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Truncation Error in differentiation I dont understand how truncation error ie, the error caused because the definition of h tends to zero, but we take the value(during differentiation) as a non zero really small number; gets reduced because of taking centre difference.
That is, take $\frac{f(x+h)−f(x−h)}{2h}$ instead of $\frac{f(x+h)−f(x)}{h}$. How does this reduce the truncation error ?
|
By taking symmetric steps around the centre, we're able to cancel out the first-order term to achieve a second-order formula. Indeed, consider their Taylor expansions:
\begin{align*}
f(x + h) &= f(x) + f'(x)h + \frac{1}{2}f''(x)h^2 + \frac{1}{6}f'''(x)h^3 + \cdots \\
f(x - h) &= f(x) - f'(x)h + \frac{1}{2}f''(x)h^2 - \frac{1}{6}f'''(x)h^3 + \cdots
\end{align*}
Subtracting the two equations and dividing through by $2h$, we obtain:
$$
\frac{f(x + h) - f(x - h)}{2h}
= f'(x) + \frac{1}{6}f'''(x)h^2 + \cdots \\
\implies \boxed{f'(x) = \frac{f(x + h) - f(x - h)}{2h} \color{red}{+ O(h^2)}}
$$
On the other hand, if we took the first equation, subtracted $f(x)$, then divided by $h$, we would obtain:
$$
\frac{f(x + h) - f(x)}{h}
= f'(x) + \frac{1}{2}f''(x)h + \frac{1}{6}f'''(x)h^2 + \cdots \\
\implies \boxed{f'(x) = \frac{f(x + h) - f(x)}{h} \color{red}{+ O(h)}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+11$ . Where did I go wrong ? Thanks !!
|
you did wrong in the first step, the derivate of (x+3) is equal to 1.
|
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|
If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$ If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$
I considered the function $f(x)=ax^2+\frac{b}{x}$ and I put $f'(x)=0$ to find $x^3=\frac{b}{2a}$
I am stuck here.
|
By AM-GM $$ax^2+\frac{b}{x}=ax^2+2\left(\frac{b}{2x}\right)\geq3\sqrt[3]{ax^2\cdot\left(\frac{b}{2x}\right)^2}=3\sqrt[3]{\frac{ab^2}{4}}.$$
Thus, we need $$3\sqrt[3]{\frac{ab^2}{4}}\geq c,$$
which gives which you wish.
|
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|
Compute $\Sigma _{n=0}^{\infty} \frac{1}{n^2+a^2}$ with Fourier series of $e^{ax}$ given $a>0$. As exercise, the teacher asks "Let $g(x)$ be the Fourier series of $e^{ax} x\in[-\pi,\pi]$. Look $g(x)$ up and use it to compute $\Sigma _{n=0}^{\infty} \frac{1}{n^2+a^2}$".
Since $g(x)=e^{ax}=\frac{2\sinh (a\pi)}{\pi}(\frac{1}{2a}+\Sigma _{n=1}^{\infty} \frac{(-1)^n(a\cos(nx)-n\sin(nx))}{n^2+a^2})$
My idea is to set $a=0$, in order to get $1=\frac{2\sinh (a\pi)}{\pi}(\frac{1}{2a}+\Sigma _{n=1}^{\infty} \frac{(-1)^n a}{n^2+a^2})$. I don't know how to eliminate $(-1)^n$ (maybe with Parseval formula??).
Any idea how to continue? Is even $g(x)$ correct?
When I try to compute it on Wolfram $\Sigma _{n=0}^{\infty} \frac{1}{n^2+a^2}$ the answer is$\frac{1+a\pi cotanh (a\pi) }{2a^2}$
|
The Fourier coefficients of $f(x)=e^{ax}$ are given by:
$$\widehat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}e^{at}e^{-int}\,dt=\frac{e^{2(a-in)\pi}-1}{2\pi(a-in)}=\frac{e^{2a\pi}-1}{2\pi(a-in)}$$
so
$$|\widehat{f}(n)|^2=\frac{(e^{2a\pi}-1)^2}{4\pi^2}\frac{1}{a^2+n^2}$$
By Parseval's identity:
$$\sum_{-\infty}^{\infty}|\widehat{f}(n)|^2=\frac{1}{2\pi}\int_0^{2\pi}e^{2at}\,dt=\frac{e^{4a\pi}-1}{4a\pi}$$
Therefore:
$$\frac{1}{a^2}+2\sum_{n=1}^{\infty}\frac{1}{n^2+a^2}=\sum_{-\infty}^{\infty}\frac{1}{a^2+n^2}=\frac{\pi(e^{2a\pi}+1)}{a(e^{2a\pi}-1)}=\frac{\pi}{a}\coth(a\pi)$$
so
$${2\over a^2}+2\sum_{n=1}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{a}\coth(a\pi)+\frac{1}{a^2}$$
which implies:
$$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{1}{2}\left(\frac{\pi}{a}\coth(a\pi)+\frac{1}{a^2}\right)$$
Remark
As it stands, the equality does not allow to take the limit of both sides as $a\to 0$. Indeed, both sides tend to infinity. But if we rewrite it (by taking the first term of the series to the r.h.s), we get:
$$\sum_{n=1}^{\infty}\frac{1}{n^2+a^2}=\frac{1}{2}\left(\frac{\pi}{a}\coth(a\pi)-\frac{1}{a^2}\right)$$
and now the limit as $a\to 0$ exists on both sides, which gives - yet another -- proof of the familiar fact that $\sum_{n=1}^{\infty}1/n^2=\pi^2/6$.
|
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|
No. Of possible triangles If there are n line segments in a plane of lengths (1,2,3,....,N) then find the no. of triangles that can be formed from these.
Attempt: let the sides be X,Y,Z such that X < Y< Z then the cardinality of required set boils down to finding out pairs such that X + Y > Z other two inequalities are automatically satisfied. Then I drew a table for values of X,Y and Z such that the above conditions hold.And now the solution becomes clumsy and lengthy. Is there a better way? And even in this approach for Z = n , I have (n-2)*(n-3)/2 solutions. And now this n can be anything from 4 to N. So I summed this again. But the problem is I am missing many things answer is different for odd and even values of N. What's wrong with my solution. It doesn't match with the correct answer.
|
If $X<Y<Z$ and $X+Y>Z$,
$$Z+1\le X+Y\le Y-1+Y$$
and hence
$$Y\ge \frac{Z}{2}+1$$
The number of triangles is
\begin{align}
\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^{Z-1}\sum_{X=Z-Y+1}^{Y-1}1&=\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^{Z-1}(Y-1-Z+Y)\\
&=\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^{Z-1}(2Y-Z-1)\\
\end{align}
Note that
\begin{align}
\sum_{Y=a}^{Z-1}(2Y-Z-1)&=2\left(\frac{Z-a}{2}\right)(a+Z-1)-(Z-a)(Z+1)\\
&=(Z-a)(a-2)\\
\end{align}
So, the number of triangles is
\begin{align}
\sum_{Z=4}^N(Z-\lfloor \frac{Z}{2}\rfloor-1)(\lfloor \frac{Z}{2}\rfloor+1-2)=\sum_{Z=4}^N(Z-\lfloor \frac{Z}{2}\rfloor-1)(\lfloor \frac{Z}{2}\rfloor-1)\\
\end{align}
If $N$ is even,
\begin{align}
&\;\sum_{Z=4}^N(Z-\lfloor \frac{Z}{2}\rfloor-1)(\lfloor \frac{Z}{2}\rfloor-1)\\
=&\;\sum_{k=2}^{N/2}(2k-\lfloor \frac{2k}{2}\rfloor-1)(\lfloor \frac{2k}{2}\rfloor-1)+\sum_{k=2}^{(N-2)/2}(2k+1-\lfloor \frac{2k+1}{2}\rfloor-1)(\lfloor \frac{2k+1}{2}\rfloor-1)\\
=&\;\sum_{k=2}^{N/2}(k-1)(k-1)+\sum_{k=2}^{(N-2)/2}k(k-1)\\
=&\;\sum_{i=1}^{(N-2)/2}i^2+\sum_{i=1}^{(N-4)/2}i^2+\sum_{i=1}^{(N-4)/2}i\\
=&\;2\left(\frac{1}{6}\right)\left(\frac{N-4}{2}\right)\left(\frac{N-4}{2}+1\right)(N-4+1)+\left(\frac{N-2}{2}\right)^2\\
&\qquad+\frac{1}{2}\left(\frac{N-4}{2}\right)\left(\frac{N-4}{2}+1\right)\\
=&\;\frac{1}{12}(N-2)(N-3)(N-4)+\frac{1}{4}(N-2)^2+\frac{1}{8}(N-2)(N-4)\\
=&\;\frac{1}{24}(N-2)[2(N-3)(N-4)+6(N-2)+3(N-4)]\\
=&\;\frac{1}{24}N(N-2)(2N-5)
\end{align}
If $N$ is odd, $N-1$ is even. So,
\begin{align}
&\;\sum_{Z=4}^N(Z-\lfloor \frac{Z}{2}\rfloor-1)(\lfloor \frac{Z}{2}\rfloor-1)\\
=&\;\frac{1}{24}(N-1)(N-1-2)[2(N-1)-5]+(N-\lfloor \frac{N}{2}\rfloor-1)(\lfloor \frac{N}{2}\rfloor-1)\\
=&\;\frac{1}{24}(N-1)(N-3)(2N-7)+(N-\frac{N-1}{2}-1)(\frac{N-1}{2}-1)\\
=&\;\frac{1}{24}(N-1)(N-3)(2N-7)+\frac{1}{4}(N-1)(N-3)\\
=&\;\frac{1}{24}(N-1)(N-3)(2N-7+6)\\
=&\;\frac{1}{24}(N-1)(N-3)(2N-1)
\end{align}
|
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|
How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta)}}{5 \sin\theta}\,d\theta $
Factor out 5 and cancel, simplified to $\int\frac { {5(1-( \sin\theta))}}{5 (\sin\theta)}\, d\theta $
Divide by $\sin\theta$ to get $\int\frac { (1-( \sin\theta))}{ (\sin\theta)}\, d\theta $
Simplified to $\int\frac {1}{ \sin\theta}d\,\theta - 1 $
I am stuck here and I can't figure it out. Supposedly the answer simplifies to $\int-5 \sin x \tan x \,dx$
|
Let us begin by $x=5t $ and $dx=5\,dt $.
it becomes
$$\int\frac {\sqrt {25 (1-t^2)}}{5t}5\,dt $$
$$=5\int \frac {\sqrt {1-t^2}}{t}\,dt $$
now put $t=\sin (u) $ and $dt=\cos (u)\,du $.
we get
$$\pm 5\int \frac {\cos(u)}{\sin (u)}\cos (u)\,du $$
$$=\pm 5 \int \frac {\cos^2 (u)}{1-\cos^2 (u)}\sin (u)\,du $$
put $v=\cos (u) $ and you can finish.
|
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|
Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration to solve this problem? and if I should use spherical or double polar or cylindrical coordinate system?
Also how to find the limits of integration in this particular problem as an example?
This is what the cross section looks like
It is clearly visible that $\theta$ goes from $0$ to $\pi$
|
$\textbf{Second method}$ (I presume that this is the method you want):
Show that the volume enclosed by the paraboloid $z=x^2+y^2$, the $xy$-plane $z=0$, and the cylinder $x^2+(y-1)^2=1$ is $\frac{3\pi}{2}$.
This time, use the change of coordinates:
$$
\begin{align*}
x &= r\cos \theta, \\
y &=r\sin\theta, \\
z &=z, \\
\end{align*}
$$
where
$$
\begin{align*}
0 \leq &z\leq r^2, \\
0\leq &r \leq 2\sin\theta, \mbox{ and }\\
0 \leq &\theta \leq \pi. \\
\end{align*}
$$
Let $E$ be the solid defined in this prompt.
Then the volume of the enclosed object is
$$
\begin{align*}
V&=\iiint_E 1 \: dV \\
&= \int_{0}^{\pi}\int_{0}^{2\sin\theta} \int_{0}^{r^2} r\: dzdrd\theta \\
&= \int_{0}^{\pi}\int_{0}^{2\sin\theta} r^3 \: dr d\theta \\
&= \int_{0}^{\pi} \frac{r^4}{4}\Bigg|_0^{2\sin\theta} \: d\theta \\
&= \int_{0}^{\pi} 4\sin^4\theta\: d\theta \\
&= \int_{0}^{\pi} 4\left( \frac{1-\cos(2\theta) }{2}\right)^2 \: d\theta \\
&= \int_{0}^{\pi} \left( 1-\cos(2\theta) \right)^2 \: d\theta \\
&= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\cos^2(2\theta) \right) \: d\theta \\
&= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\frac{1}{2}+\frac{1}{2}\cos(4\theta) \right) \: d\theta \\
&= \theta +\sin(2\theta)+\frac{1}{2}\theta +\frac{1}{8}\sin(4\theta) \Bigg|_0^{\pi} \\
&=\frac{3}{2}\pi + 0 + 0 \\
&=\frac{3}{2}\pi.
\end{align*}
$$
|
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|
How to sum an infinite matrix? Given this matrix that stretches to infinity to the right and up:
$$
\begin{matrix}
...&...&...\\
\frac{1}{4}& \frac{1}{8}& \frac{1}{16}&... \\
\frac{1}{2} & \frac{1}{4}& \frac{1}{8}&... \\
1 & \frac{1}{2}& \frac{1}{4}&... \\
\end{matrix}
$$
I was trying to find the total sum of this matrix. I know the answer should be $4$. I came up with a different solution and a different answer. What is wrong with that solution? Here it is:
The first row sums to $2$. The second row to $2-1$. The third row to $2-1-\frac{1}{2}$ etc... So we get:
$$
\begin{matrix}
2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}&-\frac{1}{16}\\
2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}\\
2&-1&-\frac{1}{2}&-\frac{1}{4}\\
2&-1&-\frac{1}{2}\\
2&-1 \\
2 \\
\end{matrix}
$$
Now for each "$2$" there is a diagonal that gives the sequence $2-1-\frac{1}{2}-\frac{1}{4}...=0$ (since the matrix goes on forever) Therefore, the sum of the matrix must be $0$!
Apparently that's wrong; but why? Thanks!
EDIT: I am looking for an answer to the question what is fundamentally wrong with my method plus an explanation for why that is wrong. I am not looking for an explanation of the correct method.
|
The ''triangular array'' is not really an array but a column of values:
$ 2 $
$2-1=1$
$2-1-\frac{1}{2}=\frac{1}{2}$
$2-1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
$2-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$
$\cdots$
so there is not a diagonal
and the sum of these values is clearly $=4$
In other words, this is not a matrix, but the ''infinite sum''
$$
2+\left(2-1\right)+\left(2-1-\frac{1}{2}\right)+\left(2-1-\frac{1}{2}-\frac{1}{4}\right)+ \cdots +\left(2-\sum_{i=1}^n\frac{1}{i}\right)+ \cdots
$$
and we cannot rearrange or associate the terms of the series in a different order, as adding them ''by diagonals''.
|
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|
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can anyone help me to continue from here
|
$$=\cos^4A-\cos^2A\sin^2A+\sin^4A$$
$$=\cos^4A-2\cos^2A\sin^2A+\sin^4A + \cos^2a\sin^2a$$
$$=(\cos^2A-\sin^2A)^2 + \cos^2A\sin^2A=1-3\sin^2A\cos^2A$$
$$(\cos^2A-\sin^2A)^2 + 4\cos^2A\sin^2A=1$$
$2\cos x\sin x=\sin2x$
-> $\sin^22x=4\cos^2\sin^2x$
and
$\cos^2x-\sin^2x =\cos2x$
$$\cos^22A + sin^22A=1$$
|
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|
Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides.
$$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$
|
Set $\theta=\pi/2-x$, so the left-hand side becomes
$$
\frac{1+\cos x}{\sin x}+\frac{\sin x}{1-\cos x}
$$
Now recall that
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}=
\frac{1-\cos x}{\sin x}
$$
so the left-hand side is actually
$$
2\cot\frac{x}{2}=2\tan\Bigl(\frac{\pi}{2}-\frac{x}{2}\Bigr)=
2\tan\Bigl(\frac{\pi}{2}+\frac{\theta}{2}-\frac{\pi}{4}\Bigr)
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to proceed anymore. Thanks in advance.
|
$\left|y-\frac{4}{3}\right|$ has a maximum at $x=1$ which is $k=\frac{2}{3}$
Derive $f(x)=\frac{x^2+1}{x^2+x+1}-\frac{4}{3}$ and set $f'(x)=0$. Verify that it is a maximum (for instance checking second derivative) then substitute in $\left|y-\frac{4}{3}\right|$ to get $k$.
Hope this helps.
|
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|
Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
|
Let $a$, $b$ and $c$ be sides-lengths of the triangle.
Thus, $$1=\frac{2S}{a+b+c}$$ or
$$4(a+b+c)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ or
$$4(a+b+c)=(a+b-c)(a+c-b)(b+c-a).$$
Now, it's obvious that $a+b+c$ is an even number.
Thus, $a+b-c$, $a+c-b$ and $b+c-a$ are even positive numbers.
Let $a+b-c=2w$, $a+c-b=2v$ and $b+c-a=2u$.
Thus, we need to solve
$$u+v+w=uvw$$ or
$$\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=1.$$
Let $uv=z$, $uw=y$ and $vw=x$.
Hence, $x$, $y$ and $z$ are natural numbers such that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$
This equation we can prove by the following way.
Let $x\leq y\leq z$.
Hence, $$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq\frac{3}{x},$$
which gives $x\leq3$.
$x=1$ is impossible and $x=2$ and $x=3$ give the following solutions:
$(2,3,6)$, $(2,4.4)$ and $(3,3,3)$.
Last two are impossible and $(x,y,z)=(2,3,6)$ gives $(u,v,w)=(3,2,1)$,
which gives $(a,b,c)=(3,4,5)$ and since $4^2+3^2=5^2,$ we are done!
|
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|
Solution of differential equation using Laplace transform Can someone please help me identify my mistake as I am not able to find it.
$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$
$L[y'']+L[y']=3L[x^2]$
$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$
$L[y']=pL[y]-y(0)=pL[y]$
$p^2L[y]-1+pL[y]=\frac{6}{p^3}$
$p^2L[y]+pL[y]=\frac{6}{p^3}+1$
$(p^2+p)L[y]=\frac{6}{p^3}+1$
$L[y]=\frac{6}{p^3(p^2+p)}+\frac{1}{p^2+p}$
$y=x^3-e^{-x}+1-e^{-x}$
|
$\frac{6}{p^3(p^2+p)}=\frac{6}{p^4} - \frac{6}{p^3} + \frac{6}{p^2} - \frac{6}{p} + \frac{6}{1+p}$ then reverse transform gives you $x^3-3x^2+6x-6+6e^{-x}$ add the rest which is right in your calculation we get $$y=x^3-3x^2+6x-6+6e^{-x}+1-e^{-x}=x^3-3x^2+6x-5+5e^{-x}$$
|
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|
Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality:
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$.
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)=y(let)$$
Next using AM-GM inequality
$$\cos\left(\frac{A}{2}\right)+\cos\left(\frac{B}{2}\right)+\cos\left(\frac{C}{2}\right)\geq 3 \left(\frac{y}{4}\right)^{\frac{1}{3}}.$$
I'm unable to proceed further. Please help me.
|
$$\sin A+ \sin B + \sin C=\sin A+ \sin B+ \sin (A+B)$$ $$= \sin(A)(1+ \cos (B))+ \sin(B)(1+\cos (A)) = \frac1{\sqrt3}((\sqrt3 \sin(A))(1+\cos(B))+(\sqrt3 \sin(B))(1+\cos(A))$$ Then use A.M.-G.M. inequality, $$\le \frac1{2\sqrt3}(3\sin^2(A)+1+\cos^2(B)+2\cos(B)+3\sin^2(B)+1+\cos^2(A)+2\cos(A))$$
$$=\frac1{2\sqrt3}(2\sin^2(A)+2\sin^2(B)+4+2\cos(B)+2\cos(A))=\frac1{\sqrt3}( 1-\cos^2(A)+1-\cos^2(B)+2+\cos(B)+\cos(A))=\frac1{\sqrt3}\left(1+\frac14-\left(\cos(A)-\frac12\right)^2+1+\frac14-\left(\cos(B)-\frac12\right)^2+2\right)\le\frac1{\sqrt3}\left(4+\frac12\right)=\frac9{2\sqrt3}=\frac{3\sqrt3}2$$
I multiplied and divided $\sin(A)$ by $\sqrt3$ because I know that the function maximizes when $A=B=\frac{\pi}3$ and so $1+\cos(B)=\frac32=\sqrt3\times \sin(\frac{\pi}3)=\sqrt3\times \sin(A)$, because $ab\le \frac{a^2+b^2}2$ and equality holds only when $a=b$.
|
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|
Fraction problem I have been reading Basic Math and Pre Algebra for Dummies, the question below has me stumped!
