$$
By the Fermat Little Theorem, $3^6\equiv 1 \quad \pmod 7 $, we have
$$3^{6 m+1} \equiv \left(3^6\right)^m \cdot 3\equiv 1^m \cdot 3
\equiv 3 \quad \pmod 7
$$
Therefore $y=6m+1$.
We now can conclude that there are infinitely many integer solutions $(x,y,n)$ which are
$$
\boxed{\left(3 h, 6 m+3 , \frac{2^{3 h}+3^{6 m+3}}{7}\right),
\left(3 k+1,6 m+5, \frac{ 2^{3 k+1}+3^{6 m+5}}{7} \right) \quad \textrm{ and }
\left(3 k+2, 6m+1, \frac{2^{3k+2}+3^{6k+1}}{7}\right) }
$$
where $h\in N$ and $k,m \in N \cup\{0\}.$
|
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|
Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.
My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$
where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.
I'm not getting any idea how to solve it further.
|
Hint:
As $x(2x+1)\ne0,$
$$\dfrac{x^4}{x^3}=\dfrac{3x+2}{2x+1}\iff x^2-x-1=0$$
Similarly check, $$\dfrac{x^5}{x^4}=?,\dfrac{x^6}{x^5}=?$$
Find the common root of all?
|
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|
Find the sum of $\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$
Find the sum of $$\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$$
The solution in the book is a lot different than what I tried
what they did in the book is say $k-1=t$ then they expanded from this point $\sum_{t=0}^{n-1} (t+1)(n) {{n-1} \choose {t-1}}$ and the final answer is $n[(n-1)(2^{n-2}-1)+2^n-2]$ but I did not understand their way as it complicated to me
I thought about using derivatives but I got stuck and I am not sure if it really works I just remember using derivatives for some cases in class and I tried it
$(x+b)^n=$ ${n \choose 0} x^0 b^{n-0}+ {n \choose 1} x^1 b^{n-1} +...+ {n \choose n} x^n b^{n-n} $
derivative with respect to $x$ is
$n(x+b)^n-1=$ $0+ {n \choose 1} b^{n-1} +...+ {n \choose n} nx^{n-1} b^{n-n} $
then derivative again
$n(n-1)(x+b)^n-2=$ $0+0+ {n \choose 2} 2b^{n-2} +...+ {n \choose n} n(n-1)x^{n-2} b^{n-n} $
but I could not continue from here
is it ok to use derivatives ? is there another way other than the book(just expanding)?
thanks for any tips and help!
|
\begin{align}
\sum_{k=1}^n k(k-1) \binom{n}{k-1}
&= \sum_{k=2}^n k(k-1) \binom{n}{k-1} \\
&= \sum_{k=2}^n k n \binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2+2)\binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2)\binom{n-1}{k-2} + 2n \sum_{k=2}^n \binom{n-1}{k-2} \\
&= n (n-1) \sum_{k=3}^n \binom{n-2}{k-3} + 2n \sum_{k=2}^n \binom{n-1}{k-2} \\
&= n (n-1) \sum_{k=0}^{n-3} \binom{n-2}{k} + 2n \sum_{k=0}^{n-2} \binom{n-1}{k} \\
&= n (n-1) \left(2^{n-3} - 1\right) + 2n \left(2^{n-2} - 1\right)\\
&= n (n+3) 2^{n-3} - n (n+1)
\end{align}
|
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|
Number of distinct arrangement of $(a,b,c,d,e)$
If $a<b<c<d<e $ be
positive integer such that
$a+b+c+d+e=20$.
Then number of distinct
arrangement of $(a,b,c,d,e)$ is
Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \ 1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\ 4\ 5\ 6$
Now after that solution given in book as
Total number of ways
$ \displaystyle =\binom{4}{0}+\frac{\binom{4}{1}}{4}+\frac{\binom{4}{2}}{3}+\frac{\binom{4}{3}}{2}+\frac{\binom{4}{4}}{1}$
$\displaystyle = 1+1+2+2+1=7$
I did not understand last $2$ line
i e solution given in book
Please have a look on that part
|
For when $e=10$, there is only one arrangement of numbers that satisfies $a+b+c+d+e=20$, which is $(1,2,3,4,10)$. I am going to call this solution our default solution.
