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How to construct an example where $\sum_{n=1}^{+\infty} a_n$ diverges and $\prod_{n=1}^{+\infty} (1+a_n)$ converges?
My Problem: Construct an example where
$$ \sum_{n=1}^{+\infty} a_n \quad \text{diverges}, \qquad \text{and}\qquad \prod_{n=1}^{+\infty} (1 + a_n) \quad \text{converges}$$
My thoughts:
By '$\prod _{n=1}^{+\infty} (1 + a_n)$ converges', we have $a_n \to 0$, and by '$\sum_{n=1}^{+\infty} a_n$ diverges', we know that the sign of $\{a_n\}$ is not fixed. But I have no idea how to construct the $a_n$.
Please give me some hints to do it. Thanks a lot!
And how can I construct the ${a_n}$ when it is not constant?
|
Some author says that $\prod_{n=1}^{\infty}(1 + a_n)$ is convergent only if
$$ \lim_{N\to\infty} \prod_{n=1}^{N} (1+a_n) \in \mathbb{C}\setminus\{0\}. $$
This definition has some advantages over the more lenient version that allows $0$ as a possible value of infinite product, see the related Wikipedia article for instance.
Even with this definition, we are still able to construct an example.
Construction. Define $(a_n)_{n\geq 1}$ so as to satisfy
$$ (1+a_n)_{n\geq 1} = \left( 1, \quad \frac{2}{1}, \frac{1}{2}, \quad \frac{3}{2}, \frac{2}{3}, \frac{3}{2}, \frac{2}{3}, \quad \frac{4}{3}, \frac{3}{4}, \frac{4}{3}, \frac{3}{4}, \frac{4}{3}, \frac{3}{4}, \frac{4}{3}, \frac{3}{4}, \quad \ldots \right) $$
By the construction, it is clear that $\prod_{n=1}^{\infty} (1 + a_n) = 1$. On the other hand,
$$ (a_n)_{n\geq 1} = \left( 0, \quad \frac{1}{1}, -\frac{1}{2}, \quad \frac{1}{2}, -\frac{1}{3}, \frac{1}{2}, -\frac{1}{3}, \quad \frac{1}{3}, -\frac{1}{4}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{3}, -\frac{1}{4}, \quad \ldots \right), $$
and so,
$$ \sum_{n-1}^{2^N - 1} a_n
= \sum_{k=2}^{N} \sum_{n=2^{k-1}}^{2^k - 1} a_n
= \sum_{k=2}^{N} 2^{k-2} \left( \frac{1}{k-1} - \frac{1}{k} \right)
= \sum_{k=2}^{N} \frac{2^{k-2}}{k(k-1)} $$
and this diverges as $N \to \infty$. Using this, it is then not hard to check that $\sum_{n=1}^{\infty} a_n = \infty$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4618127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I can prove it.
$$ab^2aab^3a=ab^5a$$ so I'm thinking something here. we get $b^5=ab^5a$.
|
We are given that $b^3 = ab^2a$. If we cube this equation, we obtain that $b^9 = ab^6a = ab^3 \cdot b^3a$. Since from the given equation we deduce that $ab^3 = b^2a$ and $b^3a = ab^2$, we can make these substitutions to obtain $b^9 = b^2a \cdot ab^2 = b^4$. Hence, $b^5 = 1$, as desired.
Edit: The idea above shows that if $b^n$ and $b^m$ are conjugated in a group by an element of order $2$, then $b^{n^2-m^2} = 1$, which is a pretty neat and fairly general result.
|
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"url": "https://math.stackexchange.com/questions/4618547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the volume of the following region $E= (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) $
Find the volume of the following region $E= \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) \} $
I figured that the region E is formed by the points that belong in the gap between the two paraboloids
$z=\sqrt{2}(x^2 + y^2)$ $\hspace{20pt}$ and $\hspace{20pt}$ $ z= \sqrt{6}(x^2 + y^2)$
that are as well inside of the sphere $x^2 + y^2 + z^2 = 1$.
$E$ can be obtained by making a whole rotation around the $z$-axis in the plane-$xy$ then using Guldino's theorem can be appropriate.
Considering now cylindrical coordinates we've that $E$ becomes the following:
$F= \{ (\rho, \theta, z): \rho^2 + z^2 ≤1, \sqrt{2}\rho^2 ≤z≤ \sqrt{6}\rho^2, 0≤ \theta≤ 2\pi \}$
Then we've that:
$vol \hspace{2pt} E = \int _{F} \rho \hspace{2pt} d\rho d\theta dz = 2\pi \int\int _{D} \rho d\rho dz $
I'm struggling to find the region $D$. Any hint?
|
Sticking to cylindrical coordinates, you solve a quadratic to deduce that $z=\sqrt6r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt3}$. Similarly, $z=\sqrt2r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt2}$.
When $0\le r\le \dfrac1{\sqrt3}$, we have $\sqrt2 r^2\le z\le \sqrt6 r^2$. Then when $\dfrac1{\sqrt3}\le r\le \dfrac1{\sqrt2}$, we have $\sqrt 2r^2\le z\le \sqrt{1-r^2}$.
Thus, your volume is given by the sum of two iterated integrals:
$$\text{vol}(E) = \int_0^{2\pi}\int_0^{1/\sqrt3}\int_{\sqrt2 r^2}^{\sqrt6 r^2}r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_{1/\sqrt3}^{1/\sqrt2}\int_{\sqrt2 r^2}^{\sqrt{1- r^2}}r\,dz\,dr\,d\theta.$$
|
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"url": "https://math.stackexchange.com/questions/4621942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x$ How can we get a closed expression for this integral,
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x$$
where an complete elliptic integral $K(x)$ defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$ for $|x|<1$ appears and $x$ is the elliptic modulus?
*
*Observation 1: We can make a use of
$$
\int_{0}^{1}x^n K\left ( \sqrt{1-x^2} \right )\text{d}x
=\frac{\pi}{4} \frac{\Gamma\left ( \frac{n+1}{2} \right )^2 }{
\Gamma\left ( \frac{n+2}{2} \right )^2}.
$$
However, I don't know exactly the behaviour of function $K\left ( \sqrt{1-x^4} \right)$ around $x=0$. Expanding him at other points seems to also be a messy calculation.
*Observation 2: We may notice,
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x=2\int_{0}^{1} \frac{K\left ( \frac{x}{\sqrt{1+x^2} } \right )
K\left ( \sqrt{1-x^4} \right ) }{
\sqrt{1+x^2} }\text{d}x.
$$
I am grateful for all your help.
|
We have
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x
=\frac{\Gamma\left ( \frac14 \right )^4}{16}
{}_4F_3\left ( \begin{array}{c|}
\frac14,\frac14,\frac14,\frac14\\
\frac12,\frac12,1
\end{array}\text{ }1 \right )-\frac{\Gamma\left ( \frac34 \right )^4}{4}
{}_4F_3\left ( \begin{array}{c|}
\frac34,\frac34,\frac34,\frac34\\
1,\frac32,\frac32
\end{array}\text{ }1 \right ).
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4622179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How fast does the sequence $a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx$ decays as $n \to \infty$? The question is as in the title.
The sequence $\{a_n \}$ define by
\begin{equation}
a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx
\end{equation}
decays to $0$ as $n \to \infty$, which is a direct consequence of the Dominated Convergence Theorem.
But, when I run the Wolfram Alpha to evluate each $a_n$, I obtain the hypergeometric functions:
\begin{equation}
a_n=2^{-n-7/2}\Bigl(4 {_2}{F}_1\bigl(\frac{1}{2}, -n -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{2}\bigr)-{_2}{F}_1\bigl(\frac{1}{2}, -n, -\frac{1}{2}; \frac{3}{2} ; -\frac{1}{32}\bigr) \Bigr)
\end{equation}
But I cannot find information on the asymptotics of the above hypergeometric functions as $n \to \infty$.
Could anyone please help me?
|
If you perform the substitution $u = -\ln(1/2+x^2)$ you have that
\begin{equation}
a_n = \int_{1/8}^{1/2} \left( \frac{1}{2} + x^2\right)^{1/2+n} = \frac{1}{2} \int_{\ln(4/3)}^{\ln(64/33)} \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} e^{-nu} du
\end{equation}
This can be solved by Laplace's method. The minimum of $p(u)=u$ is at $\ln(4/3)$. Hence giving that at this point,
\begin{equation}
q(u) = \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} = \frac{3\sqrt{3}}{4} + \frac{9\sqrt{3}}{8}\left( u-\ln(4/3) \right)^2 +...
\end{equation}
and $p(u) = u \Rightarrow p(u) = \ln(4/3) + (u - \ln(4/3))$,
the two first terms are given by
\begin{equation}
a_n = \frac{1}{2} \int_{\ln(4/3)}^{\ln(64/33)} \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} e^{-nu} du = \left(\frac{3}{4}\right)^{n+1}\frac{\sqrt{3}}{2} \left(\frac{1}{n} + \frac{3}{n^3} + O(n^{-4}) \right)
\end{equation}
when $n \rightarrow \infty$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4624792",
"timestamp": "2023-03-29T00:00:00",
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|
Finding Taylor Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ in $1<|z|<2$. I have to find Taylor series of
$f(z)=\frac{z^2-2z+5}{(z-2)(z^2+1)}$, for $1<|z|<2$.
I have started with
\begin{eqnarray*}
f(z)&=&\textstyle\frac{z^2 -2z +5}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{1}{z-2}-\frac{2}{z^2+1}\\[4pt]
&=&\textstyle\frac{1}{-2\left(1-\frac{z}{2}\right)}-\frac{2}{z^2\left(1+\frac{1}{z^2}\right)}\\[4pt]
&=&-\sum\limits_{n=0}^{\infty}\frac{z^n}{2^{n+1}}-2\sum\limits_{n=0}^{\infty}(-1)^nz^{-2n-2}.
\end{eqnarray*}
But, this looks like Laurent series. Am I right about that?
|
Since the poles of $f$ are $\pm i$ and $2$, the radius of convergence of the Taylor series of $f$ around $0$ is $1$ (the smallest distance from $0$ to the poles).
It is then divergent for $1<|z|<2$.
For $|z|<1$ we have $f(z)=\sum\limits_{k\geqslant0}a_kz^k$,
where $a_k=\begin{cases}
\displaystyle -\frac1{2^{k+1}}+2(-1)^{\frac k2+1}&\textrm{if $k$ is even,}\\[8pt]
\displaystyle -\frac1{2^{k+1}}&\textrm{if $k$ is odd.}\end{cases}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4625086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
A sphere of constant radius $r$ passes through the origin $O$ and cuts the axes in $A,B,C.$ Find the locus of the foot of the perpendicular from $O$ A sphere of constant radius $r$ passes through the origin $O$ and cuts the axes in $A,B,C.$ Find the locus of the foot of the perpendicular from $O$ to the plane $ABC$.
My solution goes like this:
We consider the intercepts by the sphere as $(A,0,0),(0,B,0),(0,0,C)$ respectively. The equation of the plane $ABC$ is $\frac xA+\frac yB+\frac zC=1$ and the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$. Substituting, $(A,0,0),(0,B,0),(0,0,C),(0,0,0)$ in the equation of the sphere, we get $A=-2u,B=-2v,C=2w$.
We can write the equation of the plane as $\overrightarrow{r}(\frac {1}{A}\overrightarrow {i}+\frac {1}{B}\overrightarrow {j}+\frac {1}{C}\overrightarrow {k})=1 $ and hence, the the coordinates of the foot of perpendicular are $$\left(\frac{\frac {1}{A}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\mathbin,
\frac{\frac {1}{B}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\mathbin,
\frac{\frac {1}{C}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\right)\cdot$$
Now, the equation of the normal is $$\frac{x-0}{\frac 1A}=\frac{y-0}{\frac 1B}=\frac{z-0}{\frac 1C}=r_1.$$
Thus, $x=\frac {r_1}{A},y=\frac {r_1}{B},z=\frac {r_1}{C}.$
Now, since
$$\left(\frac{\frac {1}{A}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2},\frac{\frac {1}{B}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2},\frac{\frac {1}{C}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\right)
=(\alpha,\beta,\gamma)(\text{say}),$$
hence $\alpha^2+\beta^2+\gamma^2=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}.$
Also, we know that, $u^2+v^2+w^2=r^2$ which implies $A^2+B^2+C^2=4r^2.$
Now, $(\alpha,\beta,\gamma)$ lies on the line
$$\frac{x-0}{\frac 1A}=\frac{y-0}{\frac 1B}=\frac{z-0}{\frac 1C}=r_1.$$
Hence, $\alpha=\frac {r_1}{A}$, $\beta=\frac {r_1}{B}$, $\gamma=\frac {r_1}{C}$, due to which $r_1=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\cdot$
Also, $r_1=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}
=(\alpha^2+\beta^2+\gamma^2)
=\alpha A=\beta B=\gamma C.$
So, $A=\frac{\alpha^2+\beta^2+\gamma^2}{\alpha}$,
$B=\frac{\alpha^2+\beta^2+\gamma^2}{\beta}$,
$C=\frac{\alpha^2+\beta^2+\gamma^2}{\gamma}\cdot$
Again, $A^2+B^2+C^2=4r^2$ thus
$$\begin{eqnarray*}
&&\left(\frac{\alpha^2+\beta^2+\gamma^2}{\alpha})^2\right)
+\left(\frac{\alpha^2+\beta^2+\gamma^2}{\beta}\right)^2
+\left(\frac{\alpha^2+\beta^2+\gamma^2}{\gamma}\right)^2\\
&&\qquad{}={}4r^2\\
&&\qquad{}={}\left(\frac{x^2+y^2+z^2}{x}\right)^2
+\left(\frac{x^2+y^2+z^2}{y}\right)^2
+\left(\frac{x^2+y^2+z^2}{z}\right)^2\\
&&\qquad{}={}4r^2,
\end{eqnarray*}$$ is the required locus.
Is the above solution correct? If not, where is it going wrong?...
|
I checked your computations. I agree with implicit equation that can be written as well under the form:
$$(x^2+y^2+z^2)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=4r^2\tag{1}$$
I will take now on WLOG the value $r=\frac12$ in order that the RHS in (1) is $1$.
(1) can be given the equivalent form avoiding "divisions by zeros" :
$$(x^2+y^2+z^2)^2\left(x^2y^2+y^2z^2+z^2x^2\right)=x^2y^2z^2\tag{2}$$
which has a representation looking like orbitals in chemistry :
For those who are interested, here is the Matlab program that I have written for generating the above image :
clear all;close all;hold on;axis equal off;
a=0.3;I=-a:0.01:a;[x,y,z]=meshgrid(I,I,I);
axis([-a,a,-a,a,-a,a]);
f=((x.^2+y.^2+z.^2).^2).*((x.*y).^2+(y.*z).^2+(z.*x).^2)-(x.*y.*z).^2;
[faces,verts,colors] = isosurface(x,y,z,f,0,abs(z));
patch('Vertices',verts,'Faces',faces,'FaceVertexCData',colors,...
'FaceColor','interp','EdgeColor','none');
view([20,24]);
plot3([0,a,0,0,0,0,0],[0,0,0,a,0,0,0],[0,0,0,0,0,0,a],'color','k')
Edit : In fact, the form of the coordinates of the projection point $$(\frac{\frac{1}{A}}{D},\frac{\frac{1}{B}}{D},\frac{\frac{1}{C}}{D})
\text{ where } D:=(\frac1A)^2+(\frac1B)^2+(\frac1C)^2)\tag{3}$$
evokes a possible connection with the geometric transform called inversion. Indeed, we can establish such a relationship. Let us consider the simplest inversion, with the origin as its center and with "power" $1$, wich means that point $M$ has image $M'$ if and only if :
$$\vec{OM'}=\frac{\vec{OM}}{\|\vec{OM}\|^2}$$
which is equivalent to
$$O,M,M' \ \text{are aligned and } \vec{OM'}.\vec{OM'}=1$$
Let $A(a,0,0), B(0,b,0), C(0,0,c)$ (please note the slight notational changes). The image of the plane $A,B,C$ is the sphere passing through the origin $O$ and the three points
$A'(\frac{1}{a},0,0), B'(0,\frac{1}{b},0), C'(0,0,\frac{1}{c})$
This sphere passing through these 4 points is easily proven to have equation :
$$x^2+y^2+z^2-\frac{1}{a}x-\frac{1}{b}y-\frac{1}{c}z=0$$
in which we "read" the coordinates of its center
$$D(\frac{1}{2a},\frac{1}{2b},\frac{1}{2c}).$$
Now consider line $OD$ : it intersects the sphere in a point $H'$ where line $OD$ is a diameter ; therefore $OD$ is orthogonal to the sphere in $H'$ but it means that $H'$ is the image of the foot of the altitude because orthogonality is preserved by inversion. Therefore (3) is established by unicity of a point $M$ such that $OM$ is orthogonal to the sphere.
|
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|
Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
|
A solution proceeding from where the OP has stopped:
Let's assume $a=x^2, b=y^2, c=z^2$. We will show:
$$(y^2x+z^2y+x^2z)^2+18 \geq9(x^2+y^2+z^2).$$
To do so, consider the function $$f(x)=(y^2x+z^2y+x^2z)^2+18-9(x^2+y^2+z^2),$$
where $x,y,z \geq 1$. We have:
$$f^{\prime}(x)=2(y^2x+z^2y+x^2z)(y^2+2xz)-18x \\ \geq 2(x+1+x^2)(1+2x)-18x\\ \geq 2(3x) \times 3-18x =0.$$
So, $f(x)$ is an increasing functions. Therefore, we will be done if we prove that $f(1) \geq 0$; in other words, $f(1)=(y^2+z^2y+z)^2+9-9(y^2+z^2) \geq 0.$
Similarly, let's consider the function:
$$g(y)=(y^2+z^2y+z)^2+9-9(y^2+z^2).$$
Then,
$$g^{\prime} (y)=2(y^2+z^2y+z)(2y+z^2)-18y \\ \geq 2(y^2+y+1)(2y+1)-18y\\ \geq 2(3y) \times 3-18y=0;$$
hence $g(y)$ is an increasing function as well. As a result:
$$f(x) \geq f(1)=g(y) \geq g(1)=(1+z^2+z)^2-9z^2 \\ \geq (3z)^2-9z^2=0.$$
We are done.
PS:How did the OP get to that inequality?
By the Cauchy-Schwarz inequality:
$$(\frac{ab^2}{a+c}+\frac{bc^2}{b+a}+\frac{ca^2}{b+c})((a+c)+(b+a)+(a+c))\\ \geq (b \sqrt a +c \sqrt b+a \sqrt c)^2 \\ \implies \frac{ab^2}{a+c}+\frac{bc^2}{b+a}+\frac{ca^2}{b+c} \geq \frac{(b \sqrt a +c \sqrt b+a \sqrt c)^2}{2(a+b+c)}$$
Similarly,
$$(\frac{1}{a+c}+ \frac{1}{b+a}+\frac{1}{b+c})((a+c)+(b+a)+(a+c))\geq (1+1+1)^2 \\ \implies \frac{1}{a+c}+ \frac{1}{b+a}+\frac{1}{b+c} \geq \frac{9}{2(a+b+c)},$$
therefore:
$$\frac{ab^2+2}{a+c}+\frac{bc^2+2}{b+a}+\frac{ca^2+2}{b+c} \geq \frac{(b \sqrt a +c \sqrt b+a \sqrt c)^2+18}{2(a+b+c)}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Can we evaluate $\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$ After investigating the integral
$$\boxed{\quad \int \frac{1}{\sin ^5 x+\cos ^5x} d x \\=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}-2}}\right)\\ \qquad\ -\frac{1}{\sqrt{\sqrt{5}+2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sqrt{5}+2}}\right)\right]+C}$$
in my post, I try to evaluate
$$\int \frac{1}{\sin ^ {7} x+\cos ^ {7} x} d x$$
using the factorisation
\begin{aligned}
\sin ^{7} x+\cos ^7 x&=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^5 x+\cos ^5 x\right)-\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\
&=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^3 x+\cos ^3 x\right) -\sin ^2 x \cos ^2 x(\sin x+\cos x) \\
&\quad -\sin ^2 x \cos ^2 x\left(\sin ^3 x+\cos ^3 x\right) \\
&=(\sin x+\cos x)\left(1-\sin x \cos x-2 \sin ^2 x \cos ^2 x+\sin^3x\cos^3x\right)
\end{aligned}
Let $t=\sin x-\cos x$, then $d t=(\cos x+\sin x) d x$ and $\sin x\cos x= \frac{1-t^2}{2}$ and yields
$$
\begin{aligned}
I
=\int \frac{8}{\left(t^2-2\right)\left(t^6+t^4-9 t^2-1\right)} d t
\end{aligned}
$$
where I was stuck in the last integrand with power $6$. Your help and comments are highly appreciated.
My Question:
Is it difficult go further with
$$\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$$
where $n\geq 3.$
|
Here is a generalized approach, which decomposes the integrand with $a_k=\frac{\pi k}{2n+1}$
\begin{align}
\frac{2n+1}{\sin^{2n+1}x+\cos^{2n+1}x}
= &\ \frac{2^n}{\sin x+\cos x}\\ &+2^{n+1}\sum_{k=1}^n(-1)^{k}\frac{\cos^{n}2a_k}{\sec a_k }\frac{\sin x+\cos x}{1+\cos 2a_k\sin2x} \\
\end{align}
Then, the decomposed terms on the RHS can be integrated individually. For example
\begin{align}
&\int \frac1{\sin^{7} x+\cos^{7} x}dx\\
=&\ \frac{8}7 \int \frac{1}{2-t^2}
- \frac{2\cos\frac\pi7 \cos^2\frac{2\pi}7}{{\sec\frac{2\pi}7+1-t^2}}
+\frac{2\cos\frac{2\pi}7 \cos^2\frac{4\pi}7}{{\sec\frac{4\pi}7+1-t^2}}
-\frac{2\cos\frac{3\pi}7 \cos^2\frac{6\pi}7}{{\sec\frac{6\pi}7+1-t^2}}\ dt
\end{align}
where the substitution $t=\sin x-\cos x$ is made.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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|
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$
I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.
|
Write \begin{align}&\frac{r^2-1}{r^4+r^2+1} = \frac{Ar+B}{r^2-r+1}+\frac{Cr+D}{r^2+r+1}\\ \iff &r^2-1 = (Ar+B)(r^2+r+1)+(Cr+D)(r^2-r+1)\tag{1}\end{align}
You can now expand the RHS as a polynomial in $r$ and by direct comparison of coefficients on the LHS and the RHS get a $4\times4$ linear system.
