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$y'=\frac{y^2}{2x(y-x)}$ I'm trying to solve the following differential equation:
$$y'=\frac{y^2}{2x(y-x)}$$
It is supposed to have a relatively easy general solution, but I can't find it.
I've tried several things, the most promising of which is the change $z(x)=y(x)-x$, which yields (if I haven't made any mistake):
$$z'=\frac{z}{2x}+\frac{x}{2z}$$
Which I guess I can solve as a Bernoulli equation, with $p=-1$ (which means solving a linear equation, then changing the variables back).
*
*My question is, is there a simpler way of solving this equation?
|
$$y'=\frac { y^{ 2 } }{ 2x(y-x) } \\$$
Solve respect to the $x $
$$\\ \\ { y }^{ 2 }{ x }^{ \prime }-2x(y-x)=0\\ \\ { x }^{ \prime }-2\frac { x }{ y } +2\frac { { x }^{ 2 } }{ { y }^{ 2 } } =0\\ x=zy\\ { x }^{ \prime }=z+y{ z }^{ \prime }\\ z+y{ z }^{ \prime }-2z+2{ z }^{ 2 }=0\\ y{ z }^{ \prime }-z+2{ z }^{ 2 }=0\\ y{ z }^{ \prime }=z\left( 1-2z \right) \\ \int { \frac { dz }{ z\left( 1-2z \right) } } =\int { \frac { dy }{ y } } \\ \frac { 1 }{ z\left( 1-2z \right) } =\frac { A }{ z } +\frac { B }{ 1-2z } \\ 1=A-2Az+Bz=\left( B-2A \right) z+A\\ B-2A=0,A=1\Rightarrow B=2\\ \int { \left( \frac { 1 }{ z } +\frac { 2 }{ 1-2z } \right) dz=\ln { C\left| y \right| } } \\ \ln { \left| z \right| } -\ln { \left| 1-2z \right| =\ln { C\left| y \right| } } \\ \ln { \left| \frac { z }{ 1-2z } \right| =\ln { C\left| y \right| } } \\ \frac { z }{ 1-2z } =Cy\\
\frac { \frac { x }{ y } }{ 1-2\frac { x }{ y } } Cy\\$$
$$ \frac { x }{ y-2x } =Cy\\ \\ \\ \\ $$ and note $y=0$ is a particular solution
|
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"timestamp": "2023-03-29T00:00:00",
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|
Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $
$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$
I've noted that, since
$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$
we will have the cosine series at the principal diagonal for $\cos(2)$. But couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$.
Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for
$$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?
Thanks!
|
For the general formula I get it to $e^A = I \cosh(\sqrt{\alpha\beta}) + A\sinh(\sqrt{\alpha\beta})/\sqrt{\alpha\beta}$ where the same square root is selected everywhere (the one under $\cosh$ doesn't matter due to symmetry).
You get this by checking the powers of $A$:
*
*$A^0 = I$
*$A^1 = A$
*$A^2 = (\alpha\beta)I$
*$A^3 = (\alpha\beta)A$
and so on. Expanding it results in
$e^A = I/0! + A/1! + (\alpha\beta)I/2! + (\alpha\beta)A/3! + ...$
suppose $\gamma^2 = \alpha\beta$ we get
$e^A = I\gamma^0/0! + (A/\gamma)\gamma^1/1! + I\gamma^2/2! + (A/\gamma)\gamma^3/3!+...$
then just separating the parts you get
$e^A = I\cosh(\gamma) + (A/\gamma)sinh(\gamma)$
|
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|
If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,could not prove it. Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal. How should i prove it?Please help me.
|
For brevity write $a=\cos A$, $b=\cos B$, $c=\cos C$. Then we have
$$C=\pi-A-B$$
and so
$$c=-\cos(A+B)=-ab+\sin A\sin B$$
and so
$$(c+ab)^2=(1-a^2)(1-b^2)\ .$$
Expanding, rearranging and using the identity you are given,
$$a^2+b^2+c^2=1-2abc=ab+bc+ca\ .$$
Now
$$(a-b)^2+(b-c)^2+(c-a)^2=2a^2+2b^2+2c^2-2ab-2bc-2ca$$
and so
$$(a-b)^2+(b-c)^2+(c-a)^2=0\ ,$$
which shows that $a,b,c$ are all equal; hence $A,B,C$ are all equal.
|
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|
simultaneous equation help Can someone help me with to solve this system of equations ?
$$
\left\{
\begin{array}{c}
y=x+1 \\
y^2+2x^2=2
\end{array}
\right.
$$
|
We know from equation one that $y = x+1$. Plugging this value into equation two we get
\begin{align*}
y^2+2x^2 &= (x+1)^2+2x^2 \\
&=x^2+2x+1+2x^2\\
&=3x^2+2x+1 \\
\end{align*}
We know that $3x^2+2x+1 = 2$ from equation two. Simplifying we get $$3x^2+2x-1 = 0$$$$(x+1)(3x-1) = 0 \tag{factoring}$$
$$x = -1, \frac{1}{3}$$Plugging in these values of $x$ into $y = x+1$ we get that $y=0, \dfrac43$ So the solutions are $$\boxed{\left(-1,0\right)\ \& \left(\frac13, \frac43\right)}$$
|
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|
Polar to Rectangular Coordinates Problem:
Transform the following equation from polar to rectangular coordinates.
\begin{eqnarray*}
\rho &=& \frac{2}{1 - \cos \theta} \\
\end{eqnarray*}
Answer:
Recall that:
\begin{eqnarray*}
\rho &=& {(x^2+y^2)} ^ {\frac{1}{2}} \\
\cos \theta &=& \frac{x}{\rho} =\frac{x}{(x^2+y^2)^{\frac{1}{2}}} \\
\end{eqnarray*}
This gives us:
\begin{eqnarray*}
{(x^2+y^2)}^{\frac{1}{2}} &=& \frac{2}{1 - \frac{x}{(x^2+y^2)^{\frac{1}{2}}}} \\
2 &=& {(x^2+y^2)^\frac{1}{2}} - x \\
x + 2 &=& (x^2+y^2)^\frac{1}{2} \\
(x + 2)^2 &=& x^2+y^2 \\
x^2 + 4x + 4 &=& x^2+y^2 \\
4x + 4 &=& y^2 \\
y^2 &=& 4x + 4
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
y^2 &=& 4(x + 2)
\end{eqnarray*}
What am I missing?
|
Let's check with $\theta=60^\circ$ so that $\rho=4$. Then: $x=2$, $y=2\sqrt3$, $y^2=12$. But $4(x+2)=16$ and $4x+4=12$. So you are right and the book is wrong.
|
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|
Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is $8^k-3^k$ is divisible by 5.
Step 3: Prove that (*) is true for $n=k+1$, that is $8^{k+1} - 3^{k+1}$ is divisible by 5. We have
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k$$
Can anyone explain the next logical expansion?
Update:
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k = 5*8^k + 3*8^k - 3*3^k = 5*8^k + 3(8^k - 3^k)$$
Now we can say that $5*8^k$ is divisible by 5 since it has the form $5*p$ where $p$ is an integer $\geq 1$
And it is assumed that $8^k-3^k$ is divisible by $5$, then $3(8^k-3^k)$ is divisible by $5$ which means it has the form $5p$
so we reduce the expression to
$$5p+5p = 5(p+p)$$
which is of the form $5p$, which is divisible by $5$.
Is my proof correct?
|
By the binomial theorem, $8^n=(5+3)^n=5a+3^n$.
Induction is used to prove the binomial theorem.
|
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|
Simple Combination of balls I encounter this problem but I cannot figure out the solution. Please help.
Given that there are 4 white balls, 2 red balls and 1 black balls of the same size. 4 balls are drawn at random. Find the number of ways this can be done.
The given answer is 1 + 4C1 + 4C1 + 4C2 + 2 × 4! ÷ 2! = 39.
Thank in advanced.
|
For the answer where balls of the same color are indistinct and order does matter, look above to Dylan's answer.
Assuming all balls are distinct (even those of the same color) and assuming order does not matter
You can break this into cases:
$\begin{array}{|c|c|c||c|}
\hline
\text{White}&\text{Red}&\text{Black}&\text{Total number of cases}\\
\hline
\hline
4&0&0&\binom{4}{4}\cdot\binom{2}{0}\cdot\binom{1}{0}=1\\
3&1&0&\binom{4}{3}\cdot\binom{2}{1}\cdot\binom{1}{0}=8\\
3&0&1&\binom{4}{3}\cdot\binom{2}{0}\cdot\binom{1}{1}=4\\
2&2&0&\binom{4}{2}\cdot\binom{2}{2}\cdot\binom{1}{0}=6\\
2&1&1&\binom{4}{2}\cdot\binom{2}{1}\cdot\binom{1}{1}=12\\
1&2&1&\binom{4}{1}\cdot\binom{2}{2}\cdot\binom{1}{1}=4\\
\hline
\end{array}$
Adding these together gives the answer. (note: this agrees with the answer $\binom{7}{4}=35$ if we were to simply treat all balls as different and ignore the colors)
Assuming all balls of the same color are indistinct and order does not matter
There will be as many possibilities as there are cases listed above, for a total of 6 cases.
|
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|
Get number of additions that lead to a specific sum with given summands Suppose you have $n$ variables. Each of these variables (e.g. $a$) have their own interval between $0$ and $a_i$ (whole numbers). The only rule is that all these variables have to add up to a given value $s$.
An example:
$n = 3$, so the three variables are $a$, $b$ and $c$
$a_i = 2$, so $0\le$ $a$ $\le2$
$b_i = 3$, so $0\le$ $b$ $\le3$
$c_i = 4$, so $0\le$ $c$ $\le4$
$s = 6$, so $a+b+c=6$
The 9 unique possible solutions for these given variables are the following (the order of the variables is still important, so solution 7 and solution 9 are not the same):
$a=0$ $b=2$ $c=4$
$a=0$ $b=3$ $c=3$
$a=1$ $b=1$ $c=4$
$a=1$ $b=2$ $c=3$
$a=1$ $b=3$ $c=2$
$a=2$ $b=0$ $c=4$
$a=2$ $b=1$ $c=3$
$a=2$ $b=2$ $c=2$
$a=2$ $b=3$ $c=1$
What is the correct formula or program to only get the number of unique, possible solutions for any value of $n$, $s$ and the other variables, which would be 9 in this case?
|
Suppose you have $k$ variables, and you want to add it up to $n$:
$$f_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}\prod_{i=1}^k \frac{1-x^{a_i+1}}{1-x}$$
Then your solution is $f(0)$.
To come up with this, start with the generating functions for your constraints:
$$(1+x+...x^{a_i}) = \frac{1-x^{a_i+1}}{1-x}$$
Suppose we compute part of this product with your example:
$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+x^3+x^4)$$
Then multiply the $x$ from the first by the $x^2$ in the second and the $x^3$ in the last. This gives $x^6$, and corresponds to the $a=1,b=2,c=3$ solution.
Then coefficient of $x^n$ in that product is the number of solutions. I used the derivative and factorial, then evaluation at $x=0$ to get this, but if you're using a CAS it probably has a more efficient coefficient function.
|
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|
Find all real functions $f$ such that $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x,y$ we have:
$f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$
I have solved this problem but the solution is "bruteforce", so I wanted to ask if there is a more elegant way of approaching this.
|
I claim that the only possible solutions to the functional equation
$$f\big(xf(y)\big)=(1-y)f(xy)+x^2y^2f(y)\tag0\label0$$
are $f(x)=0$ and $f(x)=x-x^2$.
To show that, first let $y=1$ in \eqref{0} and you'll get $f\big(xf(1)\big)=x^2f(1)$. Now assuming $f(1)\neq0$ we'll have $f(x)=\frac{x^2}{f(1)}$ for every real number $x$, whcih leads to a contradiction using \eqref{0}. So we have:
$$f(1)=0\tag1\label1$$
Next, let $x=1$ in \eqref{0} and you'll get:
$$f\big(f(y)\big)=\big(1-y+y^2\big)f(y)\tag2\label2$$
Now, substituting $f(y)$ for $y$ in \eqref{0} and using \eqref{2} we get:
$$f\Big(xf\big(f(y)\big)\Big)=\big(1-f(y)\big)f\big(xf(y)\big)+x^2f(y)^2f\big(f(y)\big)\\
\therefore f\Big(x\big(1-y+y^2\big)f(y)\Big)=\big(1-f(y)\big)f\big(xf(y)\big)+x^2f(y)^3\big(1-y+y^2\big)$$
So, if $f(y)\neq0$, letting $x=\frac1{f(y)}$ in the last equation and using \eqref{1}, we get:
$$f\big(1-y+y^2\big)=\big(1-y+y^2\big)f(y)\ne0$$
Thus by \eqref{2}, $f\big(f(y)\big)=f\big(1-y+y^2\big)\ne0$ and so $f\Big(f\big(f(y)\big)\Big)=f\Big(f\big(1-y+y^2\big)\Big)$. Hence by \eqref{2}:
$$\big(1-f(y)+f(y)^2\big)f\big(f(y)\big)=\Big(1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2\Big)f\big(1-y+y^2\big)\\
\therefore1-f(y)+f(y)^2=1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2\\
\therefore\Big(f(y)-\big(1-y+y^2\big)\Big)\Big(f(y)+\big(1-y+y^2\big)-1\Big)=0$$
$$\therefore f(y)=1-y+y^2\quad or\quad f(y)=y-y^2\tag3\label3$$
Now, if $f(y)=1-y+y^2$ then $f\big(1-y+y^2\big)=\big(1-y+y^2\big)^2$, but because $1-y+y^2\neq0$ we must have $f\big(1-y+y^2\big)=1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2$ or $f\big(1-y+y^2\big)=\big(1-y+y^2\big)-\big(1-y+y^2\big)^2$ by \eqref{3}. It's easy to check that the latter case leads to a contradiction and the former case leads to $y=0$ or $y=1$. But letting $x=y=0$ in \eqref{0} we have $f(0)=0$, and by \eqref{1} we have $f(1)=0$. So this case leads to a contradiction, too, since we assumed $f(y)\neq0$. So $f(y)$ cannot be equal to $1-y+y^2$ and hence by \eqref{3}, $f(y)=y-y^2$.
Finally, if there is a real number $y$ such that $y\neq0$, $y\neq1$ and $f(y)=0$, then by \eqref{0} we get $f(0)=(1-y)f(xy)+0$ for every real number $x$, which means that $f$ is the constant zero function. This yields our first claim. It's easy to check that indeed the two mentioned functions satisfy \eqref{0}.
|
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|
Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-2)^2}$ (first of all I don't get why that is?).
But in this exercise, will $\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64}$ be equal to $\frac A {x-2} + \frac B {(x-2)^2} + \frac C {x^2+2x+4} + \frac D {(x^2+2x+4)^2}$?
Thanks in advance.
|
The question is not entirely clear. Your expression can be also writen as
$$
\frac{3 x}{\left(x^2+2 x+4\right)^2}+\frac{x+2}{x^2+2 x+4}+\frac{1}{(x-2)^2}
$$
hope this helps
|
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|
Probability that the equation $x^2 + k_1 x + k_0 = 0$ has real solutions
$k_1$, $k_0$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $x^2 + k_1 x + k_0 = 0$ has real solutions?
This is a subproblem of another problem, and I do not know how to approach it without brute force, hope some of you will propose a fresh idea.
The answer should be devised without using computer, if possible.
|
It has real solutions iff $\Delta=k_1^2-4k_0\ge 0$.
$$P=\frac{\sum_{i=1}^{20}\lfloor\frac{i^2}{4}\rfloor+80\cdot 100}{10000}$$
$(2k+1)^2\equiv 1\pmod{4},\, (2k)^2\equiv 0\pmod{4}$. Therefore (see here and here):
$$P=\frac{\frac{1^2+3^2+5^2+\cdots+19^2-10}{4}+\frac{2^2+4^2+6^2+\cdots+20^2}{4}+8000}{10000}$$
$$=\frac{\frac{10(19)(21)}{3}-10+4\cdot \frac{10(11)(21)}{6}+32000}{40000}=0.8715$$
|
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|
Find solution to system of differential equations with initial conditions I have a system
$$
\begin{cases}
x_1'(t) = -x_2(t) \\
x_2'(t) = -x_1(t)
\end{cases}
$$
I have found the linearly independent solutions
$$
\begin{split}
\vec{x}(t) &= e^{\lambda_1 t} \vec{v}_1 = e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}, \\
\vec{x}(t) &= e^{\lambda_2 t} \vec{v}_2 = e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}.
\end{split}
$$
and the general solution
$$
\vec{x}(t) = c_1 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}
$$
Is it correct that both all the linearly independent solutions and the general solution are all named $\vec{x}(t)$ or should the linearly independent solutions be named $\vec{x}_1(t)$ and $\vec{x}_2(t)$?
Now I have to find the solution to which $x_1(0) = -1$ and $x_2(0) = 1$. How can I do this? I know I have to choose the correct values of $c_1$ and $c_2$ in order to fulfil the conditions. But I don't know where to substitute $x_1(0) = -1$ and $x_2(0) = 1$.
In my system, I have $x_1'(t)$ and $x_2'(t)$ but these are not vector function although I only have found vector functions as solutions. Is it because my general solution $\vec{x}(t)$ consists of the functions $x_1(t) = -c_1 e^t + c_2 e^{-t}$ and $x_2(t) = c_1 e^t + c_2 e^{-t}$, so I just have to find $c_1$ and $c_2$ such that $x_1(0) = -c_1 e^0 + c_2 e^0 = -c_1 + c_2 = -1$ and $x_2(0) = c_1 e^0 + c_2 e^0 = c_1 + c_2 = 1$, resulting in $c_1 = 1$ and $c_2 = 0$?
|
Apart from some confusing notation with indices $(1,2)$ - better replace that with e.g. $(a,b)$ - you've got everything OK. About the initial conditions:
$$
\vec{x}(t) = c_1 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}
\quad \Longrightarrow \quad
\vec{x}(0) = c_1 \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 \begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}-1 & 1\\ 1 &1 \end{pmatrix}\begin{pmatrix}c_1 \\ c_2\end{pmatrix}
$$ $$
\begin{pmatrix}-1 & 1\\ 1 &1 \end{pmatrix}\begin{pmatrix}c_1 \\ c_2\end{pmatrix}
= \begin{pmatrix} x_1(0) \\ x_2(0) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}
\quad \Longrightarrow \quad
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} =
\begin{pmatrix}-1 & 1\\ 1 & 1 \end{pmatrix}^{-1} \begin{pmatrix} -1 \\ 1 \end{pmatrix}
$$
Inverse of an almost orthogonal matrix is almost the transpose, so:
$$
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} = \frac{1}{2}
\begin{pmatrix}-1 & 1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix}
\quad \Longrightarrow \quad
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
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How do you prove $37^{100} - 37^{20}$ is a multiple of $10$ using induction? I've tried making it $37^{5n} - 37^n$ is a multiple of $10.$ Then I made the base case be $n = 0,$ so $1 - 1 = 0$ which is a multiple of $10.$ I assumed $37^{5n} - 37^n$ is a multiple of $10$ for all $n \geqslant0.$ The inductive step is really hard however. I did $37^{5(n + 1)} - 37^{n+1}$ which I simplified to $37^{5n + 5} - 37^{n+1}.$ I tried factoring and made it $37^5\cdot37^{5n} - 37\cdot(37^n).$ I don't know where to go from here though.
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I know OP requested induction, but I would like to post this anyway on the off-chance that someone finds it interesting:
This is the same as proving that $10 {\large\mid} 37^{80}-1$, since $37^{100}-37^{20} = 37^{20}(37^{80}-1)$ and $37^{20}$ has only powers of $37$ as divisors since $37$ is prime.
