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Compute 100th derivative A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of
$$
f(x)=\frac{x^2+1}{x^3-x}.
$$
I have computed the zeroth derivative
$$
\frac{x^2+1}{x^3-x}
$$
and the first derivative
$$
\frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2}
$$
but I don't see any obvious structure.
|
$$\frac{x^2+1}{x^3-x}=\frac{x^2-1+2}{x^3-x}=\frac{1}{x}+\frac{2}{x(x-1)(x+1)}=$$
$$=\frac{1}{x}+\frac{2}{x+1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}+2\left(\frac{1}{x+1}-\frac{1}{x}\right)=$$
$$=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}.$$
Can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof by induction - stuck on simple question! Question:
(Part 1) Show that the inequality
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} < \frac{1}{2}
$$
works for all natural numbers $n > 2$.
(Part 2) Deduce that for all natural numbers $n$, the following inequality holds:
$$
\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} < \frac{7}{4}.
$$
This is a problem I found when rummaging around through old proof by induction questions and has been quite an issue for me. You see, I struggled with the first part, I proved it works for $n=3$ and then wrote out the result for $n=k$ and $n=k+1$ but I couldn't progress from there.
What I did:
*
*When $n = 3$:
$$
\frac{1}{2\cdot 3} = \frac{1}{6}
$$
and since $1/6 < 1/2$, the proposition is true for $n=3$.
*Assuming the truth of the proposition when $n = k$ i.e.
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k-1)\cdot k} < \frac{1}{2}
$$
and considering $n = k+1$, prove that
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(k+1)\cdot k} < \frac{1}{2}
$$
From there I couldn't seem to find a way to link the assumption into the $n=k+1$ inequality in order to prove that that is also $< 1/2$.
Can't solve the first part in order to link to the second part in order to fulfill the "deduce" part.
Thanks in advance and sorry if it's just a stupid error I've made.
|
Hints:
$1.\qquad\dfrac1{2\cdot 3}=\dfrac12-\dfrac 13,\quad \dfrac1{3\cdot 4}=\dfrac 13-\dfrac14,\;\dots,\;\dfrac1{(n -1)n}=\dfrac1{n-1}-\dfrac 1n$.
$2.\qquad \dfrac1{3^2}<\dfrac 1{2\cdot 3},\quad \dfrac1{4^2}<\dfrac1{3\cdot4},\;\dots\;,\dfrac1{n^2}<\dfrac1{(n-1)n}.$
|
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|
Inverse of $A^T Y A$ Is there a formula for the inverse of a matrix
$$ X = A^\top Y A $$
in terms of $A$ and $Y$, given that $A \in F^{m \times n}$ is full rank with $m >n$, and $Y$ is positive definite?
|
Even with the restriction $m>n$, it need not be the case that $X=A^\top Y A$ be invertible. As a counterexample, suppose $m=2,n=1$. Then the choice $$Y=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\; A=\begin{pmatrix} 1 \\ 1\end{pmatrix}$$
yields $$X = A^\top Y A = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=0$$ which is plainly not invertible.
|
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|
If $a + b = \frac{1}{4}$, what is $a^3 + b^3$? I just want to know how to solve this problem.
If $a + b = \dfrac{1}{4}$, what is $a^3 + b^3$ equals to?
My work - but still doesn't give me an answer.
$a^2 + b^2 = (a + b)^2 - 2ab = \dfrac{1}{16} - 2ab$
$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = \dfrac{1}{64} - \dfrac{3ab}{4}$.
How can I get the value for $ab$ here?
|
The value of $y = a^3+b^3$ contains a family of solutions (like a curve). You cannot establish a value for $y$ unless something else is specified.
For example besides $a+b = \tfrac{1}{4}$, let us use some artbitrary value $a-b = \lambda$. The solve for $a$ and $b$ in terms of $\lambda$ and put them into $y$
$$ y = \left( \frac{1+4 \lambda}{8} \right)^3 + \left( \frac{1-4 \lambda}{8} \right)^3=\frac{1}{256} + \frac{3 \lambda^2}{16} = \frac{1}{256} + \frac{3 (a-b)^2}{16}$$
so the minimum value of y exists when a=b, and thus λ=0.
|
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|
$S_n = \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{n^4 +r^4+2r^3+2n^2r^2+2n^2r+n^2+r^2}})$ find$S_{100}$ $$S_n = \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2}})$$ find$S_{100}$
Now the denominator of this expression is to big and confusing.I don't know how can we resolve it, because of such big expression in limited amount of time. It needs to be telescopic of some kind, but I don't know how to make it. Help please
|
Notice that $\displaystyle n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2 = (n^2 + r^2 + r)^2 + n^2$
Hence
$\displaystyle S_n=$
$\displaystyle \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{(n^2 + r^2 + r)^2 + n^2}})$
$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{n}{n^2+r^2+r} $
$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{\frac{1}{n}}{1 + \frac{r^2+r}{n^2}} $
(Divide numerator and denominator by $n^2$)
$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{\frac{r+1}{n} - \frac{r}{n}}{1 + \frac{r+1}{n}\frac{r}{n}}$
$\displaystyle =\sum_{r=0}^{n-1} \left (\tan^{-1} \frac{r+1}{n} - \tan^{-1} \frac{r}{n} \right)$
$\displaystyle = \frac{\pi}{4}$
Please check the calculations.
|
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|
Solve the initial value problem $xyy' + xy' = 1$ and $y(1) = 0$ I can solve the differential equation, which is $y + y^{2}/2 = \ln(x) + C$.
But I cannot solve the IVP because I can't isolate for $y$ and find the value of $C.$
|
We have (just reproducing your work for completeness, and combining some of the comment ideas):
\begin{align*}
xyy'+xy'&=1\\
xy'(y+1)&=1\\
(y+1)\,dy&=\frac{dx}{x}\\
\frac{y^2}{2}+y&=\ln(x)+C\\
0&=\ln(1)+C\\
0&=C\\
\frac{y^2}{2}+y&=\ln(x)\\
y^2+2y&=2\ln(x)\\
y^2+2y+1&=2\ln(x)+1\\
(y+1)^2&=2\ln(x)+1\\
y+1&=\pm\sqrt{2\ln(x)+1}\\
y&=-1\pm\sqrt{2\ln(x)+1}.
\end{align*}
But notice that we can't allow the negative square root, because it doesn't actually satisfy the initial condition. Also notice that we threw out the $\ln|x|$ for $\ln(x),$ because we knew the initial condition would be for positive $x.$ So the final solution is
$$y=-1+\sqrt{2\ln(x)+1}. $$
|
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|
If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without using calculus?
I know if $a , b,c,d \in \mathbb R$ then $a^2+b^2+c^2+d ^2$ will be minimum when $a =b= c =d = \frac{4m+1}{4}$..So the minimum value would have been $4m^2 +2m +1/4$..But what to do in that case?
|
The nearest integer greater than $4m^2 +2m +1/4$ is $4m^2 +2m +1$..And if we consider $a=b=c =m $ and $d=m+1$.We can get this minimum value.
|
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|
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$
So we chose $\delta=\sqrt{\epsilon}$
|
In two variables the epsilon-delta definition for $\lim_{\substack{x\to a\\ y\to b}}f(x,y)=L$ means that for every $\epsilon >0$ there exists a $\delta>0$ such that $\big|f(x,y)-L\big|<\epsilon$ whenever $0<\sqrt{(x-a)^2+(y-b)^2}<\delta$.
In your case, you want to show that $\big|f(x,y)-0\big|<\epsilon$ whenever $0<\sqrt{x^2+y^2}<\delta$. You did this by showing that
\begin{align}x^2y^2\leq (x^2+y^2)^2\implies\bigg|\frac{x^2y^2}{x^2+y^2}-0\bigg|\le\bigg|\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}\bigg|=\bigg|x^2+y^2\bigg|=x^2+y^2\end{align}
so that you could choose $\delta=\sqrt{\epsilon}$ and then get the required form of $\big|f(x,y)-0\big|<\epsilon$.
|
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|
If $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ , find the maximum and minimum value of $u^2$.
If $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ , find the maximum and minimum value of $u^2$.
This problem was bothering me for a while. The minimum value of $u $ seemed relatively easy to find by using AM-GM followed by Cauchy-Schwarz but both equalities don't hold simultaneously. It seems a bit tempting to assume that the maximum occurs at $x= {\pi/4}$ but I couldn't find a proof to the conjecture. Would someone please help me to find the maximum and minimum with a short relevant proof?
Note: I discovered that the function $u $ is periodic with period ${\pi/2} $.
|
$u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$
Let
$p = a \cos^2 x + b \sin^2 x$
$q = b \cos^2 x + a \sin^2 x$
and
$u = \sqrt{p} + \sqrt{q}$
Then $u^2 = p + q + 2 \sqrt{pq}$
Now
$p + q = a + b$
and
$\displaystyle pq = \frac{(a+b)^2}{4} - \frac{(a-b)^2}{8} - \frac{(a-b)^2}{8} \cos 4x$
If $\cos 4x = -1$ then $u^2$ is maximum and is equal to
$2(a+b)$
If $\cos 4x = 1$ then $u^2$ is minimum and is equal to
$a+b + 2 \sqrt{ab}$
Please check the calculations.
|
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|
Find $\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$
$$\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$$
I tried using the Series Expansion of the $\ln(x)$ but then got stuck in between. I also tried using the L'Hopital but the expression got quickly messy.
After applying L'Hopital for the first time, I got
$$\lim_{x \to \infty} \frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}x\bigg)$$
The expression is still in the undefined form. Unless the question maker wants to torture the problem solver, this method would not be the way to go.
I have got no other clue for solving this problem.
Any help would be appreciated.
|
Note that as $x\to +\infty$,
$$\cot^{-1}(x)=\arctan(1/x)=\frac{1}{x}-\frac{1}{3x^3}+o(1/x^3).$$
Hence, from your work,
$$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)=
\frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\
=\frac{-x^3}{2}\bigg(-\frac{1}{x}+\frac{1}{x^3} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\to -\frac{1}{3}.$$
|
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|
System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Thank you for your time
|
Subtract the first row from the last row of the system: then you get
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1 -16 a_2 -64 &=0
\end{split}
$$
Then multiply by 8 the second row and subtract it again from the third row:
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1 -16 a_2 -64 &=0
\end{split}\implies
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
56a_1 -8a_3+64 &=0
\end{split}
$$
Then you can proceed by substitution and do the following steps
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
a_1 &=\frac{1}{7}(a_3-8)
\end{split}\implies
\begin{split}
\frac{1}{7}a_3^2-\frac{8}{7}a_3-a_2^2 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1&=\frac{1}{7}(a_3-8)
\end{split}
$$
and finally we have
$$
\begin{split}
9a_3^2-424a_3-3600 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1 &=\frac{1}{7}(a_3-8)
\end{split}
$$
where the first equation is a quadratic equation respect to the single $a_3$.
In sum, a nice way to proceed for such system is to see if you can subtract a multiple of one row from another, in order to simplify the structure and possibly arrive to a system where all the equations are linear except one, which however contains only one of the variables.
|
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|
Prove this Ramanujan series
Prove how this series: $\frac{1}{1^3}\cdot\frac{1}{2} + \frac{1}{2^3}\cdot\frac{1}{2^2} + \frac{1}{3^3}\cdot\frac{1}{2^3} + \frac{1}{4^3}\cdot\frac{1}{2^4} +...= \frac{1}{6}(\log 2)^3+\frac{\pi ^2}{12}(\log 2) + (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+..)$
I don't know how to begin but I know that the left-hand side is of the form $$\sum\frac{1}{n^3}\cdot\frac{1}{2^n}$$
|
This is by no means a complete answer but really a long comment. We could also ask if for a fixed number $k$ is it true that
$$
\sum_{n=1}^{\infty}\frac{1}{n^3k^n}=\text{Li}_{3}\left( \frac{1}{k}\right);
$$
where $\text{Li}_{3}()$ is the trilogarithm. Really we are computing $\text{Li}_{3}(\frac{1}{2}).$
We have the following functional equation from Euler
$$
\text{Li}_{3}\left(\frac{-z}{1-z}\right)+\text{Li}_{3}(z)+\text{Li}_{3}(1-z)\\
=\zeta(3)+\zeta(2)\log(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{1}{3}\log^3(1-z)\\
$$
|
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|
Throwing a die three times - check my proof "A teacher rolls a 6-sided die (numbers from 1 to 6) three times. The lowest number will be the grade of the student.
Calculate the probability of each grade."
My approach is:
Firstly we determine the sample space $\Omega$. So $\Omega:= \{\{\omega_1, \omega_2, \omega_3\},~with~\omega_i \in \{1,...,6\}\}$. The 3-sets $\{\omega_1, \omega_2, \omega_3\}$ are non-ordered and as repitition is allowed we have $\vert\Omega\vert={6+3-1\choose 3}$.
1.) Let be $A_1$ the set of all 3-sets which at least contain one $1$. We have to consider the cases in which the 3-set contains one $1$, two $1's$ and three $1's$. Hence, $\vert A_1 \vert= {5+2-1 \choose 2}+5+1$. The probabilty of $A_1$ is $\frac{21}{56}= 0.375$.
2.) Let be $A_2$ the set of all 3-sets which at least contain one $2$ and no $1$. We have to consider the cases in which the 3-set contains one $2$, two $2's$ and three $2's$. Hence, $\vert A_2 \vert= {4+2-1 \choose 2}+4+1$. The probabilty of $A_2$ is $\frac{15}{56} \approx 0.268$.
In this manner I would proceed for the remaining 3-sets.
However, the solution comes up with slightly different probabilities as it uses the combinatorial model of "ordered sample" and "repitition allowed"
What is wrong with my approach?
|
The grade $1$ occurs when $1,2$ or $3$ dice show $1$.
Only one die showing $1$:
$${3\choose 1}\cdot 5^2$$
Two dice showing $1$ each:
$${3\choose 2}\cdot 5$$
All three dice showing $1$:
$${3\choose 3}$$
Hence:
$$P(1)=\frac{{3\choose 1}5^2+{3\choose 2}5+{3\choose 3}}{6^3}\approx 0.421$$
Similarly:
$$P(2)=\frac{{3\choose 1}4^2+{3\choose 2}4+{3\choose 3}}{6^3}\approx 0.282\\
P(3)=\frac{{3\choose 1}3^2+{3\choose 2}3+{3\choose 3}}{6^3}\approx 0.171\\
P(4)=\frac{{3\choose 1}2^2+{3\choose 2}2+{3\choose 3}}{6^3}\approx 0.088\\
P(5)=\frac{{3\choose 1}1^2+{3\choose 2}1+{3\choose 3}}{6^3}\approx 0.032\\
P(6)=\frac{{3\choose 1}0^2+{3\choose 2}0+{3\choose 3}}{6^3}\approx 0.005\\$$
|
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|
Value of $(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}$ if $\beta$ is the root of $x^3-x-1=0$
If $\beta$ is the root of the equation $x^3-x-1=0$, find the value of
$$(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}.$$
This is what I tried:
$x=\beta$ is a root of $x^3-x-1=0,$
so getting $\displaystyle \beta^3-\beta-1=0\Rightarrow \beta^2=\frac{\beta+1}{\beta}.$
Now $$3\beta^2-4\beta = 3\bigg(\frac{\beta+1}{\beta}\bigg)-4\beta.$$
Don't know how to continue.
|
The trick is that if we reduce $(x + a)^3$ modulo $x^3 - x - 1$, we'll get $3 a x^2 + (3 a^2 + 1) x + (a^3 + 1)$. Therefore, if $\beta$ is a root of $x^3 - x - 1$,
$$(1 + \beta)^3 = 3 \beta^2 + 4 \beta + 2, \\
(1 - \beta)^3 = 3 \beta^2 - 4 \beta.$$
Then we need to define the cube root in such a way that $(z^3)^{1/3} = z$ for $z = 1 \pm \beta$.
|
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|
$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)= -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$?
Solution:
If $\frac{\pi}{2} < x < \pi$ then right triangle with acute angle $x$ facing west at second quadrant, the hypotenuse must be positive and the only negative is the adjacent.
$\cos(x)= -\frac{24}{25}$ then $adjacent=-24$ and $front=7$ so $\sin(x) = \frac{7}{25}$.
If $\pi < y <\frac{3\pi}{2}$ then right triangle with acute angle $x$ facing west at third quadrant, the front and adjacent are negative. If $\tan(y) = \frac{9}{40}$ then the front is $-9$, the adjacent is $-40$, and the hypotenuse is $41$. So $\sin(y) = -\frac{9}{41}$ and $\cos(y)=-\frac{40}{41}$
The right triangle that we consider is the one with the adjacent side on the $x$-axis right..?
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Your method is correct. Indeed, refer to the graphs:
$\hspace{1cm}$
$$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$
Also note:
$$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos (\pi -x)=\frac{24}{25}; \quad \sin (\pi -x)=\frac{7}{25}=\sin x.\\
\tan y=\tan(y-\pi)=\frac{9}{40}=\frac{-9}{-40} \Rightarrow \begin{cases}\sin y=-\sin(y-\pi)=-\frac{9}{41} \Rightarrow \sin (y-\pi)=\frac9{41}\\ \cos y=-\cos(y-\pi)=-\frac{40}{41} \Rightarrow \cos (y-\pi)=\frac{40}{41}\end{cases}$$
Alternatively, you can remember the signs of sine, cosine, tangent and cotangent functions in the four quarters: sine (++--); cosine (+--+); tangent and cotangent (+-+-).
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|
A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from this question here we have
\begin{eqnarray*}
\int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1.
\end{eqnarray*}
It is easy to show
\begin{eqnarray*}
\int_0^1 \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z).
\end{eqnarray*}
So we have (with a little tad of algebra)
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 &=& \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\
&=& \frac{1}{\pi} \int_0^1 \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\
&=& \frac{2}{\pi} \int_0^1 \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
\end{eqnarray*}
This suggests we should consider the integral (sub $t=\sin(\theta)$)
\begin{eqnarray*}
\frac{2}{\pi} \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
= \int_0^{\pi/2} \frac{d \theta }{1- x^2 y^2 \sin^2(\theta)}.
\end{eqnarray*}
Now it is well known (Geometrically expand, integrate term by term & sum the familiar plum) that
\begin{eqnarray*}
\frac{2}{\pi} \int_0^{\pi/2} \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}}
\end{eqnarray*}
and using this we have
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}} .
\end{eqnarray*}
The above double integral reminds of
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2} = \frac{\pi^2}{8}
\end{eqnarray*}
which can be evaluated using the substitution $x= \frac{\sin u}{\cos v}$, $y= \frac{\sin v}{\cos u}$.
My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?
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Through series expansions:
$$ \iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1-x^2 y^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\iint_{(0,1)^2}x^{2n}y^{2n}\,dx\,dy=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}=\int_{0}^{1}\frac{\arcsin(x)}{x}\,dx $$
and this is
$$ \int_{0}^{\pi/2}x\cot(x)\,dx\stackrel{\text{IBP}}{=}\int_{0}^{\pi/2}\log(\sin x)\,dx $$
which is well known to be $\frac{\pi}{2}\log(2)$. It is possible to exploit symmetry, derivatives of the Beta function, Fourier (or Fourier-Legendre) series and probably much more. For instance, the identity
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n} $$
and just Riemann sums.
