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Finding conjugates of all z $\in$ C that satisfy $z^3$ $$ z^3 = \frac{16e^{i\frac{3\pi}4}}{(1-\sqrt3)+\sqrt6e^{i\frac\pi4}} $$
Anyone know of a good way to simplify this expression?
|
$\sqrt{6}e^{i\pi/4} = \sqrt{6}\left(\cos(\pi/4) + i\sin(\pi/4)\right) = \sqrt{6}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = \sqrt{3} + \sqrt{3}i \Rightarrow 1 - \sqrt{3} + \sqrt{6}e^{i\pi/4} = 1 +\sqrt{3}i = 2e^{i\pi/3} \Rightarrow z^3 = \dfrac{16e^{i3\pi/4}}{2e^{i\pi/3}} = 8e^{i5\pi/12} = 8e^{i5\pi/12+2n\pi} \Rightarrow z = 2e^{i5\pi/36 + 2n\pi/3} \Rightarrow \bar{z} = 2e^{-i5\pi/36 - 2n\pi/3}, n \in \{0,1,2\}$.
|
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|
Find bases for eigenspaces of A $$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$
Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.
I don't really know where to start on this problem.
|
First step: find the eigenvalues, via the characteristic polynomial $$\det(A - \lambda I) = \begin{vmatrix} 6 - \lambda & 4 \\ -3 & -1-\lambda \end{vmatrix} = 0 \implies \lambda^2 - 5\lambda + 6 = 0.$$
One of the eigenvalues is $\lambda_1 = 2$. You find the other one.
Second step: to find a basis for $E_{\lambda_1}$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda_1 I)\mathbf{v} = \mathbf{0}$, in this case, we go for:
$$(A-2I)\mathbf{v} =
\begin{pmatrix}
4 & 4 \\ -3 & -3
\end{pmatrix}
\begin{pmatrix} v_1 \\ v_2
\end{pmatrix} =
\begin{pmatrix}
0 \\ 0
\end{pmatrix} \implies
\begin{cases}
4v_1 + 4v_2 = 0 \\ -3v_1 -3v_2 = 0
\end{cases}
\implies v_1 = -v_2.$$
So, $\mathbf{v} = (v_1,v_2) = (v_1,-v_1) = v_1(1,-1)$, so $(1,-1)$ is a basis for that eigenspace with eigenvalue $\lambda_1$. Try to find a basis for the other one.
|
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|
Equation $a^5+15ab+b^5=1$ What are the integer solutions of $a^5+15ab+b^5=1$?
The equation is symmetric in $a$ and $b$, so let's assume $a\geq b$. When $a=b$, we have $2a^5+15a^2=1$, which has no solution by the Rational Root Theorem. Else, $a>b$. It is degree-five, so we cannot use the quadratic formula.
|
Clearly $a \gt 0, b \le 0, a \gt |b|$ because the $15ab$ is negative. We can see that $a=1,b=0$ is a solution. If $a =1-b,$ the left side becomes $a^5+15a(1-a)-(a-1)^5=5a^4-10a^3-5a^2+10a+1$, which is greater than $1$ when $a \gt 2$ so we only need to check $a=2$ We find $a=2,b=-1$ is the only other solution.
|
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|
Proving the inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$ If $a,b,c$ are non-negative real numbers prove the following inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$.
|
It's obviously true for all $a,b,c$ real numbers.
|
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|
Simplify - Help I am supposed to simplify this:
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
The answer is supposed to be this, but I can not seem to get to it:
$$x(x^2-1)(x^3+1)(5x^3-3x+2)$$
Thanks
|
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)\\=x(x^2-1)(x^3+1)\left[3x(x^2-1)+2(x^3+1)\right] \color{blue}{\text{(Factor out common factors)}}\\=x(x^2-1)(x^3+1)\left[3x^3-3x+2x^3+2\right] \color{blue}{\text{ (Distribute inside brackets)}}\\=x(x^2-1)(x^3+1)(5x^3-3x+2) \color{blue}{\text{ (Simplify)}}$$
|
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|
Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
|
Apolozies for any typos:
My solution:
$$
\int \frac{dx}{\sin^4x+\cos^4x\:+\:\sin^2\left(x\right)\cos^2\left(x\right)}\\
=\int\frac{dx}{(\sin^2x+\cos^2x)^2-\sin^2x\cos^2x}\\
=\int\frac{dx}{1-\sin^2x\cos^2x}\\
=\int\frac{(1+\tan^2x)\sec^2xdx}{1+\tan^4x+\tan^2x}\\
=\int\frac{(1+y^2)dx}{1+y^2+y^4}\quad y:=\tan x\\
=\frac12\int\frac{1}{1+y^2+y}+\frac1{1-y+y^2}\\
=\frac12\left(\frac2{\sqrt3}\left(\arctan\frac{\tan x+0.5}{\sqrt3/2}+\arctan\frac{\tan x-0.5}{\sqrt3/2}\right)\right)+c
$$
|
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|
Prove that one of x,y,z is smaller than 3 and one is bigger than 5 if... If $x+y+z=12$ and $x^2+y^2+z^2=54$ then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.
So I got that $xy+yz+zx=45$ and with that I had a function with x,y,z as zeros: $f(a)=a^3-12a^2+45a-k$, where k=xyz.
I get that $f'(a)=3(a-3)(a-5)$, but I can't prove what I need with that.
|
A sketch that needs fleshing out: We might as well require $x \ge y \ge z$. If $z \gt 3$, let $w=3-z$ and we are thinking of $w$ as small. The largest $x^2+y^2+z^2$ can be is $(6-2w)^2+2(3+w)^2=54-12w+6w^2$, so $z \lt 3$ If $x \lt 5$, Let $x=5-w$, then the largest $x^2+y^2+z^2$ can be is $2(5-w)^2+(2+2w)^2=54-16w+6w^2$. In fact there is a solution $5,5,2$
|
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|
Fourier series for $f(x)=\begin{cases} 0 & -\pi\leq x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$
Find the Fourier series for $$f(x)=\begin{cases} 0 & -\pi\leq x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$$
I found an answer, I'm not completly sure if it's right.
The solution would be given by $$f(x)=a_0+\sum_n\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right)$$
$$a_0=\frac{1}{2\pi}\int_0^\pi\sin x dx = - \frac{1}{2\pi}\cos x|_0^\pi=\frac{1}{\pi}.$$
$a_n=\displaystyle{\frac{1}{\pi}\int_0^\pi\sin x \cos nx dx}$, I compute then the indefinite integral first:
$$\int\sin x \cos nx dx=-n\cos x\cos nx - n\int\sin nx \cos x dx\;\;\;[1]\\=-n\cos x\cos nx-n\left(-\sin x \sin nx - n\int\sin x \cos nx\right)\;\;\;[2]\\=-n\cos x \cos nx + n\sin x \sin nx+ n\int \sin x \cos nx dx\\\implies \int sin x \cos nx = \frac{n}{1-n}(\sin x \sin nx - \cos x \cos nx)$$
The last integral evaluated from $0$ to $\pi$ is $-2$, then $a_n = -\displaystyle\frac{2}{\pi}$
$[1]:$ By parts with $u=\cos nx,dv=\sin x\;dx$
$[2]:$ Parts again with $u=\cos x, dv=\sin nx\;dx$
$b_n=\displaystyle{\frac{1}{\pi}\int_0^\pi\sin x \sin nx dx}$, as before I compute then the indefinite integral first:
$$\int \sin x \sin nx dx = -\frac{1}{n}\sin x \cos nx + \frac{1}{n}\int \cos x \cos nx\;\;\; [3]\\=-\frac{1}{n}\sin x \cos nx + \frac{1}{n}\left(\frac{1}{n}\cos x \sin nx + \frac{1}{n}\int \sin x \sin nx\right) \;\;\;[4]\\=-\frac{1}{n}\sin x \cos nx + \frac{1}{n^2}\cos x \sin nx + \frac{1}{n^2}\int \sin x \sin nx dx\\ \implies \int \sin x \sin nx dx = \frac{\frac{1}{n^2}cos x \sin nx - \frac{1}{n}\sin x \cos nx}{\left(1-1/n^2\right)}.$$
Then $b_n=0$ due to $\sin n\pi = \sin 0 = 0$.
$[3]:$ Parts with $u=\sin x, dv = \sin nx$
$[4]:$ Parts with $u=\cos x, dv=\cos nx$
This together gives the Fourier series
$$f(x) = \frac{1}{\pi}-\frac{2}{\pi}\sum_{n=1}^{\infty}\cos nx$$
Is this solution correct?
|
To avoid partial integration, a complex approach:
\begin{align}
a_n&=\frac{1}{\pi}\int_0^\pi\sin(x)\cos(nx)dx\\
&=\frac{1}{4\pi i}\int_0^\pi(e^{ix}-e^{-ix})(e^{inx}+e^{-inx})dx\\
&=\frac{1}{4\pi i}\int_0^\pi e^{i(n+1)x}+e^{i(1-n)x}-e^{i(n-1)x}-e^{i(-1-n)x}dx\\
&=-\frac{1}{4\pi}\left[\frac{e^{i(n+1)x}}{n+1}+\frac{e^{i(1-n)x}}{1-n}-\frac{e^{i(n-1)x}}{n-1}-\frac{e^{i(-1-n)x}}{-1-n}\right]_0^\pi\qquad\text{$n\neq 1$}\\
&=-\frac{1}{4\pi}\left[\frac{e^{i(n+1)\pi}-1}{n+1}+\frac{e^{i(1-n)\pi}-1}{1-n}-\frac{e^{i(n-1)\pi}-1}{n-1}-\frac{e^{i(-1-n)\pi}-1}{-1-n}\right]\\
&=\cases{-\frac{2}{\pi(n^2-1)}\qquad&\text{if $n\geq 2$ even}\\
0\qquad&\text{if $n\geq3$ odd}}
\end{align}
For $a_1$, note that $\int_0^\pi\sin(x)\cos(x)dx=2\int_0^\pi\sin(2x)dx=0$.
Now we calculate $b_n$, using a product-to-sum-identity[Link]:
\begin{align}
b_n&=\frac{1}{\pi}\int_0^\pi\sin(x)\sin(nx)dx\\
&=\frac{1}{2\pi}\int_0^\pi\cos((n-1)x)-\cos((n+1)x)dx\\
&=\frac{1}{2\pi}\left[\frac{\sin((n-1)x)}{n-1}-\frac{\sin((n+1)x)}{n+1}\right]_0^\pi dx\\
&=0
\end{align}
The fact that $b_n=0$ corresponds with your solution, because $\sin(n\pi)=0$, with $n$ integer. Actually, using known identities is much more easier than partial integrating or a complex approach.
EDIT:
If this answer above was true, then $f(x)$ would be an even function, because it consists of only cosine terms. However, this is not the case. Because in the last calculation for $b_n$, there is a division by $n-1$, it should be noted that $n\geq2$. I completely forgot this. An additional calculation of $b_1$ is needed:
$$b_1=\frac{1}{\pi}\int_0^\pi\sin^2(x)dx=\frac{1}{2}$$
|
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|
A second order problem on recurrence relation equals 3^n I had this Recurrence Relation problem:
$a_{n+2} + a_{n+1} - 12a_n = 0$
And I solved in a form like this $a_n = A(r_1)^n + B(r_2)^n$
$r^{n+2} + r^{n+1} - 12r^n = 0$ -> Divided the hole eqation by the lowest $r$
Result of this:
$r^2+r -12 = 0$
And then i found the roots: $r_1 = 3$ and $r_2=-4$
And the solution of this problem is: $a_n = 2/7*(3)^n -7/2*(-4)^n $
The next one, is what i don´t get. It is exact like this one in stead it is equal to $3^n$, like is: $a_{n+2} + a_{n+1} - 12a_n = 3^n$
How do I solve this one ?
|
You didn't mentioned the initial conditions of the recurrence relation, but by the solution that you obtained
$$a_n = \frac{2}{7}(3)^n − \frac{7}{2}(−4)^n$$
one can infer the initial conditions
$$a_0 = -\frac{45}{14} \;\;\;\;\;\;\;\; a_1 = \frac{104}{7}$$
That made me think that maybe there is a typo and you intended
$$a_n = \frac{2}{7}(3)^n − \frac{2}{7}(−4)^n$$
so
$$a_0 = 0 \;\;\;\;\;\;\;\; a_1 = 2$$
I'll solve both cases just in case.
What you are trying to do is to solve a non-homogeneous linear recurrence relation.
$$ a_{n+2} + a_{n+1} - 12a_{n} = 3^n = f(n)$$
The first step is to solve the the associated homogeneous linear recurrence:
$$ a_{n+2} + a_{n+1} - 12a_{n} = 0 $$
The associated characteristic polynomial is
$$r^2 + r - 12$$
with roots $r_1 = 3$ and $r_2 = -4$.
Hence the homogeneous solution is of the form
$$h_n = A 3^n + B(-4)^n$$
Instead to plug-in the initial conditions, we now need to find a particular solution to the non-homogeneous equation.
As $f(n) = 3^n$, we would guess a solution of the form $g_n = C 3^n$, where $C$ is a constant.
But $3$ is a root of characteristic polynomial of the homogeneous part, hence the correct guess is $g_n = C n 3^n$.
We now find $C$ by plugging it on the recurrence:
$$ C(n+2)3^{n+2} + C(n+1)3^{n+1} - 12Cn3^n = 3^n$$
$$ 9C(n+2) + 3C(n+1) - 12Cn = 1 $$
have to hold for all $n$, in particular for $n = 0$, hence:
$$ 18 C + 3 C = 1 \Rightarrow C = \frac{1}{21} $$
Hence we have the particular solution $g_n = \frac{1}{21} n 3^n$.
Therefore, the final solution is of the form:
$$ a_n = h_n + g_n = \left( A + \frac{n}{21} \right) 3^n + B(-4)^n$$
where the constants $A$ and $B$ are to be determined from the initial conditions through the linear system:
$$\begin{cases}
a_0 = A + B \\
a_1 = 3A - 4B + \frac{1}{7}
\end{cases}$$
Using $a_0 = -\frac{45}{14}$ and $a_1 = \frac{104}{7}$ we have: $A = \frac{13}{49}$ and $B = - \frac{341}{91}$ hence the solution:
$$ a_n = \left( \frac{13}{49} + \frac{n}{21} \right) 3^n - \frac{341}{91}(-4)^n $$
Using $a_0 = 0$ and $a_1 = 2$ we have: $A = \frac{13}{49}$ and $B = - \frac{13}{49}$ hence the solution:
$$ a_n = \left( \frac{13}{49} + \frac{n}{21} \right) 3^n - \frac{13}{49}(-4)^n $$
If you had another set of initial values in mind, you can just plug in and find $A$ and $B$.
|
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Find a value for "c"
For what value of $c$ is $$\lim_{x\to\infty} \left(\frac{x+c}{x-c}\right)^x = e?$$
I am unsure of how to start this question in any sense.
|
You also have a solution using Taylor series.
Let $$A=\left(\frac{x+c}{x-c}\right)^x$$ $$\log(A)=x \log \left(\frac{x+c}{x-c}\right)=x \log \left(\frac{1+\frac cx}{1-\frac cx}\right)$$ For small values of $y$, we also have $$\log \left(\frac{1+y}{1-y}\right)=2 \Big(\frac{y}{1}+\frac{y^3}{3}+\cdots\Big)$$ Replace $y$ by $\frac{c}{x}$ and so $$\log(A)=2x \Big(\frac{ c}{x}+\frac{ c^3}{3 x^3}+O\left(\left(\frac{1}{x}\right)^4\right))=2 c+\frac{2 c^3}{3 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$ $$A=e^{2 c}\Big(1+\frac{2 c^3 }{3 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)\Big)$$ which shows the limit and how the expression goes to the limit.
|
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Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
|
Here's another way without using fancy formulas
$$ (x-5)^2-4 $$
$$=x^2-5x-5x+25-4 $$
$$=x^2-10x+25-4 $$
$$=x^2-10x+21 $$
$$=(x-3)(x-7) $$
$$=(x-5+2)(x-5-2) $$
|
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|
Factorising quadratics - coefficient of $x^2$ is greater than $1$ In factoring quadratics where the coefficient of $x^2$ is greater than $1$, I use the grouping method where we multiply the coefficient and constant together and then factor.
My question is can someone explain the math behind that?
Example:
$5x^2+11x+2,\quad 5\cdot2=10$
$5x^2+10x+x+2$
$5x(x+2)+1(x+2)$
$(5x+1)(x+2)$
|
Consider the factorization of $9x^2 - 18x - 16$. To split the linear term, we must find two numbers with product $9 \cdot -16 = -144$ and sum $-18$. They are $-24$ and $6$. Hence,
\begin{align*}
9x^2 - 32x - 16 & = 9x^2 - 24x + 6x - 16 && \text{split the linear term}\\
& = 3x(3x - 8) + 2(3x - 8) && \text{factor by grouping}\\
& = (3x + 2)(3x - 8) && \text{extract the common factor}
\end{align*}
Observe that when you split the linear term, the product of the quadratic and constant coefficients is equal to the product to the two linear coefficients, that is, $9 \cdot -16 = -24 \cdot 6 = -144$.
Suppose that $a, b, c, k, l, m, n$ are integers such that
$$ax^2 + bx + c = (kx + l)(mx + n) \tag{1}$$
First, we show that
\begin{align*}
a & = km \tag{2}\\
b & = kn + lm \tag{3}\\
c & = ln \tag{4}
\end{align*}
Since equation 1 is an algebraic identity, it holds for each value of the variable. In particular, it holds for $x = 0, 1, -1$.
If we substitute $0$ for $x$ in equation 1, we obtain
$$c = ln$$
Thus, equation 4 holds. If we substitute $1$ for $x$ in equation 1, we obtain
\begin{align*}
a + b + c & = (k + l)(m + n)\\
& = k(m + n) + l(m + n)\\
& = km + kn + lm + ln
\end{align*}
Since $c = ln$, we can cancel $c$ from the left hand side and $ln$ from the right hand side of the equation to obtain
$$a + b = km + kn + lm \tag{5}$$
If we substitute $-1$ for $x$ in equation 1, we obtain
\begin{align*}
a - b + c & = (-k + l)(-m + n)\\
& = -k(-m + n) + l(-m + n)\\
& = km - kn - lm + ln
\end{align*}
Since $c = ln$, we cancel $c$ from the left hand side and $ln$ from the right hand side of the equation to obtain
$$a - b = km - kn - ln \tag{6}$$
Adding equations 5 and 6 yields
$$2a = 2km$$
Dividing each side of the equation by $2$ yields
$$a = km$$
so equation 2 holds. Since $a = km$, we can cancel $a$ from the left hand side and $km$ from the right hand side of equation 5 to obtain
$$b = kn + ln$$
so equation 3 holds.