Alternatively, let $\alpha$ be a root of $x^2-x+1$. In particular, we have $\alpha^2 = \alpha - 1$. Evaluating $(1)$ at $\alpha$ gives us
\begin{align}\alpha^2-1 &= (A\alpha + B)(\alpha^2+\alpha+1)\\
(\alpha-1)-1 &= (A\alpha + B)((\alpha-1)+\alpha+1)\\
\alpha - 2 &= 2\alpha(A\alpha+B)\\
\alpha - 2 &= 2A\alpha^2+2B\alpha\\
\alpha - 2 &= 2A(\alpha - 1)+2B\alpha\\
\alpha - 2 &= 2(A+B)\alpha-2A \implies 2(A+B)=1,\ -2A = -2
\end{align}
Note that it's important to get rid of any higher powers of $\alpha$ besides $\alpha$ and $1$ before you compare coefficients. Similarly, if you take $\beta$ a root of $x^2+x+1$, you can calculate $C$ and $D$ analogously.
Now that you can calculate partial fractions decomposition, we have
\begin{align}
\sum_{r = 1}^n \frac{r^2-1}{r^4+r^2+1} &=\sum_{r = 1}^n\left( \frac{2r-1}{2(r^2-r+1)}-\frac{2r+1}{2(r^2+r+1)} \right) \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2r+1}{2(r^2+r+1)}\\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2(r+1)-1}{2((r+1)^2-(r+1)+1)} \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 2}^{n+1} \frac{2r-1}{2(r^2-r+1)} \\
&= \frac 12 - \frac{2n+1}{2(n^2+n+1)}
\end{align}
Taking a limit as $n\to \infty$ gives us $$\sum_{r = 1}^\infty \frac{r^2-1}{r^4+r^2+1} = \frac 12.$$
|
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"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
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|
Solving $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$ using complex numbers My teacher gave this question to me while teaching complex numbers:
If $\;\sin A\!+\!\sin B\!+\!\sin C=\cos A \!+\!\cos B\!+\!\cos C=0$, then find:
*
*$\cos 2A + \cos 2B + \cos 2C$
*$\cos 3A + \cos 3C + \cos 3C$
*$\cos(A\!-\!B)+\cos(B\!-\!C)+\cos(C\!-\!A)$
Now, I tried it simplifying using this formulae derived from Euler form of complex number: $\;2\cos x=e^{ix}+e^{-ix}\;$,
but nothing seems to work.
I do not understand, what is the significance of solving this question with complex numbers ?
|
As noted by copper.hat, the conditions on cosine and sine imply that $$e^{iA} + e^{iB} + e^{iC} = 0.$$ These vectors all lie on the unit circle $|z| = 1$ in the complex plane. Thus, the sum of the first two vectors must be of length one, so as to be cancellable by the third. This can only occur if the angle between them is $2\pi/3$, or 120 degrees; see below for more detail. Thus $B = A \pm 2\pi/3$. Similarly, $C = A \pm 2\pi/3$, but $C$ cannot equal $B$ or the identity will not hold. Thus, we must have either $B = A + 2\pi/3$ and $C = A - 2\pi/3$, or the opposite. Since the expressions to be evaluated do not change when we swap $B$ and $C$, we may as well use the first choice.
To see in more detail why the angle between $A$ and $B$ must be $2\pi/3$, note that
\begin{align}
1 &= |-e^{iC}|^2 \\&= |e^{iA} + e^{iB}|^2 \\&= 2 + 2 \Re (e^{i(A - B)}),
\end{align}
from which it follows that $\cos(A - B) = -1/2.$ ($\Re$ means "real part.")
We can now answer the questions:
*
*\begin{align}
\cos{2A}+ \cos{2B} + \cos{2C} &= \Re (e^{2iA} + e^{2iB} + e^{2iC})\\ &= \Re(e^{2iA}(1 + e^{i4\pi/3} + e^{-i4\pi/3})) \\&= 0.
\end{align}
*\begin{align}\cos{3A}+ \cos{3B} + \cos{3C} &= \Re (e^{3iA} + e^{3iB} + e^{3iC}) \\&=
\Re (e^{3iA} + e^{i(3A + 2\pi)} + e^{i(3A- 2\pi)}) \\&=
3\cos{3A}.\end{align}
*\begin{align}\cos(A-B) + \cos(B-C) + \cos(C-A) &= \cos(-2\pi/3) + \cos(4\pi/3) + \cos(-2\pi/3) \\&= -3/2.\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4635521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Inverse related inverse trigonometric functions Prove that for $x<0$,
$\sin^{-1}\sqrt{1-x^2}=\cos^{-1}(-x)$
This question is related to Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ but want to solve it differently.
I checked it on
If I use $x=\cos\theta$ then i get $\sin^{-1}|\sin\theta|$ but not able to use the sign
|
Let $\arccos(-x) = \theta$. Then $\theta$ is the unique angle in the interval $[0, \pi]$ such that $\cos\theta = -x$. Since $x < 0$, $-x > 0$, so $\arccos(-x) = \theta \in [0, \pi/2)$.
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\sin^2\theta & = 1 - \cos^2\theta\\
& = 1 - (-x)^2\\
& = 1 - x^2\\
|\sin\theta| & = \sqrt{1 - x^2}\\
\sin\theta & = \sqrt{1 - x^2}
\end{align*}
since $\sin\theta \geq 0$ in the interval $[0, \pi/2)$.
Let $\arcsin \sqrt{1 - x^2} = \varphi$. Then $\varphi$ is the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\varphi = \sqrt{1 - x^2}$. The only such angle is $\theta$. Hence, $\varphi = \theta$. Thus, $\arcsin \sqrt{1 - x^2} = \arccos(-x)$.
|
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|
Closed form of $\sum_{x=1}^{n} \dfrac{x-1}{e^x}$ Is there a closed form for this series?
$$\sum_{x=1}^{n} \dfrac{x-1}{e^x}$$
Ultimately, I want to find its convergence i.e.
$$\lim_{n \to \infty} \sum_{x=1}^{n} \dfrac{x-1}{e^x}$$
By the integral test, the series converges and for arbitrary large $n$ my code tells me it's approaching $\dfrac{1}{(e-1)^2}$
|
Define
$$f(y)=\sum_{x=1}^n y^x=\frac{y(1-y^n)}{1-y}$$
Differentiate $f$ with respect to $y$ and manipulate as follows:
$$\begin{aligned}
\frac{\text{d}f}{\text{d}y}=\sum_{x=1}^nxy^{x-1}&=\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}\\
\sum_{x=1}^nxy^x&=y\cdot\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}\\
\sum_{x=1}^n(x-1)y^x&=y\cdot\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}-f(y)\\
&=y\cdot\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}-\frac{y(1-y^n)}{1-y}\\
&=\frac{y(y-ny^n+(n-1)y^{n+1})}{(1-y)^2}
\end{aligned}$$
From here, substitute in $y=e^{-1}$ to get
$$\sum_{x=1}^n\frac{x-1}{e^x}=\frac{e^{-n}(n-1-en)+1}{(e-1)^2}$$
|
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|
Coefficient of a term in an expanded polynomial I came across this question while solving problems on functions. Find the coefficient of $x^{203}$ in the expansion of the following expression: $(x-2)((x)(x+1)(x+2)(x+3)...(x+202))$.
The solution given in the text is:
$(x-2)((x)(x+1)(x+2)(x+3)...(x+202)) = (x-2)(x^{203} + x^{202}(1+2+3+4...+202)+....)$
So the coefficient of $x^{203} = -2 + (1 + 2 + .... + 202) = 20501$
The part I don't understand is how the coefficient of $x^{202}$ can be written directly as (1+2+3+....+202).
*
*Is there any theorem or a method for the expansion of a polynomial in this way?
*Is there a way I can get the coefficient of any term I want($x^{199}, x^{46}$, etc.)
|
Observe that $$x(x+1)=x^2+x\\x(x+1)(x+2)=x^3+3x^2+2x\\x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x\\x(x+1)(x+2)(x+3)(x+4)=x^5+10x^4+35x^3+50x^2+24x\\\dots$$
There seems to be a pattern forming regarding all coefficients, don't you agree?
But we are primarily interested in the coefficients of the second highest power of $x$.
We notice that these follow the sequence $1,3,6,10,\dots$ which is called the Triangular Number Sequence and whose $n$-th term is given by the formula $$P(n)=\frac{n(n+1)}{2}$$
Thus, for $n=202$, we obtain $P(202)=20503$, from which (because in this particular problem we have the term $(x-2))$ we get the desired $20503-2=20501$.
|
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|
2023 MIT Integration Bee Regular Season Problem 6 How do we solve $$\int\left(\frac{x^6+x^4-x^2-1}{x^4}\right)e^{x+\frac{1}{x}}dx$$? This is an integral in the 2023 MIT integration bee. I thought that maybe this is a consequence of the chain rule, but that is not the case. Then I thought there is a product rule hiding in the integrand, but it doesn't seem like it.
|
First, factor the integrand:
$$\frac{x^6+x^4-x^2-1}{x^4}=\frac{(x^4-1)(x^2+1)}{x^4}=\left(\frac{x^2+1}{x}\right)^2\left(1-\frac1{x^2}\right).$$
Now, let $y=x+1/x$; we have $dy=(1-1/x^2)dx$, so our integral becomes
$$\int \left(\frac{x^2+1}{x}\right)^2e^ydy=\int y^2e^ydy.$$
A standard integration by parts gives
$$\int y^2e^ydy=(y^2-2y+2)e^y+C,$$
so we have the answer
$$\left(\left(x+\frac1x\right)^2-2\left(x+\frac1x\right)+2\right)e^{x+1/x}+C.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4639264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Generating Functions - Counting how many solutions for an exact $x^n$ Having trouble with this problem:
$$ x_1+x_2+x_3+x_4+x_5+x_6 = 13 $$
$$ x\ne 3 $$
Thanks in advance for any further help.
Being able to get to this point:
$$\begin{split}
f(x) &= (x^0+x^1+x^2+x^4+x^5+\cdots)^6\\
&= [(-x^3)+{x^0+x^1+x^2+(x^3)+x^4+x^5+\cdots}]^6\\
&= \left[{(-x^3)} + \frac{1}{(1-x)}\right]^6\\
&=\left[ \frac{-x^3(1-x)+1}{(1-x)}\right]^6\\
&=\left[\frac {x^4-x^3+1}{1}\cdot\frac {1}{1-x}\right]^6\\
&=(x^4-x^3+1)^6 \frac {1}{(1-x)^6}\\
\end{split}$$
Now if I long divide $(x^4-x^3+1)$ with $(1-x)$ I get $[(-x^3 + \frac {1}{1-x}) (1-x)]^6 $ thus:
$$\begin{split}
f(x)&=\left(\frac {1}{1-x} -x^3\right)^6 \frac {(1-x)^6}{(1-x)^6}\\
&=\left(\frac {1}{1-x} -x^3\right)^6\\
\end{split}$$
That's where I get stuck, honestly.
|
It seems the coefficient needs to be manually extracted. Using the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ we obtain
\begin{align*}
\color{blue}{[x^{13}]}&\color{blue}{\left(\frac{1}{1-x}-x^3\right)^6}\\
&=[x^{13}]\sum_{k=0}^6\binom{6}{k}\frac{(-1)^kx^{3k}}{(1-x)^{6-k}}\\
&=\binom{6}{0}[x^{13}]\frac{1}{(1-x)^6}-\binom{6}{1}[x^{10}]\frac{1}{(1-x)^5}+\binom{6}{2}[x^7]\frac{1}{(1-x)^4}\tag{1}\\
&\qquad-\binom{6}{3}[x^4]\frac{1}{(1-x)^3}+\binom{6}{4}[x^{1}]{(1-x)^2}\\
&=\binom{6}{0}\binom{-6}{13}(-1)^{13}-\binom{6}{1}\binom{-5}{10}(-1)^{10}+\binom{6}{2}\binom{-4}{7}(-1)^7\\
&\qquad-\binom{6}{3}\binom{-3}{4}(-1)^4+\binom{6}{4}\binom{-2}{1}(-1)^1\tag{2}\\
&=\binom{18}{5}-6\binom{14}{4}+15\binom{10}{3}-20\binom{6}{2}+15\binom{2}{1}\tag{3}\\
&=8\,568-6\,006+1\,800-300+30\\
&\,\,\color{blue}{=4\,092}
\end{align*}
Comment:
*
*In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (2) we select the coefficient $[x^k]$ from the binomial series expansion $[x^k](1-x)^{-\alpha}=[x^k]\sum_{n=0}^{\infty}\binom{-\alpha}{n}=\binom{-\alpha}{k}(-1)^k$.
*In (3) we use the binomial identity $\binom{-p}{q}(-1)^q=\binom{p+q-1}{p-1}$.
|
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|
Putnam, A3 (2016) Let $f: \mathbb R\rightarrow \mathbb R$ satisfying \begin{equation*} f(x) + f\bigg(1-{1\over x}\bigg) = \arctan x\end{equation*}
for all $x\neq 0$. I want to evaluate \begin{equation*} \int_0^1 f(x) dx \end{equation*}
I know that $\tan\bigg(f(x) + f\bigg(1-{\cfrac 1x}\bigg)\bigg) = x$ and $-\pi/2 < f(x) + f\bigg(1-{1\over x}\bigg) < \pi/2$, but none of this seems useful yet. Since $x\neq 0$, the answer should be the solution to
\begin{equation*}
\lim_{n\rightarrow 0}\int_n^1f(x)dx\end{equation*}
But again, I'm still unsure where to go from here. Any hint would be appreciated.
|
Answer being revised
Noting that on $]0,\pi/2[$ :
$$\arctan(\frac{1}{x})+\arctan(x)=\dfrac{\pi}{2} $$
Then your relation can be writen
$$ f(x)+f(1-1/x)=\arctan(x) $$
$$ f(1-1/x)+f(x) = \dfrac{\pi}{2}-\arctan(\frac{1}{x})=\dfrac{\pi}{2}-(f(1/x)+f(1-x))$$
First
\begin{align*}
\int_0^1 f(1-1/x)+f(1/x)dx& = \dfrac{\pi}{2}-\int_0^1 f(1-x)+f(x)dx \\
\int_1^\infty \dfrac{f(1-u)+f(u)}{u^2}du&= \dfrac{\pi}{2}-2\int_0^1f(x)dx
\end{align*}
Second
Composing two times in addition, as suggested in comments by @Daniel Schepler $ x \to 1-1/x$ :
\begin{align*} f(\frac{1}{1-x}) + f(x) & =\arctan(\frac{1}{1-x})\\
&=\text{sgn}(1-x)\dfrac{\pi}{2}-\arctan(1-x)\\
& =\text{sgn}(1-x)\dfrac{\pi}{2}-(f(1-\frac{1}{1-x})+f(1-x))\\
\end{align*}
Taking $x \in ]0,1[$,
\begin{align*}\int_0^1 f(1-\frac{1}{1-x})+f(\frac{1}{1-x})dx&=\dfrac{\pi}{2}-\int_0^1 f(1-x)+f(x)dx \\
&= \dfrac{\pi}{2}-2\int_0^1f(x)dx \\
\end{align*}
So
$$ -\int_1^\infty \dfrac{f(1-u)+f(u)}{u^2}du=\dfrac{\pi}{2}-2\int_0^1f(x)dx $$
Conclusion by summing the two part recognizing opposite integrals, if everything is ok.
$$ \int_0^1 f(x)dx=\dfrac{\pi}{4}$$
|
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|
Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)}
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) $$
Proof:
For $a,b>0$,
Using $\ln \left(a^2+b^2\right)=2 Re(\ln (a+b i))$, we can reduce the power $4$ to $2$.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x
& = 2\int_{0}^{\infty} \frac {Re\left[\ln \left(x^2+a ^2i\right)\right]}{b^2+x^2} d x \\
& =2 Re\left(\int_0^{\infty} \frac{\ln \left(x^2+\left[\left(\frac{1+i}{\sqrt{2}}\right) a\right]^2\right)}{b^2+x^2} d x\right)
\end{aligned}
$$
Using (*), we have $$
\begin{aligned}\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x&=2 Re\left[\frac{\pi}{b} \ln \left(\frac{1+i}{\sqrt{2}} a+b\right)\right] \\&=\frac{2 \pi}{b} R e\left[\ln \left(\frac{a}{\sqrt{2}}+b+\frac{a}{\sqrt{2}}i\right)\right] \\&= \boxed{\frac{\pi}{b} \ln \left(a^2+b^2+a b \sqrt{2}\right)}\end{aligned}
$$
In general, for any $a, b \in \mathbb{R} \backslash\{0\}$, replacing $a$ and $b$ by $|a|$ and $|b|$ yields
$$\boxed{\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) }$$
For example, $$
\int_0^{\infty} \frac{\ln \left(x^4+16\right)}{9+x^2} d x= \frac{\pi}{3} \ln (13+6 \sqrt{2})
$$
Comments and alternative methods are highly appreciated.
|
To make it more general, consider the case of
$$I_n=\int_0^{\infty} \frac{\log \left(x^n+a^n\right)}{x^2+b^2} \,d x$$ where $a$ and $b$ are positive. Let $x=a t$ to get
$$I_n=\frac{\pi \log (a)}{2 b}n+\frac 1a\int_0^{\infty} \frac{\log \left(t^n+1\right)}{t^2+c^2} \,d x \qquad\text{with} \qquad c=\frac ba$$ Using the roots of unity, this is the summation of integrals looking like
$$J=\int_0^{\infty} \frac{\log \left(t+r\right)}{t^2+c^2} \,d x $$The antiderivative exists (polylogarithms appear, for sure) and, in terms of the Lerch transcendent function
$$4c^2\,J=r \,\Phi \left(-\frac{r^2}{c^2},2,\frac{1}{2}\right)+c \left(\pi \log \left(c^2+r^2\right)+4 \log
\left(\frac{c}{r}\right) \tan
^{-1}\left(\frac{r}{c}\right)\right) $$ Summing over the roots, it is "just" a matter of simplifications.
|
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"url": "https://math.stackexchange.com/questions/4645421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Using Integration by Parts to find leading order approximation of exponential integral I'm trying to understand how finding leading order approximation of exponential integrals.
Here is my integral: $$\int^{+\infty}_0 e^{-xt}\ln(1+t^2)dt$$
I need to use Integration by parts to then find the leading order asymptotic approximation and using the remainder in integral form, showing that the approximation is in fact asymptotic.
(EDIT) This is what I have tried:
Integration by Parts gives:
$$[\frac{e^{-xt}\ln(1+t^2)}{-x}]^{+\infty}_0 + x^{-1}\int^{+\infty}_0e^{-xt}\frac{2t}{1+t^2}dt$$
The boundary term vanishes and the Remainder is the second term.
Then I show that the Remainder is Asymptotic:
$$|R(x)| \leq x^{-1}\int^{+\infty}_0|e^{-xt}||\frac{2t}{1+t^2}|dt \leq x^{-1}[\ln(1+t^2)]^{+\infty}_0$$
But I find a divergence.
Any ideas?
Many thanks!
|
You correctly derived that
$$
f(x) = \int_0^\infty e^{-xt} \ln(1+t^2) \, dt
= \frac 2x \int_0^\infty e^{-xt} \frac{t}{1+t^2} \, dt
$$
for $x > 0$. This does not yet reveal the asymptotic behavior for $x \to \infty$, so let's integrate by parts again:
$$
f(x) = -\frac{2}{x^2} e^{-xt} \frac{t}{1+t^2} \big|_{t_0}^{t=\infty}
+ \frac{2}{x^2} \int_0^\infty e^{-xt} \frac{1-t^2}{(1+t^2)^2} \, dt \, .
$$
Again the first part vanishes, and we have
$$
f(x) = \frac{2}{x^2} \int_0^\infty e^{-xt} \frac{1-t^2}{(1+t^2)^2} \, dt \, .
$$
But we don't give up and integrate once more:
$$
\begin{align}
f(x) &= -\frac{2}{x^3} e^{-xt} \frac{1-t^2}{(1+t^2)^2} \big|_{t_0}^{t=\infty}
+ \frac{2}{x^3} \int_0^\infty e^{-xt} \frac{2t(t^2-3)}{(1+t^2)^3} \, dt \\
&= \frac{2}{x^3} + \frac{4}{x^3} \int_0^\infty e^{-xt} \frac{t(t^2-3)}{(1+t^2)^3} \, dt \\
&=: \frac{2}{x^3} + R(x) \, .
\end{align}
$$
That looks promising! $\frac{t(t^2-3)}{(1+t^2)^3}$ is bounded on $\Bbb R$ (the denominator has no zeros and a higher degree than the numerator). Therefore, with some constant $M > 0$
$$
|R(x)| \le \frac{4M}{x^3} \int_0^\infty e^{-xt} \, dt
= \frac{4M}{x^4}
$$
and we have shown that
$$
f(x) = \frac{2}{x^3} + O\left( \frac{1}{x^4} \right)
$$
for $x \to \infty$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How to evaluate $\int \frac1{(x^2+a^2)^{m}}dx$ The integral $$\int\frac{1}{(x^2+a^2)^m}dx$$ can be expressed by a recursive formula of $$\frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{dx}{(x^2+a^2)^{m-1}}$$ I do not understand how integration by part leads to this result. Specifically, since $$\int u dv = uv - \int v du$$ I want to know what is $u$, $du$, $v$, $dv$.
|
Integrate by parts as follows
\begin{align}
&\int\frac{1}{(x^2+a^2)^m}dx\\
= &\int \frac1{2a^2(m-1)x^{2m-3}}\ d\left[\bigg(\frac{x^2}{x^2+a^2} \bigg)^{m-1}\right]\\
=&\ \frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{1}{(x^2+a^2)^{m-1}}dx
\end{align}
Thus
$$u= \frac1{2a^2(m-1)x^{2m-3}},\>\>\>\>\>
v= \left(\frac{x^2}{x^2+a^2} \right)^{m-1} $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Taylor series estimation for the first four non-zero terms Question:
Find the Taylor series expansion (up to the first 4 non-zero terms) of the function $f(x) = \sqrt[3]{x+27}$ about the point $a=0$. Use the series to estimate $\sqrt[3]{27.1}$, up to 4 significant figures.
My Attempt:
Given the function $f(x) = \sqrt[3]{x+27}$ and the point $a=0$, we want to find the Taylor series expansion of $f(x)$ up to the first 4 non-zero terms.