David bought a cake for himself and his friends. He cut a piece for himself that was $1/6$ of the total cake. The Sharon cut a piece that was $1/5$ of what was left. Then Armand cut a piece that was $1/2$ of what was left. How much of the cake was left after all three friends had taken their pieces?
The answer in the book is $1/3$, is this a mistake ?
When he ate $1/6$ there was $5/6$ left, so $5/6 - 1/5 = 19/30$
$19/30 - 1/2$? am I on the wrong path here?
|
$\frac{1}{6}$ is taken off the bat, so we're left with $\frac{5}{6}$ of a cake. Then $\frac{1}{5}$ of that is taken, so $\frac{1}{5}\times\frac{5}{6}=\frac{1}{6}$ of the original cake is taken. Now we have $1-2\times\frac{1}{6}=\frac{2}{3}$ of the original cake left. Finally, half of the remaining cake is taken, so $\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}$ of the original cake is taken. Thus, we are left with $\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$ of the original cake left.
|
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|
Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$
I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
|
It's a bit of overkill, but since $f(x) = \frac{1}{1+x^{2}}$ is a monotonically decreasing function on $(0, \infty)$, the Riemann-like sum $$\frac{1}{n}\sum_{k=0}^{{\color{red}{\infty}}} f \left(\frac{k}{n} \right)= \frac{1}{n} \sum_{k=0}^{\infty} \frac{1}{1+ \left(\frac{k}{n} \right)^{2}} = n\sum_{k=0}^{\infty} \frac{1}{n^{2}+k^{2}} $$ converges to the value of $\int_{0}^{\infty} \frac{dx}{1+x^{2}}$ as $n \to \infty$.
And the partial fraction expansion of $\coth (\pi z)$ is $$\coth(\pi z) = \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=1}^{\infty} \frac{1}{z^{2}+k^{2}} = - \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=0}^{\infty} \frac{1}{z^{2}+k^{2}} . $$
Therefore, $$ \begin{align}\int_{-\infty}^{\infty} \frac{dx}{1+x^{2}} &= 2 \int_{0}^{\infty} \frac{dx}{1+x^{2}} \\ &= 2 \lim_{n \to \infty} \left(\frac{1}{2n} + \frac{\pi}{2} \, \coth(\pi n) \right) \\ &= 2 \left(0+ \frac{\pi}{2}(1) \right) \\ &= \pi. \end{align}$$
|
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|
Find the value of $x$ in the triangle $ABC$ A triangle $ABC$ has angle $A=2y$ and angle $C=y$. Furthermore, $AB=x,BC=x+2$, and $AC=5$. Solve for $x$.
I tried this question whole day but could not find solution of it . Please help me
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Let $AD$ be a bisector of $\Delta ABC$ and $AD=y$.
Thus, $AD=DC=y$ and $BD=x+2-y$.
Now, since $\Delta{ABD}\sim\Delta{CBA}$, we obtain
$$\frac{y}{5}=\frac{x}{x+2}=\frac{x+2-y}{x},$$ which gives
$xy=5x-2y$ and $x^2=(x+2)^2-xy-2y$, which is $xy=4x+4-2y$
and from here $4x-2y+4=5x-2y$ or $x=4$ and we are done!
|
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Laplace's equation to solve a system in a square Solve the system in the square $S = \{0 < x < 1, 0 < y < 1\}$:
$$\begin{cases}
\Delta u = 0 \ \ \ &\text{for} \ \ (x,y)\in S\\
u(x,0) = 0, \ u(x,1) = x \ \ \ &\text{for} \ \ 0 < x < 1\\
u_x(0,y) = 0, \ u_x(1,y) = y^2 \ \ \ &\text{for} \ \ 0 < y < 1
\end{cases}$$
Attempted solution - We know that the general solution is $$u(x,y) = \sum_{n=0}^{\infty}A_n \sin \beta_n x(\beta_n \cosh \beta_n y - \sinh \beta_n y)$$
Applying the initial condition $$u(x,1) = x = \sum_{n=0}^{\infty}A_n \sin \beta_n x(\beta_n \cosh \beta_n - \sinh \beta_n)$$
Then I think $$A_n = \frac{2}{\beta_n\cosh \beta_n - \sinh\beta_n}\int_{0}^{1} x\sin \beta_n x dx = \frac{2\sin(\beta_n) - 2\beta_n \cos(\beta_n)}{\beta_n^{3}\cosh \beta_n - \beta_n^{2}\sinh \beta_n}$$ But I am really not understanding how to solve this problem. Please provide steps on how to solve this. Assuming I get $A_n$ then how do I apply the condition that $u_x(1,y) = y^2$? It seems to me that we would have
$$u_x(x,y) = \sum_{n=1}^{\infty}A_n \beta_n \cos \beta_n x(\beta_n \cosh \beta_n y - \sinh \beta_n y)$$
Then $$u_x(1,y) = y^2 = \sum_{n=0}^{\infty} A_n \beta_n \cos \beta_n(\beta_n \cosh \beta_n y - \sinh \beta_n y)$$
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Given that $\Delta u = 0$ this implies that $\nabla^2 u = 0$ or
\begin{equation}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial^2 y} = 0
\end{equation}
Let $u = X(x)Y(y)$ then
\begin{equation}
Y \frac{d^2 X}{d^2 x^2} + X \frac{d^2 Y}{d^2 y} = 0
\end{equation}
Now we have two cases for boundary conditions
$$\text{Case 1:} \ u_1(x,0) = 0, \ \ u_1(x,1) = 0, \ \ u_{1x}(0,y) = 0, \ \ u_{1x}(1,y) = y^2$$
$$\text{Case 2:} \ u_2(x,0) = 0, \ \ u_2(x,1) = x, \ \ u_{2x}(0,y) = 0, \ \ u_{2x}(1,y) = 0$$
Now using equation $(2)$ with case $1$ we have
$$\frac{1}{Y}\frac{d^2 Y}{d^2 y} = -\frac{1}{X}\frac{d^2 X}{d^2 x^2} = -p^2 \ \ (\text{say})$$
Then solving for this we have
$$Y = c_1 \cos(p y) + c_2 \sin(py) \ \ \text{and} \ \ X = c_3 e^{px} + c_4 e^{-p x}$$
Hence $$u = XY = (c_1 \cos(py) + c_2 \sin(py))(c_3 e^{px} + c_4 e^{-px})$$
Now apply case $1$ we get
$$c_1 = 0, \ \ 0 = (c_3 e^{px} + c_4 e^{-px})(c_2 \sin(p))$$
Then
$$\sin(p) = 0 = \sin(n\pi) \Rightarrow p = n\pi$$
Also $u_{1x}(0,y) = 0$ then
$$0 = (c_3(n\pi) - c_4(n\pi))(c_2 \sin(n\pi y)) \Rightarrow (c_3 - c_4)c_2\sin(n\pi y) = 0$$
Now $u_{1x}(1,y) = y^2$ then
\begin{align*}
y^2 &= (n\pi c_3 e^{n\pi} - n\pi c_4 e^{-n\pi} )(c_2 \sin(n\pi y))\\
&\Rightarrow y^2 = \sum_{n=1}^{\infty}(A_n e^{n\pi} - B_n e^{-n\pi})\sin(n \pi y)
\end{align*}
where
$$A_n = \frac{2}{e^{n\pi}}\int_{0}^{1}y^2 \sin(n\pi y) dy = \ldots = \frac{2e^{\pi(-n)}((2 - \pi^2 n^2)\cos(\pi n) + 2\pi n(\sin(\pi n) - 2 ) }{\pi^3 n^3}$$
and
$$B_n = -\frac{2}{e^{-n\pi}}\int_{0}^{1}y^2 \sin(n\pi y)dy = \ldots = -\frac{2e^{\pi n}((2 - \pi^2 n^2)\cos(\pi n) + 2\pi n(\sin(\pi n) - 2 ) }{\pi^3 n^3}$$
Hence the solution is
$$u = \sum_{n=1}^{\infty}(A_n e^{n\pi} - B_n e^{-n \pi})\sin(n\pi y)$$
Similarly for case $2$, we can find the solution to
$$\frac{1}{Y}\frac{d^2 Y}{d^2 y} = -\frac{1}{X}\frac{d^2 X}{d^2 x^2} = -p^2$$
$$\Rightarrow u = (c_1 \cos(py) + c_2 \sin(py))(c_3 e^{px} + c_4 e^{-px})$$
using case $2$, we get
$$u_2(x,0) = 0 \Rightarrow c_3 + c_4 = 0 \Rightarrow c_3 = -c_4$$
So
$$ u = (c_1 \cos(py) + c_2 \sin(py))(c_3 e^{px} + c_4 e^{-px})$$
applying $u_{2x}(0,y) = 0$ then
\begin{align*}
0 &= c_3(e^{py} - e^{-py})p c_2 \cos(p)\\
\Rightarrow c_2 &= 0
\end{align*}
So we have
$$u = c_1 c_3 \cos(px)(e^{py} - e^{-py})$$
Now, applying $u_{2x}(1,y) = 0$ then
$$\sin(p) = 0 = \sin(n\pi) \Rightarrow p = n\pi$$
So
$$u = c_1 c_3 \cos(n\pi x)(e^{n\pi y} - e^{-n\pi y})$$
We have the general solution
$$u = \sum_{n=1}^{\infty}A_n \cos(n\pi x)(e^{n\pi y} - e^{-n\pi y})$$
Applying $u_2(x,1) = x$ we have
$$x = \sum_{n=1}^{\infty}A_n \cos(n\pi x)(e^{n \pi} - e^{-n\pi})$$
where
$$A_n = \frac{2}{e^{n\pi} - e^{-n\pi}}\int_{0}^{1} x \cos(n\pi x) dx = \ldots = \frac{2(\pi n \sin(\pi n) + \cos(\pi n) - 1 )}{\pi^2 (e^{\pi n} - e^{\pi(-n)})n^2 }$$
|
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MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$ Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:
*
*$\dfrac{2n}{(1-x)^{n+1}} $
*$\dfrac{2(n!)}{(1-x)^{2n}} $
*$\dfrac{2(n!)}{(1-x)^{n+1}} $
by Leibniz formula
$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (1+x)^{(k)}\ \left(\dfrac{1}{1-x}\right)^{(n-k)}}$$
using the hint
*
*$\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$ and
*$\left(\dfrac{1}{x}\right)^{n}=\dfrac{(-1)^{n}n!}{x^{n+1}}$
so
$${\displaystyle \left( \dfrac{1+x}{1-x} \right)^{(n)} = \left( \dfrac{2}{1-x}-1 \right)^{(n)}=2\dfrac{ (-1)^{n}n! }{ (1-x)^{n+1} } } $$ but this result isn't apear in any proposed answers
what about the method of Lord Shark the Unknown
tell me please this way holds for any mqc question contain find the n th derivative so it's suffice to check each answer in y case i will start with first
*
*let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f_{n+1}$$
*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f_{n+1}$$
*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f_{n+1}$$
|
You can do it with Leibniz's formula (not that it's easier than without it); just consider that
$$
(1+x)^{(k)}=
\begin{cases}
1+x & k=0 \\
1 & k=1 \\
0 & k>1
\end{cases}
$$
so the formula gives
$$
\left( \frac{1+x}{1-x}\right)^{\!(n)}=
(1+x)\left(\dfrac{1}{1-x}\right)^{\!(n)}+
n\left(\frac{1}{1-x}\right)^{\!(n-1)}
$$
Now make a conjecture about
$$
\left(\frac{1}{1-x}\right)^{\!(n)}
$$
|
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|
Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but clearly this will not work because for $n=3$ we have $3\cdot2^3+1=3\cdot8+1=24+1=25$ which is clearly not divisible by $3$.
Any ideas how to evaluate a formula that generates all the $n$'s for which $n2^n+1$ is divisible by $3$?
|
It should also be clear that if $n$ is divisible by $3$, then $n\cdot 2^n$ is divisible by $3$ so $n\cdot 2^n + 1$ is not divisible by $3$.
Therefore, the only candidates are $n=3k+1$ and $n=3k+2$.
Now, look at the remainders of $2^n$ when divided by $3$:
$$2^1\equiv 2\mod 3\\
2^2\equiv 1\mod 3\\
2^3\equiv 2\mod 3\\
2^4\equiv 1\mod 3\\
2^5\equiv 2\mod 3\\\vdots$$
and so on. You can clearly see that $2^k\equiv 2\mod 3$ for odd $k$ and $2^k\equiv 1\mod 3$ for even $k$.
So, let's look at the whole expression depending on how $n$ is divisible by $6$:
*
*If $n\equiv 0\mod 6$, then $n$ is divisible by $3$, so $n\cdot2^n+1$ is not divisible by $3$.
*If $n\equiv 1\mod 6$, then $n$ is odd, so $2^n\equiv 2\mod 3$. Since $n\equiv 1\mod 3$, this means $$n\cdot 2^n\equiv 1\cdot 2\equiv 2\mod 3$$ meaning that $n\cdot 2^n+1\equiv 2+1=3\equiv 0\mod 3$ so $n\cdot2^n+1$ is divisible by $3$
*If $n\equiv 2\mod 6$, then $n$ is odd and $n\equiv 2\mod 3$. Because $n$ is odd, $2^n\equiv 1\mod 3$, and therefore $n\cdot 2^n\equiv 2\cdot 1\equiv 2\mod 3$ meaning $n\cdot 2^n+1\equiv 2+1\equiv 0\mod 3$, so again, $n\cdot 2^n +1$ is divisible by $3$.
I trust you can finish the rest of the cases yourself...
|
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|
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
|
Doing rationalization you get:
$$a=\frac{x+\sqrt{x^2-4}}{2}\to a-x=\frac{\sqrt{x^2-4}-x}{2}$$
so,
$$a(a-x)=\frac{(\sqrt{x^2-4})^2-x^2}{4}=-1$$
|
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|
Proving $a^{-2}bcd+b^{-2}cda+c^{-2}dab+d^{-2}abc\geqslant a+b+c+d$
If $a,b,c,d$ are positive real numbers, then prove that $$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2}\geqslant a+b+c+d$$
Attempt:
$$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2} = \frac{abcd}{a^3}+\frac{abcd}{b^3}+\frac{abcd}{c^3}+\frac{abcd}{d^3}$$
Using AM-GM inequality:
$$abcd\big[a^{-3}+b^{-3}+c^{-3}+d^{-3}\big]\geqslant4abcd(abcd)^{-\frac{1}{4}}$$
Could anyone help me?
|
We'll replace $a\rightarrow\frac{1}{a}$ and similar.
Thuse, we need to prove that
$$a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd,$$
which is AM-GM or Muirhead.
For example
$$\sum_{cyc}a^3=\frac{1}{3}\sum_{cyc}(a^3+b^3+c^3)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{a^3b^3c^3}=\sum_{cyc}abc$$
and we are done!
|
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|
What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$? Let $A,B\in M_2(\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\det A= \det B$. Our goal is to compute $\det(A^2 - B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true:
*
*$\det(A^2+B^2)+\det(A^2-B^2)=2(\det A^2+\det B^2)$. (Is this well known?)
*(1) $\implies \det(A^2-B^2)=0$.
*If $A,B\in M_2(\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$.
*$(A+B)^2=0 \implies \det(A^2-B^2)=0$.
Can someone help me with justifying these statements?
Edit:
Doug M provided an explanation for (1) in the answers. Here is an explanation for (2):
$A^2+B^2+2AB=O_2 \implies A^2+B^2=-2AB$. So $\det(A^2+B^2)=4\det(AB)$. Now using (1), $\det(A^2-B^2)= 2(\det(A^2)-2\det(AB)+\det(B^2)) = 2((\det(A)^2-\det(B)^2) = 0$.
|
If $A = \pmatrix {a&b\\c&d}, B = \pmatrix {e&f\\g&h}$
$\det (A+B) + \det (A-B)\\(a+e)(c+h) - (b+f)(c+g) + (a-e)(c-h) - (b-f)(c-g)\\ 2ac+2eh - 2bc-2fg\\2(\det A + \det B)$
$\det (A+B) + \det (A-B) = 2(\det A + \det B)$
In which case the initial premise indeed holds.
I am pretty sure this only holds in the $2\times2$ case.
|
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|
How to find $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots$? Find the sum of series $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots,$$ where the terms are the reciprocals of positive integers whose only prime factors are two's and three's:
|
Hint: factor
$$\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{1}{2^i 3^j}$$
as the product of a sum over $i$ and a sum over $j$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the maximum number of distinct positive integer's square that sums up to $2002$?
What is the maximum number of distinct positive integer's square that sums up to $2002$ ?
My tries:
$$\frac{n(n+1)(2n+1)}{6} = 2002$$
$$\implies n\approx 17 $$ but am clueless as to how to proceed any further.
|
From $\frac {17(17+1)(2\cdot 17+1)}6 \lt 2002 \lt \frac {18(18+1)(2\cdot 18+1)}6$ you can conclude that at most $17$ distinct squares can be chosen to sum to $2002$ because the smallest $18$ add up to too much. Now you need to show that you can find $17$. If we were asked the same question for $34$ we could conclude from $1^2+2^2+3^2+4^2$ that it was at most $4$, but in fact the best we can do is $3^2+5^2$ for two. We have $\frac {17(17+1)(2\cdot 17+1)}6=1785$ so you need to increase the sum by $217$. If we delete $a^2$ from the list and add $b^2$ we want $b^2-a^2=217=(b-a)(b+a)=7\cdot 31$, which gives $b=19,a=12$ and our list is all the numbers from $1$ to $17$ except $12$ plus $19$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Formula for Natural logarithm of $\pi$ Does any formula or expansion exist that gives $\ln \pi$ ?
The expansion should not just be any formula of $\pi$ with a $\ln$ before it. For example $\ln \pi$ = k + $\sum f(x)$ or something of this type.
|
We can also derive that
$$1-\sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)=ln(\pi)$$
If we use $\int_0^{\frac{\pi}{2}} ln(sin(x)) dx = -\frac{\pi}{2}ln(2)$ and replace $sin(x)$ by it's Euler product, we get
$$\int_0^{\frac{\pi}{2}} ln(x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2})) dx = -\frac{\pi}{2}ln(2)$$
We can replace the product by a summation to arrive at
$$\int_0^{\frac{\pi}{2}} ln(x) + \sum_{n=1}^\infty ln(1-\frac{x^2}{n^2\pi^2}) dx$$
Which equals
$$\left| xln(x)-x \right|_{0}^{\frac{\pi}{2}}+\sum_{n=1}^\infty \int_0^{\frac{\pi}{2}} ln(1-\frac{x^2}{n^2\pi^2}) dx$$
We see
$$\lim_{x \rightarrow 0} xln(x) = \lim_{x \rightarrow 0} \frac{\ln(x)}{\frac{1}{x}} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x \rightarrow 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}} =\lim_{x \rightarrow 0} -x = 0$$
For the integral we see
$$\int_0^{\frac{\pi}{2}} ln(1-\frac{x^2}{n^2\pi^2}) dx =\int_{0}^{\frac{\pi}{2}} ln(1-\frac{x}{n\pi}) + \ln(1+\frac{x}{n\pi}) dx\\
= \left|(x-\pi n)(ln(1-\frac{x}{\pi n})-1)+(x+\pi n)(ln(1+\frac{x}{\pi n})-1)\right|_0^{\frac{\pi}{2}}\\
=(\frac{\pi}{2}-\pi n)ln(1-\frac{1}{2n})+\pi n -\frac{\pi}{2}+(\frac{\pi}{2}+\pi n)ln(1+\frac{1}{2n})-\pi n -\frac{\pi}{2}\\
=\frac{\pi}{2}ln(1-\frac{1}{4n^2})+\pi n(ln(1+\frac{1}{2n})-ln(1-\frac{1}{2n}))-\pi\\
=\frac{\pi}{2}(ln(1-\frac{1}{4n^2})+4n\coth^{-1}(2n)-2)$$
So we see
$$-\frac{\pi}{2}ln(2)=\frac{\pi}{2}ln(\frac{\pi}{2})-\frac{\pi}{2} +\sum_{n=1}^\infty (\frac{\pi}{2}(ln(1-\frac{1}{4n^2})+4n\coth^{-1}(2n)-2))$$
Dividing by $\frac{\pi}{2}$ and swapping some terms yields
$$ln(\pi)=ln(2)+ln(\frac{\pi}{2}) = 1 - \sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)$$
|
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|
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!
|
For $a=b=c=\frac{1}{2}$ we get a value $\frac{15}{2}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\left(a+\frac{1}{a}-\frac{5}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-2)(2a-1)}{a}\geq0$$ or
$$\sum_{cyc}\left(\frac{(2a-1)(a-2)}{a}+3(2a-1)\right)+6\left(\frac{3}{2}-a-b-c\right)\geq0$$ or
$$\sum_{cyc}\frac{2(2a-1)^2}{a}+6\left(\frac{3}{2}-a-b-c\right)\geq0,$$
which is obvious.