Clearly $2{\large\mid}37^{80}-1$ since $37^{80}-1$ is even, so we must only show that $5{\large\mid}37^{80}-1$. Observe that $37 \equiv 2 \pmod{5}$, therefore $37^{80} \equiv 2^{80} \pmod{5}$. Thus we have $2^{80} \equiv (2^{10})^8 \equiv (1024)^8 \equiv (4)^8 \equiv (4^2)^4 \equiv 1^4 \equiv 1 \pmod{5}$.
Thus $37^{80} \equiv 1 \pmod{5}$, and so $37^{80} - 1 \equiv 0 \pmod{5}$, as desired.
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
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Since :$$\left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) =x-1\\ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) =x-1$$
so we have
$$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt [ 3 ]{ x } -1 }{ 2\sqrt { x } -2 } \right) = } \frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) \left( \sqrt { x } +1 \right) }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) } = } \\ =\frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( x-1 \right) \left( \sqrt { x } +1 \right) }{ \left( x-1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) } } =\frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } =\frac { 1 }{ 3 } $$
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Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
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A little deus ex machina but this seems to be related to the golden ratio $\phi = \frac{1+\sqrt 5}{2}$, which is the solution of $x+\frac1x = 1$, and to the Fibonacci sequence $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657, \ldots$
In particular $x^3+\frac1{x^3} = 18$ has the solutions $x^3=5+8\phi$ and its reciprocal $-8\phi+13$, which requires $x=1+\phi$ or its reciprocal $-\phi+2$, and so $x^{11}=10946+17711\phi$ or its reciprocal $-17711\phi + 28657$ which gives $$x^{11} + \frac{1}{x^{11}} = 10946+17711\phi -17711\phi + 28657 = 39603.$$
|
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|
Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?
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Notice, $$x^4-2x^3-7x^2+8x+12=0$$ substituting $x=-1$ we get $$(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12=0\iff 0=0$$ Hence, $x=-1$ is a root of given quartic equation i.e. $(x+1)$ is a factor of $x^4-2x^3-7x^2+8x+12$, now we have
$$(x+1)(x^3-3x^2-4x+12)=0$$
further factorizing $x^3-3x^2-4x+12=(x-3)(x^2-4)$
$$(x+1)(x-3)(x^2-4)=0$$
$$(x+1)(x-3)(x-2)(x+2)=0$$
Hence, the roots are $$x=\color{red}{-2, -1, 2, 3}$$
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|
Summation of simple sequence This is a fairly simple question, I'm sure, but I appear to be having trouble. What is the result of the following sequence:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ .... + \frac{n}{2^{n}}.$$ ?
Thanks
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$$2S_n-S_n=\left(\frac11+\frac{2}{2}+\frac{3}{4}+\cdots+ \frac{n}{2^{n-1}}\right)-\left(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ \cdots+ \frac{n}{2^{n}}\right)\\
=1+\frac{1}{2}+\frac{1}{4}+ \cdots+ \frac{1}{2^{n-1}}-\frac n{2^n}=2-\frac{n+2}{2^n}.$$
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Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any ideas?
Thanks a bunch in advance.
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Suppose $7+5\sqrt{2}=(x+y\sqrt{2})^3 = (x^3+6 x y^2)+(3 x^2 y +2 y^3)\sqrt2 $.
If we're looking for integer solultions, then we must have $7=x^3+6 x y^2$ and $5=3 x^2 y +2 y^3$, because $\sqrt2$ is irrational.
Consider $5=3 x^2 y +2 y^3=(3x^2+2y^2)y$. Since $5$ is prime, $y$ must be $1$ or $5$ because $3x^2+2y^2 \ge 0$. Now $y$ cannot be $5$ because $3x^2+2y^2$ could not be $1$. Thus, $y=1$ and $3x^2+2y^2=5$, which gives $x=\pm 1$.
It is easy to verify that $7+5\sqrt{2}=(1+\sqrt{2})^3$.
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Proof that $3\mid n^3 − 4n$ Prove that $n^3 − 4n$ is divisible by $3$ for every positive integer $n$.
I am not sure how to start this problem. Any help would be appreciated
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Well, one can start off by factoring $n^3 - 4n$:
$n^3 - 4n = n(n^2 - 4)$
$= n(n + 2)(n - 2); \tag{1}$
next we note that every $n \in \Bbb Z$ is of one of the three forms
$n = 3q, \tag{2}$
$n = 3q + 1, \tag{3}$
or
$n = 3q + 2; \tag{4}$
this follows from the basic properties of Euclidean division applied to $n$ with $3$ as the divisor; the only possible remainders $r$ be $0$, $1$, and $2$, since such integers $r$ must satisfy $0 \le r \le 2$. In case (2), clearly $3 \mid n$; in case (3), we have
$n + 2 = 3q + 1 + 2$
$= 3q + 3 = 3(q + 1), \tag{5}$
showing that $3 \mid n + 2$; finally, case (4) yields
$n - 2$
$= 3q + 2 - 2 = 3q, \tag{6}$
whence $3 \mid n - 2$. For evey possible $n$, $3$ divides one of the factors of $n^3 - 4n$ occurring in (1); thus we have $3 \mid n^3 - 4n$ for all $n \in \Bbb Z$. QED.
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integrate $\int\sqrt{x^2-1}dx$ unproven step In order to find $\int\sqrt{x^2-1}dx$ one makes substitutions $x=\sec(\theta)$ , $dx=\sec(\theta)\tan(\theta)d\theta$ and $\sqrt{x^2-1}$ = $\sqrt{\tan^2(\theta)}$. Then you find $\int{\tan^2(\theta) \sec(\theta)}d\theta$ directly as easy as $1+1=2$ it seems from many online sources and even WolframAlpha. But I tried for hours but every way I try, $\sqrt{\tan^2(\theta)}\neq\tan(\theta)$. I already taken into account $\theta$ is a substitution for $\sec^{-1}(x)$, does not make a difference. What do I think wrong? (edit: and also $\sqrt{\tan^2(\theta)} * \tan(\theta)$ != $\tan^2(\theta)$ )
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$\int\sqrt (x^2-1)dx$
(1)Use a substitution of $x=cosh \theta$, $dx=sinh\theta$
=$\int\sqrt(cosh^2-1).(sinh\theta)d\theta$
=$\int\sqrt sinh^2\theta.(sinh\theta)d\theta$
=$\int\sinh^2\theta d\theta$
Using double angle formulae: $cosh 2\theta=1+2sinh^2\theta$
Hence, $sinh^2\theta=\frac{1}{2}(cosh 2\theta-1)$
=$\frac{1}{2}\int (cosh2\theta-1)d\theta$
=$\frac{1}{2}[\frac{sinh 2\theta}{2}-\theta]$
Need to otain final answer in terms of $x$,
so from (1), substitute $cosh\theta=x$
in $cosh^2\theta-sinh^2\theta=1$. Hence $sinh\theta=\sqrt(x^2-1)$
Therefore $sinh 2\theta=2cosh\theta sinh\theta=2x\sqrt(x^2-1)$
$\frac{1}{2}[\frac{2x\sqrt(x^2-1)}{2}-cosh^{-1}x]$+C
$\frac{x\sqrt(x^2-1)}{2}-\frac{cosh^{-1}x}{2}$+C
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Finding $\lim_{x\to 1 }\left( \frac{x-1-x\ln x }{x(x-1)^{2}}\right)$ without L'Hopital's rule I would like to calculate this limit but without using L'Hopital's rule
$$\lim_{x\to 1 }\left( \frac{x-1-x\ln(x) }{x(x-1)^{2}}\right)=-\dfrac{1}{2}$$
My thoughts:
Note that $\lim\limits_{x\to 1 }\dfrac{\ln(x)}{x-1}=1$ so I tried :
\begin{align*}
\lim_{x\to 1 }\left( \dfrac{x-1-x\ln(x) }{x(x-1)^{2}}\right)&=\lim_{x\to 1 }\left( \dfrac{1}{x(x-1)}-\dfrac{\ln(x)}{(x-1)^2}\right) \\
&=\lim_{x\to 1 }\left( \dfrac{1}{x(x-1)}\right)-\lim_{x\to 1 }\left(\dfrac{1}{(x-1)}\cdot\dfrac{\ln(x)}{(x-1)}\right)
\end{align*}
but with no luck
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Let us first make $x=y+1$. So, $$\dfrac{x-1-x\ln(x) }{x(x-1)^{2}}=\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$ Now, consider Taylor expansion around $y=0$ $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ So, the numerator write $$y-(y+1) \log (y+1)=-\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$
I am sure that you can take from here.
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how can I prove it is a convergent sequence and its limit is $\sqrt{a}$ Let we have the following sequence $$x_1=1$$
$$x_{n+1}=\frac{x_n^2+a}{2x_n}$$ such that $$a>0$$ how can I prove it is a convergent sequence and its limit is $\sqrt{a}$
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Calculate $ x_{n+1} - \sqrt{a} $,
$$ x_{n+1} - \sqrt{a} = \frac{(x_n - \sqrt{a})^2}{2x_n},$$
and
$$ x_2 = \frac{a + 1}{2} \geqslant \sqrt{a}.$$
Thus by induction we can prove that $\forall n \geqslant 2,\,\, x_n \geqslant \sqrt{a}$.
Now that
$$ x_{n+1} - \sqrt{a} = \frac{x_n - \sqrt{a}}{2x_n} \cdot (x_n - \sqrt{a}) = \frac12\left(1 - \frac{\sqrt{a}}{x_n}\right) \cdot (x_n - \sqrt{a})$$
and we know that
$$ 0 < \frac{\sqrt{a}}{x_n} \leqslant 1.$$
So
$$ 0 \leqslant \frac12\left(1 - \frac{\sqrt{a}}{x_n}\right) < \frac12$$
and the sequence $\{x_n - \sqrt{a}\}$ converges to $0$ because
$$ \forall n > 2, \,\, 0 \leqslant x_n - \sqrt{a} < \frac{1}{2^{n-2}} (x_2 - \sqrt{a}).$$
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Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating the radical on one side...
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Note that $\sqrt{c+1}\geq \sqrt{c-1}$ and so $\frac{1}{\sqrt{c}+\sqrt{c-1}}\geq \frac{1}{\sqrt{c}+\sqrt{c+1}}$. Write $\sqrt{c}-\sqrt{c-1}=\frac{1}{\sqrt{c}+\sqrt{c-1}}$ and $\sqrt{c+1}-\sqrt{c}=\frac{1}{\sqrt{c}+\sqrt{c+1}}.$ Thus you have the result.
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solving ODE : $xy'=\sin(x+y)$ May I ask some hints or solution for solving $xy'=\sin(x+y)$ ??
My idea was substitution : $x+y=u$ , then it becomes $x(u'-1)=\sin u$
and still I can't approach further..
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Let $u=x+y$ ,
Then $y=u-x$
$y'=u'-1$
$\therefore x(u'-1)=\sin u$
$x\dfrac{du}{dx}-x=\sin u$
$x\dfrac{du}{dx}=\sin u+x$
$\dfrac{du}{dx}=\dfrac{\sin u}{x}+1$
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=223:
Let $v=\tan\dfrac{u}{2}$ ,
Then $\dfrac{dv}{dx}=\dfrac{v^2}{2}+\dfrac{v}{x}+\dfrac{1}{2}$
Let $v=-\dfrac{2}{w}\dfrac{dw}{dx}$ ,
Then $\dfrac{dv}{dx}=-\dfrac{2}{w}\dfrac{d^2w}{dx^2}+\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2$
$\therefore-\dfrac{2}{w}\dfrac{d^2w}{dx^2}+\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2=\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2-\dfrac{2}{xw}\dfrac{dw}{dx}+\dfrac{1}{2}$
$\dfrac{2}{w}\dfrac{d^2w}{dx^2}-\dfrac{2}{xw}\dfrac{dw}{dx}+\dfrac{1}{2}=0$
$4x\dfrac{d^2w}{dx^2}-4\dfrac{dw}{dx}+xw=0$
$w=C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)$ (according to http://www.wolframalpha.com/input/?i=4xw''-4w'%2Bxw%3D0)
$\therefore v=-\dfrac{2\dfrac{d}{dx}\left(C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)\right)}{C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)}=-\dfrac{C_1xJ_0\left(\dfrac{x}{2}\right)-C_1xJ_2\left(\dfrac{x}{2}\right)+4C_1J_1\left(\dfrac{x}{2}\right)+C_2xY_0\left(\dfrac{x}{2}\right)-C_2xY_2\left(\dfrac{x}{2}\right)+4C_2Y_1\left(\dfrac{x}{2}\right)}{2C_1xJ_1\left(\dfrac{x}{2}\right)+2C_2xY_1\left(\dfrac{x}{2}\right)}=\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}$
$u=2\tan^{-1}\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}$
$y=2\tan^{-1}\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}-x$
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Maths Challenge IMO I saw this problem on a maths challenge book.
Given any set $A=\{a_1 ,a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $S_A$. Let $n_A$ denote the number of pairs $(i,j)$ with $1\leq i\leq j\leq 4$ for which $a_i +a_j$ divides $S_A$. Find all sets of four distinct positive integers which achieve the largest value of $n_A$.
I was thinking that $4\choose 2$ is $6$. So there should be six possible ways to pair all the elements of set $A$ that will divide $S_A$ provided that $a_i + a_i$ is not allowed. Well this is only my first thought, other ideas will be appreciated .
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WLOG $a_1<a_2<a_3<a_4$. For every permutation $a,b,c,d$ of $a_1,a_2,a_3,a_4$ we have $$a+b\mid (a+b+c+d)-(a+b)=c+d\implies \boxed{a+b\mid c+d}$$
Since $a_3+a_4>a_1+a_2$ and $a_2+a_4>a_1+a_3$ we can not have $a_3+a_4\mid a_1+a_2$ and $a_2+a_4\mid a_1+a_3$, so $n_A\leq 4$.
Since $a_1+a_4\mid a_2+a_3\;\;(*)$ only iff $$ka_4<k(a_1+a_4)\leq a_2+a_3 <2a_4$$
we get $k=1$, so $(*)$ is possible only if $\boxed{a_1+a_4= a_2+a_3}$
Now $$a_1+a_3\mid a_2+a_4 \implies a_1+a_3\mid (2a_2+a_3-a_1)-(a_1+a_3)=2(a_2-a_1)$$
So $$la_3<l(a_1+a_3) = 2(a_2-a_1) <2a_2 \implies l=1$$
so we have now $\boxed{ a_3=2a_2-3a_1}$ and $\boxed{ a_4=3a_2-4a_1}$.
Let $x=a_1$ and $y=a_2$. Finaly we have
$$a_1+a_2\mid a_3+a_4 \implies a_1+a_2\mid (5a_2-7a_1)$$ and $$a_1+a_2\mid 5(a_1+a_2)-(5a_2-7a_1) =12a_1 \implies 12x = m(x+y)$$ for some natural number $1\leq m<6$, since $x<y$. So $$\boxed{(12-m)x=my}$$
*
*For $m=1$ we get $y=11x$ and so $a_3=19x$ and $a_4= 29x$ and $x$ can be arbitrary positive integer.
*For $m=2$ we get $y=5x$ and so $a_3=7x$ and $a_4= 11x$ and $x$ can be arbitrary positive integer.
*For $m=3$ we get $y=3x$ and so $a_3=3x=a_2$ and so it is not OK.
*For $m=4$ we get $y=2x$ and so $a_3=x=a_1$ and so it is not OK.
*For $m=5$ we get $5y=7x$ and so $a_1=5t$, $a_2=7t$, $a_3=-t$ and so it is not OK.
So $$(a_1,a_2,a_3,a_4)\in \{(x,11x,19x,29x),(x,5x,7x,11x);\;x\in\mathbb{N}\}$$
|
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|
Show that $p^3+4$ is prime If $p$ and $p^2 +8$ are both prime number, prove that $p^3 +4$ is also prime.
I was thinking using two cases for even and odd. So $p$ is even thus there is no prime, the statement does not hold. So, $p$ is odd, $p$ has to be of the form of $2k+1$.
let $p=2k+1$ which is prime
$(2k+1)^2 + 8 = 4k^2+4k+1+8 = 2(2k^2+2k+4)+1$ which is prime
so $(2k+1)^3 +4 = 8k^3+12k^2+6k+1 +4= 2(4k^3+6k^2+3k+2)+1$ which is also prime.
is that all that i need to show
|
Hint: if you don't know the theory of quadratic residue, you can note that every prime $p>3$ can be written in the form $6n\pm1$. It has remainder $\pm1$ when divided by $3$. Note therefore that for $p>3$, $p^2+8$ is divisible by $3$.
Indeed $$p^2+8=(6n\pm1)^2+8\equiv (\pm1)^2\!+8\equiv 0\pmod{\!3}$$
|
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|
A given polynomial equation of 5 degree has three equal roots . Given equation is $x^5-10a^3x^2+b^4x+c^5=0$ which has 3 equal roots. What I know is that since its a 5th degree equation therefore it must have 5 roots out (of which 3 are equal).
Aim is to establish the relationship between the constants $a,b$ and $c$.
Options given are:
1) $6a^5+c^5=0$.
2) $b^4=15a^4.$
I have to find which one is correct out of the two, any help?
|
Let $$f(x) = x^5 - 10a^3x^2 + b^4x + c^5 = (x - m)^3g(x)$$ where $m$ is the repeated root, and $g(x)$ is some second-order polynomial.
Then, differentiating and substituting $x = m$,
$$5x^4 - 20a^3x + b^4 = (x - m)^3g'(x) + 3(x - m)^2g(x)$$
$$5m^4 - 20a^3m + b^4 = 0$$
Differentiating another time, and similarly substituting $x = m$,
$$20x^3 - 20a^3 = (x - m)^3g''(x) + 3(x - m)^2g'(x) + 3(x - m)^2g'(x) + 6(x - m)g(x)$$
$$20m^3 - 20a^3 = 0$$
$$m = a$$
Substituting this back into our first result,
$$5a^4 - 20a^3\cdot a + b^4 = 0$$
$$b^4 = 15a^4$$
Note: The fact that $(x - m)$ appears in every term of $f'(x)$ and $f''(x)$ is in fact a usable basis for proving what Gerry Myerson mentioned in the comments - that $f(x), f'(x), f''(x)$ all have a common factor.
|
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|
Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$?
$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$
By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?
|
Taking the derivative of
$$
\sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1}
$$
and multiplying by $x$ yields
$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2}
$$
Therefore,
$$
\hspace{-1cm}\begin{align}
\frac83\sum_{n=1}^\infty\frac{n}{2^{2n}}\zeta(2n)
&=\frac83\sum_{n=1}^\infty\frac{n}{4^n}\sum_{k=1}^\infty\frac1{k^{2n}}\tag{3a}\\
&=\frac83\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{n}{\left(4k^2\right)^n}\tag{3b}\\
&=\frac83\sum_{k=1}^\infty\frac{4k^2}{\left(4k^2-1\right)^2}\tag{3c}\\
&=\frac23\sum_{k=1}^\infty\left[\frac1{(2k-1)^2}+\frac1{(2k+1)^2}+\frac1{2k-1}-\frac1{2k+1}\right]\tag{3d}\\
&=\frac23\sum_{k=1}^\infty\left[\frac2{(2k-1)^2}-\left(\frac1{(2k-1)^2}-\frac1{(2k+1)^2}\right)+\left(\frac1{2k-1}-\frac1{2k+1}\right)\right]\tag{3e}\\
&=\frac23\sum_{k=1}^\infty\frac2{(2k-1)^2}\tag{3f}\\[6pt]
&=\zeta(2)\tag{3g}
\end{align}
$$
Explanation:
$\text{(3a)}$: expand $\zeta(2n)$
$\text{(3b)}$: change the order of summation
$\text{(3c)}$: apply $(2)$
$\text{(3d)}$: partial fractions
$\text{(3e)}$: unify the sum of the odd squares by introducing a telescoping series
$\text{(3f)}$: the telescoping series cancel: $-1+1=0$
$\text{(3g)}$: the sum of the reciprocals of the odd squares is $\frac34$ of the sum of the reciprocals of all the squares
|
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|
About Integration $\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx$ What i want to prove is following integral
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx=\frac{1}{2} \sqrt{\left(1+\sqrt{2}\right) \pi }
\end{align}
can you give some explicit method to obtain this result?
|
So let us follow through on Eclipse Sun's suggestion. In doing so we will find that while the double integral can indeed be separated into two parts (two double integrals), neither of them turns out to be equal to zero.