First, let's find the derivatives of $f(x)$ at $a=0$:
$$f'(x) = \frac{1}{3(x+27)^{2/3}}, \quad f'(0) = \frac{1}{27}$$
$$f''(x) = -\frac{2}{9(x+27)^{5/3}}, \quad f''(0) = -\frac{2}{2187}$$
$$f'''(x) = \frac{10}{27(x+27)^{8/3}}, \quad f'''(0) = \frac{10}{177147}$$
Now, we can write the Taylor series expansion of $f(x)$ around $a=0$:
$$f(x) = f(0) + xf'(0) + \frac{x^2f''(0)}{2!} + \frac{x^3f'''(0)}{3!} + \cdots$$
Substituting the values we found earlier, we get:
$$f(x) = 3 + \frac{x}{27} - \frac{x^2}{2187} + \frac{5x^3}{531441} + \cdots$$
To estimate $\sqrt[3]{27.1}$ using the series, we can plug in $x = 0.1$:
$$\sqrt[3]{27.1} \approx 3 + \frac{0.1}{27} - \frac{(0.1)^2}{2187} + \frac{5(0.1)^3}{531441} \approx 3.0037$$
Therefore, $\sqrt[3]{27.1}$ is approximately $3.0037$ up to 4 significant figures.
Am I right?
|
Yes you are correct. You can use a CAS software such as the open source and free software maxima to check your computation.
(%i1) taylor((x+27)^(1/3), x, 0, 5);
(%o1) $3+\frac{x}{27}-\frac{x^2}{2187}+\frac{5\,x^3}{531441}-\frac{10\,x^4}{43046721}+\frac{22\,x^5}{3486784401}+\cdots $
Note that you can start by factoring 3 as
$$\sqrt[3]{x+27} = 3 \sqrt[3]{1+ {x\over 27}}$$
Then the binomial formula works for non-integer exponents as long as $|x|<1$.
$(1+x)^\alpha = 1 + \pmatrix{1 \cr \alpha}x + \pmatrix{2 \cr \alpha}x +\pmatrix{3 \cr \alpha}x ...$
|
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|
Non-differentiability at $x=0$ Question:
Let $f(x) =\left\{ \begin{array}{ll} x^2 + x \sin \left(\frac{1}{x}\right) & x\neq0, \\ 0& x=0. \end{array} \right. $
Discuss whether $f(x)$ is differentiable at $x=0$. If yes, find $f'(0)$.
Attempt:
To determine whether $f(x)$ is differentiable at $x=0$, we need to compute the limit of the difference quotient:
$$\lim_{x\to 0} \frac{x^2 + x\sin\left(\frac{1}{x}\right)}{x} = \lim_{x\to 0} x+ \sin\left(\frac{1}{x}\right) = 0 + \lim_{x\to 0}\sin\left(\frac{1}{x}\right) = \infty$$
Note that for $x\neq 0$, we have
$$\begin{aligned} \frac{f(x)-f(0)}{x-0} &= \frac{x^2 + x \sin \left(\frac{1}{x}\right) - 0}{x-0} \ &= x + \sin \left(\frac{1}{x}\right) \ &\leq |x| + 1. \end{aligned}$$
Based on the squeeze theorem, we conclude that $f(x)$ is not differentiable at $x=0$.
Therefore, there is no value for $f'(0)$.
Updated Attempt:
To determine whether $f(x)$ is differentiable at $x=0$, we compute the limit of the difference quotient:
$$\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{x^2 + x\sin\left(\frac{1}{x}\right)}{x} = \lim_{x\to 0} x+ \sin\left(\frac{1}{x}\right).$$
Since the limit of $\sin(1/x)$ as $x$ approaches 0 does not exist, the limit of the difference quotient also does not exist, and we can conclude that $f(x)$ is not differentiable at $x=0$.
However, we can show that $f(x)$ is continuous at $x=0$ by noting that $\lim_{x\to 0} f(x) = f(0) = 0$ and $f(x)$ is well-defined at $x=0$.
Thank you for your kind comments; am I correct?
|
Since there exist two sequences
$a_n=\dfrac1{\frac{\pi}2+2n\pi}>0\;,\quad n\in\Bbb N\;,$
$b_n=\dfrac1{\frac{3\pi}2+2n\pi}>0\;,\quad n\in\Bbb N\;,$
such that $\;a_n\to0\;,\;b_n\to0\;$ and
$\lim\limits_{n\to\infty}\dfrac{f(a_n)-f(0)}{a_n-0}=\lim\limits_{n\to\infty}\left[a_n+\sin\left(\dfrac1{a_n}\right)\right]=1\;,$
$\lim\limits_{n\to\infty}\dfrac{f(b_n)-f(0)}{b_n-0}=\lim\limits_{n\to\infty}\left[b_n+\sin\left(\dfrac1{b_n}\right)\right]=-1\;,$
it follows that it does not exist the limit
$\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}\;,$
consequently ,
the function $\,f(x)\,$ is not differentiable at $\,x=0\,.$
|
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|
Prove that the area of a circumscribed hexagon is twice the area of a triangle Let $\Delta ABC$ be an acute triangle and denote the circumscribed circle of $\Delta ABC$ $\Gamma$ with midpoint $O$. Let $A_1, B_1, C_1$ be the points on $\Gamma$ where the lines $AO, BO, CO$ intersect $\Gamma$. Show that the area of the hexagon $AC_1BA_1CB_1$ is twice as big as the area of $\Delta ABC$.
My first approach was to realize that the hexagon $AC_1BA_1CB_1$ contains $\Delta ABC$, which means that we have to prove that $|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A| = |\Delta ABC|$. This could equivalently be written as:
$
\begin{equation}
\frac{|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A|}{|\Delta ABC|} = 1.
\end{equation}
$
But since the three "outer" triangles each share a side with $\Delta ABC$, the ratio between the areas of $\Delta ABC$ and the respective "outer" triangle" will be $\frac{h_i}{H_i}$, where $h_i$ is the height of the outer triangle and $H_i$ is the height of $\Delta ABC$ (where both heights are perpendicular to the shared side). This means that we can write the above equation as:
$
\begin{equation}
\frac{h_1}{H_1} + \frac{h_2}{H_2} + \frac{h_3}{H_3} = 1
\end{equation}
$
Now, label the points where the lines $AO, BO, CO$ intersect the opposite side of $\Delta ABC$ $P, R, Q$ respectively ($P$ on $AB$, $R$ on $AC$, $Q$ on $AB$). Since we have angles in the same segment of $\Gamma$, we get the following similar triangles: $\Delta ABC_1\sim\Delta PBC, \Delta CQA_1\sim\Delta ABQ, \Delta ARB_1\sim\Delta RBC$, which means that we can write:
$
\frac{h_1}{H_1} = \frac{|AP|}{|BP|} \\
\frac{h_2}{H_2} = \frac{|CQ|}{|BQ|} \\
\frac{h_3}{H_3} = \frac{|AR|}{|CR|}
$
Which means that we have to prove that the sum of these (above) ratios equals $1$. But from here, I can't seem to make much progress...
|
Let's assume $R$ is the circumradius of $\triangle ABC$. Then, we have:
$$S_{\triangle ABC}=\frac{1}{2} |AC||AB| \sin A=\frac{1}{2} \times 2R \sin B \times 2R\sin C \times \sin A \\ = 2R^2 \sin A\sin B \sin C.$$
Now, let's compute the area of $\triangle A_1BC$. We have:
$$S_{\triangle A_1BC}=\frac{1}{2}|A_1B||A_1C| \sin A =\frac{1}{2} \times 2R \cos C \times 2R \cos B \times \sin A \\= 2R^2 \cos C \cos B \sin A.$$
Similarly,
$$S_{\triangle B_1AC}=2R^2 \cos A \cos C \sin B \\ S_{\triangle C_1AB}=2R^2 \cos A \cos B \sin C.$$
Thus, as you observed, we need to show that:
$$\cos C \cos B \sin A+\cos A \cos C \sin B+\cos A \cos B \sin C =\sin A\sin B \sin C,$$
while $\sin A=\sin(B+C)$ and $\cos A= - \cos (B+C).$
So, we must show that:
$$\cos C \cos B \sin(B+C)- \cos (B+C) \cos C \sin B- \cos (B+C)\cos B \sin C \\= \sin(B+C)\sin B \sin C \\ \iff sin(B+C) (\cos C \cos B - \sin B \sin C )= \cos (B+C) (\cos C \sin B + \cos B \sin C),$$
which obviously holds.
We are done.
|
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|
We can divide $7^{17} - 7^{15}$ by? We can divide $7^{17} - 7^{15}$ by?
The answer is $6$, but how?
Thanks in advance.
|
HINT $\rm\quad X^{n+2} - X^n \;=\; (X^2 - 1)\: X^n \;=\; (X-1)\: (X+1)\: X^n$
Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\
\frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\
\end{array}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:
$$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$
Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $\sin x$:
$$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$
but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
|
This one I found in The American Mathematical Monthly from 1951 in the article 'A simple evaluation of an improper integral' written by Waclaw Kozakiewicz.
Theorem (Riemann). If $f(x)$ is Riemann integrable in the interval $a \leq x \leq b$, then:
$$\lim_{k \to +\infty} \int_a^b f(x) \sin kx \; dx = 0 \;.$$
Next, notice that:
$$\int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{2 \sin \frac{x}{2}}\; dx = \frac{\pi}{2} \; ,n = 0,1,2,\ldots \quad (1)$$
and let:
$$\phi(x) = \begin{cases} 0 & , \;x = 0 \\ \frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} =\frac{2 \sin \frac{x}{2} - x}{2x \sin \frac{x}{2}} & ,\; 0 < x \leq \pi \; . \end{cases}$$
Then $\phi(x)$ is continuous and satisfies Riemann theorem, so choosing $k = n + \frac{1}{2}$ we write:
$$\lim_{n \to +\infty}\int_0^{\pi} \left(\frac{1}{x} - \frac{1}{2 \sin \frac{x}{2}} \right) \sin \left(n+\frac{1}{2}\right)x \; dx = 0 \;.$$
But taking $(1)$ into account we have:
$$\lim_{n \to +\infty} \int_0^\pi \frac{\sin \left(n+\frac{1}{2}\right)x}{x} \; dx = \frac{\pi}{2}\;.$$
Using substitution $u = \left(n+\frac{1}{2}\right)x$ and knowing that $\int_0^{+\infty} \frac{\sin x}{x} \; dx$ converges we finally have:
$$\int_0^{+\infty} \frac{\sin x}{x} \; dx = \lim_{n \to +\infty} \int_0^{\left(n+\frac{1}{2}\right)\pi}\frac{\sin u}{u} \; du = \frac{\pi}{2}\;.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
|
$$ \sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$
$$ \cos \theta \cdot \cos \bigl(60^\circ - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$
$$ \tan \theta \cdot \tan \bigl(60^\circ - \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta $$
|
{
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"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
Let $\int \frac{1}{\sin x \cos x}dx=\int \frac{2}{2 \sin x \cos x}dx$
=$\int \frac{2}{\sin 2x}dx$
Let $\tan x=t$
$\sin 2x=\frac{2t}{1+t^2}$
$\tan^{-1} t=x$
$\frac{1}{1+t^2}dt=dx$
$2\int \frac{dx}{\sin 2x} =2\int \frac{1+t^2}{2t}\cdot\frac{dt}{1+t^2}$
=$\int\frac{dt}{t}$
=$\ln t+C$
=$\ln (\tan x)+C$
|
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|
Floor of Square Root Summation problem I have problem calculating the following summation:
$$
S = \sum_{j=1}^{k^2-1} \lfloor \sqrt{j}\rfloor.
$$
As far as I understand the mean of that summation it will be something like $$1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+\cdots$$
and I suspect that the last summation number will be $(k-1)^2$, but I really can't find the pattern of the equal simpler summation.
|
Let $\lfloor \sqrt{n} \rfloor = a$, then the following sum holds:
$$\sum_{0\le k < n} \lfloor \sqrt{k} \rfloor = (n+1)a - \frac{a^3}{3} -
\frac{a^2}{2} - \frac{a}{6}.$$
(Edited.)
You might also enjoy the slightly more tricky sum:
$$\sum_{0\le k < n} \lfloor k^{1/3} \rfloor = nb - \frac{b^2}{4} - \frac{b^3}{2} - \frac{b^4}{4},$$
where this time $b =\lfloor n^{1/3} \rfloor$
One way to obtain these answers is to merely apply the summation identity that I mentioned here
|
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|
What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
|
The cosine formula states
$$a^2 = b^2+ c^2 – 2bc \cos A$$
and since
$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$
we have
$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$
which reduces to $a^4 = (b^2 – c^2)^2.$
And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.
Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.
And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.
|
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|
Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this
|
After $(x-4)(x+2)=16$, you get $x^2-2x-24=0$ (the coefficient of $x$ is $-2$ not $2$). So $x=\frac{2\pm \sqrt{100}}{2}$ by the quadratic formula. So $x=6$ or $x=-4$
|
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|
Evaluation of the sum $\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor$ Let $a, b$ and $c$ be positive integers. Recall that the greatest common divisor (gcd) function has the following representation:
\begin{eqnarray}
\textbf{gcd}(b,c) = 2 \sum_{k = 1}^{c- 1} \left \lfloor \frac{kb}{c} \right \rfloor + b + c - bc
\end{eqnarray}
as shown by Polezzi, where $\lfloor \cdot \rfloor$ denotes the floor function. In trying to generalize the formula I came across the following summation
\begin{eqnarray}
\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor.
\end{eqnarray}
I can prove the following identity for real $x$,
\begin{eqnarray}
\sum_{k = 0}^{c-1} \left \lceil \frac{x - kb}{c} \right \rceil = d \left \lceil \frac{x}{d} \right \rceil - \frac{(b-1)(c-1)}{2} - \frac{d-1}{2},
\end{eqnarray}
where $\lceil \cdot \rceil$ denotes the ceiling function and $d = \text{gcd}(b,c)$. (Note that in the first summation the upper index is in general independent of $c$.) Ideas or reference suggestions are certainly appreciated. Thanks in advance!
Update I can prove the identity
\begin{eqnarray}
\sum_{k = 0}^{c-1} \left \lfloor \frac{x - kb}{c} \right \rfloor = d \left \lfloor \frac{x}{d} \right \rfloor - \frac{(b+1)(c-1)}{2} + \frac{d-1}{2},
\end{eqnarray}
where $d = \text{gcd}(b,c)$. There is another identity which might be useful. If $n = c \ell +r$ with $0 \leq r < c$, then
\begin{eqnarray}
\sum_{k = 1}^{n} \left \lfloor \frac{k}{c} \right \rfloor = c \binom{\ell}{2} + (r + 1) \ell.
\end{eqnarray}
Update 2 Ok, so I can prove that for real $x, y > 0$,
\begin{eqnarray}
\sum_{k = 0}^{\lfloor y \rfloor} \left \lfloor x + \frac{k}{y} \right \rfloor = \lfloor xy + (\lceil y \rceil - y) \lfloor x + 1 \rfloor \rfloor + \chi_{\mathbb{N}}(y)(\lfloor x \rfloor + 1),
\end{eqnarray}
where $\chi_{\mathbb{N}}$ denotes the characteristic function of the positive integers. My original problem (and a nice generalization of it) will be in hand if I can evaluate the following minor generalization: For real $x, y > 0$ and $n \in \mathbb{Z}_{\geq 0}$,
\begin{eqnarray}
\sum_{k = 0}^{n} \left \lfloor x + \frac{k}{y} \right \rfloor.
\end{eqnarray}
Again, any help is certainly appreciated!
|
This isn't a complete solution but rather a reformulation for a large number of cases involving the problem above. Write $n = m \lfloor y \rfloor + r$, where $0 \leq r < \lfloor y \rfloor$ and $n, m , r \in \mathbb{Z}_{\geq 0}$. Then for real $x, y > 0$, we have
\begin{eqnarray}
\sum_{k = 0}^{n} \left \lfloor x + \frac{k}{y} \right \rfloor & = & \lfloor x \rfloor + \sum_{\ell = 0}^{m-1} \left \lfloor x y + \ell \lfloor y \rfloor + ( \lceil y \rceil - y) \left \lfloor x + \frac{\ell \lfloor y \rfloor}{y} + 1 \right \rfloor \right \rfloor +
\end{eqnarray}
\begin{eqnarray} & & \sum_{k = 1}^{r} \left \lfloor x + \frac{m \lfloor y \rfloor}{y} + \frac{k}{y} \right \rfloor + \chi_{\mathbb{N}}(y) \sum_{\ell = 0}^{m-1} \left \lfloor x + \frac{\ell \lfloor y \rfloor}{y} + 1 \right \rfloor - \sum_{\ell = 0}^{m-1} \left \lfloor x + \frac{ \ell \lfloor y \rfloor }{y} \right \rfloor .
\end{eqnarray}
This was derived by partitioning the interval $[0,n]$ into subintervals and using the last identity in the description of the problem.
For the original problem, write
\begin{eqnarray}
\sum_{k = 0}^{\lfloor a /b \rfloor} \left \lfloor \frac{a - k b}{c} \right \rfloor = \sum_{k = 0}^{\lfloor a /b \rfloor} \left \lfloor \frac{\text{mod}(a,b) + k b}{c} \right \rfloor
\end{eqnarray}
where $\text{mod}(a,b) = a - b \lfloor \frac{a}{b} \rfloor$. If $\lfloor \frac{a}{b} \rfloor \gg c$, then use the identity above to vastly reduce the number of terms in the sum.
|
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|
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:
$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$
There is a slight difference in case of $\sin$, which is:
$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)$$
How do we prove the above two identities?
|
This is similar to the currently accepted answer, but more straightforward. You can use the trig identity
\begin{equation*}
\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\sin \beta \cos \alpha.
\end{equation*}
Let $a_n = a + 2dk$ be an arithmetic sequence of difference $2d$, and set $b_n
= a_n - d = a + d(2k - 1)$. Note that $\{b_n\}$ is also an arithmetic sequence
of difference $2d$, hence $a_n + d = b_n + 2d = b_{n + 1}$. Therefore
\begin{equation*}
2 \sin d \cos a_n = \sin(a_n + d) - \sin(a_n - d) = \sin b_{n + 1} - \sin b_n.
\end{equation*}
Summing both sides from $0$ to $n$ yields
\begin{align*}
2 \sin d \sum_{k = 0}^n \cos a_k &= \sin b_{n + 1} - \sin b_0 \\
&= \sin(a + d(2n + 1)) - \sin(a - d).
\end{align*}
From our original trig identity,
\begin{equation*}
2\sin((n + 1)d) \cos(a + nd) = \sin(a + d(2n + 1)) - \sin(a - d).
\end{equation*}
Thus, if $\sin d \neq 0$, we can rewrite our result as
\begin{equation*}
\sum_{k = 0}^n \cos (a + 2dk) = \frac{\sin((n + 1)d) \cos(a + nd)}{\sin d}.
\end{equation*}
This is OP's formula with $2d$ and $n$ instead of $d$ and $n - 1$. A similar process will yield the formula for $\sum_{k = 0}^n \sin(a + 2dk)$.
|
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|
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
|
Hint: write $\tan(x) + \cot(x)$ in terms of $\sin(x)$ and $\cos(x)$, with a common denominator. Square $\sin(x) + \cos(x) = {\sqrt{3} + 1 \over 2}$ and compare.
|
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|
Basic Recursion Im trying to write recursive formulas for sequences but it seems like there are different techniques depending on what type of sequence I'm dealing with. for example I want to the sequence:
$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.... \frac{1}{n}$
What I see when I look at this is all the next terms are $\frac{1}{2}$ the previous term. So, I wrote the following recursive formula - which is incorrect.
$f(1) = 1, f(n) = f(n-1) + \frac{f(n-1)}{2}$ Why is this wrong?
Why is this correct? $f(1) = 1, f(n) = f(n-1) +\frac{1}{2}^{n-1}$
Intuitively I thought that my original attempt looked correct.
|
As Leonidas suggests, your formula doesn't work (let's try it for $n=3$):
\begin{align*}
f(2) & = f(1) + \frac{f(1)}{2} = 1 + \frac{1}{2} = \frac{3}{2} \\
f(3) & = f(2) + \frac{f(2)}{2} = \frac{3}{2} + \frac{\frac{3}{2}}{2} = \frac{9}{4} \neq 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}.
\end{align*}
Your last summand in $f(n) = 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}}$ should be $\frac{1}{2^{n−1}}$ instead of $\frac{1}{n}$ (in the denominator there always is a power of $2$). If you write it this way then the "recursion" $f(n) = \left(1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-2}}\right) + \frac{1}{2^{n-1}} = f(n-1) + \frac{1}{2^{n-1}}$ should be clear. But writing the parentheses in a different way yields
\begin{align*}
f(n) & = 1 + \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}}\right) \\
& = 1 + \frac{1}{2}\left(1 + \frac{1}{2} + \cdots + \frac{1}{2^{n-3}} + \frac{1}{2^{n-2}}\right) \\
& = 1 + \frac{1}{2} f(n-1).
\end{align*}
This shows that you can define $f(1)=1$ and $f(n)=1+\frac{1}{2}f(n−1)$ recursively, and this seems to be nicer and closer to what you tried to do.
One thing to take away from this: If you postulate a formula, you should check it for small values of $n$. If there's an obvious mistake, you'll catch it quickly and don't waste time trying to prove it. Playing with the formulas for small $n$ often gives you a feeling of what could or should be true.
|
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|
How to simplify this to the given answer? I'm trying to simplify:
$$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$$
but I can't get to the answer:
$$\frac{1}{x-2}$$
How to do it?
|
First, do the operation by finding the least common denominator. Since $x^2-4 = (x-2)(x+2)$, the least common denominator is already $x^2-4$. Then do some simple algebra:
\begin{align*}
\frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2 - (x+1)(x-2)}{x^2-4} = \frac{x^2-(x^2-x-2)}{x^2-4}\\
&= \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}.
\end{align*}
If you didn't realize that $x+2$ already divides $x^2-4$, you probably would get
\begin{align*}
\frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2(x+2) - (x+1)(x^2-4)}{(x^2-4)(x+2)}\\
&= \frac{x^3 + 2x^2 - (x^3 +x^2 - 4x - 4)}{(x^2-4)(x+2)}\\
&= \frac{x^2 +4x + 4}{(x^2-4)(x+2)} = \frac{(x+2)^2}{(x^2-4)(x+2)}\\
&= \frac{x+2}{x^2-4} = \frac{1}{x-2},
\end{align*}
with the extra work for not noticing.
|
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|
Striking applications of integration by parts What are your favorite applications of integration by parts?
(The answers can be as lowbrow or highbrow as you wish. I'd just like to get a bunch of these in one place!)
Thanks for your contributions, in advance!
|
My favorite example of integration by parts (there are other nice tricks as well in this example but integration by parts starts it off) is this:
Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$I_n = (n-1) (\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx) = (n-1) I_{n-2} - (n-1) I_n$.