Thus, if $a, b, c, k, l, m, n$ are integers such that equation 1 holds, then
$$ac = (km)(ln) = klmn = (kn)(lm)$$
where $b = kn + ln$, so the product of the quadratic and constant coefficients is equal to the product of the two coefficients into which the linear coefficient splits.
We then obtain the factorization
\begin{align*}
ax^2 + bx + c & = kmx^2 + (kn + lm)x + ln && \text{by equations 2, 3, and 4}\\
& = kmx^2 + knx + lmx + ln && \text{split the linear term}\\
& = kx(mx + n) + l(mx + n) && \text{factor by grouping}\\
& = (kx + l)(mx + n) && \text{extract the common factor}
\end{align*}
Note that to check the factorization is correct, we multiply $(kx + l)(mx + n)$, in which we perform the steps of the factorization by splitting the linear term and grouping in reverse order.
|
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How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.
|
Here's the easiest way:
$$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \frac{dx}{x^3 \sqrt{1 - \frac{1}{x^2}}}$$
Now substitute $u = 1 - \frac{1}{x^2}$
|
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Taylor series of a definite integral Consider the function of a parameter $\alpha > 0$, given by
$$f(\alpha) = \frac{2}{\sqrt 2\pi} \int_0^\infty e^{\dfrac{-x^2}{2\alpha^2}}\cosh(x)\log\cosh(x) dx.$$
From Wolfram-alpha, it seems that for small values of $\alpha$,
$$f(\alpha) = \frac{\alpha^3}{2}+\frac{\alpha^5}{2} + o(\alpha^5).$$
How can one establish this rigorously?
Edit: Per the comment below, I used dfrac to make it more visible.
|
Considering $$f(\alpha) = \frac{2}{\sqrt {2\pi}} \int_0^\infty e^{\frac{-x^2}{2\alpha^2}}\cosh(x)\log\big(\cosh(x)\big) dx$$ start developing $\cosh(x)\log\big(\cosh(x)\big)$ as an infinite Taylor series built at $x=0$. We then get $$\cosh(x)\log\big(\cosh(x)\big)=\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^6}{720}+\frac{x^8}{630}+O\left(x^9\right)$$ and so we are let with a linear combination of integrals of the type $$I_k=\int e^{\frac{-x^2}{2\alpha^2}}x^{2k}dx=-2^{k-\frac{1}{2}} \alpha ^{2 k+1} \Gamma \left(k+\frac{1}{2},\frac{x^2}{2 \alpha
^2}\right)=-\frac{1}{2} x^{2 k+1} E_{\frac{1}{2}-k}\left(\frac{x^2}{2 \alpha ^2}\right)$$ where appears the incomplete gamma function or the elliptic integral.
Concerning the corresponding definite integral $$J_k=\int_0^{\infty} e^{\frac{-x^2}{2\alpha^2}}x^{2k}dx=2^{k-\frac{1}{2}} \left({\alpha ^2}\right)^{k+\frac{1}{2}} \Gamma
\left(k+\frac{1}{2}\right)$$ So, replacing, we get $$f(\alpha)=\frac{\alpha ^3}{2}+\frac{\alpha ^5}{2}+\frac{\alpha ^7}{48}+\frac{\alpha
^9}{6}+\cdots$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0$ A limit I had to compute recently boiled down to the following limit:
$$\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0\quad\mbox{for $c\ge0$}$$
How can I show that this limit is correct?
|
Note that \begin{gather*}
\sqrt{x^2+c}-x=(\sqrt{x^2+c}-x)\frac{\sqrt{x^2+c}+x}{\sqrt{x^2+c}+x}\\
=\frac{x^2-x^2+c}{\sqrt{x^2+c}+x}\\
=\frac{c}{\sqrt{x^2+c}+x}
\end{gather*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$ In my textbook they asked me to show that
$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$
but this is not true, I think.
I put down
$$\begin{align}
\frac12(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)&=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right)]\\
&=\frac12\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]\\
&=4\\
\end{align}$$
I checked it twice and I got $4$ what am I doing wrong?
|
Now you know your mistake!
But there is an alternative way to show that
Consider this cubic
$$x^3-1=0$$
$$(x-1)(x^2+x+1)=0$$
By applying quadratic formula to
$x^2+x+1=0$
$$x=\frac{-1- i\sqrt{3}}{2} \quad\text{or}\quad x=\frac{-1+ i\sqrt{3}}{2}$$
Since $x=\dfrac{-1+ i\sqrt{3}}{2}$ is root of $x^3-1=0$ we have
$$\left(\dfrac{-1+ i\sqrt{3}}{2}\right)^3=1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$ Question :
Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$
I know that $|z_1+z_1| \leq |z_1|+|z_2|$
Also if I consider two complex numbers say $z_1 =x+iy$ and $z_2 =a+ib$
Now $|z_1|^2 =x^2+y^2 $
and $|z_2|^2 =a^2+b^2$
Also using A.M.-GM. inequality we have
$\frac{x^2+y^2}{2}\geq \sqrt{2}xy$
$\Rightarrow x^2+y^2 \geq xy 2\sqrt{2}$
Also $a^2+y^2 =2\sqrt{2}ab$
Please suggest whether we can proceed in this manner also guide further on this thanks.
|
Let $z_{1} = a+ib$ and $z_{2} = c+id$. Thus, $|z_{1}|^2 = a^2+b^2, |z_{2}|^2 = c^2+d^2, Re(z_{1}z_{2}) =ac-bd$ and $Im(z_{1}z_{2}) =ad+dc$.
The problem can be posed as
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} a^2+b^2+ c^2+d^2 + ac-bd \\
s.t. \hspace{3cm} ad+bc = 1. \\
\end{array}
\end{equation}
Let $\mathbf{x} = [a \hspace{1mm} b \hspace{1mm} c \hspace{1mm} d]^{T}$. The previsous optimization problem can be rewrite as
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} \mathbf{x}^{T}A\mathbf{x} \\
s.t. \hspace{1cm} \mathbf{x}^{T}B\mathbf{x}= 1, \\
\end{array}
\end{equation}
where
\begin{equation}
A = \left [ \begin{array}{cccc}
1 & 0 & 1/2 & 0 \\
0 & 1 & 0 &-1/2 \\
1/2 & 0 & 1 & 0 \\
0 & -1/2 & 0 & 1 \\
\end{array} \right]
\end{equation} and
\begin{equation}
B = \left [ \begin{array}{cccc}
0 & 0 & 0 & 1/2 \\
0 & 0 & 1/2 &0 \\
0 & 1/2 & 0 & 0 \\
1/2 & 0 & 0 & 0 \\
\end{array} \right].
\end{equation}
Solving the problem for lagragian function $\mathcal{L} = \mathbf{x}^{T}A\mathbf{x} + \lambda (\mathbf{x}^{T}B\mathbf{x}-1)$, we have
$\frac{\partial \mathcal{L}}{\partial \mathbf{x}} = \mathbf{0} \Rightarrow (A+\lambda B)\mathbf{x} = \mathbf{0}$.
The linear system $(A+\lambda B)\mathbf{x} = \mathbf{0}$ has non-trivial solutions ($\mathbf{x} \neq \mathbf{0}$) only if $\lambda = \pm \sqrt{3}$ (The trivial solution $\mathbf{x} = \mathbf{0}$ does not satisfy the constraint $\mathbf{x}^{T}B\mathbf{x}=1$). In this case, $rank(A+\lambda B) = 2$.
For $\lambda = \pm \sqrt{3}$ , $\mathbf{x} = \alpha_{1}\mathbf{u}_{1} + \alpha_{2}\mathbf{u}_{2}$, where $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$ are non-zero vectors in $Ker\{A+ \lambda B\}$, and $ \alpha_{1}$ and $ \alpha_{2}$ are any real parameters.
By substituting $\mathbf{x}$ in the previous optimization problem, we have the equivalent quadratic problem
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} \mathbf{\alpha}^{T}C\mathbf{\alpha} \\
s.t. \hspace{1cm} \mathbf{\alpha}^{T}D\mathbf{\alpha}= 1, \\
\end{array}
\end{equation} where $\mathbf{\alpha} = [\alpha_{1} \hspace{1mm} \alpha_{2}]^{T}$, and
\begin{equation}
C = \left [ \begin{array}{cc}
\mathbf{u}_{1}^{T}A\mathbf{u_{1}} & \mathbf{u}_{1}^{T}A\mathbf{u_{2}} \\
\mathbf{u}_{2}^{T}A\mathbf{u_{1}}& \mathbf{u}_{2}^{T}A\mathbf{u_{2}} \\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
\mathbf{u}_{1}^{T}B\mathbf{u_{1}} & \mathbf{u}_{1}^{T}B\mathbf{u_{2}} \\
\mathbf{u}_{2}^{T}B\mathbf{u_{1}}& \mathbf{u}_{2}^{T}B\mathbf{u_{2}} \\
\end{array} \right].
\end{equation}
For $\lambda = \sqrt{3}$, we can choose $\mathbf{u}_{1} = [ 1 \hspace{1.7mm} 0 \hspace{1.7mm} -1/2 \hspace{1.7mm} -\sqrt{3}/2]^{T}$, and
$\mathbf{u}_{2} = [ 0 \hspace{1.7mm} 1 \hspace{1.7mm} -\sqrt{3}/2 \hspace{1.7mm} 1/2]^{T}$. Thus,
\begin{equation}
C = \left [ \begin{array}{cc}
3/2 & 0 \\
0& 3/2\\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
-\sqrt{3}/2 & 0 \\
0& -\sqrt{3}/2\\
\end{array} \right].
\end{equation}
Here, there is no solution because the constraint becomes $\alpha_{1}^2+\alpha_{2}^2 = -2/\sqrt{3}$.
For $\lambda = -\sqrt{3}$, we can choose $\mathbf{u}_{1} = [ 1 \hspace{1.7mm} 0 \hspace{1.7mm} -1/2 \hspace{1.7mm} \sqrt{3}/2]^{T}$, and
$\mathbf{u}_{2} = [ 0 \hspace{1.7mm} 1 \hspace{1.7mm} \sqrt{3}/2 \hspace{1.7mm} 1/2]^{T}$. Thus,
\begin{equation}
C = \left [ \begin{array}{cc}
3/2 & 0 \\
0& 3/2\\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
\sqrt{3}/2 & 0 \\
0& \sqrt{3}/2\\
\end{array} \right].
\end{equation}
Here, the minimum value of the cost function is $\sqrt{3}$, for any $\alpha_{1}$ and $\alpha_{2}$ such that $ \alpha_{1}^2+\alpha_{2}^2 = 2/\sqrt{3}$.
|
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|
What is the value of $a+b+c$?
What is the value of $a+b+c$?
if $$a^4+b^4+c^4=32$$
$$a^5+b^5+c^5=186$$
$$a^6+b^6+c^6=803$$
How to approach this kind of problem. Any help.
UPDATE: Thank you all for answers. Now I realize there is no integer solution. But is there any real number solution? I am curious to know.
|
If $x$ is an integer then $x^4$ and $x^6$ have the same parity. Therefore $a^4+b^4+c^4$ and $a^6+b^6+c^6$ are either both odd or both even.
Therefore there is no solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Non linear effect in differential equation Suppose I want to study the non linear ODE
$\frac{d^2}{dx^2}f(x) + a f^2(x) + b f(x) + c=0$
I know the solution for $a=0$ and I know $a$ is a small parameter. How can I study the effect of the non linear term?
|
Multiply by $f'$ to yield a integrable equation. After it depends on $f$ if you can do the last integral.
$\textbf{Edit:}$
$$
f'' f' + \left[af^2+bf + c\right]f' = f'\dfrac{d}{dx}f' + \frac{d}{dx}\left[\frac{1}{3}af^3 + \frac{1}{2}bf^2+cf\right] = 0
$$
now we can re-write the last but one equation as
$$
\dfrac{d}{dx}\left[\frac{f'^2}{2} + \frac{1}{3}af^3 + \frac{1}{2}bf^2+cf\right] = 0
$$
or
$$
\frac{f'^2}{2} + \frac{1}{3}af^3 + \frac{1}{2}bf^2+cf = \text{Constant} = \lambda
$$
then we have a separable equation
$$
\dfrac{df}{dx} = \pm \sqrt{\lambda -\left(\frac{1}{3}af^3 + \frac{1}{2}bf^2+cf \right)}
$$
now to solve this (maybe) heinous equation..we require the $f(x)$ to play nice ;).
|
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|
Computing High Powers of a Matrix With Polynomial Entries I have a matrix where the terms are from a polynomial with two variables. Specifically it's
$C = \left(\begin{array}{cccc}
1 & 0 & 1 & 0 \\
s & 0 & s & 0 \\
0 & 1 & 0 & 1 \\
0 & st & 0 & st
\end{array}\right)$
Is there a good method for computing (exactly) the n-th power of this matrix?
|
As noted by @Simon S, the matrix can be diagonalized. By the spectral theorem
$$
%
\mathbf{A} = \mathbf{P}^{-1} \mathbf{\Lambda} \, \mathbf{P} \\
%
$$
Instead of multiplying with the full matrix $\mathbf{A}$, work with the diagonal matrix of eigenvalues. For $k\in\mathbb{N}$:
$$
\mathbf{A}^{k} = \mathbf{P}^{-1} \mathbf{\Lambda}^{k} \, \mathbf{P} \\
$$
The eigenvalue spectrum is
$$
\lambda \left( \mathbf{A} \right) = \left\{
\frac{1}{2} \left(1 + st \pm \sqrt{s^2 t^2-2 s t+4 s+1} \right), 0, 0
\right\}
$$
The matrix of eigenvalues
$$
\Lambda =
\left(
\begin{array}{cccc}
\lambda_{+} & 0 & 0 & 0 \\
0 & \lambda_{-} & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
The matrix of eigenvectors
$$
\begin{align}
%
\mathbf{P} &=
\left(
\begin{array}{cccc}
-\frac{\lambda_{+}}{2 s^2 t} & -\frac{\lambda_{-}}{2 s^2 t} & -1 & 0 \\
-\frac{\lambda_{+}}{2 s t} & -\frac{\lambda_{-}}{2 s t} & 0 & -1 \\
\frac{1}{s t} & \frac{1}{s t} & 1 & 0 \\
1 & 1 & 0 & 1 \\
\end{array}
\right), \\[4pt]
%
\mathbf{P}^{-1} &=
\left(
\begin{array}{cccc}
\frac{s t (2 s t-\lambda_{-})}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{t (\lambda_{-}-2 s)}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{s t (2 s t-\lambda_{-})}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{t (\lambda_{-}-2 s)}{(t-1) (\lambda_{-}-\lambda_{+})} \\
\frac{s t (\lambda_{+}-2 s t)}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{t (2 s-\lambda_{+})}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{s t (\lambda_{+}-2 s t)}{(t-1) (\lambda_{-}-\lambda_{+})} & \frac{t (2 s-\lambda_{+})}{(t-1) (\lambda_{-}-\lambda_{+})} \\
\frac{1}{t-1} & \frac{1}{s-s t} & \frac{t}{t-1} & \frac{1}{s-s t} \\
\frac{s t}{t-1} & \frac{t}{1-t} & \frac{s t}{t-1} & \frac{1}{1-t} \\
\end{array}
\right)
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
|
Put:
$$f(x) = \sum_{k=0}^{n-1}\exp\left[(3 k +1)x\right]=\exp(x)\frac{\exp(3 n x)-1}{\exp(3 x)-1}$$
If we expand $f(x)$ in a series expansion around $x=0$, then the coefficient of $x$ will give us the desired summation. We can do this as follows, we put:
$$g(x) = \log\left[f(x)\right] = x + \log\left[\exp(3 n x)-1\right]-\log\left[\exp(3x)-1\right]$$
Series expansion around $x = 0$ to order $x$ yields:
$$g(x) = x + \log\left(3 n x + \frac{9 n^2x^2}{2}+\cdots\right) -\log\left(3 x + \frac{9 x^2}{2}+\cdots\right)$$
$$=x + \log(n) + \log\left(1+\frac{3nx}{2}+\cdots\right)- \log\left(1+\frac{3x}{2}+\cdots\right)$$
$$=\log(n) + \frac{3 n -1}{2} x+\cdots$$
Therefore:
$$f(x) = \exp\left[g(x)\right] = n\exp\left(\frac{3 n -1}{2} x+\cdots\right)=n + \frac{n(3 n -1)}{2} x+\cdots$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 9
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|
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ = \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ = \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ = \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.
|
An idea for your fourth line:
$$\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4}=\frac{(n+1)^2}4\left(n^2+4n+4\right)=\frac{(n+1)^2}4(n+2)^2$$
|
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|
Writing an alternating series as two non-alternating series? How does one calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
if given this information: $$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6}.$$
How does one account for the $(-1)^{n+1}$ in the first series?
Note: in this earlier post, user David Hodden (thanks) pointed out that the following simplification can be made:
$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac1{n^2} = \sum_{n=1}^{\infty} \frac1{n^2} -2 \cdot \frac14 \sum_{n=1}^{\infty} \frac1{n^2} \\
= \frac{\pi^2}{12}$$
What I do not understand is where this simplification came from. Why can the alternating series be broken up into two non-alternating series?
|
Notice that
$$\frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \ ... \ = \ \frac{1}{1^2} + \left( \frac{1}{2^2} - 2 \frac{1}{2^2}\right) + \frac{1}{3^2} + \left( \frac{1}{4^2} - 2 \frac{1}{4^2}\right) + \ ...$$
See where this is going?
|
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|
Prove that there are infinitely many $m,n$ for which $\frac{m+1}{n}+\frac{n+1}{m}$ is an integer Prove that there are infinitely many pairs of positive integers (m,n) such that
$\frac{m+1}{n}+\frac{n+1}{m}$
is a positive integer.
I tried following:
clearly $(1,1)$ satisfies the condition. we assume that $(a,b)$ satisfies the condition. If we can find some $f(a)$ and $f(b)$ that satisfy the condition and they are increasing functions then we are done. But i didn't get such thing.
|
Apply the ideas of Vieta's root jumping to $ \frac{m+1}{n} + \frac{n+1}{m} = k $.
OP found $(1,1)$ as a solution, which give us the equation $ \frac{m+1}{n}+ \frac{n+1}{m} = 4$.
This simplifies to the quadratic $m^2 + (1-4n)m + n^2 + n = 0$.
Divide by $ m \neq 0$ to get: $ m - (4n-1) + \frac{n^2 + n}{m} = 0$.
So, if we have an integer solution $(m, n)$ , then
*
*$ \frac{n^2+n}{m}$ is an integer.
*$m+ \frac{n^2+n}{m} = 4n-1$.