Let
$$I = \int^\infty_{-\infty} e^{-x^2} \cos (x^2) \, dx = 2 \int^\infty_0 e^{-x^2} \cos (x^2) \, dx.$$
on squaring the integral we have
\begin{align*}
I^2 &= 4 \left(\int^\infty_0 e^{-x^2} \cos(x^2) \, dx\right)\times\left(\int_{0}^{\infty} e^{-y^2} \cos(y^2) \, dy\right)\\
&= 4 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2) \cos (y^2) \, dx dy.
\end{align*}
After making use of $2\cos x\cos y= \cos(x-y)+\cos(x+y)$ we have
$$I^2 = 2 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2 + y^2) \, dx dy + 2 \int^\infty_0 \int^\infty_0 e^{-(x^2 + y^2)} \cos (x^2 - y^2) \, dx dy.$$
Converting to polar coordinates $(x,y) \mapsto (r \cos \theta, r \sin \theta)$ we have
$$I^2 = 2 \int^{\frac{\pi}{2}}_0 \int^\infty_0 r e^{-r^2} \cos (r^2) \, dr d\theta + 2 \int^{\frac{\pi}{2}}_0 \int^\infty_0 r e^{-r^2} \cos (r^2 \cos 2\theta ) \, dr d\theta = 2 I_1 + 2 I_2.$$
For the first of these integrals, the $r$-integral can be found by preforming integration by parts twice before performing the $\theta$-integration. The result is $I_1 = \pi/8$.
For the second of these integrals, finding the $r$-integral first by performing integration by parts twice, leads to
$$I_2 = \frac{1}{2} \int^{\frac{\pi}{2}}_0 \frac{d\theta}{1 + \cos^2 (2\theta)} = \frac{\pi}{4\sqrt{2}}.$$
Thus
$$I^2 = 2 \left [\frac{\pi}{4} + \frac{\pi}{2\sqrt{2}} \right ] = \frac{\pi}{4} (1 + \sqrt{2}),$$
or
$$I = \frac{\sqrt{\pi}}{2} \sqrt{1 + \sqrt{2}},$$
as required.
|
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|
What is the following limit? $\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$ What is the following limit?
$$\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$$ Should I calculate the exact value of $(1+x)^5$?
|
Using $$\displaystyle (1+x)^5 = \binom{5}{0}+\binom{5}{1}x+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5$$
and Using $$\displaystyle \binom{n}{r} = \frac{n!}{r!\times (n-r)!}\;,$$ Where $n!=n\times (n-1)\times (n-2)\times ...2\times 1$
So we get $$\displaystyle (1+x)^5 = 1+5x+10x^2+10x^3+5x^4+x^5$$
So $$\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^5-1-5x}{x^2+x^5} = \lim_{x\rightarrow 0}\frac{10x^2+10x^3+5x^4+x^5}{x^2+x^5} = \lim_{x\rightarrow 0}\frac{10+10x+5x^2+x^3}{1+x^3} =10$$
|
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|
Power series expansion of ODE So I have this equation:
$$(1-x^2)y''-2xy'+\lambda y=0$$
Where it is given that the solution can be written as the following convergent power series:
$$y(x) =\sum_{n=0}^\infty a_n x^n$$
And my goal is to arrive to this relationship of the coefficients:
$$a_{n+2} = \frac{n(n+1)-\lambda}{(n+1)(n+2)}a_n$$
My attempt:
$$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$$
$$y''=\sum_{n=0}^\infty (n+2)a_{n+2}x^n$$
Substituting in the above equation and considering the coefficients of $x^n$ we get the following equation:
$$ 0 = (n+2)a_{n+2} - na_n -2na_n+\lambda a_n$$
Which does not seem to match the above relationship. What am I missing?
|
$y′=\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n$
$y′'=\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n$
$\lambda y = 2xy'-(1-x^2)y''$
$\lambda y = 2x[\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n]-(1-x^2)[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n]$
$\lambda y = [\sum_{n=-1}^{\infty}2a_{n+1}(n+1)x^{n+1}]-[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n]+[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^{n+2}]$
$\lambda y = \sum_{n=0}^{\infty}[2a_{n}n-a_{n+2}(n+2)(n+1)+a_{n}n(n-1)]x^n$
$\lambda a_{n} = a_{n}n(n+1)-a_{n+2}(n+2)(n+1)$
$[\lambda-n(n+1)] a_{n} = -a_{n+2}(n+2)(n+1)$
$a_{n+2} = \frac{n(n+1)-\lambda}{(n+2)(n+1)}a_{n}$
|
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|
Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
|
Let $a_n = x^n + \frac{1}{x^n}$. Then $x^2 = 3x - 1$ implies $a_{n+2} = 3a_{n+1} -a_n$ for all $n$ and also $a_1=3$. Since $a_0=2$, we get
$a_2 = 3a_1 -a_0 = 7$
$a_3 = 3a_2 -a_1 = 18$
$a_4 = 3a_3 -a_2 = 47$
|
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|
Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$
Evaluate the indefinite integral $$\int \frac{x-1}{9x^2-18x+17} \, dx .$$
This is an exercise from a book I'm studying. It gives the answer as:
$$\ln(9x^2 -18x+17)^\frac{1}{18} +C .$$
This is an easy integral. You just notice that the numerator is the derivative of the denominator.
But I didn't notice exactly that at first, so I solved it in a slightly different way. I did:
$$\int \frac{x-1}{9x^2-18x+17}\, dx=\frac{1}{18}\int \frac{x-1}{\frac{x^2}{2}-x+\frac{17}{18}}\, dx$$ so $$z=\frac{x^2}{2}-x+\frac{17}{18}, \quad \frac{dz}{dx}=x-1 $$
I find the answer in the usual way as: $$\frac{1}{18} \int\frac{1}{z} dz= \frac{1}{18}\ln\left(\frac{x^2}{2}-x+\frac{17}{18}\right)+C , $$
which is (I believe) different from the answer the book gives, because the arguments of $ln$ are different. What is the problem here? Where am I wrong?
|
Despite appearances to the contrary, there's no problem here: The two expressions actually differ by a constant, but this equivalence is buried in a few special identities involving the logarithm function, namely $$\log (ab) = \log a + \log b$$ and $\log (a^b) = b \log a$ (both for appropriate $a, b$).
More specifically, note that
\begin{align}
\frac{1}{18} \log \left(\frac{x^2}{2} - x + \frac{17}{18}\right) + \color{#0000ff}{\frac{1}{18} \log 18}
&= \frac{1}{18}\log\left[\left(\frac{x^2}{2} - x + \frac{17}{18}\right) (18)\right] \\
&= \frac{1}{18} \log\left(9x^2 - x + \frac{17}{18}\right) \\
&= \log\left[\left(9x^2 - x + \frac{17}{18}\right)^{\frac{1}{18}}\right] .
\end{align}
By absorbing the constant $\color{#0000ff}{\frac{1}{18} \log 18}$ into $C$, we see that your general antiderivative and the one the text gives, regarded as families of functions (all equal up to addition of a constant), are actually the same.
|
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|
Proving a property of a prime number (Elementary number theory) Prove the following statement.
For all prime numbers $a$, $b$, and $c$, $a
^2 + b^2
\neq c^2$
.
What i tried
Proving by contradiction
Assume the negation of the statement
There exists prime numbers $a$, $b$, and $c$ such that $a
^2 + b^2
= c^2$
Since we know that all prime number except $2$ is also an odd number
Then we let $a=2n+1$ $b=2l+1$
Then we have
$4l^{2}+4l+1+4k^{2}+4k+1=c^{2}$
$4l^{2}+4l+4k^{2}+4k+2=c^{2}$
$2(2l^{2}+2l+2k^{2}+2k+1)=c^{2}$
From this expression,since $c^{2}$ is even then $c$ is even. But we have a case where $c=2$, so if $c=2$, then $c^{2}=4$ and
$2l^{2}+2l+2k^{2}+2k+1=2$. Then
$2l^{2}+2l+2k^{2}+2k=1$ and
$l^{2}+l+k^{2}+k=0.5$
Which is not possible since the sum of integers have to also be an integer
Which contradicts with the face that $c$ is either odd or has a value of $2$ for
it to be a prime number
Is my proof correct.Could anyone explain. Thanks
|
A more elementary approach: since the only quadratic residues $\pmod{3}$ are $0$ and $1$, if $a^2+b^2=c^2$ at least one number between $a$ and $b$ has to be a multiple of three. With a similar argument $\pmod{4}$ and $\pmod{5}$, we have that $a^2+b^2=c^2$ implies that at least one number between $a$ and $b$ is even, and at least one number among $a,b,c$ is a multiple of five. It follows that no triple of primes is a pythagorean triple.
|
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|
What do I do next when trying to find the derivative of this fraction? I'm trying to find the derivative of this equation: $-\frac{3(x-6)}{2\sqrt{9-x}}$
The quotient rule: $\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ where $g(x)$ and $f(x)$ are functions.
So I take out the constants and I'm left with $\frac{-3}{2}\frac{d}{dx}[\frac{(x-6)}{\sqrt{9-x}}]$.
$f(x)$ is $(x-6)$ and $g(x)$ is $(9-x)^{1/2}$. This is what it expands to:
$$\frac{-3}{2} \left[\frac{(9-x)^{1/2}\frac{d}{dx}[x-6]-(x-6)\frac{d}{dx}[(9-x)^{1/2}]}{((9-x)^{1/2})^2}\right]$$
After simplifying, I get:
$$\frac{-3}{2}\left[\frac{(9-x)^{1/2}(1)-(x-6)[(\frac{1}{2})(9-x)^{-1/2}(-1)]}{9-x}\right]$$
$$\frac{-3}{2}\left[\frac{\sqrt{9-x}-(x-6)(-1)}{2(9-x)\sqrt{9-x}}\right]$$
$$\frac{-3}{2}\left[\frac{\sqrt{9-x}+x-6}{2(9-x)\sqrt{9-x}}\right]$$
$$\frac{-3\sqrt{9-x}-3x+18}{4(9-x)\sqrt{9-x}} \text{ ??}$$
What did I do wrong? Wolfram Alpha says the answer is: $\frac{3(x-12)}{4(9-x)^{3/2}}$
|
Until here: $$\frac{-3}{2}[\frac{(9-x)^{1/2}(1)-(x-6)[(\frac{1}{2})(9-x)^{-1/2}(-1)]}{9-x}]$$ you're good. This simplifies to: $$-\frac{3}{2}\left[\frac{\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}}{9-x}\right] = -\frac{3}{2}\left[\frac{\color{red}{(9-x)}+ \frac{1}{2}(x-6)}{(9-x)^{3/2}}\right] = -\frac{3}{4} \frac{18-2x+x-6}{(9-x)^{3/2}},$$which is readily seen to agree with Wolfram's answer. Where I pointed in red, you kept $\color{blue}{\sqrt{9-x}}$.
More detailed:
$$\begin{align}\frac{\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}}{9-x} \cdot\frac{\color{red}{\sqrt{9-x}}}{\color{blue}{\sqrt{9-x}}} &= \frac{\left(\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}\right)\color{red}{\sqrt{9-x}}}{(9-x)\color{blue}{\sqrt{9-x}}} \\ &= \frac{\sqrt{9-x}\color{red}{\sqrt{9-x}}+\frac{1}{2}(x-6)\require{cancel}\cancel{\frac{1}{\sqrt{9-x}}}\cancel{\color{red}{\sqrt{9-x}}}}{(9-x)^{3/2}} \\ &= \frac{9-x+\frac{1}{2}(x-6)}{(9-x)^{3/2}}\end{align}$$
|
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|
$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
|
This is a not so elegant approach.
Let $a_n=x^n+\frac1{x^n}$ then
$$a_n\left(x+\frac1x\right)=a_{n+1}+a_{n-1}$$ or $$a_{n+1}=a_{n}-a_{n-1}$$
We have $a_1=1$ and $a_2=-1$, hence solving recursively - plugging - we obtain \begin{align}
a_{6n-4}=a_{6n-2}&=-1\\
a_{6n-3}&=-2\\
a_{6n}&=2\\
a_{6n-1}=a_{6n+1}&=1
\end{align}
Now since $2000=6\times 334 -4$, we have that $a_{2000}=-1$.
|
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|
Solution to Complex Equation $\cos z = 2i$ I tried solving the equation $\cos{z}=2i$ where $z$ is a complex number. The solution $i$ have ended up with is $z$, $= (4k+1)\frac{\pi}{2} - i\ln{(2+\sqrt{5})}$. However the text book solution is $z=(2k+1)\frac{\pi}{2} - (-1)^k\ln{(2+\sqrt{5})}i$. Are the results geometrically same or my solution is erroneous.
|
$$\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right).$$ Solve first for $w=e^{iz}$.
$$w^2-4iw + 1=0$$
so:
$$w=\frac{4i\pm\sqrt{-16-4}}{2} = (2\pm\sqrt{5})i$$
So we have to deal with two cases with care. If $w=(2+\sqrt{5})i$, you get your solution:
$$z = (4k+1)\frac{\pi}{2} - i\ln(2+\sqrt{5})$$
If $w=(2-\sqrt{5})i$, then using $2-\sqrt{5}=-(2+\sqrt{5})^{-1}$, we have:
$$iz = -\log(2+ \sqrt{5}) +\left(\frac{3\pi}{2}+2\pi k\right)i$$
So:
$$z = \left(\frac{3pi}{2}+2\pi k\right) + \log(2+ \sqrt{5})i$$
In the first case, let $m=2k$ and you have:
$$z = (2m+1)\frac{\pi/2} - (-1)^{m}\ln(2+\sqrt{5})i$$
and the second case $m=2k+1$ then:
$$ z = (2m+1)\frac{\pi}{2} - (-1)^{2m+1}\ln(2+\sqrt{5})$$
So the full answer merges the two answers.
You can probably write this carefully as one case. The two values of $w$ can be written as $$w=i(-1)^k(2+\sqrt{5})^{(-1)^k}$$. Then:
$$iz = \log w = (-1)^k \log(2+\sqrt{5}) + \frac{\pi}{2}i + k\pi i$$
So:
$$z = \left(k+1/2\right)\pi - (-1)^k \log(2+\sqrt{5}) i$$
|
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|
I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red?}}{=}}\{a+b\sqrt2+c\sqrt 3:a,b,c\in \Bbb Q\}$
But from my last question it seems that we need to check this for multiplicative closure, $(a+b\sqrt2 + c\sqrt3)(d+e\sqrt 2 + f\sqrt 3)=ad+ae\sqrt 2+ af\sqrt 3 + 2eb+bf\sqrt 3+cd\sqrt 3+ ce\sqrt2\sqrt3+ 3cf$
So we actually find I believe that:
$\Bbb Q(\sqrt2,\sqrt3)=\{a+b\sqrt 2+c\sqrt 3+ d\sqrt2\sqrt3:a,b,c,d\in \Bbb Q\}$?
I believe that $\Bbb Q(\sqrt 2 +\sqrt 3)\overset{\huge{\color\red ?}}=\{a+b(\sqrt 2+ \sqrt 3):a,b\in \Bbb Q\}$
So let's verify multiplicative closure $(a+b\sqrt 2+ b\sqrt 3)(c+d\sqrt 2+ d\sqrt 3)$
$$=ac+ad\sqrt 2+ad\sqrt 3+bc\sqrt2+2bd+bd\sqrt3+bc\sqrt3+bd\sqrt2\sqrt3+3bd$$
$$=(ac+3bd+2bd)+(ad+bc)\sqrt 2 + (ad+bd+bc)\sqrt 3 +(bd)\sqrt2\sqrt3$$
Since $\sqrt2$ and $\sqrt 3$ and $\sqrt2\sqrt3$ don't share common coefficients, they are linearly independent and hence $\Bbb Q(\sqrt2+\sqrt 3)=\Bbb Q(\sqrt2,\sqrt3)$(from above, i.e we have deduced that:) $\Bbb Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt2+c\sqrt3:a,b,c\in\Bbb Q\}$.
Is that the correct way to show this? Also this means we have:
$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$
|
Your belief that
$$\def\Q{\mathbb{Q}}
\Q(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+ \sqrt{3}):a,b\in \Q\}
$$
is wrong, as well as
$$
\Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt{2}+ c\sqrt{3}:a,b,c\in \Q\}
$$
The second claim can be immediately dismissed, because this would mean that $[\Q(\sqrt{2}+\sqrt{3}):\Q]=3$, but clearly $\Q(\sqrt{2}+\sqrt{3})\subseteq\Q(\sqrt{2},\sqrt{3})]$ and this extension has degree a divisor of $4$, because
$$
[\Q(\sqrt{2},\sqrt{3}):\Q]=
[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2})]\,
[\Q(\sqrt{2}):\Q]
$$
and both extensions in the right-hand side are of degree either $1$ or $2$ (both $2$, actually, but it's not important).
The first belief is wrong, too: this would mean that $\sqrt{2}+\sqrt{3}$ has a degree two minimal polynomial and this contradicts the fact you have to prove. Indeed
$$
[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2})]=2
$$
because no element of $\Q(\sqrt{2})$ is the square root of $3$:
$$
(a+b\sqrt{2})^2=3
$$
means
$$
a^2+2b^2+2ab\sqrt{2}=3
$$
hence either $a=0$ or $b=0$, leaving either $2b^2=3$ or $a^2=3$; neither equation has a solution in $\Q$.
Now this is a proof that $\Q(\sqrt{2}+\sqrt{3})=\Q(\sqrt{2},\sqrt{3})$. Indeed, we know that $[\Q(\sqrt{2},\sqrt{3}):\Q]=4$ and that
$$
[\Q(\sqrt{2}+\sqrt{3}):\Q]>2
$$
(well, assuming we know that $\sqrt{2}+\sqrt{3}$ is not rational, which is almost obvious). Therefore $[\Q(\sqrt{2}+\sqrt{3}):\Q]$ is a divisor of $4$, but greater than $2$, which leaves just one possibility.
Why doesn't $\sqrt{2}+\sqrt{3}$ have degree $\le2$ over $\Q$? Since $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$, we see that
$$
\Q\subset\Q(\sqrt{6})\subseteq\Q(\sqrt{2}+\sqrt{3})
$$
and, since $\sqrt{6}$ is clearly irrational (a standard proof based on Eisenstein's criterion applies), we have that the degree of $\sqrt{2}+\sqrt{3}$ (which we already know is a divisor of $4$) can be either $2$ or $4$. If it is $2$, we find that $\sqrt{2}+\sqrt{3}=a+b\sqrt{6}$, for some $a,b\in\Q$. Squaring gives
$$
5+2\sqrt{6}=a^2+6b^2+2ab\sqrt{6}
$$
that implies $ab=1$, so
$$
a^2+\frac{6}{a^2}-5=0
$$
that is,
$$
a^4-5a^2+6=0
$$
or
$$
(a^2-2)(a^2-3)=0
$$
a contradiction.
Of course, the above is much more complicated than the direct observation that
$$
(\sqrt{2}+\sqrt{3})+(\sqrt{2}+\sqrt{3})^{-1}=
\sqrt{2}+\sqrt{3}+\frac{1}{\sqrt{3}+\sqrt{2}}=
\sqrt{2}+\sqrt{3}+\sqrt{3}-\sqrt{2}=2\sqrt{3}
$$
so $\sqrt{3}\in\Q(\sqrt{2}+\sqrt{3})$ and therefore also $\sqrt{2}\in\Q(\sqrt{2}+\sqrt{3})$, ending the proof. However, the main object was to show the issues in your argument.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of the perpendicular distance of origin from L is ?