Rearranging we get $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3}$$
$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2}$$
|
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|
How can I find the roots of this complex equation? I have to find the domain of
$\frac{z^2-1}{z^2+z+1}$
So, there can't be any $z$ that makes null the denominator,
$z^2+z+1$ = 0
and after decomposing $z$,
$x^2-y^2+x+2xyi+yi=0$
How to solve that quadratic complex equation?
|
Three points:
*
*You did the computation wrong. Writing $z=x+iy$, then
$$z^2+z+1 = (x+iy)^2 + (x+iy) + 1 = (x^2-y^2 + x + 1) + i(2xy+y).$$
*If $(x^2-y^2+x+1) + i(2xy+y) = 0$, with $x,y$ real numbers, then you need $x^2-y^2+x+1=0$ and $2xy+y=0$. You solve these the way you usually solve equations for real numbers.
So, for example, you have $0=2xy+y = (2x+1)y$. So either $y=0$ or $2x+1=0$. If $y=0$, then the first equation reduces to $x^2+x+1=0$, which has no real solutions, so there are no solutions with $y=0$. If $y\neq 0$, then $2x+1=0$, so $x=-\frac{1}{2}$. Plugging into the first equation, we get
$$0 = \frac{1}{4}-y^2 -\frac{1}{2} + 1 = -y^2 +\frac{3}{4},$$
so you get that $y^2 = \frac{3}{4}$, or $y = \pm\frac{\sqrt{3}}{2}$. So the two solutions are $z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
*The acrobatics from step 2 are unnecessary. You don't have to decompose into real and imaginary parts, because the quadratic formula works for complex numbers! (Provided you take complex square roots). Since
$$z^2 + z + 1 = \left(z+\frac{1}{2}\right)^2 + \frac{3}{4}$$
(by completing the square), then this is zero if and only if
$$z+\frac{1}{2} = \sqrt{-\frac{3}{4}},$$
if and only if
$$z = -\frac{1}{2} + \frac{\sqrt{-3}}{2}.$$
You may recognize this as exactly what you get from the quadratic formula applied to $z^2+z+1$, and you may also recognize them as the solutions you get if you go through the contorsions of step 2 above. So just find the two complex square roots of $-3$, and rejoice! (The quadratic formula works even if the coefficients of the quadratic are complex numbers, instead of real numbers).
|
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|
Proving the value of a limit using the $\epsilon$-$\delta$ definition I'm trying to solve the problem of showing that
$$\lim_{x\to6}\left(\frac{x}{4}+3\right) = \frac{9}{2}$$
using the $\epsilon$-$\delta$ definition of a limit.
|
Given an $\epsilon\gt 0$, you want to show that provided $x$ is close enough to $6$, without being equal to $6$, then $\frac{x}{4}+3$ will be $\epsilon$-close to $\frac{9}{2}$. Well, first thing is to figure out how close $\frac{x}{4}+3$ is to $\frac{9}{2}$:
$$\left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right| = \left|\frac{x}{4}+3-\frac{9}{2}\right| = \left|\frac{x + 12 - 18}{4}\right| = \frac{|x-6|}{4}.$$
So, how can you make sure that $\displaystyle \left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right|$ is smaller than $\epsilon$, by placing conditions on how close $x$ is to $6$, that is, on the value of $|x-6|$?
|
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|
Which is bigger: $9^{9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}}}$ or $9!!!!!!!!!$? In my classes I sometimes have a contest concerning who can write the largest number in ten symbols. It almost never comes up, but I'm torn between two "best" answers: a stack of ten 9's (exponents) or a 9 followed by nine factorial symbols. Both are undoubtedly huge, but I haven't been able to produce an argument that one is larger (surely they aren't equal). Any insight into which of these two numbers is bigger would be greatly appreciated.
|
I find it quite sad that none of the answers has actually formally shown that
$$9^{9^{9^{9^{9^{9^{9^{9^{9^9}}}}}}}} > 9!!!!!!!!!$$
The proces is a bit long, but it it worth it in my opinion. First, a simple calculation shows $9! < 9^8$. Then $9!! < (9^8)^{9^8} = 9^{8\cdot9^8}$. Then $$9!!!< \left(9^{8\cdot9^8}\right)^{9^{8\cdot9^8}} < \left(9^{9^9}\right)^{9^{8\cdot9^8}}=9^{9^{8\cdot9^8}\cdot9^9}=9^{9^{8\cdot9^8+9}}$$
$$9!!!! < \left(9^{9^{9^9}}\right)^{9^{9^{8\cdot9^8+9}}}=9^{9^{9^{8\cdot9^8+9}}\cdot 9^{9^9}}=9^{9^{9^{8\cdot9^8+9}+9^9}}<9^{9^{9^{8\cdot9^8+10}}}$$
Now, some notation. Let us use $f(x)$ for $9^x$ and $f^k(x)=f(f(... f(f(x))...))$. Let $x!^k$ denote $x!!!...!!!$ with $k$ factorials.
Clearly, $f(x)+1<f(x+1)$. By induction, $f^t(x)+1<f^t(x+1)$ for any $t$.
Now we proceed by induction to show that $9!^k < f^{k-1}(8\cdot9^8+k+6)$. The base case has been given above. Assume $4 \leq k \leq 9^8-7$.
$$9!^{k+1} < \left(f^{k-1}(8\cdot9^8+k+6)\right)^{f^{k-1}(8\cdot9^8+k+6)}=
9^{f^{k-2}(8\cdot9^8+k+6)f^{k-1}(8\cdot9^8+k+6)}<9^{9^{f^{k-3}(9^9)+f^{k-2}(8\cdot9^8+k+6)}}<9^{9^{9\cdot f^{k-2}(8\cdot9^8+k+6)}}<9^{9^{9^{f^{k-3}(8\cdot9^8+k+6)+1}}}<9^{9^{9^{f^{k-3}(8\cdot9^8+k+7)}}}=f^k(8\cdot9^8+k+7)$$
This completes the induction hypothesis. Hence we have shown the asked.
Now for some fun: The largest number using ten symbols. We can also use shortened Knuth's up arrow notation, $9 \uparrow^{9 \uparrow^{9 \uparrow^{9} 9} 9} 9$ has ten symbols.
Conway chained arrow notation, $9 \rightarrow 9 \rightarrow 9 \rightarrow 9 \rightarrow 9^9$.
Rado's sigma function, $\Sigma(\Sigma(\Sigma(9)))$.
|
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Why does the sum of a division applied to individual items not equal the division applied to the sum of those items? When $a_2/a_1 = b_2/b_1$, $a_1 \neq b_1$, we have
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}=
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}.$$
So why when $a_2/a_1 \neq b_2/b_1 , a_1 \neq b_1$ we don't have a similar equality?
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$$
|
Because for $a_{1},a_{2},b_{1},b_{2}\neq 0$ we have the following equivalent inequalities:
$$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$$
This can be shown as follows:
$$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}$$
$$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{%
\left( a_{1}+b_{1}\right) ^{2}}{a_{1}+b_{1}+(a_{2}+b_{2})-(a_{1}+b_{1})}$$
$$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{%
\left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$$
$$\Leftrightarrow \frac{a_{1}^{2}b_{2}+a_{2}b_{1}^{2}}{a_{2}b_{2}}\neq \frac{%
\left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$$
$$\Leftrightarrow \left( a_{1}^{2}b_{2}+a_{2}b_{1}^{2}\right) \left(
a_{2}+b_{2}\right) \neq \left( a_{1}+b_{1}\right) ^{2}a_{2}b_{2}$$
$$\Leftrightarrow
a_{1}^{2}b_{2}^{2}+a_{2}^{2}b_{1}^{2}-2a_{2}b_{2}a_{1}b_{1}\neq 0$$
$$\Leftrightarrow \left( a_{1}b_{2}-a_{2}b_{1}\right) ^{2}\neq 0$$
$$\Leftrightarrow a_{1}b_{2}\neq a_{2}b_{1}$$
$$\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$$
Your numerical case seems to be for $a_{1}=b_{1}=a_{2}=1,b_{2}=1.1$ for
which we have
$$\frac{1}{1/1}+\frac{1}{1.1/1}\neq \frac{2}{1+\frac{(1+1.1)-\left( 1+1\right) }{2%
}}=\frac{2}{1.05}\Leftrightarrow \frac{1}{1}\neq \frac{1.1}{1}.$$
|
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|
Find limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$, with the initial conditions: $a>b>0$ With the initial conditions: $a>b>0$;
I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$
I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.
|
Another way: Note that $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$. Since $b < a$, each term in the sum is bounded by $a^{n-1}$ and we have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} < na^{n-1}$$
Since the sum is at least the first term, we also have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} > a^{n-1}$$
Combining we have
$$a^{n-1}(a - b) < a^n - b^n < na^{n-1}(a - b)$$
Taking $n$th roots we get
$$a^{1 - {1 \over n}}(a - b)^{1 \over n} < \sqrt[n]{a^n - b^n} <n^{1 \over n} a^{1 - {1 \over n}}(a - b)^{1 \over n} $$
This can be rewritten as
$$a \bigg({a - b \over a}\bigg)^{1 \over n} < \sqrt[n]{a^n - b^n} < a n^{1 \over n} \bigg({a - b \over a}\bigg)^{1 \over n}$$
As $n$ goes to infinity both the left and right sides of the above go to $a$. Thus by the squeeze theorem so does the middle and we are done.
|
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|
Simple formula for integer polynomial with $2\sin(2\pi/n)$ as a root? Is there a simple formula an integer polynomial that $2\sin(2\pi/n)$ satisfies?
For $2\cos(2\pi/n)$ the answer is relatively nice. For any given $n$, we have $2\cos(2\pi/n)= z + z^{-1}$ where $z = e^{2 \pi i/n}$ satisfies a cyclotomic polynomial of degree $\varphi(n) = 2k$,
$$
0 = a_{2k}z^{2k} + a_{2k-1}z^{2k-1} + \ldots + a_{1}z + a_0,
$$
where $a_{i} = a_{2k-i}$. Dividing by $\zeta^k$ gives
$$
0 = a_{2k}z^k + \ldots + a_k + a_0z^{-k}
$$
Using the symmetry of the coefficients lets us write this as
$$
0 = a_0(z^k+z^{-k}) + \ldots + a_k.
$$
Then
$$
(z+z^{-1})^2 = z^2 + z^{-2} + \binom{2}{1}
$$
$$
(z+z^{-1})^3 = z^3 + z^{-3} + \binom{3}{1}(z+z^{-1})
$$
$$
(z+z^{-1})^4 = z^4 + z^{-4} + \binom{4}{1}(z^2+z^{-2}) + \binom{4}{2}
$$
and so on, and in the end we get something fairly nice.
What happens with $2\sin(2\pi/n)$?
EDIT: I am aware that $2\sin(2\pi/n) = -i(z - z^{-1})$.
EDIT: Arturo's comment made me realize that dividing by $(-z)^k$ or $(iz)^k$ may be the way to go.
|
I'll work with $\sin \frac{2 \pi}n$ instead of $2 \sin \frac{2 \pi}n$.
Since $\sin x = \cos (\frac \pi 2 - x)$, $\sin \frac{2 \pi}n = \cos (\frac {\pi}2 - \frac {2\pi}n) = \cos \frac{n-4}{2n}\pi$.
The Chebyshev polynomials of the first kind $T_k(x)$ have the property that $T_k(\cos x) = \cos kx$. They have integer coefficients since they satisfy the recurrence $T_0(x)=1, T_1(x)=x, T_{k+1}(x) = 2xT_k(x)-T_{k-1}(x)$.
$T_{4n}(\sin \frac {2 \pi}n) = T_{4n}(\cos \frac{n-4}{2n}\pi) = \cos((n-4) \times 2\pi) = 1$.
Therefore, $\sin \frac{2\pi}n$ is a root of $T_{4n}(x)-1$.
$2 \sin \frac{2 \pi}n$ is a root of $T_{4n}(x/2)-1$, but you may have to clear denominators to get a polynomial over the integers. Of course, this is far from the minimal polynomial.
|
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|
Multiplication in Permutation Groups Written in Cyclic Notation I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if
$$
a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4),
$$
then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?
|
I didn't find this method written as an answer here, so I am just adding it.
You might convert these cyclic notations to Cauchy's two-line notations. Carefully observing your $a$ and $b$, the permutation groups can be written in the following form:
$$a=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 1 & 5 & 4 & 2 & 6
\end{pmatrix} \\
b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 5 & 3 & 4 & 6 & 2
\end{pmatrix}
$$
Now, under the multiplication operation, we will have:
$$a\cdot b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 1 & 5 & 4 & 2 & 6
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 5 & 3 & 4 & 6 & 2
\end{pmatrix}
$$
As @Arturo explained in his answer, "you perform the permutation in $b$ first and then in $a$".
So, in $b$, $1\to1$ and then in $a$, $1\to3$. Thus, in $a\cdot b$, $1\to3$.
Similarly, in $b$, $2\to5$ and then in $a$, $5\to2$. Thus, in $a\cdot b$, $2\to2$ itself. Continuing in a similar process, you'll find that
$$a\cdot b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 2 & 5 & 4 & 6 & 1
\end{pmatrix}$$
Now convert this final form to the cyclic form. Note that we have, in $a\cdot b$, $1\to3$. Thus $$(1,3$$
Again, $3\to5$. Thus $$(1,3,5$$
Note again that $5\to6$ and $6\to1$, making our cyclic notation as $$\left(1,3,5,6\right)$$
Also, note that the rest of the elements in the group map to themselves. Thus we can ignore their presence in our cyclic notation. This makes
$$a\cdot b=\left(1,3,5,6\right)$$
Now can you similarly proceed for $a\cdot c$ also?
|
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|
Infinitely many integer solutions for the equations $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$ How do you show that the equation $x^3+y^3+z^3=1$ has infinitely many solutions in integers? How about $x^3+y^3+z^3=2$?
|
You can reduce the first equation to $$x^3 = -y^3, z = 1$$ with obvious infinite solutions.
This paper details other families of solutions.
The second equation has solutions $(x,y,z)\equiv (6t^3+1, 1-6t^3, -6t^2)$ which (AFAIK) you find by construction (i.e you have to guess it).
|
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|
If $\alpha$ is an acute angle, show that $\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$ If $\alpha$ is an acute angle, show that $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
My attempt:
Write $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\alpha} = (x+\cos{\alpha})^2+\sin^2{\alpha}$, we have:
$\displaystyle \begin{aligned}\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} & = \int_0^1 \frac{dx}{(x+\cos{\alpha})^2+\sin^2{\alpha}} = \bigg[\frac{1}{\sin{\alpha}}\tan^{-1}\left(\frac{x+\cos{\alpha}}{\sin{\alpha}}\right)\bigg]_{x=0}^1 \\ & = \frac{1}{\sin{\alpha}}\bigg[\color{blue}{\tan^{-1}\left(\frac{1+\cos{\alpha}}{\sin{\alpha}}\right)-\tan^{-1}\left(\frac{\cos{\alpha}}{\sin{\alpha}}\right)}\bigg] \end{aligned}$
I'm not sure, however, how the blue bit reduces to $\frac{1}{2}\alpha$. Any hints/suggestions? Thanks.
|
Use $1 + \cos \alpha = 2\cos^2 (\frac{\alpha}{2})$
and $\sin \alpha = 2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})$
|
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|
Integral of $\frac{1}{(1+x^2)^2}$ I am in the middle of a problem and having trouble integrating the following integral:
$$\int_{-1}^1\frac1{(1+x^2)^2}\mathrm dx$$
I tried doing partial fractions and got:
$$1=A(1+x^2)+B(1+x^2)$$
I have no clue how to solve this since it is obvious there is no way to cancel out either $A$ or $B$ to get the other variable. Please guide me.
Thank you.
|
$$
\begin{aligned}
\int \frac{1}{\left(1+x^{2}\right)^{2}} d x\stackrel{IBP}{=} &-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^{2}}\right) \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{1}{x^{2}}\right) d x \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C\\
=&\frac{1}{2}\left(\frac{x}{1+x^{2}}+\tan ^{-1} x\right)+C
\end{aligned}
$$
|
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|
Infinite limits on integration: acceptable notation So you have an integral like
$$
\int_{-\infty}^\infty{ \frac{dx}{1+4 x^2} }
$$
Schaum's Calculus 5e recommends you write this as
$$
\lim_{a \to -\infty} \int_a^b{ \frac{dx}{1+4 x^2} } + \lim_{c \to \infty} \int_b^c{ \frac{dx}{1+4 x^2} }
$$
Where b is chosen as a point where f(x) is defined.
Choosing b=0, you then get
$$
\lim_{a \to -\infty} \frac{1}{4} \int_a^0{ \frac{dx}{\frac{1}{4}+x^2} } + \lim_{c \to \infty} \frac{1}{4} \int_0^c{ \frac{dx}{\frac{1}{4}+x^2} }
$$
$$
\lim_{a \to -\infty} \frac{1}{2} \tan^{-1}{ 2x } |_a^0 + \lim_{c \to \infty} \frac{1}{2} \tan^{-1}{ 2x } |_0^c
$$
$$
0 - \lim_{a \to -\infty} \frac{1}{2} \tan^{-1}{ 2a } + \lim_{c \to \infty} \frac{1}{2} \tan^{-1}{ 2c } - 0
$$
$$
= \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}
$$
Would it be correct to shortcut this as
*
*Take the indefinite integral with no limits
$$
\frac{1}{4} \int{ \frac{dx}{ \frac{1}{4} + x^2} } = \frac{1}{2} \tan^{-1}{ 2x }
$$
*Evaluate
$$
= \frac{1}{2} \lim_{a \to -\infty} \lim_{b \to \infty} \tan^{-1}{ 2x } |_a^b
$$
$$
= \frac{1}{2} \lim_{a \to -\infty} \lim_{b \to \infty} \left( \tan^{-1}{ 2b } - \tan^{-1}{ 2a } \right)
$$
$$
= \frac{1}{2} \left( \frac{\pi}{2} + \frac{\pi}{2} \right) = \frac{ \pi }{2}
$$
So my question is surrounding limit notation and evaluation. Is the above notation ok, or is it completely necessary to break it up into 2 limits?
|
To see why they have this approach, try
$$\int_{-\infty}^\infty{ \frac{8x \;dx}{1+4 x^2} } $$
using the recommended method [the indefinite integral is $\log(1+4 x^2) + \text{constant}$], and then using your suggestion, and alternatively what would happen if you had to work out
$$\lim_{c \to +\infty} \int_{-c}^c { \frac{8x \;dx}{1+4 x^2} }.$$
You would be looking with the recommended method at something like $\infty-\infty$, with your suggestion of doing the upper limit first at $\lim_{a \to -\infty} +\infty = +\infty$, and with the alternative method at $\lim_{c \to +\infty} 0 = 0$. In fact you want the alarm bells of the difference between two infinities.
|
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|
Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
|
You want to show that $$(n-1)^3+n^3+(n+1)^3 \equiv 0 \pmod 9.$$ There are $9$ cases to check,
*
*$(0-1)^3+0^3+(0+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
*$(1-1)^3+1^3+(1+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
*$(2-1)^3+2^3+(2+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
*$(3-1)^3+3^3+(3+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
*$(4-1)^3+4^3+(4+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
*$(5-1)^3+5^3+(5+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
*$(6-1)^3+6^3+(6+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
*$(7-1)^3+7^3+(7+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
*$(8-1)^3+8^3+(8+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
|
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|
What is the relation? The function $x^2 = y\quad$ limits two areas $A$ and $B$:
$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.
$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.
What are the relations between $a$ and $b$, when $Vb = Va$?
..................
I have come to the solution that:
$Vb = (\pi b^2 )/ 2$
$Va = (\pi a^5) / 5$
so the relation between them is:
$2.5b^2 = a^5$
Is that the final solution or is it more?
|
If we take the region bounded by the $y$-axis, the $x$-axis, the line $x=a$ (with $a\gt 0$), and the parabola $y=x^2$, and rotate it about the $x$-axis, the volume of the resulting solid of revolution is easily computed (using, for example, discs perpendicular to the $x$-axis) to be
$$\text{Volume A} = \int_0^a \pi(x^2)^2\,dx = \frac{\pi}{5}x^5\Bigm|_0^a = \frac{\pi a^5}{5}.$$
If the region bounded by the $y$-axis, the $x$ axis, the line $y=b$ (with $b\gt 0$), and the parabola $y=x^2$ is revolved around the $y$-axis, then using discs perpendicular to the $y$-axis we obtain the volume to be:
$$\text{Volume B} = \int_0^b \pi (\sqrt{y})^2\,dy = \frac{\pi}{2}y^2\Bigm|_0^b = \frac{\pi b^2}{2}.$$
So your computations are correct there.
If the two volumes are the same, then we must have
$$\text{Volume A} = \frac{\pi a^5}{5} = \frac{\pi b^2}{2} = \text{Volume B};$$
there are many ways to express this: you can solve for one of $a$ or $b$ in terms of the other:
$$b = \sqrt{\frac{2a^5}{5}} = a^{5/2}\sqrt{\frac{2}{5}},$$
or, if you want to express $a$ in terms of $b$ instead,
$$ a = \sqrt[5]{\frac{5}{2}b^2} = b^{2/5}\sqrt[5]{\frac{5}{2}}.$$
Or you can simply express this relation by saying, say
$$2a^5 = 5b^2.$$
Note. If $a\lt 0$, then the volume of $A$ can be computed the same way, but the integral would go from $a$ to $0$, so that the volume would be $-\frac{\pi a^5}{5}$; to account for both possibilities, both $a\gt 0$ and $a\lt 0$, you can simply write that the volume is $\frac{\pi|a|^5}{5}$. For solid $B$, however, it makes no sense to talk about $b\lt 0$, because then we don't have a finite area "enclosed" by the curves in question.
|
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|
Problem: Sum of absolute values of polynomial roots Can you please give me some hints as to how I might approach this problem? Thanks!
Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$
Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$
|
We first have the following factorization,
$ 2x^3 - ax^2 - ax + 2 = (x+1)(2x^2 - (2+a)x + 2).$
Suppose $x_1 = -1$, then $|x_2| + |x_3| = 2$ and they are solutions to the quadratic equation $2x^2 - (2+a)x + 2 = 0$. Hence,
$ x_2 + x_3 = 1 + a/2 $ and
$ x_2 x_3 = 1$.
By observing the second equation, $x_2$ and $x_3$ are either both positive or both negative, so we have
$1 + a/2 = x_2 + x_3 = 2$ or $-2$.