Consider when $e=9$, there are four different ways to make sure that the numbers sum to 20. You can choose one number out of $a,b,c,d$ from the default solution and add one to it. The solutions for $e=9$ are the following: $(2,2,3,4,9)$, $(1,3,3,4,9)$, $(1,2,4,4,9)$, $(1,2,3,5,9)$. There is only one solution here which fulfills $a<b<c<d<e$ and that solution is $(1,2,3,5,9)$.
Note that for $e=9$, there are $4\choose1$ possibilities since we pick any one out of $a,b,c,d$ to increase, but there is only one correct answer.
For $e=8$, there are two different ways to increase numbers in the default solution. You could either (a) pick two numbers out of $a,b,c,d$ and increase them both by 1, or you could (b) pick one number out of $a,b,c,d$ and increase that number by 2.
For (a), the possibilities are:
$(2,3,3,4,8)$
$(2,2,4,4,8)$
$(2,2,3,5,8)$
$(1,3,4,4,8)$
$(1,3,3,5,8)$
$(1,2,4,5,8)$.
For (b), the possibilities are:
$(3,2,3,4,8)$
$(1,4,3,4,8)$
$(1,2,5,4,8)$
$(1,2,3,6,8)$.
From (a), the only solution is $(1,2,4,5,8)$. From (b), the solution is $(1,2,3,6,8)$. Notice that for $e=8$, there are $4\choose2$ and $4\choose1$ possibilities for (a) and (b) respectively.
Following this pattern, for $e=7$ there will be ${4\choose3}+{4\choose2}+{4\choose1}$ different possibilities corresponding to how many numbers you are changing. For $e=6$ there are ${4\choose4}+{4\choose3}+{4\choose2}+{4\choose1}$ possibilities.
To reconstruct the equation you gave, I believe you have to count the instances $n$ which each $4\choose k$ occurs and divide by $n$. For example, $4\choose{k=1}$ occurs $n=4$ different times so that term in the equation is $\frac{{4\choose1}}{4}$. In total, there are 4 instances of $4\choose1$, 3 instances of $4\choose2$, 2 instances of $4\choose3$, and 1 instance of $4\choose4$. This results in the equation $\frac{{4\choose1}}{4}+\frac{{4\choose2}}{3}+\frac{{4\choose3}}{2}+\frac{{4\choose4}}{1}$. For the final $4\choose0$ term, I think that would be the scenario where $e=10$ since that is the default solution and we pick 0 numbers to change.
|
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|
A magic basis of $\mathbb{C}^5$ This is a a small $n$ restriction of another question.
Find a $5\times 5$ matrix of unit vectors $\xi_{ij}\in \mathbb{C}^5$ such that:
*
*Entries along rows and columns are orthogonal, that is for $1\leq i,j,k,l\leq 5$:
$$\delta_{i,k}+\delta_{j,l}=1\implies \langle \xi_{ij},\xi_{kl}\rangle=0.$$
*Entries not on a common row or column are neither parallel, anti-parallel, nor orthogonal, again for $1\leq i,j,k,l\leq 5$::
$$\delta_{i,k}+\delta_{j,l}=0\implies 0<|\langle \xi_{ij},\xi_{kl}\rangle|<1$$
This is proving a real wicked problem. Motivation and an $n=4$ example in the original questions.