Let $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ be the required line.
Then $2l+3m-10n=0,4l-3m-2n=0$
So $l:m:n=1:1:\frac{1}{2}$
So the line is $\frac{x-x_1}{1}=\frac{y-y_1}{1}=\frac{z-z_1}{\frac{1}{2}}$.
But i could not solve further.Please help me.
|
Hint.
I give you a hint that you can apply to find the distance you're looking for.
So you have two lines $$\begin{cases} L_1 \equiv \frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10} \\
L_2 \equiv \frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}\end{cases}$$
For $L_1$, you can find a point $A_1 \in L_1$, for example $A_1=(-2,-6,34)$ and a direction vector of $L_1$, for example $u_1=(2,3,-10)$. You can do the same for $L_2$ to get $A_2$ and $u_2$.
Now the cross product $u = u_1 \times u_2$ is perpendicular to $u_1$ and $u_2$. The plane $P_1$ containing the line $L_1$ and the direction $u$ intersects $L_2$ at $B_2$. Similarly, The plane $P_2$ containing the line $L_2$ and the direction $u$ intersects $L_1$ at $B_1$. The line $B_1 B_2$ is the perpendicular to your orginal two lines.
Finally you can find its distance to the origin.
|
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|
Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels like nothing simple fits. Next I noticed we can normalise: restricting $a+b+c=1$ it can be made to look like this:
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
Another idea is to let $x=a/b,y=b/c, z=c/a:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{1}{x(y+1)}}+\sqrt{\frac{1}{y(z+1)}}+\sqrt{\frac{1}{z(x+1)}}\right)\leq \sqrt{x+y+z}$$
But I can't see where to go from here.
|
Using C-S inequality, we can obtain
$$\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)^2\leq3\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).$$
Since $(b+c)\left(\frac1{b}+\frac1{c}\right)\geq4$, we have $\frac{a}{b+c}\leq\frac1{4}\left(\frac{a}{b}+\frac{a}{c}\right)$, similarly we can have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right).$$
So $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ must be less than or equal to either $\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$ or $\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)$. If $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right),$$
then we may exchange $b$ and $c$, hence we can always get the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right).$$ Therefore
$$\sqrt{\frac2{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq\sqrt{2}\sqrt{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\\\leq\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}.$$
|
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|
Limit Infinity Minus Infinity form without using infinite series Expansions Evaluate $$L=\lim_{x \to 0} \frac{1}{\sin^2x}-\frac{1}{\sinh^2x}$$ If we take L.C.M and use LHopital's Rule it becomes quite Tedious. I tried in this way.
$$
\sinh x=\frac{e^x-e^{-x}}{2}
$$
Now
$$\lim_{x \to 0}\frac{e^x-e^{-x}}{x}=2 \implies \lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}=\frac{1}{4}
$$
Now
$$
\sinh x=\left(\frac{e^x-e^{-x}}{x}\right)\left(\frac{x}{2}\right)
$$
So
$$
\lim_{x \to 0}\frac{1}{\sinh^2x}=\lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}\:\lim_{x \to 0}\frac{4}{x^2}=\lim_{x \to 0}\frac{1}{x^2}
$$
Hence
$$
L=\lim_{x \to 0}\frac{1}{\sin^2x}-\frac{1}{x^2}=\lim_{x \to 0}\frac{(x-\sin x)(x+\sin x)}{(x\sin x)(x\sin x)}=\lim_{x \to 0}\frac{x-\sin x}{x\sin x}\:\lim_{x \to 0}\frac{x+\sin x}{x\sin x}$$
Individually if i evaluate above limits I am getting Zero. But answer is not correct. Can anyone spot out where I am going wrong
|
Write the original expression as
\begin{eqnarray*}
\frac{1}{\sin ^{2}x}-\frac{1}{\sinh ^{2}x} &=&\frac{\sinh ^{2}x-\sin ^{2}x}{%
\sin ^{2}x\sinh ^{2}x} \\
&=&\left( \frac{\sinh x-\sin x}{x^{3}}\right) \times \left( \frac{\sinh
x+\sin x}{x}\right) \times \left( \frac{x}{\sin x}\right) ^{2}\times \left(
\frac{x}{\sinh x}\right) ^{2}.
\end{eqnarray*}
Using L'Hospital rules successively one obtains
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sinh x-\sin x}{x^{3}}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x-\cos x}{3x^{2}}\right)
=\lim_{x\rightarrow 0}\left( \frac{\sinh x+\sin x}{6x}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x+\cos x}{6}\right) =\frac{1+1}{6}=%
\frac{1}{3}
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sinh x+\sin x}{x}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x+\cos x}{1}\right) =\frac{1+1}{1}=2
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2}=\left(
\lim_{x\rightarrow 0}\frac{x}{\sin x}\right) ^{2}=\left( \lim_{x\rightarrow
0}\frac{1}{\cos x}\right) ^{2}=\left( 1\right) ^{2}=1
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x}{\sinh x}\right) ^{2}=\left(
\lim_{x\rightarrow 0}\frac{x}{\sinh x}\right) ^{2}=\left( \lim_{x\rightarrow
0}\frac{1}{\cosh x}\right) ^{2}=\left( 1\right) ^{2}=1
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin ^{2}x}-\frac{1}{\sinh ^{2}x}%
\right) =\left( \frac{1}{3}\right) \times \left( 2\right) \times \left(
1\right) \times \left( 1\right) =\frac{2}{3}
\end{equation*}
|
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|
verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.
I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity laws to solve this problem. But i get a fraction:
$8x^2 - 2x = 2((2x - \frac{1}{2})^{2} - \frac{1}{4}) $. So the previous statement is equivalent to proving:
$(2x - \frac{1}{2})^2 \equiv 3\frac{1}{2} \cdot \frac{1}{2} \mod 75$. which obviously does not help me further.
Any tips hints on this problem?
|
Hint:
$$
8x^2-2x-3=(4x-3)(2x+1)
$$
Is $2$ invertible mod $75$?
Since a full answer has been given in another post, I will add the following.
Indeed $2\cdot38\equiv1\pmod{75}$, so there are solutions.
Although not required by the question, we can find all the solutions as follows.
$$
\begin{align}
8x^2-2x-3
&=(4x-3)(2x+1)\\
&\equiv8(x-57)(x-37)\pmod{75}
\end{align}
$$
Besides the obvious
$$
x\equiv57\pmod{75}\\
x\equiv37\pmod{75}
$$
we also have
$$
\left.\begin{align}
x&\equiv57\pmod{3}\\
x&\equiv37\pmod{25}
\end{align}\right\}x\equiv12\pmod{75}
$$
$$
\left.\begin{align}
x&\equiv57\pmod{25}\\
x&\equiv37\pmod{3}
\end{align}\right\}x\equiv7\pmod{75}
$$
However, all of the preceding are contained in
$$
\left.\begin{align}
x&\equiv57\pmod{15}\\
x&\equiv37\pmod{5}
\end{align}\right\}x\equiv12\pmod{15}
$$
$$
\left.\begin{align}
x&\equiv57\pmod{5}\\
x&\equiv37\pmod{15}
\end{align}\right\}x\equiv7\pmod{15}
$$
Therefore, all solutions are either $x\equiv7\pmod{15}$ or $x\equiv12\pmod{15}$.
|
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|
How to solve combinatorics with variable set sizes? The question: How many ways to put $15$ students into groups, such that each group has $3$~$5$ students?
If the group was $3$ equal groups of $5$, then the answer would be $\cfrac{15!}{5! \cdot 5! \cdot 5! \cdot 3!}$, and if it was $5$ equal groups of $3$, then the answer would be $\cfrac{15}{3! \cdot 3! \cdot 3! \cdot 3! \cdot 3! \cdot 5!}$. How do you deal with various sized groups?
Thanks.
|
Since it is unclear whether the groups are labelled or unlabelled,
I'll treat them as labelled, and indicate the correction needed if they are unlabelled.
Groups can be$\;$ 5-5-5, $\;$ 5-4-3-3, $\;$ 4-4-4-3, $\;$ or$\;$ 3-3-3-3-3 , so # of ways for labelled groups is:
$$\frac{15!}{5!\cdot 5! \cdot5!}+\frac{15!}{5!\cdot 4!\cdot 3! \cdot 3! }+\frac{15!}{4! \cdot 4! \cdot 4! \cdot 3!}+\frac{15!}{3!\cdot 3!\cdot 3! \cdot 3! \cdot 3!}$$
If the groups are unlabelled, divide respectively by $3!,\;2!,\;3!\;$ and $5!$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Maximizing $3 x^2+2 \sqrt{2} x y$ with $x^4+y^4=1$ I want to have maxizing value of $3 x^2+2 \sqrt{2} x y$ when $x^4+y^4=1$, $x>0,y>0$.
How can I solve it.
|
here is one way to do this: let us parametrize the constraint $$x^4 + y^4 = 1, x > 0, y > 0 \text{ by } x = \sqrt{\cos t}, y = \sqrt{\sin t}, 0 < t < \pi/2.$$ we need to maximise $$ f = 3x^2 + 2\sqrt 2xy=3\cos t + 2\sqrt 2\sqrt{\cos t \sin t} = 3\cos t + 2\sqrt{\sin 2t}, 0 < t < \pi/2$$
taking the derivative of $f$ and seeting it to zero, you have $$f' = -3\sin t + 2\frac{\cos 2t}{\sqrt{\sin 2t}}$$
only critical number of $f$ is $t = 0.4636$ and this is a global max $4.4721$ for $f(t),\, 0 \le t \le \pi/2.$
|
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|
Show that $\lim_{n \longrightarrow \infty} \frac{3n^2+2n}{4+3n^2}=1$ I need show that $\forall \, \epsilon > 0$, exist $M \in \mathbb N$ such that, $\left|\frac{3n^2+2n}{4+3n^2}-1\right| < \epsilon$ if $n \geq M$.
Let's consider
$\left|\frac{3n^2+2n}{4+3n^2}-1\right|=\left|\frac{2n-4}{4+3n^2}\right|<\left|\frac{2n-4}{3n^2}\right|=\left|\frac{2}{3n}-\frac{4}{3n^2}\right|\leq \left|\frac{2}{3n^2}-\frac{4}{3n^2}\right|=\frac{2}{3n^2}<\epsilon \Leftrightarrow n>\sqrt{\frac{2}{3\epsilon}}$
Then, I must take $M = \left\lceil \sqrt{\frac{2}{3\epsilon}}\right\rceil+1 \in \mathbb{Z}$?
Thanks.
|
If $n \geq 1$, then
$$
\bigg| \frac{3n^{2}+2n}{4+3n^{2}} - 1 \bigg|
=
\bigg| \frac{2n-4}{4+3n^{2}} \bigg|
\leq
\frac{2n}{4+3n^{2}} + \frac{4}{4+3n^{2}}
<
\frac{2}{n} + \frac{4}{n^{2}};
$$
taking any $\varepsilon > 0$, if $n \geq \max \{ \lceil \frac{4}{\varepsilon} \rceil + 1, \lceil \sqrt{\frac{8}{\varepsilon}} \rceil + 1 \},$
then
$$
\frac{2}{n} + \frac{2}{n^{2}} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
$$
|
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|
Prove that if 2 divides $x^2-5$ then 4 divides $x^2-5$ so I have to prove this and I use two different types of proof and I came to a contradicting result. Can someone point out an error I made?
Using a direct proof:
If 2 divides $x^2-5$ than $x^2-5=2k$ from that $x^2=2(k+2)+1$. From that we know x is odd. since x is odd x=2y+1. From that $(2y+1)^2-5=4y^2+4y-4=4(y^2+y+1)$ so the implication is true.
Using a contrapositive proof.
If 4 does not divide $x^2-5$ then we have 3 cases.\
$x^2-5=4k+1$\
$x^2-5=4k+2$\
$x^2-5=4k+3$\
Looking at the second case we have $4k+2=2(2k+1)$. Therefore the implication is false as from truth we have false. What am I doing wrong? Any help would be appreciated.
|
It is clear that $x^2-5$ is not divisible by $2$ when it is of the form $4k-1$ or $4k-3$. So suppose $x^2-5 = 4k-2$. This quantity is certainly divisible by $2$ if there is such an integer $k$, which is why we need to do a bit more analysis to make sure something fishy isn't going on. From this we have $$x^2 = 4k+3$$ which makes $x$ odd, so let $ x= 2y+1$ to get $$\begin{align}(2y+1)^2 = 4k+3 \\ \implies 4y^2+4y+1 = 4k+3 \\ \implies 4y^2+4y = 4k+2 \\ \implies 2(y^2+y) = 2k+1\end{align}$$ Now the LHS is necessarily even while the RHS is necessarily odd, a contradiction. Hence no such integer $y$ exists, meaning no such integer $x$ exists such that $x^2-5$ can take the form $4k+2$ for any integer $k$. This is enough to complete the proof by contrapositive.
|
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|
How to prove this $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$ Let $x,y>0$, and $x+y=2$, show that
$$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$$
I tried using Minkowski inequality
$$\sqrt{x^2+3}+\sqrt{y^2+3}\ge\sqrt{(x+y)^2+(\sqrt{3}+\sqrt{3})^2}=\sqrt{4+12}=4$$
But
$$\sqrt{xy+3}\le\sqrt{\dfrac{(x+y)^2}{4}+3}=2$$
so i'm not sure how it would work with this problem,Thanks
|
Let $\sqrt{x^2+3}+\sqrt{y^2+3}=t$.
We need to prove that
$$\sqrt{x^2+3}+\sqrt{y^2+3}-4\geq2-\sqrt{xy+3}$$ or
$$\frac{x^2+3+y^2+3+2\sqrt{(x^2+3)(y^2+3)}-16}{t+4}\geq\frac{1-xy}{2+\sqrt{xy+3}}$$ or
$$\frac{2(x^2+3)+2(y^2+3)-16-\left(\sqrt{x^2+3}-\sqrt{y^2+3}\right)^2}{t+4}\geq\frac{4-4xy}{4(2+\sqrt{xy+3})}$$ or
$$\frac{(x-y)^2-\frac{(x-y)^2(x+y)^2}{t^2}}{t+4}\geq\frac{(x-y)^2}{4(2+\sqrt{xy+3})}$$ or
$$\frac{4(t^2-4)(2+\sqrt{xy+3})}{t^2(t+4)}\geq1$$
and since $xy\geq0$, it's enough to prove that $f(t)\geq\frac{2-\sqrt3}{4}$ , where $f(t)=\frac{t^2-4}{t^2(t+4)}$.
But by Mikowski $t=\sqrt{x^2+3}+\sqrt{y^2+3}\geq\sqrt{(x+y)^2+3(1+1)^2}=4$.
In another hand,
$$t=\sqrt{x^2+y^2+6+2\sqrt{(x^2+3)(y^2+3)}}=\sqrt{10-2xy+2\sqrt{x^2y^2+3(x^2+y^2)+9}}=$$
$$=\sqrt{10-2xy+2\sqrt{x^2y^2-6xy+21}}\leq\sqrt{10-2xy+2(xy+\sqrt{21})}=\sqrt3+\sqrt7.$$
Also we have
$$f'(t)=\frac{-t^3+12t+32}{t^3(t+4)^2}>0$$
for all $t\in\left[4,\sqrt3+\sqrt7\right)$.
Thus, $f(t)\geq f(4)=\frac{3}{32}>\frac{1}{4}(2-\sqrt3)$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving binomial summation identity using generating functions An exercise for class requires me to prove the following identity using generating functions:
$$\sum_{k=0}^{m/2} (-1)^k {n \choose k} {n+m-2k-1 \choose n-1} = {n \choose m}$$ for all $m \leq n$ and $m$ is even.
I've tried a bunch of things but can't wrap my head around this one.
I assume one would start with
$${\sum_{k=0}^{n} {n \choose m}} x^m = (1+x)^n
= \frac{(1-x^2)^n}{(1-x)^n}$$
and then interpreting the expression on the right as the product of generating functions to somehow arrive at
$$\sum_{m=0}^{n} \left( \sum_{k=0}^{m/2} (-1)^k {n \choose k} {n+m-2k-1 \choose n-1} \right) x^m$$
but now idea how to actually do it.
Some help would be greatly appreciated! :-)
|
Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose n-1}$$
where $n\ge m.$
Re-write this as
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose m-2k}.$$
Now introduce
$${n+m-2k-1\choose m-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2k+1}} (1+z)^{n+m-2k-1} \; dz.$$
Observe that this is zero when $2k\gt m$ so we may extend the range of
$k$ to infinity.
We thus get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n+m-1}
\sum_{k\ge 0} {n\choose k} (-1)^k \frac{z^{2k}}{(1+z)^{2k}}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n+m-1}
\left(1-\frac{z^2}{(1+z)^2}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1+z)^{n-m+1}}
\left(1+2z\right)^n \; dz.$$
Extracting the residue we get
$$\sum_{q=0}^m {n\choose q} 2^q {m-q+n-m\choose m-q}
(-1)^{m-q}
\\ = \sum_{q=0}^m {n\choose q} 2^q {n-q\choose m-q}
(-1)^{m-q}.$$
To conclude introduce the integral
$${n-q\choose m-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-q+1}} (1+z)^{n-q} \; dz.$$
Observe that this is zero when $q\gt m$ so we may extend the range of
$q$ to infinity to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n}
\sum_{q\ge 0} {n\choose q} 2^q (-1)^{m-q} \frac{z^q}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^m}{z^{m+1}} (1+z)^{n}
\left(1-2\frac{z}{1+z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^m}{z^{m+1}}
\left(1-z\right)^n \; dz
= (-1)^m {n\choose m} (-1)^m = {n\choose m}.$$
Alternative solution.
Introduce
$${n+m-2k-1\choose n-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2k+1}} \frac{1}{(1-z)^n} \; dz.$$
This is again zero when $2k\gt m.$ We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^n}
\sum_{k\ge 0} {n\choose k} (-1)^k z^{2k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^n}
(1-z^2)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}}
(1+z)^n \; dz
\\ = {n\choose m}.$$
This is probably what the problem had in mind.
|
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|
Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x + y + z = z^2 + y^2 + 2$
But I cannot get much farther!
EDIT:
$x -z = z^2 - x^2 = (z-x)(z+x) \implies z + x = -1$.
$y - x = x^2 = y^2 = (x-y)(x+y) \implies x + y = -1$
Thus, $z - y = 0$ and $z = y$.
$2x + 7y$ is to be found then, and $x = -1 -y $ so: $2(-1 - y) - 7y = -2 - 9y$ is to be found.
Any ideas here?
|
Hint:
If you add up the first equation and the second times $-1$, you get:
$$x-z=z^2-x^2=-(x-z)(x+z)$$
Edit:
Therefore, $x=y=z$, because otherwise you can divide by $x-y$ to get $x+y=-1$ and, using the first set of equations, $z^2+1=-1$ and $z^2=-2$, which is not possible. The same goes for y and z. Using this in the first equation gives:
$$
x+x=x^2+1 \\
x^2-2x+1=0\\
x=y=z=1\\
2x+3y+4z=2+3+4=9
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1482063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically. How would I prove this? I suspect I want to use the seventh roots of unity in some way but I am not sure how to proceed.
|
Since $ \sin\left(\frac{6\pi}7\right) = \sin\left(\frac{\pi}7\right) $, then the expression is positive because $ \sin\left(\frac{2\pi}7\right) - \sin\left(\frac{\pi}7\right) > 0 $.