Therefore, $a = 2$ or $-6$.
|
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|
How many function are there? Let $f_k \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear map defined by
$$
f_k \begin{pmatrix} 1 \\ 2 \\ k \end{pmatrix}=
\begin{pmatrix} 2+k \\ 3 \\ 0 \end{pmatrix},
\quad
f_k \begin{pmatrix} 2 \\ k+1 \\ -1 \end{pmatrix}=
\begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix},
\quad
f_k \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix}=
\begin{pmatrix} 1 \\ k \\ 2 \end{pmatrix},
$$
How many functions $f_k$ are there?
|
First, consider the three vectors,
$$\left(\begin{array}{c}1\\2\\k\end{array}\right),\qquad \left(\begin{array}{c}2\\k+1\\-1\end{array}\right),\qquad \left(\begin{array}{c}-3\\1\\5\end{array}\right).$$
If they are linearly independent, then we can define a linear transformation from $\mathbb{R}^3$ to any vector space simply by saying what we want the value to be at those three vectors (since any specification of what happens to a basis gives a linear transformation). If the three vectors are linearly dependent, then any linear dependency between them must also be satisfied by their images.
So, when are they linearly independent? The three vectors are linearly independent if and only if the determinant of the $3\times 3$ matrix that has them as columns is nonzero. Since
$$\begin{align*}
\det\left(\begin{array}{ccc}1 & 2 & -3\\
2 & k+1 & 1\\
k & -1 & 5
\end{array}\right) &= \left|\begin{array}{cc}k+1 & 1\\ -1 & 5\end{array}\right| -2\left|\begin{array}{cc}2 & 1\\k & 5\end{array}\right| -3\left|\begin{array}{cc}2 & k+1\\k & -1
\end{array}\right|\\
&= 5(k+1) + 1 -2\Bigl(10 - k\Bigr) -3\Bigl(-2 -k^2-k\Bigr)\\
&= 5k + 5 +1 - 20 + 2k +6 +3k^2 + 3k\\
&= 3k^2 +10k -8 = (k+4)(3k-2).
\end{align*}$$
So the three vectors are linearly independent if $(k+4)(3k-2)\neq 0$; that is, if $k\neq -4$ and $k\neq\frac{2}{3}$. In those cases, such a linear transformation exists, and since we've specified what happens at a basis, there is one and only one such linear transformation.
If $k=-4$, then the vectors are
$$\left(\begin{array}{r}1\\2\\-4\end{array}\right),\qquad \left(\begin{array}{c}2\\-3\\-1\end{array}\right),\qquad \left(\begin{array}{c}-3\\1\\5\end{array}\right).$$
Notce that if we add all three vectors we get $\mathbf{0}$. That means that we must have that when we add the three images we get $\mathbf{0}$ as well, otherwise, we don't have a linear transformation; this holds, since the images are
$$\left(\begin{array}{r}-2\\3\\0\end{array}\right),\qquad \left(\begin{array}{r}1\\1\\-2\end{array}\right),\qquad \left(\begin{array}{r}1\\-4\\2\end{array}\right).$$
In fact, this is (up to multiplying by a scalar) the only linear relation between the three vectors: they span a 2-dimensional subspace. Add a single vector to, say, the first two, to get a basis. Then you can define a linear transformation to be what we want at these three, and anything whatsoever on the third basis basis vector. Each choice will give you a linear transformation; so there are infinitely many possible linear transformations when $k=-4$.
Finally, when $k=\frac{2}{3}$, we have that the three vectors are
$$\left(\begin{array}{r}1\\2\\\frac{2}{3}\end{array}\right),\qquad \left(\begin{array}{r}2\\\frac{5}{3}\\-1\end{array}\right),\qquad \left(\begin{array}{r}-3\\1\\5\end{array}\right).$$
The three are linearly dependent; if you do a bit of work, you will find that
$$3\left(\begin{array}{c}1\\2\\ \frac{2}{3}\end{array}\right) - 3\left(\begin{array}{r}2\\ \frac{5}{3}\\-1\end{array}\right) -\left(\begin{array}{r}-3\\1\\5\end{array}\right) = \left(\begin{array}{c}3 -6 +3\\ 6-5-1\\2+3-5
\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right).$$
That means that we need the images to satisfy the same linear relation; but
$$3\left(\begin{array}{c}2+\frac{2}{3}\\3\\0\end{array}\right) -3\left(\begin{array}{c}1\\1\\-2\end{array}\right) - \left(\begin{array}{c}1\\\frac{2}{3}\\2\end{array}\right)\neq\left(\begin{array}{c}0\\0\\0\end{array}\right),$$
so there can be no linear transformation with those values when $k=\frac{2}{3}$.
a
|
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Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the next section. (Simmons' Calculus)
The hint given with the problem is to integrate another series. This series must be $$1-\frac{ 1}{2} x^2+\frac{ 1\cdot 3}{2\cdot 4} x^4-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^6+\cdots$$
But where does it come from? [How would I know to integrate that series, if it weren't for the hint?]
The derivative of $\ln(x+\sqrt{1+x^2})$ is $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$,
but I can't see how to get from $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$ to that series.
The derivative can be written as $\displaystyle \frac{ 1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, but that looks even worse.
|
The first part of the derivate you calculated can be transformed to
$$\frac{ 1}{x+\sqrt{1+x^2}} =
\frac{ 1}{x+\sqrt{1+x^2}} \frac{ x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}} =
\sqrt{1+x^2}-x$$
and therefore the derivate is
$$\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}$$
The term $\frac{x}{\sqrt{x^2+1}}$ is the derivate of $\sqrt{x^2+1}$.
The expansion of
$$\sqrt{1+x}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^n}$$
You can try to guess the coeffizients of the series:
Let $w(x)=\sqrt{x+1}$ Then
$$(x+1)^{\frac{3}{2}}=w(x)(1+x)$$
Differentiating this equation gives
$$\frac{3}{2}w(x)=w(x)+(1+x)w'(x)$$
$$w(x)=2(x+1)w'(x)$$
If $p(x)=\sum_{n=0}^{\infty}a_n x^n$ then we get
$$a_n=2(na_n+(n+1)a_{n+1})$$
and
$$a_{n+1}=(-1)\frac{2n+1}{2(n+1)}a_n$$
if a powerseries $p(x)$ should fullfill this equation.
From this we get
$$a_n=\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}$$
by induction. So if w(x) there is a power series expansion of w(x) these are the coefficients. Both w(x) and the power series p(x) satisfy the equation
$$\frac{f'(x)}{f(x)}=\frac{1}{x+1}$$
and therefore
$$(\log{(f(x)})'=\frac{1}{x+1}$$
Therefore
$$(\log{(w(x)})'=(\log{(p(x)})'=\frac{1}{x+1}$$
$$\log{(w(x))}=\log{(p(x))}+C$$
$$w(x)=p(X)*D$$
where $D=e^C$. From $w(0)=p(0)=1$ we conclude that $D=1$ and $w(x)=p(x)$.
We have now proved that this is the power series expansion of $\sqrt{x+1}$.
Substitute $x$ by $x^2$ in $\sqrt{x+1}$ and you get $$\sqrt{1+x^2}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^{2n}}$$
differentiate the equation and get
$$\frac{x}{\sqrt{1+x^2}}=\sum_{n=1}^{\infty}{\frac{(-1)^n(2n)!(2n)}{(1-2n)(n!)^2 4^n}x^{2n-1}}$$
and therefore
$$\sqrt{x^2+1} - x*\frac{x}{\sqrt(1+x^2)}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n}x^{2n}}$$
This is the series of your hint.
Integration now gives
$$C+\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n (2n+1)}x^{2n+1}}$$
For x=0 the function is 0, therefore C=0
If you calculate $x$ from $y=\ln{(x+\sqrt{x^2+1})}$ you get $$x=\frac{(e^y)^2-1}{2e^y}=\frac{e^y-e^{-y}}{2}=\sinh(y)$$ Therefore your function is the inverse of the sinh, the arsinh.
You can chek your results at http://en.wikipedia.org/wiki/List_of_mathematical_series
|
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|
Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as:
$$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$
But how is this possible?
|
All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get
$$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$$
We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$.
Comment: If you find this not obvious, let's look at the first term, that is, at
$$\frac{ab+ca}{a^2+ab+ca}$$
The "top" is $a(b+c)$. The "bottom" is $a(a+b+c)$. So the fraction is
$$\frac{a(b+c)}{a(a+b+c)}$$
Divide top and bottom by $a$, or equivalently, "cancel" the $a$'s. We are using the "algebra" version of the familiar fact that
$$\frac{2\cdot 3}{2 \cdot 5}=\frac{3}{5}$$
Additional comment The hint is strange. True, we can rewrite the expression as
$$\left(1-\frac{a^2}{a^2+ab+ca}\right) +\left(1-\frac{b^2}{b^2+ab+bc}\right)+\left(1-\frac{c^2}{c^2+bc+ca}\right)$$
which is equivalent to what was given. And then we could divide top and bottom by $a$ in the first fraction, by $b$ in the second, by $c$ in the third, and end up with
$$3-\frac{a+b+c}{a+b+c}$$
But it sure seems like a lot of work when we can cancel immediately! You are sure that you quoted the problem correctly?
|
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|
Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
|
Hint: use $a^6 - b^6 = (a - b)(a^5 + a^4 b + a^3 b^2 + a^2 b^3 + a b^4 + b^5)$ with $a =(x^6 + x^5)^{1/6} = x (1 + 1/x)^{1/6}$ and $b = (x^6 - x^5)^{1/6} = x (1 - 1/x)^{1/6}$.
|
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|
Laplacian of a Smooth Function $f(\vec x) = 1/\|\vec x \|$ for $\| \vec x \| \geq 1$ This is a multivariable calculus problem from a past prelim exam. I have an answer for this written up (posted below), but it seemed rather time-intensive. If there is a slicker way to approach this problem, I'd appreciate seeing it. Thanks!
Recall that for a smooth function $f: \mathbb{R}^3 \to \mathbb{R}$, the Laplacian of $f$ is defined by
$$ \Delta f = \nabla \cdot ( \nabla f). $$
Suppose that $f: \mathbb{R}^3 \to \mathbb{R}$ is a smooth function satisfying $f(\vec{x}) = 1/\|\vec{x}\|$ for $\|\vec{x}\| \geq 1$.
*
*Verify that $\Delta f(\vec{x}) = 0$ for $\|\vec{x}\| \geq 1$.
*Compute $\int_{\mathbb{R}^3} \Delta f \, dV$.
|
*
*In rectangular coordinates, for $\|\vec{x}\| \geq 1$,
$
\begin{align*}
f(\vec{x}) &= \frac{1}{\sqrt{x^2+y^2+z^2}} \\
\nabla f &= \left\langle \frac{-x}{(x^2+y^2+z^2)^{3/2}}, \frac{-y}{(x^2+y^2+z^2)^{3/2}}, \frac{-z}{(x^2+y^2+z^2)^{3/2}} \right\rangle \\
&= \frac{-1}{(x^2+y^2+z^2)^{3/2}} \left\langle x,y,z \right\rangle \\
\nabla \cdot \nabla f &= \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-x)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2x) }{(x^2+y^2+z^2)^3} \\
&+ \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-y)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2y) }{(x^2+y^2+z^2)^3} \\
&+ \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-z)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2z) }{(x^2+y^2+z^2)^3} \\
&= \frac{-3(x^2+y^2+z^2) + 3x^2+3y^2+3z^2}{(x^2+y^2+z^2)^{5/2}} \\
&= 0
\end{align*}
$
*Since $\Delta f = 0$ for $\|\vec x\| \geq 1$, then $\int_{\mathbb{R}^3} \Delta f \, dV = \int_{\| \vec x \| \leq 1} \Delta f \, dV$. Then by the divergence theorem,
$\begin{align*}
\int_{\| \vec x \| \leq 1} \nabla \cdot ( \nabla f) dV &= \iint_{\|\vec{x}\|=1} (\nabla f) \cdot \vec n \, dS \\
&= \iint_{\|\vec{x}\|=1} \frac{-1}{(x^2+y^2+z^2)^{3/2}} \left\langle x,y,z \right\rangle \cdot \frac{ \left\langle x,y,z \right\rangle }{(x^2+y^2+z^2)^{1/2}}dS \\
&= \iint_{\|\vec{x}\|=1} \frac{-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2} dS \\
&= \iint_{\|\vec{x}\|=1}-1 dS \\
&= -4 \pi
\end{align*}$
|
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about $\operatorname{SO}(2)$ group When
$$
A =\begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad a,b,c,d \in \mathbb{R}, \quad A^2 -2aA + I = 0
$$
Is $ A \in \operatorname{SO}(2)\; $ if $A^n \in SO(2)\;$ for some $n \in \Bbb N$?
$\operatorname{SO}(2)$ is special orthogonal group.
|
I'm a bit surprised since nobody tried to find a counter-example! Let $p$ be a nonzero real number and
$$A = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -p \\ p^{-1} & 1 \end{pmatrix}.$$
Then we have $A^2 - \sqrt{2}A + I = O$ and $A^4 = -I \in SO(2)$, but still $A \notin SO(2)$ unless $p = \pm 1$.
But this is partially true if some auxiliary conditions are imposed. For example, let's denote $a = \cos \theta$, where $\theta$ is allowed to have complex value in the case $|a| > 1$. Also, let $n$ be the least positive integer satisfying $A^n \in SO(2)$. If we impose the condition that $\sin n\theta \neq 0$, then $A \in SO(2)$. (The condition $\sin n\theta \neq 0$ corresponds to the condition that $A^n \neq \pm I$.)
Let's prove this claim. Here we further assume that $a \neq \pm 1$, so that $\sin \theta \neq 0$. Later this excluded cases will be treated separately. Since
$$x^2 - 2ax + 1 = (x - e^{i\theta})(x - e^{-i\theta}),$$
we have
$$x^n = (x^2 - 2ax + 1)q_n (x) + \frac{\sin n\theta}{\sin\theta} x - \frac{\sin (n-1)\theta}{\sin \theta}.$$
Now by assumption and the fact pointed out by Michael Banaszek, we can write
$$\frac{\sin n\theta}{\sin\theta} A - \frac{\sin (n-1)\theta}{\sin \theta} I = A^n = R_{\varphi}, \ \cdots \ (1)$$
where
$$R_{\varphi} = \begin{pmatrix} \cos \varphi & -\sin\varphi \\ \sin\varphi & \cos \varphi \end{pmatrix}$$
is the rotation matrix corresponding to the angle $\varphi$. Now comparing the trace of both sides of (1), we find that
$$\sin n\theta \cos \theta - \sin(n-1)\theta = \sin \theta \cos \varphi,$$
or by addition formula for sine, we have
$$\cos\varphi = \cos n\theta.$$
In particular, $\theta$ is real and $\theta = \frac{1}{n}(\varphi + 2\pi k )$ for some $k \in \mathbb{Z}$. Plugging this back to (1) and simplifying, we obtain
$$\quad A = R_{\theta}.$$
Here we critically used the fact that $\sin n\theta = \sin \varphi$ is nonzero.
Now we treat the case when $a = \pm 1$. In this case, we hvae
$$ x^n = (x^2 - 2ax + 1)q_n (x) + n a^{n-1} x - (n-1) a^{n} .$$
Then by assumption,
$$ n a^{n-1} A - (n-1) a^{n} I = A^n = R_{\varphi}$$
for some $\varphi$. Now take $k$-power to both sides to obtain
$$ kna^{kn-1}A - (kn-1)a^{kn}I = A^{kn} = R_{k\varphi}.$$
Dividing both sides by $kn a^{nk-1}$, we have
$$ A = a \left( 1 - \frac{1}{kn}\right)I + \frac{a^{1-nk}}{nk} R_{k\varphi}.$$
But since $a^{1-nk} R_{k\varphi}$ is uniformly bounded, taking $k \to \infty$ gives
$$ A = aI,$$
which is again in $SO(2)$.
|
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$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$ Prove :
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$
For $a,b,c,d>0$
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I know many time has passed, however i came up with a solution and I hope it is correct: here you are:
Let us denote
$$\begin{array}{c}\frac{1}{4}\frac{(b+c)^3}{2}(8(a^2+d^2)^3-(a+d)^3(a+b)^3)+\\\frac{1}{4}\frac{(a+d)^3}{2}(8(b^2+c^2)^3-(c+d)^3(c+b)^3)=(\clubsuit)\end{array}$$
And we want to prove $(\clubsuit)\geq0.$
We recall $x^3-y^3=(x-y)(x^2+xy+y^2)$ $(\spadesuit)$ and so
$$\begin{array}{c}(\clubsuit)=(2(a^2+d^2)-(a+d)(a+b))C_1+2((b^2+c^2)-(c+d)(c+b))C_2\geq\\ \min\{C_1,C_2\}\cdot(2(a^2+b^2+c^2+d^2)-(a+d)(a+b)-(c+d)(c+d)).\end{array}$$
Where $C_1,C_2$ are nonnegative costants depending on $a,b,c,d$, and can be easily derived following $(\spadesuit)$
But now it is true that $$2(a^2+b^2+c^2+d^2)-a^2-ab-ad-bd-c^2-cd-cb-bd\geq 0,$$
since $$\begin{eqnarray} b^2+d^2&\geq& 2bd,\\\frac{a^2+d^2}{2}&\geq& ad,\\ \frac{c^2+d^2}{2}&\geq& cd,\\\frac{c^2+b^2}{2}&\geq&cb,\\\frac{a^2+b^2}{2}&\geq&ab,\\a^2&\geq&a^2,\\c^2&\geq&c^2.\end{eqnarray}$$
We have then estabilished $(\clubsuit)\geq 0$. But it is easy to show that this relation implies $$(a^2+d^2)^3(b+c)^3+(b^2+c^2)^3(a+d)^3\geq\frac{(a+b)^3(a+d)^3(b+c)^3}{8}+\frac{(c+d)^3(a+d)^3(b+c)^3}{8}$$ which in turn implies $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq \left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3,$$ as desired.
|
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When should the antiderivative of a rational function be defined as a piecewise function? I'm doing basic problems on antiderivatives and there seems to be an inconsistency in my book.
The instructions for these problems are:
Find the most general antiderivative of the function.
Number 11 is:
11. $f(x) = \dfrac{10}{x^9}$
So, I naturally wrote down $F(x) = -\dfrac{5}{4x^8} + C$.
The solution manual says this is wrong. The function has domain $(-\infty, 0) \cup (0, \infty)$, so $F(x) = \begin{cases} -\dfrac{5}{4x^8}+C_1 & \text{if } x < 0 \\
-\dfrac{5}{4x^8}+C_2 & \text{if } x > 0 \end{cases}$
Okay, I thought. That makes sense.
I then come to number 13, which is:
13. $f(x) = \dfrac{u^4 + 3\sqrt{u}}{u^2}$.
I figured the domain is $(-\infty, 0) \cup (0, \infty)$, so I wrote down $F(x) = \begin{cases} \frac{1}{3}u^3-6u^{-1/2}+C_1 & \text{if } u > 0 \\
\frac{1}{3}u^3-6u^{-1/2}+C_2 & \text{if } u < 0 \end{cases}$.
Well, the solution book doesn't mention the domain and just says $F(x) = \frac{1}{3}u^3-6u^{-1/2}+C$.
I later realized that the $\sqrt{u}$ in the numerator and the $u^2$ in the denominator must limit the domain to the positive numbers, so the antiderivative doesn't need to be defined for anything but positive numbers. So, okay, I think I get it.
Next number 19 is:
19. $f(x) = \dfrac{x^5-x^3+2x}{x^4} = x - \dfrac{1}{x} + \dfrac{2}{x^3}$
So, again, since the domain appears to be $(-\infty, 0) \cup (0, \infty)$, I wrote down $F(x) = \begin{cases} \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\
\frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}$.
But the book again doesn't mention the domain and just says that $F(x) = \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C$.
I am confused. This doesn't seem consistent, especially between #11 and #19. Why is my answer not right for #19?
|
The person who wrote out the solution may have thought that putting absolute value signs around $x$ takes care of the problem. It doesn't. Congratulations on reading with care.
The answer for Question $19$ ought to be something like
$$F(x) = \begin{cases} \frac{1}{2}x^2 - \ln x - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\
\frac{1}{2}x^2 - \ln(-x) - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}.$$
|
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|
Asymptotic behaviour of sums of consecutive powers Let $S_k(n)$, for $k = 0, 1, 2, \ldots$, be defined as follows
$$S_k(n) = \sum_{i=1}^n \ i^k$$
For fixed (small) $k$, you can determine a nice formula in terms of $n$ for this, which you can then prove using e.g. induction. For small $k$ we for example get
$$\begin{align}
S_0(n) &= n\\
S_1(n) &= \frac{1}{2}n^2 + \frac{1}{2}n \\
S_2(n) &= \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \\
S_3(n) &= \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 \\
S_4(n) &= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3 - \frac{1}{30}n
\end{align}$$
The coefficients of these polynomials are related to the Bernoulli-numbers, and getting arbitrary coefficients in these polynomials (i.e. the coefficient of $n^m$ in $S_k(n)$ for large $k,m$) is not so easy. However, th first two coefficients follow a simple pattern: the coefficient of $n^{k+1}$ is $\frac{1}{k+1}$ and the coefficient of $n^k$ (for $k > 0$) is always $\frac{1}{2}$. My main question now is:
How can we prove that $S_k(n) = \frac{1}{k+1}n^{k+1} + \frac{1}{2}n^k + O(n^{k-1})$ for $k > 0$?
The first coefficient can be explained intuitively, as
$$S_k(n) = \sum_{i=1}^n \ i^k \approx \int_{i=1}^n i^k di \approx \frac{n^{k+1}}{k+1}$$
Maybe you could make this more rigorous, but I don't see how you will get the term $\frac{1}{2}n^k$ with this.
Also, while the coefficient of $n^{k+1}$ can be explained intuitively, it's not clear to me why the coefficient of $n^k$ is $\frac{1}{2}$, and why this one is fixed while e.g. the coefficient of $n^{k-1}$ is different for different $k$. If someone could explain that, that would be appreciated as well.
Thanks.
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The standard way to sum $\sum\limits_{r=1}^n r^k$ gets this estimate.
We have:
$$
(r+1)^{k+1} - r^{k+1} = (k+1) r^{k} + \frac{k(k+1)}{2} r^{k-1} + O(r^{k-2}).
$$
Summing this equation for $r = 0, 1, 2, \ldots, n$, we get:
$$
(n+1)^{k+1} + O(1) = (k+1) \sum_{r=0}^nr^{k} + \frac{k(k+1)}{2} \sum_{r=0}^n r^{k-1} + \sum_{r=0}^nO(r^{k-2}).
$$
Using the inductive hypothesis that $\sum_{r=0}^n r^j = \frac{n^{j+1}}{j+1} + \frac{n^j}{2}$
for $j < k$, we get:
$$
\begin{eqnarray*}
(k+1) \sum_{r=0}^nr^{k} &=& (n+1)^{k+1} + O(1) - \frac{k(k+1)}{2} \left(\frac{n^k}{k} + \frac{n^{k-1}}{2} \right)+ \sum_{r=0}^nO(r^{k-2}).