Attempts thus far:
I have started with the first two rows :
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix}
\\ \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} \end{bmatrix}$$
To get orthogonality along the first row, we are multiplying four numbers, and along columns, just three. I started with 24th roots of unity but then this reduced to sixth roots, powers of $w=\exp(2\pi i/6)$. You can get some of what you want with this by considering what combinations of four and three sixth roots give zero. It seems possible to get all but one of:
*
*orthogonal along rows one, two, three
*appropriate orthogonality between rows one and two, and one and three
*appropriate orthogonality between rows two and three
But it seems to fail before it ever gets to rows four or five, or indeed the second condition of being non-orthogonal nor parallel, nor anti-parallel. This wasn't the best I did, but something like:
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1\\ 1\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ -1\\ -1\end{pmatrix} & \begin{pmatrix} 0 \\ w \\ w^4 \\ 1\\ -1\end{pmatrix} & \begin{pmatrix} 0 \\ 1\\ -1 \\ w^2\\ w^5\end{pmatrix}
\\ \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^2\\ w^4\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^5\\ 1\end{pmatrix} & \begin{pmatrix} w^2 \\ 0 \\ w^4 \\ w^2\\ w\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ -1 \\ w^4\\ -1\end{pmatrix}
\\ \begin{pmatrix} 0 \\ 0 \\ 2 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 1\\ w \\ 0 \\ w^5\\ -1\end{pmatrix} & \begin{pmatrix} w^2 \\ w \\ 0 \\ w^2\\ 1\end{pmatrix} & \begin{pmatrix} -1\\ w^2 \\ 0 \\ w^5\\ 1\end{pmatrix} & \begin{pmatrix} w^2 \\ w \\ 0 \\ w\\ w^2\end{pmatrix} \end{bmatrix}$$
This attempt falls down on orthogonality in row three, and orthogonality in row 2 vs row 3.
|
David Roberson has pointed me towards Figure 1 from Quantum symmetry vs nonlocal symmetry.
This is a so-called quantum latin square:
The produced quantum Latin squares are indexed by group elements and have the property that the inner product of the $i,j$-entry and $k,l$-entry depends only on $i^{-1}k$ and $j^{-1}l$. I would not be surprised if one can find a quantum Latin square meeting your requirements for all $n \geq 5$ by using this construction on the cyclic group $\mathbb{Z}_n$ (and picking the right permutation of its characters).
|
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|
Find rational solutions to $x^2 + y^2 = 6$ This question comes from Rational Points on Elliptic Curves (Silverman & Tate) Exercise $1.7$ (a).
Find rational solutions (if any) to $x^2 + y^2 = 6$
I think there exist no solutions and here is my proof:
Suppose that there is a rational point and write it as
$$x=\frac{X}{Z} \quad \text { and } \quad y=\frac{Y}{Z}$$
for some integers $X, Y$, and $Z$. Then
$$X^2 + Y^2 = 6Z^2$$
If $X, Y, Z$ have a common factor, then we may remove it, so we may assume that they have no common factor. It follows that neither $X$ nor $Y$ is divisible by $6$. This is true because if $6$ were to divide $X$, then $6$ divides $Y^2 = 6Z^2 - X^2$, so $6$ divides $Y$. But then $36$ divides $X^2 + Y^2 = 6Z^2$, so $6$ divides $Z$, contradicting the fact that $X, Y, Z$ have no common factors. Hence $6$ does not divide $X$, and a similar argument shows that $6$ does not divide $Y$.
Since $X$ and $Y$ are not divisible by $6$, we have
$$X \equiv \pm 1,4 \quad(\bmod 6) \text { and } Y \equiv \pm 1,4 \quad(\bmod 6),$$
and hence
$$X^2 + Y^2 \equiv 1+1 \equiv 2 \quad(\bmod 6)$$
or
$$X^2 + Y^2 \equiv 1+4 \equiv 5 \quad(\bmod 6)$$
or
$$X^2 + Y^2 \equiv 4+4 \equiv 3 \quad(\bmod 6)$$
However, we also have
$$X^2 + Y^2 = 3Z^2 \equiv 0 \quad(\bmod 6),$$
a contradiction.
I think this proof is sufficient. I based it on the method that the book showed there didn't exist any rational points on the curve $x^2 + y^2 = 3$.
|
The ideas in the proof are correct, but there are some minor errors that need correction to make your reasoning rigorous.