Thus, for a fixed $n$, if $m$ is a solution to $X^2 + ( 1-4n)X + n^2+n = 0$, then so is $ \frac{n^2+n}{m}$.
This allows us to establish the Vieta root jumping: Define a recurrence via $ a_ 1 = 1, a _2 = 1, a_{n+2} = 4a_{n+1} - a_{n} - 1 = \frac{a_{n+1} ^2 + a_{n+1} } { a_n}$, and the above shows that $(a_{n}, a_{n+1})$ satisfies the conditions of the problem.
Observe that if $ a_{n+1} \geq a_{n} > 0$, then $ a_{n+2} > \frac{a_{n+1}^2}{a_{n} } \geq a_{n+1}$.
Hence, we have a sequence of strictly increasing integers (apart from possibly the initial values), which gives us infinitely many solutions as desired.
Further Notes
*
*With $k=4$, this gives us the $1, 1, 2, 6, 21, \ldots$ solution found by paw.
*With $ k = 3$ (modify the above as needed), this gives us $2, 2, 3, 6, 14, 35, \ldots$ solution found by daniel.
*Furthermore, Vieta tells us that if we have any solution with $ m > n$, we can reverse engineer the sequence [IE Repeatedly apply $(m, n) \rightarrow (n, \frac{n^2+n}{m} )$] to end up with $ m=n$. Then, $ \frac{2(m+1)}{m} = 2+\frac{2}{m}$ is an integer, which happens iff $ m = 1, 2$ giving us $k=4, 3$ respectively. Hence, we've found all solutions to the condition.
*We could show that $a_i $ was a strictly increasing sequence using just the linear recurrence via $ka_{n+1} - a_n - 1 > a_{n+1}$ for $ k = 4$. With $ k =3$, we have to be slightly careful.
|
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|
The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)}{(x-a)}$
$0=(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$(x^2-a^2)^{\frac{1}{2}}(2a)=\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$2a\sqrt{x^2-a^2}=\frac{a(x+a)}{2\sqrt{x^2-a^2}}$
$2a(x+a)(x-a)=a(x+a)$
$2(x-a)=a$
$2x-2a=a$
$x=\frac{3}{2}a$
My book says x=2a. I'm not sure where I went wrong. Thanks in advance
|
For the ease of computation I use $$x=a\sec2\theta=a\cdot\frac{1+t^2}{1-t^2}$$ where $t=\tan\theta$ and $0\le\theta\le\dfrac\pi2$
So, $f(t)=\dfrac{4a^3}{t(1-t^2)}$
We need to attain the maximum positive value of $g(t)=t-t^3$ which can be achieved using Second derivative test
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
|
$\bf{Another\; Solution::}$Let
$$\displaystyle I = \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = \int\frac{\cos ^{\frac{5}{2}}x}{(1+\sin x)^{\frac{9}{2}}}dx = \int\frac{\cos^{\frac{3}{2}}x\cdot \cos x}{(1+\sin x)^{\frac{9}{2}}}dx$$
Now Let $\sin x = t\;,$ Then $\cos xdx = dt\;,$ So $$\displaystyle I = \int\frac{(1-t^2)^{\frac{3}{4}}}{(1+t)^{\frac{9}{2}}}dt = \int\frac{(1-t)^{\frac{3}{4}}}{(1+t)^{\frac{15}{4}}}dt$$
Now Let $\displaystyle (1+t) = s\;,$ Then $dt = ds$ and $$\displaystyle I = \int (2-s)^{\frac{3}{4}}\cdot s^{-\frac{15}{4}}ds = \int \left(\frac{2-s}{s}\right)^{\frac{3}{4}}\cdot s^{-3}ds$$
Now Let $\displaystyle \left(\frac{2}{s}-1\right) = y\;,$ Then $$\displaystyle -\frac{2}{s^2}ds = dy\Rightarrow ds = -\frac{1}{2}y^2dy$$
So $$\displaystyle I = -\frac{1}{4}\int y^{\frac{3}{4}}\cdot (y+1)dy = -\frac{1}{4}\int y^{\frac{7}{4}}dy-\frac{1}{4}\int y^{\frac{3}{4}}dy$$
So $$\displaystyle I = -\frac{1}{11} y^{\frac{11}{4}}-\frac{1}{7} y^{\frac{7}{4}}+\mathcal{C} = -\left[\frac{1}{11}\cdot \left(\frac{2-s}{s}\right)^{\frac{11}{4}}+\frac{1}{7}\cdot \left(\frac{2-s}{s}\right)^{\frac{7}{4}}\right]+\mathcal{C}$$
So $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = -\left[\frac{1}{11}\cdot \left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{11}{4}}+\frac{1}{7}\cdot \left(\frac{1-\sin x}{1+\sin x}\right)^{\frac{7}{4}}\right]+\mathcal{C}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
|
Using complex numbers, you can factor $x^6-1$ from the roots of unit.
$$(x^6-1)=(x-1)(x-e^{i\pi/3})(x-e^{2i\pi/3})(x+1)(x-e^{4i\pi/3})(x-e^{5i\pi/3}).$$
Dropping the factor $(x-1)$ and grouping for the conjugate roots*,
$$(x^2-x+1)(x+1)(x^2+x+1)=(x^4+x^2+1)(x+1)=x^5+x^4+x^3+x^2+x+1.$$
*Using
$(x-z)(x-z^*)=x^2-(z+z^*)x+zz^*=x^2-2\Re(z)x+|z|^2.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} = \frac{2\tan A}{\sec^2 A} = 2 \tan A\cos^2 A = 2 \sin A \cos A = \sin 2A $$
Since $x=A/2$, $\sin 2A = \sin x$
Let $t=\tan \frac{x}{2}$
$$ \int \frac{1}{\sin x} dx = \int \frac{2t}{1+t^2} dt $$
Let $u= 1+t^2$, $ du = 2t\,dt$
$$\int \frac{2t}{1+t^2} dt = \int \frac{2t}{u}\cdot\frac{1}{2t} du = \int \frac{1}{u} \,du = \ln u + C \\ = \ln(1+t^2) + C = \ln\left(1+\tan^2 \frac{x}{2} \right) + C $$
Then what do I do?
How do I show this is equal to $-\cos x + C$ ?
|
$$\int\frac{1}{\sin (x)} dx=\int\frac{\sin(x)}{\sin^2(x)}dx=\int\frac{\sin(x)}{1-\cos^2(x)} $$ Use substitution $t=\cos(x) \to dt=\sin(x)dx\to dx=\frac{dt}{\sin(x)}$
$$\int \frac{\sin(x)}{1-t^2}*\frac{dt}{\sin(x)}=\int \frac{dt}{1-t^2}=arth(t)=arth(\cos(x))+C$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent expansion with different annuli Find the Laurent expansion about $0$ of $$f(z)= \frac{1}{(z-i)(z-2)}$$ on the annuli:
*
*$0 \lt \lvert z \rvert \lt 1 $,
*$ 1 \lt \lvert z \rvert \lt 2$,
*$ 2 \lt \lvert z \rvert \lt \infty $.
So far I have put the function into partial fractions so $ \frac{i+2}{5} \left(\frac{1}{z-i} - \frac{1}{z-2} \right)$ but I'm not sure how to continue
|
First, we need to address your partial fraction decomposition since it is missing a multiplication by $-1$. You should have obtained
$$
\frac{1}{(z-i)(z-2)}=\frac{2+i}{5}\Bigl[\frac{1}{z-2}-\frac{1}{z-i}\Bigr]
$$
Recall that for $\lvert z\rvert < 1$, the geometric series $\sum_{n=0}^{\infty}z^n$ converges to $\frac{1}{1-z}$.
For $0<\lvert z\rvert < 1$, we have
\begin{align}
\frac{1}{z-2}&=\frac{-1}{2(1-z/2)}\\
&= -\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
\frac{-1}{z-i}&= \frac{1}{i(1-z/i)}\\
&=-i\sum_{n=0}^{\infty}\Bigl(\frac{z}{i}\Bigr)^n\\
&=\sum_{n=0}^{\infty}(-i)^{n+1}z^n
\end{align}
Our series is then
$$
\frac{2+i}{5}\sum_{n=0}^{\infty}z^n\biggl[(-i)^{n+1}-\Bigl(\frac{1}{2}\Bigr)^{n+1}\biggr]
$$
For $1<\lvert z\rvert < 2$, we have $1/\lvert z\rvert < 1$ and $\lvert z\rvert/2<1$ which means we can use our first geometric series again.
\begin{align}
\frac{1}{z-2}&=\frac{-1}{2(1-z/2)}\\
&= -\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\
\frac{-1}{z-i}&= \frac{-1}{z(1-i/z)}\\
&=\frac{-1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{i}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^{0}(-i)^{n+2}z^{n-1}
\end{align}
Our series is then
$$
\sum_{n=-\infty}^{0}(-i)^{n+2}z^{n-1}-\frac{1}{2}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n
$$
where if we peel of the zero term of the first series, we have what is referred to as the principal part of the Laurent series. That is,
$$
\frac{1}{z}+\underbrace{\sum_{n=-\infty}^{-1}(-i)^{n+2}z^{n-1}}_{\text{principal part}}
$$
For the final region $1<2<\lvert z\rvert$, we have $1/\lvert z\rvert < 1$ and $2/\lvert z\rvert < 1$.
\begin{align}
\frac{1}{z-2}&=\frac{1}{z(1-2/z)}\\
&= \frac{1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{2}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^0\Bigl(\frac{1}{2}\Bigr)^nz^{n-1}\\
\frac{-1}{z-i}&= \frac{-1}{z(1-i/z)}\\
&=\frac{-1}{z}\sum_{n=0}^{\infty}\Bigl(\frac{i}{z}\Bigr)^n\\
&=\sum_{n=-\infty}^0(-i)^{n+2}z^{n-1}
\end{align}
Our series is then
$$
\sum_{n=-\infty}^0\biggl[\Bigl(\frac{1}{2}\Bigr)^n+(-i)^{n+2}\biggr]z^{n-1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1060164",
"timestamp": "2023-03-29T00:00:00",
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|
Column space of a 3x3 matrix Let $A$ be a $3 \times 3$ matrix $A$ with reduced row echelon form $
\left[ {\begin{array}{c}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array} } \right]$.
Then
$\left[ {\begin{array}{c}
1 \\
0 \\
0
\end{array} } \right]$ and
$\left[ {\begin{array}{c}
0 \\
1 \\
0
\end{array} } \right]$ spans the column space of $A$.
I thought this would be a true statement but apparently it's not.
Isn't column space of A
$$span(\left[ {\begin{array}{c}
1 \\
0 \\
0
\end{array} } \right],
\left[ {\begin{array}{c}
0 \\
1 \\
0
\end{array} } \right],
\left[ {\begin{array}{c}
2 \\
0 \\
0
\end{array} } \right]) = span(\left[ {\begin{array}{c}
1 \\
0 \\
0
\end{array} } \right],
\left[ {\begin{array}{c}
0 \\
1 \\
0
\end{array} } \right]) $$
Thank you in advance!
|
The column space of the original matrix is spanned by the columns of the original matrix which, when reduced to row echelon form, contain a pivot. So for a counterexample, consider:
$$
A = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 1 & 0
\end{bmatrix}
$$
whose row reduced echelon form has the required form. Notice that:
$$
\text{Col}(A) = \text{span}\left\{
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
0 \\ 1 \\ 1
\end{bmatrix}
\right\}
\neq
\text{span}\left\{
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
\right\}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x$ $\forall x\neq -1$ Given function $y=f(x)$
such that
$$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$
find $f(x)$ and $f(2007)$.
|
$f\left(\dfrac{x-3}{x+1}\right)+f\left(\dfrac{x+3}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x+1}\right)+f\left(1+\dfrac{2}{x+1}\right)=x$
$f\left(1-\dfrac{4}{x}\right)+f\left(1+\dfrac{2}{x}\right)=x-1$
$f(1-2x)+f(1+x)=\dfrac{2}{x}-1$
$f(3-2x)+f(x)=\dfrac{2}{x-1}-1~......(1)$
$f(3-2(3-2x))+f(3-2x)=\dfrac{2}{3-2x-1}-1$
$f(4x-3)+f(3-2x)=-\dfrac{1}{x-1}-1~......(2)$
$(2)-(1)$ :
$f(4x-3)-f(x)=-\dfrac{3}{x-1}$
$f(4(4^x+1)-3)-f(4^x+1)=-\dfrac{3}{4^x+1-1}$
$f(4^{x+1}+1)-f(4^x+1)=-3\times4^{-x}$
$f(4^x+1)=\Theta(x)+4^{1-x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$f(x)=\Theta(\log_4(x-1))+\dfrac{4}{x-1}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
|
This might be irrelevant but it is equal to $2^{6n-1}+5 \cdot 9^n=2^{6n-6} \cdot 2^5+5\cdot 9^n$
Now since $2^6=64 \equiv -2 (mod 11)$, and $9^n \equiv -2(mod 11)$, the whole expression is equivalent to $(-2)^{n-1} \cdot 32+5 \cdot (-2)^n (mod 11) \equiv (-2)^n \cdot -16+5 \cdot (-2)^n (mod 11) \equiv 11 \cdot (-2)^n (mod 11)$, which is divisible by 11.
|
{
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"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.$
So $\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.$
Now $\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi$
Now How can I calculate $\sin 2\phi.$
Help me, Thanks
|
HINT:
$$\dfrac54=1+\sin x+\cos x+\sin x\cos x=1+\sqrt2\cos\left(\frac\pi4-x\right)+\dfrac{2\cos^2\left(\dfrac\pi4-x\right)-1}2$$
Write $\cos\left(\frac\pi4-x\right)=u$
$$4u^2+4\sqrt2u-3=0$$
We need $-1\le u\le1$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
|
Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$.
$b$ can be expressed as $a+x$, where $x$ is some positive integer.
to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$
Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$
$= a^a (a+x)^{a+x} - a^{a+x} (a+x)^a$
expanding the powers
$= a^a (a+x)^a (a+x)^x - a^a a^x (a+x)^a$
taking common factors out of bracket, we get
$= a^a (a+x)^a [ (a+x)^x - a^x ] $
since $x \gt0$, $(a+x)^x$ must be greater than $a^x$, therefore the above expression evaluates to a value greater than zero.
The other case of $a\gt b$ is the same as this since the order of $a$ and $b$ don't matter, while $a=b$ is trivial.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$24x^7 + 5y^2 = 15$ has no integer solutions, having none $\!\bmod 12$ Prove $5a^2\equiv k \pmod{12}$, where $k\in \{0,5,8,9\}$. Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions.
My lecturer used a table containing $a$, $a^2$, and $5a^2$ from $1$ to $11$ to which he applied $\pmod{12}$
for ex.
$a=3,a^2=9$, and $5a^2=9$
I didn't really understand what he continued on to do, does anyone have alternative method/proof?
|
Your lecturer calculated the remainder when $5a^2$ is divided by $12$, for $a=0$ to $11$. Actually we can stop at $6$, since for any $a$ we have $(12-a)^2\equiv a^2\pmod{12}$. (Indeed we could stop at $a=3$, since $(6-a)^2\equiv a^2\pmod{12}$.)
So calculating from $0$ to $6$, we find that modulo $12$, $5y^2$ can take on values $0,5,8,9, 8, 5,0$.
Now suppose that $24x^n+5y^2=15$. Then, modulo $12$, we have $5y^2\equiv 3$. That is impossible, since $3$ is not in our list of possible remainders when $5a^2$ is divided by $12$.
Remark: It is simpler to work modulo $4$. If $(x,y)$ is a solution of our equation, then $5y^2\equiv 3\pmod{4}$. That cannot happen if $y$ is even. And if $x\equiv \pm 1\pmod{4}$, then $5x^2\equiv 1\pmod{4}$, contradicting the fact that $5y^2\equiv 3\pmod{3}$.
|
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|
Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of
$$
\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}
$$
My idea was to use ratio test:
$$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$
$$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$
Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all
$\displaystyle n$ , but
$$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$
so I am not quite sure if I can apply ratio test.
|
Here's another way to prove the convergence
$$ \sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)} $$
$$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} $$
Also note that
$$ \left|\frac{1}{n^2+2n}\right|\leq \left|\frac{1}{n^2}\right| $$
And by the p-series test, we have
$$ \sum \limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right| \Rightarrow \mbox{converges} $$
Which implies that
$$ \sum \limits_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \mbox{converges absolutely} $$
Therefore, by the direct comparison test
$$ 2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} \Rightarrow \mbox{converges absolutely} $$
Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series.
|
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|
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$
Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation
$$8x^3+4x^2-4x-1=0$$
And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.
For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution
$$x\mapsto x^3$$
and getting another equation
$$8x^9+4x^6-4x^3-1=0$$
However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.
Also, it's given that the sum evaluates to a radical of the form
$$\sqrt[3]{\frac{1}{d}(a-b\sqrt[b]{c})}$$
where $a, b, c \: \text{&} \: d \in \mathbb Z$
Can somebody please help me with this question?
Thanks!
|
My solution:
Let $\alpha= \sqrt [3] {\cos \frac{2\pi}{7}}, \beta=\sqrt [3] {\cos \frac{4\pi}{7}}, \gamma=\sqrt [3] {\cos \frac{8\pi}{7}}$ so $(x-{\alpha}^3)(x-{\beta}^3)(x-{\gamma}^3)$ $=8x^3+4x^2-4x-1$.
hence we get $${\alpha}^3+{\beta}^3+{\gamma}^3=-\frac{1}{2} \tag{1}$$ $${\alpha}^3{\beta}^3+ {\beta}^3{\gamma}^3+ {\gamma}^3{\alpha}^3=\frac{-1}{2} \tag{2}$$
$${\alpha}^3{\beta}^3{\gamma}^3=\frac{1}{8} \Longrightarrow {\alpha}{\beta}{\gamma}=\frac{1}{2}. \tag{3}$$
Let $\mathcal{P}={\alpha}+{\beta}+{\gamma}$ and $\mathcal{Q}={\alpha}{\beta}+ {\beta}{\gamma}+ {\gamma}{\alpha}$.