\\ &\stackrel{(1)}{=}& \left[ n^{k+1} + (k+1)n^{k} + O(n^{k-1}) \right] + O(1) - \frac{(k+1)n^k}{2} + O(n^{k-1})+ O(r^{k-1})
\\ &=& n^{k+1} + \frac{k+1}{2} n^k + O(n^{k-1}),
\end{eqnarray*}
$$
where $(1)$ follows from the binomial theorem. This is equivalent to the claim.
|
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$a/b=c$, how to move the numerator to the right side of the equation Given an equation of the form $a/b=c$, how do you move $a$, the numerator, over to the other side so that you get $b =c?$ (where $?$ denotes my ignorance of what the right side should look like).
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$$
\begin{align}
\frac{a}{b} & = c \\ \\
b \cdot \frac{a}{b} & = c \cdot b \\
\require{cancel} \\
\cancel{b} \cdot \frac{a}{\cancel{b}} & = c \cdot b \\ \\
a & = cb \\ \\
\frac{a}{c} & = \frac{cb}{c} \\ \\
\frac{a}{c} & = \frac{\cancel{c}b}{\cancel{c}} \\ \\
\frac{a}{c} & = b
\end{align}
$$
|
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|
Proving that for reals $a,b,c$, $(a + b + c)^2 \leq 3(a^2+b^2+c^2)$
Proving that for reals $a,b,c$, $(a+b+c)^2\leq 3(a^2+b^2+c^2)$.
This is a homework question and I have no clue where to even start on this. I don't know if I am just tired or what but I can't get anywhere. I've expanding both sides and seeing if that gets me anywhere but I don't see how it helps.
|
We need to expand both sides of inequality:
$(a+b)^2+c^2+2(a+b)c\leq 3a^2+3b^2+3c^2$
$a^2+b^2+c^2+2ab+2ac+2bc\leq 3a^2+3b^2+3c^2$
$2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$
$0\leq a^2+b^2-2ab+a^2+c^2-2ac+b^2+c^2-2bc$
$0\leq (a-b)^2 +(a-c)^2
+(b-c)^2$ , what is obvious true.
|
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|
How can I differentiate this? I was trying to differentiate this: $\frac{1}{2} \sin^{-1} \frac{2x}{1+x^2}$ but I am really stuck.I obtained an answer that does not match with the one given in the book.Your help is appreciated.[Edit: I used the formula $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ to get $\frac{1+x^2}{2(1-x^2)}$]
Thank you.
|
The derivative of the arcsine function is
$$\frac{d}{du}\arcsin(u) = \frac{1}{\sqrt{1-u^2}}.$$
So, using the Chain Rule, you would have
$$\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) = \frac{1}{2}\left(\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right)\left(\frac{2x}{1+x^2}\right)'.$$
The derivative of $\frac{2x}{1+x^2}$ is a simple application of either the Quotient Rule or the Product, Power, and Chain Rules:\
$$\begin{align*}
\frac{d}{dx}\left(\frac{2x}{1+x^2}\right) &= \frac{(1+x^2)(2x)' - 2x(1+x^2)'}{(1+x^2)^2}\\
&= \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2}\\
&= \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2}\\
&= \frac{2(1-x^2)}{(1+x^2)^2}.
\end{align*}$$
The expression obtained from the derivative of $\arcsin(u)$ can use a bit of simplification too:
$$\begin{align*}
\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} &= \frac{1}{\sqrt{1 - \frac{4x^2}{(1+x^2)^2}}}\\
&=\frac{1}{\sqrt{\frac{(1+x^2)^2 -4x^2}{(1+x^2)^2}}}\\
&= \frac{1}{\frac{\sqrt{1+2x^2+x^4 - 4x^2}}{|1+x^2|}}\\
&= \frac{|1+x^2|}{\sqrt{1-2x^2+x^4}}\\
&= \frac{1+x^2}{\sqrt{(1-x^2)^2}}\\
&= \frac{1+x^2}{|1-x^2|}.
\end{align*}$$
(Remember that $\sqrt{a^2}=|a|$, not $a$; we can get rid of the absolute value bars around $1+x^2$ because it is always positive; the same is not true with $1-x^2$).
Putting it all together, we have:
$$\begin{align*}
\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) &= \frac{1}{2}\left(\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right)\left(\frac{2x}{1+x^2}\right)'\\
&= \frac{1}{2}\left(\frac{1+x^2}{|1-x^2|}\right)\left(\frac{2(1-x^2)}{(1+x^2)^2}\right)\\
&= \frac{1-x^2}{(1+x^2)|1-x^2|}.
\end{align*}$$
This can be rewritten with the "sign function",
$$\mathrm{sgn}(u) = \left\{\begin{array}{ll}
1 & \text{if }u\gt 0;\\
-1 &\text{if }u\lt 0.
\end{array}\right.$$
as
$$\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) = \frac{\mathrm{sgn}(1-x^2)}{1+x^2}.$$
It's also possible your book was not careful with the square root of the square, so that the answer given is just
$$\frac{1}{1+x^2}.$$
|
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|
Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$ Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$.
By plotting graph I have seen that there are no roots for $x$. Can somebody prove it theoretically?
|
Here's an answer, but I am sure there should be better ones.
Assume $x \in \mathbb R$ to be a root of the given equation. Rearrange the equation as
$$
(x^{35}-100) (x^{17}+2+\cos^2 x) = - 20205.
$$
Let $A := x^{35}-100$ and $B := x^{17}+2+\cos^2 x$. Since $AB < 0$, exactly one of $A$ and $B$ is positive and the other is negative.
If $B < 0$ and $A > 0$, then $x^{35} > 100 > 0$, which implies that $x > 0$. Then we can conclude that $B = x^{17}+2 + \cos^2 x > 0$, which contradicts the assumption about the sign of $B$. So this case is impossible.
We are left with the case $A < 0$ and $B > 0$.
*
*From $A < 0$, we get $x^{35} < 100$. Therefore $x^{17} < 10$. Hence, $B = x^{17}+2+\cos^2 x \leq x^{17}+3 < 13$.
*Similarly, from $B > 0$, we get $x^{17} > -2 - \cos^2 x \geq -3$. Therefore, $A = x^{35} - 100 > (-3)^{3}-100 = -127$. In other words, $|A| < 127$.
Multiplying the upper bounds on $|A|$ and $B$, we get $$|A \cdot B| = |A| \cdot B < 127 \cdot 13 = 1651 ,$$
which contradicts the equation $AB = - 20205$ we started out with.
|
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|
Digamma function integral Does anyone how to get a finite value to this integral ?
$ \int\nolimits_{0}^{\infty} dx \frac{ \Psi (1/4+ix/2) +\Psi (1/4-ix/2)}{x^{2}+1/4} $
i have tried residue theorem but i got nonsenses :( can anyone help
or give some advice
i believe that the integral should be equal to the $ \Psi (3/2) $ or similar although i can not prove it :(
anyway, is this integral equal to · $ \int\nolimits_{-\infty}^{\infty} dx \frac{ \Psi (1/4+ix/2)}{x^{2}+1/4} $ ??
|
Use $\sum_{n=1}^\infty \left( \frac{1}{n+a-1} - \frac{1}{n}\right) = -\gamma - \Psi(a)$, where $\Psi(a)$ is di-gamma function. Using this representation for the di-gamma function and integrating term-wise:
$$ \begin{multline}
\mathcal{I} = \int_0^\infty \frac{ \Psi(\frac{1}{4} + i \frac{x}{2}) + \Psi(\frac{1}{4} - i \frac{x}{2}) }{x^2 + 1/4} \, \mathrm{d} x = \\
\int_0^\infty \frac{ -2 \gamma }{x^2 + 1/4} \, \mathrm{d} x -
\sum_{n=1}^\infty \int_0^\infty \frac{1}{x^2 + 1/4} \left( \frac{8 (4 n-3)}{(3-4 n)^2+4 x^2}-\frac{2}{n} \right) \, \mathrm{d} x
\end{multline}
$$
Thus
$$
\mathcal{I} = -2 \pi \gamma - \sum_{n=1}^\infty \frac{2 \pi }{n (2 n-1)} = -2 \pi \left( \gamma + 2 \log 2 \right) = 2\pi \left( \Psi\left(\frac{3}{2}\right) - 2\right)
$$
Verify numerically:
In[55]:= NIntegrate[(
PolyGamma[1/4 + I x/2] + PolyGamma[1/4 - I x/2])/(
x^2 + 1/4), {x, 0, \[Infinity]}, WorkingPrecision -> 50] ==
2 Pi (PolyGamma[3/2] - 2)
Out[55]= True
|
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|
Compute $1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
Sol:
$P(n):\ 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
For $n=n_1 = 1$
$$P(1) = \frac{1}{3}{3} = (1)^2.$$
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$
$$P(k): 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 = \frac{1}{3}k(2k-1)(2k+1).$$
We have to prove that it is true for $P(k+1)$.
$$P(k+1) = 1^1+3^2+5^2+\cdots+(2k+1)^2 = \frac{1}{3}(k+1)(2k+1)(2k+3)\tag{A}.$$
Taking LHS:
$$\begin{align*}
1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\
&= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\
&= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\
&=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\
&=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\
&=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\
&=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\
&=\frac{1}{3}(2k+1)(k+1)\left(k+\frac{3}{2}\right) \tag{B}
\end{align*}$$
EDIT:
Solving EQ (A):
$=(1/3)(2k^2+5K+3) (2K+1) \tag{C}$
Comparing EQ(B) and EQ(C)
Hence proved that it is true for $n = k+1.$
Thus the proposition is true for all $n >= 1$.
Thanks.
|
NOTE: I am not saying anything different, before someone start commenting that my answer is not any different.
$$
\begin{align*}
1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\
&= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\
&= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\
&=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\
&=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\
&=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\
&=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\
&=\frac{1}{3}(2k+1) \hspace{3pt}\left[(k+1)(2k+3)\right] \\
&= \frac{1}{3} (k+1)(2(k+1)-1)(2(k+1)+1)
\end{align*}
$$
The last line shows that the result is true for $n=k+1$
|
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|
Find critical points of a function: $y = x \sqrt{4 - x^2}$ I am having trouble finding the critical points of this function, I was wondering if someone could help me out.
So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points.
$$\begin{align*}
y &= x \sqrt{4 - x^2}\\
y' &= x \frac{d}{dx}\sqrt{4 - x^2} + \sqrt{4 - x^2}&& \text{(product rule)}\\
&= x\left(\frac{1}{2}(4 - x^2)^{-1/2}(-2x)\right) + \sqrt{4 - x^2} &&\text{(chain rule on square root)}\\
&= \frac{\frac{1}{2}x(-2x)}{(4-x^2)^{1/2}} + \sqrt{4-x^2}\\
&\qquad\qquad\text{(moved to denominator changed the exponent sign)}
\end{align*}$$
This is where I am stuck, I cant seem to figure out how to proceed further, any help would be appreciated
Thanks
|
Put your last expression together as one fraction by getting a common denominator. The right hand expression will be
$$\frac{4-x^2}{\sqrt{4-x^2}}$$
so the overall fraction is
$$\frac{-x^2 + 4 - x^2}{\sqrt{4-x^2}} = \frac{4 - 2x^2}{\sqrt{4-x^2}} = \frac{-2(x^2 - 2)}{\sqrt{4-x^2}}$$
which is 0 when $x = \pm \sqrt 2$.
|
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|
Limit of $\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{n/2}$ I'm having a hard time characterising the behavior of the following expression:
$$\lim_{n\rightarrow\infty}\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{\frac{n}{2}}$$
with the following constraints on the parameters: $0<b<a<\infty$, and $\epsilon\in\mathbb{R}$. I am interested in the following:
*
*for $\epsilon>0$, does this limit go to zero or does it go to some
constant $C$? If it can both go to zero or to some constant $C>0$,
what are the conditions on the value of $\epsilon$ as a function of
$a$ and $b$ which leads to these outcomes, if any?
*for $\epsilon<0$, does it always go to some constant $C<1$, or can
it go to 1 for some $\epsilon$, if it's a function of $a$ and $b$?
*what happens to this limit when $\epsilon=0$?
|
Let the desired limit be denoted by $L$. On taking logarithms we get
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)^{n/2}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)^{n/2}\text{ (by continuity of log)}\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(\frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)}}{2a(\sqrt{n} + \epsilon) + b}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log\left(1 + \frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} - 2a(\sqrt{n} + \epsilon) - b}{2a(\sqrt{n} + \epsilon) + b}\right)\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\log(1 + (A/B))
\end{align}
where
\begin{align}
\frac{A}{B} &= \frac{2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} - 2a(\sqrt{n} + \epsilon) - b}{2a(\sqrt{n} + \epsilon) + b}\notag\\
&= \frac{4\{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)\} - 4a^{2}(\sqrt{n} + \epsilon)^{2} - b^{2} - 4ab(\sqrt{n} + \epsilon)}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&= -\frac{b^{2}}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&\to 0 \text{ as }n \to \infty\notag
\end{align}
Therefore we can continue the limit evaluation as
\begin{align}
\log L &= \lim_{n \to 0}\frac{n}{2}\log(1 + (A/B))\notag\\
&= \lim_{n \to 0}\frac{n}{2}\cdot\frac{A}{B}\cdot\frac{\log(1 + (A/B))}{A/B}\notag\\
&= \lim_{n \to \infty}\frac{n}{2}\cdot\frac{A}{B}\notag\\
&= -\frac{b^{2}}{2}\lim_{n \to \infty}\frac{n}{\left(2a(\sqrt{n} + \epsilon) + b\right)\left(2\sqrt{a^{2}(\sqrt{n} + \epsilon)^{2} + ab(\sqrt{n} + \epsilon)} + 2a(\sqrt{n} + \epsilon) + b\right)}\notag\\
&= -\frac{b^{2}}{2}\lim_{n \to \infty}\frac{1}{\left(2a(1 + \epsilon/\sqrt{n}) + b/\sqrt{n}\right)\left(2\sqrt{a^{2}(1 + \epsilon/\sqrt{n})^{2} + ab(1/\sqrt{n} + \epsilon/n)} + 2a(1 + \epsilon/\sqrt{n}) + b/\sqrt{n}\right)}\notag\\
&= -\frac{b^{2}}{2}\cdot\frac{1}{2a(2a + 2a)}\notag\\
&= -\frac{b^{2}}{16a^{2}}\notag
\end{align}
Thus $L = \exp(-b^{2}/16a^{2})$.
|
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|
quotient rule difficulties I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have
$$\begin{align*}
\frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac {d}{dr} \sqrt{r^2+1}} {r^2+1} \\\\\\\\
&= \frac {\sqrt{r^2+1} - r \frac{1}{2}(r^2+1)^{-1/2}2r} {(r^2+1)}\\\\\\\\
&= \frac {\sqrt{r^2+1} - r^2 (r^2+1)^{-1/2}} {(r^2+1)}\\\\\\\\
&= (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}
\end{align*}$$
However, the correct answer is just $$(r^2+1)^{-3/2} $$
I've been over it a number of times now, but I can't see the error. I'm pretty new to the quotient rule.
|
Your answer is right
$$ (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}= \frac{1}{\sqrt{r^2+1}} -\frac{r^2}{(r^2+1)^{3/2}}
= \frac{r^2+1}{(r^2+1)^{3/2}} -\frac{r^2}{(r^2+1)^{3/2}}$$
You can finish it from here....
|
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Showing that there are infinitely many integer solutions to $x^2+y^2=z^2$ Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.
|
I we use Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can see that
$$
A^2+B^2=\big((m^2-k^2)+(2mk)^2\\
=(m^4-2m^2k^2+k^4)+4m^2k^2\\
=(m^4+2m^2k^2+k^4)\\
=(m^2+k^2)^2=C^2
$$
One problem with Euclid's formula is that it generates trivial triples
like $f(1,0)=(1,0,1)\quad f(3,0)=(9,0,9)\quad \cdots\qquad$
I developed a formula that generate non-trivial Pythagorean triples for every pair of natural number $(n,k)$ and produces only the subset of triples where $GCD(A,B,C)$ is an odd square -- meaning it includes all primitives where
$GCD(A,B,C)=1^2$.
\begin{equation}
A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
|
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|
If $\gcd(a,b)=1$ , and $a$ is even and $b$ is odd then $\gcd(2^{a}+1,2^{b}+1)=1$? How to prove that:
$\gcd(a,b)=1 \Rightarrow \gcd(2^{a}+1,2^{b}+1)=1$ ,where $a$ is even and $b$ is odd natural number
For example:
$\gcd(2^8+1,2^{13}+1)=1 , \gcd(2^{64}+1,2^{73}+1)=1$
I know that Knuth showed that:
$\gcd(2^{a}-1,2^{b}-1)=2^{\gcd(a,b)}-1$
so: $\gcd(a,b)=1\Rightarrow \gcd(2^{a}-1,2^{b}-1)=1$
but I don't see whether this fact is useful.
|
We post a second answer that, instead of doing details, recycles old results. The OP mentioned the following standard result, but was not sure it was useful:
$$\gcd(2^{x}-1,2^{y}-1)=2^{\gcd(x,y)}-1.$$
It is useful! By the result, if $\gcd(a,b)=1$, then
$$\gcd(2^{2a}-1,2^{2b}-1)=2^{2}-1=3.$$
Thus
$$\gcd((2^{a}+1)(2^a-1)\;,\;(2^b+1)(2^b-1)=3.$$
It follows that any common divisor of $2^a+1$ and $2^b+1$ must divide $3$, and therefore $\gcd(2^a+1,2^b+1)$ is $1$ or $3$.
Like in the other answer, we rule out $3$ by using the fact that one of $a$ and $b$, say $a$, is even. For even $a$, $2^a\equiv 1\pmod{3}$, so $3$ does not divide $2^a+1$.
|
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|
integrate square of $\arctan x$. Tricky $$\int \left(\frac{\tan^{-1}x}{x-\tan^{-1}x}\right)^{2}dx$$
I ran across an integral I am having a time solving.
The solution merely works out to $\displaystyle\frac{1+x\tan^{-1}x}{\tan^{-1}x-x}$, but for the life of me I can not find a suitable method to tackle it.
Does anyone have any hints on a good strategy, substitution, parts, etc?. Thanks much.
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The denominator suggests that maybe an integration by parts will help where $v=\frac{1}{x-\arctan x}$. This may make for a strange looking $dv$ that is not readily apparent from the given integral.
$$\frac{d}{dx}\left(\frac{1}{x-\arctan x}\right)=\frac{-\frac{x^2}{1+x^2}}{(x-\arctan x)^2}$$
So let's try $dv=\frac{-\frac{x^2}{1+x^2}}{(x-\arctan x)^2}$ and $u =-\arctan^2x\frac{1+x^2}{x^2}$.
Note that $u = -\left(\frac{1}{x^2}+1\right)\arctan^2 x$.
Computing $du$, we have \begin{align}
du &=\left[-\left(\frac{1}{x^2}+1\right)\frac{2\arctan x}{1+x^2}+\frac{2\ \arctan^2x}{x^3}\right]dx\\
&=\frac{2\arctan x}{x^3}\left(\arctan x-x\right)\ dx\end{align}
Applying the integration by parts formula $\int u\,dv=uv-\int v\,du$, the given intergal is equal to $$-\left(\frac{1}{x^2}+1\right)\arctan^2 x\frac{1}{x-\arctan x}+\int\frac{2\arctan x}{x^3}\,dx$$
The new integral can be handled with another integration by parts where $u$ is $\arctan x$ and $dv$ is $2x^{-3}$. That will leave you with a rational function in $x$, which can be integrated using partial fraction decomposition. Once all of the integration is complete, the terms should add together to give what you know the antiderivative to be (perhaps with a constant difference). These next steps are "ugly" details, but follow the description above. We have
\begin{align}&-\frac{\left(\frac{1}{x^2}+1\right)\arctan^2 x}{x-\arctan x}-\frac{\arctan x}{x^2}+\int\frac{1}{x^2(1+x^2)}\ dx\\
=&-\frac{\left(\frac{1}{x^2}+1\right)\arctan^2 x}{x-\arctan x}-\frac{\arctan x}{x^2}+\int\frac{1}{x^2}\ dx-\int\frac{1}{x^2+1}\ dx\\
=&-\frac{\left(\frac{1}{x^2}+1\right)\arctan^2 x}{x-\arctan x}-\frac{\arctan x}{x^2}-\frac{1}{x}-\arctan x+C\\
=&\frac{1+x\arctan x}{x^2(x-\arctan x)}-\frac{\arctan x(x-\arctan x)}{x^2(x-\arctan x)}\\&-\frac{x(x-\arctan x)}{x^2(x-\arctan x)}-\frac{x^2(x-\arctan x)\arctan x}{x^2(x-\arctan x)}+C\\
\\
=&-\frac{\left(1+x^2\right)\arctan^2 x}{x^2(x-\arctan x)}-\frac{\arctan x(x-\arctan x)}{x^2(x-\arctan x)}\\&-\frac{x(x-\arctan x)}{x^2(x-\arctan x)}-\frac{x^2(x-\arctan x)\arctan x}{x^2(x-\arctan x)}+C\\
\\
=&\frac{-x^2-x^3\arctan x}{x^2(x-\arctan x)}+C\\
=&-\frac{1+x\arctan x}{(x-\arctan x)}+C\\
\end{align}
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Hard algebra problem Given
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
Now, it is necessary to find
$$\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=?$$
Is this possible and how? a,b,c are given constants.
I think, ? is probably a complicated function $F=F(a,b,c)$
|
The expression
$$\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}$$ isn't constant on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$ Notice that I get different values on the points $(x,y,z) = (a,0,0)$ and $(x,y,z)=(0,b,0)$.
|
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Determinant of a specific circulant matrix, $A_n$ Let
$$A_2 = \left[ \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]$$
$$A_3 = \left[ \begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
$$A_4 = \left[ \begin{array}{cccc} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0\end{array}\right]$$
and so on for $A_n$.
I was asked to calculate the determinant for $A_1, A_2, A_3, A_4$ and then guess about the determinant for $A_n$ in general. Of course the pattern is clear that
$$ \det A_n = (n-1)(-1)^{n-1} $$
but I was wondering as to what the proof of this is. I tried to be clever with cofactor expansions but I couldn't get anywhere.