First, while it is true that $X,Z$ and $Y,Z$ may be taken to be relatively prime without loss of generality, and that this directly follows from $x = X/Z$ and $y = Y/Z$, it is not immediately obvious why $X,Y$ need to also be relatively prime. To see why, consider the alternative equation $$X^2 + Y^2 = 40Z^2,$$ which of course admits integer solutions such as $(X,Y,Z) = (2,6,1)$. Then $X, Z$ and $Y, Z$ are relatively prime but $X, Y$ are not. So it is misleading to start off with such a claim; instead, it is better to omit it and directly proceed to your reasoning that neither $X$ nor $Y$ can be divisible by $6$.
Second, you have a few typographical errors:
$$X^2 + Y^2 \equiv 4 + 4 = 8 \equiv \color{red}{2} \pmod 6$$
and your last claim should read
However, we also have $$X^2 + Y^2 = \color{red}{6}Z^2 \equiv 0 \pmod 6.$$
|
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|
Create unique table rotation for an event I am planning to host an event with 14 people participating in it. There will be five tables in total. (4 tables with 3 people each and 1 with 2).
I want to rotate the tables 5-6 times and to ensure that everyone gets to meet as many people as possible and minimize (if not completely avoid) that some people see each other twice.
I figured out how to do tables of 2, but I struggle with tables of 3.
I would very much appreciate your help!
|
Here's a 7-round schedule with every pair covered exactly once, obtained via integer linear programming:
1 {6,8,11} {10,12,14} {4,5,7} {3,9,13} {1,2}
2 {3,7,11} {1,4,10} {5,8,13} {2,6,14} {9,12}
3 {2,7,8} {5,9,10} {1,11,12} {3,4,14} {6,13}
4 {1,3,8} {4,6,9} {2,5,12} {10,11,13} {7,14}
5 {2,9,11} {6,7,10} {4,8,12} {1,13,14} {3,5}
6 {2,3,10} {8,9,14} {1,5,6} {7,12,13} {4,11}
7 {5,11,14} {3,6,12} {2,4,13} {1,7,9} {8,10}
|
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|
Show this inequality $\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$ For any $\triangle ABC$, prove or disprove $$\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$$
by $$\sin{A}+\sin{B}=2\sin{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}$$
$$\sin{A}+\sin{C}=2\sin{\dfrac{A+C}{2}}\cos{\dfrac{A-C}{2}}$$
This leads to prove
$$\dfrac{\sin{\frac{B}{2}}}{\sin{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}}+\dfrac{\sin{\frac{C}{2}}}{\sin{\dfrac{A+C}{2}}\cos{\dfrac{A-C}{2}}}\le\dfrac{1}{\sin{A}}$$
|
Let's assume $a,b,c$ are the sides of the triangle. We may suppose that $a=x+y$, $b=y+z$, and $c=x+z.$ If $p=\frac{a+b+c}{2}$, one can show that:
$$\sin \frac{B}{2}=\sqrt {\frac{(p-a)(p-c)}{ac}} =\sqrt{\frac{yz}{(x+y)(x+z)}}\\\sin \frac{C}{2}=\sqrt {\frac{(p-a)(p-b)}{ab}}=\sqrt{\frac{zx}{(x+y)(y+z)}}.$$
Moreover, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$; hence, we just need to verify the relations below:
$$\frac{\sqrt{\frac{yz}{(x+y)(x+z)}}}{(x+y)+(y+z)}+\frac{\sqrt{\frac{zx}{(x+y)(y+z)}}}{(x+y)+(x+z)} \leq \frac{1}{2(x+y)} \\ \iff \frac{\sqrt {yz(x+y)}}{((x+y)+(y+z))\sqrt{x+z}}+\frac{\sqrt {zx(x+y)}}{((x+y)+(x+z))\sqrt {y+z}} \leq \frac{1}{2}.$$
On the other hand, we have:
$$\frac{\sqrt {yz(x+y)}}{((x+y)+(y+z))\sqrt{x+z}}+\frac{\sqrt {zx(x+y)}}{((x+y)+(x+z))\sqrt {y+z}} \\ \leq \frac{\sqrt {yz(x+y)}}{2\sqrt{(x+y)(y+z)}\sqrt {x+z}}+\frac{\sqrt {zx(x+y)}}{2\sqrt{(x+y)(x+z)}\sqrt{y+z}}=\frac{\sqrt {yz}+\sqrt{xz}}{2\sqrt{y+z}\sqrt{x+z}};$$
but:
$$\frac{\sqrt {yz}+\sqrt{xz}}{2\sqrt{y+z}\sqrt{x+z}} \leq \frac {1}{2};$$
because:
$$\sqrt {yz}+\sqrt{xz} \leq \sqrt{y+z}\sqrt{x+z} \\ \iff z(\sqrt x +\sqrt y)^2 \leq (y+z)(z+x)=z^2+zx+zy+xy \\ \iff 2z\sqrt {xy} \leq z^2+xy,$$
which is clear.
