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Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y}
$$
$$
\sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y}
\\ \therefore (x-y) \geqslant 2
$$
So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?
Thanks
|
Hint.
Making
$$
\cases{
u = x+y\\
v = x-y
}
$$
we have
$$
\sqrt{u^2-v^2}=2\frac uv
$$
so
$$
u^2 = \frac{v^4}{v^2-4}
$$
etc.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displaystyle x$, from the given equation but it will be tedious I think !.
$$\displaystyle x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$$
$$\displaystyle \Longrightarrow \ x^{2} +1/2 \ =\ \frac{7}{2\sqrt{5}}$$
$$\displaystyle \Longrightarrow \ x^2 \ =\ \ \frac{7-\sqrt{5}}{2\sqrt{5}}$$
Which is getting too much complicated to solve the expression by putting the value of $\displaystyle x$.
What could be the other way to solve the given expression?
|
Let me try it
we have putting $t=\tan\Big(\frac{x_1}{2}\Big)$
$$\cos(x_1)=\frac{1-t^2}{1+t^2}$$
It's the Weierstrass substitution
Putting $y=t^2$ we get :
$$\cos(x_1)=\frac{1-y}{1+y}$$
Now we put $x=y$ to get :
$$\Big(-\cos(x_1)\Big)^{\frac{1}{3}}+\Big(-\frac{1}{\cos(x_1)}\Big)^{\frac{1}{3}}=?$$
After that I have tried $\cos(3x)=4\cos^3(x)-3\cos(x)$.
Hope it helps (I think it's hard).
|
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|
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,
$$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$
That is,
$$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$
Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$
So
$$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$
$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$
I tried till this , then I have no idea . Can anyone help?
|
Multiply both sides of $x+y=5xy$, $y+z=6yz$ and $z+x=7zx$ by $z,x$ and $y$ respectively and add them together to obtain:
$xz+zy+xy+xz+zy+xy=2(xz+zy+xy)=18xyz$ so $xz+zy+xy=xy+z(x+y)=xy+5xyz=9xyz$.
So $xy=4xyz$ and hence $z=\frac{1}{4}$, etc.
|
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|
Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=\frac{1}{9}(a+b+c)^2(a-2b+c)^4\\ &+\frac{2}{3}(a+b+c)^2(a-2b+c)^2(b-c)(a-b)\\ &+\frac{1}{16}(a-b)^2(b-c)^2(a+4b+7c)(7a+4b+c)\\&\geqslant 0\end{align*}
However, this solution is too hard for me to find without computer. Could you help me with figuring out a better soltuion?
Thank you very much
|
We write inequality we have
$$4(a^3+b^3+c^3-3abc) \geqslant 9 |(a-b)(b-c)(c-a)|,$$
or
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] \geqslant 9 |(a-b)(b-c)(c-a)|.$$
It's easy to check $a + b \geqslant |a-b|,$ now using the AM-GM inequality, we have
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] $$
$$\geqslant \left(|a-b|+|b-c|+|c-a|\right) \left[(a-b)^2+ (b-c)^2+ (c-a)^2\right]$$
$$ \geqslant 3 \sqrt[3]{\left | (a-b)(b-c)(c-a)\right |} \cdot 3 \sqrt[3]{(a-b)^2(b-c)^2(c-a)^2}.$$
$$ =9 \left | (a-b)(b-c)(c-a)\right | .$$
Done.
SOS proof. We have
$$4(a^3+b^3+c^3-3abc) - 9(a-b)(b-c)(c-a) = \sum b \left[(2a-b-c)^2+3(a-b)^2\right] \geqslant 0.$$
Note. The best constant is
$$a^3+b^3+c^3-3abc \geqslant \sqrt{9+6\sqrt{3}} \cdot \left | (a-b)(b-c)(c-a)\right |.$$
|
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|
Prove that $\tan(x)\tan(x+\frac{\pi}{3})+\tan(x)\tan(\frac{\pi}{3}-x)+\tan(x+\frac{\pi}{3})\tan(x-\frac{\pi}{3}) = -3$ Let's assume that $\tan(x) = y$.
So, $\tan\Big(x+\dfrac{\pi}{3}\Big) = \dfrac{\tan(x) + \tan\Big(\dfrac{\pi}{3}\Big)}{1-\tan(x)\tan\Big(\dfrac{\pi}{3}\Big)} = \dfrac{y+\sqrt{3}}{1-\sqrt{3}y}$
Similarly, $\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{y-\sqrt{3}}{1+\sqrt{3}y}$
Also, $\tan\Big(\dfrac{\pi}{3}-x\Big) = -\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{\sqrt{3}-y}{1+\sqrt{3}y}$
Now, $\tan(x)\tan\Big(x+\dfrac{\pi}{3}\Big)+\tan(x)\tan\Big(\dfrac{\pi}{3}-x\Big)+\tan\Big(x+\dfrac{\pi}{3}\Big)\tan\Big(x-\dfrac{\pi}{3}\Big)$
$$ = y\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)+y\Big(\dfrac{\sqrt{3}-y}{1+\sqrt{3}y}\Big)+\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)\Big(\dfrac{y-\sqrt{3}}{1+\sqrt{3}y}\Big)$$
$$ = y\Big(\dfrac{(y+\sqrt{3})(1+\sqrt{3}y)+(1-\sqrt{3}y)(\sqrt{3}-y)}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$
$$ = y\Big(\dfrac{y+\sqrt{3}y^2+\sqrt{3}+3y+\sqrt{3}-y-3y+\sqrt{3}y^2}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$
$$ = \dfrac{2\sqrt{3}y+2\sqrt{3}y^3+y^2-3}{1-3y^2}$$
This is how much I've been able to simplify the expression but I'm unable to continue. I am familiar with the values of trigonometric functions at multiples and sub multiples of angles and I think the solution would involve their use (as the question has been taken from that very chapter).
Thanks!
|
As written, this statement is false, as per @quasi 's comment. I suspect the identity should have used $\tan(x-\pi/3)$ rather than $\tan(\pi/3-x)$. In this case, by the angle-sum formula for tangent,
$$\tan(x)\tan\left(x+\frac\pi3\right)=\tan(x)\left(\frac{\tan x+\sqrt 3}{1-\sqrt 3\tan x}\right)=\tan(x)\left(\frac{4\tan x+\sqrt 3\tan^2 x+\sqrt 3}{1-3\tan^2 x}\right)$$
and
$$\tan(x)\tan\left(x-\frac\pi3\right)=\tan(x)\left(\frac{\tan x-\sqrt 3}{1+\sqrt 3\tan x}\right)=\tan(x)\left(\frac{4\tan x-\sqrt 3\tan^2 x-\sqrt 3}{1-3\tan^2 x}\right)$$
and
$$\tan\left(x+\frac\pi3\right)\tan\left(x-\frac\pi3\right)=\left(\frac{\tan x+\sqrt 3}{1-\sqrt 3\tan x}\right)\left(\frac{\tan x-\sqrt 3}{1+\sqrt 3\tan x}\right)=\frac{\tan^2x-3}{1-3\tan^2x}.$$
The sum is then
$$\frac{\tan(x)(8\tan x)+\tan^2 x-3}{1-3\tan^2 x}=-3.$$
|
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|
Solving the recurrence relation $T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$ I have been trying to solve the following recurrence relation
$T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$
$T(0) = 0$
I've tried to use substitution which wasn't very useful as I couldn't figure out a way to simplify the resultant equation.
My next approach was to just plug in numbers as follows
$T(1) = 5 * 1$
$T(2) = \frac{1}{2}(5 * 1) + 5 * 2 = \frac{5 * 1}{2} + 5 * 2$
$T(3) = \frac{1}{3}(\frac{5 * 1}{2} + 5 * 2) + 5 * 3 = \frac{5 * 1}{6} + \frac{5 * 2}{3} + 5 * 3$
$T(4) = \frac{1}{4}(\frac{5 * 1}{6} + \frac{5 * 2}{3} + 5 * 3) + 5 * 4 = \frac{5 * 1}{24} + \frac{5 * 2}{12} + \frac{5 * 3}{4} + 5 * 4$
I can sort of see a pattern emerge here like $\sum_{i=1}^{n} \frac{5*i}{previousDenominator/ i}$ however I couldn't really see a way to get that bottom part into an actual equation.
|
Multiply the recurrence relation by $n$ to get $$nT_n = T_0+T_1+\cdots+T_{n-1}+5n^2.$$ Replacing $n$ with $n+1$ gives $$(n+1)T_{n+1} = T_0+T_1+\cdots+T_{n-1}+T_n+5(n+1)^2.$$ Subtracting the first equation from the second gives $$(n+1)T_{n+1}-nT_n = T_n+5(n+1)^2-5n^2,$$ which can be simplified to $$T_{n+1}-T_n = 10-\dfrac{5}{n+1}.$$ Now you have a closed form for the difference between consecutive values of $T_n$. So just use the equation $$T_n = T_0 + \sum_{k = 0}^{n-1}(T_{k+1}-T_k)$$ and simplify the result to get a formula for $T_n$.
|
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|
Calculating the volume under the surface $z = x^2 + y^2$ and above $D$ Given:
$$D = \left\{(x,y) | 1 \leq x^2 +y^2 \leq 100, ~~ \frac{x \sqrt{3}}{3} \leq y \leq x \sqrt{3} \right\}$$
Calculate the Volume under the surface $$z = x^2 + y^2$$
And above $D$.
My try:
$$I = \iint_D x^2 + y^2 \,dx\,dy$$
We can write $D$ as:
$$D = \left\{(x,y) | 1 \leq x^2 +y^2 \leq 100, ~~ \frac{\sqrt{3}}{3} \leq \frac{y}{x} \leq \sqrt{3} \right\}$$
And change to $u,v$ as the following:
$$\left\{\begin{matrix}
u = x^2 + y^2 \\ v = \frac{y}{x}
\end{matrix}\right.$$
So the set $D$ would be:
$$D = \left\{(u,v) | 1 \leq u \leq 100, ~~ \frac{1}{\sqrt{3}} \leq v \leq \sqrt{3} \right\}$$
Now, because we changed the variables, we need to calculate the Jacobian:
$$ J = \frac{ D(x,y)}{D(u,v)} = \frac{1}{\frac{D(u,v)}{D(x,y)}} = \frac{1}{\begin{vmatrix}
u_x & u_y\\
v_x & v_y
\end{vmatrix}} = \frac{1}{\begin{vmatrix}
2x & 2y\\
-\frac{y}{x^2} & \frac1x
\end{vmatrix}} = \frac{1}{2+ 2 (\frac{y}{x})^2}$$
$$J = \frac{1}{2(1+v^2)}$$
And we can calculate the integral as so:
$$\iint_{D_{uv}} u \cdot \frac{1}{2} \cdot \frac{1}{1+v^2} \,du\,dv = \frac{1}{2} \int_1^{100} ( \int_{ \frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{u}{1+v^2} du)dv = \frac{1}{2} \int_1^{100} \frac{1}{1+v^2} ( \frac{ (\sqrt{3})^2 - (\frac{1}{ \sqrt{3}})^2}{2})dv = \frac23 \int_1^{100} \frac{1}{1+v^2} dv = \frac23 ( \arctan{v} |_1^{100}) \approx 0.5169 \dots$$
Some people I talked with got an answer above $10,000$ and my answer is not close at all! So I ask for your help, if you can review my work, I would be so thankful!
Thanks for helping!
|
In polar coordinates, the region $D$ is equivalent to $$D = \{(r, \theta) : (1 \le r \le 10) \cap (\tfrac{\pi}{6} \le \theta \le \tfrac{\pi}{3})\}.$$ The proof is straightforward after noting that $r^2 = x^2 + y^2$ and $\tan \theta = \frac{y}{x}$. Consequently, the desired volume is easily expressed as an integral in cylindrical coordinates: $$V = \int_{r=1}^{10} \int_{\theta = \pi/6}^{\pi/3} r^3 \, d\theta \, dr.$$
|
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|
If $(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n = a_0 + {a_1}x + {a_2} x^2 + .... +{a_{4n}}x^{4n}$ , then prove that $a_r=a_{4n-r}$ I tried solving this question by attempting to prove $(a_r)=(a_{4n-r})$
Now, $a_r$ is the coefficient of $x^r$ which we can obtain by adding up the coefficients of $x^r$ from the 3 separate expressions $(1+x)^{4n}$ , $(1+x+x^2)^{2n}$ and $(1+x+x^2+x^3+x^4)^n$
In the first expression $(1+x)^{4n}$, it is fairly easy to see that the coefficient of $x^r$ can be obtained from binomial expansion and the coefficient $a_r=a_{4n-r}$ by some basic properties.
I tried finding the coefficient of $x^r$ in $(1+x+x^2)^{2n}$
I first tried doing so by simplifying the expression as
$$(1+x+x^2)^{2n} = [1+(x(1+x)]^{2n}$$
Then taking $x(1+x)$ as some $y$, I applied binomial expansion and obtained the following expansion,
$$[1+(x(1+x)]^{2n} = \binom {2n}{r}\binom{r}{0}+\binom{2n}{r-1}\binom{r-1}{1}+\binom{2n}{r-2}\binom{r-2}{2}+...+\binom{2n}{r/2}\binom{r/2}{r/2}$$
(when $r$ is even, otherwise instead of going upto $r/2$ we will go upto $(r-1)/2$)
However, after calculating this, I was unable to find any relation between $a_r$ and $a_{4n-r}$ for this expression.
I was also not able to find any such simplification for the third expression $(1+x+x^2+x^3+x^4)^n$
I tried using the identity
$$1+x+x^2+...+x^n = {(1-x^{n+1})}/{(1-x)}$$
But was unable to obtain any further simplification even on using negative binomial expansions.
Any help on how to approach this question is appreciated
Thanks in advance!
Regards
|
Let $$P(x)=(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n$$ Then it's reciprocal polynomial is $$\hat{P}(x)=x^{4n}P(1/x)=x^{4n}\left(\left(1+\frac{1}{x}\right)^{4n} +\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^{2n} +\left(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}\right)^n\right)=P(x)$$ Hence $P(x)$ is a palindromic polynomial and hence $a_r=a_{4n-r}$.
|
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|
Calculate $\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$ with Taylor expansion $L=\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$
If I do $u=\frac{1}{x} \Rightarrow x=\frac{1}{u}$
Then $L=\lim _{u\to 0 }\left(\frac{1}{u^3}\left(1+u\left(1+u\right)^\frac{1}{u}\right)^{u}-\frac{1}{u^3}-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(\left(1+u\left(1+u\right)^\frac{1}{u}\right)^{u}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{\frac{1}{u}ln(1+u)})}-1\right)-\frac{e}{u}\right)$
I know that, when $t\to0, \ln(1+t)=1-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+o(t^4)$
And also, $e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+o(t^3)$
So: $L=\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{\frac{1}{u}(u-\frac{u^2}{2}-\frac{u^3}{3}-\frac{u^4}{4})})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{(1-\frac{u}{2}-\frac{u^2}{3}-\frac{u^3}{4})})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+u*e*e^{\frac{-u}{2}}*e^{\frac{u^2}{3}}*e^{\frac{-u^3}{4}})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+u*e*(1-\frac{u}{2}+\frac{u^2}{8}-\frac{u^3}{48})*(1+\frac{u^2}{3})*(1-\frac{u^3}{4})))}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+e*(u-\frac{u^2}{2}+\frac{u^3}{8})*(1+\frac{u^2}{3})*(1-\frac{u^3}{4})))}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+e*(u-\frac{u^2}{2}+\frac{11u^3}{24})}-1\right)-\frac{e}{u}\right)
$
Note: I am not writing the term $o(u^3)$ I will do it at the end
How would I do it easily using the Taylor expansion?
|
Composing Taylor series is a piece of cake if you are patient, going from inside to outside.
$$A=x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex$$ As you properly did, let $x=\frac 1u$ to get
$$A=\frac{\left(1+u (u+1)^{\frac{1}{u}}\right)^u-e u^2-1}{u^3}$$ Now, let us work the pieces
$$a=(u+1)^{\frac{1}{u}}\implies \log(a)={\frac{1}{u}}\log(1+u)=1-\frac{u}{2}+\frac{u^2}{3}-\frac{u^3}{4}+O\left(u^4\right)$$
$$a=e^{\log(a)}=e-\frac{e u}{2}+\frac{11 e u^2}{24}-\frac{7 e u^3}{16}+O\left(u^4\right)$$
$$b=1+u a=1+e u-\frac{e u^2}{2}+\frac{11 e u^3}{24}-\frac{7 e u^4}{16}+O\left(u^5\right)$$
$$c=b^u \implies \log(c)=u \log(b)=e u^2-\frac{e (1+e)}{2} u^3+\frac{e(11 +12 e+8 e^2}{24} u^4+O\left(u^5\right)$$
$$c=e^{\log(c)}=1+e u^2-\frac{e (1+e)}{2} u^3+\frac{e(11+24 e+8 e^2)}{24}
u^4+O\left(u^5\right)$$
Finally
$$A=-\frac{e (1+e)}{2} +\frac{e(11+24 e+8 e^2)}{24} u+O\left(u^2\right)$$ which gives the limit and also how it is approached.
Try its for $u=0.01$ (quite far away from $0$). The exact result is $-4.90454$ while the above truncated expansion gives $-4.90037$.
This means that, if you have to solve for $u$ the equation $A=-5$, the above would give as an estimate
$$u_0=\frac{12 \left(-10+e+e^2\right)}{e \left(11+24 e+8 e^2\right)}\approx 0.003501$$ while the exact solution is $0.003536$
|
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Find all pairs of integers $(x, y)$ such that $x^3+y^3=(x+y)^2.$
Find all pairs of integers $(x, y)$ such that $$x^3+y^3=(x+y)^2.$$
Since $x^3+y^3 = (x+y)(x^2-xy+y^2)$ we get that $$x^2-xy+y^2=x+y$$
this can be expressed as $$x^2-(y-1)x+y^2-y=0.$$
Since we want integers we should probably look at when the discriminant is positive?
$$\Delta = (y-1)^2-4(y^2-y)=-3y^2+6y+1$$
so for $\Delta \geqslant 0$
$$-\frac{2\sqrt3}{3}+1 \leqslant y \leqslant \frac{2\sqrt3}{3}+1$$
only possible solutions are $y=0,1,2.$ However I don't see how this is helpful at all here. What should I do?
|
The solutions when $x=0,$ $y=0,$ $x=y$ and $x+y=0$ have already
been presented. I give a new argument for the remaining case.
Let $y=-x+t,$ where both $x$ and $t$ are nonzero. With this
substitution, the original equation becomes
$$3x^2-3tx+t^2-t=0\tag1$$
As a polynomial in $x,$ the discriminant for (1) is
$$ D=12t-3t^2=3t(4-t) \tag2$$
If $D=0,$ then $t=4,$ giving the solution $(x,y)=(2,2).$
The integer $D$ is positive iff $1\le t\le3.$
Since $D$ must be a square, it follows from (2) that
$3$ divides $t$ or $4-t.$ Hence, $t$ is $1$ or $3.$
For $t=1,$ (1) gives us $(x,y)=(0,1)\ \text{and}\ (1,0),$ which are not new.
For $t=3,$ (1) implies that $x$ is $1$ or $2,$ giving the solutions
$(1,2)\ \text{and}\ (2,1).$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...
|
$$
\begin{aligned}
\int \frac{1}{\sin ^4 x+\cos ^4 x} d x = & \int \frac{1}{1-\frac{\sin ^2 2 x}{2}} d x \\
= & 2 \int \frac{1}{2-\sin ^2 2 x} d x \\
= & \int \frac{\sec ^2 2 x}{2 \sec ^2 2 x-\tan ^2 2 x} d(2 x) \\
= & \int \frac{d\left(\tan ^2 x\right)}{2\left(1+\tan ^2 2 x\right)-\tan ^2 2 x}\\=& \int \frac{d(\tan 2 x)}{2+\tan ^2 2 x}
\\=&\int \frac{d(\tan 2 x)}{(\sqrt{2})^2+(\tan 2 x)^2}\\=&\frac{1}{\sqrt{2}}\arctan \left(\frac{\tan 2 x}{\sqrt{2}}\right)+C
\end{aligned}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}}$$ are always sides of a triangle.
For all $\triangle MNP$ where $m = MP, n = PM, p = MN$, we have that $$n + p > m, p + m > n, m + n > p$$
We need to obtain that $$\frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}} > \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}$$
First attempt, we have that $$\frac{(a - b)^2}{|a - b|\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{(b - c)^2}{|b - c|\sqrt{b^2 + 1}\sqrt{c^2 + 1}}$$
$$ \ge \frac{(c - a)^2}{\sqrt{b^2 + 1} \cdot \left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)}$$
and $$\left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)^2 \le \left[(b - c)^2 + (a - b)^2\right] \cdot (c^2 + a^2 + 2)$$
It is needed to prove that $$\sqrt{\left[(b - c)^2 + (a - b)^2\right] \cdot (b^2 + 1)(c^2 + a^2 + 2)} < |c - a|\sqrt{c^2 + 1}\sqrt{a^2 + 1}$$
Second attempt, it is to prove that $$|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1} > |c - a|\sqrt{b^2 + 1}$$
According to the Cauchy - Schwarz inequality, we have that $$\left(|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1}\right)^2 \ge 2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)}$$
What needs to be established is $$2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)} > (c - a)^2(b^2 + 1)$$
Third attempt, let $a = \tan\alpha, b = \tan\beta, c = \tan\gamma$ $\left(\alpha, \beta, \gamma \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\right)$, it could be easily deducted that $$\frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}} = \frac{|\tan\gamma - \tan\alpha|}{\sqrt{\tan\gamma^2 + 1}\sqrt{\tan\alpha^2 + 1}} = \frac{\left|\dfrac{\sin(\gamma - \alpha)}{\cos\gamma\cos\alpha}\right|}{\dfrac{1}{\cos\gamma\cos\alpha}} = \pm\sin(\gamma - \alpha)$$
For all of the above attempts, there need to be considered multiple cases of $a, b, c$, whether they're positive and negative, and their arrangements from littlest to greatest.
|
For complex numbers $z, w \in \Bbb C$ is
$$
d(z, w) = \frac{|z - w|}{\sqrt{|z|^2 + 1}\sqrt{|w|^2 + 1}}
$$
(apart from a constant factor) the “spherical distance” of $z$ and $w$, that is the euclidean distance of the stereographic projections of $z, w$ onto a sphere. See for example A metric in $\mathbb{C}^{\infty}$ or What is this metric called?.