Done!
|
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|
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere?
|
Before rotating (and plotting), you need to parametrize your parabola:
$$\begin{cases}
x(t) = t\\
y(t) = t^2
\end{cases},$$
where $t \in \mathbb{R}.$
You are rotating each point of the parabola, and hence:
$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\cdot\begin{bmatrix}x(t)\\y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}x(t)-y(t)\\x(t)+y(t)\end{bmatrix}.$$
At the end you get that:
$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}t(1-t)\\t(1+t)\end{bmatrix}.$$
This can be plotted in Matlab using the following code:
t=linspace(-3,3,100);
X=(sqrt(2)/2)*(t.*(1-t));
Y=(sqrt(2)/2)*(t.*(1+t));
plot(X,Y)
This is what you get:
|
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|
Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go about this.
Working
This is convergent if
$$
\lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx
$$
exists.
The indefinite integral is
\begin{equation*}
\begin{aligned}
\int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx
& = \int \left(\frac{nx^2}{x^3 + 1}
\right) - \int \left(\frac{1}{13x + 1} \right)dx \\
&= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1)
\end{aligned}
\end{equation*}
Which gives
\begin{equation*}
\begin{aligned}
&\lim_{b \to + \infty}
\int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\
&= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\
&= \lim_{b \to + \infty}
\left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right)
\\ &= \lim_{b \to + \infty}
\left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln(b^3 + 1) - \ln(2)
\right)
- \frac{1}{13}
\left(
\ln(13b + 1)
- \ln(14)
\right)
\right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln \left(\frac{b^3 + 1}{2}\right)
\right)
- \frac{1}{13}
\left(
\ln \left(\frac{13b + 1}{14}\right)
\right)
\right)
\end{aligned}
\end{equation*}
I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( +
\infty)$. But things went pretty south.
So I'm sure there's a better approach.
|
Note that we have
$$\begin{align}
\lim_{b\to \infty}\left(\frac n3\log\left(b^3+1\right)-\frac1{13}\log\left(13b+1\right)\right) &= -\frac1{13}\log(13)\\\\
&+\lim_{b\to \infty}\left(n\log(b)-\frac1{13}\log(b)\right)\\\\
&+\frac n3 \lim_{b\to \infty}\log\left(1+\frac1{b^3}\right)\\\\&-\lim_{b\to \infty}\log\left(1+\frac{1}{13b}\right)\\\\
&=-\frac1{13}\log(13)+\lim_{b\to \infty}\left((n-1/13)\log(b)\right)
\end{align}$$
which converges if and only if $n=1/13$
|
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|
How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once?
The total number of $5$ digit numbers using the digits $0,1,2,3,4,7$ and $8$ is $6\times6\times5\times4\times3=2160.$
Now I found the numbers not divisible by $3$, i.e. even numbers ending in $2,4,8.$
Even numbers from the digits $0,1,2,3,4,7$ and $8$ are $5\times5\times4\times3\times3=930.$
So the numbers divisible by $3$ are $2160-930=1230$ but the answer is $744.$ Where I am wrong?
|
Solution
We've to make $5$ digit numbers using given $7$ digits excluding $2$ digits every time.
Now,$0+1+2+3+4+7+8=25$.
As $25 \pmod 3 \equiv 1$, duplets to be excluded should have their sum $\mod 3=1$.
It's not much difficult to find such duplets. Without much trouble,
we get :$$(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7).$$
Thus, there are such $7$ duplets & hence $7$ such groups of $5$ digits.
Now, we find that out of $7$ groups, $4$ groups have $0$ in them which isn't allowed to take 5th place ( otherwise a $4$-digit number will be formed.).
Total numbers formed by these $4$ groups $=4(5!−4!)=384$.
Other $3$ groups have no zero so total numbers formed by them$=3(5!)=360$
Total possibel numbers: $384+360=744$
|
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|
What is the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis? Question: If $x \ge 0$ and $y \ge 0$, then the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis is?
Answer provided: $3$ units$^2$.
My doubt: How do we graph this relation in order to find the area?
|
Note that
$$
\left\lfloor x \right\rfloor = a\quad \Rightarrow \quad a \leqslant x < a + 1
$$
and therefore
$$
\begin{gathered}
\left\{ \begin{gathered}
\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2 \hfill \\
0 \leqslant \left\lfloor x \right\rfloor ,\left\lfloor y \right\rfloor \hfill \\
\end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered}
0 \leqslant a \leqslant x < a + 1 \hfill \\
0 \leqslant 2 - a \leqslant y < 2 - a + 1 \hfill \\
\end{gathered} \right. \hfill \\
\Rightarrow \quad \left\{ \begin{gathered}
0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\
a \leqslant x < a + 1 \hfill \\
2 - a \leqslant y < 2 - a + 1 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
Therefore the answer $3$ refers to the area of the region
defined by the condition ${\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2}$
in itself.
$$
\begin{gathered}
R = \left\{ {(x,y):\;\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2} \right\}\quad \Rightarrow \quad \int\limits_{\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2} {dxdy} = \hfill \\
= \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{x = a}^{a + 1} {\int_{y = 2 - a}^{2 - a + 1} {dxdy} } } = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} 1 = 3 \hfill \\
\end{gathered}
$$
The area bounded by the given condition and the $x$ and $y$ axes will be instead
$$
\begin{gathered}
S = \left\{ {(x,y):\;\left\{ \begin{gathered}
\;0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\
0 \leqslant x < a + 1 \hfill \\
0 \leqslant y < 2 - a + 1 \hfill \\
\end{gathered} \right.\quad } \right\}\quad \Rightarrow \quad \int\limits_S {dxdy} = \hfill \\
= \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{y = 0}^{2 - a + 1} {dy} } = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\left( {2 - a + 1} \right)} = \hfill \\
= 3 + 2 + 1 = 6 \hfill \\
\end{gathered}
$$
|
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|
Sum of $k^\text{th}$ Powers of the First $n$ Natural Numbers is a Perfect Square.
Question:
If for some $k$ and $\forall$ $n>2017, n\in \mathbb{N}$, there exist an $x\in \mathbb{N}$ such that
$$1^k+2^k+3^k+\cdots+n^k=x^2$$
then $k=3$.
Minor clarification:
The question says : $\forall n$. Now, we know that $1^3+2^3+..+n^3=\left(\frac{n(n+1)}{2}\right)^2$.
But, for all $n\in \mathbb{N}$ and $n>2017$, you won't get some positive integer $x$ such that $1^k+2^k+\cdots+n^k=x^2$. You will get such $x$ only when $k=3$. I don't think there's any particular significance of $2017$ here.
How do we even start this? It looks so weird.
How to proceed after that?
My idea is to use:
$$\sum_{x=0}^n (x+1)^{k+1}-x^{k+1}=n^k \left({k \choose 0} + {k-1 \choose 0} + \cdots + {0 \choose 0}\right) + n^{k-1} \left({k \choose 1} + {k-1 \choose 1} + \cdots + {1 \choose 1}\right) + \cdots + n^{k-k}\left(k\choose k \right)$$.
The $\text {LHS}$ equals $\left(n+1\right)^{k+1}-1$. And we have the $1^k+2^k+3^k+\cdots+n^k$ term in the $\text {RHS}$
But I don't know if his helps to show that the sum of the $k^\text {th}$ powers will be a non perfect sqaure $\forall n>2017$
We can also do something like this-
If $1^k+2^k+\cdots+n^k=x^2$, then $x^2+(n+1)^k=y^2$. And then, $y^2+(n+2)^k=z^2$ and so on.
First of all, is $a^2+n^k=b^2$ possible for all $n>2017$ (though $2017$ hasn't got any special significance) and for a particular $k$ (and needless to mention, $a,b$ are not constants since $n$ is variable)?
But, how do we show that it won't hold unless $k=3$?
|
Answer posted by OP (solved by some AoPSer)
$f(n)=1^m+...n^m$, obvious, that $m$ is odd.
$f(n)<n^{m+1}$
Let $p$ -big prime.
$f(p-1)=x^2,f(p)=y^2 \to p^m=y^2-x^2 \to m=a+b;b>a,x=\frac{p^b-p^a}{2}$
$4x^2=(p^b-p^a)^2=4f(p-1)<4(p-1)^{m+1}$
$p^{2b}+p^{2a}-2p^{a+b}<4p^{a+b}(p-1)$
$p^{2b-2a}+1-2p^{b-a}<4p^{b-a}(p-1)$
Let $p^{b-a}=q$
$q^2-q(4p-2)+1<0 \to q<4p-2<p^2 \to b-a=1$
So if $m=2k+1$ then $x=\frac{p^k(p-1)}{2}$ and so $f(p-1)=(\frac{p^k(p-1)}{2})^2$
We can prove, that for $k\geq 2$ is true, that $f(n)<\frac{(n+1)^{2k+2}}{6}$
$(\frac{p^k(p-1)}{2})^2<\frac{p^{2k+2}}{6}$
$3p^{2k}(p-1)^2<2p^{2k+2}$
$p^2-6p+3<0$ -wrong for big $p$
So $k=0,1 \to m=1,3$ and easy to check, that $m=3$ is answer
|
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|
Prove that a non-positive definite matrix has real and positive eigenvalues I have a $2\times2$ matrix $J$ of rank $2$ and a $2\times2$ diagonal positive definite matrix $Α$. Denote by $J^+$ a pseudoinverse of $J$.
I can find many counterexamples for which $J^+AJ$ is not positive definite (e.g. $J=\left(\begin{smallmatrix}2&1\\1&1\end{smallmatrix}\right)$ and $A=\left(\begin{smallmatrix}1&0\\0&2\end{smallmatrix}\right)$), but for all of them $J^+AJ$ has real and positive eigenvalues. So, I was wondering whether I can prove or disprove that indeed this is the case for this form.
What would happen if $J$ was still of rank $2$, but $3\times2$ this time? Of course, $A$ would be correspondingly a $3\times3$ real positive definite matrix.
|
$J$ is $2 \times 2$ and has rank $2$, so $J$ is invertible. Consequently, the Moore-Penrose pseudoinverse of $J$, $J^+ = J^{-1}$. Then \begin{align*}
J^+ A J = J^{-1} A J
\end{align*} is similar to $A$ and has the same eigenvalues as $A$. $A$ is a diagonal matrix, so is its own symmetric part. Therefore, since $A$ is a real positive definite matrix, $A$ and $J^+ A J$ have all positive eigenvalues.
Edit:
Assuming your $3 \times 3$ version of $A$ is also diagonal... Take \begin{align*}
J &= \begin{pmatrix} 1 & 4 \\ 0 & 1 \\ 0 & 1 \end{pmatrix} \text{,} \\
A &= \begin{pmatrix} 1/2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \text{.}
\end{align*} Then the rank of $J$ is $2$. (To see this, subtract row 3 from row 2 to get row-echelon form and note that we have two nonzero rows. Alternatively, the two columns are linearly independent.) We find \begin{align*}
J^+ &= \begin{pmatrix} 1 & -2 & -2 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \text{,} \\
B = J^+ A J &= \begin{pmatrix} \frac{1}{2} & -2 \\ 0 & 1 \end{pmatrix} \text{,}
\end{align*} and the symmetric part of $B$ is \begin{align*}
B_S &= \begin{pmatrix} \frac{1}{2} & -1 \\ -1 & 1 \end{pmatrix} \text{.}
\end{align*} The eigenvalues of $B_S$ are $\frac{1}{4}(3 \pm \sqrt{17})$ and the smaller one is negative. Thus, we have found choices for $J$ and $A$ yielding a $J^+ A J$ which is not positive definite.
(Note, these values aren't "special". They're more or less the first thing I tried after using a CAS to find the expressions for the eigenvalues. Two more: let the right column of $J$ and $A$ be $(85, -21, -42)$ and $\mathrm{diag}(15, 2173/29, 58)$, respectively, or $(-271, 0,-18)$ and $\mathrm{diag}(93, 70, 148)$, respectively. These were found with that same CAS using a random number function for several numbers and messing around with the remaining ones.)
|
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|
Evaluate the integral $\int\limits_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$. Evaluate the integral $\displaystyle \int_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$.
Hint: put $z = e^{i\theta}$.
Is there a way to solve this without using the Residue Theorem and $\tan(z)$? Is Cauchy's integral formula applicable?
|
Another approach.
If $|b|<1$ then we can write:
$$\frac{1}{1-b\cos x}=\sum_{n=0}^{\infty} b^n\cos^n x$$
The constant term of the Fourier expansion of $\cos^n x$ is zero when $n$ odd and $\frac{1}{2^n}\binom{n}{n/2}$ when $n$ is even.
Now, $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}$ is the constant of the Fourier series expansion, and, since the series converges absolutely, you get that:
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=\sum_{n=0}^{\infty}\left(\frac{b}{2}\right)^{2n}\binom{2n}{n}$$
But when $|z|<\frac{1}{4}$, $$\sum_{n=0}^{\infty} \binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}.$$
So, with $z=b^2/4$, $$\int_{0}^{2\pi}\frac{dx}{1-b\cos x}=2\pi \frac1{\sqrt{1-b^2}}$$
Setting $b=\frac{3}{5}$ we get:
we get:
$$\int_{0}^{2\pi}\frac{dx}{1-\frac{3}{5}\cos x}=\frac{5\pi}{2}$$
And thus your integral is $\frac{\pi}{2}$.
You get the general formula, when $|b|<a$ that:
$$\int_{-\pi}^{\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
|
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|
Length of Radius of Circle
This problem came very straightforward to me. Since the perimeter of the square is $32,$ the sidelengths are $8.$ Since the radius of the circle is $4.$ Constructing a smaller square in the lower right hand corner with side length $4$ I can subtract the radius of the circle $4$ from the constructed diagonal of $4\sqrt2$. Then I divide by two since it's asking for the radius, and I get $2\sqrt2-2.$ Except this cannot be the answer since the integer cannot be being subtracted
|
$r=12-8 \sqrt{2}$ so $k+w+f=22$
$$GB=\frac{8\sqrt 2-8}{2}=4\sqrt 2-4\\
GH=\frac{GB}{\sqrt 2}=4- 2\sqrt 2\\
r+r\sqrt 2=4\sqrt 2-4\\
r=\frac{4\sqrt 2-4}{\sqrt 2+1}=12-8\sqrt 2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.
Where is my mistake? Please explain very intuitively.
|
Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have
$$
\frac{1}{4} = \frac{n}{2}\cdot \frac{n/2}{n^2} \le \frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n-1}{n^2} + \frac{n}{n^2} \le n\cdot \frac{n}{n^2} = 1
$$
In fact, the answer is just the limit of $n^{-2}\binom{n+1}{2} = \frac{1}{2}$ as $n\to\infty$, but the above is why you should guess that the answer isn't zero.
|
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|
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even. Question:
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.
I am not quite sure how to prove this. My guess is proof by contradiction.
Assume the contrary, that is, $a^{2} + b^{2} = c^{2}$ for $c$ even and at least one of $a,b$ odd.
|
$c$ is even, thus $c^2=4n$ for some integer $n$.
Now if $a,b$ are both odd, then
$$a^2+b^2=(2k+1)^2 + (2m +1)^2$$
$$=4k^2+ 4k + 1 + 4m^2 + 4m+ 1$$
$$=4n' + 2 \equiv 2 \pmod4$$
Thus $4 \nmid (a^2+b^2)$, while $4 \mid c^2$ - contradiction.
Also it is obvious that if $a$ even and $b$ odd, or $a$ odd and $b$ even; then $a^2+b^2$ is odd, while $c^2$ is even - contradiction.
Thus it has to be that both $a$, $b$ are even.
|
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|
If $\cos A+\cos B+\cos C=\frac{3}{2}$, prove $ABC$ is an equilateral triangle If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$$ $\implies$
$$a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)=3abc$$ $\implies$
$$(a+b+c)(a^2+b^2+c^2)=2(a^3+b^3+c^3)+3abc$$
Any clues as to how I can take it from here?
|
A trick I often use with these sorts of problems is this. Let $A = P+Q$ and $B=P-Q$, so that we want to show that $Q=0$.
$A+B+C = 180 \Rightarrow C = 180 - (A+B) = 180 - 2P$. The condition becomes $$\cos(P+Q)+\cos(P-Q)+\cos(180°-2P) = \frac{3}{2}$$
Using standard trig identities yields
$$2\cos(P)\cos(Q)-2\cos^2(P)+1 = \frac{3}{2}$$
$$\Rightarrow 4\cos^2(P) - 4\cos(Q)\cos(P) + 1 = 0$$
This is a quadratic in $\cos(P)$, and since it must have real solutions, its discriminant must be $\ge 0$. Hence,
$$16\cos^2(Q)-16 \ge 0$$
$ \Rightarrow \cos^2(Q) \ge 1$ $ \Rightarrow \cos^2(Q) = 1$ $ \Rightarrow \cos(Q) = \pm1$ $\Rightarrow Q = 0 $ or $180$. However, $A-B = 2Q$, and since $A$ and $B$ cannot differ by $360$, $Q$ must be $0$ $\Rightarrow A=B$.
Now, if we swap $B$ with $C$ and repeat the process we will get $A=C$.
Hence, $A=B=C$, and the triangle is equilateral.
|
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|
Finding the general Taylor Series of a function I have to find the general Taylor Series Expansion, about $0$, of the following function.
$$ \sqrt{x^4 -6x^2+1} $$
I have tried to use the identity
$$ \left(1+t\right)^{1/2} = \sum_{n\ge 0}\frac{(-1)^{n+1}}{4^n (2n-1)} \binom{2n}{n}t^n $$.
and then substitute for $t$ accordingly, but to no avail, since the resulting expression becomes messy.
Any help will be appreciated.
Thanks.
|
Keep it simple, for $|x|<1$
$\sqrt{1+x}=1+\dfrac{x}{2}-\dfrac{x^2}{8}+\dfrac{x^3}{16}+O(x^4)$
Plug $x\to x^4-6 x^2$ and get
$$\sqrt{1+ x^4-6 x^2}= \dfrac{1}{16} \left(x^4-6 x^2\right)^3-\dfrac{1}{8} \left(x^4-6 x^2\right)^2+\dfrac{1}{2} \left(x^4-6 x^2\right)+1=\\=1 - 3 x^2 - 4 x^4 - 12 x^6+O(x^7)$$
The initial series had $4$ terms and so will the second one. The additional terms are wrong, I let you imagine the reason why. If you need more terms for the second one you must use more terms in the original series.
All this provided that $|x^4-6x^2|<1$ that is
$$-\sqrt{3+\sqrt{10}}<x<-1-\sqrt{2}\lor 1-\sqrt{2}<x<\sqrt{2}-1\lor 1+\sqrt{2}<x<\sqrt{3+\sqrt{10}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed can someone help me to find the way to simplify the expression.
|
$$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}=\frac{5\sqrt7-16\sqrt5}{4(\sqrt{3\sqrt5}-\sqrt{2\sqrt5})}.$$
|
{
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|
Solving a functional equation and general theory of functional equations $f(x-1)$ + $f(x+1)$ = $kf(x)$
given that $k>1$ then I need to find $f(x)$.