We are done.
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|
How to simplify this trigonometric equation I am not sure how to further simplify this expression:
$$\sec^2(\arcsin(y/r)) \times \frac{\frac{1}{r}}{\sqrt{1-(\frac{y}{r})^2}} \times \frac{r}{1000}$$
How should I further simplify the $\sec^2(\arcsin(y/r))$?
The simplified version looks like this:
$$\left(\frac r{10\sqrt{r^2-y^2}}\right)^3$$
|
Starting from: $$\sec^2\left(\arcsin\left(\frac yr\right)\right) \times \frac{\frac{1}{r}}{\sqrt{1-(\frac{y}{r})^2}} \times \frac{r}{1000}.\tag 1$$
We know that $$\sec \left(\arcsin \left(x\right)\right)=\frac{\sqrt{1-x^2}}{1-x^2}\implies \sec^2 \left(\arcsin \left(x\right)\right)=\frac{1}{1-x^2}$$ then rewrite the $(1)$ like as:
$$\frac{1}{1-\left(\frac yr\right)^2}\times \frac{\frac{1}{r}}{\sqrt{1-(\frac{y}{r})^2}} \times \frac{r}{1000} \tag 2$$
$$\frac{r^2}{r^2-y^2}\times \frac{\frac{1}{\not r}}{\sqrt{1-(\frac{y}{r})^2}}\times \frac{\not r}{1000}=\frac{r^2}{r^2-y^2}\times \frac{r}{\sqrt{r^2-y^2}}\times \frac{1}{10^3}=\left(\frac{r}{10\sqrt{r^2-y^2}}\right)^3$$
|
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|
Limit of $(|x| + |y|)\ln(x^2 + y^4)$ at $(0,0)$ I want to show that $$\lim\limits_{(x,y) \to (0,0)} (\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4) = 0$$
First I let $\lVert (x,y) \rVert = \lvert x \rvert + \lvert y \rvert\ < \delta$, and assume that $x,y < 1$ so $x^2 + y^4 < \lvert x \rvert + \lvert y \rvert$. Then $\ln(x^2 + y^4) < \ln(\lvert x \rvert + \lvert y \rvert)$. However, $\lvert \ln(x^2 + y^4)\rvert > \lvert \ln(\lvert x \rvert + \lvert y \rvert)\rvert$, which is where I am stuck because I wanted to show that $\lvert(\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4)\rvert < \lvert x \rvert + \lvert y \rvert < \delta $. It does not seem like this approach will work & I am not sure what else I can try.