$d$ is a metric on $\Bbb C$. It follows that
$$
d(a, c) < d(a, b) + d(b, c)
$$
for (all permutations of) pairwise distinct $a, b, c \in \Bbb C$, with strict inequality because three distinct points on a sphere cannot be collinear. This implies that $d(a, b)$, $d(b, c)$, $d(c, a)$ are the side length of a non-degenerate plane triangle.
In particular this holds for pairwise distinct $a,b, c \in \Bbb R$.
Your last approach works also: According to Metric $d(x,y)=\frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$ on $\mathbb{R}$ we have
$$
d(a, b) = \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} = |\sin(\arctan(a) - \arctan(b))|
$$
for $a, b \in \Bbb R$. It follows that
$$
d(a, c) = |\sin(\arctan(a) - \arctan(b) + \arctan(b) - \arctan(c))| \\
\le |\sin(\arctan(a) - \arctan(b))| + |\sin(\arctan(b) - \arctan(c))| \\= d(a, b) + d(b, c)
$$
since $|\sin(x+y)| \le |\sin(x)| + |\sin(y)|$. Equality holds only if $x=0$ or $y= 0$, that is if $a=b$ or $b=c$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does the inequality $ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2}$ have a name? This is somewhat embarrassing but if $a,b$ are nonnegative real numbers the following seems to hold
$$
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}.
$$
Does this inequality have a name?
|
Squaring both sides we obtain
$$
\sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2} \iff a+b\ge \frac{a+b+2\sqrt{ab}}2\iff \frac {a+b}2\ge \sqrt{ab}
$$
which is true by AM-GM and squaring again we obtain
$$\iff \frac{a^2+2ab+b^2}{4}\ge ab \iff a^2-2ab+b^2=(a-b)^2 \ge 0$$
which is a more foundamental result.
|
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|
Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n}\frac{1}{r!}$$ for all $n \geq 2$.
$\implies \displaystyle \lim_{n \to \infty}a_{n} \ \geq \ e $
Now, in my book it says $\displaystyle \lim_{n \to \infty}a_{n} \ = \frac{3}{2}e $
How can I attack this problem? Anyone please?
|
First, let's note: $$(\forall n\in\mathbb{N}) \ \ \ 1+2+3+...+n= \frac{(n+1)n}{2} $$ so $$\eqalign{\sum_{1 \le n \le \infty } \frac{1+2+...+n}{n!}&=\frac{1}{2} \sum_{1 \le n \le \infty } \frac{n^2+n}{n!}=\cr &= \frac{1}{2} \sum_{1 \le n \le \infty } \frac{n^2}{n!}+\frac{1}{2} \sum_{1 \le n \le \infty } \frac{n}{n!}=\cr
&=\frac{1}{2} \sum_{1 \le n \le \infty }\frac{n}{(n-1)!}+\frac{1}{2} \sum_{1 \le n \le \infty }\frac{1}{(n-1)!}=\cr &=\frac{1}{2} \sum_{0 \le k \le \infty }\frac{k+1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}=\cr &=\frac{1}{2} \sum_{0 \le k \le \infty }\frac{k}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}=\cr &= \frac{1}{2} \sum_{1 \le k \le \infty }\frac{k}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}=\cr &= \frac{1}{2} \sum_{1 \le k \le \infty }\frac{1}{(k-1)!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}=\cr &=\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le \infty }\frac{1}{k!}=\cr &=3\cdot\frac{e}{2} }$$
or
$$e^x= \sum_{n=0}^{ \infty } \frac{x^n}{n!}$$ $$\frac{\text{d}}{\text{d}x}e^x= \sum_{n=0}^{ \infty }n\frac{x^{n-1}}{n!}$$ $$x^2\frac{\text{d}}{\text{d}x}e^x= \sum_{n=0}^{ \infty }n\frac{x^{n+1}}{n!}$$
$$\frac{\text{d}}{\text{d}x}\left(x^2\frac{\text{d}}{\text{d}x}e^x\right)= \sum_{n=0}^{ \infty }n(n+1)\frac{x^{n}}{n!}$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(x^2\frac{\text{d}}{\text{d}x}e^x\right)}{2}= \sum_{n=0}^{ \infty }\frac{\frac{n(n+1)}{2}}{n!}x^{n}$$
count the left side at $x=1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a determinant of a large matrix Evaluate $$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a \\
1 & 1& \dots &\ldots&a&1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&1\\
a & 1& \dots &\ldots&1&1 \\
\end{vmatrix}$$
I tried cofactor expansion and it leads to
$$\det \begin{vmatrix}
1& \dots &\ldots&a&1 \\
\vdots& \ddots &&\vdots&\vdots \\
\vdots&&\ddots&\vdots&\vdots \\
a& \dots &\ldots&1&1\\
1& \dots &\ldots&1&1 \\
\end{vmatrix} -
\det \begin{vmatrix}
1 & \dots &\ldots&a&1 \\
\vdots & \ddots &&\vdots&\vdots \\
\vdots & &\ddots&\vdots&\vdots \\
1 & \dots &\ldots&1&1\\
a & \dots &\ldots&1&1 \\
\end{vmatrix} + \ldots +(-1)^{n+1}
\det \begin{vmatrix}
1 & 1& \dots &\ldots&a \\
\vdots & \vdots& \ddots &&\vdots \\
\vdots & \vdots&&\ddots&\vdots \\
1 & a& \dots &\ldots&1\\
a & 1& \dots &\ldots&1 \\
\end{vmatrix}$$
I think, except for last term, all the terms will be cancel each other somehow although it may depend on parity of $n$. If it does depend on parity, the possibility of cancelation is going to be opposite for the cofactor expansion of the last term, and so it does for the last term of the next expansion...
I could not find out how to deal with this branch.
|
Add all the columns to the last one : you get
$$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a \\
1 & 1& \dots &\ldots&a&1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&1\\
a & 1& \dots &\ldots&1&1 \\
\end{vmatrix} = \det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a+n-1 \\
1 & 1& \dots &\ldots&a&a+n-1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&a+n-1\\
a & 1& \dots &\ldots&1&a+n-1 \\
\end{vmatrix}$$
Now substract the first line from all the other lines :
$$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a+n-1 \\
1 & 1& \dots &\ldots&a&a+n-1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&a+n-1\\
a & 1& \dots &\ldots&1&a+n-1 \\
\end{vmatrix} = \det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a+n-1 \\
0 & 0& \dots &\ldots&a-1&0 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
0 & a-1& \dots &\ldots&0&0\\
a-1 & 0& \dots &\ldots&0&0 \\
\end{vmatrix}$$
Finally develop w.r.t. the last column. You get that
$$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a \\
1 & 1& \dots &\ldots&a&1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&1\\
a & 1& \dots &\ldots&1&1 \\
\end{vmatrix} =(-1)^{\frac{n(n-1)}{2}}(a+n-1)(a-1)^{n-1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
|
Easiest proof is actually the proof of Schur. Alternatively, you can even prove it using simple calculus.
Assume $a\geq b\geq c.$ Consider:
$$f(x) = x^3-x^2(b+c) - x(b^2+c^2) +3xbc+b^3+c^3-bc(b+c),\,\, x\geq b$$
Then, $f'(x) = 3x^2-2x(b+c) - b^2-c^2+3bc$ and $f''(x) = 6x - 2b-2c\geq 0.$
So $f'$ is increasing on $[b,\infty)$ and as such:
$$f'(x)\geq f'(b) = 3b^2 - 2b^2-2bc-b^2-c^2+3bc = c(b-c)\geq 0$$
which means $f$ is increasing on its domain. Finally,
$$f(a)\geq f(b) = b^3-b^3-b^2c-b^3-bc^2+3b^2c+b^3+c^3-b^2c-bc^2 = $$
$$ = c^3+b^2c-2bc^2 = c(c-b)^2\geq 0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\int_S z\,dS$ where $S=\left\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\right\}$ $$\int_Sz\,dS$$
$$S=\big\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\big\}$$
I've already calculate this surface integral " by hand " :
$$z=\sqrt{a^2-x^2-y^2}\text{, thus,}$$
$$\int_SzdS=\int\int_{x^2+y^2\le a^2}\sqrt{a^2-x^2-y^2}\cdot\bigg(\sqrt{1+z_x^2+z_y^2}\bigg)dxdy$$
$$=\int_SzdS=\int\int_{x^2+y^2\le a^2}\sqrt{a^2-x^2-y^2}\cdot\bigg(\sqrt{\frac{a^2}{a^2-x^2-y^2}}\bigg)dxdy=\int_SzdS=\int\int_{x^2+y^2\le a^2}adxdy=a^3\pi.$$
I want to prove it by using the Gauss divergence theorem, but im making a mistake somewhere:
$$\text{Let } F=(F_1,F_2,F_3),$$
$$\text{And i want to find the unit normal vector $\hat n$ of $S$ }$$
$$\hat n =\frac{\nabla f}{\lvert \nabla f \rvert}\text{, where } f:=x^2+y^2+z^2-a^2$$
$$\text{we get: }\hat n=\big(\frac{x}{a},\frac{y}{a},\frac{z}{a}\big)\quad \text{ we want:}$$
$$F\cdot\hat n=z \Rightarrow F_1=F_2=0,F_3=a$$
$$\Rightarrow \nabla \cdot F=0+0+0=0$$
$$\text{but obviously } \int_S zdS\ne 0$$
What am i doing wrong?
thank you.
|
With the same trick, you can use Stokes' theorem in one of two ways. First is directly:
$$\nabla \times H = (0,0,a) \to H = (0,ax,0)$$
which gives us the line integral
$$\int\limits_{x^2+y^2=a^2\:\cap\:z=0}(0,ax,0)\cdot dr = a^3\int_0^{2\pi}\cos^2 t\:dt = \pi a^3$$
Or we can use a corollary of Stokes' theorem, which says that we can shift the surface integral over to another surface that shares the same boundary as the original (Think distorting a bubble when you blow it from a hoop). In this case we can say that
$$\iint\limits_{x^2+y^2+z^2=a^2\:\cap\:z\geq 0}(0,0,a)\cdot dS = \iint\limits_{x^2+y^2\leq a^2\:\cap \:z=0}(0,0,a)\cdot dS = a\iint\limits_{x^2+y^2\leq a^2}\;dA = \pi a^3$$
|
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|
Let $x_{n}$ be the positive real root of $(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$, how to prove that $x_{n} > x_{n + 1}$? Let $x_{n}$ be the positive real root of equation
$$(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$$
How to prove that $x_{n} > x_{n + 1}$?
Actually, $x_{n} > 2$ and I get that $x_{1} = 5, x_{2} \approx 3.5973, x_{3} \approx 3.1033$
|
replace $x$ by $2x$, we have
\begin{equation}
1+\frac{x^n}{2} = \Big(1+\frac{x^{n-1}}{2}\Big)^{1+\frac{1}{n}} > 1 + \big(1+\frac{1}{n}\big)\frac{x^{n-1}}{2}
\end{equation}
hence $x> 1+1/n$. Let
\begin{equation}
f(y) = 1+\frac{y^{n+1}}{2} - \Big(1+\frac{y^n}{2}\Big)^{1+\frac{1}{n+1}}
\end{equation}
then
\begin{equation}
f'(y) = \frac{y^n}{2}\Big(n+1-\frac{n(n+2)\big(1+\frac{y^n}{2}\big)^{\frac{1}{n+1}}}{(n+1)y}\Big)
\end{equation}
$f'(y)>0$ if $y>1+1/n$. It is easy to check $f(x)>0$, hence $x>y_0$(root of $f$).
|
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|
Probability that a quadratic equation has real roots Problem
The premise is almost the same as in this question. I'll restate for convenience.
Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(-1,+1)$. What is the probability that the polynomial $Ax^2+Bx+C$ has real roots?
Note: The distribution is now $-1$ to $+1$ instead of $0$ to $1$.
My Attempt
Preparation
When the coefficients are sampled from $\mathcal{U}(0,1)$, the probability for the discriminant to be non-negative that is, $P(B^2-4AC\geq0) \approx 25.4\% $. This value can be obtained theoretically as well as experimentally. The link I shared above to the older question has several good answers discussing both approaches.
Changing the sampling interval to $(-1, +1)$ makes things a bit difficult from the theoretical perspective. Experimentally, it is rather simple. This is the code I wrote to simulate the experiment for $\mathcal{U}(0,1)$. Changing it from (0, theta) to (-1, +1) gives me an average probability of $62.7\%$ with a standard deviation of $0.3\%$
I plotted the simulated PDF and CDF. In that order, they are:
So I'm aiming to find a CDF that looks like the second image.
Theoretical Approach
The approach that I find easy to understand is outlined in this answer. Proceeding in a similar manner, we have
$$
f_A(a) = \begin{cases}
\frac{1}{2}, &-1\leq a\leq+1\\
0, &\text{ otherwise}
\end{cases}
$$
The PDFs are similar for $B$ and $C$.
The CDF for $A$ is
$$
F_A(a) = \begin{cases}
\frac{a + 1}{2}, &-1\leq a\geq +1\\
0,&a<-1\\
1,&a>+1
\end{cases}
$$
Let us assume $X=AC$. I proceed to calculate the CDF for $X$ (for $x>0$) as:
$$
\begin{align}
F_X(x) &= P(X\leq x)\\
&= P(AC\leq x)\\
&= \int_{c=-1}^{+1}P(Ac\leq x)f_C(c)dc\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P(Ac\leq x)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
\end{align}
$$
We take a quick detour to make some observations. First, when $0<c<x$, we have $\frac{x}{c}>1$. Similarly, $-x<c<0$ implies $\frac{x}{c}<-1$. Also, $A$ is constrained to the interval $[-1, +1]$. Also, we're only interested when $x\geq 0$ because $B^2\geq 0$.
Continuing, the calculation
$$
\begin{align}
F_X(x) &= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=-x}^{0}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=0}^{x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + 0 + 1 + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}\frac{x+c}{2c}dc + 0 + 1 + \int_{c=x}^{+1}\frac{x+c}{2c}dc\right)\\
&= \frac{1}{2}\left(\frac{1}{2}(-x+x(\log(-x)-\log(-1)+1) + 0 + 1 + \frac{1}{2}(-x+x(-\log(x)-\log(1)+1)\right)\\
&= \frac{1}{2}\left(2 + \frac{1}{2}(-x+x(\log(x)) -x + x(-\log(x))\right)\\
&= 1 - x
\end{align}
$$
I don't think this is correct.
My Specific Questions
*
*What mistake am I making? Can I even obtain the CDF through integration?
*Is there an easier way? I used this approach because I was able to understand it well. There are shorter approaches possible (as is evident with the $\mathcal{U}(0,1)$ case) but perhaps I need to read more before I can comprehend them. Any pointers in the right direction would be helpful.
|
We know from the quadratic formula that the polynomial $Ax^2 + Bx + C$ has real roots if $B^2 - 4AC \geq 0$. We can think of this problem in terms of volumes. To do so, it's easier if we rename the coefficients as $x \equiv A$, $y \equiv C$, and $z \equiv B$. Hence, in order to have real roots we require that $z^2 \geq 4xy$ for $x,y,z \in (-1,1)$. The probability we are after is the ratio between the volume of the region for which this inequality is true and the volume of the containing cube, which is 8. Begin by observing that if $x$ and $y$ have opposite signs then this inequality is trivially satisfied. The volume of the region for which they have opposite signs 4. Now consider the case where $x$ and $y$ have the same signs. In this case, we want to compute the volume above the surface $z^2 = 4xy$ and below the containing cube. There are four cases to consider:
*
*$-1 < x \leq -\frac{1}{4}$ and $\frac{1}{4x} \leq y \leq 0$.
*$-\frac{1}{4} \leq x \leq 0$ and $-1 < y \leq 0$.
*$0 \leq x \leq \frac{1}{4}$ and $0 \leq y < 1$.
*$\frac{1}{4} \leq x < 1$ and $0 \leq y \leq \frac{1}{4x}$.
By symmetry we can just consider cases 1 and 2 and then multiply that volume by 2. In each case we have to compute the integral:
\begin{align*}
\int_a^b\int_c^d 2 - 4\sqrt{xy}\,dy\,dx,
\end{align*}
where the limits of integration are defined above. Evaluating cases 1 and 2 we find that the volume is $5/18 + (1/6)\ln(4)$. Hence, the total volume that satisfies the inequality is
\begin{align*}
4 + 2\left(\frac{5}{18} + \frac{1}{6}\ln(4)\right) = \frac{41}{9} + \frac{1}{3}\ln(4)
\end{align*}
which leads to a probability of
\begin{align*}
\frac{1}{8}\left(\frac{41}{9} + \frac{1}{3}\ln(4)\right) \approx 0.62721
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3818919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 3,
"answer_id": 2
}
|
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl[\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\Biggl]$$
$$= \frac{1}{16} + \frac{1}{24} - \require{cancel} \cancel{\frac{3}{80}}$$
$$+ \frac{1}{48} + \require{cancel} \cancel{\frac{1}{40}} - \require{cancel} \cancel{\frac{3}{112}}$$
$$+ \require{cancel} \cancel{\frac{1}{80}} + \require{cancel} \cancel{\frac{1}{56}} - \require{cancel} \cancel{\frac{3}{144}}$$
. . .
. . .
. . .
$$+ \require{cancel} \cancel{\frac{1}{16(2n-3)}} + \require{cancel} \cancel{\frac{1}{8(2n-1)}} - \frac{3}{16(2n+1)}$$
$$+ \require{cancel} \cancel{\frac{1}{16(2n-1)}} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{1}{16} + \frac{1}{24} + \frac{1}{48} - \frac{3}{16(2n+1)} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{2n^{2}+6n+3}{4(2n+1)(2n+3)}$$
|
$$
\begin{align}
&\sum_{r=1}^n\frac{r}{(2r-1)(2r+1)(2r+3)}\\
&=\sum_{r=1}^n\left(\frac1{16(2r-1)}+\frac1{8(2r+1)}-\frac3{16(2r+3)}\right)\tag1\\
&=\frac1{16}\sum_{r=1}^n\left(\frac1{2r-1}-\frac1{2r+1}\right)+\frac3{16}\sum_{r=1}^n\left(\frac1{2r+1}-\frac1{2r+3}\right)\tag2\\
&=\frac1{16}\left(1-\frac1{2n+1}\right)+\frac3{16}\left(\frac13-\frac1{2n+3}\right)\tag3\\[3pt]
&=\frac18-\frac18\frac{4n+3}{(2n+1)(2n+3)}\tag4
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: group for telescoping
$(3)$: telescope the sums
$(4)$: collect terms
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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|
How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!
|
$b^nb^m = b^{n+m}$ and $b^{-n} = \frac 1b$ and $\frac {b^n}{b^m} = b^{n-m} = \frac 1{b^{m-n}}$. That is all that is going on.
$\frac {BLAH}{\frac 1{2^{-3}}\cdot 2^{-10}}=$
$\frac {BLAH}{\frac 1{2^{-3}}\frac 1{2^{-(-10)}}}=\frac {BLAH}{\frac 1{2^{-3 - (-10)}}}$
Frankly seems like a convoluted way to make things as hard as possible and to make the negative signs as many and as confusing as possible.
I'd have just done:
$\frac{6^6}{\frac 1{2^{-3}}\cdot 2^{-10}}=$
$\frac {(2\cdot 3)^6}{2^3\cdot 2^{-10}}=$
$\frac {2^6\cdot 3^6}{2^{-7}}=$
$2^6\cdot 3^6 \cdot 2^7 = 2^{13}\cdot 3^6$.
.....
Oh.... I now see the expression was supposed to be $\frac {\frac {6^6}1}{2^{-3}}\cdot 2^{-10}$ and not $\frac{6^6}{\frac 1{2^{-3}}\cdot 2^{-10}}$.
That doesn't change me comments and answer. But sheesh what kind of lunatic wrote this problem solely for the purpose of confusion?
We have $\frac {BLAH}{2^{-3}}\cdot 2^{-10} =$
$\frac {BLAH}{2^{-3}2^{-(-10)}} = \frac {BLAH}{2^{-3-(-10)}}$.
But I'd do:
$\frac {\frac {6^6}1}{2^{-3}}\cdot 2^{-10}=$
$\frac {6^6}{2^{-3}}\cdot 2^{-10}=$
$6^6\cdot 2^3 \cdot 2^{-10} =$
$6^6 \cdot 2^{-7}= $
$(2\cdot 3)^6 \cdot 2^{-7}=$
$2^6\cdot 3^6 \cdot 2^{-7}=$
$2^{-1}\cdot 3^6=$
$\frac {3^6}2$.
=====
Or we could simply do:
$\frac {\frac {6^6}1}{2^{-3}} \cdot 2^{-10}$
$\frac {\frac {6^6}1}{\frac 1{2^3}}\cdot \frac 1{2^{10}}=$
$\frac {\frac {6^6}1}{\frac 1{2^3}2^{10}}=$
$\frac {\frac {6^6}1}{2^7}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3827352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?
|
From where you left off:
$$\begin{align}
\sqrt{3}-\sqrt{\sqrt3+x}
&=x^2\\
\sqrt{3}-x^2
&=\sqrt{\sqrt3+x}\\
3-2\sqrt{3}x^2+x^4
&=\sqrt{3}+x\\
x^4-2\sqrt{3}x^2-x+\left(3-\sqrt3\right)&=0\\
\left(x^2+x+(1-\sqrt{3})\right)\left(x^2-x-\sqrt{3}\right)&=0
\end{align}$$
So if there is a solution to the original equation, it is a root of this 4th degree polynomial. It's easy to find its four roots since it factors. But some roots of this polynomial might not solve the original equation, since we squared a few times earlier. So each one should be checked.