I am clueless about solving functional equations please help me this . Also give some general advice about how to solve functional equations.
|
If we write $\displaystyle F(x) = \sum_{n=0}^\infty f(n)x^n$, then we have
$$f(n-1) + f(n+1) = kf(n)$$
$$\sum_{n=0}^\infty f(n-1)x^n + \sum_{n=0}^\infty f(n+1)x^n = kF(n)$$
$$\sum_{n=-1}^\infty f(n)x^{n+1} + \sum_{n=1}^\infty f(n)x^{n-1} = kF(x)$$
$$f(-1)+\sum_{n=0}^\infty f(n)x^{n+1} -f(0)x^{-1}+\sum_{n=0}^\infty f(n)x^{n-1} = kF(x)$$
$$f(-1)+xF(x) -f(0)x^{-1}+ x^{-1}F(x) = kF(x)$$
$$f(-1) - f(0)x^{-1} = F(x)(k - x-x^{-1})$$
$$F(x) = \frac{f(0) - xf(-1)}{x^2 - kx +1}$$
If we use $U_n$ for Chebyshev polynomials of the second kind, we find that $\displaystyle\sum_{n=0}^\infty U_n(y/2)x^n = \frac{1}{x^2 -yx +1}$, and
$$F(x)= \Big(f(0) - xf(-1))\Big) \sum_{n=0}^\infty U_n(k/2)x^n$$
$$\sum_{n=0}^\infty f(n)x^n = \sum_{n=0}^\infty \Big [f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)\Big ]x^n$$
$$f(n) = f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)$$
Now with
$$U_n(x) = \frac{(x + \sqrt{x^2-1})^{n+1}- (x - \sqrt{x^2-1})^{n+1}}{2\sqrt{x^2-1}}$$
You have a (messy) closed form.
|
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|
Is $\displaystyle\sum_{k=1}^n \frac{k \sin^2k}{n^2+k \sin^2k}$ convergent? Let $\displaystyle x_n=\sum_{k=1}^n\frac{k \sin^2k}{n^2+k \sin^2k}$ for all $n>0$. How can we prove that $(x_n)$ is convergent?
|
To prove that the limit is $\frac{1}{4}$, we just need to prove that
$\sum_{k=1}^n k \sin ^2 (k) = \frac{n^2}{4} + o(n^2)$. In fact :
$\sum_{k=1}^n k \sin ^2 (k) = \sum_{k=1}^n \frac{k}{2} (1 - \cos(2k)) = \frac{n(n+1)}{4} - \frac{1}{2} \sum_{k=1}^n k\cos(2k)$.
Let now $f$ be the function defined by
$f(x) = \sum_{k=0}^n e^{ikx} = \frac{1 - e^{inx}}{1 - e^{ix}} = \frac{\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}e^{i\frac{n}{2}x}$ for $x\in\mathbb{R}\backslash 2\pi\mathbb{Z}$. It is clear that
$\sum_{k=1}^n k \sin ^2 (k) = \frac{n(n+1)}{4} - \Im(f'(1)) = \frac{n^2}{4} + o(n^2)$
since $f'(1) = O(n)$
|
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|
If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)$ is
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these
My try
I take $f (x) = ( x^2+x+1)(x+1) = x^3+2x^2+2x+1$
Real root $x= -d/a =-1$
And also
$f (x) = ( x^2+x+1)(x-2) = x^3-x^2-x-2$
Real root $x = -d/a =2$
But how can I prove it in general?
|
The only way $x^2+x+1$ can be a factor of $ax^3+\cdots+d$ is if the
complementary factor is $ax+d$. As $x^2+x+1>0$ for all real $x$, the real solutions of $ax^3+\cdots+d=0$ must be the real solutions of $ax+d=0$.
|
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|
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
|
\begin{eqnarray*}
\frac{1}{x(x+1)}&=&\frac{1}{x}-\frac{1}{x+1} \\
\frac{2}{x(x+1)(x+2)}&=&\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2} \\
\frac{6}{x(x+1)(x+2)(x+3)}&=&\frac{2}{x(x+1)(x+2)}-\frac{2}{(x+1)(x+2)(x+3)}\\&=&\frac{1}{x}-\frac{3}{x+1}+\frac{3}{x+2}- \frac{1}{x+3}\\
\cdots
\end{eqnarray*}
$A_7= \color{red}{\binom{n}{7}}$.
|
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|
find pairs of real numbers $x, y$ to satisfy this equation equation is: $(x + y)^2 = (x + 3)(y − 3)$
I'm not asking for a solution, but an approach. How do I prove this kind of question? I have tried to arrange it so that it
is $x + y$ = ....
But I still get nothing, nothing intuitive at least. What is a way to tackle this problem?
|
Hint:
By expanding we have
\begin{align*}
(x+y)^2&=(x+3)(y-3)\\
\iff x^2+2xy+y^2&=xy-3x+3y-9\\
\iff x^2+xy+y^2+3x-3y+9&=0\\
\iff \tfrac12(x+y)^2+\tfrac12(x+3)^2+\tfrac12(y-3)^2&=0
\end{align*}
So we must have $$x+y=x+3=y-3=0$$
|
{
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|
Easy irreducibility over $\mathbb{F}_5$ One of the last days I was looking for help in a problem here and, somewhere, I saw someone asking something that involved the irreducibility of the polynomial $f(x)=x^4+2$ over $\mathbb{F}_5$. In that case the argment was that since $f$ has no roots in $\mathbb{F}_5$ it is enough to check if $gcd(x^4+2,x^{24}-1)=1$. This gcd was calculated as follows:$$gcd(x^4+2,x^{24}-1)=gcd(x^4+2,(-2)^6-1)=1$$ I understood that computation and then I kept my life happilly.
However, when I was solving another problem, I needed to verify the reducibility or irreducibility of the polynomial $f(x)=x^4+1$ over $\mathbb{F}_5$. I know that this polynomial is reducible since $x^4+1=x^4-4=(x^2+3)(x^2+2)$ but I would like to see that in that case the $gcd(x^4+1,x^{24}-1)$ would be different of $1$ and that is where I think I did not understand the previous computation. I know that the answer has to be $x^2+3$ or $x^2+2$ but I do not know how to get it. Should I have to divide $x^{24}-1$ by $x^4+1$ and following with this strategy to find the gcd or is there any easier way to find it?
Thank you so much!
|
More general, $x^4\equiv -1\pmod{x^4+1}$, so $x^{4k+r}\equiv (-1)^{k}x^r\pmod{x^4+1}$, so $x^{24}\equiv (-1)^6=1\pmod{x^4+1}$.
The same trick works for dividing $x^{24}-1$ by $x^4+2$:
$$x^4\equiv -2\pmod{x^4+2}\\
x^{24}\equiv (-2)^6\pmod{x^4+2}\\
x^{24}-1\equiv 3\pmod{x^4+2}$$
More generally, over $\mathbb F_q$, $$x^{q-1}\equiv a\pmod{x^{q-1}-a}\\
x^{q^2-1}=a^{q+1}=a^2\pmod{x^{q-1}-a}
$$
So $x^{q^2-1}-1$ is divisible by $x^{q-1}-a$ iff $a=\pm 1$. Otherwise, the two are relatively prime.
|
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|
A nice but somewhat challenging binomial identity
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
\begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{1}{2n+1-2m}}\tag{1}
\end{align*}
I have troubles to prove it and so I'm kindly asking for support.
Maybe the following simpler one-dimensional identity could be useful for a proof. We have for non-negative integers $n$:
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}\tag{2}
\end{align*}
The LHS of (2) can be transformed to
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{0}^1x^{2k}dx\\
&=\int_{0}^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,dx\\
&=\int_{0}^1(1-x^2)^n\,dx
\end{align*}
Using a well-known integral representation of reciprocals of binomial coefficients the RHS of (2) can be written as
\begin{align*}
\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}&=4^n\int_{0}^1x^n(1-x)^n\,dx
\end{align*}
and the equality of both integrals can be shown easily. From (2) we can derive a simple one-dimensional variant of (1).
We consider binomial inverse pairs and with respect to (2) we obtain
\begin{align*}
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}g_k \quad&\quad g_n=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} \quad&\quad\frac{1}{2n+1}=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k
\end{align*}
We conclude again with (2)
\begin{align*}
\frac{1}{2n+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&=\sum_{k=0}^n\frac{(-4)^{k}}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\\
\end{align*}
This identity looks somewhat like a one-dimensional version of (1). Maybe this information can be used to solve (1).
|
Firstly let us evaluate the inner sum on the left hand side. Using the beta function identity quoted above along with the identity $\left. \binom{k+l}{l} = d^l/dx^l x^{k+l}/l! \right|_{x=1}$ we have:
\begin{equation}
S^{(n)}_l:=\sum\limits_{k=0}^n \frac{(-4)^k}{2k+1} \binom{n}{k} [\binom{2k}{k}]^{-1} \binom{k+l}{l} = \left.\frac{1}{l!} \frac{d^l}{d x^l} x^l \int\limits_0^1 \left(1- 4 t (1-t) x\right)^n dt \right|_{x=1}
\end{equation}
Now if we take $m=0$ then $l=0$ and then:
\begin{equation}
rhs= 4^n \int\limits_0^1 \left[ (t-\frac{1}{2})^2 \right]^n dt= 4^n \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2 n} du = \frac{1}{2 n+1}
\end{equation}
as it should be.
Now let us take arbitrary $l \ge 0$ . Then by using the chain rule of differentiation and then by substituting $u := t-1/2$ we have:
\begin{equation}
S^{(n)}_l= \sum\limits_{p=0}^l \binom{l}{p} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p}(4 u^2-1)^p d u
\end{equation}
Therefore the left hand side of the identity to be proved reads:
\begin{eqnarray}
&&\sum\limits_{l=0}^m (-4)^l \binom{m}{l} [\binom{2 l}{l}]^{-1} S^{(n)}_l=\\
&&\sum\limits_{p=0}^m(-1)^{p+1} 2^{2p-1} \frac{\binom{m}{p} (m-p-3/2)!(p-1/2)!}{\sqrt{\pi} \binom{2 p}{p} (m-1/2)!} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&\sum\limits_{p=0}^m (-1)^{p+1} 4^p \frac{\binom{m}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&-4^n \sum\limits_{p=0}^m \binom{m}{p} \frac{\binom{n}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2n-2p} (1-4 u^2)^p du=\\
&&-4^n \frac{1}{2} \frac{n!}{(m-1/2)!} \sum\limits_{p=0}^m \binom{m}{p} \frac{(m-p-3/2)!}{(n-p)!} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (u^2)^{n-p} (1/4 - u^2)^p d u=\\
&& -\frac{1}{4} \frac{n! m!}{(n+1/2)! (m-1/2)!} \sum\limits_{p=0}^m \frac{(m-p-3/2)!(n-p-1/2)!}{(n-p)!(m-p)!}=\\
&& -\frac{1}{(2m-1)(2n+1)} F^{3,2}\left[\begin{array}{rrr} 1&-m&-n\\ \frac{3}{2}-m & \frac{1}{2}-n & \end{array};1\right] = \\
&& -\frac{1}{(2m-1)(2n+1)} \cdot \frac{(\frac{1}{2}-m)^{(n)} (\frac{3}{2})^{(n)}}{(\frac{3}{2}-m)^{(n)} (\frac{1}{2})^{(n)}} = \\
&&-\frac{1}{(2m-1)(2n+1)} \cdot \frac{(1- 2m)(1+2 n)}{1-2 m+2 n} = \frac{1}{2n-2 m+1}
\end{eqnarray}
where in the first line we summed over $l$ and in the second in the third and in the forth lines we simplified the result. Finally in the fifth line we evaluated the integral by substituting for $4 u^2$ and in the subsequent line we expressed the sum through hypergeometric functions. Finally, from Wolfram's site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/02/04/, we used the following identity:
\begin{equation}
F^{(3,2)}\left[ \begin{array}{rrr} a& b & -n \\ d & a+b-d-n+1 & \end{array};1 \right] = \frac{(d-a)^{(n)}(d-b)^{(n)}}{(d)^{(n)} (-a-b+d)^{(n)}}
\end{equation}
for $a=1$, $b=-m$ and $d=3/2-m$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$(m+9)-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}-(m-9)=27$$
$$-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=9$$
$$-(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=3$$
$$-(m+9)^\frac{1}{3}(m-9)^\frac{1}{3}\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)=3$$
$$-(m+9)(m-9)\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$-(m+9)(m-9)(27)=27$$
$$m^2-81=-1$$
$$m=\pm4\sqrt{5}$$
$$\lvert m\rvert=4\sqrt{5}$$
In the third line, I cubed both sides, and then in the forth line, I expanded the left side of the equation using the binomial theorem. Is there a faster or alternative way to do this type of question? If in the question there was a higher-index root (ie. instead of the cube roots there is a higher index), I don't think my method would work, because it would take too long to apply the binomial theorem. How would solve a question in this form?
|
Hint:
Let $\sqrt[3]{m+3}=a$ and $\sqrt[3]{m-3}=b$
$a-b=?$
$a^3-b^3=?$
Can you find $ab$ and $a^3+b^3=(a+b)\{(a-b)^2+ab\}$
|
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|
Area between arc and line I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.
Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ such that the length of arc $AB$ is $21$. Find the area of the region bounded by the arc and the line.
Progress: First, draw the full diagram as shown below.
We can compute the area of $\angle{ABC}$. Using the Law of Cosines, we have that
\begin{align*}
\cos C &= \frac{a^2+b^2-c^2}{2ab} \\
&= \frac{2r^2-400}{2r^2}\\
&= 1-\frac{200}{r^2}
\end{align*}Therefore, $\angle ACB = \cos^{-1}\left(1-\frac{200}{r^2}\right)$. The length of the arc in terms of $\angle ACB$ is
\begin{align*}
&\frac{\angle ACB}{360}2\pi r \\
&= \frac{\angle ACB}{180}\pi r \\
&= \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r
\end{align*}We know that
\begin{align*}
&21 = \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r
\implies \\&\frac{3780}{\pi r} = {\cos^{-1}\left(1-\frac{200}{r^2}\right)}
\implies\\ &\cos\left(\frac{3780}{\pi r}\right) = 1-\frac{200}{r^2}
\end{align*}
However, I couldn't solve this equation. Any help would be appreciated. I can also visualize a calculus approach involving finding the area between two curves, but I want to solve it in an elementary way if possible.
|
As angryavian commented, the problem reduces to : find the zero of $$f(\theta)=\frac \theta {\sin\left(\frac\theta 2\right)}- 2.1\qquad \text{or}\qquad g(\theta)=\theta-2.1{\sin\left(\frac\theta 2\right)}$$ which can only be solved using numerical methods.
Let $x=\frac\theta 2$ to make the equation $$x=1.05 \sin(x)$$ and use, for an approximation, the magnificent $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).
Skipping the trivial $x=0$, this leads to the quadratic
$$4 x^2+\left(\frac{84}{5}-4 \pi \right) x+\pi\left(5 \pi -\frac{84 }{5}\right)=0$$ the positive solution of which being $$x=-\frac{21}{10}+\frac{\pi }{2}+\frac{1}{40} \sqrt{7056+3360 \pi -1600 \pi ^2}\approx 0.537445$$ making $$\theta \approx 1.07489$$ while the exact solution would be $\approx 1.07682$.
With this first result, we could use a Taylor expansion around $\theta=\frac \pi 3$ and get $$g(\theta)=\left(\frac{\pi }{3}-\frac{21}{20}\right)+\left(1-\frac{21 \sqrt{3}}{40}\right)
\left(\theta-\frac{\pi }{3}\right)+\frac{21}{160} \left(\theta-\frac{\pi
}{3}\right)^2+O\left(\left(\theta-\frac{\pi }{3}\right)^3\right)$$ Ignoring the higher order terms, another quadratic leading to $$\theta \approx 1.07683$$
|
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|
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
|
$$\left( \frac { 1+\sin \theta +i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) ^{ n }={ \left( 1+\frac { 2i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) }^{ n }=\\ ={ \left( 1+\frac { 2i\cos \theta \left( 1+\sin \theta +i\cos \theta \right) }{ \left( 1+\sin \theta -i\cos \theta \right) \left( 1+\sin \theta +i\cos \theta \right) } \right) }^{ n }=\\ ={ \left( 1+\frac { 2i\cos \theta +2isin{ \theta \cos { \theta } }{ -2\cos ^{ 2 }{ \theta } } }{ 2+2\sin \theta } \right) }^{ n }=\\ ={ \left( 1+\frac { i\cos \theta +isin{ \theta \cos { \theta } }{ -\cos ^{ 2 }{ \theta } } }{ 1+\sin \theta } \right) }^{ n }=\\ ={ \left( 1-\frac { \cos ^{ 2 }{ \theta } }{ 1+sin{ \theta } } +i\frac { \cos \theta +sin{ \theta \cos { \theta } } }{ 1+\sin \theta } \right) }^{ n }={ \left( \sin { \theta } +i\cos { \theta } \right) }^{ n }=\\ ={ \left( \cos { \left( \frac { \pi }{ 2 } -\theta \right) +i\sin { \left( \frac { \pi }{ 2 } -\theta \right) } } \right) }^{ n }=\cos { n\left( \frac { \pi }{ 2 } -\theta \right) +i\sin { \left( n\left( \frac { \pi }{ 2 } -\theta \right) \right) } } \\ \\ \\ \\ $$
|
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|
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
|
Here is another way, which also illustrates how easy it is to make a mistake with these kinds of problems. If we take the negative fractions to the other side of the equation we get $$\frac 1{x-1}+\frac 1{x-4}=\frac 1 {x-3}+\frac 1{x-2}$$ which becomes $$\frac {2x-5}{(x-1)(x-4)}=\frac {2x-5}{(x-2)(x-3)}$$Now cancel and clear fractions to get $$(x-2)(x-3)=(x-1)(x-4)$$ or $$6=4$$
What went wrong? - Well in cancelling $2x-5$ I didn't check to make sure I was not dividing by zero - so the answer I want is $2x-5=0$ or $x=\frac 52$
|
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|
Number of partitions into $k$ parts: recursion has characteristic polynomial $(X-1)(X^2-1) \cdots (X^k-1)$? Let $p_k(n)$ denote the number of partitions of $n$ into $k$ parts. For example, $p_3(6)=3$ because $6$ admits the partitions $1+1+4$, $1+2+3$ and $2+2+2$ into $3$ parts.
For $k=1$ we have $p_1(n)=1$ and therefore $p_1(n) = p_1(n-1)$. This recursion has characteristic polynomial $X-1$.
For $k=2$ we have $p_2(n)=\lfloor n/2 \rfloor$ and therefore $p_2(n) = p_2(n-1) + p_2(n-2) - p_2(n-3)$. The characteristic polynomial of this recursion is $X^3-X^2-X+1=(X-1)(X^2-1)$.
For $k=3$ and $k=4$ we similarly find that $p_k(n)$ satisfies a recursion, with characteristic polynomials $(X-1)(X^2-1)(X^3-1)$ and $(X-1)(X^2-1)(X^3-1)(X^4-1)$, respectively.
Question: Is it true that $p_k(n)$ satisfies a recursion with characteristic polynomial $(X-1)(X^2-1)\cdots(X^k-1)$ for each value of $k$?
In particular, this would mean that $p_k(n)$ satisfies a recursion of order $\binom{k+1}{2}$.
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What you call $p_k$ I will call $r_k$, and use $p_k$ to count partitions of $n$ into at most $k$ parts. This gives the obvious recurrence $p_k=r_k+p_{k-1}$ (as functions of $n$). We may regard $X$ as the shift operator, satisfying $(Xf)(n):=f(n+1)$, in which case $(X^k-1)p_k$ equals $X^kp_{k-1}$, i.e.
$$ p_k(n+k)=p_k(n)+p_{k-1}(n+k). \tag{$\circ$}$$
This is because every partition of $n+k$ into at most $k$ parts either has $k$ parts, in which case we may subtract $1$ from each part to obtain a partition of $n$, or else it has at most $k-1$ parts.
