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equation $\log_{2x}216=x$ Solve the equation $\log_{2x}216=x$ where $x$ is real.
My attempt,
By trial and error $x=3$ as $6^3=216$
Is there any systematic way to solve it?
|
Noting that $$\log_{2x}(216) = \frac{\ln(216)}{\ln(2x)}$$ we can say \begin{align}\log_{2x}(216) &= x\\
\frac{\ln(216)}{\ln(2x)}&=x\\
\ln(216)&=x\ln(2x)\\
e^{\ln(216)}&=e^{x\ln(2x)}\\
e^{\ln(216)}&=e^{\ln((2x)^x)}\\
216&=(2x)^x
\end{align}
We can then note that $216=6^3$ and so \begin{align}6^3&=(2x)^x\\
(2\times 3)^3&=(2x)^x\end{align}
It is then trivial to see that $x=3$
We could also calculate this last step by noting that the exponent must be half the base.
We can then brute force this with integers:
Starting with an exponent of $1$, we have
\begin{align}2^1&=2\\
4^2&=16\\
6^3&=216\end{align}
And so our exponent, and thus $x$, must be $3$
|
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|
Distribute $a + b + c$ distinct balls into boxes $A, B, C$ such that $a$ balls, $b$ balls and $c$ balls go to boxes $A, B, C$
Distribute $a + b + c$ distinct balls into boxes $A, B, C$ such that $a$ balls, $b$ balls and $c$ balls go to boxes $A, B, C$, respectively, and show that the number of ways to do this is $\frac{(a+b+c)!}{a!b!c!}$
I want to do the special case just to see if I am thinking correctly about this. Please, see if that makes sense.
Show the number of ways to distribute $10$ distinct balls into boxes $A, B, C$ is $\frac{(5 + 3 + 2)!}{5!2!3!}$ if exactly five balls go to box $A$ exactly three balls go to box $B$ and exactly two balls go to box $C$.
There are $10$ ways to choose the first ball going into the box $A$, there are $9$ ways to choose the second ball going into the box $A$...in all there are $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6$ ways to put five balls into the box $A.$ For each such placement of five balls in $A$, we can place three balls in $5 \cdot 4 \cdot 3$ ways. Finally, there are two ways to place the last two balls in $C.$ In total, there are $10!$ ways to place ten distinct balls in the boxes $A, B, C.$ But we have overcounted by the factor of $5!3!2!$ so we divide them out: $\frac{(5 + 3 + 2)!}{5!2!3!}.$
|
You say that "in all there are $10⋅9⋅8⋅7⋅6$ ways to put five balls into the box A". That is only true if the order in which we put the balls into the box matters. Clearly, it doesn't over here thus the correct computation would be $\displaystyle\binom{10}{5},$ that is $\dfrac{10⋅9⋅8⋅7⋅6}{5!}$ since we divide by the number of ways the $5$ balls could be ordered. Having selected $5$ balls for box A, you can select $3$ balls for box B in $\displaystyle\binom{5}{3}$ ways and $2$ balls for box C in $\displaystyle\binom{2}{2}=1$ way. Since these choices are independent, the total number of ways to assign $10$ balls to $3$ boxes subject to those conditions is $\displaystyle\binom{10}{5}\cdot\binom{5}{3}\cdot\binom{2}{2}= \displaystyle\binom{10}{5,3,2}=\dfrac{10!}{5!\cdot3!\cdot2!}$, if you're familiar with multinomial coefficients.
|
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|
Error in Polynomial Factoring
From Ramanujan's Notebooks IV:
Let $\alpha,\beta$ and $\gamma$ be the roots of$$x^3-ax^2+bx-1=0\tag1$$Now, choose cube roots such that $(\alpha\beta\gamma)^{1/3}=1$ and then let$$z^3-\theta z^2+\varphi z-1=0\tag2$$Denote the cubic polynomial with roots $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. Cubing both sides of the equality$$z^3-1=\theta z^2-\varphi z\tag3$$We find that$$(z^3-1)^3-\theta^3z^6+\varphi^3z^3+3\theta\varphi z^3(z^3-1)=0\tag4$$Since $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$ are roots of $(2)$, they are also the root of $(4)$. As a cubic polynomial in $z^3$, $(4)$ thus has the roots $\alpha,\beta,\gamma$. Comparing $(1)$ with $(4)$, we deduce$$a=\theta^3+3-3\theta\varphi\tag5$$$$b=\varphi^3+3-3\theta\varphi\tag6$$ If we define $t$ by$$\theta^3=a+6+3t\tag7$$Then, by $(2)$ and $(7)$, $$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=\theta=(a+6+3t)^{1/3}$$And similarly, $$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=\varphi=(b+6+3t)^{1/3}$$
While the proof is fairly elementary, there is one problem that I'm not sure how Bruce got there. From $z^3-1=\theta z^2-\varphi z$, to the next step, how'd they get the $3\theta\varphi z^3(z^3-1)$. When you expand $\theta z^2-\varphi z$, you get$$(\theta z^2-\varphi z)^3=\theta^3z^6-3\theta^2\varphi z^5+3\theta\varphi^2z^4-\varphi^3z^3$$So where did that term come from?