Note that the original left-hand side is not real unless $x$ is in $\left[-\sqrt{3},3-\sqrt{3}\right]$. That should help eliminate several of the four polynomial roots.
|
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"url": "https://math.stackexchange.com/questions/3829995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$ Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 - 4)^2 = 5$, hence $\alpha$ is a root of the polynomial $f(x)=x^6 - 8 x^3 + 11$.
I tried to prove with various approaches that $f(x)$ is irreducible over $\mathbb{Q}$ without success, so I devised the following strategy.
Since $x^2 - 5$ is irreducible over $\mathbb{Q}$, we have $\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$. Now from $\alpha^3 = 4 + \sqrt{5}$ we get $\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$, so that $\mathbb{Q} \left( \sqrt{5} \right )$ is a subfield of $\mathbb{Q} \left( \alpha \right )$, $\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$, and we have
\begin{equation}
\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .
\end{equation}
Now $\alpha$ is a root of the polynomial $g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$ given by $g(x) = x^3 - 4 - \sqrt{5}$. So to prove our thesis it is enough to prove that this polynomial is irreducible in $\mathbb{Q} \left (\sqrt{5} \right) [ x ]$. Being $g(x)$ of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that $g(x)$ would have some root in $\mathbb{Q} \left (\sqrt{5} \right)$. Hence our problem boils down to show that there are no integers $m_0, m_1, n$, with $n \neq 0$, such that
\begin{equation}
\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},
\end{equation}
which gives
\begin{equation}
m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1 + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,
\end{equation}
or
\begin{equation}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1 - n^3 \right)=0,
\end{equation}
which implies, being $\sqrt{5}$ irrational,
\begin{cases}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1 - n^3 = 0.
\end{cases}
At this point I am stuck, because I do not know how to prove that this system admits the only integer solution $m_0 = m_1 = n = 0$.
Any help is welcome!
|
As requested by OP I am rewriting my comment as an answer. We will show that $[\mathbb{Q}(\sqrt[3]{4+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 3$ by showing that $f(x) = x^3 - (4+\sqrt{5})$ has no solution in $\mathbb{Q}(\sqrt{5})$.
Rather than the approach in the question we notice that $\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) = 11$. In particular suppose that $\alpha$ is a root of $f(x)$ in $\mathbb{Q}(\sqrt{5})$, then
\begin{align*}
\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha)^3 &= \operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha^3) \\ & =\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) =11
\end{align*}
a contradiction.
What's really going on under the hood here is that $f(x)$ is Eisenstein for the prime ideal $\mathfrak{p} = (4 + \sqrt{5})$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$
But I know that $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq \frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$ and so:
But I am stuck, this tells me that the sum for $n=k+1$ is always $1$ , not $S \leq 1$ I am so confused, because I can't use the geometric series sum formula.. any help would be appreciated!
|
We can instead try to prove the following
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1-\frac1{2^n}\le 1$$
then the induction step becomes
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n}+ \frac{1}{2^{n+1}} \le 1-\frac1{2^n}+ \frac{1}{2^{n+1}}= 1-\frac1{2^n}\left(1- \frac{1}{2}\right)=1-\frac1{2^{n+1}}$$
and we are done.
|
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"url": "https://math.stackexchange.com/questions/3837387",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original post hinted towards, setting $k = \frac{y+1}{x+2}$ and than writing the given equation as $$(x-3)^2 + (k(x+2) - 4)^2 = 6.$$ I than proceeded to expand and simplify, which gave me $$x^2(k^2 + 1) + x(4k^2 - 8k - 6) + (4k^2 - 16k + 19).$$ However, I am unsure where to go from here. Should I use the discriminant now?
|
The function and the constraint are
$$
f(x,y) = \frac{y+1}{x+2}\\
\quad g(x,y) = (x-3)^2 + (y-3)^2 - 6 = 0
$$
using Lagrange multipliers
$$
\nabla f = (-\frac{y+1}{(x+2)^2}, \frac{1}{x+2}), \quad \nabla g = (2x,2y)
$$
We get the equations
$$
\nabla f - \lambda \nabla g = 0\\
g(x,y) = 0
$$
in $\lambda, x$, and $y$. The first (vector) equation is the two equations
$$
-\frac{y+1}{(x+2)^2} - \lambda 2x = 0 \tag{1}
$$
$$
\frac{1}{x+2} - \lambda 2y = 0 \tag{2}
$$
Mathematica gives me the solution set
$$
x = \frac{1}{41}(93-4\sqrt{210}),\quad y = \frac{1}{41}(99+5\sqrt{210}),\quad \lambda = \frac{8400+971\sqrt{210}}{151620} \\
x = \frac{1}{41}(93+4\sqrt{210}),\quad y = \frac{1}{41}(99-5\sqrt{210}),\quad \lambda = \frac{8400-971\sqrt{210}}{151620} \\
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to shrink explicit function definitions to 1 line? Assume maximum domain for all following examples.
We know that $f(x)=|x-1|$ can be expanded(split) into branches like this: $f(x)=\left\{\begin{gather}-x+1, x\le 1\\x-1, x \gt 1\end{gather}\right\}$
or $f(x)=min(x^{2}-2, 2)$ can be written as $f(x)=\left\{\begin{gather}2, x\lt -2\\x^{2}-2, -2 \le x \le 2 \\ 2, x \gt 2\end{gather}\right\}$
I usually found these branched forms in math problems to study continuity and what not.
What i'm interested in is if i can do the reverse (write all branches in 1 line in closed form if possible) in a general case where branches appear.
Maybe something like: $f(x)=\left\{\begin{gather}-2x+7, x\lt -10\\x^{3}-2x^{2}+8, -10 \le x \le 3 \\ \ln (x+8), x \gt 3\end{gather}\right\}$ can be written as $f(x)=A*(-2x+7)+B*(x^{3}-2x^{2}+8)+C*(\ln (x+8))$ that is equivalent to the branched form? What other functions like abs or min and other tricks can i use to achieve that 1 line explicit form? Or in my example is it even possible to find such A, B, C ? My example is just that, an example. I want to apply this on other functions.
Thanks!
|
We can use the indicator function to write for instance
\begin{align*}
f(x)&=|x-1|\\
&=(1-x)\mathbf{1}_{x\le 1}+(x-1)\mathbf{1}_{x> 1}\\
\\
&\qquad\qquad\text{or}\\
\\
g(x)&=
\begin{cases}
-2x+7&\qquad x<-10\\
x^{3}-2x^{2}+8&\qquad -10 \le x \le 3 \\
\ln (x+8)&\qquad x > 3
\end{cases}\\
\\
&=(-2x+7)\mathbf{1}_{x<-10}+(x^{3}-2x^{2}+8)\mathbf{1}_{-10 \le x \le 3}+\ln (x+8)\mathbf{1}_{x> 3}
\end{align*}
We can also use Iversion brackets to do the same job:
\begin{align*}
f(x)&=|x-1|\\
&=(1-x)[[x\le 1]]+(x-1)[[x> 1]]\\
\\
&\qquad\qquad\text{or}\\
\\
g(x)&=
\begin{cases}
-2x+7&\qquad x<-10\\
x^{3}-2x^{2}+8&\qquad -10 \le x \le 3 \\
\ln (x+8)&\qquad x > 3
\end{cases}\\
\\
&=(-2x+7)[[x<-10]]+(x^{3}-2x^{2}+8)[[-10 \le x \le 3]]+\ln (x+8)[[x> 3]]
\end{align*}
Note that both, the indicator function as well as the Iverson brackets are sometimes conveniently used when doing calculations. Here we only need to focus on how to ease readability.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \leftarrow\text{(1)} \\
3a^2 + 3b + 1 &= 4 \leftarrow\text{(2)}
\end{align}
$$ \text{From (1):} $$
\begin{align}
2a^2 + 2b + 1 &= 1 \\
2a^2 &= -2b + 1 - 1 \\
2a^2 &= -2b \\
a^2 &= -b \leftarrow\text{(3)}
\end{align}
$$ \text{Substitute (3) into (2):} $$
\begin{align}
3(-b) + 3b + 1 &= 4 \\
-3b + 3b + 1 &= 4 \\
1 &= 4
\end{align}
Here is my second attempt at solving it:
\begin{align}
ax^2 + bx + 1 &= 1 \\
2a^2 + 2b + 1 &= 1 \\
a^2 + b + \frac{1}{2} &= \frac{1}{2} \\
a^2 + b &= \frac{1}{2} - \frac{1}{2} \\
a^2 + b &= 0 \\
4a^2 + 4b &= 0 \\
4a^2 + 4b + 1 &= 1 \leftarrow\text{(1)}
\end{align}
\begin{align}
3a^2 + 3b + 1 &= 4 \\
3a^2 + 3b &= 4 - 1 \\
3a^2 + 3b &= 3 \\
a^2 + b &= 1 \\
4a^2 + 4b &= 4 \\
4a^2 + 4b + 1 &= 5 \leftarrow\text{(2)}
\end{align}
$$ \text{(1) = (2): } 1 = 5 $$
Where did I go wrong?
|
Your problem lies in $(1)$
Since the equation is
$$f(x)=ax^2+bx+1$$
And the value when $x=2$ is $1$, equation one should be
$$f(2)=a2^2+b2+1\\
1=4a+2b+1\\
2a=-b\\
a=\frac{-b}{2}$$
Substituting in (2)
$$f(3)=9a+3b+1\\
4=9\left(\frac{-b}{2}\right)+3b+1\\
6=-9b+6b\\
b=-2\\
\therefore a=1$$
So when $x=4$
$$f(4)=4^2-2\cdot4+1=9$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$
Solve the following:
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$
Solution from my textbook:
Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get:
$$|c| \le 1$$
$$|a + b + c| \le 1$$
$$|a + 2b + 4c| \le 4$$
From the above three equations, we get,
$|b| \le 8 \text{ and } |a| \le 8$.
Therefore, $|a| + |b| + |c| \le 17$
However, I don't understand: how, from the three equations, can you find the values of $|a|$ and $|b|$?
|
Let $f(x)=ax^2+bx+c$.
Thus, $$a+b+c=f(1),$$
$$\frac{1}{4}a+\frac{1}{2}b+c=f\left(\frac{1}{2}\right)$$ and $$c=f(0),$$ which gives
$$b=4f\left(\frac{1}{2}\right)-f(1)-3f(0),$$ $$a=2f(1)+2f(0)-4f\left(\frac{1}{2}\right)$$ and by the triangle inequality we obtain: $$|a|+|b|+|c|\leq2+2+4+4+1+3+1=17.$$
It's interesting that for $f(x)=8x^2-8x+1$ we have an equality.
|
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|
Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers. Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation
$(xz+1)(yz+1)=az^{k}+1$
has infinitely many solutions in positive integers $x,y, z$.
To motivate the question, it can easily be shown that if $k <3$, the given diophantine equation has no solutions in positive integers $x, y ,z$ with $z>a$.
Proof: $(xz+1)(yz+1)=az^{k}+1$ may be simplified to $xyz^{2}+(x+y)z=az^{k}$. If $k=1$, this reduces to $xyz+x+y=a$. Its clear that $a>z$ therefore there are no positive integral solutions in $x$ and $y$ when $z>a$. if $k=2$, we have the reduced equation $xyz+x+y=az$. We have $z$ | $x+y$, $z \le(x+y) \le xy$. Therefore $LHS=xyz+x+y>z^{2}$. Because $RHS=az$, we must have $a>z$ thus there are no solutions in positive integers $x ,y$ when $z>a$.
I would like to prove that given two positive integers $a$ and $k>3 $, the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many positive integer solutions $x, y, z$. I do not know how to start the proof.
|
This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\
&=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right)
\end{align}$$
Where the last equation is because when $p$ is odd, $$
\begin{align}u^p+1&=(u+1)
\sum_{j=0}^{p-1} (-1)^ju^j
\\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$
Now, since $n>d,$ we can set $$
\begin{align}x&=a^{n-d}t^{k-p}\\
y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}
\end{align}$$
For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.
So this solves all cases $k>3.$
You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$
k even
So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$
Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$
k odd
Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$ Then for any integer $t$:
$$\begin{align}z&=a^{m-1}t^{2m-1}\\
x&=at^2\\
y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1}
\end{align}$$
is a solution.
In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$
|
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|
Why doesn't trig substitution work for definite integrals? In the following example, I am attempting to find the area of a semicircle using calculus, which is obviously $\frac{\pi{r}^2}{2}$. Effectively, I am trying to find
$$\int_{-r}^r\sqrt{r^2-x^2}dx$$
Here goes:
Let $x=r\sin\theta$:
$$\frac{dx}{d\theta}=r\cos\theta\implies dx=r\cos\theta d\theta$$
When:
$$x=r, ~~~~\text{Then}~~~~\sin\theta=1\implies\theta=\frac{\pi}{2}$$
$$x=-r, ~~~~\text{Then}~~~~\sin\theta=-1\implies\theta=-\frac{\pi}{2}$$
$$\therefore\int_{-r}^r\sqrt{r^2-x^2}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r\cos\theta\sqrt{r^2-r^2\sin^2\theta} ~~d\theta$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos^2\theta~~d\theta=\frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos2\theta)~~d\theta=\frac{r^2}{2}\left[\theta+0.5\sin2\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{r^2}{2}((\frac{\pi}{2}+0)-(-\frac{\pi}{2}-0))=\frac{\pi r^2}{2}$$
as required.
BUT
Back to the stage when I was working out the limits of the new integral with respect to $\theta$. Let's say I wrote something like this:
When:
$$x=r, ~~~~\text{Then}~~~~\sin\theta=1\implies\theta=\frac{5\pi}{2}$$
$$x=-r, ~~~~\text{Then}~~~~\sin\theta=-1\implies\theta=-\frac{5\pi}{2}$$
Then , I would get an eventual answer of
$$\frac{5\pi r^2}{2}$$
Where have I gone wrong? It seems to me that my logic is faultless, but the definite integral is simply ambiguous when using trig substitutions. This would apply to any integral involving trig substitutions; in short, can't all definite integrals computed using trig substitutions generate an incorrect answer?
If so, isn't the maths of definite integrals faulty? Thanks for your help.
|
Ideally, as others have pointed out, you will set up the bounds of your trigonometric substitution so that each value of $x$ you need is produced once and only once, in a continuous stream.
For $-\frac\pi2 \leq \theta \leq \frac\pi2,$
if $x = r\sin\theta$ then $x$ increases monotonically from $-r$ to $r$
as $\theta$ increases monotonically from $-\frac\pi2$ to $\frac\pi2.$
This makes a very "clean" substitution.
(Note: throughout this answer I am assuming $r > 0.$ In the case where $r<0$ then $\int_{-r}^r\sqrt{r^2-x^2}dx$ is negative and is not the answer to the initial problem, "What is the area of a semicircle?")
When you increase $\theta$ from $-\frac{5\pi}2$ to $\frac{5\pi}2,$
the value of $x$ starts at $-r,$ increases to $r,$ then decreases back to $-r$,
then increases to $r$ again, decreases to $-r$ again, and finally increases to $r.$
That's a lot of increasing and decreasing just to cover the distance from $-r$ to $r.$
But the up-and-down-and-up movement of $x$ is not really the problem.
What is the problem is that your substitution is not correct over the
entire domain $-\frac{5\pi}2 \leq \theta \leq \frac{5\pi}2.$
In particular, look at this equation on which you rely
(where I have written $a$ and $b$ as the bounds of the interval of integration,
since you propose to use the same method from $-\frac{5\pi}2$ to $\frac{5\pi}2$
as for $-\frac{\pi}2$ to $\frac{\pi}2$):
$$\int_a^b r\cos\theta \sqrt{r^2-r^2\sin^2\theta} \,d\theta
=\int_a^b r^2\cos^2\theta\,d\theta.$$
In order to justify this equation, you must show that
$\sqrt{r^2-r^2\sin^2\theta} = r\cos\theta.$
That is easily proved when $\cos\theta \geq 0,$
but it is false when $\cos\theta < 0.$
When $\cos\theta < 0,$ the correct equation is
$$\int_a^b r\cos\theta \sqrt{r^2-r^2\sin^2\theta} \,d\theta
=\int_a^b -r^2\cos^2\theta\,d\theta.$$
Alternatively, you could combine the two equations as
$$\int_a^b r\cos\theta \sqrt{r^2-r^2\sin^2\theta} \,d\theta
=\int_a^b -r^2\cos\theta \lvert\cos\theta\rvert \,d\theta,$$
but the integral of $\cos\theta \lvert\cos\theta\rvert$ is not the same as the integral of $\cos^2\theta,$ so you still have some work to do to sort things out.
Here's how the integral can be correctly integrated from
$-\frac{5\pi}2$ to $\frac{5\pi}2$:
\begin{align}
\int_{-r}^r \sqrt{r^2-x^2}\,dx
&= \int_{-5\pi/2}^{5\pi/2} r\cos\theta\sqrt{r^2-r^2\sin^2\theta} \,d\theta\\
&= \int_{-5\pi/2}^{5\pi/2} r^2(\cos\theta)\lvert\cos\theta\rvert \,d\theta\\
&= \int_{-5\pi/2}^{-3\pi/2} r^2\cos^2\theta \,d\theta\\
&\qquad + \int_{-3\pi/2}^{-\pi/2} -r^2\cos^2\theta \,d\theta\\
&\qquad + \int_{-\pi/2}^{\pi/2} r^2\cos^2\theta \,d\theta\\
&\qquad + \int_{\pi/2}^{3\pi/2} -r^2\cos^2\theta \,d\theta\\
&\qquad + \int_{3\pi/2}^{5\pi/2} r^2\cos^2\theta \,d\theta\\
&= \frac{\pi r^2}{2} - \frac{\pi r^2}{2} + \frac{\pi r^2}{2}
- \frac{\pi r^2}{2} + \frac{\pi r^2}{2} \\
&= \frac{\pi r^2}{2},
\end{align}
using the fact that $\lvert\cos\theta\rvert = -\cos\theta$
when $\cos\theta \leq 0.$
You actually get the correct answer, but only if you integrate the correct function over the entire interval.
Also notice that each time $\sin\theta$ decreases from $1$ to $-1$
(that is, each time $x$ decreases from $r$ to $-r$) you precisely wipe out the amount you integrated on the previous increase.
In effect, by allowing $x$ to go up and down all these times, you end up integrating
\begin{multline}
\int_{-r}^r \sqrt{r^2-x^2}\,dx
+ \int_r^{-r} \sqrt{r^2-x^2}\,dx
+ \int_{-r}^r \sqrt{r^2-x^2}\,dx\\
+ \int_r^{-r} \sqrt{r^2-x^2}\,dx
+ \int_{-r}^r \sqrt{r^2-x^2}\,dx,
\end{multline}
in which the first four integrals cancel each other out.
|
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|
$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma directly but it is not working. I meant that I wrote each term in the following way: $$\frac{a^3}{b+c}=\frac{a^3bc}{bc(b+c)}=\frac{a^2}{bc(b+c)}.$$ Then I applied Titu's lemma to the sum $$\frac{a^2}{bc(b+c)}+\frac{b^2}{ac(a+c)}+\frac{c^2}{ab(a+b)}\geq \frac{(a+b+c)^2}{bc(b+c)+ac(a+c)+ab(a+b)}=\frac{\sigma_1^2}{\sigma_1\sigma_2-3},$$ where $\sigma_1=a+b+c$ and $\sigma_2=ab+ac+bc$ are elementary symmetric functions.
This is what I got so far.
Would thankful if someone shows correct solution.
|
Also, we can use AM-GM and SOS:
$$\sum_{cyc}\frac{a^3}{b+c}=\frac{a^2+b^2+c^2}{3}+\sum_{cyc}\left(\frac{a^3}{b+c}-\frac{a^2}{2}\right)\geq$$
$$\geq\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(2a-b-c)}{b+c}=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(a-b-(c-a))}{b+c}=$$
$$=\frac{3}{2}+\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a^2}{b+c}-\frac{b^2}{c+a}\right)=\frac{3}{2}+\sum_{cyc}\tfrac{(a-b)^2(a^2+b^2+ab+ac+bc)}{2(a+c)(b+c)}\geq\frac{3}{2}.$$
The following stronger inequality is also true.
Let $a$, $b$ and $c$ be positive numbers such that $a^5+b^5+c^5=3.$ Prove that:
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq\frac{3}{2}.$$
|
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|
Fibonacci sequence and convergence of $t_n=\cfrac{x_{n+1}}{x_n}$ Fibonacci sequence define as: $x_1=x_2=1 , x_n=x_{n-1}+x_{n-2}$ for $n \ge 1$ prove the sequence $t_n=\cfrac{x_{n+1}}{x_n}$ is convergent.
First of all I start by calculating some of the first terms of the $t_n$ sequence:
$\begin{array}{rcc}
n:&1&2&3&4&5&6\\
t_n:&\frac11&\frac21&\frac32&\frac{5}3&\frac{8}{5}&\frac{13}{8}
\end{array}$
It seems for even values of $n$ the sequence is decreasing and for odd values of $n$ it is increasing.
To prove the first statement (where $n=2k$) , I should prove $t_{2k}- t_{2k+2}\ge 0$ :
$$\cfrac{x_{2k+1}}{x_{2k}}- \cfrac{x_{2k+3}}{x_{2k+2}} \ge0$$
$$\cfrac{x_{2k+1} \cdot x_{2k+2}- x_{2k+3} \cdot x_{2k}}{x_{2k} \cdot x_{2k+2}} \ge 0$$
$$x_{2k+1} \cdot x_{2k+2} \ge x_{2k+3} \cdot x_{2k}$$
$$x_{2k+1} \cdot (x_{2k}+x_{2k+1}) \ge x_{2k} \cdot (x_{2k+1}+x_{2k+2})$$
$$x_{2k+1} \times x_{2k+1} \ge x_{2k} \times x_{2k+2}$$
Here I don't khow how to proceed.
|
$$X_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]=\frac{a^n-b^n}{\sqrt{5}}$$
So $\lim_{n \to \infty} \frac{X_{n+1}}{X_n}=a=\frac{1|+\sqrt{5}}{2}$
|
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|
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer?