Could someone explain it to me please?
|
Here is an elementary way to compute the determinant of $A_n$:
Add row 2 to row 1, add row 3 to row 1, ..., and add row $n$ to row 1, we get
$$\det(A_n)=\begin{vmatrix}
n-1 & n-1 & n-1 & \cdots & n-1 \\
1 & 0 & 1 &\cdots & 1 \\
1 & 1 & 0 &\cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \ldots & 0 \\
\end{vmatrix}.$$
Next subtract column 2 by column 1, subtract column 3 by column 1, ..., subtract column $n$ by column 1, we get
$$\det(A_n)=\begin{vmatrix}
n-1 & 0 & 0 & \cdots & 0 \\
1 & -1 & 0 &\cdots & 0 \\
1 & 0 & -1 &\cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \ldots & -1 \\
\end{vmatrix}=(-1)^{n-1}(n-1).$$
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|
Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} = \frac{5 \times \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,
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$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} =\frac{10}{4} \sum_{i=1}^{10} \frac{i}{2^i}-\frac{5}{4}\sum_{i=1}^{10} \frac{1}{2^i}$
Second $\sum$ is obviously $1-\frac{1}{1024}$.
First $\sum$:
$\frac{1}{2} +$
$\frac{1}{4} + \frac{1}{4} +$
$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} +$
$\vdots$
$\frac{1}{1024} + \frac{1}{1024} + \dots + \frac{1}{1024}$
Summing by columns, $(1-\frac{1}{1024})+(\frac{1}{2}-\frac{1}{1024})+\dots+(\frac{1}{512}-\frac{1}{1024})=(1+\frac{1}{2}+\dots+\frac{1}{512})-\frac{10}{1024}=\frac{1023}{512}-\frac{10}{1024}$
Rest is boring and easy.
|
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How can I prove this equation: $(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}1(1-\frac{1}{n})\cdot(1-\frac{2}{n})…(1-\frac{k-1}{n})$
Possible Duplicate:
Proving sequence equality using the binomial theorem
$(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}\cdot1\cdot(1-\frac{1}{n})\cdot(1-\frac{2}{n})\cdot…\cdot(1-\frac{k-1}{n})$
This is how far I came using the binomial theorem:
$(1+ \frac{1}{n})^n = 1 + \sum\limits_{k=1}^n{\frac{1}{k!}\cdot\frac{n!}{n^k\cdot(n-k)!}}$
I don't know how to rearrange that tail to be the same as in the original equation.
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$$\frac{n!}{n^k(n-k)!}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{n\,n\,n\cdots n}=\left(1-\frac0n\right)\,\left(1-\frac1n\right)\,\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right).
$$
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If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
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If negative values of the variables are allowed then there are counterexamples. The simplest is the following:
$$a=2,\quad b=2,\quad c=\sqrt{3}-1-\sqrt{2\sqrt{3}}\doteq -1.129,\quad d=\sqrt{3}-1+\sqrt{2\sqrt{3}}\doteq2.593$$
with $a+b=4$, $c+d\doteq1.464\ .$
In the following we assume $0\leq a\leq b$ and $0\leq c\leq d$. Then icobes' conjecture is true:
Since $a^2+b^2=c^2+d^2=:r^2>0$ we may write
$$a=r\sin\bigl({\pi\over4}-\alpha\bigr), \quad
b=r\sin\bigl({\pi\over4}+\alpha\bigr), \quad
c=r\sin\bigl({\pi\over4}-\beta\bigr), \quad
d=r\sin\bigl({\pi\over4}+\beta\bigr)$$
for some $\alpha, \beta\in\ \bigl[0,{\pi\over4}\bigr]$. It follows that $$a+b=\sqrt{2}r\cos\alpha,\quad c+d=\sqrt{2}r\cos\beta$$
and therefore
$$2(a^3+b^3)=3(a+b)(a^2+b^2)-(a+b)^3=\sqrt{2}r^3(3\cos\alpha-2\cos^3\alpha)=:\sqrt{2}r^3 f(\alpha)\ ;$$
and similarly $2(c^3+d^3)=\sqrt{2}r^3f(\beta)$.
Now $f'(t)=3\sin t\cos(2t)>0$ for $0<t<{\pi\over 4}$, whence $f$ is strictly increasing on $\bigl[0,{\pi\over4}\bigr]$. It follows that $a^3+b^3=c^3+d^3$ implies $\alpha=\beta$, and this in turn implies $a=c$, $b=d$.
|
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$\frac{1+\sqrt{m}}{2}; (19)^{1/3}$ are algebraic integral
This question is from a very old exercise sheet, I do not know what the notation $\mathbb{Z}_{\mathbb{C}}$ means:
*
*$1\equiv 1 \mod 4 $ in $\mathbb{Z}$. Show that $\frac{1+\sqrt{m}}{2} \in \mathbb{Z}_{\mathbb{C}}$ 2. Let $a= (19)^{1/3}$. Show that $\frac{1+a+a^{2}}{3} \in \mathbb{Z}_{\mathbb{C}}$
1.
It will now be shown that $\frac{1+\sqrt{m}}{2}$ is an algebraic integer, that means that it fulfills a polynomial with leading coefficient 1 (monic) and rest integral. Wikipedia article on algebraic integers says that it suffises the monic polynomial: $x^{2}-x+\frac{1-m}{4}$ This is true because: $(x-\frac{1}{2}(1+\sqrt{m})) (x-\frac{1}{2}(1-\sqrt{m})) = x^{2}-x+\frac{1-m}{4}$
2. Let $b = 19$, then $\frac{1+b+b^{2}}{3} = \frac{1+19+19^{2}}{3}$
Then one examines : $(x-(\frac{1+19+19^{2}}{3}))(x-\frac{-1-19-19^{2}}{3}) = (x-127)(x+127) = x^{2}- 16129$
now subtituting $x$ with $x^{3}$ gives: $x^{6}-16129$, and this is the monic polynomial for $a= (19)^{1/3}$
Is this done correctly? Please do tell
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Your approach seems too complicated in both cases. Try these simpler approaches:
*
*Let $\alpha =\frac{1+\sqrt{m}}{2}%$. Then $2 \alpha -1 = \sqrt{m}$. Can you now find a monic equation for $\alpha$? Make sure you use the hypothesis on $m \bmod 4$.
*Let $\beta=\frac{1+a+a^{2}}{3}$. Note that $\beta = \frac{a^3-1}{3(a-1)}=\frac{6}{a-1}$. Then $a=\frac{6}{\beta}+1$. Now use that $a^3=19$ to find a monic equation for $\beta$.
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Find a couple of integers such that the third power of a given natural can be written as the difference of the squares of those integers Given a natural number $n$, find inegers $a, b$ such that $n^3=a^2-b^2$. I've tried, but I'm a bit rusty. Please Help
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Note that
$$n^3=\left(\frac{n^2+n}{2}\right)^2-\left(\frac{n^2-n}{2}\right)^2.$$
Comment: The magic identity that solved the problem in fact did not (for me) come by magic. Given a number $K$, we want to find numbers $a$ and $b$ such that $a^2-b^2=K$. So we want $(a+b)(a-b)=K$. This means that $a+b$ and $a-b$ are two integers whose product is $K$.
Suppose that $x$ and $y$ are any two integers whose product is $K$. If we set $a+b=x$ and $a-b=y$, then we will have $(a+b)(a-b)=K$. But will $a$ and $b$ be integers?
Solve the system $a+b=x$, $a-b=y$. Algebra gives $a=\frac{x+y}{2}$, $b=\frac{x-y}{2}$. In order to make sure that $a$ and $b$ are integers, $x+y$ (and therefore $x-y$) must be even. This means that $x$ and $y$ have to be of the same parity (both odd or both even).
Can we express $n^3$ as a product of two numbers of the same parity? If $n$ is odd, we can use $x=n^3$, $y=1$. That won't work if $n$ is even. But $x=n^2$, $y=n$ always works, because $n^2$ and $n$ have the same parity.
In general, the integer $K$ is a difference of two squares unless $K$ is even but not divisible by $4$. So $\pm 2$, $\pm 6$, $\pm 10$, $\pm 14$, and so on cannot be expressed as a difference of two squares, and everybody else can be.
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Deducing Markov inequality from reverse Markov inequality? Let $x$ be random variable, such that $E(x)=0,E(x^2)=1$ and $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$, where $C>0,s\geq 1 , t>0$.
Let $m<n$ and $m,n$ are natural numbers very big. Let also $L\geq 1$ .
Consider
$1-(1-\frac{n}{2}P(x^2\geq Ln))^m$.
Assume (*) $\frac{n}{2}P(x^2\geq Ln)\leq \frac{2c_0}{m}$, where $0 <c_0 <0.6$;
using inequality $(1-y)^m\leq 1-\displaystyle\frac{my}{2}$, valid for $y\in [0, c_0]$ , with natural $n$, we get
$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq\displaystyle\frac{nm}{4}P(x^2\geq Ln),$$
using assumption $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$ with $s^2=Ln$ , we get
$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq \displaystyle\frac{nm}{4}P(x^2\geq Ln)\geq \frac{mnC'}{(Ln)^{\frac{t}{2}}}.$$
How to show, that if $t\geq 4$, then (*) holds?
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Probably it is better reformulate the question I've posted. Please tell me if my solution is correct and where I have mistakes. Thank you for advice.
Let x be random variable with mean zero and variance 1. And such that for $C>0$, $t>0$ and natural n, $P(x^2\ge n)\geq \frac{C}{n^t}$.
We want to show that if $t\ge 2$, then $P(x^2\ge n)\geq \frac{2}{n^2}$.
First, we show that for any $n\geq \widetilde{n}$,
$$
\begin{align}
P(x^2\geq n)\leq \frac{2}{n^2}.
\end{align}
$$
Suppose to the contrary that
\begin{align}
P(x^2 \geq n_i) > \frac{2}{n_i^2}
\end{align}
for some infinite sequence $n_1$, $n_2$, $\ldots$, $n_i$, $\ldots$, such that
\begin{align}
P(x^2 > n_i^2) > 2P(x^2 > n_{i+1}^2).
\end{align}
By assumption
$$
1=E x^2 = \int_{0}^\infty x^2 d \mu.
$$
Here $\mu$ is a probability measure. We can break up this integral into the sum:
$$
E x^2 = \sum_{i=0}^\infty \int_{n_i}^{n_{i+1}}x^2 d \mu.
$$
Consider now
$$
\begin{align}
\int_{n_i}^{n_{i+1}} x^2 d\mu \geq n_i^2 \int_{n_i}^{n_{i+1}} d\mu
&= n_i^2(P(x^2\geq n_i^2)-P(x^2 \geq n_{i+1}^2))
&\gt n_i^2\frac 12 P(x^2\geq n_i^2)
&\gt 1.
\end{align}
$$
This is a contradiction to $E x^2=1$.
Thus, with $\widetilde{n}\leq n$
\begin{align}
P(w^2\geq Kn)\leq \frac {2}{n^2} .
\end{align}
But, within condition \begin{align}
\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2},
\end{align}
and $\displaystyle{n\geq \left(\frac{C}{2}\right)^{\frac{1}{t-2}}=\widetilde{\widetilde{n}}}$.
Thus, for any $n\geq \max\{\widetilde{n},\widetilde{\widetilde{n}}\}$, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$.
If $t\geq 2$, then for any $n \in N$ and ,$C=2$
$P(w^2\geq Kn)=\frac{2}{n^2}$. Thus, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$ for any natural n.
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Use Lagrange Remainder Theorem to Prove Inequality I'm supposed to use Lagrange Remainder Theorem to prove that
$$1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2} \text{ } \text{ if } x>0$$
Obviously, the left and right hand sides look like Taylor series, but how would you do this using Lagrange, it isn't obviously intuitive. Thanks!
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For the RHS inequality you can even use Mean Value Theorem (which, of course, is a special case of Taylor theorem - where the reminder term contains first derivative).
Let $f(x) = \sqrt{1+x}$. Then the theorem gives that there exists some $\theta \in (0,x)$ such that
$$
\frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1+x} - 1}{x} = f'(\theta) = \frac{1}{2 \sqrt{1+\theta}}.
$$
Hence $\sqrt{1+x} = 1 + \frac{x}{2 \sqrt{1+\theta}} < 1 + \frac{x}{2}$ since $x > 0$. Observe that, of course, $\frac{x}{2 \sqrt{1+\theta}}$ is the reminder term in the Taylor theorem.
For the LHS by Taylor theorem you get that there exists $\theta \in (0,x)$ such that
$$
\sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} (1 + \theta)^{-5/2} x^3.
$$
The reminder term is positive for $x > 0$, which proves the inequality.
|
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Isotropy over $p$-adic numbers Over what $p$-adic fields $\mathbb{Q}_p$ is the form $\langle3, 7, -15\rangle$ isotropic?
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Everything is in Cassels, Rational Quadratic Forms, pages 41-44, then pages 55-59. In particular Lemma 2.5 on page 59. Your forms is isotropic at the primes $\{2,5,\infty\}.$ It is anisotropic at primes $\{3,7\}.$ And isotropic at everything else.
To give the simplest looking version, what happens if $$ 3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}?$$
Well, $$ 3 x^2 - 15 z^2 \equiv 0 \pmod {7},$$
$$ x^2 - 5 z^2 \equiv 0 \pmod {7},$$
$$ x^2 \equiv 5 z^2 \pmod {7}.$$
If we assume $z$ is not divisible by 7, it has a multiplicative inverse $\pmod 7,$ so
$$ \frac{ x^2}{z^2} \equiv 5 \pmod {7}$$ and
$$ \left(\frac{ x}{z}\right)^2 \equiv 5 \pmod {7}.$$
However, 5 is a quadratic nonresidue $\pmod 7,$ written $(5|7)= -1.$
So the equation is impossible, contradicting the assumption, and, in fact, $z$ is divisible by 7. From $ x^2 \equiv 5 z^2 \pmod {7}$ we find that $x$ is also divisible by 7. So, actually, $$ x^2 - 5 z^2 \equiv 0 \pmod {49}.$$
It follows from $ 3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ that $7 y^2 \equiv 0 \pmod {49}$ which tells us that $ y^2 \equiv 0 \pmod {7}$ and $ y \equiv 0 \pmod {7}.$ In sum, $ 3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ implies that $(x,y,z)$ are all divisible by 7. It follows that we can divide out 7 from each of the variables and 49 from whatever the result was. Or, we cannot express a number divisible by an arbitrarily high power of 7 unless $(x,y,z)$ are all divisible by 7 to roughly half that exponent.
Or, put another way, the only solution to $ 3 x^2 + 7 y^2 - 15 z^2 = 0 $ in
$\mathbb Q_7$ is $(x=0, y=0, z=0).$
It is a worthwhile exercise for you to confirm things about the prime 3, as above. Also plow through Lemma 2.5 for each prime.
Finally, from Lemma 2.4 on page 58, $7 y^2 - 15 z^2$ is isotropic in $\mathbb Q_2,$ and $ 3 x^2 + 7 y^2$ is isotropic in $\mathbb Q_5.$
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Find $ n\geq1 $ such that 7 divides $n^n-3$ Find $ n\geq1 $ such that 7 divides $n^n-3$.
Here is what I found:
$ n\equiv 0 \mod7, n^n\equiv 0 \mod7,n^n-3\equiv -3 \mod7$ no solution.
$ n\equiv 1 \mod7, n^n\equiv 1 \mod7,n^n-3\equiv -2 \mod7 $ no solution.
$ n\equiv 2 \mod7, n^n\equiv 2^n \mod7, n^n-3\equiv 2^n-3 \mod7$. Is $2^n\equiv 3 \mod7 $ possible?
$ n=7k+2,..., 2^{7k+2}\equiv 2^{k+2} \mod7$. Is $2^{k+2}\equiv 3 \mod7$ possible?
Studying $k$ modulo 6: $2^{6q+2}\equiv 4 \mod7, 2^{6q+3}\equiv 1 \mod7, 2^{6q+4}\equiv 2 \mod7, 2^{6q+5}\equiv 4 \mod7, 2^{6q+6}\equiv 1 \mod7, 2^{6q+7}\equiv 2 \mod7 $
The congruence is never 3 so there is no solution for $n\equiv 2 \mod7$.
$ n\equiv -2 \mod7,..., n=42q+5 $
$ n\equiv-1 \mod7, n^n-3\equiv (-1)^n-3 \mod7 $: no solution.
$ n\equiv 3 \mod7, n^n-3 \equiv 3^n-3 \mod7$. Is $3^n\equiv 3 \mod 7$ possible?
$n=7k+3,..., 3^{7k+3}\equiv -3^k \mod7 $. Is $3^k \equiv 4 \mod 7 $ possible?
Studying $k$ modulo 6:
$ 3^{6q} \equiv 1 \mod7, 3^{6q+1} \equiv 3 \mod7, 3^{6q+2} \equiv 2 \mod7, 3^{6q+3} \equiv -1 \mod7, 3^{6q+4} \equiv 4 \mod7 , 3^{6q+5} \equiv -2 \mod7 $
So $ n=7k+3=7(6q+4)+3=42q+31 $ is a solution.
$ n\equiv -3 \mod7 $,..., there is no solution.
$42q+31 \equiv 3 \mod 7, (42q+31)^{42q+31} \equiv 3^{42q+31} \mod 7 \equiv 3 \mod7$ OK
$ 42q+5 \equiv 4 \mod7, (42q+5)^{42q+5} \equiv 5^{42q+5} \mod 7 \equiv 3 \mod7$ OK
So 7 divides $ n^n-3 $ if and only if $ n\equiv 31 \mod 42$ or $ n\equiv 5 \mod 42$
|
Since the question only asks (at least the way I read it) to find some $n\geq 1$ such that $n^n-3\equiv 0 \bmod 7$, you can exploit the fact that a power $a^b \bmod 7$ only depends on the class of $a\bmod 7$ and the class of $b\bmod 6$.
Thus, if you pick $n\equiv 3 \bmod 7$ such that $n\equiv 1 \bmod 6$, then:
$$n^n\equiv 3^n \equiv 3^{1+6k} \equiv 3\cdot (3^6)^k \equiv 3 \bmod 7,$$
as desired. Hence, by the Chinese remainder theorem, any $n\equiv 31 \bmod 42$ will work.
For instance,
$$3^{31}-3 = 617673396283944=2^3\cdot 3\cdot 7\cdot 11^2\cdot 13\cdot 31\cdot 61 \cdot 271\cdot 4561.$$
|
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|
approximation formula for the integral Get an approximation formula for the following integral:
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
|
$$\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
First observe the powers of $cos$ and $sin$
$$ cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) = cos^{2n-2k}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) =
(1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y)$$
$$
\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy
$$
If we use the substitution $t = sin(y) \Rightarrow dt = cos(y) \hspace{4pt} dy$
$t = 0$ when $y=0$ and $t=1$ when $y=\frac{\pi}{2}$
$$
\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^2)^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy
$$
$$
= \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt
$$
The expression
$$
\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
= \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right)
$$
$$
= \left( \frac{1}{35} \right)^{k-1}\int_0^1 \left( \sum_{k=1}^{n} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt
$$
Notice that the sum is a geometric series with first term $\frac{1}{35}(1-t^{2})^{n}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})^{2}}$
The expression therefore simplifies to
$$
= \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^2)^{(n-k)} \right) dt
$$
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
= \left( \frac{1}{35} \right)^{k-1} \sum_{k=1}^n \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right)
$$
$$
= \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt
$$
Notice that the sum is a geometric series with first term $(1-t^{2})^{(n-1)}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})}$
The expression therefore simplifies to
$$
= \int_0^1 \left( \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( t^{(2k-2)} (1-t^{2})^{(n-k)} \right) \right) dt
$$
This expression turns out to be the summation of constant multiple of Beta function
|
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|
Minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ I am attempting to compute the minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ over $\mathbb Q$. So far, my reasoning is as follows:
The Galois conjugates of $2^{1/3}$ are $2^{1/3} e^{2\pi i/3}$ and $2^{1/3} e^{4\pi i /3}$. We have $4^{1/3} = 2^{2/3}$, so the image of $4^{1/3}$ under an automorphism $\sigma$ fixing $\mathbb Q$ is determined by the image of $2^{1/3}$: it must equal the square of $\sigma(2^{1/3})$. Therefore, the Galois conjugates of $1 + 2^{1/3} + 4^{1/3}$ are $1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3}$ and $1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}$. Therefore, the minimal polynomial is $(x-a)(x-b)(x-c)$, where
$$\begin{align*}
a&=1 + 2^{1/3} + 4^{1/3},\\
b&=1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3},\text{ and}\\
c&=1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}.
\end{align*}$$
However, this polynomial does not seem to have coefficients in $\mathbb Q$! What am I doing wrong?
|
We can conclude that your polynomial has rational coefficients even without doing any expanding. The splitting field for the polynomial, which is $\mathbb{Q}(2^{1/3},\omega)$ (here $\omega = e^{2\pi i/3}$) has Galois group $S_3$ generated by the automorphisms $\sigma$ fixing roots of unity and sending $2^{1/3}$ to $2^{1/3}\omega$, and $\tau$ fixing $2^{1/3}$ and sending $\omega$ to $\omega^2$. When we expand $(x-a)(x-b)(x-c)$, the coefficients will be given (up to sign) by the elementary symmetric polynomials in $a,b,c$. Since each element of $S_3$ permutes the set $\{a,b,c\}$, the symmetric polynomials will all be fixed by $S_3 \cong \operatorname{Gal}(\mathbb{Q}(2^{1/3},\omega):\mathbb{Q})$, so will lie in $\mathbb{Q}$.
|
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|
Product in GF(16) i need some help with a product in GF(16), where it is seen as an extension of GF(4)={0, 1, x, x+1} (where $x^2 = x + 1$ ) with the irreducible polynom $f(y) = y^2 + y + x$
So the elements in the fields are: $\{0, 1, x, x+1, y, y + 1, y + x, y + x + 1, xy, ... \}$
I'm try to multiply $((x+1)y + x) * (xy + x + 1)$ and i obtain as result $(x + 1)$
I'm not sure that it is the correct result, can you help me?
Thank you,
greetings
|
Use the identities $x^2 = x+1$, $y^2 = y+x$, and $2=0$. After that you just have to compute. Here it goes explicitly for you to be confident :
\begin{align}
((x+1)y+x) \times (xy + x + 1) & = ((x+1)y+x)xy + ((x+1)y+x)x + (x+1)y+x \\
& = (x^2y^2+xy^2+x^2y)+(x^2y + xy + x^2) + xy + y + x \\
& = x^2 y^2 + xy^2 + x^2 + y + x \\
& = x^2 (y+x) + x(y+x) + x^2 + (y+x) \\
& = (x^2 + x + 1)(y+x) + x^2 \\
& = (2x^2)(y+x) + x^2 = x^2 = x+1.