Let $ I $ denote this expression, then $ I > 0 $. Consider
$ I^2 = \left[\underbrace{ \sin^2\left(\frac{2\pi}7\right) + \sin^2\left(\frac{4\pi}7\right) + \sin^2\left(\frac{6\pi}7\right)}_{J} \right] - \left [ \underbrace{2\sin\left(\frac{2\pi}7\right) \sin\left(\frac{4\pi}7\right) - 2\sin\left(\frac{2\pi}7\right) \sin\left(\frac{6\pi}7\right) - 2\sin\left(\frac{4\pi}7\right) \sin\left(\frac{6\pi}7\right)}_{K} \right ] $
By half angle formula, $\sin^2(A) = \frac12 (1-\cos(2A)) $, then $J = \frac32 - \frac12 \left( \cos\left(\frac{4\pi}7\right) + \cos\left(\frac{8\pi}7\right) + \cos\left(\frac{12\pi}7\right) \right) = \frac32 - \frac12\left( \cos\left(\frac{3\pi}7\right) + \cos\left(\frac{5\pi}7\right) + \cos\left(\frac{\pi}7\right)\right) $
$J = \frac32 + \frac12 \times \frac12 = \frac74$
By product to sum formula, $\cos(A) - \cos(B) = -2\sin\left( \frac{A+B}2\right)\sin\left( \frac{A-B}2\right) $, can you show that $ K = 0 $?
|
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|
Find all polynomial $f(x)$ such that $x^3-1\mid f(x)g(x)-1$ This is a problem that has haunted me for more than some days. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind:
Find all polynomial $f(x)$,such $f(x)\in Z[x],\deg{(f(x))}\le 2$,and there exist $g(x)\in Z[x]$,such
$$x^3-1\mid f(x)g(x)-1$$
I have $f(x)=x^2,x$ ,have other solution,and How to find all solution?. Thanks
|
Let $\rho = e^{2\pi i/3} = \frac{-1 + i\sqrt{3}}{2}$. In $\mathbb{C}[x]$, we have the factorisation $x^3-1 = (x-1)(x-\rho)(x-\rho^2)$, so if $x^3-1 \mid f(x)g(x) - 1$, we must have $f(1)g(1) = f(\rho)g(\rho) = f(\rho^2)g(\rho^2) = 1$. Since $f,g$ are assumed to be polynomials with integer coefficients, we know that we must have $f(1) = g(1) = \pm 1$, and the values at $\rho^k$ belong to the ring $R = \mathbb{Z}[\rho]$ of Eisenstein integers, so $f(\rho^k)$ must be a unit in $R$. From the ansatz $f(x) = ax^2 + bx + c$ we obtain the conditions
\begin{gather}
a + b + c = \pm 1, \\
a\rho^2 + b\rho + c \in R^{\times}, \\
a\rho^4 + b \rho^2 + c \in R^{\times}.
\end{gather}
With $\rho^2 = -(1+\rho)$ and $\rho^4 = \rho$, that becomes
\begin{gather}
a+b+c = \pm 1, \\
(c-a) + (b-a)\rho \in R^{\times}, \\
(c-b) + (a-b)\rho \in R^{\times}.
\end{gather}
The units of $R$ are $1,\, 1 + \rho,\, \rho,\, -1,\, -1-\rho,\, -\rho$, so we must have either $a = b$ or $a - b = \pm 1$. Since $(-f(x))(-g(x)) = f(x)g(x)$, we can ignore the case $a - b = -1$.
In the case $a = b$, we must have $c-a = \pm 1$, and thus the first condition yields $a + b + c = 3a \pm 1 = \pm 1$, so it follows that $a = 0$, and we have the trivial option $f(x) = \pm 1$, with $g(x) = \pm x^3$ as a possible companion.
In the case $a - b = 1$, it follows that $c-a = -1$ or $c-a = 0$ from the second condition, and the corresponding $c - b = c - a + 1 = 0$ or $c - b = c - a + 1 = 1$ then also satisfies the third condition. So we either have $c = b$ or $c = a$ then. In the first condition, $c = b = a-1$ leads to $3a - 2 = \pm 1$, thus $a = 1$ and $b = c = 0$, which yields $f(x) = x^2$. The possibility $c = a$ leads to $3a - 1 = \pm 1$, hence $a = 0$, and therefore $f(x) = -x$.
Picking up the negatives that we dropped by ignoring $a - b = -1$, the complete list of $f \in \mathbb{Z}[x]$ with $\deg f \leqslant 2$ such that there is a $g \in \mathbb{Z}[x]$ with $x^3 - 1 \mid f(x)g(x) - 1$ is
$$f(x) = \pm 1 \quad\text{or}\quad f(x) = \pm x \quad\text{or}\quad f(x) = \pm x^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1482417",
"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$
Induction step: $n=m+1$
$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x}
{sin\frac{x}{2}}$
How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.
|
Hint: If you want another approach you could probably do a series of deconvolutions in the Fourier domain. The euclidean algorithm should work.
|
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|
Problem related to maximum and minimum value of a trigonometric function I have used the following technique to calculate the maximum value of the function...
but I couldn't proceed with next step..
can anyone guide me please?
Find the maximum and minimum value of the function $$f(x) = \sin^2(\cos x)+\cos^2(\sin x)$$
I have written the following function as $$1+\sin^2(\cos x)-\sin^2(\sin x) = 1+\sin(\cos x+\sin x)\cdot \sin(\cos x-\sin x)$$
|
Using Calculus
Let $$f(x) = \sin^2(\cos x)+\cos^2(\sin x) = 1-\cos^2(\cos x)+1-\sin^2(\sin x)\;$$
So we get $$\displaystyle f(x) = 2-\left[\cos^2(\cos x)+\sin^2(\sin x)\right] = 2-\frac{1}{2}\left[2\cos^2(\cos x)+2\sin^2(\sin x)\right]$$
So we get $$\displaystyle f(x) = 2-\frac{1}{2}\left[1+\cos(2\cos x)+1-\cos(2\sin x)\right] = 1+\frac{1}{2}\left[\cos(2\sin x)+\sin(2\cos x)\right]$$
Now Using First Derivative test.
So $$\displaystyle f'(x) = \frac{1}{2}\left[-\sin(2\sin x)\cdot 2\cos x+\cos(2\cos x)\cdot -2\sin x\right]$$
So $$\displaystyle f'(x) = -\left[\sin(2\sin x)\cdot \cos x+\cos(2\sin x)\cdot \sin x\right]$$
Now $f'(x) =0$ at $$\displaystyle x=0\;\;,\frac{\pi}{2}\;\;,\pi\;\;,\frac{3\pi}{2}\;\;,2\pi,......$$
So $$\displaystyle \max[f(x)] = \sin^2(1)+1\;\;$$ at $$\displaystyle x = 0\;\;,\pi\;\;,2\pi,.....$$
And So $$\displaystyle \min[f(x)] = \cos^2(1)\;\;$$ at $$\displaystyle x = \frac{\pi}{2}\;\;,\frac{3\pi}{2}\;\;,\frac{5\pi}{2},.....$$
|
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|
Fourier sine series for $x^3$ It is asked to find the Fourier Sine Series for $x^3$ given that
$$\frac{x^2}{2} = \frac{l^2}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \cos\left(\frac{n \pi x}{l} \right)$$
integrating term by term. (This result was found in another exercise). As suggested, I integrated:
$$\int \frac{x^2}{2} dx = \frac{x^3}{6} = \frac{l^2x}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \int \cos\left(\frac{n \pi x}{l} \right)dx + C$$
$$ \Rightarrow \frac{x^3}{6} = \frac{l^2x}{6}+ \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \frac{l}{n \pi} \sin\left(\frac{n \pi x}{l} \right) + C $$
$$\Rightarrow x^3 = l^2x + \frac{12l^3}{\pi^3}\sum_{n=1}^\infty (-1)^n \frac{1}{n^3} \sin\left(\frac{n \pi x}{l} \right) + C$$
It looks like that wolfram gives a different answer. I don't know if the problem is the constant $C$ or if I made something wrong. Please, follow the book approach, don't try to calculate the coefficient of Fourier sine series.
Thanks!
|
Both terms have to agree at $x=0$, hence, simply, $C=0$.
|
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|
How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?
|
$$
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
(x - 1)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 1\right)^2 = 25 \\
(x + 5)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 9\right)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
\frac{25}{16}\left(x^2+4x+20\right) = 25 \\
\frac{25}{16}\left(x^2+4x+20\right) = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x^2+4x+20 = 16 \\
x^2+4x+20 = 16 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x^2+4x+4 = 0 \\
x^2+4x+4 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
(x+2)^2 = 0 \\
(x+2)^2 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -\frac{3}{4}(-2) - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
x=-2 \\
y = -5
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=-2 \\
y = -5
\end{cases}
$$
|
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|
Why test of divisible by $12$ works with $3$ and $4$ but not with $2$ and $6 ?$ Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$
Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $?$
Test of divisible by $2 ,$ last digit must be divisible by $2 ,$ since $10$ is always divisible by $2 .$
Test of divisible by $ 6 ,$ the number should be divisible by $2$ and $3 .$
Test of divisible by $12 ,$ the number should be divisible by $4$ and $3 ,$ somewhere is given as we can say if number is divisible by $2$ and $6$ then it also must be divisible by $12 ,$ for counter example $,$ if number is $18 .$
I have two questions $:$
*
*Why test of divisible by $3$ works $?$ Any proof $.$
*Why test of divisible by $12$ works with divisible by $3$ and $4 ,$ but not with divisible by $2$ and $6 ?$ Any proof $.$
|
Since we are working in a numerical system of basis $10$, an integer number $N>0$ can be written as $$N=a_n\times 10^n+a_{n-1}\times 10^{n-1}+\ldots +a_1\times 10 +a_0$$
The number $10=9+1$ leaves remainder $1$ under division by $3$, and so does the another powers of $10$, so $N$ has the form $$N=9m+a_{n}+a_{n-1}+\ldots+a_1+a_0$$
for some integer $m$, which means that $3$ divides $N$ iff $3$ divides $a_{n}+a_{n-1}+\ldots+a_1+a_0$, the sum of the digits of $N$. Also the same test works for $9$.
|
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|
Simple inequality with a,b,c I'm looking for proof of
$$\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)} \leq \sqrt{2}(a+b+c)$$
I tried using $m_g \leq m_a$, generating the permutations of $\sqrt{a(b+c)} \leq \frac{a+(b+c)}{2}$ and adding them together, but I get $\frac{3}{2}(a+b+c)$, and obviously $\frac{3}{2}>\sqrt{2}$, so I'm stuck and sure I'm missing something trivial, so thanks in advance.
|
Using Cauchy-Schwarz inequality with the vectors:
$$\begin{cases}
u=(\sqrt{a},\sqrt{b},\sqrt{c})\\
v=(\sqrt{b+c},\sqrt{c+a},\sqrt{a+b})
\end{cases}$$ you get
\begin{align}[\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)}]^2 &=(u.v)^2\\ &
\le \Vert u \Vert^2 \Vert v \Vert^2\\
&=(a+b+c)((b+c)+(c+a)+(a+b))\\
&=2(a+b+c)^2\end{align} Which enables to conclude to the desired result.
|
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|
What is $\angle AEB$? Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$
What is the angle $\widehat {AEB}$?
I can only find $|ED|=\sqrt{12}$
|
There has to be an easier way, but this works...
Use the law of cosines to get:
$s^2 = 5 - 4\cos(AEB)$
$s^2 = 13 - 12\cos(BEC)$
$2s^2 = 10 - 6\cos(AEB+BEC)$
Eliminate $s^2$, and use $\cos(AEB + BEC) = \cos(AEB)\cos(BEC) - \sin(AEB)\sin(BEC)$. Exchange the sines for cosines, eliminate $\angle BEC$, and a whole bunch of algebra later we get:
$(4\cos(AEB) - 5)(2\cos^2(AEB) - 1) = 0$
Reject the first root for being too large, and we get $\cos(AEB) = \pm \frac{1}{\sqrt{2}}$.
Reject $45^\circ$ because that implies that $s = \sqrt{5 - \frac{4}{\sqrt{2}}} \approx 1.473 < \frac{3}{\sqrt{2}}$, so $E$ would not be in the square. We can similarly reject $225^\circ$ and $315^\circ$.
Therefore, $\angle AEB = 135^\circ$.
|
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Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24. Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24.
|
Note that $n-1\equiv -1\pmod{3}$ and $n-1\equiv -1\pmod{8}$.
It is clear that $n-1$ is not a perfect square. If $ab=n-1$, where $a$ and $b$ are positive, call $a$ and $b$ a couple, or, in the more business-oriented language of today, partners.
If $\{a,b\}$ are a couple, then since $ab\equiv -1\pmod{3}$, one of $a$ and $b$ is congruent to $1$ modulo $3$, and the other is congruent to $-1$. So their sum is divisible by $3$.
Now we work modulo $8$. Either one of $a$ and $b$ is congruent to $1$, and the other to $-1$, or one is congruent to $3$ and the other to $5$. In either case their sum is divisible by $8$.
Thus the sum of the two members of any couple is divisible by $24$, and therefore the sum of the divisors of $n-1$ is divisible by $24$.
|
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|
an urn contains six ball of each of the three colors: red, blue, and green. An urn contains six balls of each of the three colors: red, blue, green. Find the expected number of different colors obtained when three balls are drawn:
a. with replacement;
b. without replacement.
The correct answers are:
a: 19/9
b: 2.1912
I just don't know how to get to get to the answers.
Thanks for any help
|
We can also use recurrences, where A,B,C are the number of balls of three colors:
a)
\begin{align*}
f(a,b,c) = \left\{\begin{matrix}
3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)& \text{if }A+B+C-(a+b+c) = 3\\
\dfrac{a\cdot f(a-1,b,c)+b\cdot f(a,b-1,c)+c\cdot f(a,b,c-1)}{a+b+c} & \text{otherwise}
\end{matrix}\right.
\end{align*}
b)
\begin{align*}
f(a,b,c) = \left\{\begin{matrix}
3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)& \text{if }A+B+C-(a+b+c) = 3\\
\dfrac{A\cdot f(a-1,b,c)+B\cdot f(a,b-1,c)+C\cdot f(a,b,c-1)}{A+B+C} & \text{otherwise}
\end{matrix}\right.
\end{align*}
|
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Given integer roots of $x^2+mx-n=0$ and $x^2-mx+n=0$. Show $6 \mid n$. Suppose that $m$ and $n$ are integers such that both the quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
I figured out the roots of the equations through quadratic formula-for the first one it is $\frac{-m \pm \sqrt{m^2-4n}}{2}$ and for the second one it was $\frac{m \pm \sqrt{m^2+4n}}{2}$ .
Then I proved that the discriminant has the same parity as that of $m$ through congruences. Therefore, it was clear that if $m^2-4n$ and $m^2+4n$ were perfect squares,then the roots would be integers. I started with $m^2-4n$ and equated it with $a^2$, where $a$ is some integer. Then I rearranged the terms and factorized things to get $(m-a)(m+a)=4n$. This shows that at least one of them should be even but if one is even then the other should also be even. This shows that $m$ and $a$ are of the same parity.
I don't know what to do next. I want some help to prove that $n$ is even. I could figure out how to prove that $n$ is divisible by 3 once it is proved that $n$ is even.
|
We will be using the following special case of the Vieta Relations: the sum of the roots of $x^2+bx+c=0$ is $-b$, and their product is $c$.
Suppose to the contrary that $n$ is odd. The product of the roots of each equation is therefore odd, so each root is odd.
The product of the roots in the first equation is $-n$, and in the second it is $n$. So in one equation, not necessarily the first, the product of the roots is congruent to $-1$ modulo $4$, and in the other equation the product of the roots is congruent to $1$ modulo $4$.
In the equation with product of roots congruent to $1$, the roots are both congruent to $1$ modulo $4$, or both are congruent to $-1$. So the sum of the roots is congruent to $2$ modulo $4$.
In the equation with product of roots congruent to $-1$ modulo $4$, the sum of the roots is congruent to $0$ modulo $4$.
So the sums of the roots in the two equations are incongruent modulo $4$.
But in either case, the sum of the roots is $m$ or $-m$. These are both congruent to $2$ modulo $4$, or both congruent to $0$, and we have reached our contradiction.
Remark: A very similar argument will show divisibility by $3$.
|
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Integration of $\frac{1}{(1+x^4)^\frac{1}{4}}$ This question has been puzzling me a lot. This is in the middle list of question(meaning its moderately tough). I have tried everything i could think of.
My try: $$\int \frac{1}{x(1+\frac{1}{x^4})^\frac{1}{4}}dx$$
$$-\frac{1}{4}\int \frac{-4x^4}{x^5(1+ \frac{1}{x^4})^\frac{1}{4}}dx$$
putting $$1 + \frac{1}{x^4} = t \iff x^4=\frac{1}{t-1} \iff dt = -\frac{4}{x^5}dx$$ and substituting in the integral,
$$-\frac{1}{4}\int \frac{1}{(t-1)t^\frac{1}{4}}dt$$
And Now I am stuck here.
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Let $$I = \int\frac{1}{(1+x^4)^{\frac{1}{4}}}dx\;,$$ Then Put
$$x=\sqrt{\tan z }\;\;\;, \;\; dx = \frac{\sec^2 z}{2\sqrt{\tan z}}dz$$
$$I=\int \frac{1}{\sqrt[4]{1+\tan^2 z }} \frac{ \sec^2 z } {2 \sqrt{\tan z}}dz=\frac{1}{2} \int \sec^{ \frac{3}{2}} z \tan^{-\frac{1}{2}} zdz$$
$$=\frac{1}{2} \int \frac{1}{\cos z \sqrt{\sin z}}dz=\frac{1}{2} \int \frac{1}{(1-\sin^2 z) \sqrt{\sin z}} \cos z dz$$
$$w=\sqrt{\sin z}, dw = \frac{\cos z}{2\sqrt{\sin z}}dz$$
$$I= \int \frac{1}{1-w^4}dw=\frac{1}{2}\int \frac{1}{1-w^2} + \frac{1}{1+w^2} dw$$
$$=\frac{1}{2} \int \frac{1}{2} \left[ \frac{1}{1-w}+ \frac{1}{1+w} \right] + \frac{1}{1+w^2} dw$$
$$= \frac{1}{4} \ln \frac{1+w}{1-w} + \frac{1}{2} \tan^{-1} w +\mathcal{C}$$
$$= \frac{1}{4} \ln \left( \frac{1+ \sqrt{\sin z} }{1- \sqrt{\sin z}}\right) + \frac{1}{2} \tan^{-1} (\sqrt{\sin z}) +\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{1+ \sqrt{ \sin ( \tan^{-1} x^2 )}}{1-\sqrt{ \sin ( \tan^{-1} x^2 )}}\right) + \frac{1}{2} \tan^{-1} \left( \sqrt{ \sin ( \tan^{-1} x^2 )} \right) +\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{1+ \sqrt{ \frac{x^2}{\sqrt{1+x^4}}}}{1- \sqrt{ \frac{x^2}{\sqrt{1+x^4}}}}\right) + \frac{1}{2} \tan^{-1} \left( \sqrt{ \frac{x^2}{\sqrt{1+x^4}}} \right)+\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{\sqrt[4]{1+x^4}+ x}{\sqrt[4]{1+x^4}-x}\right) + \frac{1}{2} \tan^{-1} \left( \frac{x}{\sqrt[4]{1+x^4}} \right)+\mathcal{C}$$
|
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|
Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$ Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$
For $n=1$ inequality holds.
For $n=k$
$$\left(\frac{1}{k+1}+\frac{1}{k+2}+...\frac{1}{2k}\right)^2<\frac{1}{2}$$
For $n=k+1$
$$\left(\frac{1}{k+2}+\frac{1}{k+3}+...\frac{1}{2k+2}\right)^2<\frac{1}{2}$$
I don't know how to prove this.
Is it better to use direct method or contradiction?
|
$$ a_n = H_{2n}-H_{n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} $$
is an increasing sequence, since:
$$ a_{n+1}-a_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}>0$$
hence:
$$ \forall n\geq 1,\qquad a_n \leq \lim_{n\to +\infty}a_n = \log(2) $$
and the problem boils down just to showing that $\log^2(2)<\frac{1}{2}$.