Here how I work, please correct me if I'm wrong
First method :
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\sin x \cdot \sin^2x}\,\mathrm{d}x
\\ &= \int{\sin x (1 - \cos^2x)}\,\mathrm{d}x
\\& = \displaystyle\int{(\sin x - \sin x\cos^2x)}\,\mathrm{d}x
\\& = \dfrac{1}{3}\cos^3x - \cos x + C
\end{align}
Second method :
First, we know that $$\sin 3x = 3\sin x - 4\sin^3x$$
Therefore, $$\sin^3x = \dfrac{3}{4}\sin x - \dfrac{1}{4}\sin 3x$$
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\sin x - \frac{1}{4}\sin 3x\right)}\,\mathrm{d}x\\
& = \frac{1}{12}\cos 3x - \frac{3}{4}\cos x + C
\end{align}
|
$$\cos 3x =4\cos^3x -3\cos x$$
So, $$\frac{1}{12}\color{green}{\cos 3x} - \frac{3}{4}\cos x=\frac{1}{12}(\color{green}{4\cos^3x -3\cos x})-\frac{3}{4}\cos x$$
$$=\frac{1}{3}\cos^3x-\cos x$$
Hence both the answers are the same.
|
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|
Evaluate $\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$
Evaluate $$\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$$
Of course L-hospitals rule kills it.The question is how to do without L-hospitals rule.The substituition $x-3\pi/4=h$ seems promising but only complicates it further.Traditional methods of multiplying by congugate don't apply here.
How do i find the limit then?
|
We try to simplify the numerator and factor out $(\sin x+\cos x)$ to avoid using L'Hospital.
$4\sin^2x \cos x-\cos x+\sin x\\
=\sin x + 3\sin^2 x\cos x+(-\cos x+\sin^2 x\cos x)\\
=\sin x + 3\sin^2 x\cos x +\cos x(\sin^2 x-1)\\
=\sin x + 3\sin^2 x\cos x -\cos^3 x+(\sin^3x-\sin^3x)\\
= \sin x (1-\sin^2x)+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\
= \sin x \cos^2 x+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\
= (2\sin x \cos^2 x-\sin x \cos^2 x)+ (2\sin^2 x\cos x +\sin^2 x\cos x) -\cos^3 x+\sin^3x\\
= (\sin^3x-\sin x \cos^2 x+ 2\sin^2 x\cos x) +(2\sin x \cos^2 x+\sin^2 x\cos x -\cos^3 x)\\
= \sin x(\sin^2x- \cos^2 x+ 2\sin x\cos x) +\cos x(2\sin x \cos x+\sin^2 x -\cos^2 x)\\
=(\sin x+\cos x)(\sin^2x- \cos^2 x+ 2\sin x\cos x)$
$\therefore \lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}=\lim_{x \to 3\pi/4}(\sin^2x- \cos^2 x+ 2\sin x\cos x)=-1$
|
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|
$\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for any real numbers. I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work:
Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$
$$
\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}
$$By squaring,
$$
\Rightarrow
\sqrt{(a^2+5b^2)(b^2+5c^2)}+\sqrt{(b^2+5c^2)(c^2+5a^2)}+\sqrt{(c^2+5a^2)(a^2+5b^2)}\geq2(a+b+c)^2
$$The $LHS$
$$=
\sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2\left(a^2 + 5b^2\right) \left(\sqrt{\left(b^2 + 5c^2\right) \left(c^2 + 5a^2\right)}\right)}}
$$$$
\geqslant
\sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2(a^2 + 5b^2)(bc + 5ca)}}
$$
Now we are only left to prove that
$$
\sum_{cyc}{5b^4 + 31a^2b^2 + 52a^2bc + 10a^3c + 10a^3c} \geqslant \sum_{cyc}{4a^4 + 16(a^3b + ab^3) + 24a^2b^2 + 48a^2bc}
$$$$
\sum_{cyc}{a^4 + 7a^2b^2 + 4a^2bc - 6(a^3b + ab^3)} \geqslant 0
$$
The last inequality is wrong for $(a,b,c) = (1,1,0)$. Cauchy Schwarz looks fine but I am not able to find a way.
I found this inequality posted by arqady on aops forum.
Please help!
|
Probably not the proof you are looking for, but a proof nonetheless.
The inequality is really sharp, and I don't think that a manual solution exists. Concretely, I don't think that one can find a lower bound on the LHS, such that we can algebraically confirm that it upper bounds the RHS. However, it is easy to numerically verify that the inequality holds, and I hope that you can find this convincing.
Specifically, divide both sides by $\sqrt{a^2 + b^2 + c^2}$, then we're left with the equivalent inequality:
$$
\sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} \geq \sqrt{10 + 8(xy + yz + xz)},
$$
where $x = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, y = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, z = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$, and $x^2 + y^2 + z^2 = 1$. Furthermore, it has been established that we can safely assume that $x,y,z\geq 0$, so it is sufficient to verify the inequality on the surface $\{(x,y,z) \in\mathbb{R}^3 ~\vert~ x^2 + y^2 + z^2 = 1, x,y,z\geq 0\}$, which can be parameterized with $$x = \sin\theta\sin\omega,\quad y = \sin\theta\cos\omega,\quad z=\cos\theta,$$ with $(\theta,\omega)\in[0,\pi/2]\times[0,\pi/2]$.
Now, if one minimizes the function
$$
h(\theta,\omega) = \sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} - \sqrt{10 + 8(xy + yz + xz)},
$$
over the square $[0,\pi/2]\times[0,\pi/2]$, one then finds that it has a unique global minimum 0 at $x=y=z=\frac{1}{\sqrt{3}}$, or at $\theta \approx 0.9554,~ \omega = \pi/4$, see the figure below which shows the level sets of $h$.
$h(\theta,\omega)$" />
This implies by homogeneity that the original inequality is equality only at $a=b=c$, and a strict inequality at all other values.
|
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|
A binomial inequality raising from an exercise on the Rademacher Process Consider the inequality below
\begin{gather*}
\frac{1}{2^{n+1}}\left[\binom{n+1}{k-1} + \binom{n+1}{k+1}\right] > \frac{1}{2^n}\binom{n}{k}
\end{gather*}
It can be reduced through the following passages
\begin{gather*}
\frac{1}{2}\left[\binom{n+1}{k-1} + \binom{n+1}{k+1}\right] > \binom{n}{k} \\
\frac{1}{2}\left[\binom{n+1}{k-1} + \binom{n}{k} + \binom{n}{k+1}\right] > \binom{n}{k}\\
\binom{n+1}{k-1} + \binom{n}{k+1} > \binom{n}{k} \\
\binom{n+1}{k-1} > \binom{n}{k} - \binom{n}{k+1}
\end{gather*}
The last is trivially true for $k\leq \lfloor{\frac{n-1}{2}}\rfloor$, might you conclude the proof for the remaining values?
The context motivating the inequality
Doing an exercise about the Rademacher Processes $\{S_n\}$ I've asked to shown that $\mathcal{F}'_n=\sigma \left<S_n\right>$ is not a filtration. Considering that $\frac{S_n - n}{2}$ has a binomial distribution $\text{Bin}(n, \frac{1}2)$, $S_n$ is a simple function and the $\sigma$-algebra generated from it is the the $\sigma$-algebra generated by the partition of $[0, 1)$ given by the preimages of the singletons in its image. Now two $\sigma$-algebra generated by finite partitions are equal if and only if the partitions are equal, morover one is contained in the other only if the latter's partition is finer than the former's one. It's easy to show that for each $x$ in the support of $S_n$, $S_n^{-1}(x)\cap S_{n+1}^{-1}(x+1) \neq \emptyset$ and $S_n^{-1}(x)\cap S_{n+1}^{-1}(x-1) \neq \emptyset$, that means that a necessary condition to have the mentioned property on the partitions is that we must have $S_{n+1}^{-1}(x+1) \cup S_{n+1}^{-1}(x-1) \subseteq S_n^{-1}(x)$. I've tried to falsify that with an argument based on measure, the inequality at the beginnig, in that context means, $\mathbb{P}(S_{n+1}^{-1}(x+1) \cup S_{n+1}^{-1}(x-1))>\mathbb{P}(S_n^{-1}(x))$. I've shown that there are some values for wich it's true, i.e. some non-trivial set in $\sigma \left<S_n\right>$ not measurable in $\sigma \left<S_{n+1}\right>$, and it's enough to finish the exercise. Concluding the proof will show that every non trivial set in $\sigma \left<S_n\right>$ is not measurable in $\sigma \left<S_{n+1}\right>$ and viceversa.
|
You used $\dbinom n k \le \dbinom n{k+1}$ for $k\le \left\lfloor\dfrac {n-1}2\right\rfloor$.
Similarly for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor$ we have $\dbinom n k > \dbinom n{k+1}$.
That is, for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor + 1$:
$$\binom {n+1}{k-1} = \binom {n}{k-2} + \binom {n}{k-1} \ge \binom {n}{k-1} > \binom nk \ge \binom nk - \binom n{k+1}$$
This leaves us with the case $k = \left\lfloor\dfrac {n+1}2\right\rfloor$, where we aim to prove:
$$\binom {n}{k-2} + \binom {n}{k-1} + \binom {n}{k+1} > \binom nk$$
For $n$ odd, write $n = 2m-1$, $k = m$. Then:
$$\binom {2m-1}{m-2} + \binom {2m-1}{m-1} + \binom {2m-1}{m+1} = \binom {2m-1}{m-2} + \binom {2m-1}{m} + \binom {2m-1}{m+1} > \binom{2m-1}m$$
as long as $0 \le m-2 < m+1 \le 2m-1$, that is, $m \ge 2$. This gives the edge case $n=k=1$.
For $n$ even, write $n = 2m$, $k = m$. Then:
$$\begin{align}\binom {2m}{m-2} + \binom {2m}{m-1} + \binom {2m}{m+1} &= \frac{(2m)!}{(m+2)!(m-2)!} + \frac{2(2m)!}{(m+1)!(m-1)!}
\\&=\frac{(2m)!}{(m+2)!(m-1)!}(m-1+2(m+2))
\\&=\frac m{(m+1)(m+2)}\frac{(2m)!}{(m!)^2}(3m+1)
\\&=\frac {m(3m+1)}{(m+1)(m+2)}\binom {2m}m\end{align}$$
Our inequality holds when $m(3m+1) > (m+1)(m+2)$, that is, when $2m^2-2m-2>0$.
Therefore it holds when $m > 1$, giving the edge case $n=2, k=1$.
Hopefully these edge cases are trivial to deal with in your proof.
|
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|
Integration - Evaluate $\int \frac{x^7+2}{(x^2+x+1)^2} \ dx$
Evaluate $$\int \frac{x^7+2}{(x^2+x+1)^2} \ dx$$
This problem is from G N Berman, no. 2056 (integrate using ostrogradsky's method). I referred to this question as well as this article but I could neither understand the method nor come up with a solution for this integral. Resorting to standard long division is very tedious and messy, so i refrained from doing that.
Any hints/solutions on solving this are appreciated.
|
Standard long division is not really that bad. I got $$\int (x^3-2x^2+x+2-\frac{4x^3+6x^2+5x}{(x^2+x+1)^2})dx$$
Integrate each terms. You can integrate the first four terms to get $\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+2x$ use Ostrogradsky's method on the last term.
Ostrogradsky's method
$Q(x)=(x^2+x+1)^2$, $Q_1(x)=Q_2(x)=x^2+x+1$, $P(x)=4x^3+6x^2+5x$, you can deduce that $P_1(x)=-x$, $P_2(x)=4x+1$. (Try to work it out yourself.) Therefore, the last term becomes $-\frac{x}{x^2+x+1}+\int( \frac{4x+1}{x^2+x+1})\ dx$. Can you continue from here?
Hint after Ostrogradsky's method:
Split the integral into 2 parts: $\int(\frac{4x+2}{x^2+x+1}-\frac{1}{x^2+x+1})\ dx$. Use substitution on the remaining integrals.
My final solution after the substitution is:
$$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+2x+\frac{x}{x^2+x+1}-2\ln(x^2+x+1)+\frac{2\sqrt{3} \arctan(\frac{2x\sqrt{3}+\sqrt{3}}{3})}{3}+C$$
|
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|
Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$ Here is my question:
Let $a,b,c\in\mathbb{R^+}$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$$
Here is my solution:
From C-S inequality, we get $$\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq3^2,$$
which is equivalent to $$2(a+b+c) \sum_{cyc}\frac{1}{a+b}\geq9$$
or $$\frac{a+b+c}{a+b}+ \frac{a+b+c}{b+c}+ \frac{a+b+c}{c+a} \geq\frac{9}2$$
or $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{9}2-3=\frac{3}2$$
QED.
Here is my problem:
My teacher said that this question could be done in a lot of ways, one of which is by AM-GM inequality. However, after observing the original inequality, I still couldn’t get a clue.
I am stuck since I completely don’t know where to apply AM-GM. I am thinking about substitute something in the original inequality with some new positive numbers and continue. Is my thought in the right direction?
In addition, other approaches without AM-GM will be appreciated as well. Thanks for help.
P.S.
I am new to Mathematics SE, so if my post can be improved, make sure to let me know through the comment section. Thanks a lot.
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Because
$$\frac{a}{b+c} - \frac{8a-b-c}{4(a+b+c)} = \frac{(2a-b-c)^2}{4(b+c)(a+b+c)} \geqslant 0,$$
so
$$\frac{a}{b+c} \geqslant \frac{8a-b-c}{4(a+b+c)}.$$
Therefore
$$\sum \frac{a}{b+c} \geqslant \sum \frac{8a-b-c}{4(a+b+c)} = \frac 32.$$
Note. In addition, you can see 45th-proof-Nesbitt.pdf (Vietnamese)
|
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Conjugating simultaneously two matrices to an integer matrix I have the matrices $$A=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{pmatrix},\quad B=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&\cos\frac{2\pi}{3}&-\sin\frac{2\pi}{3}\\0&0&\sin\frac{2\pi}{3}&\cos\frac{2\pi}{3} \end{pmatrix}.$$
Problem: I'm trying to decide if there exists some $P\in\mathsf{GL}(4,\mathbb{R})$ such that $P^{-1} A P=\begin{pmatrix} 1&0&0&0\\0&-1&0&0\\0&0&1&1\\0&0&0&-1\end{pmatrix}=:E$ and s.t $P^{-1}BP$ be an integer matrix. Note that $B^3=I_3$.
Thoughts: I know that there exists some $P\in\mathsf{GL}(4,\mathbb{R})$ s.t $P^{-1}AP=E$ and such $P$ must be of the form $P=\begin{pmatrix} a&0&2c&c\\b&0&2d&d\\0&e&0&f\\0&g&0&h\end{pmatrix}$.
But there are many equations involved if I try to set $P^{-1}BP=F$ and try to solve. Also, I know that there are three conjugacy classes of order 3 in $\mathsf{GL}(4,\mathbb{Z})$: $$\begin{pmatrix} 0&-1&0&0\\1&-1&0&0\\0&0&0&-1\\0&0&1&-1\\ \end{pmatrix},\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&0&-1\\0&0&1&-1\end{pmatrix},\begin{pmatrix} 1&0&1&0\\0&1&0&0\\0&0&0&-1\\0&0&1&-1\end{pmatrix}$$ and $B$ can be $\mathsf{GL}(4,\mathbb{R})$-conjugated (and only) to the last two.
I don't know if maybe another approach could be useful. Could you give me some help or ideas? Thanks
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No such $P$ exists. Suppose the contrary. Let
$$
X=P^{-1}AP=\pmatrix{1&0&0&0\\ 0&-1&0&0\\ 0&0&1&1\\ 0&0&0&-1},
\ Y=P^{-1}BP=\pmatrix{a&b&c&d\\ e&f&g&h\\ i&j&k&l\\ m&n&o&p}.
$$
Since $AB=BA$, we have $XY=YX$, i.e.
$$
\pmatrix{a&b&c&d\\ -e&-f&-g&-h\\ i+m&j+n&k+o&l+p\\ -m&-n&-o&-p}
=\pmatrix{a&-b&c&c-d\\ e&-f&g&g-h\\ i&-j&k&k-l\\ m&-n&o&o-p}.
$$
By comparing coefficients on both sides, we obtain
$$
Y=\pmatrix{a&0&2d&d\\ 0&f&0&h\\ i&j&2l+p&l\\ 0&-2j&0&p}.
$$
By swapping the two middle rows and the two middle columns of $Y$, we see that $Y$ is similar to
$$
Z=\left(\begin{array}{cc|cc}a&2d&0&d\\ i&2l+p&j&l\\ \hline 0&0&f&h\\ 0&0&-2j&p\end{array}\right)=\pmatrix{S&\ast\\ 0&T}.
$$
Hence $Z$ is similar to $B$, the submatrices $S$ and $T$ are integer cube roots of $I_2$ and one of them is similar to the rotation matrix $R$ for an angle $2\pi/3$. Yet this is impossible, because $R$ has trace $-1$ and determinant $1$. Every integer $2\times2$ matrix with an odd trace and an odd determinant must have an odd anti-diagonal, but $S$ has an even anti-diagonal entry $2d$ and $T$ has an even anti-diagonal entry $-2j$.
|
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Find a closed formula (not including $\sum$) for the expression $\sum_{k=0}^{n-1}\binom{2n}{2k+1}$ Find a closed formula (not including $\sum$) for the expression
$$\sum_{k=0}^{n-1}\binom{2n}{2k+1}$$
I started by using the fact that
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
to get that
$$\sum_{k=0}^{n-1}\binom{2n}{2k+1}=\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}+\binom{2n-1}{2k}$$
$$=\sum_{k=0}^{n-1}\binom{2n-2}{2k+1}+\binom{2n-2}{2k}+\binom{2n-2}{2k}+\binom{2n-2}{2k-1}$$
now letting $m=n-1$
$$\sum_{k=0}^{m}\binom{2m+2}{2k+1}=\sum_{k=0}^{m}\binom{2m}{2k+1}+2\cdot\binom{2m}{2k}+\binom{2m}{2k-1}$$
I'm not exactly sure where to go from here or even if this has been helpful. Any guidance/alternative methods would be greatly appreciated!
|
Note that
$$\sum_{k\ge0} a_{2k+1} = \sum_{k\ge0} \frac{1-(-1)^k}{2}a_k.$$
Taking $$a_k=\binom{2n}{k}$$ yields
\begin{align}
\sum_{k\ge0} \binom{2n}{2k+1}
&= \sum_{k\ge0} \frac{1-(-1)^k}{2} \binom{2n}{k} \\
&= \frac{1}{2}\sum_{k\ge0} \binom{2n}{k} - \frac{1}{2}\sum_{k\ge0} (-1)^k\binom{2n}{k} \\
&= \frac{1}{2}(1+1)^{2n} - \frac{1}{2}(1-1)^{2n} \\
&= \frac{4^n-[n=0]}{2}.
\end{align}
|
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|
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.
But I think it is a wrong method to proceed...
|
Let $x=b+c$, $y=a+c$ and $z=a+b$.
Thus, $a$, $b$ and $c$ are positives, $a+b+c=1$ and
$$xy+xz+yz-xyz=\sum_{cyc}(a^2+3ab)\sum_{cyc}a-\prod_{cyc}(a+b)=$$
$$=\sum_{cyc}(a^3+a^2b+a^2c+3a^2b+3a^2c+3abc)-\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)>$$
$$>\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=(a+b+c)^3=1.$$
For $a=b\rightarrow\frac{1}{2}$ and $c\rightarrow0^+$ we obtain equality,
which says that $1$ it's an infimum.
Now, for $a=b=c=\frac{1}{3}$ we obtain a value $\frac{28}{27}.$
But $$\sum_{cyc}(a^3+a^2b+a^2c+3a^2b+3a^2c+3abc)-\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\leq\frac{28}{27}$$ it's
$$\sum_{cyc}\left(a^3+3a^2b+3a^2c+\frac{7}{3}abc\right)\leq\frac{28}{27}(a+b+c)^3$$ or
$$\sum_{cyc}(a^3+3a^2b+3a^2c-7abc)\geq0,$$ which is true by Muirhead.
Thus, $\frac{28}{27}$ it's a maximal value.
Since our expression is continuous, we got the answer: $$\left(1,\frac{28}{27}\right].$$
|
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|
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.
If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.
So, we cannot prove as follows:
$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
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Here's another way . . .
On the domain $[-1,1]$, let $f(x)=\sqrt{1-x^2}$ and let $g=f^2$.
If $f$ was differentiable at $x=1$ then by the power rule, the equation
$$
g'=2ff'
$$
would hold at $x=1$, but at $x=1$ the $\text{LHS}$ evaluates to $-2$ (since $g'(x)=-2x$) while the $\text{RHS}$ evaluates to $0$ (since $f(1)=0$).
|
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Antiderivative to $\int\frac{1}{(\cos x+\sin x)^2} \ dx$ I have tried the following:
$\int\frac{1}{(\cos x+\sin x)^2} \ dx \ = \int \frac{\sec^2x}{(\tan x+1)^2}\ dx \ $.
After using the substitution $t=\tan x$, I got the solution: $- \frac{1}{\tan x+1} + C$.
Wolfram alpha gives the solution: $\frac{\sin x}{\sin x+\cos x}$.
At the same time $\frac{\sin x}{\sin x+\cos x} \neq - \frac{1}{\tan x+1}$.