Thus $(X^k-1)\cdots(X-1)p_k=0$. Substituting $p_k=r_k+p_{k-1}$ in $(X^k-1)p_k=X^kp_{k-1}$ and then rewriting $X^{k-1}p_{k-2}$ as $X(X^{k-1}-1)p_{k-1}$ and cancelling like terms yields
$$ (X^k-1)r_k=(1-X)p_{k-1}+X^kr_{k-1}. \tag{$\ast$}$$
If we start with the induction hypothesis that $(X^{k-1}-1)\cdots(X-1)$ anihilates $r_{k-1}$, then it is clear that $(X^{k-1}-1)\cdots(X-1)$ annihilates both sides of $(\ast)$, or in other words we have concluded that $(X^k-1)(X^{k-1}-1)\cdots(X-1)r_k=0$. The base case $(X-1)r_1=0$ is clear.
|
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|
What is the result of $\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}$ without L'Hôpital's rule. I have the limit
$$
\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}
$$
which I need to compute without L'Hôpital's rule.
(The result is $-\frac{1}{2}$ with L'Hôpital's rule).
Thanks.
|
$\lim \limits_{x \to0+}$
$\frac{e^x-xe^x-1}{\left(e^x-1\right)^2}\cdot\frac{x^2}{x^2}$=
$\lim \limits_{x \to0+}$
$\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}$
$\frac{x^2}{\left(e^x-1\right)^2}$
=
$\lim\limits_{x\to 0^{+}}\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}\left[\frac{x}{\left(e^x-1\right)}\right]^2$
=$\lim\limits_{x\to 0^{+}}\frac{e^x-x-1+x-xe^x}{x^2}\cdot1$
=$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}-\lim \limits_{x \to0+}\frac{x\left(e^x-1\right)}{x^2}$
=$\frac{1}{2}-\lim \limits_{x \to0+}\frac{e^x-1}{x}$
=$\frac{1}{2}-1=-\frac{1}{2}$
note
$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}=\frac{1}{2}$
can be proved without Lohspital rule or series
|
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|
Inequality : $ \sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$ \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}.$$
My attempt :
By Holder inequality,
$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \displaystyle\sum_{cyc}(a+2c)\geq \left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3$
$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{\displaystyle\sum_{cyc}(a+2c)}$
$\Leftrightarrow \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}$
$\Leftrightarrow \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq \sqrt{\frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}}$
|
Your idea to use Holder was very good!
By Holder $$\left(\sum_{cyc}\frac{a+b}{\sqrt{a+2c}}\right)^2\sum_{cyc}(a+b)(a+2c)\geq8(a+b+c)^3.$$
Thus, it's enough to prove that
$$8(a+b+c)^3\geq4(a+b+c)\sum_{cyc}(a+b)(a+2c)$$ or
$$2(a+b+c)^2\geq\sum_{cyc}(a+b)(a+2c),$$ which is
$$\sum_{cyc}(a-b)^2\geq0.$$
Done!
|
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|
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimum.
So, I assumed $x=3$, $y=4$ and $z=4$.
Now putting this value in $xyz+xy+yz+zx$ I got my answer $88$. But actual answer is $78$. Where am I doing it wrong?
|
For $x=5$, $y=4$ and $z=2$ we get a value $78$.
We'll prove that it's a maximal value.
Indeed, let $x<y<z$, $x=1+a$, $y=2+a+b$ and $z=3+a+b+c$,
where $a$, $b$ and $c$ are no-negative integer numbers.
Thus, the condition gives
$$1+a+2+a+b+3+a+b+c=11$$ or
$$3a+2b+c=5,$$
which says that $a\in\{0,1\}$.
Now, let $a=0$.
Thus, $x=1$, $y+z=10$ and we need to prove that
$$y+z+2yz\leq78$$ or
$$yz\leq68,$$
which follows from AM-GM:
$$yz\leq\left(\frac{y+z}{2}\right)^2=25<68.$$
Let $a=1$.
Hence, $x=2$, $y+z=9$, $z\geq5$ and we need to prove that
$$2y+2z+3yz\leq78$$ or
$$yz\leq20$$ or
$$(9-z)z\leq20$$ or
$$(z-4)(z-5)\geq0,$$
which is obvious.
By the same way we can prove that $42$ is a minimal value,
where the equality occurs for $(x,y,z)=(1,2,8)$.
Indeed, if $a=0$ then we get $x=1$, $y\geq2$, $y+z=10$, $6\leq z\leq8$ and we need to prove that
$$xy+xz+yz+xyz\geq42$$ or
$$y+z+2yz\geq42$$ or
$$yz\geq16$$ or
$$(10z)z\geq16$$ or
$$(8-z)(z-2)\geq0,$$ which is obvious.
If $a=1$ then $x=2$, $y\geq3$, $y+z=9$, $5\leq z\leq6$ and we need to prove that
$$2(y+z)+3yz\geq42$$ or
$$yz\geq8$$ or
$$(9-z)z\geq8$$ or
$$(8-z)(z-1)\geq0,$$
which is obvious.
Done!
|
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|
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:
A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$
I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.
|
$$\frac {7^2}{24^2}=\frac {1+\cos 2\alpha}{1-\cos 2\alpha} \implies$$ $$\implies 7^2(1-\cos \alpha)=24^2(1+\cos \alpha)\implies$$ $$\implies 7^2- 7^2\cos 2\alpha = 24^2+ 24^2 \cos 2\alpha\implies$$ $$\implies 7^2-24^2= (7^2+24^2)\cos 2\alpha =25^2 \cos 2\alpha\implies$$ $$\implies -527=625\cos 2\alpha .$$
The missing negative sign on the LHS of the above line is your first error.
Your second error is writing $\cos \frac {\alpha}{2}=\sqrt {\frac {1+\cos \alpha}{2}}\;.$ We have $|\cos \frac {\alpha}{2}|=\sqrt { \frac {1+\cos \alpha}{2} }\;.\;$.... If $450^o<\alpha<340^o$ then $225^o<\frac {\alpha}{2}<270^o,$ implying $\cos \frac {\alpha}{2}<0.$
In general if $\cot x=\frac {a}{b}$ then the proportion of $\cos^2 x$ to $\sin^2 x$ is $a^2$ to $b^2$, so let $\cos^2 x=a^2y$ and $\sin^2 x=b^2y$. Since $1=\cos^2 x +\sin^2 x$, we have $y=a^2+b^2$, so $\cos^2 x =\frac {a^2}{a^2+b^2}$ and $\sin^2 x =\frac {b^2}{a^2+b^2}\;$ and therefore $\;|\cos x|=\frac {|a|}{\sqrt {a^2+b^2}}$ and $|\sin x|=\frac {|b|}{\sqrt {a^2+b^2}}.$
So if $\cot \alpha =\frac {-7}{24}$ and $\cos \alpha<0$ then $\cos \alpha =-\frac {7}{\sqrt {7^2+24^2}}=-\frac {7}{25}.$
|
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|
Where the object that moves along the intersection of $x^2+y^2=1$ and $y+z=1$ needs to be if we want the sum $x+2y+z$ to be max/min?
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maximize/minimize the sum $x+2y+z$? What is the min/max sum?
I'm really not sure how to determine the constraint function. The intersection of the cylinder and the plane is:
$$
x^2+(1-z)^2=1
$$
Then:
$$
z=\frac{x^2+z^2}{2}
$$
I guess we can define new function:
$$
g(x,y)=x+2y+z=x+2y+\frac{x^2+z^2}{2}=x+2y+\frac{x^2+(1-y)^2}{2}
$$
Then we can look for critical points:
$$
g_x=1+x\\
g_y=1+y
$$
Which means that $(-1,-1)$ is the critical point. I don't see how to proceed though.
|
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maximize/minimize the sum $x+2y+z$? What is the min/max sum?
So you're looking for the extreme values of $f(x,y,z) = x+2y+z$ for points $(x,y,z)$ located on the cylinder $\color{blue}{x^2+y^2=1}$ and on the plane $\color{red}{y+z=1}$; i.e. on their intersection. Introduce two Lagrange multipliers to limit the points to those on this intersection:
$$\begin{align}F(x,y,z,\lambda,\mu) & = f(x,y,z)+\lambda \left( \color{blue}{x^2+y^2 -1} \right)+\mu \left( \color{red}{y+z -1} \right) \\
& = x+2y+z+\lambda \left( \color{blue}{x^2+y^2 -1} \right)+\mu \left( \color{red}{y+z -1} \right)\end{align}$$
Now you solve the system:
$$\left\{ \begin{array}{rcl}
F_x = 0 \\
F_y = 0 \\
F_z = 0 \\
F_\lambda = 0 \\
F_\mu = 0
\end{array}\right. \iff \left\{ \begin{array}{rcl}
F_x = 0 \\
F_y = 0 \\
F_z = 0 \\
\color{blue}{x^2+y^2=1} \\
\color{red}{y+z=1}
\end{array}\right. \iff \ldots$$
Can you proceed?
Addition after comments. In your notation with
$$\color{green}{p(x,y,z)=x+2y+z \implies \nabla p = \langle 1,2,1\rangle }$$
the function to be optimized and constraints
$$\color{blue}{j(x,y,z)=x^2+y^2-1 \implies \nabla j = \langle 2x,2y,0 \rangle}$$
and
$$\color{red}{k(x,y,z)=y+z-1 \implies \nabla k = \langle 0,1,1 \rangle}$$
the system becomes:
$$\left\{ \begin{array}{l}
\color{green}{\nabla p} = \lambda \color{blue}{\nabla j} + \mu \color{red}{\nabla k} \\
\color{blue}{j(x,y,z)=0} \\
\color{red}{k(x,y,z)=0}
\end{array}\right. \iff
\left\{ \begin{array}{l}
1=2\lambda x \\
2 = 2\lambda y + \mu \\
1 = \mu \\
\color{blue}{x^2+y^2=1} \\
\color{red}{y+z=1}
\end{array}\right.$$
|
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|
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$.
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
|
hint
Prove that
$$1-\frac {x^2}{2}\le \cos (x)\le 1-\frac {x^2}{2}+\frac {x^4}{24} $$
and integrate.
to prove this, it is easier to show that
$$-x\le -\sin (x)\le -x+\frac {x^3}{6} $$
or the easiest
$$0\le 1-\cos (x)\le \frac {x^2}{2} $$
then integrate twice between $0$ and $x $.
Taylor expansion give only local information.
|
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|
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success.
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.
|
If you know the cubic formula for solving $$x^3 + px + q = 0$$ that helps. The formula is $$x = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ Given the "$2$"s in your expression, you might be led to consider $q=-4$, so the cubic $x^3 + px -4 = 0$ with solution $$x = \sqrt[3]{2+\sqrt{4+\frac{p^3}{27}}}+\sqrt[3]{2-\sqrt{4+\frac{p^3}{27}}}$$ Now it's clear that $p=3$ gives your expression. So your expression is a solution to $$x^3+3x-4=0$$ which factors as $$(x-1)(x^2+x+4)=0$$ The quadratic factor has no real roots. So your number must be $1$.
|
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|
If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$.
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.
This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
|
Rewrite our equation in the following form.
$$(x^3+y^3-11^3+3xy)+(x+y)^3-11^3=0$$
Now, we can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$
Thus, we obtain:
$$(x+y-11)(x^2+y^2+11^2-xy+11x+11y+(x+y)^2+11(x+y)+11^2)=0$$ or
$$(x+y-11)(2x^2+2y^2+xy+22x+22y+242)=0$$
and since
$$2x^2+2y^2+xy+22x+22y+242=(x+11)^2+(y+11)^2+x^2+xy+y^2>0,$$
we obtain $$x+y=11$$
Done!
|
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|
maximize volume of triangle in given perimeter question: Find the triangle with perimeter 2a given that, when we rotate it around one of its sides, the solid obtained have the maximum volume.
suppose we have a triangle of sides x,y,z such that $x+y+z=2a$ and $h$ is the height, we rotate it around the side with size z, we want to get the solid with maximum volume. $Area = \sqrt{a(a-y)(a-y)(a-z)}=\frac{1}{2} zh$
but i was confuse why the volume is $(\frac{1}{3}\pi h^2z)$
if it is cone why not $(\frac{1}{3}\pi z^2h)$??
and is it better to use Lagrange multipliers or substitution method? thankyou
|
From $Area=\frac12 \sqrt{a (a-x) (a-y) (a-z)}$
we get $h=\dfrac{2 \sqrt{a (a-x) (a-y) (a-z)}}{z}$
Plug in the volume $V=\frac{1}{3} \pi h^2 z$ and get
$V=\dfrac{4 \pi a (a-x) (a-y) (a-z)}{3 z}$
with the constraint $x+y+z-2a=0$
We use the Lagrangian multiplier and consider
$f(x,y,z,k)=\dfrac{4 \pi a (a-x) (a-y) (a-z)}{3 z}+k (-2 a+x+y+z)$
Derivative wrt $x,y,z$ must be zero
$$
\left\{ {\begin{array}{*{20}{l}}
{ k-\dfrac{4 \pi a (a-y) (a-z)}{3 z}=0} \\
{k-\dfrac{4 \pi a (a-x) (a-z)}{3 z}=0 } \\
{k-\dfrac{4 \pi a (a-x) (a-y) (a-2 z)}{3 z^2}=0 }
\end{array}} \right.
$$
we get the solution
$$x=\frac{4 \pi a^2-2 \sqrt{3 \pi } a \sqrt{k}+3 k}{4 \pi a},\;y= \frac{4 \pi a^2-2 \sqrt{3 \pi } a \sqrt{k}+3 k}{4 \pi a},\;z= \frac{2 \pi a-\sqrt{3 \pi } \sqrt{k}}{2 \pi }$$
Now we must consider the constraint so we substitute in the equation
$x+y+z=2a$ and we get
$$\frac{4 \pi a^2+2 \sqrt{3 \pi } a \sqrt{k}+3 k}{2 \pi a}+\frac{2 \pi a+\sqrt{3 \pi } \sqrt{k}}{2 \pi }-2 a=0$$
there are two solutions $\quad k=\dfrac{\pi a^2}{3},\;k= \dfrac{4 \pi a^2}{3}$
but only the first leads to the result which is
$x=\dfrac{3 a}{4},\;y=\dfrac{3 a}{4},\;z=\dfrac{a}{2}$
and a volume $V=\dfrac{\pi }{12}\,a^3$
I would have bet that the maximum was the equilateral triangle, but solid generated has a volume smaller, namely $V'=\dfrac{2 \pi }{27}\,a^3$
hope this helps
|
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|
A question of trigonometry on how to find minimum value.
Find The minimum value of $$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}.$$
I can easily get the maximum value but minimum value is kinda tricky. Please help.
|
$f'(\alpha)=2\log 2 \sin \alpha \cos \alpha \left[2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}\right]$
$f'(\alpha)=0 \to 2\sin\alpha\cos\alpha=0$ or $2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0$
$ \sin 2\alpha=0\to 2\alpha=k\pi\to\alpha=\dfrac{k\pi}{2}$
$2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0\to 2^{\sin ^2\alpha}= 2^{\cos ^2\alpha}$
$\sin^2\alpha=\cos^2\alpha\to |\sin\alpha|=|\cos\alpha|\to \alpha=\dfrac{\pi}{4}+k\dfrac{\pi}{2}$
Now it's easy to see that $\alpha=\dfrac{\pi}{4}$ etc leads to the minimum
$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}=2^{\frac12}+2^{\frac12}=2\sqrt 2$
hope this helps
|
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|
Angle bisector problem In triangle $ABC$, $AY$ is a perpendicular to the bisector $\angle ABC$ and $AX$ is a perpendicular to the bisector of $\angle ACB$. If $AB= 9cm$ , $AC=7cm$ and $BC= 4cm$, then the length of $XY$ is?
Any help would be much appreciated.
Thank you.
|
Let $AB=c$, $AC=b$, $BC=a$, $\measuredangle BAC=\alpha$, $\measuredangle ABC=\beta$, $\measuredangle ACB=\gamma$,
$R$ be a radius of circumcircle of $\Delta ABC$, $S$ be an area of the triangle
and let be our bisectors intersects in the point $O$.
Thus, $$\measuredangle XAY=90^{\circ}-\frac{\beta}{2}+90^{\circ}-\frac{\gamma}{2}-\alpha=180^{\circ}-90^{\circ}-\frac{\alpha}{2}=90^{\circ}-\frac{\alpha}{2}.$$
By law of sines for $\Delta AOC$ we obtain:
$$\frac{AO}{\sin\frac{\gamma}{2}}=\frac{b}{\sin\frac{\alpha+\gamma}{2}}$$ or
$$AO=\frac{b\sin\frac{\gamma}{2}}{\cos\frac{\beta}{2}}.$$
But $AO$ is a diameter of the circumcircle of $AXOY$, which says
$$XY=AO\sin\measuredangle XAY=\frac{b\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}}{\cos\frac{\beta}{2}}=\frac{2R\sin\beta\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}}{\cos\frac{\beta}{2}}=4R\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\cos\frac{\alpha}{2}.$$
Now, $p=\frac{a+b+c}{2}=\frac{4+7+9}{2}=10$, which says
$$S=\sqrt{10(10-4)(10-7)(10-9)}=6\sqrt5,$$
$$R=\frac{abc}{4S}=\frac{4\cdot7\cdot9}{4\cdot6\sqrt5}=\frac{21}{2\sqrt5}.$$
Now, by law of cosines for $\Delta ABC$ we obtain:
$$\cos\alpha=\frac{7^2+9^2-4^2}{2\cdot7\cdot9}=\frac{19}{21},$$
$$\cos\beta=\frac{4^2+9^2-7^2}{2\cdot4\cdot9}=\frac{2}{3}$$ and
$$\cos\gamma=\frac{4^2+7^2-9^2}{2\cdot4\cdot7}=-\frac{2}{7}.$$
Thus, $$\sin\frac{\beta}{2}=\sqrt{\frac{1-\frac{2}{3}}{2}}=\frac{1}{\sqrt6},$$
$$\sin\frac{\gamma}{2}=\sqrt{\frac{1+\frac{2}{7}}{2}}=\frac{3}{\sqrt{14}}$$ and
$$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\frac{19}{21}}{2}}=2\sqrt{\frac{5}{21}}.$$
Id est,
$$XY=4\cdot\frac{21}{2\sqrt5}\cdot\frac{1}{\sqrt6}\cdot\frac{3}{\sqrt{14}}\cdot2\sqrt{\frac{5}{21}}=6.$$
Done!
|
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|
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
|
If $(1,y,z)$ is a solution then $(1,z,3z-y+1)$ is also a solution because \begin{align*} &\quad \Big( 1 + z + (3z - y + 1) \Big)^2 + 2 \Big( 1 + z + 3z - y + 1 \Big) - 5 \Big( z + z(3z - y + 1) + 3z - y + 1 \Big) \\ &= (1 + y + z)^2 + 2(1 + y + z) - 5(y + yz + z). \end{align*} You can use this to generate the infinite family
$(1,1,1)$, $(1,1,3)$, $(1,3,9)$, $(1,9,25)$, $(1,25,67), ...$ of solutions.
|
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|
Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$
The problem is when the substitution comes in. I end up having
$$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong.
By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.
|
Here's what I did.
$$\int\frac{\sqrt{9-4x^2}}{x}dx \\
\begin{align}
& a=3\\
& u=2x\\
\text{since} \quad& u=a\sin\theta \\
& 2x=3\sin\theta \\
& 2dx=3\cos\theta; \quad x=\frac{3}{2}\sin\theta; \quad \sin\theta=\frac{2x}{3}; \quad \theta= \sin^1\frac{2x}{3}\\
& dx = \frac{3}{2}\cos\theta \\ \\
\sqrt{a^2-u^2}& =\sqrt{9-(2x)^2} \\
& = \sqrt{9-(3sin\theta)^2} \\
& = \sqrt{9-(9sin^2\theta)}\\
& = 3\cos\theta \\
\\
\text{Going back to the problem.}\\
\int\frac{\sqrt{9-4x^2}}{x}dx &=\int\frac{3\cos\theta}{\frac{3}{2}\sin\theta}(\frac{3}{2}\cos\theta d\theta) \\
&=3\int\frac{\cos^2\theta}{\sin\theta} \\
&=3\int\frac{1}{\sin\theta}-3\int\sin\theta \\
&=3\ln(\csc\theta-\cot\theta)+3\cos\theta \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+3\frac{\sqrt{9-4x^2}}{3} \\
&=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3\ln(\frac{3-\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\
&=3(\ln(3-\sqrt{9-4x^2}-\ln(2x))+\sqrt{9-4x^2} \\
&=3\ln(3-\sqrt{9-4x^2}-3\ln(2x)+\sqrt{9-4x^2} \\
\end{align}
$$
|
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|
The easiest way to solve the improper integral $\int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t$? What is the easiest way to solve this improper integral?
$$I = \int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t $$
The value of this integral according to wolfram alpha is $-\frac{23\pi}{96}$. I could find the same result but my method was too long and complicated. I'm looking for a clear and easy method.
|
$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx +\int_1^{\infty}\frac{\ln x}{(1+x^2)^4}\ dx
\end{align}$
Perform the change of variable $y=\frac{1}{x}$ in the second integral,
$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx -\int_1^{\infty}\frac{x^6\ln x}{(1+x^2)^4}\ dx\\
&=\int_0^{1}\frac{(1-x^6)\ln x}{(1+x^2)^4}\ dx\\
&=\left[\frac{x(3x^4+5x^2+3)\ln x}{3(1+x^2)^3}\right]_0^1-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\left[\frac{-x^3+23(1+x^2)^2\arctan x+x}{24(1+x^2)^2}\right]_0^1\\
&=-\frac{23\pi}{96}\\
\end{align}$
|
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|
Convergence of sequence $\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-a\sqrt{n}+\sqrt{n-1})$ I am able to show that the sequence converge only if $a=2$, I also did a numeric analysis of the sequence, and I am sure that the value of this limit is $-\frac{1}{4}$. However, I do not know how to prove that
$\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})=-\frac{1}{4}$
|
For an analytic approach, we can use a higher order version of MVT.