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|
Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _1 \sqrt{1+[f'(x)]^2} dx
\\ = \int ^2 _1 \sqrt{1+({1\over 2}x^3 -{1 \over 2}x^{-3})^2} dx
\\ = \int ^2 _1 \sqrt{{1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}} dx$$
Let $u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}$
$du = {{3 \over 2} x^5 - {3\over 2}x^{-7} dx}$
However, I cannot seem to integrate this
How can I integrate this equation?
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Hint:
$$2+x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2$$
|
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|
Difficult integral involving trigonometric and hypertrigonometric functions This is definitely the most difficult integral that I've ever seen.
Of course, I'm not able to solve this.
Could you help me?
$$\int { \sin { x\cos { x } \cosh { \left( \ln { \sqrt { \frac { 1 }{ 1-\sin { x } } } +\tanh ^{ -1 }{ \left( \sin x \right) +\tanh ^{ -1 }{ \left( \cos { x } \right) } } } \right) } } dx } $$
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Mathematica gives:
$$-\frac{\sqrt{\frac{1}{1-\sin (x)}} \sqrt{\sin ^2(2 x)} \csc^2(x) \\ \left(-90 \sin \left(\frac{x}{2}\right)+35 \sin \left(\frac{3 x}{2}\right)-3 \sin \left(\frac{5 x}{2}\right)+15 \cos \left(\frac{3 x}{2}\right)+3 \cos \left(\frac{5 x}{2}\right)+30 \cos \left(\frac{x}{2}\right) \left(4 \sqrt{\frac{1}{\cos (x)+1}} \log \left(\tan
\left(\frac{x}{2}\right)-1\right)-4 \sqrt{\frac{1}{\cos (x)+1}} \log \left(2 \sqrt{\frac{1}{\cos (x)+1}}+\tan \left(\frac{x}{2}\right)+1\right)+1\right)\right)}{60 \left(\csc \left(\frac{x}{2}\right)+\sec \left(\frac{x}{2}\right)\right)}$$
so I doubt you'll want to work through this by hand.
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|
Simplify system of 3 equations of 3 recursive functions Is it possible to simplify this system into one equation of f only?
f(N) = f(N-1) + f(N-2) + 2*g(N) + h(N)
g(N) = f(N-2) + g(N-1)
h(N) = f(N-2) + h(N-2)
Initial values if needed:
f(1) = 2
f(2) = 5
g(2) = 1
g(3) = 2
h(2) = 1
h(3) = 1
I'm not sure it is possible. The simplification I have is:
()=(−1)+5(−2)+(−3)−(−4)
But it maybe includes some additional inputs.
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We start with
$$\begin{align}
&f(n) = f(n - 1) + f(n - 2) + 2·g(n) + h(n)
\\&-\\
&f(n - 2) = f((n - 2) - 1) + f((n - 2) - 2) + 2·g(n - 2) + h(n - 2)
\end{align}$$
And obtain
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1) - 2·g(n - 2) + 2·g(n) - h(n - 2) + h(n)
$$
We can substitute $h(n)-h(n-2)$ in the equation and get
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1) - 2·g(n - 2) + 2·g(n)+f(n-2)
$$
Now all that's left to eliminate are the summands $- 2·g(n - 2) + 2·g(n)$.
Using the recurrence $g(n) = f(n - 2) + g(n - 1)$ this becomes:
$$- 2·g(n - 2) + 2·f(n - 2) + 2·g(n - 1)$$
Which, using the same recurrence (shifted by $n\mapsto n-1$) becomes
$$ 2·f(n - 2) + 2·f(n-3)$$
And so, we obtain the recurrence:
$$
f(n) = - f(n - 4) - f(n - 3) + 2·f(n - 2) + f(n - 1)+ 2·f(n - 2) + 2·f(n-3)+f(n-2)
$$
Or, simplified
$$f(n) = f(n - 1)+ 5·f(n - 2)+ f(n - 3) - f(n - 4) $$
So, your reduction was correct.
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|
Formula for calculating the odds per user of winning in a raffle each player can win once Am trying to write a program which gives each user in a raffle contest their odds so far at winning.
The rules of the game are simple:
A predefined number of tickets are sold for example 1000 each user can at most buy 50 tickets and a user can only win once in the game after that all his tickets become invalid or pulled out and a new draw begins giving the other users a better chance of winning. The game will have 10 winners only.
With the data collected from the software I would then know how many tickets were sold, the users that bought them and the amount of tickets that each user has. Now I just need a formula or pseudocode that would give each user their statistic probability of winning based on the data acquired, so that it be can used before and after each draw in the game to show each user their odds so far.
I have looked at similar questions asked here, but no one seems to want to address the part that if a person wins the rest of their tickets become invalid. Am not good with probability or understand those fancy notations, so I don't understand is such a thing possible to calculate per user.
Thanks for the help
Update
Testing my understanding of joriki second method:
lets say 10 tickets were sold to 4 users each bought A: 1, B: 2, C: 4, D: 3
and there will be 3 prizes given to users.
I calculated the total probability of being drawn for each user to be
A = $\frac{1}{10} + \frac{2}{10}*\frac{1}{8}*\frac{1}{6} + \frac{4}{10}*\frac{1}{6}*\frac{1}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.1482
B = $\frac{1}{10}*\frac{2}{9}*\frac{2}{8} + \frac{2}{10} + \frac{4}{10}*\frac{2}{6}*\frac{2}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.3817
C= $\frac{1}{10}*\frac{4}{9}*\frac{4}{8} + \frac{2}{10}*\frac{4}{8}*\frac{4}{6} + \frac{4}{10} + \frac{3}{10}*\frac{4}{7}*\frac{4}{4}$ = 0.6603
D= $\frac{1}{10}*\frac{3}{9}*\frac{3}{8} + \frac{2}{10}*\frac{3}{8}*\frac{3}{6} + \frac{4}{10}*\frac{3}{6}*\frac{3}{2} + \frac{3}{10}$ = 0.6500
Their total sum is 1.8403 and not 3 ? also is this considered the total probability of being drawn for the 3 draws or just for the first round of the game with the tickets becoming invalid
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I don't think you'll find a closed formula for this. If you're OK with only specifying the odds up to a couple of decimal places, the easiest way to do this would be to simulate the draw a couple of billion times and use the resulting relative frequencies. If you want the exact probabilities instead, you can keep track of all possibilities of who already won and work through one draw at a time. For instance, if $A$, $B$ and $C$ bought $1$, $2$ and $3$ tickets, respectively, and $2$ prizes are being drawn, in the first step $A$ would be drawn with probability $\frac16$, $B$ with probability $\frac26=\frac13$ and $C$ with probability $\frac36=\frac12$. Then in the second step, if $A$ was drawn in the first step, there would be $5$ tickets left, and $B$ would be drawn with probability $\frac25$ and $C$ with probability $\frac35$, and so on. The total probability of being drawn would be $\frac16+\frac26\cdot\frac14+\frac36\cdot\frac13=\frac5{12}$ for $A$, $\frac16\cdot\frac25+\frac26+\frac36\cdot\frac23=\frac{11}{15}$ for $B$ and $\frac16\cdot\frac35+\frac26\cdot\frac34+\frac36=\frac{51}{60}$ for $C$. For debugging your code, you can check to see that they add up to $2$ as they should.
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|
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7},\cos\frac{5 \pi}{7}$
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7}\space \text{and}\space \cos\frac{5 \pi}{7}$.
How to even solve it ? I have no idea.
Because cubic have no formula like quadratic equations to obtain the roots and the expression like $\cos\frac{\pi}{7}$ is something that I don't think I can evaluate with my limited knowledge (I'm in highschool) of only $\cos 2x$ or $\cos 3x$ type. Since it has denominator of 7, I thought maybe the seventh root of unity might help me, but since only some of its multiples are used, I don't think so it will be of much use.
How do I solve this problem. Please help.
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You could prove it with just a couple of high-school trigonometric identities.
Let $\theta = \pi/7 $ and recognize
$$x_1=\cos \theta = -\cos 6\theta$$
$$x_2=\cos3\theta = -\cos 4\theta$$
$$x_3=\cos5\theta = -\cos 2\theta$$
(1). Evaluate their product by applying the identity $\sin 2x = 2\sin x \cos x$
$$ \cos2\theta \cos4\theta \cos6\theta = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{1}{8}$$
(2). Evaluate their sum by applying $\sin(x+y)+\sin(x-y)=2\sin x\cos y$,
$$ 2\sin 2\theta(\cos2\theta + \cos4\theta + \cos6\theta)$$
$$=\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)$$
which after some cancellation leads to,
$$\cos2\theta + \cos4\theta + \cos6\theta = -\frac{1}{2}$$
(3). Evaluate the sum of their cross products by applying $\cos(x+y)+\cos(x-y)=2\cos x\cos y$,
$$\cos4\theta \cos6\theta + \cos6\theta \cos2\theta + \cos2\theta \cos4\theta $$
$$= \cos2\theta + \cos4\theta + \cos6\theta = -\frac{1}{2}$$
So, in terms of $x_1$, $x_2$ and $x_3$,
$$x_1+x_2+x_3=\frac{1}{2}$$
$$x_1x_2+x_2x_3+x_3x_1=-\frac{1}{2}$$
$$x_1x_2x_3=-\frac{1}{8}$$
And
$$(x-x_1)(x-x_2)(x-x_3)=x^3 - \frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{8}$$
Therefore, $x_1=\cos \theta$, $x_2=\cos 3\theta$ and $ x_3=\cos 5\theta$ are the roots of
$$x^3-\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{8}=0$$
or,
$$8x^3-4x^2-4x+1= 0$$
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Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$ How to prove
$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$
Here is how I came up with this relation:
In this solution @Kemono Chen elegantly proved
$$\int_0^a\frac{\ln(1+ax)}{1+x^2}dx=\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{2}$$
and while trying to prove the identity in (2) starting from RHS, I ended up with the relation in (1). So any straightforward method to prove (1)? Plus any good applications of (1)?
The transformation of the integral in (2) was done by @Jack D'Aurizio here.
I will post my proof in the answer section soon and I am tagging "harmonic number" as the proof involves it in case you are curious. Thanks
UPDATE: If we let $\frac{1-x}{x}\mapsto x$ in (1) then combine with (2) we have
$$\int_0^1\frac{1+a^2x}{1+a^2x^2}dx=\int_0^\infty\frac{\ln x}{a^2+(1+x)^2}=\frac1{2a}\arctan a\ln(1+a^2)\tag{3}$$
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In other words we want to show that:
$$\color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$
This can be seen via the substitution:
$$\frac{1-x}{1+a^2 x}=t\Rightarrow x=\frac{1-t}{1+a^2 t}\Rightarrow dx=-\frac{1+a^2}{(1+a^2t)^2}dt$$
$$\Rightarrow \color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\int_0^1 \frac{\ln t}{1+a^2 \frac{(1-t)^2}{(1+a^2t)^2}}\frac{1+a^2}{(1+a^2t)^2}dt\overset{t=x}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$
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Evaluate $\sum\limits_{n=0}^{\infty} \frac{\cos(nx)}{2^n}$ where $\cos x = \frac15$ Evaluate
$$\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{2^n}$$
where $\cos x = \frac{1}{5}$.
This is a complex number question. But I don’t know where to start. Maybe need to use the DeMoivre’s Theorem?
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Consider the following auxiliar sum: $\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}$, and use that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ to get: $\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+\sum_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+i\dfrac{\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{\cos(nx)+i\sin(nx)}{2^{n}}=\sum_{n=0}^\infty \dfrac{e^{inx}}{2^{n}}=\sum_{n=0}^\infty (\dfrac{e^{ix}}{2})^{n}=\dfrac{1}{1-e^{ix}/2}=\dfrac{2}{2-e^{ix}}=\dfrac{2}{2-\cos(x)-i\sin(x)}=\dfrac{2}{((2-\cos(x))-i\sin(x))}\dfrac{(2-\cos(x))+i\sin(x)}{((2-\cos(x))+i\sin(x))}=\dfrac{4-2\cos(x)+2i\sin(x)}{(2-\cos(x))^{2}+\sin^{2}(x)}=\dfrac{4-2\cos(x)+2i\sin(x)}{4-4\cos(x)+\cos^{2}(x)+\sin^{2}(x)}=\dfrac{4-2\cos(x)+2i\sin(x)}{5-4\cos(x)}=\dfrac{4-2\cos(x)}{5-4\cos(x)}+i\dfrac{2\sin(x)}{5-4\cos(x)}$
and comparing imaginary and real parts with $\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+\sum\limits_{n=0}^\infty i\dfrac{\sin(nx)}{2^{n}}=\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}+i\sum\limits_{n=0}^\infty \dfrac{\sin(nx)}{2^{n}}$
to obtain: $\sum\limits_{n=0}^\infty \dfrac{\sin(nx)}{2^{n}}=\dfrac{2\sin(x)}{5-4\cos(x)}$ and $\sum\limits_{n=0}^\infty \dfrac{\cos(nx)}{2^{n}}=\dfrac{4-2\cos(x)}{5-4\cos(x)}$
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|
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\tan\theta=-5$
Ref$(\theta)$=$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}$
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\
Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\
=\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix}
$$
Why am I not getting the required solution in Method two using matrix method ?
Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?
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Translating up by $6$ units again, we find that the angle between the $x$-axis and $(4,-7)$ is $-\tan^{-1} \frac{7}{4}$. In addition, the angle between the line and the $x$-axis is $-\tan^{-1} 5$. So the reflected point must have an angle of:
$$\theta =-\tan^{-1} 5 -\tan^{-1} 5 + \tan^{-1} \frac{7}{4}$$
In addition, the distance between the origin and the original point is the same as the distance between the origin and the reflected point (since the reflected angles are equal). This gives:
$$r = \sqrt{4^2 + 7^2}$$
Now, the reflected point is at $(r \cos \theta, r \sin \theta) = (-1,-8)$. Translating down by $6$ units, we again get $(-1, -14)$.
|
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|
Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.com/user/Khansis/favorites/folder/7408635/math
|
It is $$4(a^6+b^6)-(a+b)(a^2+b^2)(a^3-b^3)=(a-b)^2(a^2+b^2)(3a^2+5ab+3b^2)\geq 0$$
|
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|
how do you differentiate x^(3/4) using first principle $$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$
I understand the process till
$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}$$
and post expansion
$$\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}$$
but beyond this i am unable to reduce to:
$\frac 34\cdot x^{\frac{-1}{4}}$
|
Why don't you expand from the start
$$\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=x^n\cdot\lim_{h\to0}\dfrac{\left(1+\dfrac hx\right)^n-1}h$$
Now use Binomial series
Alternatively, set $$(x+h)^{1/4}=a,x+h=a^4; x^{1/4}=b, x=b^4$$
$$\lim_{h\to0}\dfrac{(x+h)^{3/4}-x^{1/4}}h=\lim_{a\to b}\dfrac{a^3-b^3}{a^4-b^4}=\lim_{a\to b}\dfrac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}=\dfrac{3b^2}{4b^3}=\dfrac3{4b}=\dfrac3{4x^{1/4}}$$
|
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|
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$
Can someone help me with the second one?
I forgot to tell that solving the equation is not an option in my case.
|
Hint: What is $\tfrac{-b+\sqrt{b^2-4ac}}{2a} - \frac{-b-\sqrt{b^2-4ac}}{2a}$ ?
|
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|
Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$
Then, I want to find a geometric way to prove this identity, as $A+B+C=\pi$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!
|
For positives $A$, $B$ and $C$ there is the following way.
Let $A=\max\{A,B,C\},$ $\pi-A=\alpha,$ $\frac{\pi}{2}-B=\beta$ and $\frac{\pi}{2}-C=\gamma$.
Thus, $\alpha$, $\beta$ and $\gamma$ are measured angles of the triangle and let sides-lengths of the triangle be $a$, $b$ and $c$ respectively.
Thus, since by law of sines $$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma},$$ by law of cosine we obtain:
$$\sin^2\alpha=\sin^2\beta+\sin^2\gamma-2\sin\beta\sin\gamma\cos\alpha$$ or
$$\sin^2(\pi-A)=\sin^2\left(\frac{\pi}{2}-B\right)+\sin^2\left(\frac{\pi}{2}-C\right)-2\sin\left(\frac{\pi}{2}-B\right)\sin\left(\frac{\pi}{2}-C\right)\cos(\pi-A)$$ or
$$1-\cos^2A=\cos^2B+\cos^2C+2\cos B\cos C\cos A$$ and we are done!
|
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
|
Hint:
Let $$x^{81}+x^{49}+x^{25}+x^9+x=p(x)(x^3-x)+ax(x-1)+bx(x+1)+c(x-1)(x+1)$$
where $a,b,c$ are arbitrary constants
Put $x=0,-1,1$ to find $a,b,c$
|
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|
What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$?
If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.
What is the correct way to solve this kind of problems? I have tried to simplify the given inequality:
$$0.70<\frac{p}{q}< \approx0.73 \quad\text{ or }\quad \frac{21}{30}<\frac{p}{q}<\frac{22}{30}$$
Thank you in advance!
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The way to find the smallest denominator "from scratch" is with continued fractions.
Begin by rendering the proposed bounds thusly:
$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$
$\dfrac{11}{15}=\dfrac{1
}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$
The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue. We may now replace those entries with the smallest whole number lying between, thus
$1+\dfrac{1}{3}<2<3$
So the smallest denominator fraction meeting the betweenness criterion will be
$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$
|
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|
Sum of the series $1-2r\cos~\theta +3r^2 \cos~2\theta -4r^3\cos~3\theta+\dots$..
Sum of the series
$$1-2r\cos~\theta +3r^2 \cos~2\theta -4r^3\cos~3\theta+\dots\qquad |r|<1.$$
I don't know how to find sum other than for geometric series and this is not geometric.
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You'll need to combine$$\sum_{n\ge0}(n+1)z^n=\frac{d}{dz}\sum_{n\ge0}z^{n+1}=\frac{d}{dz}\frac{z}{1-z}=\frac{d}{dz}\frac{1}{1-z}=\frac{1}{(1-z)^2}$$with$$\sum_{n\ge0}(n+1)(-r)^n\cos n\theta=\Re\sum_{n\ge0}(n+1)(-r\exp i\theta)^n=\Re\frac{1}{(1+r\exp i\theta)^2}\\=\Re\frac{(1+r\exp -i\theta)^2}{(1+r^2+2r\cos\theta)^2}=\frac{1+2r\cos\theta+r^2\cos 2\theta}{(1+r^2+2r\cos\theta)^2}.$$
|
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|
Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?