So I'm a bit confused.
|
$$
\begin{aligned}
\int \frac{1}{(\cos x+\sin x)^{2}} d x &=\int \frac{\sec ^{2} x}{(1+\tan x)^{2}} d x \\
&=\int \frac{d(\tan x)}{(1+\tan x)^{2}} \\
&=-\frac{1}{1+\tan x}+C \\
&=-\frac{\cos x}{\cos x+\sin x}+C\\
&=-\frac{\cos x+\sin x-\sin x}{\cos x+\sin x}+C\\
&=-1+\frac{\sin x}{\cos x+\sin x}+C\\
&=\frac{\sin x}{\cos x+\sin x}+C^{\prime}
\end{aligned}
$$
$\therefore$ the answers differ by only a constant.
|
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By application of calculus of residues, can you please solve this problem? By application of calculus of residues, prove that
$$
\int_{0}^{2\pi}
\frac{\cos^{3}\left(3\theta\right)}
{1 - 2p\cos\left(2\theta\right) + p^{2}}
\,\mathrm{d}\theta = \frac{\pi\left(1 - p + p^{2}\right)}{1 - p}
$$
I have attempted the above question using the substitution
$z = \mathrm{e}^{\mathrm{i} \theta}$ and I have also used the substitution
$\cos\left(x\right) =
\mathrm{e}^{\mathrm{i}\theta}$ but I didn't get the solution correctly.
Solution
\begin{equation*}
\text{Let}\, z=e^{i\theta}\, \hspace{2mm} dz=ie^id\theta,\,\hspace{2mm} dz=izd\theta\\
\end{equation*}
\begin{equation*}
\cos\theta=\frac{z+z^{-1}}{2}\\
\end{equation*}
\begin{equation*}
\cos2\theta=\frac{z^2+z^{-2}}{2}\\
\end{equation*}
\begin{equation*}
\cos3\theta=\frac{z^3+z^{-3}}{2}\\
\end{equation*}
\begin{equation*}
\begin{split}
\int_0^{2\pi} \frac{\cos ^3{3\theta}d\theta}{1-2p\cos 2\theta+p^2} &= \oint\frac{(\frac{z^3+z^{-3}}{2})^3}{1-2p(\frac{z^2+z^{-2}}{2})+p^2}\cdot\frac{dz}{iz}\\
&= \frac{1}{8}\oint\frac{-(z^3+z^{-3})^3i}{[z-p(z^3+z^{-1})+p^2z]}\cdot dz\\
&= \frac{1}{8}\oint\frac{(z^3+1)^3i}{z^8[pz^4+(p^2+1)z^2-p]}\cdot dz
\end{split}
\end{equation*}
There is a pole at z=0 of order 8.
Kindly help, to determine the remaining poles and the residues.
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{2\pi}
{\cos^{3}\pars{3\theta} \over
1 - 2p\cos\pars{2\theta} + p^{2}}
\,\dd\theta = \require{cancel}
\cancelto{0}{{\pi\pars{1 - p + p^{2}} \over
1 - p}}}:\ {\Large ?}}$.
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}
{\cos^{3}\pars{3\theta} \over
1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta}
\\[5mm] = &\
-\int_{-\pi}^{\pi}
{\cos^{3}\pars{3\theta} \over
1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta
\\[5mm] = &\
-2\int_{0}^{\pi}
{\cos^{3}\pars{3\theta} \over
1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta
\\[5mm] = &\
2\int_{-\pi/2}^{\pi/2}
{\sin^{3}\pars{3\theta} \over
1 + 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta =
\bbx{\large 0} \\ &
\end{align}
*
*because the integrand is an $\ds{\underline{odd\ function}}$
*and the integral is evaluated between
symmetric limits $\ds{\pars{~\mbox{i.e.}\ \pm{\pi \over 2}~}}$.
|
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Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$
What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?
|
Any number that is a square mod $100$ is necessarily a square both mod $4$ and mod $5$, which is to say $0$ or $1$ mod $4$ and $0$, $1$, or $4$ mod $5$. The only number in the $70$s that satisfies both criteria is $76$.
|
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Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\infty}\frac{(x(x+1)(x+2))^{\frac{1}{3}}-2(x+1)}{-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}}=1$$
Using the Hospital rule but it doesn't help here .Moreover I have tried power series without success .
Any helps is welcome .
Thanks in advance
|
Expanding at $\infty$ with Taylor formula, we get
$$x^{\frac{x}{3 x+3}} (x+1)^{\frac{x+1}{3 x+3}} (x+2)^{\frac{x+2}{3 x+3}}\sim 1+x;\;x\to\infty$$
and
$$\sqrt[3]{x (x+1) (x+2)}\sim 1+x;\;x\to\infty$$
Therefore as $x\to\infty$ limit becomes
$$1+x-(2(x+1)-(1+x))=0$$
$$.....................................$$
To expand at $\infty$ we plug $1/x=y$ and expand at $y=0$
For instance
$$\sqrt[3]{x (x+1) (x+2)}=\frac{1}{y}\sqrt[3]{(y+1) (2 y+1)}$$
Expand at $y=0$
$$\sqrt[3]{(y+1) (2 y+1)}= 1+y+O\left(y^2\right)$$
plug again $y=1/x$ to get
$$\sqrt[3]{x (x+1) (x+2)}=x\left(1+\frac{1}{x}\right)+O\left(1/x^2\right)$$
therefore
$$\sqrt[3]{x (x+1) (x+2)}\sim 1 + x;\,x\to\infty$$
|
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Find $a$ and $b$ such that probability is the biggest Basketball player A hit the basket with probability $\frac{1}{2}$, and B with $\frac{1}{10}$. Player A can shoot $a$ times, and B $b$ times, $a+b=20$. Find $a$ and $b$ such that probability to make at least one point each is the biggest.
I just begging my course in probability and I have a problem with this. I tried to find max value of $(1-\frac{1}{2^{a}})(1-\frac{9^{20-a}}{10^{20-a}})$(complement of event that some of players doesn't make a point), and get solution a=1, b=19, but solution in book is a=5,b=15.
Sorry for my bad English, any help?
|
Denote the possible events of no hits/hits by (no hits, hits). We get the following sequence for A:
$E_A=(0,1),(1,1),(2,1),....,(a-1,1)$ and similarly for b: $E_B=(0,1),(1,1),(2,1),....,(b-1,1)$. The probablities are:
$$
P(E_A)=(\frac{1}{2})^0(\frac{1}{2})^1+(\frac{1}{2})^1(\frac{1}{2})^1+(\frac{1}{2})^2(\frac{1}{2})^1...+(\frac{1}{2})^{a-1}(\frac{1}{2})^1\\=1-\frac{1}{2^a}\\
P(E_B)=(\frac{9}{10})^0(\frac{1}{10})^1+(\frac{9}{10})^1(\frac{1}{10})^1+(\frac{9}{10})^2(\frac{1}{10})^1...+(\frac{9}{10})^{b-1}(\frac{1}{10})^1\\=1-(\frac{9}{10})^b=1-(\frac{9}{10})^{20-a}
$$
Thus we have to maximize
$$P(a)=P(E_A)P(E_B)=(1-\frac{1}{2^a})(1-(\frac{9}{10})^{20-a})$$
Numerically I found that the maximum of $P(a)$ is at $a=4.764$. So the closest integer is:
$a=5$
and we have $b=15$
|
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Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.
I've already tried to write it like this:
$$\frac{\sqrt{1}}{1} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 3$$
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 4$$
but I don't know that to do next or if it's the right way prove this.
|
Have you tried using induction?
Let's assume the following proposition is true for $n$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
By induction we must prove it works for the case $n=1$ and for all the cases $n+1$.
$P(1)$:
The base case is trivial, since
$$\frac{1}{\sqrt{1}} = 1 \leq 3\sqrt{2} -3 = 3(\sqrt2 - 1)$$
Since $\sqrt2 > 1$ it is easy to see why it is true.
Now we must bring our proposition $p(n) := \sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$ to test for all $n+1$ and see if it holds.
$P(n) \Rightarrow P(n+1)$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} + (\frac{1}{\sqrt{n+1}}) \leq 3\sqrt{n+1} -3 + (\frac{1}{\sqrt{n+1}})$$
$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3 + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2} - 3 $$
Therefore
$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+2} - 3 $$
You can check that $3\sqrt{n+1} + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2}$ is true for all $n$. Hence we proved by induction that the inequality $P(n)$ is true.
|
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prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.
The case $a\ge c\ge b$ is analogous.
Thus Proved!
Is it correct?...And any other alternative ways possible?
|
Your application of rearrangement is correct, in either case, $(a, b, c)$ and $(a^2+a, b^2+b, c^2+c)$ are similarly ordered, so
$$\sum_{cyc} \frac{a}{a^2+2a} \leqslant \sum_{cyc} \frac{a}{b^2+2b}$$
For another way, which generalises, consider
$$f(x) = \sum_{cyc} \frac{a+x}{b+x}, \quad f'(x) = \sum_{cyc} \frac{b-a}{(b+x)^2} = \sum_{cyc} \frac{b}{(b+x)^2} - \sum_{cyc}\frac{a}{(b+x)^2} \leqslant 0$$
again by Rearrangement. Hence $f$ is decreasing, and $f(0) \geqslant f(2)$
|
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Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$? Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?
$\int \sqrt{a^2-x^2}dx = \int \sqrt{a^2 - a^2 \cos^2 \theta} d\theta = \int a \sin\theta d\theta = \int a \sin\theta(-a \sin\theta)d\theta = -a^2 \int sin^2 \theta = -a^2[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}]+C$
However, the correct answer seems to be $a^2(\frac{\theta}{2} + \frac{\sin(2\theta)}{4}) + C$
|
$I(x)$ integral done by two different methods yields $I_1(x), I_2(x)$. These two differ only by a constant independent of $x$: $I_1(x)-I_2(x)=$Cosnstant, eventually.
In @lab Bhattacharjee 's answer above
$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}-\frac{a^2}{2} \cos^{-1}(x/a)$$
and$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2} \sin^{-1}(x/a)$$
So $$I_1(x)-I_2(x)=\frac{a^2}{2}[-\cos^{-1}(x/a)-\sin^{-1}(x/a)]=-\frac{a^2\pi}{4}$$
|
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The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$
So let $EC = k$ , $CD = DE = \frac{(\sqrt{3} + 1)k}{2}$ .
From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$
Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$
$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$
This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?
Thank You.
|
$DC=x$ and the altitude of $\Delta DEC$ from $E$ is $\frac{x}{2}$. This gives us
$$[\Delta DEC] = \frac{x^2}{4}$$
And the result
$$\frac{[\Delta ADE]}{[\Delta DEC]}=\sqrt{3}$$
|
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Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$
|
The sequence is
$$1=1 \ , \ 3+5 = 8 \ , \ 7+9+11 = 27 \ , \ \dots$$
So you can see that
$$a_n = \sum_{k=1}^{1+2+...+n} (2k-1) - \sum_{k=1}^{n-1} a_k =$$
$$\sum_{k=0}^{n(n+1)/2} (2k-1) - a_{n-1} =\frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} a_k$$
because you can view the $n$-th term as the sum of $1+2+\dots+n$ odd numbers minus the sum of all the other previous terms ($27 = 1+3+5+7+9+11 - (1)-(3+5)$).
So the base step is clear, now for the inductive one we use strong induction (we say it works from $1$ to $n-1$) then
$$a_{n} = \frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} k^3 = \frac{1}{4}n^2(n+1)^2 -\frac{1}{4}(n-1)^2n^2 = n^3$$
so the formula is proven.
|
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On the number of prime divisors of a product of 4 consecutive integers In PEN (Problems in Elementary Number Theory) Project, there is a problem A111 as follows
Find $n$ if $n(n+1)(n+2)(n+3)$ has exactly three distinct prime divisors.
It's easy to see that at least 2, 3 should be two of the prime divisors of $n(n+1)(n+2)(n+3)$, and
I think there are infinitely many $n$ satisfying the above property, but I haven't found out the general answer.
Please help me.
Thanks.
|
First not ethat $n=1$ is only almost a solution as $n(n+1)(n+2)(n+3)$ has only two prime divisors. Hence we ma assume $n>1$.
Let $p$ denote the third prime.
If $n$ is a multiple of $3$, then so is $n+3$. Then $n+1, n+2$ are divisible only by $2$ and $p$, and non of them is divisible by both. In other words, we have either
$$\tag1 n=3^a,\quad n+1=2^b,\quad n+2=p^c,\quad n+3=2^k3^d$$
or
$$\tag2 n=2^k3^a,\quad n+1=p^c,\quad n+2=2^b,\quad n+3=3^d.$$
Note that one of $a,d$ must be $=1$, and one of $b,k$ must be $=1$. This boils down to the following variants:
*
*$(1)$ with $a=1$, so $n=3$, $n+1=4$, $n+2=5$, $n+3=6$, which is indeed a solution
*$(1)$ with $b=1$ leads to $n=1$, so no.
*$(1)$ with $d=k=1$, so $n=3$ - see above
*$(2)$ with $d=1$ or $b=1$ leads to $n=0$, so no
*$(2)$ with $a=k=1$, so $n=6$, $n+1=7$, $n+2=8$, $n+3=9$, which is a solution.
Next, assume that $n$ is not a multiple of $3$. Then exactly one of the four numbers is a multiple of $2$, exactly two are multiples of $2$ and exactly one is a multiple of $p$. To cover four numbers this way, we must have that each of the for numbers is a prime power. In particular, the two powers of $2$ must be $2$ and $4$. This leaves us with $n=2$, $n+1=3$, $n+2=4$, $n+3=5$, which is a solution.
In summary, $n$ is a solution iff
$$ n\in\{2,3,6\}.$$
|
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Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges How would you show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?
My approach
At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\right)^n}{n}$ is a monotone sequence with $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$. So after a few algebraic manipulations I get:
$$
\frac{\frac{\left(1+\frac{1}{n}\right)^n}{n}}{\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{n+1}}=
\frac{\left(1+\frac{1}{n}\right)^n(n+1)}{\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)n}=
\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n}\cdot \frac{n+1}{\frac{n^2+2n}{n+1}}=
\cdots=\frac{n^2+2n+1}{n^2+2n}\left(\frac{n^2+2n+1}{n^2+2n}\right)^n\geq1.
$$
So the expression is monotonically decreasing. Further, we know that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^1$. Hence, $\left(1+\frac{1}{n}\right)^n$ is bounded, say $S\geq\Big| \left(1+\frac{1}{n}\right)^n\Big|$. This allows us to conclude:
$$
\Big| \frac{\left(1+\frac{1}{n}\right)^n}{n}-0\Big|\leq \frac{S}{n}<\epsilon,\text{ where n is sufficiently large enough}.
$$
Hence, $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$ and by the alternate series crterion the series $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges.
Is this correct? Is there a faster way?
|
Since
$$
2\le\left(1+\frac1n\right)^n\le3
$$
and
$$
\sum_{n\ge1}\frac{(-1)^n}n=\log2
$$
Calling $S$ the desired limit (that is, your sum), you can immediately get that
$$
2\log 2\le S\le3\log 2\;.
$$
|
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Allowed values of $x$ in $x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$ when $y=0$? Can someone please check the given explanation? I have this two-variable equation
$$x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$$
for $0\leqslant x,y<\infty$, and I want to check which values of $x$ satisfy this equation when $y=0$.
From the above equation, I see that when $y=0$, it is fulfilled for all $0\leqslant x<\infty$.
Then, I rewrite the equation as
$$\frac{x}{3 x^2-5}=\frac{1}{2} \tan (3 y),$$
from here, I see that when $y=0$, there are only two solutions for $x$, i.e. $x=0$ and $x\to \infty$.
*
*Then, what can I say in general about the solutions of $x$ when $y$ is zero?
*And what can we say about the point $x=\sqrt{\frac{3}{5}}$? Is it an acceptable solution for $3y=\frac{\pi }{2}$? Since from the first equation, it can be a solution, but I have doubts about it from the second form of the equation.
|
When you "rewrite" an equation by multiplying or dividing by some factor, the new equation is equivalent to the old when that factor is not $0$. When it is $0$, the two are generally not equivalent.
So, your first equation
$$ x \sin 6 y - (3 x^2 - 5) \sin^2 3 y $$
is indeed always true when $y = 0$, or more generally when $\sin 3 y = 0$. When you divide by $(3 x^2 - 5) \sin 6 y$ to get your second equation
$$ \frac{x}{3x^2-5} = \frac{1}{2} \tan 3y $$
this is equivalent to the first as long as $3 x^2 - 5 \ne 0$ and $\sin 6 y \ne 0$. When
$\sin 3y = 0$, the right side is $0$ but the left is not (unless $x=0$). When $3x^2 - 5 = 0$ the left side is undefined but the right side is defined (unless $\cos 3 y = 0$).
When $\cos 3y = 0$ the right side is undefined but the left side is defined (unless $3 x^2-5 = 0$).
|
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Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$
Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$
From here I don't know how to proceed...
|
Your method might work, but I couldn't really see how to proceed. A substitution that I am sure would work is as follows:
First expand your polynomial: $x^4-(m^2+2m+2)x^2+m^2+2m+1$.
Next, you can use your $a=x^2$ substitution: $a^2-(m^2+2m+2)a+m^2+2m+1$. Then, you use essentially the same $b$ substitution, but expanded: $b=m^2+2m+1$. Once you do this you get $a^2-(b+1)a+b$, you solve and get $a=1, a=b$ (verify yourself that they work). Then substitute back $a$ and $b$: $x^2=1, x^2=m^2+2m+1$, so your solutions for $x$ are: $x=1$, $x=-1$, $x=m+1$, $x=-m-1$. Therefore, since an extremely large or small value of $x$ will make the equation positive, your range of x-values are $(-\infty, -|m|]\cap[-1, 1]\cap[|m|, \infty)$. Note when $m=-2$, your range of x-values are $(-\infty, \infty)$
Edit: My steps are essentially your steps (expanded out), but I didn't use quadratic formula like on the fourth step, but factored it.
Using what you have:
$ a = x^2 $
$ b = (m + 1)^2$
$ a^2 - (1 + b)a + b \geq 0$
$a=b, a=1$ are solutions to the inequality, and solve for bounds
The rest is the same.
|
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Problem of minimum in physics.
It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.
Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.
I've called $x$ the distance of A from P.
The function is $$t(x)=\frac{x}{v_1}+\frac{\sqrt{(a-x)^2+b^2}}{v_2}$$
$$\frac{dt}{dx}=\frac{1}{v_1}+\frac{(a-x)*(-1)}{v_2* \sqrt{(a-x)^2+b^2}}=0 \Rightarrow v_2*\sqrt{(a-x)^2+b^2}=v_1*(a-x) \Rightarrow x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2)=0$$
The solutions of this equation are $x_{1/2} =a \pm \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $
Putting $\frac{dt}{dx}>0$ it is verified for $0<x<a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and for $x>a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$.
So the minimum point is for $a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$
and the minimum time is
$$t(a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b (v_1^2+v_2^2)}{v_1v_2 \sqrt{v_1^2-v_2^2}}+\frac{a}{v_1}$$
I'm not sure that I haven't done mistakes, because in the solution on the book it indicates
$a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $ as the solution of minimum.
and the minimum time
$$t(a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}+\frac{a}{v_1}$$
but, in particular , it distinguishes the case of $\frac{v_2^2}{v_1^2}\ge \frac{a^2}{a^2+b^2}$
in which the minimum time is possible walking from A to B directly.
|
Your calculations are mostly correct, you just have to interpret the results correctly.
You have a continuous function $t : \Bbb{R} \to \Bbb{R}$ and hence it must attain its minimum on the compact set $[0,a]$ in some point $x_{0} \in [0,a]$.
Moreover, $t$ is differentiable on $\langle 0,a\rangle$ and if it happens that $x_0 \in \langle 0,a\rangle$ then $x_0$ is a local extremum of $t$ so in particular $t'(x_0) = 0$. Assuming $v_1 > v_2$ (if $v_1 \le v_2$, then simply $x_0 = 0$), you correctly found that the possible zeroes of $t'$ are included in $\left\{a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+ \frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}$ (as @David K pointed out, the only actual zero of $t'$ is $a- \frac{bv_2}{\sqrt{v_1^2-v_2^2}}$ but it doesn't matter here). Therefore, we conclude that $$x_0 \in \left\{0,a, a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}},a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right\}.$$
However, we are interested only in $x_0 \in [0,a]$. Notice that $a+\frac{bv_2}{\sqrt{v_1^2-v_2^2}} > a$ so this option is certainly wrong.
Moreover, we have
$$a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a] \iff \frac{bv_2}{\sqrt{v_1^2-v_2^2}} \le a \iff \frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}.$$
Hence if $\frac{v_2}{v_1} \le \frac{a}{\sqrt{a^2+b^2}}$, then $a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}} \in [0,a]$ and this is indeed the minimum point. Namely,
$$t\left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}$$
which is smaller than both $t(0)$ and $t(a)$. We can use Cauchy-Schwarz:
$$\frac{a}{v_1}+\frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}\le \sqrt{a^2+b^2}\sqrt{\frac{1}{v_1^2}+\frac{v_1^2-v_2^2}{v_2^2}} = \frac{\sqrt{a^2+b^2}}{v_2} = t(0),$$
$$\frac{a}{v_1}+\frac{b}{v_2} \cdot \underbrace{\frac{\sqrt{v_1^2-v_2^2}}{v_1}}_{\le 1} \le \frac{a}{v_1}+\frac{b}{v_2} = t(a).$$
On the other hand, if $\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}}$, then this point is not in $[0,a]$ so it has to be simply $x_0 = 0$ or $x_0 = a$. It is not hard to see that
$$t(0) \le t(a) \iff \frac{\sqrt{a^2+b^2}}{v_2} \le \frac{a}{v_1}+\frac{b}{v_2} \iff \frac{v_2}{v_1} \ge \frac{\sqrt{a^2+b^2}-b}a$$
and the latter is true in our case since
$$\frac{v_2}{v_1} > \frac{a}{\sqrt{a^2+b^2}} \ge \frac{\sqrt{a^2+b^2}-b}a.$$
So it has to be $x_0 = 0$.