When $a = 2$, the expression
$$\sqrt{n-1} - 2\sqrt{n} + \sqrt{n+1} = \left. \sum_{k=0}^{2}(-1)^k\binom{2}{k} \sqrt{x+k}\;\right|_{x = n-1}$$
is the second order finite differences of the function $x \mapsto \sqrt{x}$ at $n-1$.
The MVT we need has the form:
For any $f(x)$ defined on $[a,a+mh]$ continuously differentiable up to $m$ times, there exists a $b \in (a,a+mh)$ such that:
$$\left.\frac{d^m f(x)}{dx^m}\right|_{x=b}
= \left.\frac{1}{h!}\Delta^n_h[f](x)\right|_{x=a}
\stackrel{def}{=} \frac{1}{h!} \sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}f(a+kh)
$$
For a proof of above, see this answer.
Apply this MVT to the function $\sqrt{x}$ at $a = n-1$ for any $n > 1$, with $h = 1$ and $m = 2$.
Notice $\sqrt{x}'' = -\frac14 x^{-3/2}$, we find:
$$n^{3/2}\left(\sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right)
= -\frac14\left(\frac{n}{b_n}\right)^{3/2}
\quad\text{ for some }\quad b_n \in (n-1,n+1)
$$
Since $\left(\frac{n}{n+1}\right)^{3/2} < \left(\frac{n}{b_n}\right)^{3/2} <
\left(\frac{n}{n-1}\right)^{3/2}$ and terms on both sides converge to $1$ as $n \to \infty$,
we can conclude
$$\lim_{n\to\infty}n^{3/2}\left(\sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right) = -\frac14$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve the following identity? $$
\sum_{i=2}^{n-k+1} \frac{{{n-i}\choose{k-1}}}{{n} \choose k} \frac{1}{i -1} = \frac{k}{n}\sum_{i=k +1}^{n} \frac{1}{i - 1}.
$$
I came across this identity while I was solving a probability problem. I was able to show that it is true for $k=1$ and $k=2$. But I haven't yet cracked how to solve it for all $k>1$.
For k = 1, it is obviously true.
$$
LHS = \frac{1}{n} \sum_{i=2}^{n} \frac{1}{i-1} = RHS
$$
For k=2, the equations are correct.
\begin{eqnarray}
LHS &=& \frac{2}{n(n-1)}\sum_{i=2}^{n-1}\frac{n-i}{i-1}\\
&=& \frac{2}{n(n-1)}\sum_{i=2}^{n-1}\left(\frac{n-1}{i-1} - 1\right)\\
&=& \frac{2}{n}\left(\sum_{i=2}^{n-1}\frac{1}{i-1} - 1 + \frac{1}{n -1} \right)\\
&=&\frac{2}{n}\sum_{i=3}^{n}\frac{1}{i-1}\\
&=&RHS
\end{eqnarray}
|
Multiplication with $\binom{n}{k}$ gives
\begin{align*}
\sum_{i=2}^{n-k+1}\binom{n-i}{k-1}\frac{1}{i-1}&=\binom{n}{k}\frac{k}{n}\sum_{i=k+1}^n\frac{1}{i-1}\tag{1}\\
&=\binom{n-1}{k-1}\left(H_{n-1}-H_{k-1}\right)
\end{align*}
In (1) we apply the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and the definition of the Harmonic numbers $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$.
Shifting the index $i$ by one gives
\begin{align*}
\sum_{i=1}^{n-k}\binom{n-i-1}{k-1}\frac{1}{i}=\binom{n-1}{k-1}\left(H_{n-1}-H_{k-1}\right)
\end{align*}
Substituting $n \rightarrow n+1$ and $k \rightarrow k+1$ gives the somewhat more convenient representation
\begin{align*}
\sum_{i=1}^{n-k}\binom{n-i}{k}\frac{1}{i}=\binom{n}{k}\left(H_n-H_k\right)\tag{2}
\end{align*}
We transform the left-hand side and obtain
\begin{align*}
\sum_{i=1}^{n-k}\binom{n-i}{k}\frac{1}{i}&=\sum_{i=0}^{n-k-1}\binom{n-i-1}{k}\frac{1}{i+1}&\qquad(i\rightarrow i-1)\\
&=\sum_{i=0}^{n-k-1}\binom{k+i}{k}\frac{1}{n-k-i}&\qquad(i\rightarrow n-k-1-i)\\
&=\sum_{i=k}^{n-1}\binom{i}{k}\frac{1}{n-i}&\qquad(i\rightarrow i+k)
\end{align*}
The identity (2) can now be written as
\begin{align*}
\color{blue}{\sum_{i=k}^{n-1}\binom{i}{k}\frac{1}{n-i}=\binom{n}{k}\left(H_n-H_k\right)}\tag{3}
\end{align*}
We show the validity of (3) for $1\leq k\leq n$ with the help of generating functions.
Expanding
\begin{align*}
\color{blue}{-\frac{\log(1-z)}{(1-z)^{k+1}}}&=\left(\sum_{i=0}^\infty\binom{-k-1}{i}(-z)^i\right)\left(\sum_{l=1}^\infty\frac{1}{l}z^l\right)\tag{4}\\
&=\left(\sum_{i=0}^\infty\binom{k+i}{k}z^i\right)\left(\sum_{l=1}^\infty\frac{1}{l}z^l\right)\tag{5}\\
&=\sum_{n=1}^\infty\left(\sum_{{i+l=n}\atop{i\geq 0,l\geq1}}\binom{k+i}{k}\frac{1}{l}\right)z^n\\
&=\sum_{n=1}^\infty\left(\sum_{i=0}^{n-1}\binom{k+i}{k}\frac{1}{n-i}\right)z^n\\
&=\sum_{n=1}^\infty\left(\sum_{i=k}^{n+k-1}\binom{i}{k}\frac{1}{n-i+k}\right)z^n\tag{6}\\
&\color{blue}{=\sum_{n=k+1}^\infty\left(\sum_{i=k}^{n-1}\binom{i}{k}\frac{1}{n-i}\right)z^{n-k}}\tag{7}\\
\end{align*}
shows $-\frac{\log(1-z)}{(1-z)^{k+1}}$ is a generating function of the left-hand side of (3).
Comment:
*
*In (4) we apply the binomial series expansion and the logarithmic series expansion.
*In (5) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^{q}
\end{align*}
*In (6) we shift the index $i$ to start from $i=k$.
*In (7) we shift the index $n$ to start from $n=k+1$.
On the other hand according to formula (1.2) in Riordan arrays and harmonic number identities by W. Wang it is also the generating function of the right-hand side of (3)
\begin{align*}
\color{blue}{-\frac{\log(1-z)}{(1-z)^{k+1}}}&=\sum_{n=1}^\infty\binom{n+k}{k}(H_{n+k}-H_k)z^n\\
&=\color{blue}{\sum_{n={k+1}}^\infty\binom{n}{k}(H_n-H_k)z^{n-k}}
\end{align*}
and the claim follows.
|
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|
multiple integral finding limit integration question :
$\iint_D \ln(x^2 + y^2)dxdy$ ,
domain : $4\le x^2+y^2 \le 9$
im having hard times to find the limit for integration because it is in polar coordinates, i got that $2 \le r\le 3$ but i dont know how to find the $\theta$ limit.
and also i dont know the graph actually, is it the same as logarithm graph? is it better to draw the graph? is there another way beside drawing the graph? thanks!
|
You make the change of variables:
$$x=r\cos{\theta},y=r\sin{\theta}.$$
Hence:
$$4\le x^2+y^2 \le 9 \Rightarrow 2^2\le r^2\le 3^2 \Rightarrow 2\le r\le 3.$$
The domain is the region between the cocentric circles with the center at the origin and the radii $2$ and $3$, therefore:
$$0\le \theta \le 2\pi.$$
Note: If the domain was $4\le x^2+y^2\le 9, x\ge 0, y\ge 0$, then it would be the region between the concentric circles with the center at the origin and the radii $2$ and $3$ that lies in the first quadrant, implying $0\le \theta \le \frac{\pi}{2}$.
The formula of double integration in polar coordinates is:
$$\iint_{D} f(x,y)dxdy = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2}f(r\cos{\theta},r\sin{\theta})\cdot Jdrd{\theta},$$
where $J$ is the Jacobian:
$$J=\begin{vmatrix}
x_r & x_{\theta} \\
y_r & y_{\theta} \\
\end{vmatrix} = \begin{vmatrix}
\cos{\theta} & -r\sin{\theta} \\
\sin{\theta} & r\cos{\theta} \\
\end{vmatrix} = r.$$
Hence the integral:
$$\iint_{4\le x^2+y^2\le 9} \ln(x^2+y^2)dxdy =\int_0^{2\pi} \int_2^3 \ln{r^2}\cdot rdrd{\theta}=$$
$$\left(\int_0^{2\pi}d{\theta}\right) \left(\int_2^3 \underbrace{\ln{r}}_{u} \cdot \underbrace{2rdr}_{dv}\right)=(2\pi)\left(\ln r \cdot r^2 \bigg{|}_2^3-\int_2^3 r^2\cdot \frac1r dr\right)=$$
$$(2\pi)\left(9\ln 3-4\ln 2-\frac52 \right)\approx 28.996.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trigonometry pre calculus level good question Here is a trigonometry problem.
Given
$$\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m$$
Show that
$$m^2+m\cos(\alpha) = 2.$$
I tried to convert $\sin^3(x)$ into $\sin(3x)$ and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it.
|
From the equation we have following:
$m \sin^3\theta = \sin\alpha \cos3\theta-\cos\alpha \sin3\theta \cdots (1)$
$m \cos^3\theta = \cos\alpha \cos3\theta+\sin\alpha \sin3\theta \cdots (2)$
$\cos3\theta \times (2)-\sin3\theta \times (1) \rightarrow m(\cos^3\theta\cos3\theta-\sin^3\theta\sin3\theta)=\cos\alpha$
$\cos3\theta \times (1) + \sin3\theta \times (2) \rightarrow m(\cos^3\theta\sin3\theta+\sin^3\theta\cos3\theta)=\sin\alpha$
Using $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta$, we have (from here I will write $\cos\theta$ and $\sin\theta$ as $c$ and $s$, respectively)
$m(4(c^6+s^6)-3(c^4+s^4))=\cos\alpha \cdots (3)$
$m(3cs(c^2-s^2))=\sin\alpha \cdots (4)$
Since $4(c^6+s^6)-3(c^4+s^4)=1-6c^2s^2$, it is enough to prove $m^2+m\times m(1-6c^2s^2)=2$, or $m^2(1-3c^2s^2)=1$.
Now $(3)^2+(4)^2$ leads to $m^2(1-6c^2s^2)^2+m^2(9c^2s^2(c^2-s^2)^2)=1$. Since $(c^2-s^2)^2=1-4c^2s^2$, we have $m^2(1-12c^2s^2+36c^4s^4+9c^2s^2(1-4c^2s^2))=1$, or $m^2(1-3c^2s^2)=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} + \frac{4^2}{4^3} + \frac{5^2}{4^4} + \frac{6^2}{4^5} + ...$
If I subtract S from 4S I get 3S = $ 1 + \frac{3}{4^1} + \frac{5}{4^2} + \frac{7}{4^3} + \frac{9}{4^4} + \frac{11}{4^5} + ...$
I cannot find a common ratio to plug in to $\frac{1}{1-r}$ in order to figure this out. I asked my professor for some assistance, and he said the answer should be $\frac{20}{27}$, but I cannot figure out what steps he took to get that. He didn't explain it very well, and all other resources I've checked only show how to compute a geometric sum when there is a common ratio like $\frac{1}{2}$.
How do I determine a common ratio to figure out the sum? I tried division by sequential terms, and that gave me a common ratio of $\frac{5}{16}$, but that left me with an answer of S = $\frac{16}{27}$, which isn't correct.
What am I doing wrong? Or what am I missing?
|
\begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^i = \frac{x(1+x)}{(1-x)^3} \mid_{x=\frac{1}{4}} = \frac{\frac{1}{4} \left( 1+ \frac{1}{4} \right)}{\left(1- \frac{1}{4} \right)^3} = \color{red}{ \frac{20}{27}} .
\end{eqnarray*}
EDIT:
You are happy with a geometric sum ?
\begin{eqnarray*}
\sum_{i=0}^{\infty} x^i = \frac{1}{1-x}
\end{eqnarray*}
Now differentiate this equation with respect to $x$ ( use $ \frac{d}{dx} x^n = nx^{n-1}$)
\begin{eqnarray*}
\sum_{i=0}^{\infty} i x^{i-1} = \frac{1}{(1-x)^2}
\end{eqnarray*}
multiply by $x$
\begin{eqnarray*}
\sum_{i=0}^{\infty} i x^{i} = \frac{x}{(1-x)^2}
\end{eqnarray*}
differentiate again (The RHS is tricky & I will supply more detail if needs be)
\begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^{i-1} = \frac{1+x}{(1-x)^3}
\end{eqnarray*}
finally multiply by $x$ and we have the result
\begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^{i} = \frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
|
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|
Generalizing the dot product to multivectors I am studying the book Linear and Geometric Algebra (Macdonald), and I've been stuck on a couple related, seemingly-elementary problems for a couple of days.
5.3.4. Suppose that $\mathbf{a} \bot \mathbf{b}$. Show that $\mathbf{a} \cdot (\mathbf{a} \land \mathbf{b}) = |\mathbf{a}|^2\mathbf{b}$.
5.3.5. Show that $\mathbf{e_1} \cdot (\mathbf{e_2} \land \mathbf{e_3}) = 0$
This early in the text, $\cdot$ has only been defined for vectors, not for multivectors. The only relevant identities seem to be:
*
*$\mathbf{a}\mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \land \mathbf{b}$ for pure vectors $\mathbf{a}$ and $\mathbf{b}$ (presented as the fundamental identity)
*The geometric product distributes over addition and is associative.
*Identities for the geometric product of vectors:
$$ \mathbf{a} \parallel \mathbf{b} \Rightarrow \mathbf{a} \mathbf{b} = \mathbf{b} \mathbf{a}$$
$$\mathbf{a} \bot \mathbf{b} \Rightarrow \mathbf{a} \mathbf{b} = - \mathbf{b} \mathbf{a}$$
*Scalar multiplication identities for $\land$ and the geometric product
I tried assuming that fundamental identity was not limited to 1-vectors, and applied to vector-bivector multiplication as well:
$$
\begin{align}
aB & = a \cdot B + a \land B && \text{unmotivated generalization} \\
Ba & = B \cdot a + B \land a && \text{unmotivated generalization} \\
aB + Ba & = a \cdot B + B \cdot a + a \land B + B \land a && \text{add and rearrange} \\
aB + Ba & = a \cdot B + a \cdot B + a \land B - a \land B && \text{assume} \cdot \text{commutes and} \land \text{anticommutes}\\
a \cdot B& = \frac 12 (aB + Ba) && \text{}\\
\end{align}
$$
But this doesn't work; as Guillermo Angeris shows in his answer, the correct form of $\cdot$ is actually $a \cdot B = \frac 12 (aB - Ba)$.
Both these problems seem to require a rule governing how $\cdot$ applies to vectors and bivectors (incorrect strawman: $\mathbf{a} \cdot (\mathbf{b}\mathbf{c}) = (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$). However, this more-general $\cdot$ is not presented until the next chapter.
These two problems seem designed to teach or motivate something, but I don't see how they're soluble given the minimal set of definitions presented, at least not without the student trying to define their own generalization of the geometric product, $\cdot$, and/or $\land$ operators.
Is a solution derivable given what's already been presented?
|
I don't have that book by MacDonald (although I recall liking the preprint of that I read eons ago), and I don't remember the lingo that he uses.
Define a k-vector as a quantity having a single grade.
Examples of a 2-vector:
$$\begin{aligned} &\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4 \\ &\mathbf{e}_1 \mathbf{e}_3\end{aligned}$$
The general dot product formula for two k-vectors $ a_r, b_s $, of grades r and s respectively, is typically defined as a grade selection of the following sort:
$$a_r \cdot b_s={\left\langle{{ a_r b_s }}\right\rangle}_{{\left\lvert {r - s} \right\rvert}}.$$
For 5.3.4 we have
$$\begin{aligned} \mathbf{a} \cdot (\mathbf{a} \wedge \mathbf{b}) &= {\left\langle{{ \mathbf{a} (\mathbf{a} \wedge \mathbf{b}) }}\right\rangle}_{1} \\ &= {\left\langle{{ \mathbf{a} (\mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b}) }}\right\rangle}_{1} \\ &= {\left\langle{{ \mathbf{a} \mathbf{a} \mathbf{b} }}\right\rangle}_{1} -{\left\langle{{ \mathbf{a} }}\right\rangle}_{1} (\mathbf{a} \cdot \mathbf{b}) \\ &= \mathbf{a}^2 {\left\langle{{ \mathbf{b} }}\right\rangle}_{1} -{\left\langle{{ \mathbf{a} }}\right\rangle}_{1} (\mathbf{a} \cdot \mathbf{b}) \\ &= \mathbf{a}^2 \mathbf{b} -\mathbf{a} (\mathbf{a} \cdot \mathbf{b}) ,\end{aligned}$$
but $ \mathbf{a} \cdot \mathbf{b} = 0 $ for this problem, proving the desired result.
The dot product of a multivector is typically defined of the dot products of all the grade components of the multivector, as in
$$\begin{aligned}A &= \sum_k {\left\langle{{A}}\right\rangle}_{{k}} \\ B &= \sum_k {\left\langle{{B}}\right\rangle}_{{k}} \\ A \cdot B&=\sum_{r, s} {\left\langle{{A}}\right\rangle}_{{r}} \cdot {\left\langle{{B}}\right\rangle}_{{s}},\end{aligned}$$
so
$$\begin{aligned} \mathbf{a} \cdot ( \mathbf{b} \mathbf{c} ) &= \mathbf{a} \cdot ( \mathbf{b} \cdot \mathbf{c} + \mathbf{b} \wedge \mathbf{c} ) \\ &= {\left\langle{{\mathbf{a} ( \mathbf{b} \cdot \mathbf{c})}}\right\rangle}_{{1-0}} + {\left\langle{{\mathbf{a} ( \mathbf{b} \wedge \mathbf{c})}}\right\rangle}_{{2-1}} \\ &= (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} + (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{b}.\end{aligned}$$
This is not generally equal to $ (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} $.
|
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|
Series $\sum_{k=1}^n\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$ Evaluate
$$\sum_{k=1}^n\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$
I am struck with calculation for this question
|
Hint:$$2\cot (2x)=\cot x-\tan x\\ \to \tan x=2\cot (2x)-\cot x$$so
$$\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}=\\
\frac{2\cot (2\frac x{2^k})-\cot \frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$
|
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|
Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$
This is what I should get:
$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$
Which I can simplify to:
$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$
However, this answer can't be found in my multiple choice question here:
Any hints on what I'm doing wrong?
|
The answer is b, which is equivalent to your answer after simplifying.