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The last term on the left multiplies out to $3\theta\varphi(z^3-1)=3\theta\varphi z^3-3\theta\varphi$ which are intended to be the two middle terms from cubing the original right side. It should be $-3\theta\varphi z^3(\theta z^2-\varphi z)$
|
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|
Linear differential equation - inseparable? I need help with solving following linear differential equation:
$$y^\prime + \dfrac{3y}{x} = \dfrac{3}{x^4}$$
I have set the right side equal to $0$ and started solving:
$$y^\prime + \dfrac{3y}{x} = 0$$
$$y^\prime = - \dfrac{3y}{x}$$
$$\dfrac{dy}{dx}\times \dfrac{1}{y} = - \dfrac{3}{x}$$
$$\int\dfrac{1}{y} dy = -3\int \dfrac{1}{x} dx$$
$$\ln(y)=-3\times \ln(x) + c$$
And finally solution with left side (lets say this $c$ is: $e^c$)
$$y = c\times(-3x)$$
Using some formulas
$$y = c\times f(x)$$
$$y = c(x)\times f(x)$$
$$y^\prime = c^\prime(x)\times f(x)+c(x)\times f^\prime (x)$$
$$y = c(x)\times (-3x)$$
$$y^\prime = c^\prime(x)\times (-3x)+c(x)\times (-3)$$
And I replaced $y$ and $y^\prime $ in original equation and get
$$c^\prime(x)\times (-3x)+c(x)\times (-3)+\dfrac{3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^4}$$
And after adjustment:
$$\dfrac{c^\prime(x)\times (-3x)+c(x)\times (-3)+3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^5}$$
And what are next steps? I know I need to get $c(x)$ but how will I obtain it and what will be the final result?
|
By multiplying both sides by the integrating factor $x^3$, we get
$$D(x^3y(x))=x^3y'(x) + 3 x^2 y(x) = \frac{3}{x}.$$
Hence
$$x^3y(x)=\int \frac{3}{x}\, dx=3\ln|x|+C,$$
that is
$$y(x)=\frac{3\ln|x|+C}{x^3}.$$
|
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|
Maximum value for $M^4$
$$M=|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+4}|$$
Find the maximum value of $M^4$.
I think it could have a geometric solution. Because it looks like the difference between two points formula. Please help.
|
$$\sqrt{x^2+4x+5}=\sqrt{(x+2)^2+1^2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$
$$\sqrt{x^2+2x+4}=\sqrt{(x+1)^2+(\sqrt{3})^2}~~~~~~~~~~~~~~~~~~~~~~~~(2)$$
$(1)$ is the distance of $(x,0)$ to $(-2,1)$ or $(-2,-1)$
$(2)$ is the distance of $(x,0)$ to $(-1,\sqrt{3})$ or $(-1,-\sqrt{3})$
|
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|
Parallel Vectors and Coefficients of $\mu$ I have a question that states the following
Given that c = 3i + 4j and d = i - 2j, find μ if μc + d is parallel to i + 3j
So I wrote the following
$$ c = \begin{pmatrix} 3\\ 4\\ \end{pmatrix}$$
$$ d = \begin{pmatrix} 1\\ -2\\ \end{pmatrix}$$
I then wrote
$$ \mu \textbf{c} + \textbf{d} =\mu\begin{pmatrix} 3\\ 4\\ \end{pmatrix} + \begin{pmatrix} 1\\ -2\\ \end{pmatrix}$$
$$= \begin{pmatrix} 3\mu + 1\\ 4\mu-2\\ \end{pmatrix}$$
Now I thought that if this is parallel to i + 3j then from above,
$$3\mu + 1 = 3(4\mu -2)$$
but the answer says that, in fact $$4(3\mu + 1) = 4\mu -2$$
I am confused. Why is this?