$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$
What is the differentiation rule used?
|
Using the product rule reveals
$$\begin{align}
\frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=m\vec u \frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)+\frac{1}{\sqrt{1-u^2/c^2}}\frac{d}{dt}\left(m\vec u\right)\\\\
&=m\vec u\left(-\frac12(1-u^2/c^2)^{-3/2}\left(\frac{-2u}{c^2}\right)\,\frac{du}{dt}\right)\\&+(1-u^2/c^2)^{-3/2}(1-u^2/c^2)\frac{d}{dt}\left(m\vec u\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(\vec uu\frac{du}{dt}+(c^2-u^2)\frac{d\vec u}{dt}\right)\tag1
\end{align}$$
Forming the inner product of the right-hand side of $(1)$ with $\vec u$, we find that
$$\begin{align}
\frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^3\frac{du}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12\frac{du^2}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12 \frac{d(\vec u\cdot \vec u)}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \vec u\cdot \frac{d\vec u}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\
&=m(1-u^2/c^2)^{-3/2}\vec u\cdot \frac{d\vec u}{dt}
\end{align}$$
as was to be shown!
|
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Is there a systematic way to derive constraints in a system of equations? I have this system of equations:
$x_1 = y_1+ y_2+ y_3$
$x_2 = y_1- y_2$
$x_3 = y_1 + y_2 - 2y_3$
And I have these constraints:
$y_1 \geq y_2 \geq 0 \geq y_3$
From the constraints I can derive these constraints for the $x$'s:
$x_2 \geq 0$
$x_3 \geq x_1$
$x_3 \geq x_2$
I "derived" these constraints by eye, but they all seem very natural. However, when I put the constraints into Mathematica, I get a different set of constraints:
$x_2 \geq 0$
$x_3 \geq x_1$
$2x_1 + x_3 \geq 3x_2$
It's straightforward to show that Mathematica's 3rd constraint is equivalent to $y_1 + y_2 \geq y_1 - y_2$, which in turn is equivalent to $y_2 \geq 0$, so it's a valid constraint. However, I don't see why these constraints are different from the ones I derive (they're definitely different - I got Mathematica to integrate an equation subject to these constraints, and the two constraints yield a different answer). I can't tell which constraint is incorrect or why, so I'm asking if there's a systematic way to derive them.
|
Your answer is not true. Indeed,
\begin{align}
\left\{\begin{array}{l}
x_2 \ge 0\\
x_3 \ge x_1\\
x_3 \ge x_2 \\
\end{array}
\right. \Leftrightarrow
\left\{\begin{array}{l}
y_1-y_2 \ge 0\\
y_1+y_2 - 2y_3 \ge y_1+y_2 + y_3\\
y_1+y_2 - 2y_3 \ge y_1 - y_2 \\
\end{array}
\right. \Leftrightarrow
\left\{\begin{array}{l}
y_1 \ge y_2\\
0 \ge y_3\\
y_2 \ge y_3 \\
\end{array}
\right. \Leftrightarrow (y_1\ge y_2\ge y_3) \land (0 \ge y_3)
\end{align}
which is different from $y_1\ge y_2\ge 0\ge y_3$.
You may get the correct answer as follows.
\begin{align}
\left\{\begin{array}{l}
x_1 = y_1 + y_2 + y_3 \\
x_2 = y_1 - y_2 \\
x_3 = y_1 + y_2 - 2y_3 \\
\end{array}
\right. \Longleftrightarrow
\left\{\begin{array}{l}
y_1 = \frac{1}{3}x_1 + \frac{1}{2}x_2 + \frac{1}{6}x_3\\
y_2 = \frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3\\
y_3 = \frac{1}{3}x_1 - \frac{1}{3}x_3
\end{array}
\right.
\end{align}
Then,
\begin{align}
y_1\ge y_2 \ge 0 \ge y_3 \Leftrightarrow \left\{\begin{array}{l}
y_1 \ge y_2\\
y_2 \ge 0\\
0 \ge y_3
\end{array}
\right.
\Leftrightarrow
\left\{\begin{array}{l}
\frac{1}{3}x_1 + \frac{1}{2}x_2 + \frac{1}{6}x_3 \ge \frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3\\
\frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3 \ge 0\\
0 \ge \frac{1}{3}x_1 - \frac{1}{3}x_3
\end{array}
\right.
\Leftrightarrow
\left\{\begin{array}{l}
x_2 \ge 0\\
2x_1 + x_3 \ge 3x_2\\
x_3 \ge x_1 \\
\end{array}
\right.
\end{align}
|
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|
How to find the number of solutions of an augmented matrix? Given a system of linear equations is represented as a augmented matrix:
$$M = \left[\begin{array}{cccc|c}1&0&8&0&1\\0&1&1&0&3\\0&0&0&1&1\end{array}\right]$$
How do I find the exact number of solutions in the system of equations?
If I'm not mistaken, it has unique solutions, therefore, it shouldn't be inconsistent or have an infinite amount of solutions.
Is it wrong to say that there are 3 solutions?
|
In case you mean to solve the under-determined system $AX=B:$
$$\begin{pmatrix} 1 & 0 & 8 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}
\begin{pmatrix} w \\x \\ y \\ z \\ \end{pmatrix}= \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}.$$ Multiply from let with $A^T$ to get
$$\begin{pmatrix} 1 & 0 & 8 & 0\\ 0 & 1 & 1 &0\\ 8 & 1 & 65 & 0 \\ 0 & 0 & 0& 1 \end{pmatrix} \begin{pmatrix} w \\x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \\ 0 \\ 1 \end{pmatrix}. $$
This simple equation can be solved to give $w=1, x=3, y=0, z=1$.
|
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|
What is the probability that the highest of the three numbers obtained from three dice will be exactly a $4$?
Question: You are going to toll three dice. What is the probability that the highest of the three numbers will be exactly a $4$?
My attempt:
The first die obtains number $4$ with probability $\frac{1}{6}.$
Then the remaining two dice can take numbers $1,2,3,4$ only. So each of them has probability $\frac{4}{6}.$
In total, the required probability is
$$\frac{1}{6}\times \frac{4}{6} \times \frac{4}{6} \times 3 = \frac{2}{9}.$$
But apparently this is not the answer.
It is $\frac{37}{216}.$
May I know where is the flaw in my reasoning?
|
You've summed the probabilities of the events that one particular die shows a $4$ and the other two show not more than $4$. Unfortunately, those events are not disjoint, so you've double-counted the cases where several $4$s appear. To fix that, it helps to separate your computations by the number of $4$s rolled:
The probability that exactly one $4$ is rolled, and the other dice show less, is $\frac 16 \times \frac 36 \times \frac 36 \times 3 = \frac{27}{216}$.
The probability that exactly two $4$s are rolled, and the third die shows less, is $\frac 16 \times \frac 16 \times \frac 36 \times 3 = \frac{9}{216}$.
The probability that all three dice show a $4$ is $\frac 16 \times \frac 16 \times \frac 16 = \frac{1}{216}$.
Together, this makes $\frac{37}{216}$.
|
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|
Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30} \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$
But these give me more complicated expressions.
|
Let $\arctan\left(\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}\right)=y,-90^\circ<y<90^\circ$
$\implies\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}=\tan y=\dfrac{\sin y}{\cos y}$
Rearranging we get $$\cos(6^\circ+y)+\cos(18^\circ+y)=\sin y=\cos(90^\circ-y)$$
$$\iff\cos(90^\circ-y)=\cos(18^\circ+y)+\cos(6^\circ+y)=2\cos6^\circ\cos(12^\circ+y)\ \ \ \ (1)$$
Like Solve equation $\tan(x)=\sec(42^\circ)+\sqrt{3}$
using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
$$\cos36^\circ=\cos72^\circ+\cos60^\circ=2\cos6^\circ\cos66^\circ\ \ \ \ (2)$$
Compare $(1),(2)$
$$\frac{\cos36^\circ}{\cos(90^\circ-y)}=\dfrac{2\cos6^\circ\cos66^\circ}{2\cos6^\circ\cos(12^\circ+y)}$$
$$\iff\dfrac{\cos(12^\circ+y)}{\cos(90^\circ-y)}=\dfrac{\cos66^\circ}{\cos36^\circ}$$
Apply Componendo et Dividendo and Prosthaphaeresis Formulas to find
$$\tan(y-39^\circ)=\tan15^\circ$$
The rest should be easy!
|
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|
Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + 1) + 18$
System
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HINT.-$$N=3^x\cdot5^y\cdot7^z\quad\text{ has (x+1)(y+1)(z+1) divisors }\\5N=3^x\cdot5^{y+1}\cdot7^z\quad\text{ has (x+1)(y+2)(z+1) divisors }\\27N=3^{x+3}\cdot5^{y}\cdot7^z\quad\text{ has (x+4)(y+1)(z+1) divisors }$$ You can get from this the system
$$(x+1)(z+1)[(y+2)-(y+1)]=8\\(y+1)(z+1)[(x+4)-(x+1)]=18$$ Then.........
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
I think that $(0,0)$ , $(1,0)$ and $(0,-1)$ are the only solutions to the above equation, but I'm unable to prove it.
I tried all sorts of things like working $\mod 9$ (but there are just too many cases ), a little bit of algebraic manipulations, tried to determine the parity of $x$ and $y$ etc. But they were to no avail for me.
I tried working modulo $9$ because $a^{3}\equiv 0,1$ or $-1 \pmod 9$.
The manipulations done by me were as follows:-
$x^2 + y^2 =(x-y)^3$ implies that by adding and subtracting $2xy$ on LHS we can rewrite the above equation as $(x-y)^2 +2xy=(x-y)^3$ . This can be rewritten as $2xy=(x-y)^3 -(x-y)^2$. This is all I could achieve here. One thing I did here was substitute $x-y=a$ and $x=a+y$ and rewrite the last equation as $2y^2 +2ay+a^2 -a^3=0$ and then I tried to find the roots of this quadratic in $y$ but this didn't work for me(I think there is something wrong with this approach, do tell me if you see it).That is all I could do.
Another question I would like to ask is do there exist integers $a,b$ and $c$, with none of them equal to zero, which satisfy $a^2 + b^2=c^3$ ? Thank you .
|
Letting $x=y+k$, we are looking for the solutions of
$$ 2y^2+2yk+k^2 = k^3 $$
$$ (2y+k)^2+k^2 = 2k^3 $$
which depend on the integer points on the elliptic curve $w^2=2z^3-z^2=z^2(2z-1)$.
We may assume that $z=\frac{q^2+1}{2}$, leading to the solution $k=\frac{q^2+1}{2},w=q\frac{q^2+1}{2},y=(q-1)\frac{q^2+1}{4},x=(q+1)\frac{q^2+1}{4}$.
Of course, in order to have $\frac{q^2+1}{4}\in\mathbb{Z}$ $q$ must be odd, $q=(2t+1)$. These leads to the solutions
$$\boxed{ x = 2t^3+4t^2+3t+1,\qquad y= 2t^3+2t^2+t. }$$
|
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|
Find the maximum of the value $f(x)=\frac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$
Let $n$ be a given positive integer, find the maximum of the value of
$$f(x)=\dfrac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$$ where $x \le \dfrac{n^2}{4}$ and $x \in \mathbb N^{+}.$
If the $x$ is real positive number, the problem also not easy, because
$$f'(x)=-\dfrac{n^4}{x^2}+4(n^2+3)-8x=0$$
This cubic equation is not easy to handle. But if $x$ be positive integer, maybe have other method can solve it.
|
Let $\mathbb{N}$ denote the positive integers. Given $n\in\mathbb{N}$ fixed, we look for
\begin{align*}
&f:\mathbb{N}\to\mathbb{R}\\
&f(x)=\frac{n^4}{x}+4\left(n^2+3\right)x-4x^2-5n^2\qquad\longrightarrow\qquad\max\tag{1}\\
\end{align*}
provided that
$x\leq \frac{n^2}{4}$
*
*The additive constant $-5n^2$ in (1) is not relevant for determining the value of $x$ which maximizes $f$ and can be ignored.
We consider therefore
\begin{align*}
&g:\mathbb{N}\to\mathbb{R}\\
&g(x)=\frac{n^4}{x}+4\left(n^2+3\right)x-4x^2\qquad\longrightarrow\qquad\max\tag{2}\\
\end{align*}
*We can do some more simplifications by letting $x=\alpha n^2$ with $\alpha\in\left(0,\frac{1}{4}\right]$ respecting this way $x\leq \frac{n^2}{4}$. We obtain
\begin{align*}
g\left(\alpha n^2\right)&=\frac{n^2}{\alpha}+4\left(n^2+3\right)\alpha n^2-4\alpha^2 n^4\\
&=n^2\left(12\alpha+\frac{1}{\alpha}\right)\qquad\longrightarrow\qquad\max
\end{align*}
provided that $\alpha\in\left(0,\frac{1}{4}\right]$.
*The problem finally boils down to analyse a function $h$
\begin{align*}
&h:\left(0,\frac{1}{4}\right]\to\mathbb{R}\\
&h(\alpha)=12\alpha+\frac{1}{\alpha}\qquad\longrightarrow\qquad\max
\end{align*}
Looking for positive extrema: $h^\prime(\alpha)=12-\frac{1}{\alpha^2}=0$ we find a minimum at $$\alpha=\frac{1}{2\sqrt{3}}$$ (whereby $h^{\prime\prime}\left(\frac{1}{2\sqrt{3}}\right)\ne 0$).
Since the minimum $\alpha=\frac{1}{2\sqrt{3}}>\frac{1}{4}$ and $h^\prime(\alpha)<0$ for $\alpha \in\left(0,\frac{1}{4}\right]$ we conclude the function $h$ is monotonically decreasing in $\left(0,\frac{1}{4}\right]$.
Conclusion: The maximum value of $x=\alpha n^2$ is given when $\alpha=\frac{1}{n^2}$ resulting in $\color{blue}{x=1}$.
We observe due to $x\leq \frac{n^2}{4}$ the solution $\color{blue}{x=1}$ is valid iff $n\geq 2$ and the empty set in case $n=1$.
|
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|
Are there any examples of $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$? I am writing a short discussion of Farey numbers and was wondering if there are any examples of when the mediant function is ever actually equal to the sum of the two fractions in the usual sense? (Not to produce Farey numbers, I just thought it might be an amusing way to introduce Farey addition).
Explicitly: Are there any examples of fractions $\frac{a}{b}$ and $\frac{c}{d}$ where
$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$
for positive integers?
Currently I have found the example
$$\frac{1}{1}+\frac{1}{i} =\frac{1+1}{1+i}$$
if we remove the integers requirement but it would be nice to find a case that didn't involve complex numbers though!
|
Seems like straight forward algebra manipulation
$\frac ab + \frac cd = \frac {a+c}{b+d}$
$(ad+bc)(b+d) = bd(a+c)$
$abd + b^2c + ad^2+bcd = abd +bcd$
$b^2c +ad^2 = 0$
$b^2c = -ad^2$
Wolog $b,d$ are positive. Wolog $c$ is positive an $a$ negative (I assume you don't want $a=c =0$.)
If we let $c = c'(M^2)$ where $c'$ is square free then $a =-c'N^2$ for some $n$. We can shuffle the square factors of $b,c,M,N$ around anyway we want.
$c =12$ and $a=-75$ and $b=5$ and $d=2$ would be a solution.
$\frac {-75}5 + \frac {12}{2} =\frac {-75+12}{5+2}=-9$.
|
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|
A property of Nesbitt's inequality I think it's not new however I think it's interesting :
Let $a,b,c>0$ and $x>0$ then the function :
$$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$
$f(x)$ is increasing
My first try :
The derivative of $f(x)$ is :
$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a^x(b^x\ln(b)+c^x\ln(c))}{(b^x+c^x)^2}$$
We multiply by $x$ it gives :
$$f'(x)=\frac{1}{x}\Big(\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\Big)$$
$x$ is positive so we have to prove :
$$\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\geq 0$$
We put $a^x=u$ , $b^x=v$ and $c^x=w$ so the inequality is equivalent to :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq 0$$
Now with the concavity of $\ln(x)$ and the inequality of Jensen's we have :
$$\ln\Big(\frac{v^2+w^2}{v+w}\Big)\geq \frac{v\ln(v)+w\ln(w)}{v+w}$$
So we have :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\ln(u)-u\Big(\ln\Big(\frac{v^2+w^2}{v+w}\Big)\Big)}{(v+w)}\geq 0$$
Or :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Now we have to prove :
$$\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Unfortunatly I'm stuck here ...
Second try :
For $a,b,c>0$ the function $g(x)=\frac{a^x}{b^x+c^x}$ is convex so the function $f(x)$ is convex as sum of convex functions .
Now we have with $a,b,c>0$ :
$$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq \sum_{cyc}\frac{a}{b+c}$$
Fact wich have been proven by Michael Rozenberg (see my questions)
Remains to apply the three chord lemma to get the increase of the function $f(x)$
My question :
Can someone could find an alternative proof or prove this last inequality ?
There are generalizations of this facts ?
|
By your assumption
$$f'(x)=\sum_{cyc}\frac{a^x\ln{a}(b^x+c^x)-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}\frac{a^x(b^x(\ln{a}-\ln{b})-c^x(\ln{c}-\ln{a}))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln{a}-\ln{b})\left(\frac{a^xb^x}{(b^x+c^2)^2}-\frac{a^xb^x}{(c^x+a^x)^2}\right)=$$
$$=\sum_{cyc}\frac{a^xb^x(\ln{a}-\ln{b})(a^x-b^x)(a^x+b^x+2c^x)}{(b^x+c^x)^2(c^x+a^x)^2}\geq0$$ because $$(\ln{a}-\ln{b})(a^x-b^x)=b^x\ln\frac{a}{b}\left(\left(\frac{a}{b}\right)^x-1\right)\geq0$$ for all $x>0$ and we are done!
|
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|
Find all n such that : $3 \mid (n2^{n}+1)$ Question :
Determine all natural numbers n such that 3
divides $n\cdot2^{n}+1$
Actually I don't have any ideas to approach but
my efforts :
I see $n=1,2,7,8,13,14$ so I think :
$n=6k+1$ and $n=6k+2$ $k\in \Bbb{N} $
If I'm not wrong but I don't know how ? I prove it ?
|
$2^n \equiv (-1)^n \pmod 3$.
So if $n$ is even then $n2^n + 1\equiv n+1 \equiv 0 \pmod 3$ and $n\equiv -1\pmod 3$.
And so $n\equiv 0 \pmod 2$ and $n\equiv 2\pmod 3$ so $n\equiv 2\pmod 6$.
And indeed $(6k+2)2^{6k+2}+1=(6k+2)4^{3k+1}+1\equiv (6k+2)*1^{3k+1} +1\equiv 6k + 2+ 1\equiv 0 \pmod 3$.
If $n$ is odd then $n2^n+1 \equiv -n+1\equiv 0 \pmod 3$ so $n\equiv 1\pmod 3$.
And so $n\equiv 1 \pmod 2$ and $n\equiv 1 \pmod 3$ so $n\equiv 1 \pmod 6$.
And indeed $(6k+1)2^{6k+1}+1\equiv 2^{6k+1}+1\equiv 4^{3k}*2 + 1\equiv 1^{3k}*2 + 1 \equiv 2+ 1\equiv 0\pmod 3$.
...
To check that $n\equiv 0,3,4,5\pmod 6$ can not be solutions, clearly $n\equiv 0,3\pmod 6$ imply $3|n$ so $3\not \mid n2^{n} + 1$.