This is because $ 9 - \sqrt{45} = 9 - \sqrt{9 * 5} = 9 - 3\sqrt{5}$. Then,$ \frac{9 - 3\sqrt{5}}{18} = \frac{3 - \sqrt{5}}{6}. $
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
|
Use equivalents:
$$\begin{aligned}
7x^4+x^2\, 3^x+2&\sim_\infty x^2 3^x\\
x^3+x\,4^x+1&\sim_\infty x\,4^x
\end{aligned}\quad\text{hence}\quad\frac{7x^4+x^2\, 3^x+2}{x^3+x\,4^x+1}\sim_\infty\frac{ x^2 3^x}{ x\,4^x}=x\Bigl(\frac34\Bigr)^{\!x}\to 0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that the function $f(z) = \sum_{n=0}^\infty \dfrac{1}{1+z^n}, \quad |z|>1,$ is holomorphic. I attempted to solve the following problem and would like feedback (corrections, suggestions for improvement, or a better way).
Prove that the series $\sum_{n=0}^{\infty} \dfrac{1}{1+z^n}$ converges for $|z|>1$ and that the function
$$f(z) = \sum_{n=0}^\infty \dfrac{1}{1+z^n}, \quad |z|>1,$$
is holomorphic.
\textbf{Solution:}
For $|z|>1$, we note that $|z^n| -1 \leq |1+ z^n| \leq |z^n| + 1 $ which implies
$$\dfrac{1}{|z|^n -1} \geq \dfrac{1}{|1+ z^n|} \geq \dfrac{1}{|z|^n + 1}.$$
Letting $r=|z|$, we note that by the comparison test for series,
\begin{eqnarray}
\lim_{n \to \infty} \dfrac{\dfrac{1}{r^{n+1}+1}}{\dfrac{1}{r^n+1}} &=& \lim_{n \to \infty} \dfrac{r^n+1}{r^{n+1}+1} \\
&=& \lim_{n \to \infty} \dfrac{r^n}{r^{n+1}} \\
&=& \dfrac{1}{r} \\
&<&1, \ \text{since} \ r>1,
\end{eqnarray}
that is converges. Also,
\begin{eqnarray}
\lim_{n \to \infty} \dfrac{\dfrac{1}{r^{n+1}-1}}{\dfrac{1}{r^n-1}} &=& \lim_{n \to \infty} \dfrac{r^n-1}{r^{n+1}-1} \\
&=& \lim_{n \to \infty} \dfrac{r^n}{r^{n+1}} \\
&=& \dfrac{1}{r} \\
&<&1, \ \text{since} \ r>1,
\end{eqnarray}
that is converges. So by the Sandwich" orSqueeze" Theorem", $$\sum_{n=1}^\infty\dfrac{1}{|1+ z^n|}$$
converges. Hence $\sum_{n=1}^\infty\dfrac{1}{1+ z^n}$ converges since it converges absolutely for $|z|>1$.
To show that $f(z)$ is holomorphic we can either show that $f(z)$ is differentiable in the region $G$ and apply Goursat's theorem or we can show that $f(z)$ is continuous in $G$ and for every closed path $C$ in $G$, $\int_C f(z) \ dz = 0$.
I'm having difficulty showing $f(z)$ is point-wise continuous. For every $\varepsilon > 0$, there exists at $\delta > 0$ such that
$$|f(z)-f(z_0)| < \varepsilon, \quad \text{ if } \ |z-z_0| < \delta.$$
\begin{eqnarray}
|f(z)-f(z_0)| &=& \left| \sum_{n=0}^\infty \dfrac{1}{1+z^n} - \sum_{n=0}^\infty \frac{1}{1+z_0^n}\right| \nonumber \\
&=& \left| \sum_{n=0}^\infty \dfrac{1}{1+z^n} - \frac{1}{1+z_0^n}\right| \nonumber \\
&=& \left| \sum_{n=0}^\infty \dfrac{z_0^n-z^n}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&\leq& \sum_{n=0}^\infty \left|\dfrac{z_0^n-z^n}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&=& \sum_{n=0}^\infty \left|\dfrac{(z_0-z)(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2} +z^{n-1})}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&=& \sum_{n=0}^\infty |z_0-z| \cdot \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2} +z^{n-1})}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&<& \sum_{n=0}^\infty \delta \cdot \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2} +z^{n-1})}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&=& \delta \cdot\sum_{n=0}^\infty \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2} +z^{n-1})}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&\leq & \delta \cdot\sum_{n=0}^\infty \dfrac{|z_0^{n-1}|+|z_0^{n-2}z|+|z_0^{n-3}z^2|+ \cdots +| z_0z^{n-2}| + |z^{n-1}|}{(|z^n|-1)(|z_0^n|-1|)} \nonumber \\
&= & \delta \cdot\sum_{n=0}^\infty \dfrac{|z_0^{n-1}|+|z_0^{n-2}||z|+|z_0^{n-3}||z^2|+ \cdots +| z_0||z^{n-2}| + |z^{n-1}|}{(|z^n|-1)(|z_0^n|-1|)} \nonumber \\
&\leq & \delta \cdot\sum_{n=0}^\infty \dfrac{|z_0^{n-1}|+|z_0^{n-2}|(|z_0|+\delta) +|z_0^{n-3}|(|z_0|+\delta)^2+ \cdots +| z_0|(|z_0|+\delta)^{n-2} + (|z_0|+\delta)^{n-1}}{((|z_0|+\delta)^{n}-1)(|z_0^n|-1|)} \nonumber \\
&< & \delta \cdot\sum_{n=0}^\infty \dfrac{2n \cdot (|z_0|+\delta)^{n-1}}{(|z_0|^n-1|)^2} \nonumber \\
&=& \delta \cdot M \nonumber \\
&<& \varepsilon
\end{eqnarray}
I call $M$ the sum to which the series converges (by the series ratio test). Is there a better way to show that $f(z)$ is point-wise continuous?
To use Morera's theorem, we need to show uniform convergence of $f(z)$. But I ran into problems with that as well.
Thanks for your assistance.
|
This is hard work! Take $R>1$ and consider $U=\{z:|z|>R\}$. On $U$,
$$\left|\frac{1}{1+z^n}\right|\le\frac{1}{R^n-1}\le\frac{R^{-n}}{1-1/R}$$
(for $n\ge1$). So on $U$ the series is uniformly convergent, and the sum
is therefore holomorphic on $U$.
|
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|
Product of consecutive natural numbers Let $n,k$ be two natural numbers, $n\ge k+2$ and
$$(n-k-1)(n-k)(n-k+1)(n-k+2)=k(k+1)n(n+1).$$
In the LHS we have product of 4 consecutive numbers and in the RHS we have the product of 2 consecutive numbers on the product of 2 consecutive numbers.
Problem. Prove that $k,k+1,n,n+1$ also are consecutive numbers.
My attempt. All factors are in increasing order. Suppose that $n-k-1=k$, then $n=2k+1$ and on we can reduce the products to $k(k+1)(k+2)(k+3)=k(k+1)n(n+1)$ or $(k+2)(k+3)=n(n+1)=(2k+1)(2k+2)$ We have equal products of two 2 consecutive numbers so it implies that $k+2=2k+1$ and we get a solution $k=1, n=3$.
How to prove that the case $n-k-1 \neq k$ is impossible?
|
Let $a:=n-k\ (\ge 2)$.
Then, we have
$$(a-1)a(a+1)(a+2)=k(k+1)(k+a)(k+a+1)$$
This can be written as
$$(2k^2+2ka+2k+a)^2=4a^4+8a^3-3a^2-8a$$
Now, for $a\gt 2$, we have
$$(2a^2+2a-2)^2\lt 4a^4+8a^3-3a^2-8a\lt (2a^2+2a-1)^2$$
from which we have that $4a^4+8a^3-3a^2-8a$ cannot be a square number for $a\gt 2$.
So, we have to have $n-k=a=2$.
It follows from this that $k,k+1,n,n+1$ are consecutive numbers.
|
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|
Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?
|
Hint: The ceiling of a real number is always an integer. So $0<\lceil x \rceil\leq 1$ is equivalent to $\lceil x\rceil = 1$, or $0<x\leq 1$.
|
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|
Vector Analysis Identity One Question given in class was to prove that:
$$\mathbf{(A \cdot B \times C)(a \cdot b \times c)} $$
is equal to
$$\begin{vmatrix} \mathbf A \cdot a & \mathbf A \cdot b & \mathbf A \cdot c \\
\mathbf B \cdot a & \mathbf B \cdot b & \mathbf B \cdot c \\ \mathbf C \cdot a & \mathbf C \cdot b & \mathbf C \cdot c \end{vmatrix}$$
I tried to write everything out in components, but that went no where fast. Anywhere to start? I am only allowed to use vector analysis and properties of the dot product and cross product. However, I do know that
$$\mathbf{(A \cdot B \times C)} $$
is equal to
$$\begin{vmatrix} \mathbf A_x & \mathbf A_y & \mathbf A_z\\
\mathbf B_x & \mathbf B_y & \mathbf B_z\\ \mathbf C_x & \mathbf C_y & \mathbf C_z \ \end{vmatrix}$$
any help? It looks like it has some cool geometric properties.
|
Just for reference in case someone want to see the actual algebraic manipulation.
In addition to some symmetry relations among dot and cross products
$$p \times q = -q \times p \quad\text{ and }\quad p\cdot(q \times r) = q \cdot( r \times p) = r \cdot( p \times q)$$
The key identities we need are:
$$\begin{align}
p \times (q \times r) &= (p\cdot r) q - (p \cdot q)r &\tag{*1}\\
(p \times q) \cdot ( r \times s) &= (p\cdot r)(q \cdot s) - (p\cdot s)(q \cdot r)& \tag{*2}
\end{align}$$
Using $(*2)$, we can rewrite the determinant at hand as
$$\left|\begin{matrix}
A\cdot a & A\cdot b & A\cdot c\\
B\cdot a & B\cdot b & B\cdot c\\
C\cdot a & C\cdot b & C\cdot c\\
\end{matrix}\right|
= \begin{array}{rl}
& A\cdot a((B\cdot b)(C\cdot c)- (B\cdot c)(C\cdot b))\\
- & B\cdot a((A\cdot b)(C\cdot c)-(A\cdot c)(C\cdot b))\\
+ & C\cdot a((A\cdot b)(B\cdot c)-(A\cdot c)(B\cdot b))\end{array}
= \begin{array}{rl}
& A\cdot a((B\times C)\cdot(b \times c))\\
-& B\cdot a((A\times C)\cdot(b \times c))\\
+& C\cdot a((A\times B)\cdot(b \times c))
\end{array}
$$
The determinant now has the form
$$u \cdot (b \times c)
\quad\text{ where }\quad
u = A\cdot a(B\times C)
- B\cdot a (A \times C)
+ C\cdot a (A \times B)$$
Using $(*1)$, we have
$$\begin{align}A\cdot a(B\times C) - B\cdot a(A\times C)
&= ((A\cdot a)B - (B\cdot a)A)\times C\\
&\stackrel{(*1)}{=} -(a \times (A \times B))\times C\\
&= C \times (a \times (A \times B))\\
&\stackrel{(*1)}{=} (C\cdot(A \times B)) a - (C\cdot a)(A\times B)
\end{align}
$$
Adding $(C\cdot a)(A \times B)$ to both sides, we obtain
$$u = (C\cdot (A \times B)) a = (A\cdot(B \times C)) a$$
and results follows.
|
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|
If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
|
Put $t = \tan \frac{x}{2}$. Then $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}$. The equation reduces to
$$3t^2 - 16t + 5 = 0$$ and hence $(3t-1)(t-5) = 0$. Hence the solutions are given by $\tan \frac{x}{2} = 5$ or $\frac{1}{3}$
Let us verify the solutions graphically. The following shows the graphs of $8 \sin(x) - \cos(x) = 5$ (green curve), $\tan \frac{x}{2} = 5$ (blue curve) and $\tan \frac{x}{2} = \frac{1}{3}$ (red curve). It is clear that we have obtained all the solutions.
|
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|
Strange inequality over the positive reals Let $a$, $b$ and $c$ be positive real numbers with $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$. Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geqslant (ab+bc+ca)^2$$
|
We need to prove that
$$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)\geq0$$ or
$$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)+3\sum_{cyc}(2a^3b-a^2b^2)\geq0$$ or
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0.$$
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence,
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)=$$
$$=6(u^2-uv+v^2)a^2+(5u^3+3u^2v-6uv^2+5v^3)a+uv(2u-v)^2\geq0.$$
Done!
Also, we can use C-S.
We need to prove that
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0$$ or
$$\sum_{cyc}(4a^3b-4a^2b^2+ab^3)\geq(a+b+c)abc$$ or
$$\sum_{cyc}ab(2a-b)^2\geq(a+b+c)abc$$ or
$$\sum_{cyc}\frac{(2a-b)^2}{c}\geq a+b+c.$$
Now, by C-S
$$\sum_{cyc}\frac{(2a-b)^2}{c}\geq\frac{\left(\sum\limits_{cyc}(2a-b)\right)^2}{\sum\limits_{cyc}c}=a+b+c.$$
|
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|
What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$
Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$
(DO NOT use trigonometric methods)
|
prove that $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}\le \frac{10}{3}$$ and the equal sign holds if $$a=\frac{2}{3}\left(\frac{1}{2\sqrt{2}}-\sqrt{2}\right),b=-\sqrt{2},c=-\frac{1}{2\sqrt{2}}$$
|
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|
Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$
Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing seem to work. Then putting the original equation in wolfram alpha I got solution $x=\frac{2}{9}(7+2 \sqrt{10})$. Can anyone suggest a method to solve it without wolfram alpha or any such computer method. Thanks in advance.
|
First, the condition is $x\geq 0$.
Now, as you did, put $t = \sqrt{x}$, then we have the equation.
$$18t^4-18t^3 -17t^2-8t-2 = 0.$$
Obviously, $t=0$ is not the solution of this equation. Then, we can divide two sides by $t^4$, $$18-\frac{18}{t}-\frac{17}{t^2}-\frac{8}{t^3}-\frac{2}{t^4} = 0$$
Let $u = \frac{1}{t}$, we have
$$18-18u-17u^2-8u^3-2u^4=0$$
$$25 - 5(u+1)^2 - 2(u+1)^4 = 0$$
Now, you can solve the quadratic equation of $(u+1)^2$:
$$(u+1)^2 = \frac{5}{2}$$
Thus, $u = \frac{\sqrt{5}}{\sqrt{2}}-1 = \frac{\sqrt{10}-2}{2}$.
So, $t = \frac{1}{u} = \frac{\sqrt{10}+2}{3}$. Finally, $x = t^2 = \frac{14+4\sqrt{10}}{9}.$
|
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|
Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.
Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{x+h}{x^2+2hx+h^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{(x+h)(x^2+1)-x(x^2+2hx+h^2+1)}{(x^2+1)(x^2+2hx+h^2+1)}}{h}\\
&= \lim_{h\to 0} \frac{\frac{hx^2-2hx-h^2+h}{(x^2+1)(x^2+2hx+h^2+1)}}{h}
\end{align}
$$
I'm currently stuck with simplify the fraction so that I can finally find the derivative. I'd really appreciate some advice on how to proceed with the problem.
|
\begin{align}
f'(x)
&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h(x^2+1)((x+h)^2+1)}\\
&= \lim_{h\to 0} \frac{h - h^2x - hx^2}{h(x^2+1)((x+h)^2+1)}
\end{align}
$$= \lim_{h\to 0} \frac{1 - hx - x^2}{(x^2+1)((x+h)^2+1)}$$
$$=\frac{1 - x^2}{(x^2+1)^2}$$
There are two common ways to compute a derivative. This is the other.
\begin{align}
\left. f'(x) \right|_{x=x_0}
&= \lim_{x\to x_0} \frac {f(x)-f(x)}
{x-x_0}\\
&= \lim_{x\to x_0} \frac {\frac{x}{x^2+1}-\frac{x_0}{x_0^2+1}}
{x-x_0}\\
&= \lim_{x\to x_0} \frac {x(x_0^2+1)-x_0(x^2+1)}
{(x^2+1)(x_0^2+1)(x-x_0)}\\
&= \lim_{x\to x_0} \frac {(x x_0^2 - x_0 x^2) + (x-x_0)}
{(x^2+1)(x_0^2+1)(x-x_0)}\\
&= \lim_{x\to x_0} \frac {-x x_0(x-x_0) + (x-x_0)}
{(x^2+1)(x_0^2+1)(x-x_0)}\\
&= \lim_{x\to x_0} \frac {-x x_0 + 1}
{(x^2+1)(x_0^2+1)}\\
&= \frac{1-x_0^2}
{1+(x_0^2)^2}\\
\end{align}
|
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|
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
|
I will use the cubic equation with roots $x$, $y$, $z$, inspired by the answer of @dxiv:.
$$X^3 - s X^2 + 12 X - (2+s) =0$$
The discriminant of the equation is $-(2s+13)(2s+15)(s-6)^2$. If $s\ge 0$ then the discriminant is positive only for $s=6$, which gives $a=b=c=2$. Therefore: the system has only one positive solution, but infinitely many negative ones, for each $s \in [-\frac{15}{2}, -\frac{13}{2}]$ we get a solution with $a+b+c = s$. For instance, for $s=-\frac{13}{2}$, we get $a=b=-3$, $c=-\frac{1}{2}$ a solution.
|
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|
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1-a)(1+a)
\end{array}
\right.$$
In order to have an infinite set of solutions, I want to have $0=0$ in the last row, meaning that $a=1\Rightarrow 0=1,$ which is not what I want. Have I made any arithmetical mistake? Did this computation times now.
|
The only way for this system not to have a unique solution is if the matrix
$$
\left[
\begin{array}{ccc}
1 & a &1 \\
a & 1 & 1 \\
1 & -1 &1
\end{array}
\right]
$$
is singular. Its determinant is $1 - a^2$. So it's singular if $a =1 $ or $a = -1$. If $a = 1$, the first two rows are inconsistent equations. If $a = -1$, the first and the third equations are identical, and the solutions are of the form $(x,y,z) = (y+1/2,y,1/2)$, where $y$ is a slack variable.
|
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|
Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$ where $a,b,c$ is the coefficients of $x^2,-x,1$. So what should be the next step I need to take?
|
You can find which coefficients are divisible by $3$ by noting that $x^2-x+1\equiv(x+1)^2\pmod{3}$ so we are seeking the coefficients of $(x+1)^{44}$ which are divisible by $3$. Now notice that:
$$(x+1)^{44}=(x+1)^{27}(x+1)^{9}(x+1)^{6}(x+1)^2\equiv (x^{27}+1)(x^9+1)(x^{6}-x^3+1)(x^2-x+1)$$
Inspection sees the zero terms are $x^{18},x^{19},x^{20},x^{24},x^{25},x^{26}$.
Of these, you need to figure out which are even. That part is a little harder.
|
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|
Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}
By using both my answer to Closed form solutions for a family of hypergeometric sums. and by generalizing the approach from my answer to About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$ I have derived the following results:
\begin{eqnarray}
{\mathfrak S}^{(m,0)}(z) &=& \frac{x \left(1-z x^{m-1}\right)}{1-m z x^{m-1}}\\
{\mathfrak S}^{(m,1)}(z) &=& \frac{m}{m-1} \cdot\frac{ x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-\frac{1}{m-1}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}\\
{\mathfrak S}^{(m,2)}(z) &=& \frac{m^2}{(m-1)^2}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}
-\frac{(m-2) }{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&
-\frac{1}{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}} \cdot \frac{((m-1) x+2) }{ x}\\
{\mathfrak S}^{(m,3)}(z) &=& \frac{m^3}{(m-1)^3}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-
\frac{(m-2)(2 m-3)}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&-\frac{m-3}{(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}}\cdot \frac{((m-1) x+2)}{x}+\\
&&-\frac{1}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{3}{m-1}}\cdot
\frac{2(m-1)^2 x^2+6(m-1)x+3(m+2)}{x^2}
\end{eqnarray}
Here $x:=x(z)$ is obtained in the following way. Out of the solutions of the equation:
\begin{equation}
1-x+z \cdot x^m=0
\end{equation}
we choose the one that is the closest to unity.