That follows from:
$$\log(2) = \int_{0}^{1}\frac{dx}{1+x}< \sqrt{\int_{0}^{1}\frac{dx}{(1+x)^2}} = \frac{1}{\sqrt{2}}.$$
|
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|
Olympiad problem about finding minimum value with $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$ Let $x,y,z$ be positive real numbers such that $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$.
Find the minimum value of
$$\frac{x^2y^2} {z^3(x^2+y^2)}+\frac {y^2z^2} {x^3(y^2+z^2)}+\frac {z^2x^2} {y^3(z^2+x^2)}$$
I'm pretty sure that the answer would be $\frac {\sqrt {3}} {2}$, when all parameters are $\sqrt {3}$. But I couldn't prove it after some hours of thinking. So can anyone help me?
Any help would be welcome.
Thanks:D.
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Let
$$a=\dfrac{1}{x},b=\dfrac{1}{y},c=\dfrac{1}{z},a^2+b^2+c^2\ge 1$$
Use Cauchy-Schwarz
$$\sum_{cyc}\dfrac{x^2y^2}{z^3(x^2+y^2)}=\sum_{cyc}\dfrac{c^3}{(a^2+b^2)}\ge \dfrac{(a^2+b^2+c^2)^2}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}$$
since
$$c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)\le\dfrac{2}{3}(a+b+c)(a^2+b^2+c^2)\le\dfrac{2}{\sqrt{3}}(a^2+b^2+c^2)^{\frac{3}{2}}$$
so
$$LHS\ge \dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\ge\dfrac{\sqrt{3}}{2}$$
|
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|
Simple integral: $\int \frac {1+r}{-r^2+r-1} dr$. I was solving a much longer exercise, and while solving an ODE, I got this integral $$\int \frac {1+r}{-r^2+r-1} dr$$
I think this must be pretty simple, but I couldn't solve it, my substitutions didn't work and the polynomial in the denumerator does not have real roots, so partial frac. decomp. looks quite messy.
How are do you solve the general integral:
$$
\int \frac {r+a}{dr^2+er+f}dr
$$
When the denumerator has no real roots?
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$r^2-r+1 = \frac{r^3+1}{r+1}$, hence $r^2-r+1=(r-\omega)(r-\omega^2)$, with $\omega=\exp\left(\frac{2\pi i}{6}\right)$.
By the residue theorem:
$$ \text{Res}\left(\frac{r+1}{r^2-r+1},r=\omega\right) = \frac{\omega+1}{2\omega},$$
$$ \text{Res}\left(\frac{r+1}{r^2-r+1},r=-\omega\right) = \frac{-\omega+1}{-2\omega},$$
hence the partial fraction decomposition is given by:
$$ \frac{r+1}{r^2-r+1} = \frac{1+\omega}{2\omega}\cdot\frac{1}{r-\omega}-\frac{1-\omega}{2\omega}\cdot\frac{1}{r+\omega}. $$
We may compute the primitive through the previous line, or through the trick:
$$ \frac{4r+4}{4r^2-4r+4} = \frac{1}{2}\frac{2r-1}{r^2-r+1}+\frac{2}{1+\frac{1}{3}(2r-1)^2}$$
that gives:
$$ \int \frac{r+1}{r^2-r+1}\,dr = \frac{1}{2}\log(r^2-r+1)+\sqrt{3}\arctan\left(\frac{2r-1}{\sqrt{3}}\right)+C.$$
|
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|
Parametrization of $a^2+b^2+c^2=2d^2$ Is a complete parametrization of primitive solutions to the equation
$$a^2+b^2+c^2=2d^2\qquad a,b,c,d \in \mathbb{Z}$$
known? A reference would be great.
Solutions to $a^2+b^2=c^2$ give solutions to the equation above, but I know that there are other solutions.
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Look for a single rational solution to $x^2+y^2+z^2=2$. Say, $(x_0,y_0,z_0)=(-1,0,1)$.
Then take any three integers $(u,v,w)$ and seek solutions of form $(-1+tu,tv,1+tw)$. You get:
$$1-2tu+t^2u^2 + t^2v^2 + 1+2tw + t^2w^2=2$$
solving, excluding $t=0$, you get:
$$t=\frac{2(u-w)}{u^2+v^2+w^2}$$
Substituting that bach in, we get:
$$\begin{align}
x&=-1+tu = \frac{-(u^2+v^2+w^2)+2(u-w)u}{u^2+v^2+w^2}&=&\frac{u^2-2uw-v^2-w^2}{u^2+v^2+w^2}\\
y&=tv &=& \frac{2(u-w)v}{u^2+v^2+w^2}\\
z&=1+tw &=&\frac{u^2+v^2-w^2 +2uw}{u^2+v^2+w^2}
\end{align}$$
Then $$(a,b,c,d)=(u^2-2uw-v^2-w^2,2(u-w)v,u^2+v^2-w^2+2uw,u^2+v^2+w^2)\tag{1}$$
When $w=0$, this gives: $(a,b,c,d)=(u^2-v^2,2uv,u^2+v^2,u^2+v^2)$, which is the result you note that if $(a,b,c)$ is a Pythagorean triple then $(a,b,c,c)$ is a solution of your equation. (I had to pick the $(x_0,y_0,z_0)$ carefully to make that work.
The formula (1) should yield all integer solutions, up to constant multiples.
To ensure a primitive solution, you need that $u+v+w$ is odd and that $u-w$ and $v^2+2u^2$ are relatively prime. That is equivalent to $\gcd(2(u-w),u^2+v^2+w^2)=1$, which is saying that $t$ above is in lowest common denominator form.
As Will noted in comments, this doesn't give all primitive solutions directly, since:
$$9^2+4^2+1^2=2\cdot 7^2$$
is a solution, and there is no way to write $7$ as the sum of three squares.
Instead, we get: $21=4^2+2^2+1^2$, and we get a solution:
$$(27,12,3,21)$$ from my formula, which is obviously not primitive.
|
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|
Summing reciprocal logs of different bases I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$$
\frac{1}{\log_{2}1 + \log_{2}2 + \log_{2}3 + \dots + \log_{2}50}
$$
How do the fractions with such irrational values become $1$? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?
|
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
$$=\log_{50!}2+\log_{50!}3+\log_{50!}4+...+\log_{50!}50$$
$$=\log_{50!}(2\cdot3\cdot4\cdot...50)$$
$$=\log_{50!}(50!)$$
$$=1$$
|
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|
Writing an integer as a sum of two square in many ways, with consecutive arguments Let $n\in{\mathbb N}$. I call $n=x_1^2+y_1^2=x_2^2+y_2^2=\ldots x_r^2+y_r^2$
(where $(x_1,y_1),(x_2,y_2),\ldots,(x_r,y_r)$ are distinct uples
in ${\mathbb N}^2$) a multi-decomposition of $n$, of length $r$. If the
$x_1,x_2,\ldots,x_r$ form an arithmetic progression of step $1$, I say
that the multi-decomposition is connected.
For example,
$$
1105=\left\lbrace\begin{array}{lcl}
31^2 &+& 12^2 \\
32^2 &+& 9^2 \\
33^2 &+& 4^2 \\
\end{array}\right.
$$
is a connected multi-decomposition (CMD in abbreviated notation) of length three
for $n=1105$.
With the help of my computer, I found out that for $n\leq 10^6$ there is
no CMD of length four, and the $n$'s having CMD of length three are exactly
the following : $$25, 1105, 12025, 21025, 66625, 252601, 292825, 751825.$$
My questions are :
*
*Are there CMD's with length larger than three ?
*Is there a simple description of all $n$'s having a CMD of length three ?
|
Although the answer is not quite on the topic.
$$x_1^2+y_1^2=x_2^2+y_2^2=x_3^2+y_3^2=x_4^2+y_4^2$$
For the equation is possible to write General standard formula - moreover it is symmetric. Interestingly, so monjo to record for any number of summands.
$$x_1=(a^2+b^2)(t^2+k^2)(n^2+q^2)p$$
$$y_1=(a^2+b^2)(t^2+k^2)(n^2+q^2)s$$
$$x_2=(a^2+b^2)(n^2+q^2)(2tkp-(t^2-k^2)s)$$
$$y_2=(a^2+b^2)(n^2+q^2)((t^2-k^2)p+2tks)$$
$$x_3=(a^2+b^2)(t^2+k^2)(2nqp-(n^2-q^2)s)$$
$$y_3=(a^2+b^2)(t^2+k^2)((n^2-q^2)p+2nqs)$$
$$x_4=(t^2+k^2)(n^2+q^2)(2abp-(a^2-b^2)s)$$
$$y_4=(t^2+k^2)(n^2+q^2)((a^2-b^2)p+2abs)$$
|
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Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
MyApproach
Since I cannot form pattern above
I simplified this as $21^3$ = $3^3$ $. $ $7^3$
Similarly I did $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)=$9$ $.$ $1072$/ $2^5$.
From solving this I get Remainder as $1$
And Similarly on solving $25^3$+$23^3$ separately on dividing by $96$.I get remainder as $73$ and $71$
On adding and Finding remainder I get Remainder as $49$.
I am getting wrong Ans.
Please correct me how to approach towards the problem.
|
This is essentially a simpler version of lab's answer:
For any $a,b$, $(a-b)^3+(a+b)^3 = 2a^3 + 6ab^2 \equiv 0 \mod a$ for all $a,b$. In you case, you have two such pairs with $a=24$ (and $b=1,3$ respectively). So the answer is $0$.
|
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|
Find the least squares solution of the linear system...... Hi I'm looking for the least squares solutions of....
$$ \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
So Assuming this goes by $ A\vec{x}=\vec{b}$
Then using.... $A^T A \vec{x} = A^T \vec{b} $
$$ \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \vec{x}= \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
Which solves to be
$$ \begin{pmatrix} 3&-3 \\ -3&5 \end{pmatrix} \vec{x} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} $$
which i solve to get
$$ \vec{x} = \begin{pmatrix} -7/3 \\ 3/2 \end{pmatrix} $$
|
The normal equations are correct:
$$
\mathbf{A}^{*}\mathbf{A} x = \mathbf{A}^{*}b.
$$
The error is in the solution which should be
$$
x = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}b.
$$
The inverse is
$$
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1} {6}
\left(
\begin{array}{cc}
5 & 3 \\
3 & 3 \\
\end{array}
\right)
$$
which provides
$$
x = -\frac{1}{2}
\left(
\begin{array}{r}
5 \\
1
\end{array}
\right).
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1510590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
|
$$
n\log\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)=1+\frac xn-\frac1{2n}+O\left(\frac1{n^2}\right)
$$
Therefore,
$$
\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)^n=e\left(1+\frac xn-\frac1{2n}\right)+O\left(\frac1{n^2}\right)
$$
Thus
$$
\begin{align}
&\frac{\overbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}^{x=1}-\overbrace{\left(1+\frac1n-\frac1{n^2}\right)^n}^{x=-1}}{2\underbrace{\left(1+\frac1n+\frac1{n^2}\right)^n}_{x=1}-\underbrace{\left(1+\frac1n-\frac1{n^2+1}\right)^n}_{x=-1}-\underbrace{\left(1+\frac1n-\frac1{n^2(n^2+1)}\right)^n}_{x=0}}\\
&=\frac{e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}{2e\left(1+\frac1n-\frac1{2n}\right)-e\left(1-\frac1n-\frac1{2n}\right)-e\left(1-\frac1{2n}\right)+O\left(\frac1{n^2}\right)}\\[6pt]
&=\frac{\frac2n+O\left(\frac1{n^2}\right)}{\frac3n+O\left(\frac1{n^2}\right)}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve the system of conguruences: $x = 2 \bmod7, x = 5 \bmod12, x = 8 \bmod 25$ Question:
Solve the system of conguruences: $x = 2 \bmod7, x = 5 \bmod12, x = 8 \bmod 25$
So far I have
The solution: $X= B_1X_1C_1 + B_2X_2C_2 + B_3X_3C_3$ and that the answer is -something- $\mod2100$
where
$B_1 = 300, B_2 = 175, B_3 = 84$
and
$C_1=2, C_2=5, C_3=8$
and
$300\times X_1 = 1 \mod7 \Rightarrow 6\times X_1 = 1 \mod7$
$175\times X_2 = 1 \mod12 \Rightarrow 7\times X_2 = 1 \mod12$
$84\times X_3 = 1 \mod25 \Rightarrow 9\times X_3 = 1 \mod25$
When I attempt to solve these for $X_1$, $X_2$ and $X_3$, the solution $X$ is either a lot bigger than 2100 or a negative number...
|
Let us first begin by trying to reduce this to a system of two congruencies.
Focus our attention first to the following two:
$\begin{cases} x\equiv 2\pmod{7}\\ x\equiv 5\pmod{12}\end{cases}$
We know that $\gcd(7,12)=1$ and so there should be some $a,b$ such that $7a+12b=1$. We can find the $a$ and $b$ necessary as a byproduct of running the Euclidean Division Algorithm:
$\begin{array}{c|c|c}
12&1&0\\
\hline
7&0&1\\
\hline
5&1&-1\\
\hline
2&-1&2\\
\hline
1&3&-5\end{array}$
Implying that $3\cdot 12 + (-5)\cdot 7 = 1$ (indeed, $36-35=1$)
Note that this implies that $3\cdot 12 \equiv 1\pmod{7}$ and that $(-5)\cdot 7\equiv 1\pmod{12}$
Now, we can use this information to get:
$x\equiv 2\cdot 12\cdot 3 + 5\cdot (-5)\cdot 7\pmod{84}$
simplifying to
$x\equiv -103\equiv 65\pmod{84}$
(check: $65+84n\equiv 2\pmod{7}$ and $65+84n\equiv 5\pmod{12}$)
Continue the problem by combining the new congruence with the remaining congruence:
$\begin{cases}x\equiv 65\pmod{84}\\
x\equiv 8\pmod{25}\end{cases}$
It will combine to give you a solution to $x\equiv y\pmod{2100}$
Begin by noting that $14\cdot 84 -47\cdot 25 = 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
|
Using analysis: define $f(x)=x^3+x$. Since $f'(x)=3x^2+1>0$, the function is strictly increasing, hence, injective, and the result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us
$0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I still cannot grasp why summing, say, 4C0+4C1+4C2+4c3+4C4=2^4.
|
This can also be proven by induction. In order to do so, it is very helpful to prove the following identity:
$$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$
This can be proven combinatorially, but for the sake of expediency,
\begin{align*}
\binom{n}{k} + \binom{n}{k+1} &= \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\
&= \frac{n!}{k!(n-k)(n-k-1)!} + \frac{n!}{(k+1)k!(n-k-1)!} \\
&= \frac{n!}{k!(n-k-1)!}\left(\frac{1}{n-k} + \frac{1}{k+1}\right) \\
&= \frac{n!}{k!(n-k-1)!}\left(\frac{k+1}{(n-k)(k+1)} + \frac{n - k}{(n-k)(k+1)}\right) \\
&= \frac{n!}{k!(n-k-1)!} \cdot \frac{n+1}{(n-k)(k+1)} \\
&= \frac{(n+1)n!}{(k+1)k!(n-k)(n-k-1)!} \\
&= \frac{(n+1)!}{(k+1)!(n-k)!} \\
&= \binom{n+1}{k+1}.
\end{align*}
Now, let's launch into the induction proof. First, note that $\binom{0}{0} = 1 = 2^0$, so the base case holds.
Next, assume that $\binom{n}{0} + \ldots + \binom{n}{n} = 2^n$. Then, using the identity,
\begin{align*}
&\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\
= \, &\binom{n+1}{0} + \left(\binom{n}{0} + \binom{n}{1} \right) + \left(\binom{n}{1} + \binom{n}{2} \right) + \ldots + \left(\binom{n}{n-1} + \binom{n}{n} \right) + \binom{n+1}{n+1}.
\end{align*}
Note that each term of the form $\binom{n}{k}$ where $1 \le k \le n-1$ occurs exactly twice in the above sum. Also, the other terms, $\binom{n+1}{0}, \binom{n}{0}, \binom{n+1}{n+1}, \binom{n}{n}$ are all $1$. Thus we have,
\begin{align*}
&\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\
= \, &2\left(\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} \right) = 2 \cdot 2^n = 2^{n+1}.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Is every integer a mixed sum of three squares? Lagrange's four-square theorem states that every natural number can be represented as the sum of four integer squares $n = a^2 + b^2 + c^2 + d^2$.
Question: Is every integer a mixed sum of three integer squares $n = \pm a^2\pm b^2 \pm c^2$ ?
Note that the signs are independently positive or negative, for example $28 = 36-9+1$.
|
You can write every number $n$ in the form $a^2+b^2-c^2$. Just pick $a$ so that $n-a^2$ is odd and then solve
$$\begin{align}
b+c&=n-a^2\\
b-c&=1
\end{align}$$
for $b$ and $c$:
$$\begin{align}
b&={n-a^2+1\over2}\\
c&={n-a^2-1\over2}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1522470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Bounds for $S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}$ Let $x,y,z\in[0,2]$ and $x+y+z=3$. What are the maximum and minimum of $$S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}?$$
When $x=y=z=1$, we have $S=3\sqrt{2}$.
For the upper bound, we can use the AM-GM inequality to get $$S\leq\frac{1}{2}(x+y+1+y+z+1+z+x+1)=\frac{9}{2},$$ but equality cannot hold since we need $x=y+1$, etc. simultaneously, which cannot happen.
|
A little numerical investigation gives that the maximum is attained for $x=y=z=1$ and the minimum for $x=1,y=2$ and $z=0$. An intuitive proof of this fact can be achieved by noticing that $F(x,y,z) = \sum_{cyc} \sqrt{x(y+1)}$ is concave in every variable and thus, the minimum should be attained at extremal points and the maximum in the interior of the domain.
To prove rigorously the upper bound apply Cauchy Schwarz inequality to see that
$$ S^2 \leq (x+y+z)(y+1+z+1+x+1) = 18.$$
This implies that $S \leq 3\sqrt{2}$ and thus the maximum is $3\sqrt{2}$ and is attained only for $x=y=z=1$.
In order to find the minimal value we could suppose that $x \leq y \leq z$ and we look what happens if we replace the three values by $x-\varepsilon\leq y \leq z+\varepsilon$. We define
$$ f(\varepsilon)=\sqrt{(x-\varepsilon)(y+1)}+\sqrt{y(z+\varepsilon+1)}$$
and we note that
$$ f'(\varepsilon) = -\sqrt{\frac{y+1}{x-\varepsilon}}+\sqrt{\frac{y}{z+\varepsilon+1}} $$
Note that since $\frac{y+1}{x}>\frac{y}{z+1}$ (simple consequence of the ordering) for small $\varepsilon$ the derivative is negative. Thus $f(\varepsilon)<f(0)$. This means that increasing the distance between $x$ and $z$ decreases the value of the function.
We can choose $\varepsilon$ as large as the domain allows it until we reach an extremal case: $x=0$ or $z=2$.
Case 1. $x=0$. Then $y+z=3$ and we want to minimize $g(y,z) = \sqrt{y(z+1)}+\sqrt{z}$. This is a one dimensional problem. It is not hard to see that the same principle applies. If we enlarge the distance between $y$ and $z$ then the value of the function decreases. Thus the minimum is attained for $y=1,z=2$.
Case 2. $z=2$. The $x+y=1$ and we want to minimize $g(x,y) = \sqrt{x(y+1)}+\sqrt{3y}+\sqrt{2(x+1)}$. The same type of analysis applies and we get that the minimum is found for $x=0,y=1$.
Thus the minimal value is equal to $3$ and is attained for $x=0,y=1,z=2$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $f(x)=x^4$ is continuous on $[0,\infty)$ I understand that we want $\varepsilon$ so that $|x^2+y^2||x+y||x-y|<\varepsilon$, and for $|x+y||x-y|<\varepsilon$ I can choose $\delta$ so that $\delta(2|y|+\delta)<\varepsilon$, but I don't know what to do with the $|x^2+y^2|$ part.
|
If $y=0$, then $|x^4-y^4|=x^4< \epsilon$ whenever $|x-y|=|x|<\epsilon^{1/4}$. So, take $\delta=\epsilon^{1/4}$.