From $(2)$, ${\alpha}^3+{\beta}^3+{\gamma}^3-3{\alpha}{\beta}{\gamma}=\mathcal{P}(\mathcal{P} ^2-3\mathcal{Q} ) \Longrightarrow \mathcal{P} ^3+2=3 \mathcal{P} \mathcal{Q} \ \tag{4}.$
From $(1), (3)$ and $\sum ( {\beta} ^2 {\gamma}+{\beta} {\gamma} ^2) =\mathcal{P} (\mathcal{P} ^2- 2 \mathcal{Q} )- \sum {\alpha}^3$ we get $\mathcal{Q} ^3=\sum {\beta}^3 {\gamma} ^3+6 {\alpha}^2 {\beta}^2 {\gamma}^2 +\frac{3}{2} ( \mathcal{P} (\mathcal{P} ^2- 2 \mathcal{Q} )+\frac{1}{2}),$ so combine with $(4)$ we get $$\mathcal{Q} ^3=\frac{7}{4}+\frac{3}{2} \mathcal{P} ^3-3 \mathcal{P} \mathcal{Q} =\frac{-1}{4}+\frac{1}{2} \mathcal{P} ^3 \Longrightarrow \mathcal{Q} ^3=\frac{1}{4} ( 2 \mathcal{P} ^3 -1). \tag{5}$$
From $(4), (5)$, $(\mathcal{P} ^3 +2)^3=27 \mathcal{P} ^3 \mathcal{Q} ^3=\frac{27}{4} \mathcal{P} ^3 ( 2 \mathcal{P} ^3 -1 ) \Longrightarrow 8 (\mathcal{P} ^3 )^3-60 (\mathcal{P} ^3)^2+150 \mathcal{P} ^3+64=0 ,$ hence we get:
$$(2\mathcal{P} ^3 -5)^3 = -189 \Longrightarrow \mathcal{P} ^3=\frac{5-3\sqrt [3]{7}}{2} \Longrightarrow \sqrt [3] {\cos \frac{2\pi}{7}}+\sqrt [3] {\cos \frac{4\pi}{7}}+\sqrt [3] {\cos \frac{8\pi}{7}}=\mathcal{P}$$
$$= \sqrt [3]{\frac{5-3\sqrt [3]{7}}{2}}.$$
Q.E.D
|
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|
Minimum value of $|z+1|+|z-1|+|z-i|$ How to find the minimum value of $|z+1|+|z-1|+|z-i|$.
I have tried geometrically etc but failed.
|
The post has received good and simple answers based on triangles.
However, if you really want to see the calculus based problem, let me start using Dr. Sonnhard Graubner's approach to the solution. Writing $$f(a,b)=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ Computing the partial derivatives lead to the equations $$\frac{df(a,b)}{da}=\frac{a}{\sqrt{a^2+(b-1)^2}}+\frac{a-1}{\sqrt{(a-1)^2+b^2}}+\frac{a+1}{\sqrt{(a+1)^2+
b^2}}=0$$ $$\frac{df(a,b)}{db}=\frac{b-1}{\sqrt{a^2+(b-1)^2}}+\frac{b}{\sqrt{(a-1)^2+b^2}}+\frac{b}{\sqrt{(a+1)^2+b^
2}}=0$$ which are extremely complex to solve except if you know (or are able to identify) that one solution corresponds to $a=0$. In such a case, the first partial is effectively equal to $0$ and the second one becomes $$\frac{df(0,b)}{db}=\frac{2 b}{\sqrt{b^2+1}}+\frac{b-1}{\sqrt{(b-1)^2}}=\frac{2 b}{\sqrt{b^2+1}}\pm 1=0$$ for which the solutions are $\mp \frac{1}{\sqrt{3}}$. To these solutions correspond $$f(0,-\frac{1}{\sqrt{3}})=1+\frac{5}{\sqrt{3}}$$ $$f(0,\frac{1}{\sqrt{3}})=1+\sqrt{3}$$ and then the result ($a=0,b=\frac{1}{\sqrt{3}},f=1+\sqrt{3}$).
You can notice that, for $b=\pm \frac{1}{\sqrt{3}}$, $$\frac{d^2f(0,b)}{db^2}=\frac{3 \sqrt{3}}{4} \gt 0$$
Please, notice that the problem with $|z+\alpha|+|z+\beta|+|z-i|$ will be extremely difficult to solve as soon as $\alpha+\beta \neq 0$.
|
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|
Rewrite $\sin(\cos^{-1}(x)-\tan^{-1}(y))$ as an algebraic function of $x$ and $y$. Rewrite the expression as an algebraic function of $x$ and $y$:
$$\sin(\cos^{-1}(x)-\tan^{-1}(y)).$$
I am unsure of how to change this into an algebraic function, yet I am able to simplify inso sin and cos.
thank you
|
Let $\theta = \arccos x$. Then $\cos\theta = x$, so we can draw a right triangle with adjacent side of length $|x|$ and hypotenuse of length $1$. By the Pythagorean Theorem, the opposite side has length $\sqrt{1 - x^2}$.
Let $\varphi = \arctan y$. Then $\tan\varphi = y$, so we can draw a right triangle with opposite side of length $|y|$ and adjacent side of length $1$. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{1 + y^2}$.
Thus,
\begin{align*}
\sin(\arccos x + \arctan y) & = \sin(\theta + \varphi)\\
& = \sin\theta\cos\varphi + \cos\theta\sin\varphi
\end{align*}
From the diagrams, we see that
\begin{align*}
\cos\theta & = x\\
\sin\theta & = \sqrt{1 - x^2}\\
\cos\varphi & = \frac{1}{\sqrt{1 + y^2}}\\
\sin\varphi & = \frac{y}{\sqrt{1 + y^2}}
\end{align*}
Make the appropriate substitutions, then simplify.
|
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|
How many distinct values of floor(N/i) exists for i=1 to N. Say we have a function $F(i)=\text{floor}(N/i)$.
Then how many distinct values of $F(i)$ will exist for all $0 \leq i \leq N$
e.g. We have $N=25$ then.
$F(1)=25$
$F(2)=12$
$F(3)=8$
$F(4)=6$
$F(5)=5$
...
...
...
$F(24)=1$
$F(25)=1$
So total distinct values of $F(i)$ are $(N=25)$ :- $25, 12, 8, 6, 5, 4, 3, 2, 1$
total distinct values are $9$: $(2 \times 5-1)$
Can anyone please help in that total number of distinct values are $O(\sqrt{N})$?
|
If $k \leqslant \sqrt{N}$, then $\lfloor N/\lfloor N/k\rfloor\rfloor = k$, since, letting $m = \lfloor N/k\rfloor$, we have $$mk\leqslant N < (m+1)k = mk + k \leqslant mk + m = m(k+1),$$ so that gives you $\lfloor \sqrt{N}\rfloor$ values. And the values of $\lfloor N/k\rfloor$ for $k \leqslant \lfloor \sqrt{N}\rfloor$ are all different, since
$$\frac{N}{k-1} - \frac{N}{k} = \frac{N}{k(k-1)} > 1$$
for $1 < k \leqslant \lfloor \sqrt{N}\rfloor$.
So you have either $2\lfloor \sqrt{N}\rfloor$ or $2\lfloor\sqrt{N}\rfloor - 1$ distinct values, depending on whether
$$N \geqslant \lfloor \sqrt{N}\rfloor(\lfloor\sqrt{N}\rfloor+1)$$
or not.
|
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|
Using trig substitution to solve for integration? So I used a trig sub for this problem:
$$\int \frac{1}{x^2\sqrt{9-x^2}}dx.$$
${x=3\sin\theta}$
${dx=3\cos\theta\ d\theta}$
${\sqrt{9-x^2}= 3\cos\theta}$
I ended up with
$$\frac19 \int \frac{ d\theta}{\sin^2\theta}$$
From there how do I solve that?
|
$${\int{1\over x^2\sqrt{9-x^2}}}dx$$
Your substitutions are fine:
$$x = 3\sin \theta \implies dx = 3\cos \theta \,d\theta$$
But be careful about $$\sqrt{9 - (3\sin \theta)^2}=\sqrt{9\cos^2 \theta} = 3|\cos\theta|$$
So we have $$\int \frac{3\cos \theta d\theta}{9\sin^2 \theta\cdot 3|\cos \theta|} \;= \;\frac{1}9\int \frac{\pm d\theta}{\sin^2 \theta}\; = \;\frac 19 \int \pm \csc^2\theta \,d\theta \;\;= \;\; \pm \frac 19 \int -\frac{d}{dx}\left(\cot \theta\right)\,dx= \;\;\mp \frac 19 \cot\theta + C$$
*
*Let me explain the use of $\pm, \mp$. When $\pi/2 \lt \theta\lt
3\pi/2, $ then $\cos \theta < 0$ and so $|\cos \theta| = -\cos
\theta$.
Now you need only back-substitute by expressing $\theta$ in terms of $x$.
|
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|
How to separate variables in this equation: $\;y\frac{dy}{dx} = (x+7)(y^2+6)\;?$ I need to solve the differential equation $$y\frac{dy}{dx} = (x+7)(y^2+6)$$
I know that the first step is to isolate both term each side and then integrate...
But I can't figure out how to isolate term on this one, I would probably be able to solve the rest of the equation, just need to know how to start.
Thanks.
|
$$y\dfrac{dy}{dx}=(x+7)(y^2+6)\implies \dfrac{y}{y^2+6}\dfrac{dy}{dx}=(x+7)\implies \dfrac{y}{y^2+6} dy=(x+7) dx\implies \int\frac{y}{y^2+6}dy=\int x+7 \,dx\implies \frac12\log|y^2+6|=\dfrac{x^2}2+7x\implies \log|y^2+6|=x^2+14x+c\implies y^2+6=e^{x^2+14x+c}\implies y=\sqrt{e^{x^2+14x+c}-6}$$
|
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|
Showing that a function diverges for large x Consider the function $$f(x) = \frac{x^{3}}{x^{2}+1}$$ Show that as $x\to\infty$ we have that $f(x)\to\infty$. ie. I want to prove that $$\lim_{x\to\infty} \frac{x^3}{x^2+1} = \infty.$$
So given $k>0$ I want to find $M>0$ such that $$x>M\implies |f(x)|>k.$$
First note that
\begin{align}
|f(x)| &= \left|\frac{x^3}{x^2+1}\right|\\
&\ge\left|\frac{x^3}{2x^2}\right|\\
&= \frac{x}{2}
\end{align}
So to ensure $|f(x)|>k$, we can let $x/2>k$, or $x>2k$, thus whenever $M=2k$ we are sure that for all $x>M$
\begin{align}
|f(x)|&=\left|\frac{x^3}{x^2+1}\right|\\
&\ge\left|\frac{x^3}{2x^2}\right|\\
&= \frac{x}{2} \\
&> k.& \text{since } x>M>0\implies \frac{x}{2}>\frac M2=k>0.
\end{align}
Is this a sufficient and correct proof?
|
Write $f(x) = x-{x \over x^2+1}$, and since $|{x \over x^2+1}| \le {1 \over 2}$, you have
$f(x) \ge x-{1 \over 2}$. Then for all $k>0$, if you let $M=k+1$, then you have the desired result.
|
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|
How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
|
I would go about this by first splitting the series up:
$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$
Letting $s=1+2x^2+3x^4$ we can do a few tricks:
$$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$
$$=1+x^2+x^4+...$$
Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, and can be done by a technique like the above). This gives
$$s -x^2s=\frac{1}{1-x^2}\Leftrightarrow s=\frac{1}{(1-x^2)^2}=$$
Thus the original series converges to:
$$(1+x)s=\frac{(1+x)}{(1-x^2)^2}$$
For $-1 < x - 1$.
|
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|
Series expansion of $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$? How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again??
|
$$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$
$$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$
|
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|
How to find LCM of this equation? Can $(x+1)(2x-1)$ be the LCM of this biquadratic equation
$$\frac{5x-1}{x+7}=\frac{3x+1}{x+5}$$
|
The lowest common denominator (LCD) of rational fractions
$$\frac{5x-1}{x+7}\quad {\rm and} \quad \frac{3x+1}{x+5}$$
is the lowest common multiple (LCM) of denominator polynomials
$$ x+7 \quad {\rm and} \quad x+5, $$
which is
$$ (x+7)(x+5).$$
So, formally, the original equality of rational fractions can be re-written as:
$$\frac{(5x-1)(x+5)}{(x+7)(x+5)}= \frac{(3x+1)(x+7)}{(x+5)(x+7)}.$$
|
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|
Proving the convergence of $\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$ How to prove that the series
$$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$
is convergent? What about finding the sum?
My attempt:
$$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$
The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.
|
$$log(1-\frac{1}{n^2})= \\log(\frac{n^2-1}{n^2})=\\log(\frac{n-1}{n})+log(\frac{n+1}{n})\\s_{1}+s_{2}\\s_{1}=log(\frac{2-1}{2})+log(\frac{3-1}{3})+log(\frac{4-1}{4})+...++log(\frac{n-1}{n})=\\+log(\frac{1}{2}\frac{2}{3}\frac{3}{4}...\frac{n-1}{n})\\=log(\frac{1}{n})\\$$ $$
s_{2}=log(\frac{2+1}{2})+log(\frac{3+1}{3})+log(\frac{4+1}{4})+...++log(\frac{n+1}{n})=\\log(\frac{3}{2}\frac{4}{3}\frac{5}{4}...\frac{n+1}{n})\\=log(\frac{n+1}{2})\\s_{1}+s_{2}=\\log(\frac{1}{n})+log(\frac{n+1}{2})=log(\frac{n+1}{2n}) \rightarrow log(\frac{1}{2})
$$
|
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|
Cube roots escape $ \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c}, $
where $ a, b $ and $ c $ are positive integers. What is the value of $ a+b+c $?
This question appeared in one of my exams
|
You can check the following expression works by squaring the right hand side
$$
\sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25} \, .
$$
So $ a + b + c = 47 $.
As for finding it: It is initially a good idea to try working in the field extension $ \mathbb{Q}(\sqrt[3]{4}, \sqrt[3]{5}) = \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, which has a basis given by
$$
\{ 1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{1\times5}, \sqrt[3]{2\times5}, \sqrt[3]{4\times5}, \sqrt[3]{1\times25}, \sqrt[3]{2\times25}, \sqrt[3]{4\times25} \} \, .
$$
Then you want to consider the equation
$$
3\sqrt{\sqrt[3]{5} - \sqrt[3]{4}} = r + s \sqrt[3]{2} + t \sqrt[3]{4} + u \sqrt[3]{5} + v \sqrt[3]{10} + w \sqrt[3]{20} + x \sqrt[3]{25} + y \sqrt[3]{50} + z \sqrt[3]{100} \, ,
$$ with $ r, \ldots, z \in \mathbb{Q} $. In particular you want $ r, \ldots, z \in \{ -1, 0, 1 \} $ with exactly three of them non-zero. Square both sides, and compare the coefficient of $ \sqrt[3]{5} $ and $ \sqrt[3]{4} $ on both sides. We get
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 2ru + 4tv + 4sw + 20yz \\
\sqrt[3]{4} &: \quad -9 = s^2 + 2rt + 10wx + 10vy + 10uz \, .
\end{align}
You've assumed that all the non-zero coefficients are $ \pm 1 $, so we must have $ x = \pm 1 $ and $ s = \pm 1 $, by parity considerations. By assumption there are only 3 non-zero coefficients, $ s, x $ and one other. So $ ru = tv = yz = 0 $ and $ rt = vy = uz = 0 $ and the equations reduce to
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 4sw \\
\sqrt[3]{4} &: \quad -9 = s^2 + 10wx \, .
\end{align}
We must have $ sw = 1 $ and $ wx = -1 $, so $ s = w = -x $. This gives us two possibilities by taking $ x = 1 $ or $ x = - 1 $. The form of the solution you are looking for has two $ +1 $ coefficients. This must come from $ x = -1 $, and be
$$
\sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25}
$$
Now you need to go back and check this actually does work as a solution because we've made many assumptions in finding it. In particular we've assumed that such a solution exists and lives in $ \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, rather than some larger field extension. And we've only used two equations which come from comparing coefficients; do the rest also work?
|
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|
Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form
$\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple algorithm to generate these patterns: simply take $(n+2^p-1)^p$ and subtract $(n+2^p-2)^p$ (using $n\to n-1$) to get a polynomial of degree $p-1$. Take this difference and subtract $(n+2^p-3)^p-(n+2^p-4)^p$ (using $n\to n-2$) to get a polynomial of degree $p-2$. Repeat this process until $n^p$ is reach. The first few examples of this are
$$
\begin{align}
n^0&=1\\
(n+1)^1-n^1&=1\\
(n+3)^2-(n+2)^2-(n+1)^2+n^2&=4\\
(n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3&=48
\end{align}
$$
Upon doing this for the next several powers and checking OEIS, it would appear the constant corresponding to the power $p$ is $$\large C_p=2^{\frac{p(p-1)}2} p!$$
However, this is an observation only, and I have no idea how to go about proving this. The only thing I notice is that $\frac{p(p-1)}{2}=\sum\limits_{k=1}^{p-1}k$, but I don't know how to use this fact. Does any one know how to prove this observation?
|
Let $X = \mathbb{R}^{\mathbb{Z}}$ be the space of real valued sequences defined over $\mathbb{Z}$. Let $R : X \to X$ be the operator on $X$ replacing the terms of a sequence by those on their right. More precisely,
$$X \ni (\ell_n)_{n\in\mathbb{Z}} \quad\mapsto\quad ( (R\ell)_n = \ell_{n+1} )_{n\in\mathbb{Z}} \in X$$
The identities you have can be rewritten as
$$\begin{array}{rcl}
(R - 1)n^1 &=& 1\\
(R^2-1)(R-1) n^2 &=& 4\\
(R^4 - 1)(R^2-1)(R-1) n^3 &=& 48\\
&\vdots&\\
(R^{2^{p-1}}-1)(R^{2^{p-2}}-1)\cdots(R^{2^0}-1) n^{p} &\stackrel{?}{=}& C_p = ???\tag{*1}
\end{array}$$
Notice for any polynomial $f(n)$ of degree $p$ and leading coefficient A,
$$f(n) = A n^p + A' n^{p-1} + ( \text{something of degree }< p-1 )$$
We have
$$\begin{align}
(R^{2^{p-1}} - 1) f(n)
&= A\left((n+2^{p-1})^p - n^p \right) + A'\left((n+2^{p-1})^{p-1} - n^{p-1}\right) + \cdots\\
&= A p2^{p-1} n^{p-1} + ( \text{mess with degree }< p-1 )\\
\end{align}
$$
This means $(R^{2^{p-1}}-1)f(n)$ will be a polynomial with degree $p-1$ and leading coefficient $A \cdot p 2^{p-1}$. Repeat applying this to the last equation in $(*1)$ $p$ times, we find the LHS of the last equation is equal to a polynomial with degree $0$. i.e.