\end{align}
If there is a way to make this shorter I haven't looked for it.
Hope that helps,
|
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|
Primes of the form : $x^2+4\cdot y^2$ How to prove following statement :
Conjecture:
Odd prime $p$ is expressible as : $p=x^2+4\cdot y^2$ , $x,y > 0$
if and only if : $p\equiv 1 \pmod {12}$ or $p\equiv 5 \pmod {12}$ .
Similar statements (without a proof) related to the Fermat's theorem on sums of two squares can be found here .
|
HINT $\rm\ (\Rightarrow)\ \ odd\ p\ =\ x^2+4\ y^2\ \Rightarrow\ x\ odd,\ so\ \ mod\ 4\!:\ x\ \equiv\: \pm1\ \Rightarrow\ p\ \equiv\ x^2\:\equiv\ 1 $
Thus $\rm\ p\equiv 1\pmod 4\ \Rightarrow\ p\equiv 1,5,9\pmod{12}\:,\:$ but not $\:9\:$ since $\rm\:3\:|\:p\ \Rightarrow\ p = 3\ne x^2 + 4\ y^2$
$\rm(\Leftarrow)\ $ By Fermat $\rm\: p \equiv 1\pmod 4\: \Rightarrow\ p\: =\: m^2 + n^2\: \Rightarrow\ 2\ |\ m\ \ or\ \:n\ ($else $\rm\: 2\ |\ p = odd^2+odd^2)$
|
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|
What type of distribution would rand()/rand() produce? If rand() is a function which produces a linearly distributed random number over a range not containing zero, then what type of distribution would rand() / rand() produce?
I know it would center at 1, and there would be a few extreme values very close to zero or very large.
Does this type of random distribution have a name?
|
Let $U_1$ be a uniform random variable with domain $[a,b]$ for $b>a>0$, and let $U_2$ be another identically distributed independent uniform variable.
Let $X= \frac{U_1}{U_2}$. It is clear that $\frac{b}{a} \geqslant X \geqslant \frac{a}{b}$ with probability 1. Let $x$ be from such an interval. Then
$$ \begin{eqnarray}
F(x) &=& \mathbb{P}\left( X \leqslant x\right) = \mathbb{P}\left( U_1 \leqslant U_2 x \right) = \mathbb{E}\left( \min\left(1, \max\left(0, \frac{x U_2 - a}{b-a} \right) \right) \right) \\
& = &
\int_0^1 \min\left(1, \max\left(0, \frac{x (a+(b-a) u) - a}{b-a} \right) \right) \mathrm{d} u \\
&=& \int_0^1 \min\left(1, \max\left(0, u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\
\end{eqnarray}
$$
Consider now two cases, $\frac{b}{a} \geqslant x \geqslant 1$ and $1 > x \geqslant \frac{a}{b}$.
*
*$\frac{b}{a} \geqslant x \geqslant 1$, which implies $u \cdot x + \frac{a}{b-a} (x-1) > 0$:
$$
\begin{eqnarray}
F(x) &=& \int_0^1 \min\left(1, \max\left(0, u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\
&=& \int_0^1 \min\left(1, u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
&=& \int_0^{u^\ast} \left( u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u + \int_{u^\ast}^1 \mathrm{d} u \\
&=& \left(\frac{u^\ast}{2} x - \frac{b - a x}{ x(b-a)}\right) u^\ast + 1 \\
&=& 1 - \frac{1}{2 x} \left( \frac{b-a x}{b-a} \right)^2
\end{eqnarray}
$$
where $u^\ast$ solves $u x + \frac{a}{b-a} (x-1) = 1 $, i.e. $u^\ast = \frac{b-a x}{x (b-a)}$.
*
*$1 > x \geqslant \frac{a}{b}$ implies $u \cdot x + \frac{a}{b-a} (x-1) < 1$:
$$
\begin{eqnarray}
F(x) &=& \int_0^1 \min\left(1, \max\left(0, u \cdot x + \frac{a}{b-a} (x-1) \right) \right) \mathrm{d} u \\
&=& \int_0^1 \max\left(0, u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
&=& \int_{u_\ast}^1 \left(u \cdot x + \frac{a}{b-a} (x-1) \right) \mathrm{d} u \\
&=& x \left( \frac{1}{2}- \frac{u_\ast^2}{2} \right) + \frac{a}{b-a} (x-1) \left( 1- u_\ast \right) \\
&=& \frac{ (a- b x)^2}{2 (b-a)^2 x}
\end{eqnarray}
$$
where $u_\ast$ solves $ u \cdot x + \frac{a}{b-a} (x-1) = 0$, i.e. $u_\ast =
\frac{a}{b-a} \frac{1-x}{x}$.
Given $F(x)$ computed above, the probability density follows by differentiation.
|
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|
If $f(x)$ and $f(x)-x$ have only one real root, then $f(f(x))-x$ has only one real root. First edition was: Let $f(x)$ be a polynomial such that $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that
$f(f(x))-x$ also has only one real root?
Second edition: Let $f(x)=a_3 x^3 + a_2 x^2 + a_1 x + a_0$. All $a_i \neq 0$. $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that
$f(f(x))-x$ also has only one real root?
|
As Jonas Meyer has pointed out, the problem is actually not true.
As suggested, let $f(x)= -x$. Then $f(x) -x = -x -x = -2x$. Considering $f(f(x)) -x= f(-x) -x = -(-x) -x = 0$. That is to say, for $f(x)=-x$, we have shown $f(x) -x$ has only one real root, namely $x=0$, while $f(f(x))-x=0$ has infinitely many real roots.
Response to comments:
Let $f(x)=x^3 - 2 x^2 + 5 x - 1$. Then, $$f(x) -x = x^3 -2x^2 +5x -1 =x^3 -2x^2 +4x -1$$ Solving for $x$ (by Wolfram Alpha) we have that $x \approx 0.284775$ and that there is only one real root. Considering $f(f(x))-x$ we have the following, $$f(f(x))-x = (x^3 -2x^2 +4x -1)^3 - 2(x^3 -2x^2 +4x -1)^2 +4(x^3 -2x^2 +4x -1) -1$$ which can be "simplifed" to, $$f(f(x))-x = x^9-6 x^8+24 x^7-61 x^6+116 x^5-156 x^4+155 x^3-102 x^2+44 x-8$$
Which has only one real root $x \approx 0.379555$. So, you have shown that there does exist a polynomial, namely $f(x) = x^3 -2x^2 +5x -1$ for which it is true. But this does not show in general that your conjecture is true, since there exists at least one counterexample.
@Jonas: I can't think of another counterexample off the top of my head, I don't want to try to hard at thinking about this.
|
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|
Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?
I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$
So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?
With the binomial theorem, I get it equal to
$$
\sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}).
$$
Can this be made nicer?
Thanks.
|
This is not a particularly elegant solution, but an alternative route is to simply note that $(1 \pm i)^2 = \pm 2i$. I will show a lot of steps, but this method involves only very easy calculations. If $n$ is even, then
$$
(1\pm i)^n = \left((1+i)^2\right)^{ \frac{n}{2}}= (\pm 2i)^{n/2} = \left\{ \begin{array}{ccl}
2^{n/2} & & n = 8k
\\
\pm i \cdot 2^{n/2}& & n = 8k +2
\\
i \cdot 2^{n/2} & & n = 8k + 4
\\
\mp i 2^{n/2} & & n = 8k + 6
\end{array} \right.
$$
Hence,
$$(1+i)^n + (1-i)^n =
\left\{
\begin{array}{ccl}
2^{\frac{n}{2}+1} & & n = 8k
\\
0 & & n = 8k+2 \text{ or } n = 8k+6
\\
i \cdot 2^{\frac{n}{2}+1} & & n = 8k + 4
\end{array}
\right.
$$
This is all you need since if $n$ is odd, then $n-1$ is even. For example, when $n = 8k + 1$, we have
$$\begin{align}
(1+i)^n + (1 - i)^n &= (1+i)^{n-1}(1+i) + (1 -i)^{n-1}(1-i)
\\
&= 2^{\frac{n-1}{2}}(1+i) + 2^{\frac{n-1}{2}} (1-i)
\\
&= 2^{\frac{n-1}{2}} + i 2^{\frac{n-1}{2}} + 2 ^{\frac{n-1}{2}} - i 2^{\frac{n-1}{2}}
\\
&= 2\cdot 2^{\frac{n-1}{2}}
\\
&= 2^{\frac{n+1}{2}}.
\end{align}$$
The cases $n = 8k+3$, $n=8k+5$, and $n = 8k + 7$ follow similarly.
|
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|
Solving $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ Could someone please explain how to solve this:
$$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$
not the answer only, but a step-by-step solution.
I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.
Thank you so much in advance!
|
First note that $x= \pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows:
$$ x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$
$$\qquad \Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$
$$ (x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0 $$
$$\qquad \Rightarrow (x^2-4)(x^2-10x+25)=0 $$
Hence we can solve it as
$$ (x^2-4)(x-5)^2 =0 $$ or $$x=\{ -2,+2,5 \}$$
|
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|
Integral( using Euler substitution) I try to solve this integral, but without success.
Can you help me please?
$$\int \frac{1}{2x+\sqrt{4x^{2}-x+1}}\,dx$$
Thanks a lot!
|
Hyperbolic substitution
Note $4 x^2 - x +1 = 4\left(x - \frac{1}{8} \right)^2 + \frac{15}{16}$. Therefore, let's perform a $u$-substitution, $x = \frac{1}{8} + \frac{\sqrt{15}}{8} \sinh(t)$. Then
$$
4 x^2 - x +1 = \frac{15}{16} \cosh^2(t)
$$
and $\mathrm{d} x = \frac{\sqrt{15}}{8} \cosh(t) \mathrm{d} t$, therefore
$$
\int \frac{\mathrm{d} x}{2x + \sqrt{ 4x^2-x+1}} = \int \frac{\frac{\sqrt{15}}{8} \cosh(t)}{\frac{1}{4} + \frac{\sqrt{15}}{4} \sinh(t) + \frac{\sqrt{15}}{4} \cosh(t)} \mathrm{d} t = \frac{\sqrt{15}}{4} \int \frac{ \left( \mathrm{e}^t + \mathrm{e}^{-t}\right)\mathrm{d} t} {1 + \sqrt{15} \mathrm{e}^t}
$$
The latter integral is easy
$$ \begin{eqnarray}
\sqrt{15} \int \frac{\mathrm{e}^t + \mathrm{e}^{-t}}{1+\sqrt{15} \mathrm{e}^t} \mathrm{d} t &=& \int \left( \frac{\sqrt{15}}{\mathrm{e}^{t} } - 15 + \frac{16 \sqrt{15} \mathrm{e}^t }{1 + \sqrt{15} \mathrm{e}^t} \right) \mathrm{d} t \\
&=& -\sqrt{15} \mathrm{e}^{-t} - 15 t + 16 \log\left( 1 + \sqrt{15} \mathrm{e}^t\right) + \color{\gray}{\text{const}}
\end{eqnarray}
$$
Back-substitution of $t = \sinh^{-1}\left(\frac{8x-1}{\sqrt{15}} \right)$, and using $\mathrm{e}^{-\sinh^{-1}(z)} = \sqrt{1+z^2}-z$ gives
$$
\int \frac{\mathrm{d} x}{2x + \sqrt{ 4x^2-x+1}} = \frac{1}{4} (8 x-1)-\frac{1}{4} \sqrt{(8 x-1)^2 + 15}+\frac{1}{4} \sinh
^{-1}\left(\frac{8 x-1}{\sqrt{15}}\right)+4 \log \left(15-(8 x-1)+\sqrt{(8
x-1)^2+15}\right) + \color{\gray}{\text{const}}
$$
Euler substitution
Let $u = 2x + \sqrt{4 x^2 - x+1}$. Notice that $(u-2x)^2 = u^2 - 4 x u + 4 x^2 = 4 x^2 - x + 1$, that makes it
$$
x = \frac{u^2-1}{4 u-1} = \frac{1}{16} + \frac{u}{4} - \frac{15}{16(4u-1)}
$$
Therefore
$$
\mathrm{d} x = \frac{\mathrm{d} u}{4} + \frac{15}{4} \frac{\mathrm{d} u}{(4u-1)^2}
$$
Using the above
$$ \begin{eqnarray}
\int\frac{\mathrm{d} x}{2x + \sqrt{4x^2-x+1}} &=& \int \frac{1}{u} \left( \frac{1}{4} + \frac{15}{4} \frac{1}{(4u-1)^2} \right) \mathrm{d} u \\
&=& \int \left( \frac{4}{u} + \frac{15}{(4u-1)^2} - \frac{15}{4u-1} \right) \mathrm{d} u \\
&=& 4 \ln(u) - \frac{15}{4(4u-1)} - \frac{15}{4} \ln(1-4u) + \color{\gray}{\text{const}}
\end{eqnarray}
$$
|
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|
How to solve the ordinary differential equation? $\displaystyle\frac{d²y}{dx^2}+ \frac{4}{y}\left(\frac{dy}{dx}\right)^2+2=0$
with $y(0) = 1$ and $\displaystyle\frac{dy}{dx} = 0$ for $x = 0$.
|
your equation is $yy''+4(y')^{2}+2y=0$
$z=y^5$
$z'=5y^{4}y'$
$z''=20y^{3}(y')^{2}+5y^{4}y''=5y^{3}(4(y')^{2}+yy'')$
$(4(y')^{2}+yy'')=\frac{z''}{5y^{3}}$
If we put it to your equation
$\frac{z''}{5y^{3}}+2y=0$
$z''=-10y^{4}=-10z^{4/5}$
$z'z''=-10z^{4/5}z'$
$\int z'z'' dx=-10\int z^{4/5}z'dx$
$ (z')^{2}/2 =-(50/9) z^{9/5}+m$
$ (z')^{2} =-(100/9) z^{9/5}+k$
$ z' =\sqrt{-(100/9) z^{9/5}+k}$
$z'=5y^{4}y'=\sqrt{-(100/9) y^{9}+k}$
if $x=0$ and
$y'(0)=0$ and $y(0)=1$ then $k=100/9$
$\frac{5y^{4}y'}{\sqrt{-(100/9) y^{9}+100/9}}=1$
$\int \frac{y^{4}y'}{\sqrt{1-y^{9}}} dx=\frac{2x}{3}+c$
I asked to wolfram that the solution is expressed by hypergeometric functions
the solution
$y^{5} \frac{_2F_1(1/2,5/9;14/9;y^{9})}{5}=\frac{2x}{3}+c$
if $x=0$ and
$y(0)=1$ then $c=\frac{_2F_1(1/2,5/9;14/9;1)}{5}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Groups of units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ On page 230 of Dummit and Foote's Abstract Algebra, they say: the units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ are determined by the integers $a,b$ with $a^2+ab+b^2=\pm1$ i.e. with $(2a+b)^2+3b^2=4$, from which is is easy to see the group of units is a group of order $6$ given by $\{\pm1,\pm\rho,\pm\rho^2\}$ where $\rho=\frac{-1+\sqrt{-3}}{2}$.
First, why change the characterization of unit from integers solutions of $a^2+ab+b^2=\pm1$ to integers solutions of $(2a+b)^2+3b^2=4$? How did they arrive at their answer?
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Let $\omega=\frac{1+\sqrt{-3}}{2}$. Notice that the norm of an element $a+b\omega$ in $\mathbb{Z}[\omega]$ is $$(a+b\omega)(a+b\overline{\omega})=a^2+ab+b^2.$$ Since the norm is multiplicative, all units must have norm $1$, otherwise they would not be invertible. This means we are looking for solutions to $a^2+ab+b^2=1$. (It can't be $-1$, I don't know why you put $\pm$ for all these quantities.)
For the second part, the author likely found it nicer to shift the quadratic form, as we need not consider different cases. Here is a direct way: To solve $a^2+ab+b^2=1$, split based on whether $ab$ negative or nonnegative.
Case 1: If $ab$ is nonnegative, then $1=a^2+ab+b^2\geq a^2 +b^2$, so we see that $a=\pm 1$, $b=0$ and $a=0$, $b=\pm 1$ are the only solutions.
Case 2: If $ab$ is negative, then both $a$ and $b$ are not zero, and then $1=a^2+ab+b^2=a^2+2ab+b^2-ab=(a+b)^2 -ab$. This then gives the only other solutions, $a=1=-b$, and $b=1=-a$.
|
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|
value of $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$? Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?
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The denominators $a_n$ are
$$
a_n = \sum_{k=1}^{n}k = \frac{n(n+1)}{2}
$$
so
$$
\sum_{n=1}^{\infty} \frac1{a_n} =
\sum_{n=1}^{\infty} \frac{2}{n(n+1)} =
2\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2
$$
If you don't have the resources to check this on a computer (out to infinity),
you will have your computer or calculator carry out the calculation for $n<N$.
You should then get a number near the partial sum
$$
s_N=\sum_{n=1}^{N-1} \frac1{a_n} = 2 \left( 1 - \frac{1}{N} \right),
$$
where whatever difference you find between the computer's answer and this expected answer will be the result of truncation or rounding errors due to the machine representation of each floating point number (which usually involve binary mantissas as in IEEE formats) encountered along the way.
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|
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
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Set $a = 2^x, b = 3^x$, then $6^x = (2 \cdot 3)^x = 2^x \cdot 3^x = ab$ and $2^{2x+2} = 2^2 \cdot 2^{2x} = 4 \cdot 2^x \cdot 2^x = 4a^2$, $9^{x+1} = 9 \cdot 9^x = 9 \cdot 3^x \cdot 3^x = 9b^2$. So we get $$12a^2 - 35ab + 18b^2 = 0.$$ I think.
What can you do with that?
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$? Polynomials of the form :
$ T_n(x) =2^{-1} \cdot ((x+\sqrt {x^2-1})^n+ (x-\sqrt {x^2-1})^n)$
are known as Chebyshev polynomials of the first kind .
Consider the polynomials of the form :
$P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$
Have these polynomials some special name ?
First few polynomials of this form are :
$P_0(x) = 2$
$P_1(x) = x$
$P_2(x) =x^2- 2$
$P_3(x) =x^3-3x$
$P_4(x) = x^4-4x^2+2$
$\vdots$
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I don't think they do have names (although they bear a superficial resemblance to the Lucas polynomials), but note that your polynomials are mere rescalings of the Chebyshev polynomial of the first kind:
$$\begin{align*}
2^{-n}((x+\sqrt{x^2-4})^n+(x-\sqrt{x^2-4})^n)&=\left(\frac{x}{2}+\sqrt{\frac{x^2}{4}-1}\right)^n+\left(\frac{x}{2}-\sqrt{\frac{x^2}{4}-1}\right)^n\\
&=\left(\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^2-1}\right)^n+\left(\frac{x}{2}-\sqrt{\left(\frac{x}{2}\right)^2-1}\right)^n\\
&=2T_n\left(\frac{x}{2}\right)
\end{align*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$ How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find:
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$
For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$
Thus, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$
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$(x^2-1)$ is negative iff $x\in(-1,1)$. Similarly, $(x^2-2)$ is negative for $x\in(-\sqrt{2},\sqrt{2})$, $(x^2-3)$ for $x\in(-\sqrt{3},\sqrt{3}), \dots,(x^2-6)$ for $x\in(-\sqrt{6},\sqrt{6})$.
Consider the open intervals $(-\infty,-\sqrt{6}), (-\sqrt{6},-\sqrt{5}),(-\sqrt{5},-2),\dots$ separately. Don't forget to look at how $f(x)$ behaves on the boundaries. E.g $x=\pm\sqrt{n}$ for $n=1,2,3,4,5,6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
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$$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$
$$=\lim_{h \to 0}
\frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h}
\cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$
$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3})}$$
$$=\lim_{h \to 0} \frac{1}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}$$
$$=\frac{1}{(x)^{2/3} + x^{1/3}(x)^{1/3} + (x)^{2/3}}$$
$$=\frac{1}{3x^{2/3}}$$
$$=\frac{x^{-2/3}}{3}$$
As obtained from the $Dx^{n} = n.x^{n-1}$
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|
Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$
Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something?
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Use the identities $$\sin 2\theta=2\cos\theta\sin\theta$$
and $$\eqalign{\sin 3\theta &=\sin(2\theta+\theta)\cr
&= \sin2\theta \cos\theta+\sin\theta\cos2\theta\cr
&= 2\sin\theta\cos^2\theta +\sin\theta(2\cos^2\theta-1)\cr
}
$$
to write the identity
$$\textstyle\sin{2\pi\over5}=\sin(\pi-{2\pi\over5})=\sin{3\pi\over5}$$ as
$$\tag{1}\textstyle
2\sin{\pi\over5}\cos{\pi\over5} = 2\sin{\pi\over5}\cos^2{\pi\over5} +\sin{\pi\over5}(2\cos^2{\pi\over5}-1).
$$
Cancelling the term $\sin{\pi\over5}\ne0$ from both sides of equation $(1)$ gives:
$$\textstyle
2 \cos{\pi\over5} = 2\cos^2{\pi\over5} + (2\cos^2{\pi\over5}-1);
$$
or,
$$\textstyle
0 = 4\cos^2{\pi\over5} -2\cos{\pi\over5} -1 ,
$$
as desired.
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|
Writing $3\sin(x)+4\cos(x)$ in the form of $r \sin(x-a)$ I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.
Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$
Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.
Getting $r$:
$$\begin{align*}
3^2 + 4^2 &= r^2 \cos(a)^2 + r^2 \sin(a)^2\\
25 &= r^2 (\cos(a)^2 + \sin(a)^2)\\
25 &= r^2\\
r &= -5, 5
\end{align*}$$
Getting $a$:
$$\begin{align*}
\frac{-r\sin(a)}{ r\cos(a)} &= \frac{3}{4}\\
\tan(a) &= -\frac{3}{4}\\
a &= \arctan(-3/4)\\
a &= -36.87, 143.13\\
\end{align*}$$
The questions:-
There are two values for both $r$ and $a$, how should I choose the values to be in the final form?
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The two values of $a$ differ by $180^{\circ}$. Since $\sin(y\pm 180^{\circ}) = -\sin(y)$, you want to pick $r$ and $a$ in such a way that $r\sin(x-a)$ has the same sign as $3\sin x+ 4\cos x$.
When $x=0$, $3\sin x + 4\cos x=4$ is positive; so you want to pick $r$ and $a$ so that $r\sin(-a)$ is positive. This happens when $r=5$ and $a=-36.87^{\circ}$; or when $r=-5$ and $a=143.13$. The two choices work, the other two possibilities do not.
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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.