If $y\ne 0$, then take $|x-y|<y/2$ so that $y/2<x<3y/2$.
Then, we have
$$\begin{align}
|x^4-y^4|&=|x-y|(x+y)(x^2+y^2)\\\\
&<|x-y|\left(\frac52 y\right)\left(\frac{13}{4}y^2\right)\\\\
&=|x-y|\frac{65}{8}y^3\\\\
&<\epsilon
\end{align}$$
whenever $|x-y|<\frac{8}{65y^3}\epsilon$.
So, let $\delta =\min\left(\frac{y}{2},\frac{8}{65y^3}\epsilon\right)$. Then, for all $\epsilon>0$, $|x^4-y^4|<\epsilon$ for any $x>0$,
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Value of the limit without (or with, but giving rigorous arguments) using the Taylor expansion of sin I'm trying to evaluate the limit as $N\to \infty.$
$$\frac{ \left(\dfrac{\sin \frac{1}{N}} {\frac{1}{N}}\right)^{N} -1 }{\frac{1}{N}}.$$
Note first that, using L'Hôpital, one can easily show that the numerator goes to $0.$
Using the Taylor series expansion for $sin$, the value of the actual limit seems to be $-\frac{1}{6}$. But I'm not fully sure how to justify the infinite series in the numerator is $-\frac{1}{6N}+O(\frac{1}{N^2})$. You could either justify that, if I was right, or may be use some other method to tell me what the limit is?
Many thanks in advance!
|
Just develop carefully the Taylor expansion.
$$\sin \frac{1}{N}=\frac{1}{N}-\frac{1}{6N^3}+O(\frac{1}{N^5})$$
Then
$$\left(\frac{\sin \frac{1}{N}}{\frac{1}{N}}\right)^N=\exp \left[N \log\left( 1-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left[N\left(-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left(-\frac{1}{6N}+O(\frac{1}{N^3})\right)=1-\frac{1}{6N}+O(\frac{1}{N^2})$$
And
$$\frac{\left(\frac{\sin \frac{1}{N}}{\frac{1}{N}}\right)^N-1}{\frac{1}{N}}
=N\left(-\frac{1}{6N}+O(\frac{1}{N^2})\right)=-\frac{1}{6}+O(\frac{1}{N})$$
Hence the limit as $N\to\infty$ is $-\frac{1}{6}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$? My integral is
$$I=\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$$
and hence
$$I=\int\frac{1}{(x-1)(x+1)}\sqrt[3] {\frac{x+1}{x-1}}dx $$
$\cos2\theta$ substitution wont be helpful here because of the cube root. Should I apply by parts ?
|
First notice that
$$\eqalign{
I &= \int {\root 3 \of {{1 \over {{{(x + 1)}^2}{{(x - 1)}^4}}}} } dx \cr
&= \int {{1 \over {\left( {x - 1} \right)\left( {x + 1} \right)}}\root 3 \of {{{x + 1} \over {x - 1}}} } dx \cr
&= \int {{1 \over {{{\left( {x - 1} \right)}^2}}}{{x - 1} \over {x + 1}}\root 3 \of {{{x + 1} \over {x - 1}}} } dx \cr} $$
Now, make the substitution
$$\eqalign{
u &= {{x + 1} \over {x - 1}} \cr
du &= - {2 \over {{{\left( {x - 1} \right)}^2}}}dx \cr} $$
to obtain
$$\eqalign{
I &= - {1 \over 2}\int {{1 \over u}\root 3 \of u du} \cr
&= - {1 \over 2}\int {{u^{ - {2 \over 3}}}du} \cr
&= - {3 \over 2}{u^{{1 \over 3}}} + C \cr
&= \boxed{ - {3 \over 2}{\left( {{{x + 1} \over {x - 1}}} \right)^{{1 \over 3}}} + C} \cr} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving a trigonometric inequality in a triangle In $\Delta ABC$, prove that $$\sin\left(\frac{\pi-A}{4}\right) \sin\left(\frac{\pi-B}{4}\right)\sin\left(\frac{\pi-C}{4}\right) \geq \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ Some ideas on how to start manipulating. I tried breaking it in sum, but it led to nothing. Thanks.
|
Hint: In $\triangle{ABC}$ we have,
\begin{align*}\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}} &=1+4\sin{\left(\frac{B+C}{4}\right)}\sin{\left(\frac{A+C}{4}\right)}\sin{\left(\frac{A+B}{4}\right)}\\
\cos{A}+\cos{B}+\cos{C}&=1+4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}.\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Basic proofs for epsilon delta limits: $\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$ Hello I am wondering if my approach and understanding of the epsilon delta definition is correct.
For example, if I wanted to use it to show that $$\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$$ for $a \in \mathbb{R}, a \neq -1$
Then here are my thoughts;
I know that we want that for all $\epsilon \gt 0$ such that $0 \lt |x-a| \lt \delta$ , then $|f(x)-L| \lt \epsilon$ for a suitable $\delta$.
My attempt:
If $|x-a| \lt a$ then $-a \lt x-a \lt a$ and $0 \lt x \lt 2a$
so $1 \lt x+1 \lt 2a+1$
so $\frac{1}{x+1} \lt 1$
and hence,
$|\frac{x}{1+x}-\frac{a}{1+a}|$=$\frac{|x-a|}{|x+1||a+1|} \lt \frac{|x-a|}{a+1}$
So then we could choose $$\delta=\min\{a,\epsilon|a+1|\}$$
and this would prove that for all $|\frac{x}{x+1}-\frac{a}{a+1}| \lt \epsilon$,
$|x-a| \lt \delta$ ?
How does this seem? Does it make sense? If there are mistakes, what should I consider?
|
You write
we want that for all $\epsilon \gt 0$ such that $0 \lt |x-a| \lt \delta$ , then $|f(x)-L| \lt \epsilon$ for a suitable $\delta$.
What you actually want is: For every $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all $x$, $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$. Also, since $a$ can be zero or negative, you cannot start with the assumption $|x - a| < a$. The $\delta$ you have chosen can be negative for the same reasons.
Using the fact that
$$\frac{x}{1 + x} = 1 - \frac{1}{1 + x}$$
for all $x\neq -1$, we have
$$\left|\frac{x}{1 + x} - \frac{a}{1 + a}\right| = \left|\frac{1}{1 + a} - \frac{1}{1 + x}\right| = \left|\frac{x - a}{(1 + a)(1 + x)}\right| = \frac{|x - a|}{|1 + a||1 + x|}.$$
for all $x\neq -1$. Now since $a \neq -1$, $|1 + a| > 0$. So we may consider first $0 < |x - a| < \frac{|1 + a|}{2}$. By the triangle inequality,
$$|1 + a| = |(1 + x) - (x - a)| \le |1 + x| + |x - a|,$$
thus
$$|1 + x| \ge |1 + a| - |x - a| > |1 + a| - \frac{|1 + a|}{2} = \frac{|1 + a|}{2}.$$
Hence if $0 < |x - a| < \frac{|1 + a|}{2}$, then
$$\frac{|x - a|}{|1+a||1 + x|} < \dfrac{|x - a|}{|1+a|\cdot \frac{|1+a|}{2}} = \frac{2}{(1 + a)^2}|x - a|,$$
which can be made less than $\epsilon$ by having $|x - a| < \frac{\epsilon(1 + a)^2}{2}$. So choose
$$\delta = \min\left\{\frac{|1+a|}{2},\frac{\epsilon(1+a)^2}{2}\right\}.$$
|
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|
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
|
Using $x=e^{\ln x}$ we take the log. So the limit is the exponential of the limit of the log, which is the limit of $\frac{n-1}{n+1}\ln(1-\frac{1}{n^2+1}$. That fraction goes to zero, so the log is asymptotic to it, hence the limit is the limit of $\frac{n-1}{n+1}\frac{1}{n^2+1}$, which is easily seen to be 0. Hence yout limit is 1.
Sorry for the wordy answer but typing the passages in MathJax from my mobile... I decided to spare myself such an ordeal :). If I remember, I will edit this tomorrow.
There is, however, a far easier way: the base and exponent both tend to 1, and $1^1$ is no indeterminate form, it is precisely 1.
Update
Here are the above passages in symbols:
$$\lim_{n\to\infty}\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}=\lim_{n\to\infty}e^{\ln\left(\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}\right)}=e^{\lim_{n\to\infty}\frac{n-1}{n+1}\ln\frac{n^2}{n^2+1}}=e^A,$$
and:
\begin{align*}
A={}&\lim_{n\to\infty}\frac{n-1}{n+1}\ln\left(1-\frac{1}{n^2+1}\right)=\lim_{n\to\infty}\frac{n-1}{n+1}\left(-\frac{1}{n^2+1}\right)={} \\
{}={}&-\left(\lim_{n\to\infty}\frac{n-1}{n+1}\right)\cdot\left(\lim_{n\to\infty}\frac{1}{n^2+1}\right)=-1\cdot0=0,
\end{align*}
hence, the desired limit is:
$$e^A=e^0=1.$$
Faster way:
$$\lim_{n\to\infty}\left(\frac{n^2}{n^2+1}\right)^{\frac{n-1}{n+1}}=\left(\lim_{n\to\infty}\frac{n^2}{n^2+1}\right)^{\lim_{n\to\infty}\frac{n-1}{n+1}}=1^1=1.$$
|
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|
Limits of finite sums I know that: $$\int_0^1 1 - x^2 dx = \frac{2}{3}$$
And that represents the area below the curve, delimited by the lines $x= 0$ and $x = 1$
But $$\int_0^1 1 - x^2 dx = \lim_{n \to \infty} \left( 1 - \frac{2n^3 + 3n^2 +n}{6n^3} \right) $$
And this limit is obtained by calculating an approximation of the inferior sum using $n$ rectangles of width $\Delta x = (1 - 0)/n$ and than making $n \to \infty$
The sum is:
$$\left[ f \left( \frac{1}{n} \right) \right] \left( \frac{1}{n} \right) + \left[ f \left( \frac{2}{n} \right) \right] \left( \frac{1}{n} \right) + \dots \left[ f \left( \frac{k}{n} \right) \right] \left( \frac{1}{n} \right) + \dots
\left[ f \left( \frac{n}{n} \right) \right] \left( \frac{1}{n} \right) =$$
$$ = \sum_{k = 1}^n f \left( \frac{k}{n} \right) \left( \frac{1}{n} \right) = \sum_{k = 1}^n \left( 1 - \left( \frac{k}{n} \right)^2 \right) \left( \frac{1}{n} \right)$$
A) Determine the superior sum.
I thought that:
$$ \sum_{k = 0}^n f \left( \frac{k - 1}{n} \right) \left( \frac{1}{n} \right) $$
Could represent this, am I correct?
B) I've tried to generalize the inferior sum for the interval $[a,b]$, and I found it:
$$\sum_{k = 1}^n f \left( \frac{k(b - a)}{n} \right) \left( \frac{b - a}{n} \right) $$
But when I calculate the sum as $n \to \infty$, it's not equal to $\int_a^b 1 - x^2 dx $.
Is this a correct sum or I'm doing something wrong?
|
You're not quite summing it right. If, for example, we use $a=2, b=3$ in:
$$\sum_{k = 1}^n f \left( \dfrac{k(b - a)}{n} \right) \left( \dfrac{b - a}{n}\right)$$
the function in the sum runs from $\dfrac1n$ to $1$.
Instead use:
$$\sum_{k = 1}^n f \left( a+\dfrac{k(b - a)}{n} \right) \left( \dfrac{b - a}{n}\right)$$
Your superior sum looks fine.
|
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|
trigonometry: why this is not a triangle?
As the photo shown, angle x doesn't nesscarry to be 90 degree.But when y=0, x=90 degree.
So here is my problem, what happen if I increase y, does x becomes larger (greater than 90) or less than 90 degree?
My attempt:
My reasoning is that when you increase y (say set y=1). Then the base (12-y) becomes shorter, and the height (5+y) becomes taller. So that you can either swing (5+y) clockwise or counter clockwise. Therefore, angle x could be either greater than 90 or less than 90. But the solution says I am wrong.
What are some intuitive way to explain this without using fancy equations?
|
Suppose that $y$ is a little more than $0$, so that the base is a little less than $12$. What height would be needed in order to have a right triangle with base $12-y$ and hypotenuse $13$? The Pythagorean theorem says that is would have to be
$$\sqrt{13^2-(12-y)^2}=\sqrt{25+24y-y^2}\;.$$
Is this more or less than $5+y$? If it’s more, $x$ must be more than $90°$: the $5+y$ side must tip over to the left as the $13$ side drops down. If it’s less, $x$ must be less than $90°$: the $5+y$ side must tip to the right as the $13$ side rises.
Now $(5+y)^2=25+10y+y^2$, so the question is whether $25+24y-y^2$ is more or less than $25+10y+y^2$, i.e., whether $14y-y^2$ is more or less than $y^2$.
*
*Clearly $14y-y^2<y^2$ if and only if $14y<2y^2$, or $7y<y^2$. We need only consider $y$ such that $0<y<12$ (why?), so this occurs if $y>7$.
*It’s also clear that $14y-y^2>y^2$ if and only if $14y>2y^2$, or $7y>y^2$, which for $0<y<12$ occurs when $y<7$.
Thus, if we increase $y$ just a bit from $0$, we’ll be in the second case above, and $x$ will be more than $90°$. However, when $y=7$ the two sides will have changed places, and we’ll have a right triangle again. And as $y$ increases from $7$ towards (but not reaching) $12$, we’ll be in the first case above, and $x$ will be less than $90°$.
You can also use the cosine law, which says that
$$\begin{align*}
13^2&=(12-y)^2+(5+y)^2-2(12-y)(5+y)\cos x\\
&=169-14y+2y^2-2(60+7y-y^2)\cos x\;,
\end{align*}$$
i.e.,
$$(60+7y-y^2)\cos x=y^2-7y\;.$$
Moreover, the geometry makes it clear that $0\le y<12$, so $60+7y-y^2=(12-y)(5+y)\ne 0$. Thus,
$$\cos x=\frac{y^2-7y}{60+7y-y^2}=\frac{y(y-7)}{(12-y)(5+y)}\;.$$
You can now do a sign analysis. We know that $x=90°$ when $y=0$, so we’ll consider only $0<y<12$. On that interval $y>0$ and $12-y>0$, so the algebraic sign of $\cos x$ is the same as that of $\frac{y-7}{y+5}$. This is negative for $0<y<7$, $0$ at $y=7$, and positive for $7<y<12$. Thus, when you increase $y$ just a little from $0$, $\cos x$ becomes negative, and that can happen only if $x>90°$. When $y$ reaches $7$, however, $\cos x=0$, and we have the original right triangle, but with the base and height interchanged. And when $7<y<12$, $\cos x>0$, and we must have $x<90°$.
|
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|
Diagonal Scaling of $ A \in {\mathbb{R}}^{2 \times 2} $ Positive Definite and Its Conditional Number Given a Positive Definite Matrix $ A \in {\mathbb{R}}^{2 \times 2} $ given by:
$$ \begin{bmatrix}
{A}_{11} & {A}_{12} \\
{A}_{12} & {A}_{22}
\end{bmatrix} $$
And a Matrix $ B $ Given by:
\begin{bmatrix}
\frac{1}{\sqrt{{A}_{11}}} & 0 \\
0 & \frac{1}{\sqrt{{A}_{22}}}
\end{bmatrix}
Now, defining the Diagonal Scaling of $ A $ given by $ C = B A B $.
One could see the main diagonal elements of $ C $ are all $ 1 $.
Actually $ C $ is given by:
$$ C = \begin{bmatrix}
1 & \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} \\
\frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} & 1
\end{bmatrix} $$
This scaling, intuitively, makes the Matrix closer to the identity Matrix (Smaller off diagonal values, 1 on the main diagonal) and hence improve the Condition Number.
Yet I couldn't prove it.
How could one prove $ \kappa \left( C \right) = \kappa \left( B A B \right) \leq \kappa \left( A \right) $?
|
Since multiplying by a constant doesn't change the Condition Number one could make the choice of:
$$ A = \begin{bmatrix}
a & 1 \\
1 & b
\end{bmatrix} $$
Hence the matrix $ B $ becomes:
$$ B = \begin{bmatrix}
\frac{1}{\sqrt{a}} & 0 \\
0 & \frac{1}{\sqrt{b}}
\end{bmatrix} $$
And the Matrix $ C $ becomes:
$$ C = \begin{bmatrix}
1 & \frac{1}{\sqrt{ab}} \\
\frac{1}{\sqrt{ab}} & 1
\end{bmatrix} $$
Moreover, defining $ {\alpha}_{1} \geq {\alpha}_{2} $ are the Eigen Values of $ A $ and $ {\gamma}_{1} \geq {\gamma}_{2} $ are the Eigen Values of $ C $.
One should notice that $ \det \left( B \right) = \frac{1}{\sqrt{ab}} < 1 $ by the definition of $ C \succ 0 $ (By Quadratic Form of PD Matrix) or $ A \succ 0 $ which implies $ a b > 1 \Rightarrow \frac{1}{ab} < 1 $.
In addition since $ \kappa \left( A \right) = \frac{{\alpha}_{1}}{{\alpha}_{2}} $ and $ \det \left( A \right) = {\alpha}_{1} {\alpha}_{2} $ it follows that $ \det \left( A \right) \kappa \left( A \right) = {\alpha}_{1}^{2} $ and $ \det \left( A \right) {\det \left( B \right)}^{2} \kappa \left( C \right) = {\gamma}_{1}^{2} $.
Assuming the inequality holds:
$$ \kappa \left( C \right) \leq \kappa \left( A \right) \Rightarrow \det \left( A \right) \kappa \left( C \right) \leq \det \left( A \right) \kappa \left( A \right) $$
Since $ \det \left( A \right) > 0 $, now this implies:
$$ \frac{{\gamma}_{1}^{2}}{\det \left( B \right) \det \left( B \right)} \leq {\alpha}_{1}^{2} \Rightarrow {\gamma}_{1} \leq \det \left( B \right) {\alpha}_{1} $$
Namely, showing the above inequality of the Eigen Values is equivalent of proving the original inequality (Because all terms are positive).
One could see, using the Characteristic Polynomial of the matrix $ A $ and $ C $ that $ {\alpha}_{1} = \frac{a + b}{2} + \frac{\sqrt{{\left( a - b \right)}^{2} + 4}}{2}, \; {\gamma}_{1} = 1 + \frac{1}{\sqrt{ab}} $.
Working on the right hand of the inequality:
$$
\begin{align}
\det \left( B \right) {\alpha}_{1} & = \frac{1}{\sqrt{a b}} \left( \frac{a + b}{2} + \frac{\sqrt{{\left( a - b \right)}^{2} + 4}}{2} \right) \\
& = \frac{a + b}{2 \sqrt{a b}} + \sqrt{\frac{{\left( a - b \right)}^{2} + 4}{4 a b}} \\
& \geq 1 + \sqrt{\frac{{\left( a - b \right)}^{2} + 4}{4 a b}} \\
& \geq 1 + \sqrt{\frac{4}{4 a b}} \\
& = {\gamma}_{1}
\end{align}
$$
Where the first inequality comes from the Inequality of Arithmetic and Geometric Means and the second by removing a non negative term.
|
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$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$ $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$
I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?
|
Hint:
\begin{equation*}
\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\left( \frac{\sin x-x}{x^{3}}%
\right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2}
\end{equation*}
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\sin x+x}{x}\right) &=&1+1=2 \\
\lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2} &=&1^{2}=1.