And $(6k + 3 + i)2^{6k+3 + i}+1\equiv i2^{6k+ 3+i}+1\equiv i4^k*2^{3+i}+1\equiv i2^32^i\equiv i8*2^i +1\equiv -i2^i+1\pmod 3$ and $-2^1+1 \equiv -1\pmod 3$ and $-2*2^2+1\equiv -7\equiv -1\pmod 3$.
|
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|
Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
|
Using the Extended Euclidean Algorithm (which can be applied to $\mathbb{Q}[x]$ as well as $\mathbb{Z}$), we get
$$
\left(36n^2+24n+16\right)\left(n^4+n+1\right)-\left(18n^3+12n^2+8n-27\right)\left(2n^3+5\right)=151
$$
if $2n^3+5\mid n^4+n+1$, we must also have $2n^3+5\mid151$. Since $151$ is prime and there is no integer $n$ so that $2n^3+5=\pm1$ or $2n^3+5=\pm151$, there is no integer $n$ so that $2n^3+5\mid n^4+n+1$.
|
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|
Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$ Additional Info
$x < 1, y<2, x+y>0$
I tried doing AM-GM (since all of the terms are positive.
$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$
this way we can cancel the x and y's
thus the maximum value is
$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$
but the answer is said to be 1(?) where $x =0,$ and $y = 1$
Can someone tell me what I did wrong.
|
A much more Logical way to solve this without algebraic manipulations:
Since we want to maximize $(1-x)(2-y)^2(x+y)$ , and all the terms are positive, we can see that we might want to maximize $(2-y)^2$ in particular as it is the biggest among all terms.
$$\max \space(2-y)^2 = (2-0)^2 = 4 \space\space\space\space\space\space\space \text for \space y = 0$$
So the equation is reduced to $$(1-x)\times 4\times(x+0)$$
$$= 4\times(1-x)\times x$$
Again to $\max$ this equation further , we want to maximize $(1-x)\times x$ , which happens at $x=\frac 12.$
Hence $$\max \space[(1-x)(2-y)^2(x+y)] = \left(1-\frac12\right) \times (2-0)^2 \times \left(\frac12 + 0 \right)$$
$$ = \frac12\times4\times\frac12 = 1$$
|
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|
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
|
You can get a faster result by using a different way to complete the square: split the middle term
$$
z^4-3z^2+1=(z^2-1)^2-z^2=(z^2-z-1)(z^2+z-1)
$$
Now the remaining two quadratic equations are easy to solve.
$$
0=z^2-z-1\implies z=\frac12(1\pm\sqrt{5})\\
0=z^2+z-1\implies z=\frac12(-1\pm\sqrt{5})\\
$$
The fifth unit roots satisfy the equation $z^5-1=0$. The non-trivial ones satisfy
$$
|z|=1\text{ and } z^2+\bar z^2 +z+\bar z + 1=0
$$
Eliminating the imaginary part leaves for the real part the equation
$$
0=2(x^2-(1-x^2))+2x+1=(2x)^2+(2x)-1
$$
so that $2x\in\{2\cos(72^\circ),2\cos(144^\circ)\}$ is one of the solutions of the second quadratic equation, and $(-2x)\in\{2\cos(36^\circ),2\cos(108^\circ)\}$ one of the solutions of the first quadratic equation.
|
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|
$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3}$ and $2+\frac{1}{3}$
My working:
$-3|x-2|-1=0$
$-3|x-2|=1$
$|x-2|=-\frac{1}{3}$
Then solve for both the negative and positive value of $\frac{1}{3}$:
Positive version:
$x-2=\frac{1}{3}$
$x=2+\frac{1}{3}$
Negative version:
$x-2=-\frac{1}{3}$
$x = 2-\frac{1}{3}$
How can I arrive at "$(0, -7)$; no x intercepts"? Where did I go wrong in my understanding?
|
For $$x\geq 2$$ we get $$3(x-2)-1=3x-6-1=3x-7$$ .For $$x<2$$ we have $$3(2-x)-1=-3x+5$$
|
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|
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary.
My attempt is as follows:-
Let $t=\frac{1}{1+x^2}$, and let's find out its range for which x is imaginary
$$t=\frac{1}{1+x^2}$$
$$(1+x^2)\cdot t=1$$
$$tx^2+t-1=0$$
$$D<0$$
$$0-4t(t-1)<0$$
$$t(t-1)>0$$
$$t\in (-\infty,0)\quad \cup \quad (1,\infty)$$
So for the equation $at^2-3at+1=0$, we have to find such values of a for which $t\in (-\infty,0) \cup (1,\infty)$. As $t$ should be real,so
$$D\geq 0\Leftrightarrow 9a^2-4a\geq 0\Leftrightarrow a(9a-4)\geq0$$
$$a\in \left(-\infty,0\right] \cup \left[\frac{4}{9},\infty\right)$$
But if we place $a=0$ in the quadratic equation in $t$, then $0+0+1=0$, which is not possible hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\infty\right)$.
Now as we know that roots of quadratic equation $at^2-3at+1=0$ should lie in $(-\infty,0)\cup (1,\infty)$. So
Case 1 : When both roots are negative
$$af(0)>0$$
$$a>0$$
$0$ is greater than both the roots, so
$0>(a+b)/2$ where a and b are roots.
$$0>\frac{3a}{2a}$$
$$0>\frac{3}{2}$$
So $a\in \phi$ for first case
Case 2: When both roots are greater than $1$
$$af(1)>0$$
$$a(a-3a+1)>0$$
$$a(2a-1)<0$$
$$a\in \left(0,\frac{1}{2}\right)$$
$1$ should lie before the roots on the x-axis, so $1<\frac{a+b}{2}$
$$1<\frac{3a}{2a}$$
$$1<\frac{3}{2}$$
So $a\in \left(0,\frac{1}{2}\right)$ for the second case
Case 3: When one root is greater than $1$ and another is negative:
$$af(0)<0\quad \cap \quad af(1)<0$$
$$a<0\quad \cap\quad a(a-3a+1)<0$$
$$a<0\quad \cap \quad a(2a-1)>0$$
$$a\in \left(-\infty,0\right)$$
Hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\frac{1}{2}\right)$
but answer is $a\in \left(-\infty,\frac{1}{2}\right)$
What mistake I am doing, I thought about it a lot but didn't get any breakthroughs. Please help me in this.
|
In your proof you have excluded the interval $(0,4/9)$. Why? For example if $a=1/3\in(0,4/9)$ then the equation becomes
$$\frac{1+3x^2+3x^4}{(1+x^2)^2}=0$$
which has not real roots because the l.h.s. is always positive. Hence $1/3$ should be included in the required set.
If $z(x)=1/(1+x^2)$ then $z(\mathbb{R})=(0,1]$. Let $p(z)=az^2-3az+1$ then
we have to find for which real $a$, $p((0,1])$ does not contain $0$.
We have $3$ cases according to the sign of $a$.
1) If $a>0$ then $p$ is decreasing in $(-\infty,3/2)$ and $p((0,1])=[-2a+1,1)$ and therefore $0\not \in [-2a+1,1)$ iff $-2a+1>0$ iff $a<1/2$.
2) If $a<0$ then $p$ is increasing in $(-\infty,3/2)$ and $p((0,1])=(1,-2a+1]$ and therefore $0\not \in (1,-2a+1]$ for all $a<0$.
3) If $a=0$ then $p$ is identically $1$ and therefore it is never zero.
We may conclude that the given equation has not real roots if and only if $a\in \left(-\infty,\frac{1}{2}\right)$.
|
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|
Taylor Series for $\dfrac{x}{x ^ 2 + x + 1}$. I'm trying to calculate the taylor series of $\dfrac{x}{x ^ 2 + x + 1}$.
Algebraic manipulation didn't get me anywhere, since the roots of $ x ^ 2 + x + 1 $ are complex.
Integrate or derive made the problem worse
Any tips on how to proceed?
|
There seem to be many ways to go about this, so here is one: put $\dfrac{x}{x ^ 2 + x + 1}=\sum_{n=0}^{\infty} a_nx^n$, then $$x=\sum_{n=0}^{\infty} a_nx^n(x^2+x+1)=a_0+(a_1+a_0)x+\sum_{n=2}^{\infty}(a_n+a_{n-1}+a_{n-2})x^n,$$
and by comparing the coefficients we get $a_0=0$, $a_1=1$, and $a_n+a_{n-1}+a_{n-2}=0$ for $n \geq 2$. You can see the coefficients repeat ($a_2=-1,a_3=0,a_4=1,\dots$), so we have $a_{3k}=0, a_{3k+1}=1, a_{3k+2}=-1$, or in other words
$$
\dfrac{x}{x ^ 2 + x + 1}=x-x^2+x^4-x^5+\dots
$$
|
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|
Compute $\lim_{N\to \infty} A_N$ where $\{A_N\}$ is a sequence of matrices
Given matrix, $$A_N = \begin{bmatrix} 1 & \frac{\pi}{2N}\\ \frac{-\pi}{2N} & 1\end{bmatrix}^N$$ compute $$\lim_{N \rightarrow \infty} A_N$$
I took the logarithm of both sides but was not able to figure out the limit. Any suggestions on how to approach this problem?
|
The characteristic polynomial of $A$ is given by:
$$ \det \begin{bmatrix} x-1 & \frac{-\pi}{2N} \\ \frac{\pi}{2N} & x-1\end{bmatrix} =(x-1)^2+\frac{\pi^2}{4N^2}$$
To find the eigenvalues of $A$, set the characteristic polynomial equal to zero, and solve for $x$. This gives us the complex conjugate eigenvalues $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$.
Now, $2$ linearly independent eigenvectors of $A$ would then be: $x=\begin{bmatrix} i \\ -1 \end{bmatrix}$ and $x=\begin{bmatrix} i \\ 1 \end{bmatrix}$, corresponding to the eigenvalues $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$ respectively ( This part should be relatively easy to obtain ).
Hence, $A$ is diagonalisable, and we have $A=PDP^{-1}$, where the diagonal matrix $D$ is such that $D = \begin{bmatrix} 1+\frac{i\pi}{2N} & 0 \\ 0 & 1-\frac{i\pi}{2N} \end{bmatrix} $ and $P= \begin{bmatrix} i & i \\ -1 & 1 \end{bmatrix}$. In addition, we may easily evaluate $P^{-1}$ to be: $P^{-1}= \frac{1}{2i} \begin{bmatrix} 1 & -i \\ 1 & i \end{bmatrix}$.
Since $A^{N}$=$(PDP^{-1})^N$=$PD^NP^{-1}$, and $D^N \rightarrow \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$ as $N \rightarrow \infty$ , we have that $A^N=P\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}P^{-1}$ as $N \rightarrow \infty$ . Furthermore, this is simply equal to: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, giving us the desired limit of the matrix.
|
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|
How to write this as a multiplication: $4 x^2 y^2 - (x^2 + y^2 - z^2)^2$? I have the following polynomial:
$$ 4 x^2 y^2 - (x^2 + y^2 - z^2)^2$$
(which comes up, for example, in computing the area of a triangle using the cosine law).
I would like to convert this to a product.
Wolfram tells me it's
$$ -(x - y - z) (x + y - z) (x - y + z) (x + y + z)$$
How can I find this form if I don't already know it? What operations should I perform?
|
$$ \begin{align} \color{magenta}{4 x^2 y^2 - (x^2 + y^2 - z^2)^2} & = \color{brown}{(2xy +x^2+y^2-z^2)(2xy -x^2-y^2+z^2)} \\ &= \color{blue}{((x+y)^2-z^2)(-(x-y)^2 -z^2)}\\
& = \boxed{\color{red}{-(x+y+z)(x+y-z)(x-y-z)(x-y+z)}} \end {align}$$
|
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|
Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
|
Not really that different, but this avoids doing most of the multiplications. It is probably a bit longer though.
First a simplification of your computations. Write $x^4=x^4-1+1$. Then
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{(x^4-1)(1-x)^4}{1+x^2}+\frac{(1-x)^4}{1+x^2}=(x^2+1)(1-x)^4+\frac{(1-x)^4}{1+x^2}$$
Now, do the substitution $u=1-x$. Then
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx= \int_0^1(x^2+1)(1-x)^4 dx +\int_0^1 \frac{(1-x)^4}{1+x^2} d x \\
= \int_0^1(u^2-2u+1)u^4 d u+ \int_0^1 \frac{(1-2x+x^2)^2}{1+x^2} d x $$
The first integral is trivial, while the second can be simplified the following way:
$$\frac{(1-2x+x^2)^2}{1+x^2} = \frac{1-2x+x^2}{1+x^2} (1-2x+x^2) =\left(1- \frac{2x}{1+x^2}\right)) (1-2x+x^2)\\=(1-2x+x^2)-(2x) \frac{1-2x+x^2}{1+x^2}=(1-2x+x^2)-(2x) \left(1- \frac{2x}{1+x^2}\right)
\\=(1-2x+x^2)-(2x) + \frac{4x^2}{1+x^2}=1-4x+x^2 + 4-\frac{4}{1+x^2}$$
|
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|
Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
Hence,
\begin{equation*}
\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1} = \frac{0+\infty}{0+1} = \infty.
\end{equation*}
Thus, $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ diverges.
Is this correct? Thanks.
|
That does work. What would be easier is to notice that $5^n$ dominates the numerator while $4^n$ dominates the denominator. Therefore $$\lim_{n \to \infty} \frac{2^n+5^n}{3^n+4^n} = \lim_{n \to \infty} \frac{5^n}{4^n} $$
This clearly diverges.
By the $n$th term test, you can see that the series also diverges.
|
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|
Compute the determinant of $A$
How to prove that the determinant of the following matrix
$$A = \begin{bmatrix}
a&-b&-c&-d\\
b&a&d&-c\\
c&-d&a&b\\
d&c&-b&a
\end{bmatrix}$$
is $\det A=(a^2+b^2+c^2+d^2)^2$?
Note that $AA^t=(a^2+b^2+c^2+d^2)I_4$, but we just have to $$|\det A|=(a^2+b^2+c^2+d^2)^2$$
Any hint would be appreciated.
|
A different approach: If $A,B,C,D$ are square matrices and $CD=DC$, then
$$\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(AD-BC).$$
Letting
$$A=\begin{pmatrix}a&-b\\b&a\end{pmatrix},B=\begin{pmatrix}-c&-d\\d&-c\end{pmatrix},C=\begin{pmatrix}c&-d\\d&c\end{pmatrix},D=\begin{pmatrix}a&b\\-b&a\end{pmatrix},$$
you can check that $CD=DC$ and thus compute
$$\det(AD-BC)=\det\begin{pmatrix}a^2+b^2+c^2+d^2&0\\0&a^2+b^2+c^2+d^2\end{pmatrix}=(a^2+b^2+c^2+d^2)^2.$$
|
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|
If $a$, $b$, $c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are all integers, then $|a|=|b|=|c|$
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.
Well this is what I have done so far:
From the fact that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, we get
$$abc \mid ab^2 + bc^2 + ca^2 \tag{1}$$
In the same way, we also have
$$abc \mid a^2b + b^2c + c^2a \tag{2}$$
so
$$\begin{align}
abc \mid a(ab^2 + bc^2 + ca^2) - b(a^2b + b^2c + c^2a) &\Rightarrow abc \mid c(a^3-b^3) \tag{3}\\
&\Rightarrow ab \mid a^3-b^3 \tag{4}\\
&\Rightarrow ab \mid b^3(a^3-b^3) \tag{5}
\end{align}$$ and so $a \mid b^6$. In the same way, we can also get $b \mid c^6$ and $c \mid a^6$.
But what should I do after this?
Any help is surely appreciated! Thanks!
|
You have the $2$ values, stated to be integers, of
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a^2c + ab^2 + bc^2}{abc} \tag{1}\label{eq1A}$$
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} = \frac{a^2b + b^2c + ac^2}{abc} \tag{2}\label{eq2A}$$
Consider any prime $p$ where $p \mid abc$. Note the $p$-adic valuation function is $v_{p}(d) = n$, which means $p^n \mid d$ but $p^{n+1} \not\mid d$. Then have
$$v_{p}(a) = i \tag{4}\label{eq4A}$$
$$v_{p}(b) = j \tag{5}\label{eq5A}$$
$$v_{p}(c) = k \tag{6}\label{eq6A}$$
This means that for \eqref{eq1A} and \eqref{eq2A} to be integers, both numerators must have at least
$$m = i + j + k \tag{7}\label{eq7A}$$
factors of $p$. Also, with the numerator in \eqref{eq1A},
$$v_{p}(a^2c) = 2i + k \tag{8}\label{eq8A}$$
$$v_{p}(ab^2) = i + 2j \tag{9}\label{eq9A}$$
$$v_{p}(bc^2) = j + 2k \tag{10}\label{eq10A}$$
and the numerator in \eqref{eq2A} has
$$v_{p}(a^2b) = 2i + j \tag{11}\label{eq11A}$$
$$v_{p}(b^2c) = 2j + k \tag{12}\label{eq12A}$$
$$v_{p}(ac^2) = i + 2k \tag{13}\label{eq13A}$$
The sum of \eqref{eq8A}, \eqref{eq9A} and \eqref{eq10A}, as well as of \eqref{eq11A}, \eqref{eq12A} and \eqref{eq13A}, is $3i + 3j + 3k$, so their average is $i + j + k = m$. If all $3$ values being summed are the same, then $i = j = k$. Otherwise, consider them to not all be the same. Then if just one is less than $m$, the other $2$ must be greater than or equal to $m$, which is not possible since it means the numerator will have less than $m$ factors of $p$. Thus, you need to have $2$ be less than $m$, with both being the same, and one greater than $m$. For the first numerator, consider \eqref{eq10A} to be the largest, with the other $2$ of \eqref{eq8A} and \eqref{eq9A} being the same. Thus, you get that
$$2i + k = i + 2j \implies i = 2j - k \tag{14}\label{eq14A}$$
$$2i + k \lt j + 2k \implies 2i \lt j + k \tag{15}\label{eq15A}$$
Substituting \eqref{eq14A} into \eqref{eq13A} gives
$$i + 2k = 2j - k + 2k = 2j + k \tag{16}\label{eq16A}$$
This is the same as \eqref{eq12A}. The value in \eqref{eq11A} can't be the same, nor as discussed earlier, can it be less, so it must be more than \eqref{eq16A}. Thus, you have
$$2i + j \gt 2j + k \implies 2i \gt j + k \tag{17}\label{eq17A}$$
However, this contradicts \eqref{eq15A}! This means the original assumption of \eqref{eq10A} being the largest can't be true. You can similarly show that neither of \eqref{eq8A} or \eqref{eq9A} can be the largest, with the other $2$ being smaller. This means $i = j = k$. Since $p$ was any prime where $p \mid abc$, this means the $p$-adic values for all these prime factors is the same for each of $a$, $b$ and $c$, and since any primes which divide $a$, $b$ or $c$ must divide $abc$, this means you get
$$|a| = |b| = |c| \tag{18}\label{eq18A}$$
which is what was requested to be proven.
|
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|
$f(x+1) + f(x-1) = x^2$ ; $f(x+4) + f(x-4) = 2\sin x$ , then $f(x) =$? Given $f$ is a complex valued function satisfying $$f(x+1) + f(x-1) = x^2 \\
f(x+4) + f(x-4) = 2\sin x$$
what is $f(x)$ ?