Therefore:
$$x_0 = \begin{cases}
a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}, &\text{ if } \frac{v_2}{v_1} \in \left\langle 0, \frac{a}{\sqrt{a^2+b^2}}\right] \\
0, &\text{ if } \frac{v_2}{v_1} \in \left[\frac{a}{\sqrt{a^2+b^2}}, 1\right\rangle \\
\end{cases}$$
If $v_1 >> v_2$ then $x_0$ approaches $a$, namely
$$\lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = \lim_{\frac{v_2}{v_1} \to 0} \left(a-\frac{b}{\sqrt{\left(\frac{v_1}{v_2}\right)^2-1}}\right) = a$$
but it is always faster to go at least somewhat diagonally than straight left and then straight up.
On the other hand, if $v_1$ approaches $v_2$ then you may wonder why
$$\lim_{\frac{v_2}{v_1} \to 1} \left(a-\frac{bv_2}{\sqrt{v_1^2-v_2^2}}\right) = -\infty$$
and not $0$ but this expression stops being relevant as soon as $\frac{v_2}{v_1} \ge \frac{a}{\sqrt{a^2+b^2}}$ as is it is faster to simply go diagonally.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating matrix exponential
Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?
I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as
$$P(\lambda)=\lambda^2+89$$
Is there an easier way to do this than trying to compute the diagonalized matrix?
|
Your matrix $M$ is diagonalizable with eigenvalues $\pm i\sqrt{89}$. This means that $e^M = p(M)$ where $p \in \Bbb{C}[x]$ is the unique polynomial of degree less than $2$ such that
$$p(i\sqrt{89}) = e^{i\sqrt{89}}, \quad p(-i\sqrt{89}) = e^{-i\sqrt{89}}.$$
Using Lagrange interpolation formula, we see that
$$p(x) = \frac{x+i\sqrt{89}}{2i\sqrt{89}}e^{i\sqrt{89}}-\frac{x-i\sqrt{89}}{2i\sqrt{89}}e^{-i\sqrt{89}} = \frac{\sin\sqrt{89}}{\sqrt{89}}x+ \cos\sqrt{89}$$
so
$$e^M = p(M) = \frac{\sin\sqrt{89}}{\sqrt{89}}M+ \cos\sqrt{89}I = \left(
\begin{array}{cc}
\frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89}& -\frac{(6+2 i) \sin \sqrt{89}}{\sqrt{89}} \\
\frac{(6-2 i) \sin \sqrt{89}}{\sqrt{89}} & -\frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89} \\
\end{array}
\right).$$
|
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|
Solving equation $a_n=1+\frac12a_{n-1}+\frac12a_{n+1}$ Let's consider sequence following for $n=0...,T$ for which $a_0=a_T=0$. I want to solve equation $$a_n=1+\frac12a_{n+1}+\frac12a_{n-1}$$
My work so far
$$a_1=1+\frac12a_2 \Leftrightarrow a_2=2a_1-2$$
$$a_2=1+\frac12a_3+\frac12a_1 \Leftrightarrow2a_1-2=1+\frac12a_3 \Leftrightarrow a_3=3a_1-6$$
Also I calculated $a_4$ to get $a_4=4a_1-12$. So my observation is that for any $n \in \{1,...,T-1\}$
we have $$a_n=na_1+n(n-1)$$
And here I stacked, I'm not sure what to do next. Could you please give me a hand solving this equation ?
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Consider the generating function $$f(x)=\sum_{n\ge0}a_nx^n,\qquad|x|<1.$$
We may write
$$2a_{n+1}x^{n+2}=2x^{n+2}+a_{n+2}x^{n+2}+a_{n}x^{n+2},$$
and sum over $n\ge 0$:
$$2x\sum_{n\ge0}a_{n+1}x^{n+1}=2x^2\sum_{n\ge0}x^n+\sum_{n\ge0}a_{n+2}x^{n+2}+x^2\sum_{n\ge0}a_nx^n,$$
which is
$$2x(f(x)-a_0)=\frac{2x^2}{1-x}+f(x)-a_0-a_1x+x^2f(x).$$
Using $a_0=0$, and treating $a_1$ as a constant,
$$(x^2-2x+1)f(x)=\frac{(a_1+2)x^2-a_1x}{x-1},$$
so that
$$f(x)=\frac{(a_1+2)x^2-a_1x}{(x-1)^3}.$$
Then let the sequences $p_n$ and $q_n$ be defined by
$$\begin{align}
p(x)=\frac{x^2}{(x-1)^3}&=\sum_{n\ge0}p_nx^n\\
q(x)=\frac{x}{(x-1)^3}&=\sum_{n\ge0}q_nx^n,
\end{align}$$
so that
$$f(x)=(a_1+2)p(x)-a_1q(x)=\sum_{n\ge0}((a_1+2)p_n-a_1q_n)x^n,$$
giving $a_n=(a_1+2)p_n-a_1q_n$.
Differentiating the series $\frac{1}{1-x}=\sum_{n\ge0}x^n$ three times gives
$$\frac{1}{(x-1)^3}=\sum_{n\ge0}x^n\left(-\frac{1}{2}(n+1)(n+2)\right),$$
so that
$$\begin{align}
p(x)=\frac{x^2}{(x-1)^3}&=\sum_{n\ge0}x^n\left(-\frac{1}{2}n(n-1)\right)\quad &\Rightarrow p_n=-\frac12n(n-1)\\
q(x)=\frac{x}{(x-1)^3}&=\sum_{n\ge0}x^n\left(-\frac{1}{2}n(n+1)\right)\quad &\Rightarrow q_n=-\frac12n(n+1).
\end{align}$$
Therefore
$$a_n=(a_1+2)p_n-a_1q_n=n(a_1+1-n).$$
To find $a_1$, we use $a_T=0$. Specifically we have
$$T(a_1+1-T)=0\Rightarrow a_1=T-1.$$
Thus we have $$a_n=n(T-n).$$
|
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|
Showing $\frac{f(a+2b)+f(a-2b)}{2f(a)}\leq\sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$ for $f(x)=\sqrt{x(1-x)}$ with some constraints My question is from a inequality that is not proved (it is just implicitly mentioned I guess) in a book.
Specifically, let $a \in (0,1)$ and $b \in (0,1)$ with $a - 2b \geq 0$ and $a + 2b \leq 1$. Then with $f(x) := \sqrt{x(1-x)}$ we are asked to show that $$\frac{f(a+2b)+f(a-2b)}{2f(a)} \leq \sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$$
I am really stuck at this, the only related inequality that comes to mind is that $$\frac{\sqrt{1+2\alpha}+\sqrt{1-2\alpha}}{2} \leq \sqrt{1-\alpha^2},\quad \forall \alpha \in [0,\tfrac{1}{2}]$$
But I still cannot prove the desired inequality.
Any help is greatly appreciated!
|
I will just bash it. Let's work on the square of the LHS first. You will actually find that:
$$f^2(a+2b) + f^2(a-2b) = 2a(1-a) - 8b^2 = 2f^2(a) - 8b^2$$
and
$$f(a+2b)f(a-2b) = \sqrt{(a^2-4b^2)((1-a)^2 - 4b^2)} = \sqrt{(f^2(a)+4b^2)^2 - 4b^2}$$
Therefore, the squared version of your inequality is:
$$LHS^2 = \dfrac{f^2(a) - 4b^2 + \sqrt{(f^2(a)+4b^2)^2- 4b^2}}{2f^2(a)}\leq 1 - \dfrac{b^2}{f^4(a)}=RHS^2$$
Now it looks like some nice substitution will simplify things:
$$\begin{cases}
f^2(a) = a(1-a) = x\in(0,\frac 14) \\
4b^2 = y\in (0,\frac 14)
\end{cases}.$$
Then, you have:
$$\dfrac{x-y+\sqrt{(x+y)^2-y}}{2x}\leq1 - \dfrac{y}{4x^2}\iff \dfrac{(x+y)^2-y}{4x^2}\leq \left(\frac 12-\frac{y}{4x^2}+\frac{y}{2x}\right)^2$$
and after a couple line of simplifying, you will get:
$$y(1-x)(yx-y+4x^2+2xy) + 4x^2y\geq 0$$
which is trivially true. There is no equality attained in the domain you specified, but $b =y= 0$ gives an equality.
P.S: I did try an naive AM-GM in the beginning, but that resulted in $a(1-a)\geq\frac 14$, which is only true when $a=\frac 12$, so some cute AM-GM is probably not gonna cut it.
|
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|
Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$ We have to prove the convergence and find the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$. I tried to find $a_{n+1}$ in terms of $a_{n}$ and got $a_{n+1}=\frac{n+2}{2(n+1)}\left(a_n+1\right)$ and could not proceed further.Any help is appreciated.
Edit : Following K.defaoite's answer limit appears to be 1 . But we still need to prove convergence.
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OK, I see what needs to be done here. Denote the sum in brackets by $S_n$
$$S_n = \left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$$
First we prove by induction (for $n \ge 4$) the following:
$$S_n \gt \frac{ 2^{n+1} (n+2)}{(n+1)n} \tag{1}$$
Then using (1) it is easy to prove by induction that:
(2) the sequence is decreasing for $n\ge4$ i.e. $a_n \gt a_{n+1}$
(3) $a_n \ge 1$ for $n\ge4$ (that is even true for every $n$ but for us it's enough that it's true for $n\ge4$)
Once these two statements are proved, the convergence follows.
And then you take limits (that's the easy part as K.defaoite showed) and you're done.
Here is how using (1) we can prove the sequence is decreasing: we have $a_n=\dfrac{n+1}{2^{n+1}}S_n$ and $a_{n+1}=\dfrac{n+2}{2^{n+2}}S_{n+1}$ $=\dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$. Now, the condition that $a_n\gt a_{n+1}$ is that $\dfrac{n+1}{2^{n+1}}S_n\gt \dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$; this is equivalent to $2(n+1)S_n\gt (n+2)(S_n+\dfrac{2^{n+1}}{n+1})$, which is equivalent to $nS_n\gt \dfrac{(n+2)2^{n+1}}{n+1}$ (subtract $(n+2)S_n$ from either side), and dividing this by $n$ gives (1).
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|
What is the solution of $x^3+x=1$? According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:
$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$
$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way. Am I right?
|
$$ \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3} $$
|
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|
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is
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The centroid $G$ belongs to the line
$\mathcal{L}:$ $y=x+1$.
Since the circle $\mathcal{C}$ must pass through
$G(2,3)$ and $D(1,1)$, it must also pass through
the point $K(4,4)$,
which is a reflection of $D$ wrt the perpendicular through $G$.
Hence, the sought circle is the circumcircle of $\triangle GDK$
with the side lengths
\begin{align}
|GD|=|GK|&=\sqrt5
,\\
|DK|&=3\sqrt2
,
\end{align}
the area
\begin{align}
S_{GDK}&=
\tfrac12\,|DK|\,\sqrt{|GD|^2-\tfrac14\,|DK|^2}
=\tfrac32
\end{align}
and the circumradius
\begin{align}
R&=\frac{|GD|^2|DK|}{4S_{GDK}}
=\tfrac52\,\sqrt2
.
\end{align}
Since in this special simple case
$GO$ is the diagonal
of the $\tfrac52\times\tfrac52$ square,
the coordinates of the center of the circle $\mathcal C$
can be found as
\begin{align}
O&=G+(\tfrac52,\, -\tfrac52)
=(\tfrac92,\, \tfrac12)
.
\end{align}
In case of general $\triangle ABC$
with the side lengths $a,b,c$,
we could find
the center $O$
using a known expression
\begin{align}
O&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{a^2\,(b^2+c^2-a^2)+b^2\,(a^2+c^2-b^2)+c^2\,(b^2+a^2-c^2)}
\\
&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{16S_{ABC}^2}
,
\end{align}
which, of course, works in this special case as well
and provides the same result.
|
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|
Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$ I want to prove that
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
My approach is the most logical one, rewrite as follows:
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $
But the remaining series are both divergent so I get to a sort of "$ \infty - \infty = 1 $".
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Rewriting the term,
$\color{blue}{\left(\sum_{n=1}^{\infty} \frac{2}{4 n^{2} - 1}\right)}=\color{black}{\left(\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)\right)}$
This is the telescoping series:
$\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)=\left(1-\color{green}{\frac{1}{3}}\right)+\left(\color{green}{\frac{1}{3}}-\color{red}{\frac{1}{5}}\right)+\left(\color{green}{\frac{1}{5}}-\color{red}{\frac{1}{7}}\right)+\left(\color{blue}{\frac{1}{7}}-\color{red}{\frac{1}{9}}\right)+...=1$
|
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|
Integrating factor for the ODE $(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$ I was asked to find an integrating factor $\mu = \mu(x+y^2)$ for the ODE
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0.$$
So the natural approach was define $P = 3x+2y+y^2$ and $Q = x+4xy+5y^2.$ Then,
$$\partial_yP = 2+2y,~~~\partial_xQ = 1+4y$$
and hence $\partial_yP - \partial_xQ = 1-2y.$ Let us consider then
$$\frac{\partial_yP-\partial_xQ}{Q} = \frac{1-2y}{x+4xy+5y^2}$$
and define $z = x+y^2.$ Thus,
$$\frac{\partial_yP - \partial_xQ}{Q} = \frac{1-2y}{(z-y^2)(1+4y)+5y^2}.$$
But from here I really don't know how to proceed. Any hint?
|
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$$
That you can rewrite as:
$$(x+y^2)dx+2(x+y)dx + (x+y^2)dy+4y(x+y)dy = 0$$
$$(x+y^2)d(x+y)+2(x+y)(dx+2ydy) = 0$$
$$(x+y^2)d(x+y)+2(x+y)d(x+y^2) = 0$$
$$\dfrac {d(x+y)}{x+y}+2\dfrac {d(x+y^2)}{x+y^2} = 0$$
Integrate.
$$(x+y)(x+y^2)^2 = C$$
|
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|
What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$? Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?
I know there exists a real number $a\in(1,2)$ such that $\frac{1}{3}x=x^{\frac {1}{3}}$, but there are three variables even though they are all positive.
Also I know how to simplify $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3-x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, but then I get stuck too.
I also tried to relate this to the roots of polynomial, but I don't quite know how.
Could anyone help please?
|
For what it's worth, you can prove this inequality without knowing the more general result: write $x_i=y_i^3$ so$$\frac{y_1^3+y_2^3+y_3^3}{3}-y_1y_2y_3=\tfrac16(y_1+y_2+y_3)((y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2).$$
|
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|
Combination of a piecewise defined function and a square function Let's say I have two functions, $f(x)=x^2$, and:
$$g(x)=
\begin{cases}
- 4 & x \leq 0\\
|x- 4| & x > 0
\end{cases}$$
Now I have to write combinations $f\circ g$ and $g\circ f$.
I think $f\circ g$ should be:
$$f\circ g(x)=
\begin{cases}
16 & x \leq 0\\
(x- 4)^2 & x > 0
\end{cases}$$
and $g\circ f$ should be $g\circ f(x)= |x^2- 4|$, considering that $x$ will always be positive, but now I don't get how would I graph these composite functions, and also how to check continuity at $x=0$, especially of $g\circ f$ since I just eliminated half part of it.
|
$f(x) = x^2$
$$
g(x) = \begin{cases}
-4 & \text{if \(x \leq 0\)}\\
|x - 4| & \text{if \(x > 0\)}
\end{cases}
$$
You correctly found that
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
& = \begin{cases}
f(-4) & \text{if \(x \leq 0\)}\\
f(|x - 4| & \text{if \(x > 0\)}
\end{cases}
\\
& = \begin{cases}
16 & \text{if \(x \leq 0\)}\\
|x - 4|^2 & \text{if \(x \geq 0\)}
\end{cases}
\\
& = \begin{cases}
16 & \text{if \(x \leq 0\)}\\
(x - 4)^2 & \text{if \(x \geq 0\)}
\end{cases}
\end{align*}
Since
\begin{align*}
\lim_{x \to 0^+} (f \circ g)(x) & = 16\\
\lim_{x \to 0^-} (f \circ g)(x) & = 16\\
\end{align*}
we obtain
$$\lim_{x \to 0} (f \circ g)(x) = 16$$
For a function to be continuous at $x = 0$, the limit as $x$ approaches $0$ must exist and the function must be equal to its limit when $x = 0$. Since $(f \circ g)(0) = 16$,
$$\lim_{x \to 0} (f \circ g)(x) = (f \circ g)(0)$$
Thus, the function $f \circ g$ is continuous at $x = 0$.
Since $x^2 > 0$ unless $x = 0$, where $f(0) = 0$, we obtain
\begin{align*}
(g \circ f)(x) & = g(f(x))\\
& = g(x^2)\\
& = \begin{cases}
-4 & \text{if \(x^2 \leq 0\)}\\
|x^2 - 4| & \text{if \(x^2 > 0\)}
\end{cases}
\\
& = \begin{cases}
-4 & \text{if \(x = 0\)}\\
|x^2 - 4| & \text{otherwise}
\end{cases}
\end{align*}
Observe that
\begin{align*}
\lim_{x \to 0^+} (g \circ f)(x) = 4\\
\lim_{x \to 0^-} (g \circ f)(x) = 4
\end{align*}
Hence,
$$\lim_{x \to 0} (g \circ f)(x) = 4$$
Thus, the limit of the function exists at $x = 0$. However, $(g \circ f)(0) = -4$. Since
$$\lim_{x \to 0} (g \circ f)(x) \neq (g \circ f)(0)$$
the function $g \circ f$ is not continuous at $x = 0$ because it is not equal to its limit. Since the continuity could be removed by defining $(g \circ f)(0) = 4$, which would require defining $g(x) = 4$ when $x = 0$, the function $g \circ f$ is said to have a removable discontinuity at $x = 0$.
|
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|
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}
For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 - 5n^2=4$ or $m^2 - 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?
Thank you very much in advance for your attention.
|
The number field $K=\Bbb Q(\sqrt{5})$ has class number one because its Minkowski bound satisfies $B_K<2$ . Hence its ring of integers $\mathcal{O}_K=\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is even a PID and hence a UFD.
On the other hand, it is enough to see that $\mathcal{O}_K/(4+\sqrt{5})$ is a field, so that the ideal $(4+\sqrt{5})$ is prime.
|
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|
How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$? I need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
|
Hint:
$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$
|
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|
What is the asymptotic expansion of $x_n$ where $x_{n+1} = x_n+1/x_n$? Let
$x_{n+1} = x_n+1/x_n,
x_0 = a \gt 0$
and
$y_n = x_n^2$.
What is the asymptotic expansion
of $x_n$ ($y_n$ will do)?
I can show that
$y_n
=2n+\dfrac12 \ln(n) + O(1)
$.
Is there an explicit form
for the constant
implied by the $O(1)$?
What is the asymptotic form
of the terms following that constant
(e.g.,
$O(\frac{\ln(n)}{n}),
O(\frac1{n}), ...
$)?
|
Not a complete answer. Let me first reproduce what I imagine is your argument. We have $y_{n+1} = y_n + \frac{1}{y_n} + 2$. In particular $y_{n+1} \ge y_n + 2$ which gives $y_n \ge 2n + a^2$ and hence $\frac{1}{y_n} \le \frac{1}{2n + a^2}$. This gives
$$y_{n+1} \le y_n + 2 + \frac{1}{2n + a^2}$$
which gives
$$y_n \le 2n + \sum_{i=0}^{n-1} \frac{1}{2i + a^2} + a^2 = 2n + \frac{1}{2} \log n + C$$
for some constant $C$. Write $y_n = 2n + \frac{1}{2} H_{n-1} + e_n$, where we now know that $e_n$ is bounded from above by a constant (and it's not hard to show that it's bounded from below by a constant also). Then the recurrence relation gives
$$y_{n+1} - y_n = 2 + \frac{1}{2n} + e_{n+1} - e_n = \frac{1}{2n + \frac{1}{2} H_{n-1} + e_n} + 2$$
and rearranging a bit gives
$$e_n - e_{n+1} = \frac{1}{2n} - \frac{1}{2n + \frac{1}{2} H_{n-1} + e_n} = \frac{ \frac{1}{2} H_{n-1} + e_n}{2n \left( 2n + \frac{1}{2} H_{n-1}+ e_n) \right)}.$$
Heuristically this gives something like $e_n = C + \frac{\ln n}{8n} + \dots$ but I don't know how to prove it off the top of my head, mostly because I can't think of a reasonable way to describe the constant $C$.
|
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|
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic
lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.
(1) $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$
(1) $868x+527y\equiv 558\pmod{41}$
(2) $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$
(2) $868x+308y\equiv 980\pmod{41}$
So finally we can subtract equation (1) from (2) we get:
$219y\equiv -422\pmod{41}$
lets check the $\gcd(219,41)$ by Euclidian algorithm :
$219 = 41\cdot 5 + 14$
$41= 14\cdot 2 + 13$
$14= 13\cdot 1 + 1$
$\gcd(219,41)=1$
Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.
$219y\equiv 1\pmod{41}$
$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.
I don't know how to continue from here .
|
$$-13x+17y\equiv 18 \pmod{41}\tag1$$
$$-10x+11y\equiv -6 \pmod{41}\tag2$$
$(1)\times 10-(2)\times 13$
$$
27y\equiv 12 \pmod{41} \implies y \equiv \frac{12}{27} \equiv \frac{4}{9} \equiv \frac{20}{45} \equiv \frac{20}{4} \equiv 5 \pmod{41}
$$
$$
x \equiv \frac{11y+6}{10}\equiv \frac{61}{10} \equiv \frac{20}{10} \equiv 2 \pmod{41}.\blacksquare
$$
|
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|
number of positive integer solutions of $x+y+z+w=24$ Determine the number of positive interger solutions of $x+y+z+w=24$ such that $x\leq 6, y \leq 7, z\leq 8, w\leq 9$
My try:
I used generating polynomial as
$$f(x)=(x+x^2+\cdots+x^6)(x+x^2+\cdots+x^7)(x+x^2+\cdots+x^8)(x+x^2+\cdots+x^9$$
$\implies$
$$f(x)=x^4(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$$
We need to collect the coefficient of $x^{20}$ in $(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$
So we can write $$(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}=(1-x)^{-4}(1-(x^6+x^7+x^8+x^9)+(x^{13}+x^{14}+2x^{15}+x^{16}+x^{17})-..)$$
Also using $$(1-x)^{-4}=\sum_{k=0}^{\infty}\binom{k+3}{k}x^k$$
We get required coefficient of $x^{20}$ as
$$\binom{23}{3}-\left(\binom{14}{11}+\binom{15}{12}+\binom{16}{13}+\binom{17}{14}\right)+\left(\binom{6}{3}+\binom{7}{3}+2\binom{8}{3}+\binom{9}{3}+\binom{10}{3}\right)=83$$
But the answer is not matching.
|
Your answer is right here is an easier way to go about this which may bring less calculation errors
set $x=6-u.y=7-v,z=8-t,w=9-p$
then with $0\le u\le 5,0\le v\le 6,0\le t\le 7,0\le p\le 8 $ we have to find the number of solutions of $$u+v+t+p=6$$ but $u\neq 6$ hence number of solutions is $\binom{6+4-1}{4-1}-1=83$
|
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|
critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$ so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2}\\ -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2} & \frac{8\,y^2\,\sin\left(x\right)}{{\left(y^2+1\right)}^3}-\frac{2\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
My first problem: What $x$ and $y$ do I plug in the hessian-matrix? If I solve $f'(x,y)$ for $(0,0)$ I get:
$x = \pi \cdot (k+1/2)$ or $x = \pi \cdot k$
Now, do I have to substitute these values for $x$ and solve for $y$? Seems like a huge case-differentiation to me.