|
I will use the following inequality in this answer :
$\ln z<z\;\;$ for any $\;z\in\Bbb R^+.\quad\color{blue}{(*)}$
For any $\,(x,y)\in\Bbb R^2\!\setminus\!\{(0,0)\}\,$ such that $\,x^2+y^2<1\,,\,$ it results that
$\begin{align}\color{blue}{-16\sqrt[4]{x^2+y^2}}&=-8\sqrt[4]{x^2+y^2}-8\sqrt[4]{x^2+y^2}\leqslant\\&\leqslant-8\sqrt[4]{x^2}-8\sqrt[4]{y^2}=-8\sqrt[8]{x^4}-8\sqrt[8]{y^4}\leqslant\\&\leqslant-8\sqrt[8]{\dfrac{x^8}{x^4+y^4}}-8\sqrt[8]{\dfrac{y^8}{x^4+y^4}}=\\&=-8\dfrac{|x|}{\sqrt[8]{x^4+y^4}}-8\dfrac{|y|}{\sqrt[8]{x^4+y^4}}=\\&=-8\big(|x|+|y|\big)\!\cdot\!\dfrac1{\sqrt[8]{x^4+y^4}}<\\&\overset{\color{brown}{(*)}}{<}-8\big(|x|+|y|\big)\ln\!\bigg(\!\dfrac1{\sqrt[8]{x^4+y^4}}\!\!\bigg)=\\&=\big(|x|+|y|\big)\ln\left(x^4+y^4\right)<\\&\color{blue}{<\big(|x|+|y|\big)\ln\left(x^2+y^4\right)}<\\&<\big(|x|+|y|\big)\ln\left(x^2+y^2\right)\color{blue}{<0}\;.\end{align}$
Since $\!\lim\limits_{(x,y)\to(0,0)}\!\left(-16\sqrt[4]{x^2+y^2}\right)=0\;,\;$ by applying the Squeeze theorem, it follows that
$\lim\limits_{(x,y)\to(0,0)}\!\big(|x|+|y|\big)\ln\left(x^2+y^4\right)=0\;.$
|
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|
Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a homogenous differential equation. Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}[v-\frac{1}{v}]$
$\Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}$
$\Rightarrow -\frac{2v}{v^2+1}dv=\frac{dx}{x}$
Integrating both sides
$\Rightarrow -\log|v^2+1|=\log x+\log c$
$\Rightarrow -\log|\frac{y^2}{x^2}+1|=\log xc$
$\Rightarrow -\log|\frac{x^2+y^2}{x^2}|=\log xc$
$\Rightarrow \frac{x^2}{x^2+y^2}= xc$
$\Rightarrow x= c(x^2+y^2)$
$\Rightarrow y=\pm \sqrt{xc-x^2}$
Kindly review my solution and let me know if there are other methods of solving such problems.
|
Your answer is correct. Here is another way to solve the DE:
$$(x^2-y^2)dx+2xydy=0$$
Divide by $x^2dx$:
$$1+\dfrac {2xyy'-y^2}{x^2}=0$$
$$1+\left (\dfrac {y^2}{x}\right)'=0$$
Integrate:
$$x+\dfrac {y^2}{x}=C$$
|
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|
Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$ Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$
Note 2 things here please: I KNOW the differentiation rules, I KNOW that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$, but I'm having a trouble proving this with a certain method, you see I was training on the basic method -I don't really know what they call it- to get back to the origins of calculus, which is $\frac{d}{dx}f(x) = \lim_{h \to 0}\frac {f(x+h)-f(x)}{h}$
I've done the following:
$$\lim_{h \to 0}\frac {3(x+h)^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}$$
which is so far correct, but I'm struggling to transform this to $6x+\frac{2}{\sqrt{x}}$, I know the problem isn't hard but I wish that someone will be kind and tell me what the next step is.
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You made a mistake. There was a $- 4\sqrt{x}$ not $+4\sqrt{x}$.