Here, only for the first part MathWolfram alpha is showing $f(x)$ to be of type $c_1(i)^x + c_2(-i)^x + \dfrac{x^2 - 1}{2}$ but how are they introducing "$i$"? I used a generating function but was still unable to get those terms containing $i$ . Please help me with this problem.
|
Let's do two change of variables. First, $u = x-1$:
$$f(u+2) = (u+1)^2 - f(u)$$
Let's apply that multiple times:
$$\begin{align}
f(u+4) &= (u+3)^2 - f(u+2) = (u+3)^2 - (u+1)^2 + f(u)\\
f(u+6) &= (u+5)^2 - f(u+4) = (u+5)^2 - (u+3)^2 + (u+1)^2 - f(u)\\
f(u+8) &= (u+7)^2 - f(u+6) = (u+7)^2 - (u+5)^2 + (u+3)^2 - (u+1)^2 + f(u)
\end{align}$$
Which can be simplified to:
$$f(u+8) = 8u + 32 + f(u)$$
However we can also do the variable substitution $u = x - 4$ in our second identity:
$$f(u+8) = 2\sin (u+4) - f(u)$$
Giving equation:
$$8x + 32 + f(x) = 2\sin (x+4) - f(x)$$
$$2f(x) = 2\sin (x+4) - 8x - 32$$
$$f(x) = \sin (x+4) - 4x - 16$$
|
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|
How can I solve $2^{x}+1=y^{2}$? I found an exponential equation with two variables $x$ and $y$, and I can't find the solutions. I want to solve it for $x$ and $y$. I tried using logarithms but I couldn't solve it. This is the equation:
$$2^{x}+1=y^{2}\qquad\text{with }\big(x,y\big)\in\mathbb{N}.$$
|
Observe that $2^x=y^2−1=(y+1)(y−1) \Rightarrow y+1 = 2^a$ and $y-1 = 2^b$, where $x=a+b$. Then, we have $a \geq b$, and $2^a-2^b= (y+1) - (y-1) = 2$, which gives us $2^b(2^{a-b}-1)=2$. So, we have the following cases:
(I) $\left\{\begin{array}[rcl]
22^b& = & 1 \\
2^{a-b}-1 & = & 2
\end{array} \right. $ $\quad$ (II) $\left\{\begin{array}[rcl]
22^b& = & 2 \\
2^{a-b}-1 & = & 1
\end{array} \right. $
Note that (I) is not possible. From (II), $b=1$ and $a=2$, and so $y=3$ and $x=3$.
|
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|
$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.
|
Before doing that substitution I might say
$\int \frac{\sin x}{1+2\sin x} \ dx\\
\int \frac 12 \frac{2\sin x + 1 - 1}{1+2\sin x} \ dx\\
\int \frac 12 - \frac{1}{1+2\sin x} \ dx\\
\int \frac 12 \ dx - \frac 12 \int \frac{1}{1+2\sin x} \ dx$
Now when we do the substitution it isn't as messy.
$\frac x2 - \int \frac{1}{u^2 + 4u + 1} \ du$
|
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|
Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$ $$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$
This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.
|
As you stated:
$$ \sum_{x=1}^{n} [(x+1)^3-x^3] = \sum_{x=1}^{n} 3x^2+3x+1 = 3\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n = $$ $$ = n(n+1)(n+2)+n = (n+1)^3-1$$
Keep in mind that:
$$ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6} $$
$$ \sum_{x=1}^{n} x = \frac{n(n+1)}{2}$$
|
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|
How to find limit of sequence $a_1=1$, $a_{n+1}=\left( 1-\frac{1}{(n+1)^2} \right)a_n$ for $n\geq 1$? I need some help finding this limit.
Define a sequence by $a_1=1$ and
$a_{n+1}=\left(1-\frac{1}{(n+1)^2}\right)a_n$ for $n\geq1$. Show that the limit exists and find the limit.
I've shown that the limit $L$ exists by showing it's decreasing and bounded below by $0$, but I'm not sure how to actually find it. I couldn't get a general formula for $a_n$. Also, the trick where I write $L=\lim_{n\to\infty}a_{n+1}$ and substitute the formula for $a_{n+1}$ doesn't seem to work here.
I calculated the first few terms and it seems like $L$ is $1/2$, but otherwise I'm stuck here and not sure how to proceed.
|
$1-\frac 1{k^2} = \frac{(k-1)(k+1)}{k^2}$, so the product is
$$
\frac{(2-1)(2+1)}{2\cdot 2}\cdot\frac{(3-1)(3+1)}{3\cdot 3}\cdot\frac{(4-1)(4+1)}{4\cdot 4}\cdot\frac{(5-1)(5+1)}{5\cdot 5}\cdot\dots
$$
i.e.,
$$
\frac{3}{2\cdot 2}\cdot\frac{2\cdot 4}{3\cdot 3}\cdot\frac{3\cdot 5}{4\cdot 4}\cdot\frac{4\cdot 6}{5\cdot 5}\cdot\dots
$$
Do you see something?
|
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For all $x$ in $\mathbb{R}$, there exists $n$ such that the below inequality holds (Proof)
Prove that for all $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ such that:
$$ \frac{3}{3x^2 - 3x + 2} < \frac{4}{n} < \frac{5}{5x^2 - 5x +2} .$$
Here's how I went about it:
For the first part, I reasoned that if we choose $n = 1$ we satisfy the equation since $\frac{3}{3x^2 - 3x + 2}$ attains a maximum of $\frac{12}{5} < 4$, and us bounded below by zero. So, we can write:
$$3x^2 - 3x + 2 > n \implies \frac{1}{3x^2 - 3x + 2} < \frac{1}{n} \implies\frac{3}{3x^2 - 3x + 2} < \frac{3}{n} < \frac{4}{n} $$
For the second part however, I am having some difficulty. I've tried the same approach as above and have not had success.
Disclaimer: I have not done any sort of math in quite a few years, and therefore feel that I am (a) not sure that my approach is even valid (b) feel that I am missing something incredibly obvious.
Any help is greatly appreciated.
|
Since $5x^2-5x+2>0$ and $3x^2-3x+2>0$, we have $$\frac{4}{5}(5x^2-5x+2)<n<\frac{4}{3}(3x^2-3x+2).$$ Now, we see that it's enough to prove that:
$$\frac{4}{3}(3x^2-3x+2)-\frac{4}{5}(5x^2-5x+2)>1.$$
Can you end it now?
|
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|
What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.
I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.
if I'm correct here then the answer should be $(16*270)+(24*243)$? but this does not appear to be the case.
any assistance appreciated.
|
\begin{array}{r}
(x+2)^4 = && \star x^4 &+ \star x^3 &+ 24x^2 &+ 32x &+ 16 \\
(x+3)^5 = &\star x^5 &+ \star x^4 &+ \star x^3 &+ 270x^2 &+ 405x &+ 243
\end{array}
\begin{array}{c|c}
\times &16 & 32x &24x^2 &\cdots \\
\hline
243 & \star & \star x & 5832x^2 &\cdots \\
405x & \star x & 12960x^2 & \star x^3 &\cdots \\
270x^2 & 4320 x^2 & \star x^3 & \star x^4 &\cdots \\
\vdots & \vdots & \vdots& \vdots \\
\hline
\end{array}
|
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|
Formula for length of the diagonal of a parallelepiped Let $a,b,c$ and $\alpha, \beta, \gamma $ are sides and angles ($\alpha$ is the angle between the sides $b$ and $c$ and so on) of a parallelepiped. By using the vector algebra it is easу to prove the formula for the length of the diagonal $d$ of this parallelepiped
$$
d=\sqrt{a^2+b^2+c^2+2ab\cos \gamma+2ac\cos \beta+2bc \cos \alpha}
$$
Question. How to prove the formula without vectors?
It is clear that we have to use two times the cosine theoren but what is the angle between one side and the diagonal of parallelogram formed by two other sides?
|
The formula can be derived from two geometric results/theorems:
*
*Parallelogram law
In a parallelogram $ABCD$ with $AB = CD = a, BC = DA = b$, one has $$AC^2 + BD^2 = 2(AB^2 + BC^2) = 2(a^2+b^2)$$
Since in a parallelogram, $\angle A = \angle C$ and $\angle B = \angle D = \pi - \angle A$, one can easily derive this result from law of cosines.
*The second result concerns parallel lines in space.
Two lines in space are parallel if either they are the same line or they lie in a common plane and didn't intersect. The results we need is "parallel-ness" among lines is transitive:
Given any three lines $a, b, c$; if $a$ is parallel to $b$ and $b$ is parallel to $c$, then $a$ is parallel to $c$.
$$a \parallel b\quad\text{ and }\quad b \parallel c\quad\implies\quad a \parallel c
$$
This can be proved from first principle using Hilbert's axioms.
For a proof, see this answer.
Back to the original problem.
Let $\mathcal{P}$ be a parallelepiped with sides $a,b,c$. Let $O$ be a vertex of $\mathcal{P}$. Let $A,B,C$ be the three vertices adjacent to $O$ such that
$$|OA| = a, |OB| = b, |OC| = c, \angle BOC = \alpha, \angle COA = \beta, \angle AOB = \gamma$$
Let $A_1,B_1,C_1,D$ be the remaining 4 vertices of $\mathcal{P}$ oppositie to $A, B, C$ and $O$ respectively.
Being a parallelepiped, it faces are parallelograms. In particular,
$OAB_1C$ and $AC_1DC$ are parallelograms. This implies
$$OC \parallel AB_1, |OC| = |AB_1| \quad\text{ and }\quad AB_1 \parallel CD_1, |AB_1| = |C_1D|$$
By second result, $OC \parallel CD_1, |OC| = |CD_1|$ and $OC_1DC$ is a parallelogram. By a similiar arguments, $OA_1DA$ and $AC_1A_1C$ are parallelograms too.
Apply parallelogram law to parallelograms $OC_1DC, OA_1DA, AC_1A_1C, OAB_1C$, we obtain
$$\begin{align}
OD^2 + CC_1^2 &= 2(OC^2 + OC_1^2)\\
OD^2 + AA_1^2 &= 2(OA^2 + OA_1^2)\\
AA_1^2 + CC_1^2 &= 2(AC^2 + AC_1^2) = 2(AC^2 + OB^2)\\
AC^2 + OB_1^2 &= 2(OA^2 + OC^2)\\
\end{align}$$
Sum the $1^{st}$ and $2^{nd}$ equation and subtract $3^{rd}$ equation from it, we obtain
$$\begin{align}
OD^2 &= OC^2 + OC_1^2 + OA^2 + OA_1^2 - AC^2 - OB^2\\
&= OC^2 + OC_1^2 + OA^2 + OA_1^2 - (2OA^2 + 2OC^2 - OB_1^2) - OB^2\\
&= OA_1^2 + OB_1^2 + OC_1^2 - OA^2 - OB^2 - OC^2
\end{align}\tag{*1}
$$
Apply parallelogram law and law of cosines to faces $OAB_1C$, $OBC_1A$ and $OCA_1B$, we find
$$\begin{align}
OA_1^2 &= b^2 + c^2 + 2bc\cos\alpha\\
OB_1^2 &= c^2 + a^2 + 2ca\cos\beta\\
OC_1^2 &= a^2 + b^2 + 2ab\cos\gamma
\end{align}$$
Substitute this back into $(*1)$, the desired formula follows:
$$d^2 \stackrel{def}{=} OD^2 = a^2 + b^2 + c^2 + 2bc\cos\alpha + 2ca\cos\beta + 2ab\cos\gamma$$
|
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|
Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $ Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $
My attempt is as follows:-
$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$
$$\left(\sin\theta-\cos\theta\right)^2+\sin\theta\left(\sin\theta-\cos\theta\right)=0$$
$$\left(\sin\theta-\cos\theta\right)(2\sin\theta-\cos\theta)=0$$
$$\tan\theta=1 \text { or } \tan\theta=\frac{1}{2}$$
$$\theta=n\pi+\frac{\pi}{4}$$
For $\tan\theta=\dfrac{1}{2} \text { how to make use of given condition } \tan71 ^{\circ}34^{\prime}=3$
|
As $1+2+3=1\cdot2\cdot3$
$\arctan1+\arctan2+\arctan 3=n\pi$
using Show $\tan(x)+\tan(y)+\tan(z) = \tan(x) \tan(y) \tan(z)$
As $\arctan(u)<\dfrac\pi2$
$n=1$
Now $\arctan1=\dfrac\pi4$
$\arctan2+\arctan\dfrac12=\arctan2+$arccot$2=\dfrac\pi2$
See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
|
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|
proof the given limit using epsilon delta definition
$$\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$$
let $$f(x)=\frac{\sqrt{x}-1}{x-1}$$ then
$\forall \epsilon>0, \exists\delta>0,\forall x\in D_f \left(\left|x-1 \right|<\delta\Longrightarrow \large\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|<\epsilon\right)$
$$\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|=\left|\frac{1}{\sqrt{x}+1}-\frac{1}{2}\right|=\left|\frac{1-\sqrt{x}}{2\left(1+\sqrt{x}\right)}\right|=\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\epsilon$$
take $\delta\le1$ implies:$$0.5<x<1.5$$$$\left(\sqrt{0.5}+1\right)^{2}<\left(1+\sqrt{x}\right)^{2}$$$$\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{\left|x-1\right|}{2\left(\sqrt{0.5}+1\right)^{2}}<\epsilon$$
hence $$\delta\le\min\left\{0.5,2\left(\sqrt{0.5}+1\right)^{2}\epsilon\right\}$$
e.g. if I take $\epsilon=1$ then $2\left(\sqrt{0.5}+1\right)^{2}=5.8$ so $\delta\le 0.5$
$$\left|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}\right|=\left|\frac{1}{\sqrt{x}+1}-\frac{1}{2}\right|=\left|\frac{1-\sqrt{x}}{2\left(1+\sqrt{x}\right)}\right|=\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{0.5}{2\left(1+\sqrt{x}\right)^{2}}<\frac{1}{2\left(1+\sqrt{x}\right)^{2}}<1=\epsilon$$
is the solution fine?
|
The proof looks fine, there is a typo for
$$\delta <\color{red}{\frac12} \implies0.5<x<1.5$$
and here, since $\sqrt x$ is an increasing function, we can take
$$\frac{\left|x-1\right|}{2\left(1+\sqrt{x}\right)^{2}}<\frac{\left|x-1\right|}{2\left(0+1\right)^{2}}<\epsilon$$
which leads to a simpler expression.
|
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|
$\cos(\pi/7)$ is a root of $8x^3-4x^2-4x+1=0$. How is the polynomial generated? According to Wolfram MathWorld, $\cos(\frac{\pi}{7})$ is a root of $8x^3-4x^2-4x+1=0$. Similarly, $\cos(\frac{2\pi}{7})$ is a root of $8x^3+4x^2-4x-1=0$. What's the procedure to generate these polynomials? I understand you can solve the cubic equations and check the cosines are indeed roots. My question is how does one arrive at those polynomials in the first place?
|
By DeMoivre's theorem,
$$(\cos x+i\sin x)^7=\cos7x+i\sin7x\quad\quad(*)$$
With $x=\frac\pi7$, the right hand side of $(*)$ reduces to $-1$. Expand the binomial on the left hand side and match up the real and imaginary parts (the latter of which is $0$), leaving us with
$$\cos^7x-21\cos^5x\sin^2x+35\cos^3x\sin^4x-7\cos x\sin^6x=-1$$
Rewrite each instance of $\sin^2x$ as $1-\cos^2x$ and simplify the result; you should end up with
$$64\cos^7x-112\cos^5x+56\cos^3x-7\cos x+1=0$$
the left hand side of which can be factorized as
$$(1+\cos x)(8\cos^3x-4\cos^2x-4\cos x+1)^2=0$$
but $\cos\frac\pi7\neq-1$,
$$8\cos^3\frac\pi7-4\cos^2\frac\pi7-4\cos\frac\pi7+1=0$$
Similarly, with $x=\frac{2\pi}7$, the right hand side of $(*)$ reduces to $1$, and so on.
|
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|
Division of 2 different Infinite Nested square roots Has anyone come across the following type of nested square roots problem?
$\sqrt{2-{\sqrt{2+{\sqrt{2+...n times {\sqrt{2}}}}}}}$ divided by
$\sqrt{2-{\sqrt{2+{\sqrt{2+...(n+1)times {\sqrt{3}}}}}}}$
Converging towards 3 as the 'n' increases
Are there any theorem or formulas to calculate multiplication or division of infinite nested square roots?
Note: 2nd sum done in calculator has same $\sqrt3$ at its end which is not visible. Just one term of nested square root is increased which is shown in picture
$\sqrt{2}$ = $2cos(\frac{\pi}{4})$
$\sqrt{2+\sqrt{2}}$ = $2cos(\frac{\pi}{8})$
$\sqrt{2+\sqrt{2+\sqrt{2}}}$ = $2cos(\frac{\pi}{16})$
.
.
.
$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$ = $2sin(\frac{\pi}{32})$ or $2sin(\frac{\pi}{2^5})$
$\sqrt{3}$ = $2cos(\frac{\pi}{6})$
$\sqrt{2+\sqrt{3}}$ = $2cos(\frac{\pi}{12})$
$\sqrt{2+\sqrt{2+\sqrt{3}}}$ = $2cos(\frac{\pi}{24})$
$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$ = $2cos(\frac{\pi}{48})$
.
.
.
$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$ = $2sin(\frac{\pi}{96})$ = $2sin(\frac{\pi}{3*2^5})$
This looks more interesting to note
$\frac{2sin(\frac{\pi}{2^5})}{2sin(\frac{\pi}{3*2^5})}$
For very small values. "x" and "sin(x)" are almost the same
(as long as "x" is in Radians!)
It is true that $\frac{(\frac{\pi}{2^5})}{{(\frac{\pi}{3*2^5})}}$ simplifying to 3?
Is there any other means like, using limits or integration to solve such problems?
Please throw light me someone
Thanks in advance
|
Well, you seem to have basically proven it. We want to show that
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}} = 3
$$
We would like to use that $\sin x \approx x$ for small $x$. To be rigorous, we can write $\sin x = x + xo(1)$, where $o(1)$ is a function that goes to $0$ as $x\to0$. Then
$$
\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
= \frac{\frac{\pi}{2^n} + \frac{\pi}{2^n}o_1(1)}{\frac{\pi}{3\cdot 2^n} + \frac{\pi}{3\cdot 2^n}o_2(1)}
= \frac{3+3o_1(1)}{1+o_2(1)}
\underset{n\to\infty}\to 3
$$
Alternatively, we can use the well known fact that $\frac{\sin x}{x} \to 1$ for $x\to0$. Thus:
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}\left(\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
\cdot \frac{\frac{\pi}{2^n}}{\sin\frac{\pi}{2^n}}
\cdot\frac{\sin\frac{\pi}{3\cdot 2^n}}{\frac{\pi}{3\cdot 2^n}} \right)
= \lim_{n\to\infty}\frac{\frac{\pi}{2^n}}{\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}3 = 3
$$
|
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|
How do I find all solutions of $X^2 = I_2$ in $M_2(\mathbb{N})$? I know that the solutions of the equation:
$$X^2 = I_2$$
in $M_2(\mathbb{N})$ are the $2$x$2$, natural, involutory matrices:
$$ X_1 =
\begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}
$$
and
$$ X_2 =
\begin{pmatrix}
0 & 1\\
1 & 0\\
\end{pmatrix}
$$
How can I arrive at these solutions by calculations?
|
Take $$X=\begin{bmatrix}a& b \\ c &d
\end{bmatrix} $$
with $a,b,c,d \in \mathbb{N}$
Now
$$X^2=\begin{bmatrix}a^2+bc & ab+bd \\ ac+cd & bc+d^2
\end{bmatrix} $$
and we have a system of equations in $\mathbb{N}$:
$$\begin{cases}
a^2+bc=1\\
ab+bd=0 \\
ac+cd=0 \\
bc+d^2=1
\end{cases}$$
For $b(a+d)0=0$, we have either $b=0$ or $a=d=0$ since we are in $\mathbb{N}$.