Now the question is how does the result look like for arbitrary $j \ge 2$.
|
Hint: The series ${\mathfrak S}^{(m,j)}(z)$ is strongly related with the generalized binomial series $B_m(z)$
\begin{align*}
B_m(z)=\sum_{i=0}^\infty\binom{mi+1}{i}\frac{1}{mi+1}z^i
\end{align*}
defined in (5.58) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
The series $B_m(z)$ satisfies the identities (see (5.60), (5.61)):
\begin{align*}
B_m(z)^j&=\sum_{i=0}^\infty \binom{mi+j}{i}\frac{j}{mi+j}z^i\\
\frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}&=\sum_{i=0}^\infty\binom{mi+j}{i}z^i
\end{align*}
so that the following holds
\begin{align*}
\color{blue}{{\mathfrak S}^{(m,j)}(z)=\frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}}
\end{align*}
|
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|
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
|
You could use formal manipulations with a truncated power series:
$$\frac{1-x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n \\
\frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = 1 + 2x + \cdots + n x^{n-1} \\
x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = x + 2x^2 + \cdots + n x^n \\
\frac{d}{dx} \left( x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) \right) = 1 + 4x + \cdots + n^2 x^{n-1}
$$
Now evaluate the derivatives on the left hand side, and then substitute $x := -1$.
|
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|
Integrating a Rational Function with a Square Root : $ \int \frac{dx}{\sqrt{x^2-3x-10}} $
Integrate $$ \int \frac{dx}{\sqrt{x^2-3x-10}}. $$
I started off with the substitution $\sqrt{x^2-3x-10} = (x-5)t$. To which I got
$$ x = \frac{2+5t^2}{t^2-1} \implies \sqrt{x^2-3x-10} = \left(\frac{2+5t^2}{t^2-1} - 5\right)t = \frac{7t}{t^2-1} $$
and
$$ dx = \frac{-14t}{(t^2-1)^2} dt. $$
Substituting back in
\begin{align*}
\int \frac{t^2-1}{7t} \frac{-14t}{(t^2-1)^2}dt &= -2 \int \frac{dt}{t^2-1} \\
&= - \ln\left| \frac{t-1}{t+1} \right| +c \\
&= - \ln\left| \frac{\frac{\sqrt{x^2-3x-10}}{x-5} - 1}{\frac{\sqrt{x^2-3x-10}}{x-5} + 1} \right| +c \\
&= - \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| + c.
\end{align*}
Is this a valid answer? Wolfram Alpha's answer is slightly different. I know this can come from a difference in method and/or the functions differ by a constant, but with an expression like this it's hard to check.
|
Use: $\int \frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C$
Hence $$\int \frac{dx}{\sqrt{x^2-3x-10}} = \int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{49}{4}}}=\int \frac{dx}{\sqrt{(x-\frac{3}{2})^2-\frac{49}{4}}} $$
$$=\{x-\frac{3}{2}=t; dx=dt\}= \int \frac{dx}{\sqrt{t^2-\frac{49}{4}}}=\log|t+\sqrt{t^2-\frac{49}{4}}|+C$$
$$=\log|x-\frac{3}{2}+\sqrt{x^2-3x-10}|+C=\log|2x-3+2\sqrt{x^2-3x-10}|+ const$$
|
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|
Compute $2\arctan{2}+\arcsin{\frac{4}{5}}.$ Proceeding as you guys have tought me on my previous posts by setting the whole thing equal to $X$ taking the tangent of the whole expression and then using the additon formula for tangent:
$$\tan{X}=\tan{\left(2\arctan{2}+\arcsin{\frac{4}{5}}\right)}=\frac{\tan{(2\arctan{2})}+\tan{\left(\arcsin{\frac{4}{5}}\right)}}{1-\tan{(2\arctan{2})}\cdot\tan{\left(\arcsin{\frac{4}{5}}\right)}}.$$
So lets split this into two sub-computations like this:
i): $$N=\tan{(2\arctan{2})}=\tan{(\arctan{2}+\arctan{2})}=\frac{2\tan{(\arctan{2})}}{1-\tan^2{(\arctan{2})}}=\frac{2\cdot 2}{1-2^2}=-\frac{4}{3}.$$
ii):
$$M=\tan{\left(\arcsin{\frac{4}{5}}\right)}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\cos{\left(\arcsin{\frac{4}{5}}\right)}}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\sqrt{1-\sin^2{\left(\arcsin{\frac{4}{5}}\right)}}}=\frac{\frac{4}{5}}{\sqrt{1-\left(\frac{4}{5}\right)^2}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.$$
So,
$$\tan{X}=\frac{N+M}{1-MN}=\frac{-\frac{4}{3}+\frac{4}{3}}{1+\frac{4}{3}\cdot\frac{4}{3}}=0\Longrightarrow X=\arctan{0}=0.$$
The book says $X=\pi.$
|
The last implication is false: $\tan{X} = 0 \implies X = k\pi$ for some $k \in \mathbb{Z}$. To find out which value, it's probably simplest to estimate the terms:
*
*$\arctan{2}>\arctan{1}=\pi/4$, so the sum is certainly bigger than $\pi/2$.
*On the other hand, $0<\arcsin{(4/5)} < \pi/2$ and $\arctan{2}<\pi/2$, so the result is bounded above by $3\pi/2$.
The only multiple of $\pi$ between these is $\pi$ itself, so the value of the sum must be $\pi$.
|
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|
Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
$$a^3 -(b^3)i + 3(a^2)bi -3(b^2)a$$
Equating real and imaginary parts, we get:
$$ a^3 -3(b^2)a = 1 \tag{1}$$
$$ b^3 -3(a^2)b = 1 \tag{2}$$
Factoring $(1)$, $(2)$ to get some roots in the process:
$$a(a^2 - 3(b^2)) = 1 \tag{1'}$$
$$b(b^2 - 3(a^2)) = 1 \tag{2'}$$
From $(1')$, the roots are possibly given by equations below:
$$a = 1 \tag{3}$$
$$ - \text{or} - $$ // 'or' is used in logical or sense, i.e. either or both
$$ a^2 - 3(b^2) = 1 \tag{4}$$
Substituting $a = 1$ in $(4)$, we get:
$b = 0$; which cannot be a possible solution, as the right side of
question has $b = -1$. Hence, $a \ne 1$ also.
Next, trying for the possible value from $(2')$.
$$ b = 1 \tag{5} $$
$$ -\text{or}- $$ // 'or' is used in logical or sense, i.e. either or both
$$ b^2 - 3(a^2) = 1 \tag{6} $$
Substituting $b = 1$ in $(6)$, we get:
$$ a = 0; $$
which cannot be a possible solution, as the right side of question
has $a = 1$. Hence, $b \ne 1$ also.
I am unable to solve it further, as no solution emerges from the
two equations - $(1')$, $(2')$.
|
Since $1-i= \sqrt 2 (\cos315^\circ + i\sin315^\circ),$ the cube roots of $1-i$ must be
$$
2^{1/6} (\cos(105^\circ+n120^\circ) + i\sin(105^\circ+n120^\circ))
$$
where the only values of $n$ we need to consider are $0,$ $1,$ and $2.$
If we can believe the tables on this page, we have
$$
\cos105^\circ = -\frac 1 4 (\sqrt 6 - \sqrt 2\,) \quad \text{and} \quad \sin(105^\circ) = \frac 1 4 (\sqrt6 + \sqrt 2\,).
$$
(The page gives $\cos15^\circ = \dfrac 1 4 ( \sqrt 6 + \sqrt 2\,)$ and $\sin15^\circ = \dfrac 1 4 (\sqrt 6 - \sqrt 2\,).$ If you consider $\dfrac 1 2 = \sin30^\circ = 2\sin15^\circ\cos15^\circ,$ then this appears to make sense. And then $\cos105^\circ = -\sin15^\circ$ and $\sin105^\circ = \cos15^\circ$ since $105 = 15 +90.$)
So we should have
$$
\sqrt 2\,\left( -\frac 1 4(\sqrt6-\sqrt2\,) + i \frac 1 4( \sqrt6+\sqrt 2\,) \right)^3 = 1-i.
$$
|
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|
Density of multiplication table Is there any easy way to show $$\lim_{N \to \infty} \frac{1}{N^2}\#\{ab : 1 \le a,b \le N\} = 0$$ A quick calculation I did shows that the number of positive integers $\le N^2$ with a prime divisor $p > N$ is at most the order of $(\log 2) \cdot N^2$, so just getting rid of the numbers with a high prime divisor is not sufficient.
|
For $N \ge 1$, let $$E_N := \{n \le N : \omega(n) \in [\frac{9}{10}\log\log N,\frac{11}{10}\log\log N]\},$$ where $\omega(n) := \sum_{p \mid n} 1$. An easy double-counting / [second moment method] argument shows that $|E_N| = N-o(N)$ as $N \to \infty$. So, $$\frac{1}{N^2}\#\{ab : 1\le a,b \le N\} = \frac{1}{N^2}\#\{ab : a,b \in E_N\}+o(1).$$ Now, $$\omega(ab) = \omega(a)+\omega(b)-\sum_{p \mid a, p \mid b} 1,$$ so $a,b \in E_N$ and $ab \in E_{N^2}$ imply $\sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N$. Since $|E_{N^2}| = N^2-o(N^2)$, we have $$\frac{1}{N^2}\#\{ab : 1 \le a,b \le N\} \le \frac{1}{N^2}\#\{ab : a,b \in E_N, \sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N\}+o(1).$$ Fix $a \in E_N$. Suppose $b$ shares at least $\frac{7}{10}\log\log N$ prime factors with $a$. The number of $b \le N$ with prespecified $p_1,\dots,p_{\frac{7}{10}\log\log N}$ dividing $b$ is at most $\frac{N}{\prod_{i=1}^{\frac{7}{10}\log\log N} p_i}$, which, if $p_1,\dots,p_{\frac{7}{10}\log\log N}$ denote the first $\frac{7}{10}\log\log N$ primes, is at least $\exp(\frac{6.99}{10}\log\log N\log\log\log N)$ by the prime number theorem. Therefore, since $a$ has at most $\frac{11}{10}\log\log N$, the number of $b \le N$ with $\sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N$ is at most $${\frac{11}{10}\log\log N \choose \frac{7}{10}\log\log N}\frac{N}{\exp(\frac{6.99}{10}\log\log N\log\log\log N)},$$ which is $o(N)$, since ${\frac{11}{10}\log\log N \choose \frac{4}{10}\log\log N} \le (\frac{11}{10}\log\log N)^{\frac{4}{10}\log\log N} \le \exp(\frac{4.01}{10}\log\log N\log\log\log N)$. We conclude $$\frac{1}{N^2}\#\{ab : a,b \in E_N, \sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N\} = o(1).$$
|
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|
For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ =
\ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.
|
$$E=\gcd(3^{16}2a+10,3^{17}a+66)=\gcd(b,c)=\gcd(b,c,3c-2b)=\gcd(b,c,102)$$
$$E=\gcd(b \mod 102,b\mod 102)=\gcd(36 a+10,3a+66)=\gcd(d,e)$$
$$E=\gcd(d-12e \mod 102,e)=\gcd(34,e)=\gcd(34,3(a+22))=\gcd(34,a+22)$$
because $34 \mod 3 \neq 0$
so $E=\gcd(a,2)\times \gcd(a+5,17)$
|
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|
How to find values such that the curves $y=\frac{a}{x-1}$ and $y=x^2-2x+1$ intersect at right angles? Problem:
Find all values of $a$ such that the curves $y = \frac{a}{x-1}$ and $y = x^2-2x+1$ intersect at right angles.
My attempt:
First, I set the two curves equal to each other:
$ \frac{a}{x-1} = x^2 - 2x + 1 $
$ \frac{a}{x-1} = (x-1)^2 $
$ \frac{a}{x-1}(x-1) = (x-1)^2(x-1) $
$ a = (x-1)^3 $
$ \sqrt[3] a = x-1 $
$ \sqrt[3] a + 1 = x $
I found the derivative of the first curve:
$ y = \frac{a}{x-1} $
$ y = a(x-1)^{-1} $
$ y' = -a(x-1)^{-2} $
$ y' = \frac{-a}{(x-1)^{2}} $
Then I found the derivative of the second curve:
$ y = x^2 - 2x + 1 $
$ y' = 2x - 2 $
Next, I multiplied them together and set them equal to -1:
$ \frac{-a}{(x-1)^{2}} ⋅ 2x - 2 = -1 $
$ \frac{-2ax + 2a}{(x-1)^2} = -1 $
$ \frac{-2a(x-1)}{(x-1)^2} = -1 $
$ \frac{-2a}{x-1} = -1 $
$ \frac{-2a}{x-1} ⋅ (x-1) = -1(x-1) $
$ -2a = -x+1 $
$ a = \frac{-x+1}{-2} $
Now I am unsure how to finish the problem. Do I substitute what I found for $x$ into $ a = \frac{-x+1}{-2} $?
But my problem shows that there are two possible answers for $a$ so I am confused.
Any help would be appreciated!
Thank you in advance!
|
hint: $\dfrac{a}{x-1} = x^2- 2x+1 \implies a = (x-1)^3 = \dfrac{x-1}{2}$. Can you solve for $x$ and then $a$?
|
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Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant.
Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant.
my attempt
I tried writing a family through line $x=y$ and $(1,1)$ and $(2,2)$ as
$$(x-1)(x-2)+(y-1)(y-2)+p(x-y)=0$$
where $p$ is a variable parameter.
I tried to make $g^2<c$ and $f^2<c$; so as to have no $x,y$ intercepts. But I am having trouble maximising the radius. I was looking for any suggestions or any better solutions.
|
The circles passing through $A$ and $B$ are a linear combination of the equation of the line $AB:x-y=0$ and the circle having diameter $AB$, $\mathscr{C}:x^2+y^2-3x-3y+4=0$
Therefore its general equation is $x^2+y^2-3x-3y+4+k(x-y)=0$
$x^2+y^2-3x-3y+4+k(x-y)=0$ intersect $x-$axis if $y=0$. Substitute in the circle and get
$x^2-(3-k)x+4=0$
This equation has non real roots if discriminant is less than zero
$\Delta=(3-k)^2-16<0 \to k^2-6k-7<0\to -1<k<7$
The circle intersects the $y-$axis if $x=0$ and with the same procedure as above we get no intersection when $k^2+6 k-7<0$ that is $-7<k<1$
So the condition for $k$ to have circles non intersecting axes is $-1<k<1$
Therefore the maximum radius is achieved when $k=\pm 1$
Collect $x$ and $y$ in the previous equation we get
$x^2+y^2+(k-3) x-(k+3) y+4=0$
radius is $r=\sqrt{\left(\dfrac{k-3}{2}\right)^2+\left(\dfrac{k+3}{2}\right)^2-4}$
for $k=\pm 1$ we have $r=1$ which is the maximum radius
Requested circles have equations
$x^2+y^2-4 x-2 y+4=0$
$x^2+y^2-2 x-4 y+4=0$
Hope this helps
|
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|
Find the solutions to $\left\lfloor\left(\frac{5}{3} \right)^n\right\rfloor = 3^m$
Find all positive integer solutions to $\left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor = 3^m$.
Let $a_n = \left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor$. Then $$a_n = 1,2,4,7,12,21,35,59,99,165,275,459,765,1276,2126,3544,5907,9846,16410,\ldots.$$ Since a power of $3$ is odd, we only need to look at the odd terms of $a_n$: let these be $b_n$. Then $b_n = 1,7,21,35,59,99,165,275,459,765,5907,\ldots$. There don't seem to be powers of $3$ with a positive exponent in $b_n$ in the first few terms.
Using the Binomial Theorem, we have $$\left(\dfrac{5}{3}
\right)^n = \left(1+\dfrac{2}{3}\right)^n = 1+\dfrac{2}{3} \binom{n}{1}+\left(\dfrac{2}{3}\right)^2 \binom{n}{2}+\cdots+\left(\dfrac{2}{3}\right)^n\binom{n}{n}.$$ How can we continue from here?
|
If we write
$$
\left(\frac{5}{3}\right)^n = \left[ \left(\frac{5}{3} \right)^n \right] + \left\{ \left(\frac{5}{3}\right)^n \right\},
$$
then supposing that
$$
\left[ \left(\frac{5}{3} \right)^n \right]=3^m
$$
implies the inequality
$$
0 < 5^n - 3^{n+m} < 3^n.
$$
Applying lower bounds for linear forms in (two complex) logarithms (say from Laurent-Mignotte-Nesterenko) to
$$
\Lambda = n \log 5 - (n+m) \log 3
$$
leads, after a wee bit of work, to the conclusion that $n \leq 1$.
I'm not sure if there's an elementary way to prove this. I can't see one off the top of my head.
|
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Showing a sequence converges from the definition of limit I am having some trouble proving that this series converges to zero
$b_n=\dfrac{n+5}{n^2-n-1}$, $n\geq 2$
directly from the definition of limit. My attempts have only left me with a lengthy, complicated form for $n$ in terms on $\epsilon$ so I must be missing some simple step.
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Notice that $n^2 - n - 1 = \dfrac{1}{4} \left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)$ so we can rewritten:
\begin{align}
b_{n} & = \dfrac{4n+20}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& = \dfrac{4n - 2 + 2\sqrt{5} + 22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& = \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)}
\end{align}
In addition, $22 - 2\sqrt{5} > 0$ and $2n - 1 + \sqrt{5} \geq 1, 2n - 1 - \sqrt{5} \geq 0, \forall n \geq 2$, we have further:
\begin{align}
\lvert b_{n} \rvert & \leq \left\lvert \dfrac{2}{2n - 1 - \sqrt{5}} \right\rvert + \left\lvert \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \right\rvert \\
& = \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& \leq \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{2n - 1 - \sqrt{5}} = \dfrac{24 - 2\sqrt{5}}{2n - 1 - \sqrt{5}}
\end{align}
The last term less than $\varepsilon$ when:
$$ n > \dfrac{1 + \sqrt{5}}{2} + \dfrac{24 - 2 \sqrt{5}}{\varepsilon} . $$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2443784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
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Find limit of trigonometric function $$\lim_{x\rightarrow0} \frac{\tan^3(3x)-\sin^3(3x)}{x^5}$$
I think it should be decomposed with $$\lim_{x\to0} \frac{\sin x}{x}=1$$ but I'm always getting indefinity $\frac{0}{0}$.
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Take $3x=a \to x=\frac{a}{3}$
$$\quad{\lim_{x\rightarrow0} \frac{tg^3(3x)-\sin^3(3x)}{x^5}=\\
\lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(\frac a3)^5}=\\3^5\lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(a)^5}=\\
3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^5}=\\
3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\frac{(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^2}=\\
3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\lim_{a\rightarrow0}\frac{(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^2}=\\
3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}.\lim_{a\rightarrow0}\underbrace{\frac{\tan^2(a)+\sin^2(a)+\tan (a).\sin (a)}{(a)^2}}_{3}=\\
3^6\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))}{(a)^3}=\\
3^6\lim_{a\rightarrow0} \frac{(\frac{\sin
(a)}{\cos (a)}-\sin(a))}{(a)^3}=\\
3^6\lim_{a\rightarrow0} \frac{(\frac{\sin
(a)-\sin(a)\cos (a)}{\cos (a)})}{(a)^3}=\\
3^6\lim_{a\rightarrow0} \frac{(\frac{\sin
(a)-\sin(a)\cos (a)}{1})}{\cos (a).(a)^3}=\\
3^6\lim_{a\rightarrow0} \frac{\sin
(a)-\sin(a)\cos (a)}{(a)^3}=\\
3^6\lim_{a\rightarrow0} \frac{\sin
(a)(1-\cos (a))}{\cos (a).(a)^3}=\\
3^6\lim_{a\rightarrow0} \underbrace{\frac{\sin
(a)}{\cos (a).(a)}}_{1}.\lim_{a\rightarrow0}\frac{(1-\cos (a))}{(a)^2}=\\
3^6\lim_{a\rightarrow0}\frac{(1-\cos (a))}{(a)^2}=\\
3^6\underbrace{\lim_{a\rightarrow0}\frac{2\sin^2(\frac a2)}{(a)^2}}_{\frac 12}=\\\frac {3^6}{2}}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2444868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
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Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $
Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $
Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)
=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$
Separating the real & imaginary parts:
$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)
Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);
Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'),
dividing both sides by $b^4$, and having c = a/b, $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),
having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$.
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$
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Trigonometrical way: One has
$$\sin 36^\circ = \cos 54^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ\cos 18^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ = 4\cos 18^\circ -3 = 1-4\sin 18^\circ$$
So, $\sin 18^\circ$ is positive root of equation $4x^2+2x-1=0$, or $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2445926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
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Fractions in Questions and Answers
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