$C_p$ is indeed a constant. Furthermore,
$$C_p = 1 (p 2^{p-1} )((p-1) 2^{p-2}) \cdots (2 \cdot 2^{1}) (1 \cdot 2^{0})
= p! 2^{(p-1)+(p-2)+\cdots + 1} = p!2^{\frac{p(p-1)}{2}}$$
|
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}
|
How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?
|
$$\sqrt{x^2+6x}=\sqrt{x^2+6x+9-9}=\sqrt{(x+3)^2-9}\approx |x|+3,\text{ as }x\to-\infty$$
$$\sqrt{x^2-2x}=\sqrt{x^2-2x+1-1}=\sqrt{(x-1)^2-1}\approx |x|-1,\text{ as }x\to-\infty$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1080479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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|
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
|
Rewrite the given term as
$$x\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}=\frac{x}{\sqrt{x^2+1}+x}$$
Assuming $x>0$, we have
$$\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}$$
This clearly tends to $\frac{1}{2}$ as $x\rightarrow\infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
How to integrate $f(x) = \frac{1}{a + b \cos x + c \sin x }$ over $x \in (0,\pi/2)$
Conjecture 1
$$ \begin{align*} I_{T}=\int_0^{2\pi} \frac{\mathrm{d}x}{a + b \cos x + c
\sin x} & = \frac{2\pi}{\rho} \tag{1} \\ I_{T/4} = \int_0^{\pi/2} \frac{\mathrm{d}x}{a +
b \cos x + c \sin x} & = \frac{2}{\rho} \arctan \left(\frac{\rho}{a+b+c}\right)\tag{2}
\end{align*} $$
Where $\rho^2 = a^2-b^2 -c^2$.
*
*Is this result correct, and does it cover all cases?
*Does the result above cover all cases - what if the denominator becomes a perfect square?
*When does the integral converge?
I am mainly posting this so that i can close all the duplicates in
the future. Integrals related to these seems to pop up regularly.
|
First we need a lemma or two.
Lemma 1:
$ \hspace{3.75cm} \displaystyle
\int \frac{\,\mathrm{d}x}{1+x^2} = \arctan x+\mathcal{C}
$
Proof: Use the substitution $x \mapsto \tan u$. Then we have
$$
\mathrm{d}x
= \left( \frac{\sin u}{\cos u} \right)'\mathrm{d}u
= \frac{(\sin u)' \cdot \cos u - \sin u \cdot (\cos u)'}{\cos^2u}\mathrm{d}u
= (1+\tan^2u)\,\mathrm{d}u
$$
Which means that $\mathrm{d}u = \cfrac{\mathrm{d}x}{1+(\tan u)^2} = \cfrac{\mathrm{d}x}{1+x^2}$. Hence
$$
\int \frac{\,\mathrm{d}x}{1+x^2}
= \int \mathrm{d}u
= u + \mathcal{C}
= \arctan x + \mathcal{C}
$$
Since $u \mapsto \tan x$, this means $x = \arctan u$. $\square$
Lemma 2:
Assume that $t \in [0,\infty]$ and $\lambda \in \mathbb{R}\backslash\{0\}$. Then
$$
\int_0^t \frac{\mathrm{d}x}{\lambda^2+x^2} = \frac 1\lambda \arctan \frac t \lambda
$$
Proof: Using Lemma 1 this can be done quickly. Use $x \mapsto \lambda v$. Then $\mathrm{d}x = \lambda \mathrm{d}v$, and so
$$
\int_0^t \frac{\mathrm{d}x}{\lambda^2+x^2}
= \int_0^{t/\lambda} \frac{\lambda \mathrm{d}v}{\lambda^2(1+v^2)}
= \frac{1}{\lambda} \int_0^{t/\lambda} \frac{\mathrm{d}v}{1+v^2}
= \frac 1\lambda \arctan \frac{t}{\lambda}
$$
Corollary 1:
Assume that $\lambda \in \mathbb{R}\backslash\{0\}$. We have
$$
\int_0^1 \frac{\,\mathrm{d}x}{\lambda^2+x^2} = \frac 1\lambda \arctan \frac 1\lambda \ \ \text{and} \ \ \int_{-\infty}^\infty \frac{\mathrm{d}x}{\lambda^2+x^2} = \frac{\pi}{\lambda}
$$
Proof: The first equation follows directly from setting $t=1$ in Lemma 1. For the next equation note that $\int_{-t}^t \frac{\mathrm{d}x}{1+x
2} = 2 \int_0^t \frac{\mathrm{d}x}{1+x^2}$. Now the rest follows directly since $\arctan t \to \pi/2$ as $t \to \infty$. Proving $\int_{-t}^0 \frac{\mathrm{d}x}{1+x^2} = \int_{-t}^0 \frac{\mathrm{d}x}{1+x^2}$ I will leave as an excercise for the reader. $\square$.
Now for the last part of the proof, we need the following proposition.
Proposition 1 (Weierstrass substitution):
Assume $a,b \in[-\pi,\pi]$. Then we have
$$
\int_a^b R(\sin x,\cos x,\tan x)\,\mathrm{d}x
= \int_\varphi^\psi R \left( \frac{2t}{1+t^2} , \frac{1 - t^2}{1+t^2},\frac{2t}{1-t^2}\right) \frac{2 \mathrm{d}t}{1+t^2}
$$
where $\varphi = \tan(a/2)$ and $\psi = \tan(b/2)$.
The proof for this is omitted but can be proven by using $x \mapsto \tan u/2$. It is not hard but requires some work. For a lengthier discussion see Wikipedia.
Now we can finally start our proof.
Using the above proposition one has
$$
\int \frac{\mathrm{d}x}{a + b \cos x + c \sin x}
= \int \frac{1}{a + b \frac{1-t^2}{1+t^2} + c \frac{2t}{1+t^2}} \frac{\mathrm{d}t}{1+t^2}
= \int \frac{2\,\mathrm{d}t}{(a-b)t^2+2ct+a+b}
$$
Now we can complete the square. This gives
$$
\int \frac{\mathrm{d}x}{a + b \cos x + c \sin x}
= \frac{1}{a-b} \int \frac{\mathrm{d}t}{\left(t+\frac{c}{a-b}\right)^2+\frac{a+b}{a-b}-\frac{c^2}{(a-b)^2}}
$$
Alas not much more can be done before inserting the limits. The simplest case is the $I_T$ case. Here we have
$$
\begin{align*}
I_T & = \int_0^{2\pi} \frac{\mathrm{d}x}{a + b \cos x + c \sin x} \\
& = \frac{1}{a-b} \int_{-\infty}^\infty \frac{\mathrm{d}t}{\left(t+\frac{c}{a-b}\right)^2+\frac{a+b}{a-b}-\frac{c^2}{(a-b)^2}} \\
& = \frac{2}{a-b}\int_{-\infty}^\infty \frac{\mathrm{d}y}{y^2+\frac{a^2-b^2-c^2}{a-b}}
\end{align*}
$$
where we used the obvious substitution $y \mapsto t + c/(a-b)$. Using Corollary 1 we have
$$
I_T = \frac{2}{a-b} \pi \Big/\left( \frac{a^2-b^2-c^2}{a-b} \right)
= \frac{2\pi}{\lambda}
$$
where $\lambda^2 = a^2-b^2-c^2$ was used in the last equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1081516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
What is wrong with the sum of these two series? Could anyone help me to find the mistake in the following problem? Based on the formula of the sum of a geometric series:
\begin{equation}
1 + x + x^{2} + \cdots + x^{n} + \cdots = \frac{1}{1 - x}
\end{equation}
\begin{equation}
1 + \frac{1}{x} + \frac{1}{x^{2}} + \cdots + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - 1/x} = \frac{x}{x-1}
\end{equation}
Adding both equations
\begin{equation}
2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = \frac{1}{1 - x} + \frac{x}{x-1} = \frac{1-x}{1-x} = 1
\end{equation}
So,
\begin{equation}
2 + x + \frac{1}{x} + x^{2} + \frac{1}{x^{2}} + \cdots + x^{n} + \frac{1}{x^{n}} + \cdots = 1
\end{equation}
And the left side is always bigger than $2$ for $x>0$.
What is wrong?? Thanks in advance
|
The first series only applies when $|x| < 1$ whereas the second series only applies when $\left|\frac{1}{x}\right| < 1$ (i.e. $|x| > 1$). By adding them, you are assuming that they both apply simultaneously, but they don't (for any $x$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1081892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
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|
Find Min Values Of $P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$ Given $x,y,z>0$ and $y+z=x(y^2+z^2)$
Find Min Values Of
$P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$
Could Someone give me an idea ?
|
let $y+z=2t\implies y^2+z^2\ge 2t^2 \implies xt\le 1$
$P\ge \dfrac{1}{(1+x)^2}+\dfrac{2}{(1+y)}\dfrac{1}{(1+z)}+\dfrac{4}{(1+x)(1+y)(1+z)}=\dfrac{1}{(1+x)^2}+\dfrac{2}{(1+y)(1+z)}(1+\dfrac{2}{(1+x)})\ge \dfrac{1}{(1+x)^2}+\dfrac{2*4}{(1+y+1+z)^2}(1+\dfrac{2}{(1+x)})={(1+x)^2}+\dfrac{2}{(1+t)^2}(1+\dfrac{2}{(1+x)})\ge {(1+x)^2}+\dfrac{2x^2}{(1+x)^2}(1+\dfrac{2}{(1+x)})=f(x) $
note all "=" above will hold when $y=z$,so you can find min of $f(x)$ and $P$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1082914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate Complex Integral with $\frac{\Gamma(\frac{s}{2})} {\Gamma\big({\beta +1\over 2} - {s\over 2}\big)}$ I am proving this integral:
$$ \int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\beta^{1 \over 2}\,\right)^{s}\ \Gamma\left(\,s \over 2\,\right) \Gamma\left(\,{\beta +1 \over 2} - {s \over 2}\,\right)\,{\rm d}s = \Gamma\left(\,{\beta +1 \over 2}\,\right)\left(\,1+{1 \over \beta}\left(x \over \sigma\right)^2\,\right)^{-{\beta +1 \over 2}}$$
where $\beta>0$, $\sigma>0$ and $x$ is real number
The clue I have is that Cauchy's residue theorem is applicable for the evaluation but I cant figure out how can the simplification be made
|
The equation should actually be
$$\frac{1}{4\pi j} \int_{c - j\infty}^{c + j\infty} (x^{-1}\sigma \beta^{\frac{1}{2}})^s\, \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{\beta + 1}{2} - \frac{s}{2}\right)\, \mathrm{d}s = \Gamma\left(\frac{\beta + 1}{2}\right) \left[1 + \frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-\frac{\beta + 1}{2}}$$
for $0 < c <\beta + 1$. Let $F(x;\sigma,\beta)$ be the expression on the left-hand side. Then
\begin{align}F(x;\sigma,\beta) &= \frac{1}{2\pi j}\int_{\frac{c}{2} - j\infty}^{\frac{c}{2} + j\infty} (x^{-1}\sigma\beta^{\frac{1}{2}})^{2s}\, \Gamma(s) \Gamma\left(\frac{\beta + 1}{2} - s\right)\, \mathrm{d}s\\
&= \frac{\Gamma\left(\frac{\beta+1}{2}\right)}{2\pi j}\int_{\frac{c}{2} - j\infty}^{\frac{c}{2} + j\infty} \left[\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-s} B\left(s, \frac{\beta+1}{2}-s\right)\, \mathrm{d}s\\
&= \Gamma\left(\frac{\beta+1}{2}\right) \mathcal{M}^{-1}\left\{B\left(s,\frac{\beta+1}{2}-s\right)\right\}\left[\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right],
\end{align}
where $B(u,v)$ represents the Beta function and $\mathcal{M}^{-1}(f(s))[r]$ represents the inverse Mellin transfrom of $f(s)$ evaluated at $r$. Now, for $0 < \text{Re}(s) < \frac{\beta + 1}{2}$, $$B\left(s,\frac{\beta+1}{2}-s\right) = \int_0^\infty \frac{r^{s-1}}{(1 + r)^{s + \left(\frac{\beta + 1}{2} - s\right)}}\, \mathrm{d}r = \int_0^\infty r^{s-1} (1 + r)^{-\frac{\beta+1}{2}}\, \mathrm{d}r,$$
that is, $B\left(s, \frac{\beta+1}{2}-s\right)$ is the Mellin transform of $(1 + r)^{-\frac{\beta + 1}{2}}$ evaluated at $s$. Hence,
$$F(x;\sigma,\beta) = \Gamma\left(\frac{\beta+1}{2}\right)(1 + r)^{-\frac{\beta+1}{2}}\bigg|_{r = \frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2} = \Gamma\left(\frac{\beta+1}{2}\right)\left[1+\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-\frac{\beta+1}{2}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1083360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Interesting Surd Problem If $ x $, $ y $ and $ z $ are rational numbers such that
$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then find $ x,y,z $
|
According to Srinivasa Ramanujan,
$\sqrt[3]{\sqrt[3]{2}-1}
=\sqrt[3]{\frac{1}{9}}
-\sqrt[3]{\frac{2}{9}}
+\sqrt[3]{\frac{4}{9}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1083434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Double integral $ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy$, $D$=$\{(x,y): x^2+y^2 \le1 , y\ge0\}$
Solve $$ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy \ \ \ \ . . . \ (*)$$
where $D$=$\{$$(x,y): x^2+y^2 \le1 , y\ge0 $$\}$
$$
$$
Here is my attempt.
$$\begin{align}
&(1).\ \ \ (*)=\int_{-1}^1 \int_{0}^{\sqrt{1-x^2}}\frac{y}{x^2+(y+1)^2}dydx \\
&(2).\ \ \ (*)= \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\
&(3). \ \ \int\frac{y+1}{x^2+(y+1)^2}dx = \arctan\left(\frac{x}{y+1}\right) + C \\
&(4). \ \ \ (*)=\int_{0}^{\pi} \int_{0}^{1}\frac{r^2sin\theta}{r^2+2rsin\theta+1}drd\theta \\\\
\end{align}$$
I used $(1)$, $(4)$ and $(2)$ with $(3)$,
but didn't solve yet.
$$$$
Did I make a mistake?
Could you give me some advice, please?
How can I solve this integral...
Thank you for your attention to this matter.
$$$$
P.S.
Here is result of wolframalpha
$$$$
$$ $$
Additionally... I did like this.. maybe useless :-(
$$\begin{align}
(*)
& = \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y+1}{x^2+(y+1)^2}dxdy + \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \left(\arctan\left(\frac{\sqrt{1-y^2}}{y+1}\right) - \arctan\left(\frac{-\sqrt{1-y^2}}{y+1}\right)\right)dy \\
& \ \ \ \ + \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1 \left(\arctan\left(\sqrt\frac{1-y}{1+y} \ \right) - \arctan\left(-\sqrt\frac{1-y}{1+y} \ \right)\right)dy \\
& \ \ \ \ +\int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&= terrible?! \\
\end{align}$$
$$
$$
$$ $$
---------------------------------------------------------------------------
This picture is for asking to Christian Blatter
(I am really sorry, if I bother you guys for this picture.)
|
I shall introduce new coordinates (again denoted by $x$, $y$) such that the point $(0,-1)$ becomes the origin, and your vertical axis is my horizontal axis. Your integral then appears as
$$J:=\int_H{x-1\over x^2+y^2}\>{\rm d}(x,y)\ ,$$
where $H$ is the right half of the unit disk with center $(1,0)$. Introducing polar coordinates we obtain
$$J=\int_{-\pi/4}^{\pi/4}\int_{1/\cos\phi}^{2\cos\phi}{r\cos\phi-1\over r^2} r\>dr\ d\phi\ .$$
Here the inner integral evaluates to
$$(2\cos^2\phi-1)-(\log 2+2\log\cos\phi)\ .$$
We therefore get
$$J=1-{\pi\over2}\log 2-4\int_0^{\pi/4}\log\cos\phi\ d\phi=1+{\pi\over2}\log 2-2 \>{\tt Catalan}\doteq0.256862\ ,$$
where ${\tt Catalan}$ is Catalan's constant ($\doteq0.915966$)
|
{
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"url": "https://math.stackexchange.com/questions/1085319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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|
Arithmetic Error in Calculation of the Limit of a Given Function I consider a function $f(x)$ which is equal to $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6}$
While trying to evaluate the $\lim_{x \to 6} f(x)$
It is true that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6}$
It is also true that $\dfrac{a}{\frac{b}{c}} = \dfrac{a\cdot c}{b}$
I apply this property to $f(x)$ and come to the conclusion that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6} = \dfrac{3x-18}{x}-\dfrac{x-6}{2}$, which is of course incorrect.
Additionally, I believe I'm taking an improper approach to solving this limit.
|
$$
\frac 3 x\cdot 2x = 6\text{ since $x$ cancels, and }\frac 1 2\cdot 2x = x\text{ since $2$ cancels.}
$$
Therefore
$$
\frac{\frac{3}{x}-\frac{1}{2}}{x-6}=\frac{\left(\frac{3}{x}-\frac{1}{2}\right)2x}{(x-6)2x} = \frac{6-x}{(x-6)2x} = \frac{-1(x-6)}{(x-6)2x} = \frac{-1}{2x}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1086930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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|
Computing the exponential of the operator $ H(\textbf{x}) = \alpha \textbf{n} \times\textbf{x}$ Define the vector operator: $$ H(\textbf{x}) \equiv \alpha \textbf{n} \times\textbf{x}$$ For unit vector $\textbf{n}$ and some constant $\alpha$. We define further the operator: $$G \equiv I + H + \frac{H^2}{2!} + \frac{H^3}{3!} + ... $$ Where powers of $H$ represent iterations and $I$ is defined as the identity such that $I(\textbf{x}) = \textbf{x}$
We are to prove $$ G\textbf{x} = \textbf{x} +(\textbf{n}\times\textbf{x})\sin{\alpha} + \textbf{n}\times(\textbf{n}\times\textbf{x})(1-\cos{\alpha}) $$
My attempt at the proof involved noting: $$ G = e^{H} $$ And thus $$G\textbf{x} = e^H\textbf{x}$$
But I can't seem to see where to go from here, or even if the last step is a legitimate one. Many thanks in advance.
|
$H x = \alpha \, n \times x$ gives the matrix:
$$
H = \alpha \left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right) = \alpha N
$$
Then for your proof
$$
G = I + \sin(\alpha) N + (1-\cos\alpha) N^2
$$
needs to be shown.