Suppose $b=0$, then
$$X^2=\begin{bmatrix}a^2 & 0 \\ ac+cd & d^2
\end{bmatrix} $$
It follows that $a=d=1$ and $c=0$. Thus $$X=\begin{bmatrix}1& 0 \\ 0 &1
\end{bmatrix} $$
Suppose now $a=d=0$, then
$$X^2=\begin{bmatrix}bc & 0 \\ 0 & bc
\end{bmatrix} $$
and it follows that $b=c=1$. Thus $$X=\begin{bmatrix}0& 1 \\ 1 &0
\end{bmatrix} $$
|
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|
Inverse of polynomial over a finite field The question is the following:
Can you deduce if ${2x + 1}$ is invertible in $\mathbb{Z}_3[x]/(x^2 + 2x + 2)$? In case of a positive answer, give its inverse.
Following the Wilson's theorem, for $K[X]/(f)$, any polynomial of degree $1 \leq deg < n$ will admit an inverse of degreee $1 \leq deg \leq n$ mod $f$. Since $n = 3$ because I'm working on $\mathbb{Z}_3$ and $deg = 2$, I guess $2x+1$ is irreductible in this case.
Once I want to find the inverse I've been following this post. The table using Euclidean algorithm is the following:
\begin{array}{r|r|r|r}
& & (x+\frac{3}{2})/2 & (2x+3)/5 \\\hline
1 & 0 & 1 & -(2x+3)/5\\\hline
0 & 1 & -\big(\dfrac{x}{2} + \dfrac{3}{4}\big) & \dfrac{x^2}{5} + \dfrac{3x}{5} + \dfrac{29}{20}\\\hline
x^2+2x+2 & 2x+1 & 5/4 & 0
\end{array}
So, finding the lineal combination I obtain this result:
$\dfrac{5}{4} = 1\times(x^2 +2x +2) - \big(\dfrac{x}{2}+\dfrac{3}{4}\big)\times(2x+1)\xrightarrow{}\dfrac{5}{4} = f(x) - \big(\dfrac{x}{2}+\dfrac{3}{4}\big)g(x)$,
Here is where I get stuck. I let 1 on the left side so I have the following result:
$1 = \bigg(\dfrac{4}{5}\bigg)(x^2+2x+2)- \bigg(\dfrac{2x}{5}+\dfrac{3}{5}\bigg)(2x+1)$
But once I arrive here I don't know how to get the value of the inverse.
Can anyone help me? Thank you very much.
My result, which I'm not absolutely sure is that the inverse of $2x+1$ is:
$-\big(\dfrac{2x}{5}+\dfrac{3}{5}\big)$ mod $x^2+2x+2$.
In the remote case it's correct, is there any way to get the final value instead of letting it in function of mod $f$?
Thank you again.
Bernat
|
Hint: $0=x^2 + 2x + 2$ implies $0 = 2x^2+4x+4 = 2x^2+x+1 = x(2x+1)+1$
|
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|
From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$
I recognize that the locus of a chord whose midpoint is at $(h,k)$ is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.
Here I am getting the equation in terms of $x,y,h,k$, but to find the locus I need the equation entirely in the form of $h,k$, right? So how do I eliminate $x,y$ from the equation of the locus of the midpoint?
|
This may not be the solution you are looking for:
Let $\mathcal C=\{x^2+y^2=1\}$ be the unit circle. Let $O' = (\alpha, \beta)$ be any point outside of this circle. Let $O'P'$ and $O'Q'$ be two tangent line to $\mathcal C$. One can check that the midpoint $m' = (x, y)$ of $P'Q'$ is given by (why?)
$$(x, y) = m' = \frac{1}{\alpha^2+ \beta^2} (\alpha, \beta).$$
Now assume that $O = (\alpha, \beta)$ is on the ellipse $\{ (ax)^2 + (by)^2 = d^2\}$. Thus $(a\alpha)^2 + (b\beta)^2 = d^2$. Then
$$ d^2(x^2 + y^2)^2 = \frac{d^2}{(\alpha^2 + \beta^2)^2}$$
and
$$(ax)^2 +(by)^2 = \frac{(a\alpha)^2 + (by)^2}{(\alpha^2 + \beta^2)^2}=\frac{d^2}{(\alpha^2 + \beta^2)^2}$$
Thus the locus of the midpoint $m'$ is given by
$$ \tag{1} (ax)^2 + (by)^2 = d^2 (x^2+ y^2)^2. $$
The above is related to your question in the following way: Consider the transformation: $$(x, y) \mapsto (x/a, y/b).$$
Under this transformation, the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ is sent to the unit circle $\mathcal C$, while the circle $x^2 + y^2 = d^2$ is sent to the ellipse $(ax)^2 + (by)^2 = d^2$. The crucial observation is that tangent lines $OP, OQ$ are also sent to tangent lines $O'P', O'Q'$, and the midpoint $m$ of $PQ$ are sent to the midpoint $m'$ of $P'Q'$ (see here). Thus if you take the inverse transformation
$$ (x, y) \mapsto (ax, by)$$
Then the locus of $m'$ will be sent to the locus of $m$. This implies your equation: if you change $x$, $y$ to $x/a$, $y/b$ respective in (1), you get
$$x^2 + y^2 = d^2 \left(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\right)^2.$$
|
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|
Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16 Prove that if p and q are both prime numbers, with p > q > 2, then $p^4 − q^4$ is divisible by 16.
This is my attempt so far:
Since p and q are both prime numbers greater than 2, then they must be odd and hence can be written:
$p=2m+1$, for some integer n,
And $q=2n+1$, for some integer m.
Consider, $p^4-q^4$ $\implies$ $(p^2+q^2)(p+q)(p-q)$
$p-q=2m+1-(2n+1)=2(m-n),$ hence is a multiple of 2.
$p+q=2m+1+2n+1=2(m+n+1),$ hence is a multiple of 2.
$p^2+q^2=(2m+1)^2+(2n+1)^2=$
$2(2m^2+2m+2n^2+2n+1),$ hence is a multiple of 2.
However, this is where I am stuck. Doesn’t this just prove the statement is a multiple of 8, not 16? Could anyone help with this proof, or maybe suggest an alternative method. Thanks!
|
Alternatively you may show that any odd integer power $4$ leaves the same remainder $1$ when divided by $16$:
$(2n+1)^4 \\
= (2n)^4 + \binom{4}{1}(2n)^3 + \binom{4}{2}(2n)^2+\binom{4}{3}(2n)+1
\\= 16(n^4+2n^3+n^2) + 8n(n+1)+1\\\equiv 1\pmod{16}$
|
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|
Determine all prime numbers $p,q,r$ such that : $p^{2}+1=r^{2}+q^{2}$ Problem :
find all prime numbers such that :
$$p^{2}+1=q^{2}+r^{2}$$
My attempt :
equation equivalent :
$p^{2}-q^{2}=r^{2}-1$
so :
$(p+q)(p-q)=(r+1)(r-1)$
now :
$r=2$ we find : $p+q=2$ or $p-q=1$ $×$
$r=3$ we find : $p+q=3,2$ or $p-q=2,3$ $×$
$r=5$ we find : $p+q=10,2,4,5,20$ or $p-q=2,20,10,5,4,$
so I have many case how I find it ?
see :
$13^{2}+1=7^{2}+11^{2}$
$17^{2}+1=11^{2}+13^{2}$
$23^{2}+1=13^{2}+19^{2}$
$31^{2}+1=11^{2}+29^{2}$
I need generalized this solution ?
I think ?
$(5x+13)^{2}+1=(3x+7)^{2}+(4x+11)^{2}$
I think we have infinity prime numbers
but how I prove it ? and how I find this case ??
|
COMMENT.- One way could be to consider the general solution of the equation $x^ 2 + y^ 2 = z^ 2 + w^ 2$ that gives a parameterization with four independent coprime parameters with which one could look for prime values for $r$ and $q$. For example by making $at-bs = 1$ and $as + bt = p$, try to see if primes $r$ and $q$ are obtained with appropriate values of the parameters $a, b, t, r$. But this march is very laborious.
A curious fact is that for the (type Pell-Fermat) equation $p^2+1=2q^2$ (trying with $r=q$) all the infinitely many integer solutions given by Wolfram does not appear the solution $(p,q)=(7,5)$ but it appears, for example $(p,q)=(41,29)$. Again this way is very laborious to find primes as solutions.
Anyway, it seems in fact as the O.P. suggests that there are an infinite number of solutions of which we give a few here.
$$7^2+1=5^2+5^2\\41^2+1=29^2+29^2\\13^2+1=7^2+11^2\\17^2+1=11^2+13^2\\23^2+1=13^2+19^2\\37^2+1=23^2+29^2\\47^2+1=29^2+37^2\\53^2+1=31^2+43^2$$
|
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|
Finding a Lyapunov function for a specific problem I have tried for a long time to find a Lyapunov function for the specific problem
$$\begin{align} x' &= -x - 2y + xy^2 \\ y' &= 3x - 3y + y^3 \end{align}$$
Do you know any Lyapunov function what is going to work to determine the stability of the origin? According to a phase portrait produced by pplane the stability of the origin is asymptotically stable.
Thanks in advance!
|
Hint.
We have
$$
\cases{
3x \dot x = -3 x^2-6 x y + 3x^2y^2\\
2y \dot y = 6x y-6 y^2+2y^4
}
$$
and after addition
$$
\frac 12\frac{d}{dt}(3x^2+2y^2) = -3x^2-6y^2+3x^2y^2+2y^4
$$
now there exist $\rho_m$ such that $x^2+y^2 < \rho_m^2$ makes
$$
-3x^2-6y^2+3x^2y^2+2y^4 < 0
$$
in fact, $\rho_m = 1$ suffices.
|
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|
Proving a Difficult Definite Integral in One Variable Let $t > 0 $, $N \in \mathbb{N}$, $q \geq 1 + \frac{1}{N}$.
Also let $r > 0 $, $r' = \frac{r}{r-1}$, such that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$.
We wish to prove the following integral equation:
$ \displaystyle \int^{t}_{0} \large (t-s)^{ -\frac{N}{2}(1 - \frac{1}{r}) - \frac{1}{2} } (t+s)^{ - \frac{N}{2} (q - \frac{1}{r'}) } \text{d}s = \large C t^{ \frac{1}{2} - \frac{N}{2} q } $,
where $C > 0$ is a constant, possibly just $1$.
I'm afraid I do not even know where to begin with this monster. Integral calculators have not been of much help. I presume it is a key point that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$, as the negative of this figure appears as the exponent of $(t-s)$.
Any hints/suggestions as to how I can calculate this integral are much appreciated.
EDIT
After substituting $s = tx$, we arrive at the integral:
$ \large t^{\frac{1}{2} - \frac{N}{2}q } \int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x $.
Thus, it remains only to show that
$\int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x = C$.
From our setting of $r$, as explained above, we have the following bounds on each of the exponents, which seem to be important for the existence of this integeral:
$ \large -1 < -\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} < 0 $
$ \large -\frac{1}{2} -\frac{N}{2}q < -\frac{N}{2} (q - \frac{1}{r'}) < \frac{1}{2} -\frac{N}{2}q $
In particular, by our definition of $q$, both exponents are always negative.
|
Only a partial answer:
For $a>-1,b>-1,x\in[0,1]$, let $$B(x,a+1,b+1):=\int_0^x u^a \cdot (1-u)^b \,\mathrm du$$ denote the Incomplete Beta function.
For $a,b>0$ and $c>0$, we can give an anti-derivative on $]0,c[$ of the function $$f:[0,c]\to \mathbb R, x\mapsto(c+x)^a\cdot(c-x)^b$$ in terms of the incomplete Beta function: In your case, I would like to choose $a=- \frac{N}{2} (q - \frac{q}{r'})$ and $b=-\frac{N}{2}(1 - \frac{1}{r})$. Also, note that my $c$ is your $t$. I don't see why $a,b>0$ though.
\begin{split}\int f&=\int(c+x)^a\cdot(c-x)^b\,\mathrm dx\\
&\overset{x=2 c y -c}{=}2c\cdot\int (2 c y)^a\cdot(2c\cdot(1-y))^b\,\mathrm dy \\
&=2^{a+b+1}\cdot c^{a+b+1}\cdot B(y,a+1,b+1) +const.\\
&=2^{a+b+1}\cdot c^{a+b+1}\cdot B\left(\frac{c+x}{2c},a+1,b+1\right) +const.
\end{split}
Put more formally, we have for all $x\in]0,c[$, $F'(x)=f(x)$, where $$F: [0,c]\to\mathbb R, x\mapsto 2^{a+b+1}\cdot c^{a+b+1}\cdot B\left(\frac{c+x}{2c},a+1,b+1\right).$$
(Note that, since $c\geq x$, we have $\frac{c+x}{2c}\in[0,1]$.)
By the fundamental Theorem of calculus, we thus have $$\int_0^c f(x)\,\mathrm dx = F(c)-F(0).$$
EDIT: If $a,b$ are not positive you might use that (this was determined by WolframAlpha) $F'(x)=f(x)$ for $x\in]0,c[$, where
$$F:[0,c]\to\mathbb R, \\x\mapsto \frac{2^b (c - x)^b (c + x)^{1 + a} (1 - x/c)^{-b} {}_2F_1(1 + a, -b, 2 + a, \frac{c + x}{2 c})}{1 + a}.$$
Here, $${}_2F_1(x_1,x_2,x_3,x_4)$$ denotes the Hypergeometric function.
A word of caution: I know almost nothing about the Hypergeometric function and it seems that you can run into trouble for $x_4=1$ (i.e. when $x=c$ so it is not clear if $F(c)$ is well-defined).
|
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|
Trigonometric determinants I've recently obtained my University entrance papers from 1967 (yes,52 years ago!) and I found the question below difficult. I presume the answer is a symmetric expression in the differences between alpha,beta and gamma.Am I missing some obvious trick? Any help would be appreciated.
Simplify and evaluate the determinant
and show that its value is independent of theta.
|
To make it a bit better readable, you may substitute
*
*$u = \theta + \alpha, v= \theta + \beta, w = \theta + \gamma$
So, the first row of your determinant looks like
$$\begin{pmatrix} \sin^2 u & \sin u \cos u & \cos^2 u
\end{pmatrix}$$
Now, you may use
*
*$\cos^2 u - \sin^2 u = \cos 2u$ and
*$2 \sin u \cos u = \sin 2u$ and
*$\sin^2 u= \frac{1}{2}(1-\cos 2u)$
Applying this to the columns of the determinant you get (here only for the first row)
$$\begin{pmatrix} \frac{1}{2}(1-\cos 2u) & \frac 12\sin 2u & \cos 2u
\end{pmatrix}$$
So, your determinant looks now like this
$$\left|
\begin{pmatrix}
\frac{1}{2} & \frac 12\sin 2u & \cos 2u \\
\frac{1}{2} & \frac 12\sin 2v & \cos 2v \\
\frac{1}{2} & \frac 12\sin 2w & \cos 2w \\
\end{pmatrix}\right| = \frac 14 \left|
\begin{pmatrix}
1 & \sin 2u & \cos 2u \\
1 & \sin 2v & \cos 2v \\
1 & \sin 2w & \cos 2w \\
\end{pmatrix}\right|$$
Now, expand along the first column and use
*
*$\sin a\cos b- \cos a \sin b = \sin (a-b)$
$$\sin(2(v-w)) - \sin(2(u-w)) + \sin(2(u-v))$$
This also shows the independence from $\theta$.
|
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|
Prove that the equation has only one real root. Prove that $(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only one real root.
It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root.
But it doesn't guarantee that the real root is unique. If we calculate the root then it can be shown that it has only one real root.
|
Here is an elementary way that uses only
*
*$(1)$: $a^3+b^3 = (a+b)(a^2-ab+b^2)$ and
*$(2)$: $a^2+b^2 >ab$ for $|a|+|b| > 0$
Note, that $|x-1|+|x-4|\geq |x-1 +(4-x)| =3 > 0$ and $|x-2|+|x-3|\geq |x-2 +(3-x)| =1 > 0$
Let's call $p(x) = (x-1)^3+ (x-2)^3 + (x-3)^3 + (x-4)^3$.
Now, using $(1)$ write
$$(x-1)^3 + (x-4)^3 = (x-1 + x-4)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$
$$= (2x-5)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$
$$(x-2)^3 + (x-3)^3 = (x-2 + x-3)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$ $$= (2x-5)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$
Hence,
$$p(x) = $$
$$(2x-5)\color{blue}{\left(\underbrace{(x-1)^2 + (x-4)^2 - (x-1)(x-4)}_{\stackrel{(2)}{>}0} + \underbrace{(x-2)^2 + (x-3)^2 - (x-2)(x-3)}_{\stackrel{(2)}{>}0}\right)}
$$
So,
$$p(x) = (2x-5)\color{blue}{q(x)} \mbox{ with } \color{blue}{q(x)} > 0 \mbox{ for all } x \in \mathbb{R}$$
|
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|
Use the epsilon-delta definition to show that $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$ $\lim_{x\to\sqrt2} \frac{1}{2}(\frac{2}{x}+x) = \sqrt2$
by using the epsilon-delta method.
I have been trying so solve this for the last hour, but I'm completely stuck.
Help, anyone?
|
Fix $\epsilon > 0$.
Find $\delta_1$ such that $|\frac{2}{x} - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x-\sqrt{2}| < \delta_1$.
Find $\delta_2$ such that $|x - \sqrt{2}| < \epsilon$ for any $x$ satisfying $|x - \sqrt{2}| < \delta_2$.
Then let $\delta = \min\{\delta_1, \delta_2\}$.
We then have $$|\frac{1}{2} (\frac{2}{x} + x) - \sqrt{2}| \le \frac{1}{2} |\frac{2}{x} - \sqrt{2}| + \frac{1}{2} |x - \sqrt{2}| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ for any $x$ satisfying $|x - \sqrt{2}| < \delta$.
Response to comment:
If $|x| \ge 1$, then
$$|\frac{2}{x} - \sqrt{2}|
= \sqrt{2} \left| \frac{\sqrt{2} - x}{x}\right|
\le \sqrt{2} |\sqrt{2} - x| < \sqrt{2} \delta_1.$$
So, choose $\delta_1 = \min\{\sqrt{2}-1, \epsilon/\sqrt{2}\}$.
(The $\sqrt{2}-1$ is to ensure $|\sqrt{2}-x|<\delta_1 \le \sqrt{2}-1 \implies |x| \ge 1$.)
|
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|
Prove that for $f: [0, 1] \to \mathbb{R} , f_n(x) = \frac{nx}{1+n^2 x^2}$ is uniform convergent. Prove that for $f: [0, 1] \to \mathbb{R} , f_n(x) = \frac{nx}{1+n^2 x^2}$ is uniform convergent.