Wouldn't that imply that it is parallel to 3i + j?
The answer, strangely is given as such
|
Using your notation, we have a vector
$$\mathbf a= \begin {pmatrix}a_1\\a_2 \end{pmatrix}=\begin {pmatrix}1\\3 \end{pmatrix}$$ so a vector parallel to $\mathbf a$ is of the form
$$
\mathbf b =\begin {pmatrix}b_1\\b_2 \end{pmatrix}=\begin {pmatrix}b_1\\3b_1 \end{pmatrix}
$$
Apply this to
$$ \mathbf b=\begin {pmatrix}b_1\\b_2 \end{pmatrix}=
\begin {pmatrix}3\mu+1\\4\mu -2 \end{pmatrix}
$$
we have
$$
b_2=3b_1 \iff 4\mu -2=3(3\mu +1)
$$
|
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|
Find integer part of the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$ for $x+y=3$
If $M$ is the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$. Subject to $x+y=3$. Find the value of $\lfloor M \rfloor$.
I think the maximum of $M$ occurred when $x=y=3/2$. And I just put the value of that in M and found 34. What is the original answer with good solution ?? Somebody please help me.
|
After substitution $y=3-x$ we get:
$$M=-11x^4+66x^3-139x^2+120x$$
$$M'(x)=2(3-2x)(11x^2-33x+20),$$
which gives a maximal value $\frac{400}{11}$ for $x=\frac{3+\sqrt{\frac{19}{11}}}{2}$.
|
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|
$\frac{(10^4+324)(22^4+324)\cdots(58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$
From AIME 1987, compute $$\frac{(10^4+324)(22^4+324)\cdots (58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$$
So basically the way used to solve this is by Sophie Germain's Identity which is $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$
My question is , how is possible a student to solve this question without knowing this identity? Is there another way to solve this?
|
(Without using Sophie Germain's identity, per OP's request.)
The given expression is $\,f(7, 5)\,$ where $\,\displaystyle
f(x, n) = \prod_{k=0}^{n-1} \frac{(x+12k+3)^4+324}{(x+12k-3)^4+324}\,$.
Taking each term in turn, let $\,y=x+12k\,$ then the fraction can be written as $\,\displaystyle\frac{(y+3)^4+324}{(y-3)^4+324}\,$. Looking to simplify the fraction, it turns out that $\,\gcd\left((y+3)^4+324, (y-3)^4+324\right)=y^2+9\,$ (n.b. by polynomial Euclidean algorithm, not using any Sophie Germain's insight). After canceling out the common factor of $\,y^2+9\,$, the term becomes:
$$\require{cancel}
\displaystyle\frac{(y+3)^4+324}{(y-3)^4+324} = \displaystyle\frac{\cancel{(y^2+9)}(y^2+12y+45)}{\cancel{(y^2+9)}(y^2-12y+45)}=\frac{(y+6)^2+9}{(y-6)^2+9}=\frac{\big(x+12k+6\big)^2+9}{\big(x+12(k-1)+6\big)^2+9}
$$
It follows that the product telescopes nicely:
$$
\prod_{k=0}^{n-1} \frac{\big(x+12k+6\big)^2+9}{\big(x+12(k-1)+6\big)^2+9} \;=\; \frac{\big(x+12(n-1)+6\big)^2+9}{\big(x+12(0-1)+6\big)^2+9} \;=\; \frac{(x+12n-6)^2+9}{(x-6)^2+9}
$$
So in the end $\,\displaystyle f(7,5) = \frac{(7+5 \cdot 12 - 6)^2+9}{(7-6)^2+9} = \frac{61^2+9}{10}=\frac{3721+9}{10}=373\,$.
|
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|
Cubic equation with unknown coefficients given roots I was given this equation. $x^3 + 3px^2 + qx + r=0$. The roots are $1, -1$, and $3$.
Ive tried dividing the equation by $(x-1)$ to get a quadratic to make it easier for me. But that ended up really badly.
I also inputted the different roots into the equation to get different equations that I could solve. When I tried to prove my answer it turned out to be a flop.