So we calculate $N^2$ and $N^3$:
\begin{align}
N^2
&=
\left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
-(n_2^2+n_3^2) & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & -(n_1^2 + n_3^2) & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & -(n_1^2 + n_2^2)
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
n_1^2-1 & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & n_3^2 - 1
\end{matrix}
\right)
\end{align}
where we used $1 = n_1^2 + n_2^2 + n_3^2$ for the unit vector $n$ and then
\begin{align}
N^3
&=
\left(
\begin{matrix}
n_1^2-1 & n_1 n_2 & n_1 n_3 \\
n_1 n_2 & n_2^2 - 1 & n_2 n_3 \\
n_1 n_3 & n_2 n_3 & n_3^2 - 1
\end{matrix}
\right)
\left(
\begin{matrix}
0 & -n_3 & n_2 \\
n_3 & 0 & -n_1 \\
-n_2 & n_1 & 0
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
0 & n_3 & -n_2 \\
-n_3 & 0 & n_1 \\
n_2 & -n_1 & 0
\end{matrix}
\right) = -N
\end{align}
This means all $N^k$ can be expressed with $I$, $N$ and $N^2$:
$$
N^0 = I \quad N^{2k} = (-1)^k N^2 \quad N^{2k+1} = (-1)^k N
$$
for $k > 0$ and we can calculate
\begin{align}
G = e^H
&=
\sum_{k=0}^\infty \frac{1}{k!} H^k \\
&=
\sum_{k=0}^\infty \frac{1}{(2k)!} H^{2k} +
\sum_{k=0}^\infty \frac{1}{(2k+1)!} H^{2k+1} \\
&=
\sum_{k=0}^\infty \frac{1}{(2k)!} \alpha^{2k} N^{2k} +
\sum_{k=0}^\infty \frac{1}{(2k+1)!} \alpha^{2k+1} N^{2k+1} \\
&=
I + \sum_{k=1}^\infty \frac{(-1)^k}{(2k)!} \alpha^{2k} N^2 +
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \alpha^{2k+1} N \\
&=
I + (1-\cos \alpha) N^2 + \sin(\alpha) N
\end{align}
|
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|
A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$? The inequality
$$
e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n,
$$
where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.
Are there any more elementary proofs of this inequality?
|
You can use induction.
For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,
$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \right) \left(1 - \frac{1}{n} \right) = 3 - \frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)}.$$
It is left to show
$$-\frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)} \leq -\frac{1}{n}$$
or equivalently
$$ -3n+3-n+1 \leq -n+1,$$
i.e. $n \geq 1$. Since $n-1 \in \mathbb{N}$, we have $n \geq 2$ by choice, so the inequality holds.
|
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|
How to find sum of this series? $\sum_{n=1}^{+\infty}(-1)^{n+1}\prod_{k=1}^n\frac{3k-2}{5k}$ How to find sum of this series?
$$\frac{1}{5}-\frac{1\cdot 4}{5\cdot 10}+\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}-\ldots = \sum_{n=1}^{+\infty}(-1)^{n+1}\prod_{k=1}^n\frac{3k-2}{5k}$$
Any hint is welcome.
|
The sum of the series is: $$1-\frac{1}{2}\sqrt[3]{5}.$$
As suggested by Abhishek Bakshi, this comes from considering the Taylor series of
$$ f(x) = \left(1+\frac{3}{5}x\right)^{1/3} $$
around $x=0$.
|
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|
$y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\frac{dx}{dt}$$
but the correct answer is apparently
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\left({\frac{dx}{dt}}\right)^2$$
|
You are misapplying the product rule. Note that
$$\begin{aligned}
\frac{d}{dt} \left( - \frac{1}{x^2} \frac{dx}{dt} \right) & = \left( \frac{d}{dt} \frac{(-1)}{x^2} \right) \frac{dx}{dt} - \frac{1}{x^2} \left( \frac{d}{dt} \frac{dx}{dt} \right) \\
& = \left( \frac{2}{x^3} \cdot \frac{dx}{dt} \right) \frac{dx}{dt} - \frac{1}{x^2} \frac{d^2x }{dt^2} \\
& = \frac{2}{x^3} \left( \frac{dx}{dt} \right)^2 - \frac{1}{x^2} \frac{d^2 x}{dt^2}.
\end{aligned}$$
You forgot to apply the chain rule to $- \frac{1}{x^2}$.
|
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|
Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me a formal proof based on this animation?
|
Notice that the base of the $n^\text{th}$ cube is the diagonal square prism $n^2\times 1$, and that each cube is build out of the diagonal square prism and rectangular prisms above and to the left of the diagonal square prism. Since adding another term to the left hand sum will give us one more diagonal square and set of prisms above and left of it, this suggests we can use induction.
For $n=1$, we trivially have $\left(\sum_{m=1}^n m\right)^2 = \sum_{m=1}^n m^3$.
Now, let's suppose $\left(\sum_{m=1}^n m\right)^2 = \sum_{m=1}^n m^3$ for some $n$. Then,
$$\begin{align*}\left(\sum_{m=1}^{n+1} m\right)^2 &= \left(n+1 + \sum_{m=1}^{n} m\right)^2\\
&= (n+1)^2 + 2(n+1)\left(\sum_{m=1}^{n} m\right) + \left(\sum_{m=1}^{n} m\right)^2 \tag{*}\\
&= (n+1)^2 + 2(n+1)\frac{n(n+1)}{2} + \sum_{m=1}^{n} m^3 \tag{**}\\
&= (n+1)^2(n+1) + \sum_{m=1}^{n} m^3\\
&= \sum_{m=1}^{n+1} m^3
\end{align*}$$
where $(**)$ comes from the well known identity $\sum_{m=1}^n m = \frac{n(n+1)}{2}$. In $(*)$, the first term is the volume of the diagonal square prism, the middle term is the volume of the rectangular prisms left of and above the diagonal prism, and the last term is the volume of the remaining shapes.
|
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|
How to solve $\tan x =\sin(x+45^{\circ})$? How do I solve $\tan x = \sin(x +45^{\circ})$?
This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$
|
we will show that the only solution to $\tan x = \sin(x+45^\circ)$ is $x = 45^\circ$
here is another way to solve $\tan t = \sin (t + 45^\circ).$ i have changed the independent variable to $t$ because i want to use $x = \cos t$ and $y = \sin t.$
using the addition formula for $\sin$ we can rewrite the equation as two equations
in $x,y$ in the following way: $${y \over x} = {x + y\over \sqrt 2}, x^2 + y^2 = 1$$ this simplifies to the unit circle and a hyperbola $$x^2 + y^2 = 1, x^2 + xy -\sqrt 2y = 0 $$ the unit circle and hyperbola may cut at most four points. one such point is $(\sqrt 2/2,\sqrt 2/2)$ using quadratic formula we can solve write the equation for hyperbola explicitly as $$2x = -y \pm \sqrt{y^2 + 4\sqrt 2y} $$ which shows that $-\infty < y \le -4\sqrt2$ or $0 \le y <\infty$ therefore the lower branch of the hyperbola cannot cut the unit circle.so we are only looking for the intersection of
$$ x^2 + y^2 = 1, 2x = \sqrt{y^2 + 4\sqrt 2 y}-y, y>0$$ the only solution for this is one we already guessed that
$$x = y = {\sqrt 2 \over 2} \text{ which corresponds to } t = 45^\circ$$
i wish i could add the graphs of the unit circle and the graphs of $2x = -y \pm \sqrt{y^2 + 4\sqrt 2y} $
|
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|
How does polynomial long division work?
I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$
|
As said, "it works the same way as euclidean division works for integers" (quoting Bernard). So, if you take $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}$$ The division of $3x^3$ by $x^2$ (the highest terms) gives $3x$. So $$A=3x+\Big(\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}-3x\Big)=3x -\frac{11 x^2+5x+3}{x^2 + 3x + 3}=3x-B$$ Now, consider $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}$$ The division of $11x^2$ by $x^2$ (the highest terms) gives $11$. So $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}=11+\Big(\frac{11 x^2+5x+3}{x^2 + 3x + 3}-11\Big)=11-\frac{28 x+30}{x^2 + 3x + 3}=11-C$$ Now, the highest degree in the numerator of $C$ is smaller that the highest degree in its denominator so you stop and have the remainder. So, by the end $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}=3x-11-\frac{28 x+30}{x^2 + 3x + 3}$$
|
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|
$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}=\prod_{k=1}^{n}\bigl(1-\frac{1}{2k}\bigr)$
i cant see why we have :
*
*$$\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$$
*$$\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$$
Even i see the notion of Double factorial
this question is related to that one : Behaviour of the sequence $u_n = \frac{\sqrt{n}}{4^n}\binom{2n}{n}$
*
*For $\frac{(2n)!}{4^n n!^2} = \frac{(2n-1)!!}{(2n)!!}$
$\dfrac{(2n)!}{4^n n!^2}=\dfrac{(2n)!}{2^{2n} n!^2}=\dfrac{(2n)\times (2n-1)!}{2^{2n} (n\times (n-2)!)^2}$
*
*For $\dfrac{(2n-1)!!}{(2n)!!} =\prod_{k=1}^{n}\left(1-\dfrac{1}{2k}\right),$
note that $n!! = \prod_{i=0}^k (n-2i) = n (n-2) (n-4) \cdots$
$\dfrac{(2n-1)!!}{(2n)!!}=\dfrac{\prod_{i=0}^k (2n-1-2i)}{\prod_{i=0}^k (2n-2i)}$
|
Note first that $(2n)!=(2n)!!(2n-1)!!$, because $(2n)!!$ gives you the even factors in $(2n)!$, and $(2n-1)!!$ gives you the odd factors. Now
$$\begin{align*}
(2n)!!&=(2n)(2n-2)(2n-4)\ldots(4)(2)\\
&=\big(2n\big)\big(2(n-1)\big)\big(2(n-2)\big)\ldots\big(2(2)\big)\big(2(1)\big)\\
&=2^nn!\;,
\end{align*}$$
so
$$\frac{(2n)!}{4^nn!^2}=\frac{(2n)!!(2n-1)!!}{2^{2n}n!^2}=\frac{2^nn!(2n-1)!!}{2^{2n}n!^2}=\frac{(2n-1)!!}{2^nn!}=\frac{(2n-1)!!}{(2n)!!}\;.$$
Now
$$\begin{align*}
\frac{(2n-1)!!}{(2n)!!}&=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdot\ldots\cdot\frac34\cdot\frac12\\\\
&=\left(1-\frac1{2n}\right)\left(1-\frac1{2n-2}\right)\left(1-\frac1{2n-4}\right)\ldots\left(1-\frac14\right)\left(1-\frac12\right)\\\\
&=\prod_{k=1}^n\left(1-\frac1{2k}\right)\;.
\end{align*}$$
|
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|
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.
|
Proof of the Identities
$$
\begin{align}
n^2&=\sum_{k=1}^nn\tag{1}\\
n^2+\sum_{k=1}^n\left(n^2+k\right)&=\sum_{k=1}^n\left(n^2+n+k\right)\tag{2}\\
\sum_{k=0}^n\left(n^2+k\right)&=\sum_{k=n+1}^{2n}\left(n^2+k\right)\tag{3}
\end{align}
$$
Explanation:
$(1)$: multiplication written as a sum
$(2)$: add $\sum\limits_{k=1}^n\left(n^2+k\right)$ to both sides
$(3)$: include $n^2$ in the sum on the left and reindex the sum on the right
Derivation of the Formula for Each Side
Using the formula
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{4}
$$
and equation $(2)$, we can compute the sum of each side of $(3)$ as
$$
\begin{align}
n^2+n^3+\frac{n(n+1)}2
&=\frac{2n^3+3n^2+n}2\\
&=\frac{n(n+1)(2n+1)}2\tag{5}
\end{align}
$$
|
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|
Trigonometric Integration with Negative Exponents How do you integrate $\csc^4 x/\cot^2 x$? I know that this is the same as $\csc^4 x \cot^{-2} x$ and when you use techniques in trig integrals you end up with
$$\int \csc^2 x \csc^2 x \cot^{-2} x \,dx = \int \left(1 + \cot^2 x\right)\left(\cot^{-2} x\right)\csc^2 x \,dx.$$
Making the substitution $u = \cot x \Rightarrow du = -\csc^2 x \,dx$,
$$\begin{align}
\ldots{} &= -\int \left(1 + u^2 \right) u^{-2} \,du \\
&= -\int \left( u^{-2} + u^0 \right) \,du \\
&= -\int \left( u^{-2} + 1 \right) \,du \\
&= -\left[ -u^{-1} + u \right] \\
&= (1/\cot x) - \cot x \,.
\end{align}$$
Is this the final answer or did I just mess up?
|
Another approach is to use
$\displaystyle\int\frac{\csc^{4}x}{\cot^{2}x}dx=\int\frac{1}{\sin^{4}x}\cdot\frac{\sin^{2}x}{\cos^{2}x}dx=\int\frac{1}{\sin^{2}x\cos^{2}x}dx=\int\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}dx$
$\displaystyle=\int\left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx=\int(\sec^{2}x+\csc^{2}x)dx=\tan x-\cot x+C.$
|
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|
Evaluate $ \int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx$ I want to evaluate the integral:
$$\int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx$$
but I have no clue of how to start in order to attack it.
I cannot even split that root because it does not make sense otherwise.
Any hints on how to begin?
Edit: Typos were corrected...
|
Hint:
Notice that
$$\int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx = \int_{-2}^{1} (x+2)\sqrt{\frac{1-x}{x+3}}\,dx, \text{because $ x\geq -2$}.$$
Take $t^2 = \frac{1-x}{x+3} \implies x= \frac{1-3t^2}{t^2+1}\implies dx = \frac{-8t}{(t^2+1)^2}\,dt.$
Hence,
$$\int_{-2}^{1}\sqrt{\frac{\left ( 1-x \right )\left ( x+2 \right )^2}{x+3}}\,dx = 8\int_{0}^{\sqrt{3}}\frac{(3-t^2)t^2}{(t^2+1)^3}\,dt.$$
Now, let $f(t) = t^3$ and $g(t) = (t^2+1)^2$. Notice that
$$\frac{(3-t^2)t^2}{(t^2+1)^3} = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}.$$
I think you can finish now!
|
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|
How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\dfrac{1}{8}$$
My some idea: since
$$1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n=\dfrac{(n!)^n}{1!\cdot 2!\cdot 3!\cdots (n-1)!}$$
and it is well know
$$v_{5}(n!)=\lfloor \dfrac{n}{5}\rfloor+\lfloor\dfrac{n}{5^2}\rfloor+\lfloor\dfrac{n}{5^3}\rfloor+\cdots+\lfloor\dfrac{n}{5^k}\rfloor+\cdots=\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor$$
so
$$v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=n\sum_{i=1}^{\infty}\lfloor\dfrac{n}{5^k}\rfloor-\sum_{i=1}^{n-1}\sum_{k=1}^{\infty}\left(\lfloor\dfrac{i}{5^k}\rfloor\right)$$
then I can't it.
Thank you
|
Exploiting:
$$\sum_{k=1}^{+\infty}\left\lfloor\frac{n}{5^k}\right\rfloor = \frac{n}{4}+O(1)$$
there is left very little to do.
|
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|
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
|
$$\begin{align}
\text{Your sum} &= \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \dots \\
\\
{\rm e} &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}\dots
\end{align}$$
|
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|
Numbers that are the sum of the squares of their prime factors A number which is equal to the sum of the squares of its prime factors with multiplicity:
*
*$16=2^2+2^2+2^2+2^2$
*$27=3^2+3^2+3^2$
Are these the only two such numbers to exist?
There has to be an easy proof for this, but it seems to elude me.
Thanks
|
Maple confirms that $A$ and $B$ are prime, and that $N$ is a solution to the question. $N$ has 179 digits.
$$N=45AB\\
A=(C^{107}+D^{107})/(C+D)\\
B=(C^{109}+D^{109})/(C+D)\\
C=(\sqrt{47}+\sqrt{43})/2\\
D=(\sqrt{47}-\sqrt{43})/2$$
The general solution to $3^2+3^2+5^2+A^2+B^2=3×3×5×A×B$ is a sequence with recursion
$$a_{n+1}=45a_n-a_{n-1}$$
The basic equation is a quadratic in any of the prime factors. The roots of the quadratic add up to the product of the rest of the prime factors. That is an easy way to generate solutions to $\sum x_i^2=\prod x_i$, although the $x_i$ might not be prime.
The starting point for this solution was $1,1,3,3,4$. Replace $4$ by $1×1×3×3-4=5$, then alternately replace the first factor $A$ by $B×3×3×5-A$ and the second factor $B$ by $A×3×3×5-B$.
Another starting point might be $1,1,1,2,2,2,3$ but I haven't found a prime solution from that.
|
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|
How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? I got stuck on this question:
How many roots does $(1+z^3)^8 = (1 + z^4)^6$ have? (including complex roots and roots with multiplicity)
My attempt at a solution:
First we can write the equation as:
$$(1+z^3)^8 - (1 + z^4)^6 = 0$$
We can then factor $(1+z^3)$
$$(z+1)^8(z^2-z+1)^8 - (1 + z^4)^6 = 0$$
But I am stuck here, the expression $(1+z^4)$ cannot easily be factored, but it should have 4 complex roots. I think there should be a simpler way to solve this question...
|
$(z^3+1)^8 = (z^4+1)^6$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 = z^{24} + 6z^{20} + 15z^{16} + 20z^{12} + 15z^8 + 6z^4 + 1$
$z^{24} + 8z^{21} + 28z^{18} + 56z^{15} + 70z^{12} + 56z^9 + 28z^6 + 8z^3 + 1 - z^{24} - 6z^{20} - 15z^{16} - 20z^{12} - 15z^8 - 6z^4 - 1 = 0$
$z^3(8z^{18} - 6z^{17} + 28z^{15} - 15z^{13} + 56z^{12} + 50z^9 + 56z^6 - 15z^5 + 28z^3 - 6z + 8) = 0$
$z^3(4z^6 - 3z^5 + 6z^3 - 3z + 4)(2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2)$
Cannot be factored further, but we can try to solve it from there:
*
*$z^3 = 0 \Rightarrow \sqrt[3]{z^3} = \sqrt[3]{0} \Rightarrow z = 0$
*$4z^6 - 3z^5 + 6z^3 - 3z + 4 = 0$ has 6 complex roots (three pairs of complex conjugates)
*$2z^{12} + 4z^9 + 3z^8 + 6z^6 + 3z^4 + 4z^3 + 2 = 0$ has 12 complex roots (six pairs of complex conjugates)
The equation has 19 roots.
|
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|
Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
|
$$\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}} =$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 2x^2-x^2 + 4}-(x -2)}{(x - 2)(x+2)}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^2(x-2)-(x-2)(x+2)}}{(x - 2)(x+2)}}-\lim \limits_{x \to 2}{\frac{1}{x+2}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x^2-x-2)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x-2)(x+1)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{|x-2| \cdot \sqrt{(x+1)}}{(x-2)(x+2)}}-\frac{1}{4}$$
Next , find limit when $x \to 2^{-}$ and when $x \to 2^{+}$
|
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|
Find a polynomial given the remainders of the division of that polynomial with 3 other polynomials
A polynomial from $ \mathbb{C}[x]$ divided by $ x - 1$, $x + 1$, $ x -2$ has the remainders 2, 6 and 3. Find the remainder of the division of that polynomial by $(x-1)(x+1)(x-2)$
The degree of the expression $(x-1)(x+1)(x-2)$ is 3, so the degree of the polynomial that I am looking for is 4. So the polynomial should be of the form: $ax^4 + bx^3 +cx^2 + dx + e$. I have tried to use Horner's scheme and I got the following expressions: $a+b+c+d+e=2$, $a+c-b-d+e=6$ and $16a+8b+4c+2d+c=3$. But this is not enough in order to find the polynomial. Is this the right approach?
|
Lets call this polynomial $P(x)$ than by the conditions you have that the polynomial $P(x)$ can be written as $$P(x)=(x-1)Q_1(x)+2\\P(x)=(x+1)Q_2(x)+6\\P(x)=(x-2)Q_3(x)+3$$ From this you can see that $P(1)=2,P(-1)=6,P(2)=3$
Now you can also write your polynomial as
$$P(x)=(x-1)(x+1)(x-2)Q_4(x)+ax^2+bx+c$$
Where $ax^2+bx+c$ is reminder when $P(x)$ is divided by $(x-1)(x+1)(x-2)$ plugging in $x=1,-1,2$ you can find $a,b,c$
|
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|
Determinant equation. let
A = [[x,-1],
[3,(1-x)]]
and
B = [[1,0,-3],
[2,x,-6],
[1,3,(x-5)]]
solve for x;
det(A) = det(B)
on expanding both the determinants and equating them, i obtain;
$-2x^2+3x+3 = 0$
completeing the square
$2[x^2-\frac{3x}{2}] -3 = 0$
$2[(x-\frac{3}{4})^2 - \frac{9}{16}] -3 = 0$
$2(x-\frac{3}{4})^2 - \frac{18}{16} -3 = 0$
$(x-\frac{3}{4}) = \pm \sqrt{\frac{33}{16}}$
$x = \frac{3}{4} \pm \sqrt{\frac{33}{16}}$
The answer in the back of the book is not this, could someone shed some light please.
|
You have properly solved this exercise. Actually you can simplify the result:
$$x = \frac{3 \pm \sqrt{33}}{4}$$
|
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|
How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1.$$
How to prove it by trigonometric method ?