$$\begin{aligned}\lim_{h \to 0}\frac {3(x+h)^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}& = \lim_{h \to 0}\frac {3x^2 + 6xh+3h^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}\\& = \lim_{h \to 0}\frac {6xh+3h^2+4\sqrt{x+h}-4\sqrt{x}}{h}\\&= \lim_{h \to 0}\frac {6xh+3h^2}{h}+\frac{4\sqrt{x+h}-4\sqrt{x}}{h}\\& = \lim_{h \to 0}{6x+3h}+\lim_{h\to 0}\frac{4(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+ \sqrt{x})}{h\sqrt{x+h}+ \sqrt{x})}\\&= 6x+\lim_{h\to 0}\frac{4(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+ \sqrt{x})}{h(\sqrt{x+h}+ \sqrt{x})}\\& = 6x+\lim_{h\to 0}\frac{4}{(\sqrt{x+h}+ \sqrt{x})}\\& = 6x+\frac{2}{\sqrt{x}}\end{aligned}$$
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|
How much information one root can give us about the other roots of a polynomial Suppose $a,b \in \mathbb{Q}$ and $f(X) \in \mathbb{Q}[X]$. In order to prove if $a+\sqrt{2}b$ is a root of $f(X)$, then $a-\sqrt{2}b$ is also a root, I did as follows.
I showed if $(a + \sqrt{2}b)^n=A_n+\sqrt{2}B_n$ then $(a - \sqrt{2}b)^n=A_n-\sqrt{2}B_n$ by induction on $n$, where $A_n, B_n \in \mathbb{Q}$. Then $f(a+\sqrt{2}b)=(c_0A_0+\cdots+c_nA_n)+\sqrt{2}(c_0B_0+\cdots+c_nB_n)$ implies $f(a-\sqrt{2}b)=(c_0A_0+\cdots+c_nA_n)-\sqrt{2}(c_0B_0+\cdots+c_nB_n)$, and the result is followed.
Now, my question is what can we say about roots of $f(X)$ if $a+\sqrt{\sqrt{2}}b$ is a root of $f(X)$?
Or, more generally, let $a,b \in R$, where $R$ is a domain, and $f(X) \in R[X]$. If $a+zb$ is a root of $f(X)$ where $z^i \not \in R$ for $1\le i<m-1$ and $z^m \in R$, then what can we say about roots of $f(X)$?
|
The minimal polynomial of $\sqrt{\sqrt{2}}$ is $X^4-2$. If $f(a + \sqrt{\sqrt{2}} b) = 0$ where $f(X) \in \mathbb Q[X]$ and $b \ne 0$, that says $\sqrt{\sqrt{2}}$ is a root of
$f(a + X b)$, so $X^4 - 2 \mid f(a + X b)$, i.e. $\left(\frac{Y-a}{b}\right)^4 - 2 \mid f(Y)$. Now $a - \sqrt{\sqrt{2}} b$ is a root of $\left(\frac{Y-a}{b}\right)^4 - 2$, so that $f\left(a - \sqrt{\sqrt{2}} b\right) = 0$.
This still leaves the case $b = 0$, but that's trivial because $a + \sqrt{\sqrt{2}} b = a = a - \sqrt{\sqrt{2}} b$.
|
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|
Complex numbers - solving for smallest positive value of $n$ Given that $z_1=2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)$ and $z_2=2\operatorname{cis}\left(\frac{2\pi}{3}\right)$ find the smallest positive value of $n$ such that $\left(\frac{z_1}{z_2}\right)^n \in \Bbb{R}^+$.
My attempt:
$$\frac{z_1}{z_2}=\frac{2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)}{2\operatorname{cis}
\left(\frac{2\pi}{3}\right)}=\sqrt{3}\operatorname{cis}\left(\frac{5\pi}{6}\right)$$
Hence,
$$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^n\operatorname{cis}\frac{5n\pi}{6}$$
Since $(\frac{z_1}{z_2})^n \in \Bbb{R}^+$, so
$$\sqrt{3}^ni\sin\frac{5n\pi}{6}=0$$
$$5n\pi=0,6\pi$$
$$n=0,1.2$$
At the same time, I also know that
$$\sqrt{3}^n\cos\frac{5n\pi}{6}>0$$
Solving for the above will lead to $n>\frac{3}{5}$
However, the textbook answer is $n=12$.
Somehow, I could not reconcile the my two solutions to $n=12$.
May I know where did I did wrongly? Thank you in advance.