\end{eqnarray*}
|
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|
Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$. $$
\int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x.
$$
What I have tried:
I tried writing denominator as
$ \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $
so the integral becomes,
$$
\int\frac{\sin x+\cos x}{1-\frac{\sin^2(2x)}{2}}\,\text{d}x.
$$
Anyone, how do I solve this further?
|
Let $$\displaystyle I = \int\frac{\sin x+\cos x}{\sin^4x +\cos^4 x}dx = \int\frac{\sin x+\cos x}{1-2\sin^2 x\cos^2 x}dx$$
So we get $$I = 2\int\frac{\sin x+\cos x}{2-(\sin 2x)^2}dx = 2\int\frac{\sin x+\cos x}{\left(\sqrt{2}-\sin 2x\right)\cdot \left(\sqrt{2}+\sin 2x\right)}dx$$
So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{\sqrt{2}+\sin 2x}+\frac{1}{\sqrt{2}-\sin 2x}\right]\cdot (\sin x+\cos x)dx$$
So we get $$I = \frac{1}{\sqrt{2}}\int \left[\frac{1}{1+\sqrt{2}-(\sin x-\cos x)^2}+\frac{1}{\sqrt{2}-1+(\sin x-\cos x)^2}\right]\cdot(\sin x+\cos x)dx$$
Now Put $(\sin x-\cos x) =t\;,$ Then $(\sin x+\cos x)dx = dt$
|
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|
Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. I know it is supposed to be $\Re z <{1\over 2}$, but unfortunately I've been trying to show this for too long and my sanity along with my concentration drops monotonically. I would really appreciate any help with this.
|
Let $z=x+iy$ with $x,y\in\mathbb{R}$. We have
$$\frac{z}{z-1}=\frac{x+\mathrm{i}y}{(x-1)+\mathrm{i}y}=\frac{x(x-1)+y^2}{(x-1)^2+y^2}-\mathrm{i}\frac{y}{(x-1)^2+y^2}$$
so
\begin{align}
\Re\frac{z}{z-1} & =\frac{x(x-1)+y^2}{(x-1)^2+y^2}\\
& =\frac{x^2+y^2-x}{x^2+y^2-2x+1}\\
& < \frac{1-x}{2-2x}\\
& =\frac{1}{2}\frac{1-x}{1-x}\\
& =\frac{1}{2}
\end{align}
whence $f\left(D\right)\subset\{z\in\mathbb{C}\mid \Re z<\frac{1}{2}\}$.
Now take $\omega=\left(\frac{1}{2}-\varepsilon\right)+\mathrm{i}y$ with $\varepsilon>0$ and $y\in\mathbb{R}$. We can check that $f$ is an involution, that is $f$ is bijective and $f^{-1}=f$, so we see that we can choose $z=\frac{\omega}{\omega-1}$ to have $f(z)=\omega$. Moreover, this choice is licit since
\begin{align}
|z| & =\left|\frac{\omega}{\omega-1}\right|\\
& =\frac{\left|\left(\frac{1}{2}-\varepsilon\right)+\mathrm{i}y\right|}{\left|-\left(\frac{1}{2}+\varepsilon\right)+\mathrm{i}y\right|}\\
& = \frac{\left(\frac{1}{2}-\varepsilon\right)^2+y^2}{\left(\frac{1}{2}+\varepsilon\right)^2+y^2}\\
& = \frac{\frac{1}{4}+\varepsilon^2+y^2-\varepsilon}{\frac{1}{4}+\varepsilon^2+y^2+\varepsilon}\\
& <1.
\end{align}
Hence, we have $\{z\in\mathbb{C}\mid \Re z<\frac{1}{2}\}\subset f\left(D\right)$.
|
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What is the splitting field of $(X^3-2)(X^3-3)(X^2-2)$ What is the splitting field of $$(X^3-2)(X^3-3)(X^2-2)\ \ ?$$
The roots are $$\{\sqrt[3]2,\ \ j\sqrt[3]2,\ \ j^2\sqrt[3]2,\ \ \sqrt[3]3,\ \ j\sqrt[3]3,\ \ j^2\sqrt[3]3,\ \ \sqrt 2,\ \ -\sqrt 2\}$$
where $j=e^{\frac{2i\pi}{3}}$. So it splitting field is included in $$\mathbb Q(j,\sqrt[3]2,\sqrt[3]3,\sqrt 2)$$
but is all the field ?
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Yes. It is clear that $(X^3-2)(X^3-3)(X^2-2)$ splits over $\mathbb Q(j,\sqrt[3]2,\sqrt[3]3,\sqrt 2)$. On the other hand, any splitting field of $(X^3-2)(X^3-3)(X^2-2)$ over $\Bbb{Q}$ must contain $\sqrt[3]2,\sqrt[3]3,\sqrt 2$ and $\frac{j\sqrt[3]2}{\sqrt[3]2}=j$.
|
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|
Evaluation of $ \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$ $\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$
Now Here $$\displaystyle I_{2} = \int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx.$$ Put $\displaystyle x=\frac{1}{t}.$ Then $\displaystyle dx = -\frac{1}{t^2}dt$ and changing limit
We get $$\displaystyle I_{2} = \int_{1}^{0}\frac{-t^3\cdot \ln t}{(1+t^2)^2}\cdot -\frac{1}{t^2}\,dt = \int_{1}^{0}\frac{t\ln t}{(1+t^2)^2}\,dt = -\int_{0}^{1}\frac{t\ln t}{(1+t^2)^2}\,dt$$
So we get $$\displaystyle I_{2} = -\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}dt = -I_{1}\Rightarrow I_{1}+I_{2} = 0$$
So we get $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx = I_{1}+I_{2} =0$$
My question is, can we solve it using another method? If so then please explain.
|
$$\int\frac{x\ln(x)}{(x^2+1)^2}\space\text{d}x=$$
Integrate by parts $\int f\text{d}g=fg-\int g\text{d}f$: $f=\ln(x),\text{d}g=\frac{x}{(x^2+1)^2}\space\text{d}x, \text{d}f=\frac{1}{x},g=\frac{1}{2(x^2+1)}$:
$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\frac{1}{x(x^2+1)}\space\text{d}x=$$
Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:
$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\int\frac{1}{u(u+1)}\space\text{d}u=$$
$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\int\left(\frac{1}{u}-\frac{1}{u+1}\right)\space\text{d}u=$$
$$-\frac{\ln(x)}{2(x^2+1)}-\frac{1}{4}\int\left(\frac{1}{u+1}-\frac{1}{u+1}\right)\space\text{d}u+\frac{1}{4}\int\frac{1}{u}\space\text{d}u=$$
Substitute $s=u+1$ and $\text{d}s=\text{d}u$:
$$-\frac{\ln(x)}{2(x^2+1)}-\frac{1}{4}\int\left(\frac{1}{u+1}-\frac{1}{s}\right)\space\text{d}s+\frac{1}{4}\int\frac{1}{u}\space\text{d}u=$$
$$-\frac{\ln(x)}{2(x^2+1)}-\frac{\ln(s)}{4}+\frac{\ln(u)}{4}+\text{C}=$$
$$-\frac{\ln(x)}{2(x^2+1)}-\frac{\ln(x^2+1)}{4}+\frac{\ln(x^2)}{4}+\text{C}=$$
$$\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)+\text{C}$$
Setting the boundaries:
*
*Zero:
$$\lim_{x\to 0}\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=$$
$$\frac{1}{4}\lim_{x\to 0}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=$$
$$\frac{1}{4}\left(2\lim_{x\to 0}x^2\ln(x)-\lim_{x\to 0}\ln(x^2+1)\right)=$$
$$\frac{1}{4}\left(2\cdot 0-0\right)=0$$
*Infinity:
$$\lim_{x\to\infty}\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=0$$
So:
$$\int_{0}^{\infty}\frac{x\ln(x)}{(x^2+1)^2}\space\text{d}x=\left[\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)\right]_{0}^{\infty}=0-0=0$$
|
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|
How to calculate this infinite sum? $$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$
I am not sure how to calculate the value of this summation. My working so far is as follows:
Let $S=\sum_ {n=0}^\infty \frac {1}{(4n+1)^2}$.
$\Longrightarrow S=\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{9^2}+...$
$\Longrightarrow S=(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...)-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{6^2}+...) $
$\Longrightarrow S=\zeta(2)-[(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}...)-(\frac{1}{5^2}+\frac{1}{9^2}+...)]$
$\Longrightarrow S=\zeta(2)-[(\zeta(2)-1)-(S-1)]$
$\Longrightarrow S=\zeta(2)-\zeta(2)+1+S-1$
$\Longrightarrow 0=0$
Does anyone have a better way of evaluating this that does not involve a cyclical answer as mine eventually does?
|
Step 1: The sum of all the reciprocal squares is $\zeta(2)=\frac{\pi^2}{6}$.
Step 2: The sum of all the reciprocal even squares is $\frac{1}{4}\zeta(2)=\frac{\pi^2}{48}$.
Step 3: Their difference, the sum of all the reciprocal odd squares, is $\frac{\pi^2}{8}$.
Step 4: Catalan's constant is $G=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}\cdots$
Step 5: Adding the results from steps 3 and four, and dividing by 2, gives the desired series. It also gives the sum $$\frac{1}{2}\left(\frac{\pi^2}{8}+G\right)$$
There is not much known about $G$, but it's also $\beta(2)$, the Dirichlet beta function. Hence if you wanted an answer in terms of special functions, the sum is $$
\frac{3}{8}\zeta(2)+\frac{1}{2}\beta(2)$$
|
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|
Differentiate an exponential integral Would you guide me differentiating this integral for $m$:
$$\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x$$
noting that $a , b >0$
|
$$I_1=\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x=$$
$$I_1=\frac{e^{-\frac{\left(4a^2-b^2c^2\right)(a-m)^2}{4a^2b^2}}\cdot\sqrt{\pi}\left(1+c\sqrt{\frac{1}{c^2}}\cdot\text{Erf}\left(\frac{a}{c}-\frac{c}{2}+\frac{cm}{2a}-\frac{n}{c}\right)\right)}{2\sqrt{\frac{1}{c^2}}}=$$
$$I_1=\frac{c\sqrt{\pi}e^{-\frac{\left(4a^2-b^2c^2\right)(a-m)^2}{4a^2b^2}}\left(1+\text{Erf}\left(\frac{2a^2-ac^2-2an+c^2m}{2ac}\right)\right)}{2}$$
$$I_1=\frac{\text{d}}{\text{d}n} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x=$$
|
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|
estimating absolute value of complex function Let $f(z) = \frac{1}{z^2 - 2z + 4 }$ where $|z| = R > 1 $. How can I estimate $|f(z)|$ to get an upper bound of the form $\frac{ M}{|z|^p }$ where $M > 0$ and $p>1$ ? I know
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|z^2 - 2z|} = \frac{1}{|z||z-2|}$$
and here is where I got Stuck. Any ideas?
|
Clearly $|z^2 -2z+4|=|z^2-2(z-2)|=|2(z-2)-z^2|\geq ||2z-4|-|z^2||$.
$\Rightarrow$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z-4|-|z^2| |}$$
So by triangle inequality we have that $|2z-4|\geq |2z|-|4|$
Since $|z|=R>1$ , there exists $a \in \mathbb R^+$ such that $4=a|z|$ and $a>1$ .
Therefore $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z|-|4|-|z^2| |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{\left | |2z|-a|z|-|z^2| \right |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }$$
So $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }=\frac{1}{||z^2|-(2-a)|z|| }$$
Since $|z|>1$ , $|z|^2 >|z|$.
So $||z^2|-(2-a)|z|| \geq ||z^2|-(2-a)|z^2||=||z^2|(a-1)||$
Now put $M= \frac{1}{a-1} >0$ as $a>1$ .
Then $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{M}{|z|^2 }$$
|
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|
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2+4}}=\\
& = \lim_{x \to 2}\sqrt{(2-x)(2+x)}=0.
\end{align}
|
You messed up in the first step when you said $3-\sqrt{x^2+5}=\sqrt{9-x^2-5}$. You seem to be trying to use $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$ even though this is almost always false (as an example, take $a=b=1$).
|
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|
Solving a recurrence relation without the characteristic equations I am new to solving recurrence relations and I am presented with the following two problems
(1) $$a_n = (3n-1)a_{n-1}, a_0 = 3$$
and
(2) $$a_{n+3}-6a_{n+2} + 11a_{n+1}-6a_{n} = 0, a_0 =1, a_1 =1, a_2 = -2$$
For (1), I try to find a pattern so $a_n = (3n-1)(3(n-1)-1)a_{n-2} = (3n-1)(3n-4)a_{n-2}$ Now do this one more time so we have $a_n = (3n-1)(3n-4)(3n-7)a_{n-4}$ and so we see a pattern emerging
So I just conclude that $$a_n =(3n-1)(3n-4)(3n-7) \dots \times 3$$ But I feel there is something wrong here.
For (2) Let $a_n = cr^n$ where $c,r \neq 0$and then I use the characteristic equation $$r^3 -6r^2 + 11r - 6 = 0 =(r-3)(r-2)(r-1)=0$$
So I have $r=1,2,3$ as the characteristic roots
Now we have $a_n = A(1)^n + B(2)^n + C(3)^n$
$1 = A + B + C$
$1 = A + 2B + 3C$
$-2 = A + 4B + 9C$
And after solving this linear system, I get $$A= \frac{-1}{2}, B=3, C=\frac{-3}{2}$$
And so finally we have that $$a_n = \frac{-1}{2} + 3(2)^n -\frac{3^{n+1}}{2}, n \geq 0$$
I think I got the second recurrence right, But the first one I don't know how to find it
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$$a_n = (3n-1)a_{n-1}=(3n-1)(3n-4)a_{n-2}=...=a_{n-m}\prod_{k=1}^m(3n-3k+2)\\
=a_{n-n}\prod_{k=1}^n(3n-3k+2)=a_0\prod_{k=1}^n(3n-3k+2)$$
|
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|
Find the integral part of $S$. It is given that $S=\sqrt{2012\sqrt{2013\sqrt{2014\sqrt{\cdots \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}}}}$. Find the integral part of $S$.
Thanks in advance!
|
This is a partial answer!
Define
$$
S(n) := \sqrt{n \sqrt{(n+1) \dotsm \sqrt{n^2}}} = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k-1}}.
$$
So far I can only prove that
$$
n^2 \leq S(n)^2
$$
for every $n \geq 1$. Indeed, first observe that this is trivial for $n = 1$, so let's assume $n \geq 2$. Then note that
$$
S(n)^2 = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k}} \geq n \prod_{k=1}^{n(n-1)} (n+1)^{2^{-k}} = n (n+1)^{e_n}
$$
where
$$
e_n = \sum_{k = 1}^{n(n-1)} \frac{1}{2^k} = 1 - 2^{-n^2 + n}.
$$
Thus we need to prove that
$$
\log(n) \leq e_n \log(n+1)
$$
or, in other words
$$
2^{-n^2+n} \log(n+1) \leq \log(n+1) - \log(n)
= \log\left(1 + \frac{1}{n}\right)
$$
and rearranging the terms this becomes
$$
0 \leq \left(2^{n^2-n} - 1\right) \log\left(1 + \frac{1}{n}\right) - \log(n).
$$
The derivative of the rhs is
$$
2^{n^{2} - n} {\left(2 \, n - 1\right)} \log\left(2\right) \log\left(\frac{1}{n} + 1\right) - \frac{2^{n^{2} - n} - 1}{n (n+1)} - \frac{1}{n}
$$
which (squinting a bit) can be seen to be positive for every $n \geq 2$. Finally
$$
\left(2^{2^2-2} - 1\right) \log\left(1 + \frac{1}{2}\right) - \log(2) = 3\log(3) - 4\log(2) = \log\left(\frac{27}{16}\right) > 0
$$
is enough to conclude that $n^2 \leq S(n)^2$ for every $n \geq 2$.
|
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|
Why is $\left( {\begin{array}{*{20}{c}} B & B \\ B & B \\ \end{array}} \right)$ positive semidefinite? Let $B\in M_n$ be positive semidefinite .
Why is $\left( {\begin{array}{*{20}{c}}
B & B \\
B & B \\
\end{array}} \right)$ positive semidefinite?
|
Let $x = \left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) \in \mathbb{C}^{2n}$ with $x_1 ,x_2 \in \mathbb{C}^n$. Then we have
\begin{align}
\langle x, \left( \begin{matrix} B & B \\ B & B \end{matrix} \right)x \rangle &= \langle \left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) , \left( \begin{matrix} Bx_1 + Bx_2 \\ Bx_1 + Bx_2 \end{matrix}\right) \rangle \\
&= \langle x_1 , Bx_1 +Bx_2\rangle + \langle x_2, Bx_1 + Bx_2 \rangle \\
&= \langle x_1 + x_2 , B(x_1 + x_2 ) \rangle \\
&\ge 0
\end{align}
As $x_1 + x_2 \in \mathbb{C}^n$ and $B$ is positive semidefinite.
Also, $$\left( \begin{matrix} B & B \\ B& B \end{matrix} \right)^\dagger = \left( \begin{matrix} B^\dagger & B^\dagger \\ B^\dagger & B^\dagger \end{matrix} \right) = \left( \begin{matrix} B & B \\ B& B \end{matrix} \right).$$
So $\left( \begin{matrix} B & B \\ B& B \end{matrix} \right)$ is positive semidefinite.
|
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|
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
|
$$I=\int_{0}^{+\infty}\frac{dx}{(x+1)^3+1}=\int_{1}^{+\infty}\frac{dx}{x^3+1}=\int_{0}^{1}\frac{x}{1+x^3}\,dx $$
The roots of $1+x^3$ lie at $-1,\xi=\frac{1+i\sqrt{3}}{2},\bar{\xi}=\frac{1-i\sqrt{3}}{2}$ and they are simple. Since:
$$ \text{Res}\left(\frac{x}{1+x^3},x=-1\right) = -\frac{1}{3}$$
the partial fraction decomposition of $\frac{x}{1+x^3}$ is given by:
$$ \frac{x}{1+x^3} = -\frac{1}{3(x+1)}+\frac{1+x}{3(1-x+x^2)} $$
and:
$$ I = \frac{1}{2}-\frac{1}{5}+\frac{1}{8}-\frac{1}{11}+\ldots = \color{red}{-\frac{\log(2)}{3}+\frac{\pi}{3\sqrt{3}}}.$$
|
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|
Integrate $\int \frac{dx}{3^x-8}$ I have a problem with solving this Integrate
$$\int \frac{dx}{3^x-8}$$
At first I use substitute $$3^x =t $$
$$\log_3{t} = x$$
$$\frac{1}{t\ln{3}}dt = dx$$
Next:
$$\int \frac{dt}{t^2\ln3} - \int\frac18dx$$
$$=\frac{1}{\ln3}\int \frac{dt}{t^2} - \int\frac18dx$$
$$=\frac{1}{\ln3} \left(\frac{-1}{t}\right) - \frac18$$
$$=\frac{1}{\ln3} \left(\frac{-1}{3^x}\right) - \frac18$$
But Wolfram alpha gives me different result. Where did I go wrong?
|
Notice $\frac{1}{3^x - 8} = -\frac{1}{8 - 3^x} = - \frac{1}{8} \frac{8 - \mathbf{3^x} + \mathbf{3^x} }{(8-3^x)} = -\frac{1}{8} \left( 1 + \frac{3^x}{8-3^x} \right)$ and so
$$ \int \frac{1}{3^x - 8} = -\frac{1}{8} \left( x + \int \frac{3^x dx}{8 - 3^x} \right) = -\frac{x}{8} +\frac{1}{8 \ln 3} \int \frac{d(8-3^x)}{8 - 3^x} = -\frac{x}{8} + \frac{\ln (8-3^x)}{8 \ln 3} + C $$
|
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|
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
|
The error is in the very first implication. If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$. There is no reason to believe that we can do this division and get a number.
Therefore you have a proof that $0/0$ cannot equal any nonzero $x$. Combine this with a proof that $0/0$ cannot equal zero, and you have proved that $0/0$ is not a number.
|
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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.