Thank in advances.
|
I'd try using that
$$\frac{1}{5} - \frac{1}{2} + \frac{1}{3} = \frac{1}{30},$$
$$\frac{4}{3} - \frac{6}{5} = \frac{2}{15}.$$
And the formula:
$$
\tan(a\pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a\tan b}.
$$
|
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|
Show that x is independent of this trig expression For some reason I cannot get the solution.
Show that $x$ is independent of: $\sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z)$
I have used all the identities but seem to be missing something.
|
$\begin{align}
& \sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z) \\
&=\dfrac{1}{2}(1-\cos(2x+2y)) + \dfrac{1}{2}(1-\cos(2x+2z)) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\
&=1-\cos(2x+y+z)\cos(y-z) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\
&=1-\cos(y-z)\left[\cos(2x+y+z) + 2\sin(x+y)\sin(x+z) \right]\\
&=1-\cos(y-z)\left[\cos(2x+y+z) + \cos(y-z) - \cos(2x+y+z)\right]\\
&=1-\cos^2(y-z)\\
\end{align}$
which is independent of $x.$
|
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|
Evaluate the double integral $\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)} \,dx\,dy$ We know that
$$\int _0^1\int _0^1\frac{x-1}{(1+x y) \ln (x y)} \, dx\,dy=\gamma$$
$$\int _0^1\int _0^1\frac{x+1}{(1-x y) \ln (x y)}\,dx\,dy=\ln \frac4\pi$$
I wonder what would be
$$\int _0^1\int _0^1\frac{x+i}{(1-ix y) \ln (x y)}\,dx\,dy$$
Mathematica fails.
|
It is quite easy to prove that:
$$\iint_{(0,1)^2}\frac{(xy)^n}{\log(xy)}\,dx\,dy = -\frac{1}{n+1},\tag{1}$$
$$\iint_{(0,1)^2}\frac{x^{n+1}y^n}{\log(xy)}\,dx\,dy = -\log\frac{n+2}{n+1},\tag{2}$$
hence:
$$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{(x+i)}{(1-i xy)\log(xy)}\,dx\,dy = \sum_{n\geq 0}\iint_{(0,1)^2}\frac{i^n(x+i)(xy)^n}{\log(xy)}\,dx\,dy\\&=&-\sum_{n\geq 0}i^n \left(\frac{i}{n+1}+\log\frac{n+2}{n+1}\right)=\frac{1}{2}\log 2-\frac{\pi}{4}i+\sum_{n\geq 0}i^n\left(\log(n+1)-\log(n+2)\right)\\&=&\frac{1}{2}\log 2-\frac{\pi}{4}i+\color{red}{S_1}+i\color{blue}{S_2}\tag{3}\end{eqnarray*} $$
where:
$$\color{red}{S_1} = \sum_{n\geq 0}(-1)^n\left(\log(2n+1)-\log(2n+2)\right),$$
$$\color{blue}{S_2} = \sum_{n\geq 0}(-1)^n\left(\log(2n+2)-\log(2n+3)\right),\tag{4}$$
from which:
$$ \color{red}{S_1} = \log\prod_{n=0}^{+\infty}\frac{4n+1}{4n+2}\cdot\frac{4n+4}{4n+3},$$
$$ \color{blue}{S_2} = \log\prod_{n=0}^{+\infty}\frac{4n+2}{4n+3}\cdot\frac{4n+5}{4n+4}.\tag{5}$$
Using now the Euler product for the $\Gamma$ function it is not difficult to check that:
$$ \color{red}{S_1}=\log\frac{\sqrt{2\pi^3}}{\Gamma\left(\frac{1}{4}\right)^2},\qquad \color{blue}{S_2}=\log\frac{4\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}.\tag{6}$$
|
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|
Convert the Polar Equation to Cartesian Coordinates $$
r^2=\sec 4\theta
$$
I graphed this equations using Wolfram Alpha and found it to be 2 hyperbolas. I'm having difficulty showing this using the standard equations
$$
x=r\cos\theta \;, \; y=r\sin\theta \;, and \; x^2 +y^2 =r^2
$$
My work so far:
$$
r^2 = \sec4\theta=\frac{1}{\cos4\theta}=\frac{1}{\cos(2\theta+2\theta)}=\frac{1}{\cos2\theta \cos2\theta - \sin2\theta\sin2\theta} \\ \\ =\frac{1}{1-8\sin^2\theta + 8\sin^4\theta}
$$
I'm getting nowhere from here. I've tried using a few other trig identities, but no luck! Can one please point me in the right direction? I would appreciate any help. Thank you!!!
|
Hint
$$\dfrac{y^2}{x^2}=\tan^2{\theta}=\dfrac{\sin^2{\theta}}{1-\sin^2{\theta}}\Longrightarrow \sin^2{\theta}=\dfrac{y^2}{x^2+y^2}$$
and you have
$$x^2+y^2=r^2=\dfrac{1}{1-8\sin^2{\theta}+8\sin^4{\theta}}=\dfrac{1}{1-\dfrac{8y^2}{x^2+y^2}+8\left(\dfrac{y^2}{x^2+y^2}\right)^2}$$
so
$$x^2+y^2=\dfrac{(x^2+y^2)^2}{(x^2+y^2)^2-8y^2(x^2+y^2)+8y^4}$$
so
$$\Longrightarrow x^4+y^4-6x^2y^2=x^2+y^2$$
|
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|
How to show $\lceil ( \sqrt3 +1)^{2n}\rceil$ where $n \in \mathbb{ N}$ is divisible by $2^{n+1}$
Show that $ \left\lceil( \sqrt3 +1)^{2n}\right\rceil$ where $n \in \mathbb{N}$ is divisible by $2^{n+1}$.
I wrote the binomial expansion of $ ( \sqrt3 +1)^{2n}$ and $( \sqrt3 -1)^{2n}$ and then added them to confirm that the next integer is even. Afterwards I applied $AM \ge GM$ on the two terms to get $ ( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} \ge (2^{n+1})$.
Now I'm unable to figure out the next step. Any help would be appreciated. :)
|
Let:
$$ A_n = (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n} = (4+2\sqrt{3})^n+(4-2\sqrt{3})^{n}.\tag{1} $$
Since $0<4-2\sqrt{3}<\frac{2}{3}$, we have that $A_n$ is the integer giving the ceiling of $(\sqrt{3}+1)^{2n}$.
Now we have:
$$ A_0 = 2, \quad A_1 = 8,\qquad A_{n+2} = 8 A_{n+1} - 4 A_{n}\tag{2} $$
and if we take:
$$ \nu_2(n) = \max\{m\in\mathbb{N}: 2^m\mid n\} \tag{3}$$
it happens that:
$$ \nu_2(A_n)\geq n+1\tag{4} $$
can be proved by induction from $(2)$. We can also factor a $2^n$ from the RHS of $(1)$ then just study the parity of the sequence given by:
$$ B_n = (2+\sqrt{3})^n+(2-\sqrt{3})^n,\tag{5}$$
for which:
$$ B_0=2,\quad B_1=4,\quad B_2=14,\quad B_{n+2}=4B_{n+1}-B_n\equiv -B_n\pmod{4}.\tag{6}$$
That gives:
$$ \nu_2(A_n) = \left\{\begin{array}{rl}n+1 & \text{if }n\text{ is odd,}\\n+2 &\text{if }n\text{ is even.}\end{array}\right.\tag{7}$$
|
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|
How to find the x intercepts
$\frac{4}{3} e^{3x} + 2 e^{2x} - 8 e^x$
I have some confusion especially because of the e
how can I approach the solution?
The solution of the x-intercept is 0.838
Many thanks
|
\begin{align}
0&=\frac{4}{3}e^{3x}+2e^{2x}-8e^x\tag{1} \\[1em]
& = \frac{\frac{4}{3}e^{3x}+2e^{2x}-8e^{x}}{2e^x}\tag{2} \\[1em]
& = \frac{2}{3}e^{2x}+e^x-4\tag{3} \\[1em]
\end{align}
Now let $\xi=e^x,\therefore e^{2x}=\left(e^x\right)^2=\xi^2.$ This gives us
\begin{align}
0&=\frac{2}{3}\xi^2 +\xi-4\tag{4} \\[1em]
\therefore \xi & = \left\{\frac{-1+\sqrt{1-4\left(\frac{2}{3}\right)\left(-4\right)}}{2},\:\frac{-1-\sqrt{1-4\left(\frac{2}{3}\right)\left(-4\right)}}{2}\right\}\tag{5} \\[1em]
\end{align}
And I'm sure you can do the rest...
|
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|
the general solution of $y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$ I have some trouble finding the correct solution for the difference equation
$$y(n+3)-\frac{2}{3}y(n+1)+\frac{1}{3}y(n) = 0$$
I've found that the characteristic equation of the difference equation is $\lambda^3-\frac{2}{3}\lambda+\frac{1}{3}$.
By computation I then have;
\begin{array}{lcl}
\lambda^3-\dfrac{2}{3}\lambda+\dfrac{1}{3}\\
(\lambda+1)(\lambda^2-\lambda+\dfrac{1}{3})\\
\lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{\frac{1}{3}}}{2}, \lambda_3 = \dfrac{1+i\sqrt{\frac{1}{3}}}{2}
\end{array}
I checked my answer using wolframalpha and it gave me the eigenvalues
$$\begin{array}{lcl}
\lambda_1 = -1, \lambda_2 = \dfrac{1-i\sqrt{3}}{2}, \lambda_3 = \dfrac{1+i\sqrt{3}}{2}
\end{array}$$
Could someone please tell me what I did wrong, or what I should do differently?
|
Wolfram Alpha gives the roots as
$-1$ and
$1/6 (3\pm i \sqrt(3))$.
This last is
$\frac{1\pm i\sqrt{1/3}}{2}$
which is your answer.
My guess is
either you misread WA's answer
or you entered the equation incorrectly.
|
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|
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$
and we know that:
$$\sum_{n=1}^\infty b_n$$
doesnt converge cause
$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$
doesnt converge
so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge
but i know the final answer is that it does converge
so what am i doing wrong?
|
Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1}$$
NB that your estimate contains an error in the last step, [fixed now]
$$2 \sqrt n - 2 \sqrt{n+1} = 2 ( \sqrt n - \sqrt{n+1}) \ne 2 (\sqrt n + \sqrt {n+1})$$
Your estimate only proves $S_n > 0$, wich doesn't show anything on its own.
|
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|
Find smallest $n \in \mathbb{N}$ s.t $A^n=I$ Let $A$ be $2 \times 2$ matrix:
$$
\left(
\begin{matrix}
\sin\frac{\pi}{18} \\ \sin\frac{4\pi}{9}
\end{matrix}
\begin{matrix}
-\sin\frac{4\pi}{9} \\ \sin\frac{\pi}{18}
\end{matrix} \right) $$
Then find smallest $n \in \mathbb{N}$ s.t $A^n=I$.
Looking at this question I seems I am missing some concept of of linear algebra without which it cannot be solved. How should one approach this problem?
|
Computing directly gives that the eigenvalues of $A$ are $\lambda_{\pm} := \exp \left(\pm \frac{4 \pi i}{9}\right)$, so there is a some matrix $P$ such that
$$A := P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix} P.$$
So, for any integer $n$,
$$A^n = (P^{-1} A P)^n = P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix}^n P = P^{-1} \begin{pmatrix} \lambda_+^n & 0 \\ 0 & \lambda_-^n \end{pmatrix} P.$$ Now, $\lambda_{\pm}^n = \exp\left(\pm \frac{4 \pi i n}{9}\right)$, so the smallest $n$ for which $\lambda_{\pm}^n = 1$ is $n = 9$, and for this $n$, we thus have $A^n = P^{-1} I^n P = I$. The only matrix similar to $I$ is $I$ itself, so no smaller positive $n$ satisfies the equation.
|
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|
Modular Arithmetic, Pythagorean triples I'm not sure if this question should be under Modular Arithmetic, but that's where it was in my book.
Show that, if $x$, $y$, and $z$ and integers such that $x^2 + y^2 = z^2$, then at least one of $\{x, y, z\}$ is divisible by $2$, at least one of $\{x, y, z\}$ is divisible by $3$ and at least one of $\{x, y, z\}$ is divisible by $5$.
What I did so far is to write them as:
$$x = 2k , \quad y = 3m , \quad z = 5n$$
Then the equation becomes $4k^2 + 9m^2 = 25n^2$
This would work for $k=2$, $m=1$, $n=1$, which would give $x=4$, $y=3$, $z=5$
I don't know how to show that it would only work for these numbers, or multiples of them.
|
If $x^2$ and $y^2$ are both odd $z^2$ is even.
If $x^2$ and $y^2$ are not multiples of $3$ they are $1 \bmod 3$ so $z^2$ is $2\bmod 3$, which is not a quadratic residue.
If $x^2$ and $y^2$ are not multiple of $3$ they are either $1$ or $-1 \bmod 5$. So $z^2$ is $0,2$ or $3\bmod 5$. $2$ and $3$ are impossible because they are not quadratic residues.
|
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|
Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+\cdots+1 )$
$\implies n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n^2$
$\implies \dfrac {1}{ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) } \geq \dfrac {1}{n^2}$
I guess this result is of not much use. Can somebody please tell me a direction to move ahead in this problem?
Thank you very much for your help .
|
Using $GM\ge HM$ $$\left( 1\cdot\frac{1}{2}\cdot \frac{1}{3}\cdots \frac{1}{n}\right)^{\frac{1}{n}}\ge \frac{n}{\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}}$$ so that $$\frac{1}{n}\cdot\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}\ge (n!)^{\frac{1}{n}}$$ and as $(n!)^{\frac{1}{n}}>1 \forall n\ge2$ we see that $\sum (n!)^{1/n}$ cannot converge and thus $$\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)}$$ cannot converge.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1119733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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|
If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
|
$0\leq(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)\implies ab+ac+bc\geq-\dfrac{1}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1119879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Working Out Easy Equations does anyone know how to do this equation? I know it's easy but I can't work out what the question means.
When I expanded the first equation:
$(y+4)-(y-3)$
$y^2 -3y +4y - 12$
$y^2-1y-12$
Not sure what I should do after.
Can someone explain how would you work it out in easy terms?
Thank you
|
$$(y + 4) - (y - 3) = (y + 4) - (y) - (-3) = (y+4) + (3 - y) = (y - y) + (3 + 4) = 7$$
$$(y-2) - (y - 3) = (y - 2) - y - (-3) = (y-2) + (3 - y) = (y - y) + (3 - 2) = 1$$
In general,
$$(y-a) - (y - b) = b - a$$
In the same way that $x + (y + z) = x + y + z$, $x - (y + z) = x - y - z$. The minus sign distributes to all terms and reverses their sign. Mainly because
$$z - (x + y) = z + (-1)(x + y) = z + (-1)x + (-1)y = z - x - y$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1122375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Combinatorics, surjective functions with conditions Question: $A=\left\{ 1,2,3,4,5\right\} $
, $B=\left\{ 1,2,3\right\} $
. How many surjective functions are there such that $ f(1)\neq1$
,$f(2)\neq2$
,$ f(3)\neq2$
.
Solution: Overall we have $3^{5}-{3 \choose 1}2^{5}+{3 \choose 2}1^{5}=150$
functions that is surjective. We will denote $A_{i,j}=\left\{ f:A\rightarrow B\,|\, f\,\text{is surjective and }f(i)=j\right\} $
We will calculate $\left|A_{1,1}\cup A_{2,2}\cup A_{3,2}\right|$
then subtract it from 150.
So $\left|A_{1,1}\right|=\left|A_{2,2}\right|=\left|A_{3,2}\right|=3^{4}-{3 \choose 1}2^{4}+{3 \choose 2}1^{4}=36\Rightarrow\left|A_{1,1}\right|+\left|A_{2,2}\right|+\left|A_{3,2}\right|=3.36=108$
and $\left|A_{1,1}\cap A_{2,2}\right|=\left|A_{1,1}\cap A_{3,2}\right|=\left|A_{2,2}\cap A_{3,2}\right|=3^{3}-{3 \choose 1}2^{3}+{3 \choose 2}1^{3}=6\Rightarrow\left|A_{1,1}\cap A_{2,2}\right|+\left|A_{1,1}\cap A_{3,2}\right|+\left|A_{2,2}\cap A_{3,2}\right|=3.6=18$
lastly $\left|A_{1,1}\cap A_{2,2}\cap A_{3,2}\right|=3.2=6$
because we have 4 or 5 has to go to 3, otherwise we wouldn't have a surjective functions.