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For $n\in\Bbb Z$ we have
$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^{\,n}\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $(\,\cos \frac{5n\pi}{6}>0 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $(\,\cos \frac{5n\pi}{6}=1 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $\frac {5n}{6}\,$ is twice an integer
iff $\,12|n$.
|
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|
How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(\frac{1}{x}+x\right)^2} d x\tag*{} $
Averaging them gives the exact value of the integral
$\displaystyle \begin{aligned}I & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+4} d x \\& =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{2}\right)\right]_0^{\infty} \\& =\frac{1}{4}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\& =\frac{\pi}{4}\end{aligned}\tag*{} $
I guess that we can similarly evaluate the general integral
$$
I_n=\int_0^{\infty} \frac{d x}{\left(x+\frac{1}{x}\right)^{2 n}}
$$
by mapping $x\mapsto \frac{1}{x}$ and then averaging.
$$
I_n=\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2 n}} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+4\right]^n}
$$
Letting $x-\frac{1}{x}=\tan \theta$ yields
$$
\begin{aligned}
I_n & =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \sec ^2 \theta d \theta}{4^n \sec ^{2 n} \theta} =\frac{1}{4^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta= \boxed{\frac{\pi(2 n-3) ! !}{4^n(2 n-2) ! !}}
\end{aligned}
$$
where the last answer comes from the Wallis cosine formula.
My question: How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$?
|
Writing the integral on the form
$$
I_n = \int_0^{\infty} \frac{x^{2n}}{(x^2+1)^{2n}}dx
$$
can be seen as a Mellin transform of $f(x)=(1+x)^{-\rho}$. The Mellin transform is given by
$$
\mathcal{M}[f(x);s]=\frac{1}{\Gamma(\rho)}\Gamma[s,\rho-s]$$
Which then gives that
$$
\mathcal{M}[f(x^2);s]=\frac{1}{2\Gamma(\rho)}\Gamma[s/2,\rho-s/2]
$$
With $\rho=2n$, $s=2n+1$, this gives
$$
I_n = \frac{1}{2\Gamma(2n)}\Gamma[n+1/2,n-1/2] = \frac{1}{2}B(n+1/2,n-1/2)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral with cosine in the denominator and undefined boundaries I am trying to solve the integral
$$ \int_0^{\frac{3}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx $$
And I get the primitive function to be
$$ \frac{2\arctan\left(\sqrt{\frac{7}{3}}\tan{x}\right)}{\sqrt{21}} $$
But $\tan(\frac{3}{2}\pi)$ is not defined. Mathematica gets the same primitive function and can solve it when using the given boundaries. How do I handle the boundaries?
|
Since the function is $\pi$-periodic, $$\int_{\pi}^{\frac{3}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx=\int_{0}^{\frac{1}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx.$$
It is also symmetric about $x=\frac{1}{2}\pi$-line and therefore by $x\rightarrow \pi-x$ substitution
$$\int_{\frac{1}{2}\pi}^{\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx=\int_{0}^{\frac{1}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx.$$
Hence,
$$\int_{0}^{\frac{3}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx=3\int_{0}^{\frac{1}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx.$$
Finally, by using OP's antiderivative, the result will be
$$3\left(\frac{2\arctan\left(\sqrt{\frac{7}{3}}\tan{x}\right)}{\sqrt{21}}\bigg{|}_0^{\frac{\pi}{2}}\right)=3\left(\frac{2\arctan\left(\infty\right)}{\sqrt{21}}\right)=\frac{\sqrt3}{\sqrt7}\pi.$$
|
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|
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$. I have the following question:
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.
Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx-1)^{1.5}$ for $x^3$ in the second equation. Ideas?
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$$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)\\=(x+\frac{1}{x})[(x+\frac{1}{x})^2-3]$$
|
{
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|
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
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Let:
$$S = 1 + 2 + \ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \ldots + 2 + 1.$$
Add the two equations term by term; each addition results in $n+1.$ So:
$$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1).$$
Divide by 2: $$S = \frac{n(n+1)}{2}.$$
|
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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.