I found the pointwise limit to be 0.
Taking the first derivative, $$f_n '(x) = \frac{n(1-n^2 x^2)}{(1+n^2 x^2)^2} $$ $f_n '(x) = 0$ at $x = \frac{1}{n}, 0, $ and $ \frac{-1}{n}$. $f_n(\frac{1}{n}) = \frac{1}{2}, f_n(\frac{-1}{n}) = \frac{-1}{2}, $ and $ f_n(0) = 0 $. Thus $f_n(\frac{1}{n}) = \frac{1}{2}$ is the max of the function.
Thus, $sup_{x \in [0,1]} |f_n(x) - 0| < 1/2 $
I do not know how to proceed in order to prove uniform convergence.
|
Clearly, the convergence is not uniform on $[0,1]$ since $1/n \in [0,1]$ and as $n \to \infty,$
$$\sup_{x \in [0,1]} |f_n(x)-0| = \sup_{x \in [0,1]} \frac{nx}{1 + n^2x^2} \geqslant \frac{n \cdot \frac{1}{n}}{1 + n^2 \cdot \frac{1}{n^2}} = \frac{1}{2} \not\to 0$$
|
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|
How to calculate this integral using Cauchy's Formula?
Calculate $\int_{C} \sinh(z+1)/(z^2+1)dz$ where $C$ is $x^{2/3}+y^{2/3}=3^{2/3}$
My try is redefine $\int_{C} \sinh(z+1)/(z^2+1)dz$ as
$
\begin{align}
\int_{C} \sinh(z+1)/(z^2+1)dz &= \int_{C} \sinh(z+1)/((z-i)(z+i))dz \\
&= \int_{C} \sinh(z+1)/(2i(z-i))dz-\int_{C} \sinh(z+1)/(2i(z+i))dz \\
&= \frac{1}{2i} \left( \int_{C} \sinh(z+1)/(z-i)dz-\int_{C} \sinh(z+1)/(z+i)dz \right)
\end{align} $
I know that $\sinh(z+1)$ is analytic inside and on the contour $C$ (because is analytic in $\mathbb{C}$) and $i,-i$ lie in $C$, then using the Cauchy's Formula
$$\int_{C} \sinh(z+1)/(z-i)dz=2 \pi i\cdot \sinh(i+1)$$
and
$$\int_{C} \sinh(z+1)/(z+i)dz=2 \pi i\cdot \sinh(1-i)$$
then
$
\begin{align}
\int_{C} \sinh(z+1)/(z^2+1)dz &= \frac{1}{2i} \left( 2 \pi i\cdot \sinh(i+1)-2 \pi i\cdot \sinh(1-i) \right) \\
&= \pi \cdot \sinh(1+i)- \pi\cdot \sinh(1-i) \\
\end{align} $
I've do it well?. I don't know how to get the answer $i2 \pi \sin(1)\cosh(1)$
|
Your answer is actually equivalent after applying some identities!
There is a formula for the difference of hyperbolic sines (Wikipedia article)
$$\sinh(x) - \sinh(y) = 2 \cosh \left( \frac{x+y}{2} \right) \sinh \left( \frac{x-y}{2}\right)$$
Another identity of importance is
$$\sinh(x) = -i \sin(ix) \;\;\; \text{or, equivalently,} \;\;\; \sinh(ix) = -i \sin(-x)$$
Finally, note that, in particular, the above holds if $x=1$, and that $\sin(x)$ is an odd function ($\sin(-x) = -\sin(x)$). You could skip a bit of a middleman by using this in the previous fact to conclude
$$\sinh(ix) = i \sin(x)$$
Apply all of these facts to your answer:
$$\begin{align}
\pi \sinh(1+i) - \pi \sinh(1-i) &\overset{(1)}{=} \pi \left( \sinh(1+i) - \sinh(1-i) \right) \\
&\overset{(2)}{=} 2\pi \cosh \left( \frac{1+i+1-i}{2} \right) \sinh \left( \frac{1+i-(1-i)}{2}\right) \\
&\overset{(3)}{=} 2\pi \cosh \left( 1 \right) \sinh \left( i \right) \\
&\overset{(4)}{=} 2i\pi \cosh \left( 1 \right) \sin(1)
\end{align}$$
Each equality follows as:
*
*$(1):$ Factor out $\pi$
*$(2):$ Apply the difference formula for hyperbolic sine
*$(3):$ Simplify
*$(4):$ Apply $\sinh(ix) = i \sin(x)$ for $x=1$
|
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|
Evaluate the following limit: $=\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $ I am studying MIT OpenCourseware 18.01 Single Variable Calculus on my own and am stuck on a final exam question.
Evaluate the following limit:
$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} \left(\frac{2}{n}\right)$$
$$=\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$
We can do this using the Riemann sum, which states that if the interval [a,b] is divided into $n$ equal pieces of length, where $\Delta x = \frac{b-a}{n}$, then the sum of all the areas of the rectangle is $ \sum_{i = 1}^{n} f(x_{i-1}) \Delta x $. Also, in the limit as $n$ goes to infinity, the Riemann sum approaches the value of the definite integral:
$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} f(x_{i-1}) \Delta x =\int_a^b f(x)\,dx$$
In this case, $\Delta x = \frac{b-a}{n} = \frac{2}{n}$, and therefore, $b-a = 2$. Also $f(x_0) = \sqrt{1+\frac{2}{n}}$, $f(x_1) = \sqrt{1+\frac{4}{n}}$, $f(x_2) = \sqrt{1+\frac{6}{n}}$, and so on and so forth until we reach $n$.
How do we convert $$\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$ to a definite integral?
|
Just for your curiosity (without Riemann sum).
Assuming that you know about generalized harmonic numbers, you could approximate quite well the partial sums since
$$ S_n= \frac{2}{n} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}}=2 \sqrt{2}\,\frac{ H_{\frac{3
n}{2}}^{\left(-\frac{1}{2}\right)}-H_{\frac{n}{2}}^{\left(-\frac{1}{2}\right)}
}{n^{3/2}}$$
Now, using the asymptotics
$$H_p^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{p^{1/2}}{2}+\zeta
\left(-\frac{1}{2}\right)+\frac{1}{24p^{1/2}}-\frac{1}{1920p^{5/2}}+O\left(
\frac{1}{p^{9/2}}\right)$$
$$S_n=\left(2 \sqrt{3}-\frac{2}{3}\right)+\frac{\sqrt{3}-1}{n}+\frac{\sqrt{3}-3}{18
n^2}+\frac{27-\sqrt{3}}{3240 n^4}+O\left(\frac{1}{n^6}\right)$$ which, for sure, shows the limit and also how it is approached.
Moreover, it is a quite good approximation. For example, computing $S_5\approx 2.9410395$ while the above truncated formula would give $\frac{4459499 \sqrt{3}-1768473}{2025000}\approx 2.9410399$.
|
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|
CNF or DNF, $p(x)=0 \implies f=1$ Write either a $CNF$ or a$\;DNF$ of the expression $f$ such that
$f=0\iff (x_2x_1x_0)_{(2)}$ is a zero point of the polynomial:
$$p(x)=x^3-8x^2+12x=x(x-2)(x-6)$$
$$p(x)=0\iff x\in \{0,2,6\}$$
$$\begin{array}{|c|c|c|c|}
\hline
x_2& x_1 & x_0 & f=1 \\ \hline
0 &0 &0 & 1\\ \hline
0 &0 &1 &0\\ \hline
0 &1 &0 &1\\ \hline
0 & 1 &1 &0\\ \hline
1 &0 &0 &0\\ \hline
1 &0 &1 &0\\ \hline
1 &1 &0 &1\\ \hline
1 & 1 &1 &0\\ \hline
\end{array}$$
$DNF:\overline{x_2}\cdot\overline{x_1}\cdot\overline{x_0}+\overline{x_2}\cdot x_{1}\cdot\overline{x_0}+x_2\cdot x_1\cdot \overline{x_0}$
Is this correct?
|
It's correct, and if you want to find the Minimal form, you can use a k-map:
$$\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&\color{orange}1&\color{orange}1&1&0\\
x_0&0&0&0&0\end{array}}
\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&\color{red}1&\color{red}1&0\\
x_0&0&0&0&0\end{array}}$$
$$\color{orange}{x_2'x_0'}+\color{red}{x_1x_0'}\tag*{Minimal DNF form}$$
$$\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&1&1&0\\
x_0&\color{orange}0&\color{orange}0&\color{orange}0&\color{orange}0\end{array}}
\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&1&1&\color{red}0\\
x_0&0&0&0&\color{red}0\end{array}}
$$
$$(\color{orange}{x_0})'(\color{red}{x_2x_1'})'=x_0'(x_2'+x_1)\tag*{Minimal CNF form}$$
|
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|
In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$
$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$
$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$
$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$
$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$
$$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$
$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$b^2-a^2-c^2+2ac=ba+bc-b^2$$
$$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$
$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$
$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$
$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$
Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$
$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$
Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$
$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$
$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$
$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A+\cos C=2\cos B$$
Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.
|
Given
\begin{align}
\tan\tfrac12A&=u-d
,\quad
\tan\tfrac12B=u
,\quad
\tan\tfrac12C=u+d
,\quad
u,d\in\mathbb{R}
\tag{1}\label{1}
.
\end{align}
We can express $d$ in terms of $u$ using
known identity for triangles
\begin{align}
\tan\tfrac A2\tan\tfrac B2+
\tan\tfrac B2\tan\tfrac C2+
\tan\tfrac C2\tan\tfrac A2&=1
,\\
(u-d)u+u(u+d)+(u+d)(u-d)=3u^2-d^2 &= 1
,\\
d&=\sqrt{3u^2-1}
\tag{2}\label{2}
,
\end{align}
and \eqref{1} becomes
\begin{align}
\tan\tfrac12A&=u-\sqrt{3u^2-1}
,\quad
\tan\tfrac12B=u
,\quad
\tan\tfrac12C=u+\sqrt{3u^2-1}
\tag{3}\label{3}
.
\end{align}
Since all the tangents must be positive,
we have a condition
$u\in(\tfrac{\sqrt3}3,\,\tfrac{\sqrt2}2)$,
the endpoints correspond to degenerate solutions:
one corresponds to the equilateral triangle, $d=0$,
and the other corresponds
to the degenerate triangle with $\tan\tfrac12A=0$.
We also know that
\begin{align}
\cos x&=\frac{1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2}
,
\end{align}
so the corresponding three cosines
\begin{align}
\cos A&=
\frac{u-u^3+\sqrt{3u^2-1}}{u(1+u^2)}
,\\
\cos B&=
\frac{1-u^2}{1+u^2}
,\\
\cos C&=
\frac{u-u^3-\sqrt{3u^2-1}}{u(1+u^2)}
\end{align}
are indeed in arithmetic progression,
\begin{align}
u'-d',\quad &u',\quad u+d'
,\\
u'&=\frac{1-u^2}{1+u^2}
,\qquad
d'=-\frac{\sqrt{3u^2-1}}{u(u^2+1)}
.
\end{align}
Bonus:
a suitable example
of the triangle $ABC$ with given properties
has integer side lengths
$a=27$,
$b=32$ and
$c=35$ units,
and
\begin{align}
u&=\tfrac{4\sqrt{47}}{47}
,\quad
d=\tfrac{\sqrt{47}}{47}
,\\
u'&=\tfrac{31}{63}
,\quad
d'=-\tfrac{47}{252}
,\\
\alpha&=
2\arctan(\tfrac{3\sqrt{47}}{47})=\arccos(\tfrac{19}{28})
\approx 47.2679^\circ
,\\
\beta&=
2\arctan(\tfrac{4\sqrt{47}}{47})=\arccos(\tfrac{31}{63})
\approx 60.5237^\circ
,\\
\gamma&=
2\arctan(\tfrac{5\sqrt{47}}{47})=\arccos(\tfrac{11}{36})
\approx 72.2084^\circ
.
\end{align}
|
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"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solving limit for $\lim_{n \to \infty} \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13}$ Trying to solve:
$$\lim_{n \to \infty} \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13}$$
I tried to divide with $n^3$ and $n^2$, but the limit of $(-1)^{n}$ is not existed, so i am wondering if the limit of $(-1)^{n}\cdot n$ does exist ?
|
We have indeed that $\lim_{n \to \infty} (-1)^{n}\cdot n$ doesn't exist but we have that
$$ \frac{3n^3 + 5n^2 + (-1)^n \cdot n}{5n^2 - 13} = \frac{n^3}{n^2}\cdot\frac{3 + \frac 5 n +\frac{(-1)^n}{n^2} }{5- \frac{13}{n^2}}$$
and
$$\left|\frac{(-1)^n}{n^2}\right|= \frac{1}{n^2}\to 0 \implies \frac{(-1)^n}{n^2}\to 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3453401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the family of functions given an equation I played with the solution for the problem $$\text{if }\; x + \frac{1}{x} = a$$ what is $$x^5 + \frac{1}{x^5}$$
I tried different exponents other than 5 and tried finding the solution to it. I defined $f(x) = a^x + \frac{1}{a^x}$. I got $f(x + y)=f(x)f(y) - f(x-y).$ I tried reversing the equation I got to get $f(x)$ but I only got these:
$$f(0) = 2$$ by substituting $b=0$
$$f(x)=f(-x)$$
$$(f(a)^2 - 4)(f(b)^2 - 4) \geq 0$$
$$f'(0)=0$$
Can this be solved using the given information? Is $f(x) = a^x + \frac{1}{a^x}$ the only solution? Thanks in advance!
Edit: I already got the solution for $x^5 + \frac{1}{x^5}$, I'm asking if how can I get the family of functions $f(x)$ from $f(x + y)=f(x)f(y) - f(x-y)$, sorry for the unclear question
|
Hint:
$$\left(x^3+\dfrac1{x^3}\right)\left(x^2+\dfrac1{x^2}\right)=x^5+\dfrac1{x^5}+x+\dfrac1x$$
Now $x^3+\dfrac1{x^3}=\left(x+\dfrac1{x}\right)^3-3\left(x+\dfrac1{x}\right)$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3454313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solution attempt $xuu_x+yuu_y=u^2-1$ Solve $$
\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}
$$
I have got using Lagrange method:
$$F\left(\frac{x}{y}\right)=\frac{x^2}{u^2-1}$$
Applying $u(x,x^2)=x^3$:
$$u^2=\frac{y^6-x^6}{x^2y^2}+1$$
But plug in it to the PDE show that there is a mistake
|
For the given PDE, Lagrange auxiliary equation $$\dfrac{dx}{xu}=\dfrac{dy}{yu}=\dfrac{du}{u^2-1}\tag1$$
From the first two ratio, $$\dfrac{dx}{xu}=\dfrac{dy}{yu}\implies \dfrac{dx}{x}=\dfrac{dy}{y}$$
Integrating we have $$\log x~=~\log y~+~\log c_1\implies\dfrac xy=c_1$$where $~c_1~$is a constant.
Again from the first and the last ratio, $$\dfrac{dx}{xu}=\dfrac{du}{u^2-1}\implies \dfrac{dx}{x}=\dfrac{u~du}{u^2-1}$$
Integrating we have$$2\log x~=~\log(u^2-1)~+~\log c_2\implies \dfrac{x^2}{u^2-1}~=~c_2$$where $~c_2~$is a constant.
Hence the general solution is $$F\left(\dfrac xy\right)=\dfrac{x^2}{u^2-1}\tag2$$where $~F~$is an arbitrary function.
Now given that $~u(x,x^2)=x^3~$ i.e., when $~y=x^2~$, then $~u=x^3~$. So from $(2)$, we have$$F\left(\dfrac 1x\right)=\dfrac{x^2}{x^6-1}$$
Hence $$F\left(\dfrac xy\right)=\dfrac{y^2/x^2}{y^6/x^6-1}=\dfrac{x^4y^2}{y^6-x^6}$$
From equation $(2)$ we have $$\dfrac{x^4y^2}{y^6-x^6}=\dfrac{x^2}{u^2-1}$$
$$\implies u^2-1=\dfrac{y^6-x^6}{x^2y^2}$$
$$\implies u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1$$This is the solution of the given PDE.
Cross-check : If possible let $~ u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1~\tag3$ is the solution of the given PDE $$\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}~.$$
Now putting $~y=x^2~$ in the solution we have
$$~ u^2(x,x^2)=\dfrac{x^{12}-x^6}{x^2x^4}~+~1~=~x^{6}-1~+~1~=x^6\implies u(x,x^2)=x^3$$Hence the given condition is satisfied.
Now we check whether the value satisfy the PDE or not ?
Differentiating the equation $(3)$ partially with respect to $~x~$, we have
$$2uu_x=\dfrac{-6x^5}{x^2y^2}~-~2~\dfrac{y^6-x^6}{x^3y^2}=-2\dfrac{2x^6+y^6}{x^3y^2}$$ Again differentiating the equation $(3)$ partially with respect to $~y~$, we have
$$2uu_y=\dfrac{6y^5}{x^2y^2}~-~2~\dfrac{y^6-x^6}{x^2y^3}=2\dfrac{x^6+2y^6}{x^2y^3}$$
Now $$xuu_x+yuu_y=-~\dfrac{2x^6+y^6}{x^2y^2}~+~\dfrac{x^6+2y^6}{x^2y^2}=\dfrac{-x^6+y^6}{x^2y^2}=u^2-1$$
Hence it is clear that
$$~ u^2(x,y)=\dfrac{y^6-x^6}{x^2y^2}~+~1$$ is the solution of the given PDE $$\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}~.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$
then:
$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
and I can solve for A by setting $x=-2$, which yields $A=1$.
I know the final answer is this:
$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$
But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.
|
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}= \frac {(3x^2+3x +3)-(x+2)}{(x+2)(x^2+x+1)^2}=$$
$$ \frac {3}{(x+2)(x^2+x+1)}-\frac { 1}{(x^2+x+1)^2} $$
Now you may proceed with the first fraction.
$$ \frac {3}{(x+2)(x^2+x+1)}=\frac {1}{x+2} -\frac {x-1}{x^2+x+1}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \Rightarrow x^2 + x = 0 \Rightarrow x_3= 0 \text{, (but doesn't fulfill } x+1 < 0), x_4 = -1$$
$$\Rightarrow L = \{2,-1\}$$
What I don't get is how $x+1 < 0 \Rightarrow x^2 + x = 0$. How do we get $x^2 + x = 0$?
|
When $x + 1 < 0$ then the initial equation becomes $$\begin{align*} && -(x+1) &= x^2 - 1 \\ &\iff&-x -1 &= x^2 - 1 \\ &\iff& 0 &= x^2 + x\end{align*}$$
|
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|
How to prove that $ \arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2} $ I want to prove that: $$\arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2}, \forall x \in \mathbb{R} $$
I know that $$ \arctan x + arctan \frac{1}{x} = \frac{\pi}{2} $$
But that doesn't seem to be helping. How should I proceed?
|
You may set $x=\tan\theta$ (for some $\theta\in\left(-\pi/2,\pi/2\right)$) and check that
$$2\arctan\left(\frac{1-\sin\theta}{\cos\theta}\right) = \frac{\pi}{2}-\theta $$
holds, or that
$$ 2\arctan\left(\frac{1-\cos\theta}{\sin\theta}\right)=\theta $$
holds for any $\theta\in(0,\pi)$. Since $1-\cos\theta=2\sin^2\frac{\theta}{2}$ and $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ this is equivalent to
$$ 2 \arctan\tan\frac{\theta}{2} = \theta $$
for any $\theta\in(0,\pi)$, which is fairly obvious.
|
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|
$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=d$. Also notice that $(a+b+c+d)^{2}$ must be odd and only have three factors: $1, p, p^{2}$.
$$ p^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ab+ac+ad + bc+bd + cd) $$
$$ = 2(c^{2}+d^{2}+ac+ad++bc+bd) + (ab + 3cd) $$
$$ =2(c^{2}+d^{2}+ac+ad++bc+bd + cd) + (ab + cd) $$
So $(ab+cd)$ must be odd.