So my answer was $150-108+18-6=54$
. However I was told that the solution was 45. I would appreciate if someone can tell me where my mistake is. Thank you!
|
Your calculation for $|A_{11}|$ gives the number of surjective maps $f:\{1,2,3,4,5\}\to\{1,2,3\}$ such that $f(1)=1$ and $f(\{2,3,4,5\})=\{1,2,3\}$. However, because $f(1)=1$, you do not need $f(\{2,3,4,5\})=\{1,2,3\}$, you only need $f(\{2,3,4,5\})\supseteq\{2,3\}$. You need to make similar corrections to various other parts of your calculation.
|
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|
Rationality and triangles
Consider a triangle with angles $\alpha, 5\alpha, 180-6\alpha$. What is the minimum perimeter of that triangle, if it has integer sides and $5\alpha<90$?.
Let's call the sides that face each angle $x,y,z$ respectively, and set $\cos \alpha =t$. Using the law of sines , $x=2R\sin\alpha, y=2R\sin5\alpha, z=2R\sin6\alpha$. Setting $2R\sin\alpha=k$, we have the parametrized forms
$$x=k$$
$$y=k(16t^4-12t^2+1)$$
$$z=k(32t^5-32t^3+6t)$$
I got stuck here. If $t$ is rational, I think we could make an argument for concluding that $t=20/21$(Final update: Nope, I was wrong). But that would depend on this lemma:
If $32t^5-32t^3+6t$ and $16t^4-12t^2+1$ are rational, then so is $t$.
Is this assertion correct? If not, is there an easier way to solve the original question?
Edit:
I managed to prove that my later proposition is false when $t=\pm\frac{\sqrt{3}}{2}, t=\frac{1\pm\sqrt5}4$, but true otherwise. $16t^4-12t^2+1$ is rational iff $16t^4-12t^2$ is, and then we set:
$$16t^4-12t^2=p$$
$$32t^5-32t^3+6t=q$$
$$\implies 2tp-\frac{p}{2t}=q$$
Solving for $t$, we get
$$t=\frac{q\pm\sqrt{4p^2+q^2}}{4p}$$
Substituting in the first equation with the positive case(it is analogous in the negative), we get
$$\frac{q^4}{2 p^4}+\frac{q^2}{2 p^2}-\frac{q\sqrt{4 p^2+q^2}}{2 p^2}+\frac{q^3\sqrt{4 p^2+q^2}}{2 p^4}-2=p$$
If $4p^2+q^2$ is not a perfect square, then $p=\pm q$ or $q=0$. Wolframalpha sledgehammering both cases (it is only necessary to check one case, since the other one corresponds to inverting the sign of $t$), we see that that all solutions yield rational $p,q$.
Going back to the original problem, we can forget those solutions since we wanted $0<5\alpha<90$
Second update: Substituting $2t=a/b$ with coprime $a,b$, we see that we have to minimize $x+y+z$ subject to $x,y,z$ being integers and $2\cos18<a/b<2$. Therefore, we want to minimize:
$$k\left(\left(\frac ab\right)^5+\left(\frac ab\right)^4-4\left(\frac ab\right)^3-3\left(\frac ab\right)^2+3\left(\frac ab\right)+2\right)$$
It is easy to see that when reducing to the common denominator $b^5$. the numerator will be coprime due to the leading coefficint being $1$. Therefore, $k$ must be $b^5$ in order to have integers(and not another multiple of $b^5$ since we want to minimize the perimeter). So the problem is reduced to:
Minimize $$a^5+a^4b-4a^3b^2-3a^2b^3+3ab^4+2b^5$$
Subject to $a,b$ positive integers, $2\cos18<a/b<2$
|
Finally solved the problem! for shortness, let's set $2\cos18=x$. From the inequalities $a\le2b$ and $2\cos(18)b\le a$, we see the trivial inequalities:
$$a^5\ge(xb)^5,a^4\ge(xb)^4, -a^3\ge-8b^3,-a^2\ge-4b^2,a\ge xb$$
So
\begin{align*}
P(a,b)&=a^5+a^4b-4a^3b^2-3a^2b^3+3ab^4+2b^5\\
&\ge b^5(x^5+x^4-32-12+3x+2)\\
&\approx 1.69549b^5
\end{align*}
If $b\ge15, P>1287512.71875\implies P\ge1287513$. We now have an upper bound on the minimum $P$. Since we want $2\cos18<\frac ab\le \frac{2b-1}{b}<2$, we get that $b>(2-2\cos18)^{-1}>10$, so $b\ge 11$. Finally, we are left to try the pairs $a,b$ that meet the requirements and compare the $P$ that corresponds to each pair. So we calculate
$$P(21,11)=1224640$$
$$P(23,12)=1972355$$
$$P(25,13)=3046536$$
$$P(27,14)=4543825$$
So the minimum perimter of the proposed triangle is $1224640$, with $t=21/22$ and not $20/21$ as initially hypothesised.
|
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|
Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.
I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or is there a more general solution?
|
Another way:
$$a^2 + b^2 + c^2 = \frac{a^2 + b^2}{2} + \frac{b^2 + c^2}{2} + \frac{c^2 + a^2}{2} \ge ab + bc + ca$$
The last part follows from the fact that $x^2 + y^2 \ge 2xy$ for all $x, y$ (you can get this by observing $x^2 + y^2 - 2xy = (x - y)^2 \ge 0$ for all $x,y$).
|
{
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"url": "https://math.stackexchange.com/questions/1125709",
"timestamp": "2023-03-29T00:00:00",
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|
Trig substitution fails for evaluating $ \int \frac{\cos x \sin x}{\sin^2{x} + \sin x + 1} dx$? Evaluate the integral
\begin{equation}
\int \frac{\cos x \sin x}{\sin^2{x} + \sin x + 1} dx
\end{equation}
Basically I could substitute: $t = \sin x$ and get:
$$\int \frac{t}{t^2 +t + 1} dt$$
But, although it seems a reasonable integral to solve, it doesn't.
So I tried to utilize some trigonometric identities but it didn't work out so well.
I'd be glad for help.
|
$$\int \frac{t}{t^2 +t + 1} dt = \int \frac{t}{\left(t+\frac 12\right)^2 + \frac 34}\,dt$$
$$\frac 34 = \left(\frac{\sqrt 3}2\right)^2$$
So , in therms of substitution, $$t+\frac 12 = \frac {\sqrt 3}2 \tan\theta\implies dt = \frac {\sqrt 3}2 \sec^2\theta\,d\theta,\quad \text{and}\;t = \frac{\sqrt 3}2\tan\theta - \frac 12$$
We can also express $\theta$ as a function of $t$: $$\tan\theta = \frac 2{\sqrt 3}\left(t+\frac 12\right) \implies \theta = \arctan\left(\frac 2{\sqrt 3}\left(t+\frac 12\right)\right)$$
$$
\begin{align}\int \frac{t}{\left(t+\frac 12\right)^2 + \frac 34}\,dt &= \int \frac{\left(\frac{\sqrt 3}2\tan\theta - \frac 12\right)\cdot \frac{\sqrt 3}2 \sec^2\theta}{\frac 34(\tan^2 \theta + 1)}\\ \\
& =\int \frac{\left(\frac 34\right)\tan\theta\sec^2\theta - \frac{\sqrt 3}4\sec^2\theta\,d\theta}{\frac 34(\tan^2\theta + 1)}\\ \\
&= \int \frac{\tan\theta\sec^2\theta}{\sec^2 \theta}\,d\theta -\frac 1{\sqrt 3}\int \frac{\sec^2\theta\,d\theta}{\sec^2 \theta} \\ \\
&= \int \frac {\sin\theta}{\cos\theta}\,d\theta - \frac 1{\sqrt 3}\int \,d\theta
\end{align}$$
Now all that remains is readily evaluating the integrals, then back-substituting in terms of $t$, then again for the substitution $t = \sin\theta$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$ Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$
I tried to redo this problem and got to this point
$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$
Let $z=x+iy$ then
$$2\sqrt{x^2+y^2}>1+(x+iy)(x-iy)$$
$$2\sqrt{x^2+y^2}>1+(x^2+y^2)$$
$$0>1-2\sqrt{x^2+y^2}+(x^2+y^2)$$
$$0>(1-\sqrt{x^2+y^2})^2$$
since $x,y$ are real number , so there is no solution for this inequality?
|
Let $z=re^{i\theta}$. Then after squaring both sides we have
$$
4r^2 > r^4 + 2r^2\cos(2\theta) + 1\Rightarrow
0>(r^2-2r\sin(\theta)-1)(r^2+2r\sin(\theta)-1)\tag{1}
$$
where I used the following identity $\cos(2\theta) = 1 -2\sin^2(\theta)$. Let's write $(1)$ in Cartesian coordinates so recall that $r^2 = x^2 + y^2$ and $r\sin(\theta) = y$. Then we have
$$
0>(x^2+y^2-2y-1)(x^2+y^2+2y-1)=[x^2+(y-1)^2-2][x^2+(y+1)^2-2]
$$
Then
\begin{align}
x^2+(y-1)^2&< 2\\
x^2+(y+1)^2&< 2
\end{align}
That is, we have two disc of radius $\sqrt{2}$ centered at $(0,\pm i)$. So the solution set is
$$
S=\bigl\{z\in\mathbb{C}\colon \lvert z-i\rvert < \sqrt{2}\text{ or }\lvert z+i\rvert < \sqrt{2}\text{ but not in the intersection}\bigr\}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1126953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Evaluating $\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$ without L'Hopital I need to calculate the following limit (without using L'Hopital - I haven't gotten to derivatives yet):
$$\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$$
We have $\sin$ function in the numerator so it looks like we should somehow make this similair to $\lim_{x\rightarrow 0} \frac{\sin x}{x}$. When choosing $t=x^2-\pi ^2$ we get $\lim_{t\rightarrow 0} \frac{\sin \sqrt{t+\pi ^2}}{t}$ so it's almost there and from there I don't know what to do. How to proceed further? Or maybe I'm doing it the wrong way and it can't be done that way?
|
\begin{align}
\frac{\sin(x)}{x^2 - \pi^2} = \frac{x}{x^2(1-\pi^2/x^2)} \prod_{n=1}^{\infty}\left(1 -\frac{x^2}{n^2\pi^2}\right) = \frac{1}{x} \frac{1-x^2/\pi^2}{1-\pi^2/x^2} \prod_{n=2}^{\infty} \left(1 -\frac{x^2}{n^2\pi^2}\right)
\end{align}
Since no L'Hospital available, we consider of the sign of numerator and denominator as $x \to \pi^{+}$ and $x \to \pi^-$ to obtain
\begin{align}
\lim_{x\to \pi}\frac{1-x^2/\pi^2}{1-\pi^2/x^2} = -1
\end{align}
and hence
\begin{align}
\lim_{x\to \pi }\frac{\sin(x)}{x^2 - \pi^2}
=
-\frac{1}{\pi} \prod_{n=2}^{\infty} \left(1 -\frac{1}{n^2}\right)
= -\frac{1}{2\pi}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculation of determinant of an arrowhead matrix Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?
$$\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & 0 &\cdots & 1
\end{vmatrix} = n$$
I figured it with a smaller dimension and it indeed produces the determinant that is the size of dimension. I tried to do a cofactor expansion with the first row, and each term produces the determinant of $1$ and if you sum them up, then the total determinant will be $n$. But the sign change for each cofactor is confusing, and it is not easily seen that each cofactor term is actually positive $1$.
|
$$
\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 & 1
\end{vmatrix} =
\begin{vmatrix}
n & 0 & 0 & 0 & \cdots & 0 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & & 0 \\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0 & 1
\end{vmatrix} =
n
$$
Note that in a triangular matrix its determinant equals the product of the diagonal entries.
|
{
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"url": "https://math.stackexchange.com/questions/1128330",
"timestamp": "2023-03-29T00:00:00",
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|
Transformation Matrix for cube in 2D My task is to transform the cube from the left corner to the big cube in the middle:
What I did was:
First i scale the cube:
$$
\begin{pmatrix}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Next i rotate it $45^\circ$:
$$
\begin{pmatrix}
\cos(45^\circ) & -\sin(45^\circ) & 0 \\
\sin(45^\circ) & \cos(45^\circ) & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
And last i translate the cube:
$$
\begin{pmatrix}
1 & 0 & 10 \\
0 & 1 & 7 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
When i combine this matrices i get my transformation matrix:
$$
\begin{pmatrix}
4\cos(45^\circ) & -4\sin(45^\circ) & 40\cos(45^\circ) - 28\sin(45^\circ) \\
4\sin(45^\circ) & 4\cos(45^\circ) & 40\cos(45^\circ) + 28\sin(45^\circ) \\
0 & 0 & 1 \\
\end{pmatrix}
$$
But then i tried to apply this matrix to the point $(0,1,1)$ and i got an wrong result:
$$(5.6, 50.9116 ,1)$$
As you can see it should be around:
$$(7.1, 9.9, 1)$$
What did I wrong? Thanks!
|
There is a mistake in your multiplication.
$$
T_{total}=\begin{pmatrix}
4\cos(45^\circ) & -4\sin(45^\circ) & 40\cos(45^\circ) - 28\sin(45^\circ) \\
4\sin(45^\circ) & 4\cos(45^\circ) & \color{red}{28\cos(45^\circ) + 40\sin(45^\circ)} \\
0 & 0 & 1 \\
\end{pmatrix}
$$
But, since $sin (45^\circ)$ and $cos (45^\circ)$ have the same value, the problem is not from this point.
The problem is related to the order of the Transforms. In matrix multiplication, you should care about matrix order.
$T1$ is scale transform
$T2$ is rotation transform
$T3$ is translation transform
you computed:
$$X_{new}=(T1 \times T2 \times T3) \times X_{old}$$
while, you need to scale first then rotation then translation:
$$X_{new}=(T3 \times T2 \times T1) \times X_{old}$$
$$T3 \times T2 \times T1=\begin{pmatrix}
4\cos(45^\circ) & -4\sin(45^\circ) & 10 \\
4\sin(45^\circ) & 4\cos(45^\circ) & 7 \\
0 & 0 & 1 \\
\end{pmatrix}$$
$$T1\times T2 \times T3 \times \begin{pmatrix}
0 \\
1 \\
1 \\
\end{pmatrix}=\begin{pmatrix}
5.66 \\
50.91 \\
1 \\
\end{pmatrix}$$
While
$$T3\times T2 \times T1 \times \begin{pmatrix}
0 \\
1 \\
1 \\
\end{pmatrix}=\begin{pmatrix}
7.17 \\
9.83 \\
1 \\
\end{pmatrix}$$
|
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|
Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{x}{(1+x^n)\sqrt[n]{1+x^n}} \Bigg|_0^1-\int_0^1 x \ d(1+x^n)^{-1-\frac{1}{n}} \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1 x\left(-1-\frac{1}{n}\right)(1+x^n)^{-2-\frac{1}{n}}(nx^{n-1}) \ dx \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1(-n-1)\frac{x^n}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx\\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)\int_0^1\frac{x^n+1-1}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx \\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)I_1-(n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}.\end{align} $$
So that,
$$ (n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{1}{2\sqrt[n]{2}}+nI_1$$
But how to find the following integral ?
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
|
Given $$I = \int_{0}^{1}\frac{1}{(1+x^n)^2\sqrt[n]{1+x^n}}dx\;,$$ Now Put $\displaystyle x = (\tan \theta )^{\frac{2}{n}}\;,$ Then $\displaystyle dx = \frac{2}{n}(\tan \theta)^{\frac{2}{n}-1}\cdot \sec^2 \theta d\theta$ and changing limit, We get
$$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\tan \theta)^{\frac{2}{n}-1}\cdot \sec^2 \theta}{\sec^4 \theta \cdot \sec ^{\frac{2}{n}}\theta}d\theta = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\tan \theta)^{\frac{2}{n}-1}}{(\sec \theta)^{\frac{2}{n}+2}}d\theta$$
So we get $$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\frac{(\sin \theta)^{\frac{2}{n}}}{\tan \theta \cdot \sec^2 \theta}d\theta = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\cos^3 \theta \cdot (\sin \theta)^{\frac{2}{n}-1}d\theta$$
So we get $$I = \frac{2}{n}\int_{0}^{\frac{\pi}{4}}\left[(\sin \theta)^{\frac{2}{n}-1}-(\sin \theta)^{\frac{2}{n}+1}\theta)\right]\cdot \cos \theta d\theta $$
So $$I = \frac{2}{n}\left[\frac{(\sin \theta)^{\frac{2}{n}}}{\frac{2}{n}}-\frac{(\sin \theta)^{\frac{2}{n}+2}}{\frac{2}{n}(n+1)}\right]_{0}^{\frac{\pi}{4}} = \left(\frac{1}{2}\right)^{\frac{1}{n}+1}\cdot \frac{2n+1}{n+1}$$
|
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}
|
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$
So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate
both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$
Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$
We can write $f'(x)$ as $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$
So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$
So $$\displaystyle f'(x) = = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$
So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.
In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it
will Cross $\bf{X-}$ axis at least one time.
My question is can we solve it any other way, i. e without using Derivative test.
Help me , Thanks
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$f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$ So, as you note,
$f'(x)=1+x+x^2+x^3+..........+x^6$. Now consider $x$ in three ranges
*
*For $x \ge 0$ then clearly $f'(x) >0$
*For $ -1 \le x < 0$, $f'(x)=(1+x) + (x^2+x^3) + (x^4+x^5) + x^6$. Each bracketed term is non-negative, and $x^6$ is positive, so $f'(x) >0$
*For $x < -1$, $f'(x)=1 +(x + x^2) +(x^3 + x^4) +( x^5 + x^6)$. Again, each bracketed term is positive so $f'(x) >0$
From that conclude that $f(x) $ is a monotonic increasing function and can therefore have at most 1 zero.
Finally, as $x \to -\infty$, $f(x) \to -\infty$ and as $x \to +\infty$, $f(x) \to +\infty$ so that by continuity $f(x) $ must have at least one zero, and from the previous it therefore has exactly one zero.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to variables.
How to approach such question?
|
$$
(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab + bc + ca)
$$
The LHS of the above identity is a perfect square, hence it is always positive or 0.
Thus,
$$
a^2 + b^2 +c^2 + 2(ab + bc + ca) ≥ 0
$$
It is given that $a^2 + b^2 +c^2 =1 $ .
$$
1 + 2(ab + bc + ca) ≥ 0
$$
Therefore,
$$
ab + bc + ca ≥ -1/2
$$
It is possible to determine the upper bound of the above expression by applying the AM-GM inequality as follows: (as already done by Arian)
$$
a^2+b^2≥2ab\\
b^2+c^2≥2bc\\
a^2+c^2≥2ac
$$
Adding the above three inequalities, we get
$$
a^2 + b^2 +c^2 ≥ ab + bc + ca
$$
which implies
$$
ab + bc + ca ≤ 1
$$
Thus, the value of $ab + bc + ca$ lies in the interval $[-1/2, 1]$ .
Note: It is not possible to find an exact value for this expression, as already noted by the other answers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1129070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
}
|
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.