...There must be an easier way, right?
|
Note that for:
$$f'(x,y) = (0,0) \iff \frac{cos(x)}{y^2+1} = 0 \quad \wedge \quad \frac{y\operatorname{sin}(x)}{(y^2+1)^2} = 0 $$
The first condition can only be achieved if $x\in \frac{\pi}{2}+\pi\mathbb{Z}$, and since cos and sin are linearly independent, then the only way they can both be 0 at the same time is for y to be equal to 0. Thus, the critical points are of the form $\lbrace \pi/2 + k\pi: k\in\mathbb{N}\rbrace\times\lbrace 0\rbrace$. Now you can just plug them in the Hessian
|
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|
(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$ Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).
Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in \mathbb Q[\sqrt[3]{7}]$$
Note: Application of the Euclidean algorithm to the polynomials $x^2 + 3x + 1$ and $x^3 - 7$ could help.
Attempt:
I know there is a multiplcative inverse $\beta$ with $\alpha\beta=1$, and that one should exist in $\mathbb Q[\sqrt[3]{7}]$, but do not know how I would go about expressing it in any form simpler than $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$.
How can I determine a simpler way to express the value of $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$?
|
Alternatively, $\alpha$ is a root of $x^3 - 3 x^2 - 60 x - 176$ and so its inverse is $\frac{1}{176}(\alpha^2-3\alpha-60)$.
|
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|
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos\beta)^n\int_0^{\frac{\pi}{2}} \cos^n\alpha{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos \beta) ^n\frac{\sqrt{\pi}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2\Gamma\left(\frac{n}{2}+1\right)}\\ &=\frac{\pi-2\arcsin\cos \beta}{2\sqrt{1-\cos^2\beta}}\\ &=\frac{\beta}{\sin \beta}. \end{align*}
Is it correct? How to obtain the fourth equlity?
|
$\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ = $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+k\cos\alpha}$
then use $cos\alpha$ = ($cos$ ^2)$\alpha/2$ - $(sin$^2)$\alpha/2$.
divide numerator and denom by ($cos$ ^2)$\alpha/2$
then take tan$\alpha/2$ = t
then u can get a quadratic in denominator which can be factorized, get two fraction.
|
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|
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$. Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.
Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?
|
Notice that
$$\frac{1}{\sqrt{an+b}} \cdot \frac{2n+1}{2n+2} \le \frac{1}{\sqrt{a(n+1)+b}} \\
\iff (a(n+1)+b)(2n+1)^2 \le (2n+2)^2 (an+b) \\
\iff an+a-4bn-3b \le 0$$
Hence if $a=3$, then $b=1$ would work. Of course, you need to prove the initial case ($n$=1).
BTW: how amazing it is that the first two answers got $e$ and $\pi$, respectively.
|
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|
Simplification of multiplying fractions Simplify $(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})$
I noticed the denominator of the five fractions are all in the form of:$$\frac{...}{(n-1)(n+1)}$$
Which can be simplified to:$$\frac{...}{n^2-1}$$
I futher attempted to move the integral part into the fraction, but it just got more complicated. How should I start from there?
|
$$m+\frac{m}{(n-1)(n+1)}=m\times(1+\frac{1}{(n-1)(n+1)})=m\times (\frac{n^2-1+1}{(n-1)(n+1)})=m\times\frac{n^2}{(n-1)(n+1)}$$
$$(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})=\frac{21^2}{20\times 22}\times 2\times \frac{22^2}{21\times 23}\times 2\times\frac{23^2}{22\times 24}\times 2\times \frac{24^2}{23\times 25}\times 13 \times \frac{25^2}{24\times 26}=105$$
|
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|
How to deduce the Vieta's substitution $y = z-\frac {p}{3z}$ for cubic equation $y^3+py +q=0$ In deriving the formula of cubic equations, Vieta substituted the following
$$y = z-\frac {p}{3z}$$
for the depressed cubic equation
$$y^3+py +q=0$$
and transformed it into a quadratic one.
My question: How did he get that substitution, or how did he know that by substituting $y = z-\frac {p}{3z}$, he could turn cubic into quadratic?
Please help me!
|
This is a sligntly different way of looking at your problem and getting the same results.
You have an equation of the form $$t^3 + pt + q=0 \tag{A.}$$
Let $t=z-w$. Then
$\begin{align}
t^3 &= z^3 - 3z^2V + 3zw^2 - w^3 \\
t^3 &= 3zw(z - w) + (z^3 - w^3) \\
t^3 &= 3zwt + (z^3 - w^3)
\end{align}$
Hence
$$t^3 - 3zwt - (z^3-w^3) = 0 \tag{B.}$$
Comparing (A.) and (B.), we get
\begin{align}
p &= -3zw \\
q &= -z^3 + w^3 \\
\end{align}
Rewriting the second equation as
$$z^3 + q - w^3 = 0$$
And substituting $w = -\frac{p}{3z}$, we get
$z^3 + q - \left(\dfrac{p}{3z}\right)^3 = 0$, which becomes
$$z^6 + qz^3 - \left(\dfrac{p}{3}\right)^3 = 0$$
|
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Show factorization $ x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1) $ I'm interested in how to show that
$$
x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1)
$$
I've seen this equality too often, but have no idea how to derive it. I've tried the following:
$$
x^{2n}-1=(x^n-1)(x^n+1)
$$
We all know that $x^n-1=(x-\xi^0)(x-\xi^1) \dots (x-\xi^{n-1})$, where $\xi=\exp(i \cdot \frac{2 \pi}{n})$.
Here $x-\xi^0$ gives us the desired $x-1$(for $x^2-1$ in RHS).
But the problem is, we can not do the same with $x^n+1$, because as I understand, there is no such general factorization of $x^n+1$. Firstly, it depends whether $n$ is even or not. Secondly, if for the sake for simplicity $n$ is odd, then we can decompose the polynomial such a way:
$$x^n+1=(x+1)(1-x+x^2-x^3 \dots -x^{n-2}+x^{n-1})$$
I don't see here anything to continue with. Any suggestions?
|
Note
\begin{align}
x^{2n}-1
&=\prod_{k=0}^{2n-1} (x-e^{i\frac{\pi k}n})
=\prod_{k=0}^{n-1} (x-e^{i\frac{\pi k}n})\cdot
\underset{k=2n-j}{ \prod_{k=n}^{2n-1} (x-e^{i\frac{\pi k}n})}\\
&=(x-1)\prod_{k=1}^{n-1} (x-e^{i\frac{\pi k}n})
\cdot \prod_{j=1}^{n-1} (x-e^{-i\frac{\pi j}n}) (x-e^{-i \pi})\\
&=(x-1)(x+1)\prod_{k=1}^{n-1} (x-e^{i\frac{\pi k}n})
(x-e^{-i\frac{\pi k}n})\\
&=(x^2-1)\prod_{k=1}^{n-1} (x^2 - 2x\cos\frac{\pi k}n+1)
\\
\end{align}
|
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|
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.
Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any textbooks with similar problems?
Thank you so much!
|
You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 + \cdots \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then
$$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$
which completes our inductive argument and we conclude that $x_n$ converges.
Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$
|
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|
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.
One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)
|
$$I_n=\int_{0}^{1}\frac{dx}{1+x+x^2+\ldots+x^n}=\underbrace{\int_{0}^{1}\frac{1-x}{1-x^{n+1}}dx}_{x\rightarrow y^{\frac{1}{n+1}}}=\frac{1}{n+1}\int_{0}^{1}\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y(1-y)}dy$$
$$=\frac{1}{n+1}\int_{0}^{1}\left(\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y}+\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{1-y}\right)dy$$
$$1-\frac{n+1}{2(n+2)}+\frac{1}{n+1}\int_{0}^{1}\left(\frac{y^{\frac{1}{n+1}-1}\log{\left(1-y\right)}}{n+1}-\frac{y^{\frac{2}{n+1}-1}\log{\left(1-y\right)}}{n+2}\right)dy$$
$$=\frac{n+3}{2(n+2)}+\frac{1}{n+1}\left[\frac{\mathfrak{B}\left(\frac{1}{n+1},1\right)\left(\psi^{\left(0\right)}\left(1\right)-\psi^{\left(0\right)}\left(1+\frac{1}{n+1}\right)\right)}{n+1}-\frac{\mathfrak{B}\left(\frac{2}{n+1},1\right)\left(\psi^{\left(0\right)}\left(1\right)-\psi^{\left(0\right)}\left(1+\frac{2}{n+1}\right)\right)}{n+2}\right]$$
$$=\frac{n+3}{2(n+2)}+\frac{\psi^{\left(0\right)}\left(1+\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(1+\frac{1}{n+1}\right)}{n+1}=\frac{1}{2\left(n+2\right)}+\frac{\psi^{\left(0\right)}\left(\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(\frac{1}{n+1}\right)}{n+1}$$
Therefore:
$$I_n=\int_{0}^{1}\frac{dx}{1+x+x^2+\ldots+x^n}=\frac{1}{2\left(n+2\right)}+\frac{\psi^{\left(0\right)}\left(\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(\frac{1}{n+1}\right)}{n+1}$$
Notes:
$\mathfrak{B}(x,y)$ stands for the Beta Function and $\psi^{(0)}(z)$ stands for the Digamma Function.
|
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|
How to solve the equation $x^2-2y=z^2$? Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 \Rightarrow 2y=(x-z)(x+z)\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).
|
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $-10\le x\le10$, $-10\le y\le10$, $-10\le z\le10$, and $x$, $y$, $z$ are all integers with $xy\ne0$.
I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
FullSimplify[
Solve[{x^2 - 2*y == z^2,
x*y != 0 && -10 <= x <= 10 && -10 <= y <= 10 && -10 <= z <=
10}, {x, y, z}, Integers]]
Running the code gives:
Out[1]={{x -> -6, y -> 10, z -> -4}, {x -> -6, y -> 10, z -> 4}, {x -> -5,
y -> 8, z -> -3}, {x -> -5, y -> 8, z -> 3}, {x -> -4, y -> -10,
z -> -6}, {x -> -4, y -> -10, z -> 6}, {x -> -4, y -> 6,
z -> -2}, {x -> -4, y -> 6, z -> 2}, {x -> -4, y -> 8,
z -> 0}, {x -> -3, y -> -8, z -> -5}, {x -> -3, y -> -8,
z -> 5}, {x -> -3, y -> 4, z -> -1}, {x -> -3, y -> 4,
z -> 1}, {x -> -2, y -> -6, z -> -4}, {x -> -2, y -> -6,
z -> 4}, {x -> -2, y -> 2, z -> 0}, {x -> -1, y -> -4,
z -> -3}, {x -> -1, y -> -4, z -> 3}, {x -> 1, y -> -4,
z -> -3}, {x -> 1, y -> -4, z -> 3}, {x -> 2, y -> -6,
z -> -4}, {x -> 2, y -> -6, z -> 4}, {x -> 2, y -> 2,
z -> 0}, {x -> 3, y -> -8, z -> -5}, {x -> 3, y -> -8,
z -> 5}, {x -> 3, y -> 4, z -> -1}, {x -> 3, y -> 4,
z -> 1}, {x -> 4, y -> -10, z -> -6}, {x -> 4, y -> -10,
z -> 6}, {x -> 4, y -> 6, z -> -2}, {x -> 4, y -> 6,
z -> 2}, {x -> 4, y -> 8, z -> 0}, {x -> 5, y -> 8,
z -> -3}, {x -> 5, y -> 8, z -> 3}, {x -> 6, y -> 10,
z -> -4}, {x -> 6, y -> 10, z -> 4}}
So, using these bounds I get $36$ number of solutions. Extending the bounds to $-10^3$ and $10^3$ gives $12716$ number of solutions. Extending the bounds, again, to $-10^4$ and $10^4$ gives $173364$ number of solutions.
|
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|
Cannot find limit using epsilon delta definition Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.
I try as below.
Let $\varepsilon>0$. We choose $\delta>0$.
Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}
Now I don't know how to continue this answer. I confused with $6x$.
Anyone can help me?
EDIT: I have tried as below as JC12's answer.
Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$
Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$
Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
\vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}
Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
\frac{\varepsilon}{19} = \varepsilon. \end{align}
So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.
|
Let $\epsilon > 0$ be arbitrary and start off with $\delta_{1} := 1$ so that we have $|x-2| < \delta_{1} = 1$.
Then $x>1$ and $x < 3$ so that $(x-2)^{2} < 1$ and $6x < 18$.
Then $|(x-2)^{2} + 6x|<19$ and set $\delta _{2}$ so that $|x-2| < \delta_{2}$ where $\delta_{2} := \frac{\epsilon}{19}.$
Hence, if $\delta := \min\{\delta_{1},\delta_{2}\}$, we have the result.
The reason why it is necessary to end with a single $\delta$ is that, for this argument to work, we need $|x-2| < 1$ and $|x-2| < \frac{\epsilon}{19}$.
Then to find a fixed value that will work all the way through, choose the smaller one.
|
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|
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!
|
$$45x^2-30x+1=(-9x^2+6x+7)^2$$
$$-5(-9x^2+6x+7)+36=(-9x^2+6x+7)^2$$
Substitute $u=-9x^2+6x+7$ and solve the quadratic equation.
|
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|
Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only) , $b(3) = 3$
$a(2) = 2$, because $2=1+1=2$ ,and $b(4)=2$, because $4=2+2=4$
$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$
$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$
$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$
By this way i am able to show that $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which i can understand.
Background:-This problem is from pathfinder for Olympiad mathematics.
|
For compositions into ones and twos we have the OGF
$$A(z) = \sum_{q\ge 1} (z+z^2)^q =
\frac{z+z^2}{1-z-z^2}
= -1 + \frac{1}{1-z-z^2}.$$
Compositions into parts at least two have OGF
$$B(z) = \sum_{q\ge 1} (z^2+z^3+\cdots)^q
= \frac{z^2/(1-z)}{1-z^2/(1-z)}
= \frac{z^2}{1-z-z^2}.$$
We then have for $n\ge 1$
$$[z^n] A(z) = [z^n] \left(-1 + \frac{1}{1-z-z^2}\right)
= [z^n] \frac{1}{1-z-z^2}
\\ = [z^{n+2}] \frac{z^2}{1-z-z^2} = [z^{n+2}] B(z)$$
as claimed.
|
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|
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...
|
With difference of squares and difference of cubes, you get that the limit simplifies to
$$\lim_{x\to\infty} \frac{(x+1)-(x-1)}{(x+2)-(x-2)}\cdot \frac{f(x)}{g(x)} = \frac{1}{2}$$
because $f$ and $g$ both grow at the rate $\sim 6 x$ for large $x$.
|
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|
Research question - prove inequality for $n \log n$. The following inequality has come up in some research and appears true when plotted, but I'm not sure how to to prove it. If $g(z) = z \log z$, then for $x>3$ ($x$ real),
$$
g(x) < g(x-1)\cdot \left(1 + \frac{g(x-1) - g(x-2)}{g(x-1)-1} \right).
$$
|
Alternative solution:
Fact 1: $\frac{2u}{2+u} \le \ln (1 + u) \le \frac{2u + u^2}{2 + 2u}$ for all $u \ge 0$.
(The proof is easy. Hint: Take derivative.)
Now, first, by Fact 1, we have
$$g(x) - x\ln (x-1) = x \ln(1 + \tfrac{1}{x-1})
\le x \cdot \frac{2x-1}{2x(x-1)} = \frac{2x-1}{2(x-1)}$$
which results in
$$g(x) \le x\ln(x-1) + \frac{2x-1}{2(x-1)}.$$
Second, by Fact 1, we have
$$(x-2)\ln(x-1) - g(x-2) = (x-2)\ln(1 + \tfrac{1}{x-2})
\ge (x-2) \cdot \frac{2}{2x-3} = \frac{2(x-2)}{2x-3}$$
which results in
$$g(x - 2) \le (x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}.$$
Third, it suffices to prove that (note: $g(x-1) - 1 > 0$)
$$x\ln(x-1) + \frac{2x-1}{2(x-1)}
< g(x-1) \cdot \left(1 + \frac{g(x-1) - [(x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}]}{g(x-1)-1} \right)$$
that is
$$\frac{1}{2(2x-3)[(x-1)\ln(x-1)-1]}\left(4(x-1) - \frac{1}{x-1} - \ln(x-1)\right) > 0$$
which is true since $x-1 \ge \ln(x-1)$ and $x - 1 \ge \frac{1}{x-1}$.
We are done.
|
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|
How to do integral $\int\frac1{x^3+x+1}dx$? I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$
|
$x^3+x+1$ has one real and a pair of complex roots. The real root is $r= -0.6823$, given analytically by the Cardano’s formula. Decompose the integrand in terms of $r$
\begin{align}
\frac1{x^3+x+1}&= \frac1{(x-r)(x^2+rx-1/r)}
=\frac r{2r +3}\left(-\frac1{x-r}+\frac{x+2r}{x^2+rx-1/r}\right)
\end{align}
and then integrate to obtain
$$\int \frac{dx}{x^3+x+1}
=\frac r{2r +3}\left(-\ln\frac{x-r}{\sqrt{ x^2+rx-1/r}}
+\frac{ 3r \tan^{-1}\frac{2x+r}{\sqrt{1-3/r} } }{\sqrt{1-3/r}}\right)
$$
where $r^3=-r-1$ is recognized in above derivation.
|
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|
Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).
How would I work this out?
|
From your expansions of $e^x$ and $e^{-x}$, we have $e^x+e^{-x}=2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots$,
so sech$( x)=\dfrac2{e^x+e^{-x}}=\dfrac{2}{2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots}=\dfrac1{1+\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots}$
$=1-\left(\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots\right)+\left(\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots\right)^2\cdots$
$=1-\dfrac{x^2}2-\dfrac{x^4}{24}+\dfrac{x^4}{4}\cdots=1-\dfrac{x^2}2+\dfrac{5x^4}{24}\cdots$
|
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Indefinite integral $\int \frac {\mathrm d x} {p^2 + q^2 \cosh^2 a x}$ Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.584$ gives:
$$\int \dfrac {\mathrm d x} {p^2 + q^2 \cosh^2 a x} = \begin{cases} \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \left({\dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } }\right) + C \end{cases}$$
(As usual he glosses over negative argument to the $\ln$, but all you need to do is modulus it.)
He does not give the conditions under which either case holds. This is crucial.
First I multiplied top and bottom by $\operatorname {csch}^2 a x$, then used $\operatorname {csch}^2 a x = \coth^2 a x - 1$ for the bottom, then substituted $u = \coth a x$ to convert it into the form:
$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\frac {p^2} {p^2 + q^2} - u^2}$$
I can't see there is any other case than that $\dfrac {p^2} {p^2 + q^2}$ is positive, hence:
$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {\left({\frac p {\sqrt {p^2 + q^2} } }\right)^2 - u^2}$$
This is a standard integral, giving:
$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} }\right| + C$$
which leads to:
$$\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }\right) + C$$
And that all works fine, no worries.
BUT, as far as I can see, there is no way to get to the arctangent result. To get there, we need $u^2 + k^2$ for some constant $k$, and I can't see how to get there from the given integrand.
|
Note
$$ \frac12\ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }
= \frac12\ln {\frac {\frac {p \tanh a x }{\sqrt {p^2 + q^2} } + 1}{\frac{p \tanh a x }{\sqrt {p^2 + q^2} }-1}}
=\tanh^{-1}\frac {p \tanh a x }{\sqrt {p^2 + q^2} }\\
$$
So, there is a typo in the second expression.
|
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|
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$ Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.
That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?
I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.
But from there I am stuck, I am not able to see how to link all the concepts.
Could someone give me a hint?
Thank you and Happy New Year 2021 :)
|
Using a result from Some Logarithmic Inequalities, E.R.Love :
$$
\forall x>0 \quad ; \quad \ln (1+x) > \frac{2x}{2+x}
$$
Making the transformation $x \to \frac 1x$ gives us :
$$
\forall x > 0 \quad ; \quad \ln\left(1+\frac 1x\right) > \frac{\frac 2x}{2+\frac 1x} = \frac 2{2x+1}
$$
Thus if $g(x) = x^2 + \frac {2}{2x+1}$ then $g(x) \leq f(x)$ for $x>0$.
(Note : there may be easier bounds, but for this one the minima was the easiest to find)
The minimum of $g(x)$ is $\frac 54$ at $x = \frac 12$. We easily have $\frac {1+ 2 \ln 2}{4} \leq 1$ by using $\ln 2 \leq 1$. Thus, we are done.
|
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|
Find all positive integer solutions for the following equation: Find all positive integer solutions for the following equation:
$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$
I've tried simplifying the equation and then refactoring but I can't find any solutions.
|
Using wolfram alpha, all the solutions are given by the expression : $$\begin{cases}y=\frac{x-1}{x+1}, x\neq -1 \\y = \frac {x+3}{x+1}, x\neq -1 \end{cases}$$
For example, the solutions for $x=1$ are $y=0$ and $y=2$.