Now if $a+b+c+d$ is prime $>2$ then either 3 of them is odd and 1 is even, or 3 of them is even and one is odd.
$WLOG$, let $a,b,c$ be even and $d$ is odd, then $a^{2} + b^{2}+ ab$ is even and $c^{2} + d^{2}+ cd$ is odd, so we can't have 3 even and 1 odd.
But it is possible for 3 odd and 1 even.
|
If $ \{ a, b \} = \{c, d \}$ then $a+b+c+d = 2(a+b)$ is not a prime.
Henceforth, we have distinct pairs. If $ a = c$, then $b \neq d$ are roots to $x^2 + ax + a^2 = 0$, so $b+d = -a$ contradicting the requirement that all of the terms are positive.
WLOG, $ a > c \geq d > b$.
Observe that $$ -(a-b+c-d)(a-b-c+d) = -\left((a-b)^2 - (c-d)^2\right) = 3(ab-cd) = 3\left( (a+b)^2 - (c+d)^2\right) = 3 (a+b-c-d)(a+b+c+d).$$
Since $ a+b+c+d > a-b+c-d \geq a-b-c+d > 0$, we can rewrite the above as a product of positive integers,
$$ (a-b+c-d)(a-b-c+d)= 3 (c+d-a-b)(a+b+c+d)$$
If $a+b+c+d$ is prime, then the LHS can never be a multiple of $a+b+c+d$ since both terms are smaller, hence we have a contradiction. So $ a+b+c+d$ is composite.
Notes:
*
*This problem/solution is very reminiscent of 2001/6 IMO. There we had $a^2+ab+b^2 = c^2 -cd+d^2$.
*I do not know a direct way to show that $cd > ab$. Any thoughts? This seems obvious.
*This being a contest problem suggests that there is a non "abstract algebra approach", even though it is very tempting. E.g. The equation in my solution can also be derived from WhatsUp's characterization (which is a stronger requirement).
|
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|
Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$ Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$
I have tried using Taylor series. On a side not, I am not sure what to write when expanding both the $\sqrt{x^6+1}$ and the $e^{1/x}$. Do I write both $o(x^n)$ and $o(1/x^m)$? Seems like an abuse of notation.
Here is my attempt:
$(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}=$
$(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+\frac{1}{4!x^4}+o(\frac{1}{x^5}))-(1+\frac{x^6}{2}+o(x^6))$
But when I distribute terms there is no way for me to kill off the $x^n$ terms for positive $n$ and which makes me believe the limit is infinite. According to wolfram, the limit is $\frac{1}{6}$.
What am I missing here? Is this notation acceptable?
|
Let $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}=(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6(1+\frac{1}{x^6})}$
This is $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -x^3\sqrt{1+\frac{1}{x^6}}$
$(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+\frac{1}{4!x^4}+o(\frac{1}{x^5}))-x^3(1+\frac{1}{2x^6}+o(\frac{1}{x^{12}}))$
Expand your brackets and you're sorted.
|
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|
Help with this multi variable limit. $ \lim_{x,y\to (1,0)} \frac{2x^4+x^2y^2-2x^2-y^2}{2x^2y^2-x^2-2y^2+1} $
By just plugging in $(1,0)$ into the limit you get $-2$, but I don't know if this enough to solve the question
Ive tried approaching along $x=y+1$ but it got extremely messy.
|
You have
$$
\frac{2x^4+x^2y^2-2x^2-y^2}{2x^2y^2-x^2-2y^2+1}= \frac{(2x^2+y^2)(x^2-1)}{(2y^2-1)(x^2-1)}= \frac{2x^2+y^2}{2y^2-1}\to\frac2{-1}
$$
|
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|
Solve $x^{2}y''-3xy'+4y=x^{2}lnx$.
Solve $x^{2}y''-3xy'+4y=x^{2}lnx$.
I wanted to try and follow my professor's notes but I was confused when she wrote $\eta=lnx$ and $\frac{dy}{dx}=\frac{1}{x}\frac{dy}{d\eta}$. Something about changing the equation to have constant coefficients.
Is there a step-by-step way of solving such a differential equation?
|
Let $D \equiv \frac{d}{dx}, D^2 \equiv \frac{d^2}{dx^2}$ and $\theta \equiv\frac{d}{d\eta}, \theta^2 \equiv\frac{d^2}{d\eta^2}$.
Once you substitute $\ln x = \eta \implies xD = \theta, x^2D^2 = \theta (\theta -1)$
In general
$x^nD^n = \theta(\theta-1)\cdots(\theta -n +1)$
This works only in case you have differential equations with coefficients of the form $kx^nD^n$, where $k$ is a constant.
So the equation reduces to,
$\theta(\theta-1)y - 3\theta y + 4y = \eta e^{2\eta} \implies (\theta^2-4\theta+4)y =\eta e^{2\eta} \implies (\theta -2)^2y = 2 \eta e^{2\eta}$
You'll get a complementary solution:$C_y = (A\eta+b)e^\eta = (A\ln x + b)x$
and a particular solution, $$P_y = \frac{1}{(\theta-2)^2}\eta e^{2\eta} = e^{2\eta}\frac{1}{(\theta+2-2)^2}\eta = e^{2\eta}\frac{1}{\theta^2}\eta = \frac{1}{6}e^{2\eta}\eta^3= \frac{1}{6}x^2(\ln x)^3$$
So, $$ y = (A\ln x + b)x + \frac{1}{6}x^2(\ln x)^3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3470595",
"timestamp": "2023-03-29T00:00:00",
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|
MixColumn AES manually with GF(2^8) In the step of MixColumn in AES, calculate d0,0 if given c0,0 = c2,0 =
1100 0101, and c1,0 = c3,0 = 0100 1100.
Converting to polynomials:
c0,0 = c2,0 = x^7 + x^6 + x^2 + 1
c1,0 = c3,0 = x^6 + x^3 + x^2
now using the Galois field to get d0,0 we will get 4 results and then XOR all of them
c0,0 * 10 = (x^7 + x^6 + x^2 + 1)*(x) mod (x^8 + x^4 + x^3 + x + 1) mod 2 = 10010001
c1,0 * 11 = (x^6 + x^3 + x^2) * (x+1) mod (x^8 + x^4 + x^3 + x + 1) mod 2 = 11100100
c2,0 * 01 = (x^7 + x^6 + x^2 + 1)*(1) mod (x^8 + x^4 + x^3 + x + 1) mod 2 = 11000101
c3,0 * 01 = (x^6 + x^3 + x^2) * (1) mod (x^8 + x^4 + x^3 + x + 1) mod 2 = 01001100
Now XOR all of those 8 bit binary values I get 11111100 but the answer should be 11001100 where am I making my mistake?
Any help is appreciated
|
The second product is incorrect. There is no overflow to degree $8$ in that product, and it comes out as
$$
\begin{aligned}
c_{1,0}\cdot 11&=\overline{(x^6+x^3+x^2)(x+1)}\\
&=\overline{x^7+x^6+x^4+2x^3+x^2}\\
&=\overline{x^7+x^6+x^4+x^2}=11010100.
\end{aligned}
$$
I use overline to denote the coset of a polynomial because iterated mods give me pimples.
Anyway, this differs exactly in those two bit positions your end result differs from the given answer, so I'm optimistic about it being the only error.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the $\int \frac{dx}{\sin^5x}$ $\int \frac{dx}{\sin^5x}$
I did $t= \cos(x)$, $-\frac{dt}{\sin(x)}=dx$:
$$\int \frac{dx}{\sin^5x} = \int \frac{dt}{\sin^6x} = -\int \frac{dt}{(1-t^2)^3}=-\int (1-t^2)^{-3} = \frac{(1-t^2)^{-3}}{-3} = -\frac{1}{3(1-\cos^2x)}+C$$
Is this correct?
|
With $t=\cos x$
\begin{align}
\int \frac{1}{\sin^5x} dx= &-\int \frac{1}{(1-t^2)^3}dt
=-\int \frac1{4t^3}\ d\left( \frac{t^4}{(1-t^2)^2}\right)\\
\overset{ibp}= & -\frac t{4(1-t^2)^2}-\frac38\int \frac1{t}\ d\left( \frac{t^2}{1-t^2}\right)\\
\overset{ibp}= & - \frac t{4(1-t^2)^2}-\frac38 \left(
\frac t{1-t^2}+\tanh^{-1}t \right)+C\\
= & -\frac {\cos x}{4\sin^4 x}-
\frac {3\cos x}{8\sin^2x}-\frac38\tanh^{-1}\cos x+C\\
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
The set of differences of square rationals At first, we observe that $A:=\{ p^2-m^2 : p,m\in \Bbb{Z}\}=\mathbb{Z}\setminus (4\mathbb{Z}+2)$ (because an integer $a$ can be written
as the form $a=p^2-m^2$ if and only if $a\neq 4k+2$, for every integer $k$) and so $A^c=4\mathbb{Z}+2$. Hence, $B:=\left\{ p^2-m^2 : p,m\in \Bbb{Z}\setminus \{0\}\right\}=A\setminus \left\{\pm k^2: k\in \Bbb{Z} \mbox{ and } k^2\neq s^2-t^2, \mbox{ for all } s,t\in \Bbb{Z}\right \}$
Now, put
$$
D:=\left\{ \frac{p^2}{q^2}-\frac{m^2}{n^2}: p,q,m,n\in \Bbb{Z}\setminus \{0\} \text{ and} \gcd(p,q)=
\gcd(m,n) =1\right\}
$$
We would like to determine this set and its complement $D^c$ in $\Bbb{Q}$ exactly.
It is clear that $B\subseteq D$. Is it true that $D\neq \Bbb{Q}$?
|
We have $4m = (m+1)^2-(m-1)^2$ for all $m \in \mathbb Z$. Therefore,
$$
\frac{a}{b}
=
\frac{4ab}{4b^2}
=
\frac{(ab+1)^2-(ab-1)^2}{(2b)^2}
=
\left(\frac{a b + 1}{2 b}\right)^2 - \left(\frac{a b - 1}{2 b}\right)^2
$$
When $ab=\pm 1$, one of the terms is zero. In this case, $\frac{a}{b}=\pm1$ and we can use
$$
1 = \left(\frac{5}{3}\right)^2 - \left(\frac{4}{3}\right)^2,
\quad
-1 = \left(\frac{4}{3}\right)^2 - \left(\frac{5}{3}\right)^2
$$
Thus, all rationals are the difference of two nonzero rational squares.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve a system by putting new variables
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$$
The first step is to determine the domain: $\begin{array}{|l} x \ne 0 \\ y \ne 0 \end{array}$
We can simplify the first equation of the system, and we get: $\begin{array}{|l} \dfrac{x^2+y^2}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$
$\begin{array}{|l} \dfrac{25}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} 7xy=625 \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} xy=\dfrac{625}{7} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ y^4-25y^2+\dfrac{390625}{49}=0 \end{array}$
The equation: $$y^4-25y^2+\dfrac{390625}{49}=0$$ has no solutions, so the whole system does not have a solution. In the text of the problem is said I should solve by "putting a new variable". I don't know how this method is called in English, and I would be grateful if you tell me. Let me give you a basic example of a system that can be solved using this method:
$$\begin{array}{|l} (x+2y)^2-(y-2x)^2=168 \\ (x+2y)^2+(y-2x)^2=12 \end{array}$$
Let $$\begin{array}{|l} (x+2y)^2=a \\ (y-2x)^2=b \end{array}...$$
|
Rewrite the first equation as
$$
\frac{x^2}{xy}+\frac{y^2}{xy}=\frac{7}{25}.
$$
The second equation then gives $\frac{625}{7}=xy$. Substituting gives
$$
49x^4 - 1225x^2 + 390625=0,
$$
which is a biquadratic equation with no real solution. This seems short enough.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{n^2+1} \sim \frac { 2^{\ln n }}{n^2+1} \rightarrow 0$$
Is it right?
I have another exercize that ends similarly with
$$ \frac { 10^{\ln n }}{n^2+1} \rightarrow 0$$
But the book says that the result is $+\infty$.
|
Your use of $\sim$ is not correct although this small mistake does not change the limit. You have
$$2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}} \stackrel{n\to\infty}{\sim}\color{blue}{2\cdot}2^{\ln n}$$
This is so because
$$\frac{2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{2^{\ln n}}= 2^{\ln n\left(\sqrt{ 1+ \frac {2}{\ln n}}-1\right)} = 2^{\ln n \frac{\frac{2}{\ln n}}{\sqrt{ 1+ \frac {2}{\ln n}}+1}}$$ $$=2^{\frac{2}{\sqrt{ 1+ \frac {2}{\ln n}}+1}}\stackrel{n\to\infty}{\longrightarrow}2^1=2$$
At the end you may add another step (which is also useful for your second question):
*
*$2^{\ln n} = e^{\ln 2\cdot \ln n} = n^{\ln 2}$ and similarly for your second question $10^{\ln n} = n^{\ln 10}$
Putting this together, you get
$$0\leq \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} \ldots \sim \frac { 2\cdot 2^{\ln n }}{n^2+1}=\frac{2n^{\ln 2}}{n^2+1}\stackrel{\color{green}{\ln 2 < 1}}{\leq}\frac{2n}{n^2+1}\stackrel{n\to\infty}{\longrightarrow}0$$
Similarly for your second question
$$\frac { 10^{\ln n }}{n^2+1} = \frac{n^{\ln 10}}{n^2+1}\stackrel{\color{green}{\ln 10 > \frac 94}}{\geq}\frac{n^{\frac 94}}{n^2+1}$$ $$\geq\frac{1}{2}\cdot\frac{n^{\frac 94}}{n^2}=n^{\frac 14}\stackrel{n\to\infty}{\longrightarrow}+\infty$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
In triangle $ABC$, find maximum value of $\sin A \cos B + \sin B \cos C + \sin C \cos A$ In triangle $ABC$, find maximum value of $$\sin A \cos B + \sin B \cos C + \sin C \cos A$$
We could make $\cos C = - \cos(A+B)$ and $\sin C = \sin(A+B)$.
But then we have a rather awkward expression that doesn't share the same power
$$ \sin A \cos B + \sin A \sin^{2}B + \sin B \cos^2A $$
The answer btw is not hard to be guessed, $\frac{3}{4} \sqrt{3}$, but not sure how to prove it.
Actually looks like we can solve like below:
if $A < B < C$, then $cosA > cosB > cosC$, $sinC > sinB > sinA$
Then $sinA cosB + sinB cosC + sinC cosA \leq cosAsinC + sinBcosB + sinAcosC = sinBcosB + sinB = sinB \sqrt{1-sin^2 B} + sinB$
did I do this correctly?
|
You can use Lagrange multipliers. You are maximizing
$$
\sin A\cos B+\sin B\cos C+\sin C\cos A
$$
under
$$
A+B+C=\pi.
$$
Using Lagrange multipliers and the identity $\cos (x+y)=\cos x\cos y-\sin x\sin y$, you easily get
$$
\cos(A+B)-\lambda=\cos(B+C)-\lambda=\cos(A+C)-\lambda=0,
$$
from where it follows that $\cos C=\cos A=\cos B=-\lambda$ and then necessarily (given that at least two of them are acute) $A=B=C$ and the maximum is achieved for the equilateral triangle.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int \frac{dx}{x\sqrt{x^2-2}}$ $\int \frac{dx}{x\sqrt{x^2-2}}$
My book says to solve this using $x = 1/t$ so I did $dx = -1/t^2$
$$-\int\frac{\frac{1}{t^2}}{t\sqrt{1-2t^2}} = -\int \frac{1}{\sqrt{1-2t^2}} = -\arcsin(\sqrt{2t})=-\arcsin(\sqrt{2}/x)$$
My book says the solution is $\frac{1}{\sqrt{2}}\arccos(\frac{\sqrt{2}}{x}).$ What went wrong?
|
First, remember integration constants: $\arcsin y=\frac{\pi}{2}-\arccos y$. Secondly, be careful with coefficients: $\sqrt{1/t^2-2}=\sqrt{2}\sqrt{1/(2t^2)-1}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.
|
Assuming $n\neq 0$ (to make the expression under limit meaningful) the given expression can be written as $$\dfrac{\dfrac{x} {\sin x} - 1}{\left(\dfrac{x}{\sin x}\right)^n-1}\cdot\frac{x^3}{x-\sin x} \cdot x^{n-2}$$ The first factor tend to $1/n$ and second one tends to $6$ and hence the limiting behavior of the expression crucially depends on that of $x^{n-2}$. The desired limit is thus $6/n=3$ if $n=2$, is $0$ if $n>2$ and does not exist if $n<2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$? What do we know about the rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$ in five variables $x,y,z,u,v$?
I am looking for references/papers related to this equation.
|
$ax^2+by^2+cz^2+du^2=v^2$
Let assume $a+b+c+d=r^2.$
$p,q$ are arbitrary.
Substitute $x=pt+1, y=qt+1, z=pt-1, u=qt-1, v=t+r$ to above equation, then we get $$t = \frac{2(ap-qr^2+qa+2bq+qc-cp-r)}{(-ap^2-q^2r^2+q^2a+q^2c-cp^2+1)}.$$
Thus, we get a parametric solution below.
\begin{eqnarray}
&x& = (a-3c)p^2+(2qc-2qr^2+2qa+4bq-2r)p+q^2a+q^2c+1-q^2r^2 \\
&y& = (-a-c)p^2+(2qa-2qc)p+3q^2c-3q^2r^2+3q^2a+4bq^2+1-2qr \\
&z& = (3a-c)p^2+(2qc-2qr^2+2qa+4bq-2r)p-q^2a-q^2c-1+q^2r^2 \\
&u& = (c+a)p^2+(2qa-2qc)p+q^2c-q^2r^2+q^2a+4bq^2-1-2qr \\
&v& = (-ra-rc)p^2+(2a-2c)p+2qc-2qr^2+2qa+4bq-q^2r^3+rq^2a-r+rq^2c.
\end{eqnarray}
Example for $(a,b,c,d,r)=(1,2,3,3,3).$
\begin{eqnarray}
&x& = -8p^2+(-2q-6)p+1-5q^2\\
&y& = -4p^2-4qp-7q^2-6q+1\\
&z& = (-2q-6)p+5q^2-1\\
&u& = 4p^2-4qp+3q^2-6q-1\\
&v& = -12p^2-4p-2q-3-15q^2
\end{eqnarray}
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.