You could try to find these solutions by hand solving the quadratic equation in respect to $y$. If you expand the original equation you get:
$$x^2 y^2 - 2 x^2 y + x^2 + 2 x y^2 + 2 x + y^2 - 2 y + 1 = 4 x y + 4$$ $$\Leftrightarrow (x^2 +2x+1) y^2 + (-2x^2-4x-2)y + (x^2+2x+1-4)=0 $$
Which you can solve using the quadratic formula.
|
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|
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not help me anywhere. Perhaps I have to multiply something with $a$ only so that the expression becomes usable, what what to multiply?
Next, the expression we need to find is $\frac{3a^2 + 3a + 1}{a^3}$ , which I did not find any cool factorizations, and do not think anything will help here.
Can anyone help me?
|
Note that
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{(a+1)^3-a^3}{a^3}.
\end{eqnarray*}
Note also that $a+1=\sqrt[3]{2}a$ so ...
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{2a^3-a^3}{a^3}=1.
\end{eqnarray*}
|
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|
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?
|
$\sqrt{3}f(x) = \frac{1}{\sqrt{3^{2x-1}}+1}$ and $\frac{1}{\sqrt{3^{a}}+1}+\frac{1}{\sqrt{3^{-a}}+1}=1$.
You can pair like $(-11,11),(-9,9),(-7,7)\cdots(-1,1)$ then answer is $6$.
|
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|
Sums of integer powers similar to Prouhet–Tarry–Escott problem Recall Prouhet–Tarry–Escott problem. Its solution shows that certain sums of powers of integers can be made to vanish simultaneously if their signs are chosen to follow the Thue–Morse sequence, e.g.
$$\sum_{k=0}^{2^4-1}(-1)^{\sigma_2(k)}\,k^n=0\quad\text{for}\;n<4,\quad\text{i.e.}\\
\small
\begin{array}{l}
0^1+3^1+5^1+6^1+9^1+10^1+12^1+15^1=1^1+2^1+4^1+7^1+8^1+11^1+13^1+14^1 \\
0^2+3^2+5^2+6^2+9^2+10^2+12^2+15^2=1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2 \\
0^3+3^3+5^3+6^3+9^3+10^3+12^3+15^3=1^3+2^3+4^3+7^3+8^3+11^3+13^3+14^3, \\
\end{array}$$
where $\sigma_2(k)$ is the sum of digits of $k$ in base-2.
I am interested in sums of powers of a similar form, but with larger exponents (such that the sum does not vanish anymore). I was able to make a few conjectures:
$$\begin{align}
&\sum _{k=0}^{2^n-1} (-1)^{\sigma _2(k)}\, k^{n+2}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{36}\, (-1)^n\, \left(2^n-1\right)\, \left(5\times2^n-4\right)\, (n+2)!\\
\\
&\sum _{k=0}^{2^n-1} (-1)^{\sigma _2(k)}\,k^{n+3}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{72}\,(-1)^n \,\left(2^n-1\right)^2\,\left(2\times2^n-1\right)\,(n+3)!\\
\\
&\small\sum _{k=0}^{2^n-1} (-1)^{\sigma_2(k)}\,k^{n+4}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{32400}\,(-1)^n\,\left(2^n-1\right)\,\left(143\!\times\! 8^n-307\!\times\!4^n+193\!\times\! 2^n-32\right)\,(n+4)!\end{align}$$
This pattern seems to continue within increasingly complicated polynomials of $2^n$ on the right-hand side. Can you suggest an approach to prove these, and to find a more general formula for an arbitrary exponent $k^{n+p}$?
|
The exponential generating function for $S_{n,m}=\displaystyle\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}k^m$ (w.r.t. $m$) is $$\sum_{m=0}^\infty S_{n,m}\frac{x^m}{m!}=\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}e^{kx}=\prod_{k=0}^{n-1}(1-e^{2^k x})=2^{n(n-1)/2}(-x)^n\exp\sum_{k=0}^{n-1}f(2^k x),$$ where $$f(x)=\log\frac{e^x-1}{x}=\sum_{m=1}^\infty\frac{B_m^+}{m}\frac{x^m}{m!}$$ using the Bernoulli numbers (look at $f'(x)$ to see the expansion). Thus,
\begin{align*}
S_{n,n+p}&=(-1)^n 2^{n(n-1)/2}(n+p)!\ A_p(2^n),
\\A_p(a)&=[x^p]\exp\sum_{m=1}^\infty\frac{a^m-1}{2^m-1}\frac{B_m^+}{m}\frac{x^m}{m!}.
\end{align*}
This gives an algorithmic way to compute $S_{n,n+p}$ for any fixed $p$. Continuing your table,
\begin{align*}
A_0(a)&=1
\\2A_1(a)&=a-1
\\36A_2(a)&=(a-1)(5a-4)
\\72A_3(a)&=(a-1)^2(2a-1)
\\32400A_4(a)&=(a-1)(143a^3-307a^2+193a-32)
\\64800A_5(a)&=(a-1)^2(2a-1)(19a^2-24a+2)
\\85730400A_6(a)&=(a-1)(5765a^5-19372a^4+22670a^3-10405a^2+1355a+32)
\\342921600A_7(a)&=(a-1)^2(2a-1)(1166a^4-2850a^3+1715a^2+60a-1)
\end{align*}
Observe that $A_p(a)$ is always divisible by $a-1$, and even by $(a-1)^2(2a-1)$ if $p>1$ is odd (because of $B_m^+=0$ for odd $m>1$, and [the g.f. of] $A_p(1/2)$ in closed form).
|
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|
Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ and thus $u\leq 1$:
\begin{align}
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
&=-2\int_{-1}^{-2} \frac{(u^2+2u-1)(u-1)}{u}\, du\\
&= -2 \left( \int_{-1}^{-2} u^2 d u + \int_{-1}^{-2} u\, du +\int_{-1}^{-2} -3\, du + \int_{-1}^{-2} \frac{1}{u}\, du \right) \\
&= -2\left[\frac{u^3}{3}+\frac{u^2}{2}-3u+\ln{|u|}\right]_{-1}^{-2}\approx -18
\end{align}
Can someone please help me pinpoint the issue?
|
exactly, when you have $u=1-\sqrt{2+x}\ => x=u^2-2u-1$
Then $du=2(u-1)dx$
Plugging it into your formula we get
$=2\int\ \frac{(u^2-2u-1)(u-1)}{u}$
|
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|
For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $? For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?
This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-frac-1-1-xy-3-mathrm-d-x-mathrm-d-y
where it is shown that
$I(1, 3)
=\dfrac34\left(\ln(3)+\dfrac{\pi\sqrt{3}}{9}\right)
$.
Here's what I've done
so far.
$\begin{array}\\
I(a, b)
&=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}\\
&=\int_0^1 \int_0^1 dx\,dy\sum_{n=0}^{\infty} (x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 dx\,dy(x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 x^{an}dx \int_0^1 y^{bn}dy\\
&=\sum_{n=0}^{\infty} \dfrac{x^{an+1}}{an+1}\big|_0^1 \dfrac{y^{bn+1}}{bn+1}\big|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{1}{an+1} \dfrac{1}{bn+1}\\
&=ab\sum_{n=0}^{\infty} \dfrac{1}{abn+b} \dfrac{1}{abn+a}\\
\\
I(a.a)
&=\sum_{n=0}^{\infty} \dfrac{1}{(an+1)^2}\\
&=\dfrac1{a^2}\sum_{n=0}^{\infty} \dfrac{1}{(n+1/a)^2}\\
&=\dfrac1{a^2}\psi^{(1)}(1/a)\\
\text{If } a \ne b\\
I(a, b)
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \left(\dfrac{1}{abn+b} -\dfrac{1}{abn+a}\right)\\
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \int_0^1 \left(x^{abn+b-1} -x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\sum_{n=0}^{\infty}x^{abn+b-1} -\sum_{n=0}^{\infty}x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\sum_{n=0}^{\infty}x^{abn} -x^{a-1}\sum_{n=0}^{\infty}x^{abn}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\dfrac1{1-x^{ab}} -x^{a-1}\dfrac1{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\dfrac{x^{b-1} -x^{a-1}}{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 x^{b-1}\left(\dfrac{1 -x^{a-b}}{1-x^{ab}}\right)dx\\
\end{array}
$
But,
in general,
I can't go further.
Special cases can be handled.
For example,
if $b=1$
and $a$ is a positive integer,
we can show that
$I(a, 1)
=\dfrac{a}{(a-1)^2} \left(\ln(a)-\int_0^1 \left(\dfrac{\sum_{k=0}^{a-3}(a-2-k)x^k}{\sum_{k=0}^{a-1}x^k}\right)dx\right)
$.
This gives,
as above,
$I(3, 1)
=\dfrac34\left(\ln(3)-\int_0^1 \left(\dfrac{1}{1+x+x^2}\right)dx\right)
$
and completing the square
gives the answer above.
Also
$I(2, 1)
=2\ln(2)
$.
Similarly,
$\begin{array}\\
I(4, 1)
&=\dfrac49\left(\ln(4)-\int_0^1 \left(\dfrac{2+x}{1+x+x^2+x^3}\right)dx\right)\\
&=\dfrac49\left(\ln(4)- \dfrac{3 π + \ln(4)}{8}\right)
\quad\text{(According to Wolfy)}\\
&=\dfrac49\left(\dfrac{7\ln(4)-3 π}{8}\right)\\
\end{array}
$
We could get
$I(5, 1)$
since Wolfy gives
$$\dfrac1{1+x+x^2+x^3+x^4}
=\dfrac{-2 x + \sqrt{5} - 1}{\sqrt{5} (2 x^2 - (\sqrt{5}-1) x + 2)} + \dfrac{2 x + \sqrt{5} + 1}{\sqrt{5} (2 x^2 + (\sqrt{5}+1) x + 2)}
$$
but I'm not going to bother
to work it out.
|
The following is trivial due to expansion of digamma (formula 14 of this page) $$\sum _{n=0}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{\psi ^{(0)}(a)-\psi ^{(0)}(b)}{a-b}$$ Thus OP's integral equals to $$I(a,b)=\sum _{n=0}^{\infty } \frac{a b}{(a b n+b) (a b n+a)}=\frac{\psi ^{(0)}\left(\frac{1}{b}\right)-\psi ^{(0)}\left(\frac{1}{a}\right)}{a-b}$$ Due to Gauss's digamma theorem (formula 11 of link above) for all $a,b\in \mathbb Q$ the integral can be expressed in terms of log and trig functions (even square roots, when $a,b$ are small). For instance $$I(3,5)=\frac{1}{2} \pi \sqrt{\frac{1}{6} \left(-\sqrt{3+\frac{6}{\sqrt{5}}}+\frac{3}{\sqrt{5}}+2\right)}+\frac{1}{8} \log \left(\frac{3125}{729}\right)+\frac{1}{4} \sqrt{5} \coth ^{-1}\left(\sqrt{5}\right)$$
|
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|
Find $\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$ Find $\displaystyle \lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$
My attempt:
\begin{align*}
A&=\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
\ln A&=\lim_{k\to 2}\ln\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
&=\lim_{k\to 2}(\ln(2^{2^k}-2^{k^2})-\ln(2^k-k^2))
\end{align*}
I stuck at $\ln(0)$.
|
Applying L'Hopital's rule you get
$$
\lim_{x\to 2}\dfrac{2^{x+2^x} \log ^2 2-2^{x^2+1} x \log (2)}{2^x \log 2-2 x} = 16 \log 2.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Singular Value Decomposition gives a row permutated matrix I want to find the SVD of the following matrix.
$$A = \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}$$
The singular values of the above matrix are $\sqrt{17}$,$1$. The following results are posted here in order to make this question shorter.
$$AA^T=\begin{bmatrix}5&6&4\\6&8&6\\4&6&5\end{bmatrix}$$
Eigen values of $AA^T$=$\{0,1,17\}$
Eigen vectors of $AA^T$=$\{\begin{bmatrix}-2\\3\\-2\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix},
\begin{bmatrix}3\\4\\3\end{bmatrix}\}$
$$A^TA=\begin{bmatrix}9&8\\8&9\end{bmatrix}$$
Eigen values of $A^TA$=$\{1,17\}$
Eigen vectors of $A^TA$=$\{\begin{bmatrix}1\\-1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\}$
Therefore, we can write the U,V,W matrices as follows.
$$U=\begin{bmatrix}\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}&\frac{-2}{\sqrt{17}}\\
\frac{4}{\sqrt{34}}&0&\frac{3}{\sqrt{17}}\\
\frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}&\frac{-2}{\sqrt{17}}\end{bmatrix}$$
$$W=\begin{bmatrix}\sqrt{17}&0\\0&1\\0&0\end{bmatrix}$$
$$V=V^T=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$
My problem is that why in here, $$A\neq UWV^T$$
$$UWV^T=\begin{bmatrix}2&1\\2&2\\1&2\end{bmatrix}$$
which is a row permuted version of A.
This problem won't occur if I multiplied the eigenvector of $AA^T$ corresponding to the eigenvalue of $1$ by $(-1)$.(i.e; by using $\begin{bmatrix}-1\\0\\1\end{bmatrix}$ instead of $\begin{bmatrix}1\\0\\-1\end{bmatrix}$). Both of those eigenvectors are correct for the eigenvalue $1$ of that matrix. So, my problem is how do we identify such occasions when obtaining the SVD?
|
With the help of Lutz Lehmann's comment above I figured out the way to obtain the SVD without worrying about the problem I previously faced. I am posting my answer here so that it would be a help to anyone who has the same problem.
In the SVD, we express the matrix A as a product of 3 matrices which I am denoting as U, W, $V^T$. In this less problematic method, what we have to first do is obtaining the $A^TA$.
$$A = \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}$$
$$\implies A^TA=\begin{bmatrix}9&8\\8&9\end{bmatrix}$$
Then we obtain the eigenvalues and corresponding eigenvectors of $A^TA$ as follows.
$det(A^TA-\lambda I)=det(\begin{bmatrix}9-\lambda&8\\8&9-\lambda\end{bmatrix})=0$
$\implies (1-\lambda)(17-\lambda)=0$
$\therefore \lambda=\{17,1\}$
When $\lambda = 17,$
eigenvector = $\begin{bmatrix}1\\1\end{bmatrix}$
$\therefore$ normalized eigenvector when ($\lambda = 17$) = $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}=v_1$
When $\lambda = 1,$
eigenvector = $\begin{bmatrix}1\\-1\end{bmatrix}$
$\therefore$ normalized eigenvector when ($\lambda = 1$) = $\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\end{bmatrix}=v_2$
The corresponding singular value would be $\{\sigma_1=\sqrt{17},\sigma_2=1\}$.
Using the above results, we can write W and V as follows.
$$W=\begin{bmatrix}\sqrt{17}&0\\0&1\end{bmatrix}$$
$$V=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$
Now we can obtain the columns of U without explicitly calculating the eigenvectors of $AA^T$ as follows.
Simply use the following relation,
$$Av_i=\sigma_i u_i, \forall i$$
$$\therefore Av_1=\sigma_1 u_1$$
$$\begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{bmatrix}= \sqrt{17} u_1$$
$$\therefore \begin{bmatrix}\frac{3}{\sqrt{2}}\\\frac{4}{\sqrt{2}}\\\frac{3}{\sqrt{2}}\end{bmatrix}=\sqrt{17} u_1\implies u_1=\begin{bmatrix}\frac{3}{\sqrt{34}}\\\frac{4}{\sqrt{34}}\\\frac{3}{\sqrt{34}}\end{bmatrix}$$
Also, $Av_2=\sigma_2 u_2$
$$\therefore \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\end{bmatrix}= 1 u_2 \implies u_2 = \begin{bmatrix}\frac{-1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}$$
$$\therefore U = \begin{bmatrix} \frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}\\
\frac{4}{\sqrt{34}}&0\\
\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}\end{bmatrix}$$
So, the SVD of A is,
$$\begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}=
\begin{bmatrix} \frac{3}{\sqrt{34}}&\frac{-1}{\sqrt{2}}\\
\frac{4}{\sqrt{34}}&0\\
\frac{3}{\sqrt{34}}&\frac{1}{\sqrt{2}}\end{bmatrix}
\begin{bmatrix}\sqrt{17}&0\\0&1\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{bmatrix}$$
The above is the reduced SVD. In any case if you want to obtain the full SVD, all you have to do is obtain the normalized eigenvector of $AA^T$ corresponding to the eigenvalue of $0$ and, put that as the third column of the above U matrix. Also, remember to add another row of zeros to the W matrix as its third row.
The above method can be applied to any given matrix to obtain the singular value decomposition.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3978753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Number of solutions of $ax+by+cz=n$ in nonnegative integers Let $a,b,c,n\in\mathbb{Z}_0^+$. The number of solutions of
$$ax+by+cz=n$$
(i.e. the number of ordered triplets $(x,y,z)$ satisfying the equation) in $\mathbb{Z}_0^+$ as a function of $n$ is
$$\frac{1}{n!}\lim_{w\to 0}\frac{d^n}{dw^n}\frac{1}{(1-w^a)(1-w^b)(1-w^c)}=\frac{1}{2\pi i}\oint \frac{dw}{(1-w^a)(1-w^b)(1-w^c)w^{n+1}}$$
by the theory of generating functions and Cauchy's integral formula.
If $a,b,c$ are small, then the $n$th derivative/contour integral is amenable for computation using partial fraction decomposition. But for large $a,b,c$, the partial fraction decomposition becomes very tedious.
Are there any faster methods of computing this particular integral, or faster methods of obtaining the number of solutions in general?
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The topic is standardly known as the Frobenius coin problem, and
it is quite complicated in general.
In the 2D case $ax+by=n$ there is an interesting theorem, the Popovicius' theorem
which tells that the number of non-negative solutions $ p_{\left\{ {a,b} \right\}} (n)$ is given by
$$
\eqalign{
& p_{\left\{ {a,b} \right\}} (n) = \left| {\,\left\{ \matrix{
0 \le x,y,a,b,n \in Z \hfill \cr
\gcd (a,b) = 1 \hfill \cr
ax + by = n \hfill \cr} \right.\;} \right| = \cr
& = {n \over {ab}} - \left\{ {{{b^{\,\left( { - 1} \right)} n} \over a}} \right\}
- \left\{ {{{a^{\,\left( { - 1} \right)} n} \over b}} \right\} + 1 \cr}
$$
where
$$
\left\{ x \right\} = x - \left\lfloor x \right\rfloor
\quad b^{\,\left( { - 1} \right)} b \equiv 1\;\left( {\bmod a} \right)
\quad a^{\,\left( { - 1} \right)} a \equiv 1\;\left( {\bmod b} \right)
$$
For the 3D case, like the present, we can apply the above to $ax+by=n-cz$ and sum over $z$.
$$
\eqalign{
& p_{\left\{ {a,b,\,c} \right\}} (n)\quad \left| {\; \gcd(a,b) = 1} \right.\quad
= \sum\limits_{z = 0}^{\left\lfloor {n/c} \right\rfloor } {p_{\left\{ {a,b} \right\}} (n - cz)} = \cr
& = \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {{{n - ck} \over {ab}}
- \left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
- \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\} + 1} \right)} = \cr
& = {n \over {ab}}\left( {\left\lfloor {{n \over c}} \right\rfloor + 1} \right)
- {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor \left( {\left\lfloor {{n \over c}} \right\rfloor
+ 1} \right) + \left( {\left\lfloor {{n \over c}} \right\rfloor + 1} \right) + \cr
& - \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {\left\{ {{{b^{\,\left( { - 1} \right)}
\left( {n - ck} \right)} \over a}} \right\} + \left\{ {{{a^{\,\left( { - 1} \right)}
\left( {n - ck} \right)} \over b}} \right\}} \right)} = \cr
& = \left( {{n \over {ab}} - {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor + 1} \right)
\left( {\left\lfloor {{n \over c}} \right\rfloor + 1} \right)
- \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor }
{\left( {\left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
+ \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\}} \right)} \cr}
$$
Example:
$$n=13, \, a=3, \, b=4, \, c=5$$
the solutions are
$$
\left( {x,y,z} \right) \in
\left\{ {\left( {0,2,1} \right),\left( {1,0,2} \right),\left( {3,1,0} \right)} \right\}
$$
The modular inverses of $a$ and $b$ are
$$
\eqalign{
& 3 \cdot 3 \equiv 1\left( {\bmod 4} \right) \Rightarrow a^{\,\left( { - 1} \right)} = 3 \cr
& 1 \cdot 4 \equiv 1\left( {\bmod 3} \right) \Rightarrow b^{\,\left( { - 1} \right)} = 1 \cr}
$$
and $p_{\left\{ {a,b,\,c} \right\}} (n)$ becomes
$$
\eqalign{
& p_{\left\{ {3,4,5} \right\}} (13) = \cr
& = \left( {{{13} \over {12}} - {5 \over {24}}\left\lfloor {{{13} \over 5}} \right\rfloor + 1} \right)
\left( {\left\lfloor {{{13} \over 5}} \right\rfloor + 1} \right)
- \sum\limits_{k = 0}^2 {\left( {\left\{ {{{\left( {13 - 5k} \right)} \over 3}} \right\}
+ \left\{ {{{3\left( {13 - 5k} \right)} \over 4}} \right\}} \right)} = \cr
& = 5 - \sum\limits_{k = 0}^2 {\left( {\left\{ {{{13} \over 3}} \right\}
+ \left\{ {{{39} \over 4}} \right\} + \left\{ {{8 \over 3}} \right\} + \left\{ {{{24} \over 4}} \right\}
+ \left\{ {{3 \over 3}} \right\} + \left\{ {{9 \over 4}} \right\}} \right)} = \cr
& = 5 - \sum\limits_{k = 0}^2 {\left( {{1 \over 3} + {3 \over 4} + {2 \over 3} + {1 \over 4}} \right)} = 3 \cr}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3980471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.