And I also tried multiplying the three factors and comparing coefficients. It didnt seem right.
|
We can compare coefficients when we multiply all the roots in factor form:
$$\begin{align}(x-1)(x+1)(x-3) &= x^3-3x^2-x+3\\ &=x^3 + \underset{p}{3(-1)}x^2+\underset{q}{(-1)}x+\underset r{(3)}.\end{align}$$
We then check the roots:
1: $$(1)^3 - 3(1)^2 - 1 + 3 = 1-3-1+3 = 0$$
3: $$(3)^3 - 3(3)^2 - 3 + 3 = 3^3 - 3^3 + 0 = 0$$
I leave it to you to check $-1$.
|
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|
Find the minimum value of $xy+yz+xz$ Question:
Find the minimum value of $xy+yz+xz$, given that $x,y,z$ are real and $x^2+y^2+z^2=1$
My attempt,
Since $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$=1+2(xy+xz+yz)$
$(x+y+z)^2\geq0$
So that $1+2(xy+xz+yz)\geq0$
$(xy+xz+yz)\geq -\frac{1}{2}$
So the minimum value is $-\frac{1}{2}$.
My question is am I correct ? Because I don't have the solution. And is there another way to solve this? Thanks in advance.
|
You can also solve this problem using the method of Lagrange multipliers:
Note that $f(x,y,z) = xy + yz + xz$ is a $C^{\infty}$ function. We will use the method of Lagrange multipliers to determine the maximum/minimum of $f(x,y,z)$ given the restriction $g(x,y,z)= x^2 + y^2 + z^2 = 1$.
Therefore, because of Lagrange multipliers Theorem we know that if $x,y,z,\lambda \in R$ are solutions of the following system, then $(x,y,z)$ is either a max/min of $f(x,y,z)$ under the restriction $g(x,y,z) = 1 $.
\begin{equation}
\begin{cases}
\nabla f(x,y,z) = \lambda \nabla g(x,y,z) , \quad \lambda \neq 0 \Rightarrow \begin{cases} y + z = 2\lambda x \\
x + z = 2\lambda y \\
y+x = 2\lambda z
\end{cases}\\
g(x,y,z) = x^2 + y^2 + z^2 = 1
\end{cases}
\end{equation}
Solving the system we get the following values for $(x,y,z,\lambda)$:
\begin{equation}
(x,y,z,\lambda) =\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x), -\frac{1}{2}\right)
\end{equation}
\begin{equation}
(x,y,z,\lambda) = \left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x), -\frac{1}{2} \right)
\end{equation}
\begin{equation}
(x,y,z,\lambda) = \left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}} ,1 \right)
\end{equation}
\begin{equation}
(x,y,z,\lambda) = \left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}} ,1 \right)
\end{equation}
In order to know whether the points obtained solving the system are max or min, we look evaluate them:
\begin{equation}
f\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x)\right) = -\frac{1}{2}
\end{equation}
\begin{equation}
f\left(x, \frac{1}{2} (\sqrt{2 - 3 x^2} - x), \frac{1}{2} (-\sqrt{2 - 3 x^2} - x)\right) = -\frac{1}{2}
\end{equation}
\begin{equation}
f \left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right) = 1
\end{equation}
\begin{equation}
f \left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}}\right) = 1
\end{equation}
Therefore we have two minimums at $\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x)\right)$ and $\left(x, \frac{1}{2} (\sqrt{2 - 3 x^2} - x), \frac{1}{2} (-\sqrt{2 - 3 x^2} - x)\right)$ $\forall x\in R$, and a maximum at $\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right)$ and $\left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$.
Now, to prove that the minimum is indeed global you can perfectly use your steps and conclude that $f(x,y,z) \geq -\frac{1}{2}$ and so, the minimum is $-\frac{1}{2}$.
In fact, proving that $f(x,y,z) \geq -\frac{1}{2}$ and then finding some $x_0,y_0,z_0 \in \mathbb{R}$ for which $f(x_0,y_0,z_0) = -\frac{1}{2}$ is sufficient and necessary to state that $-\frac{1}{2}$ is the minimum. However, by using Lagrange multipliers you will certainly find those $x_0,y_0,z_0$ without "making them up".
|
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Proving $\int_{0}^{\pi/2}\left[2\cos\left({x\over 2}\right)-x\sin\left({x\over 2}\right)\right]\ln\left[2\cos^2\left({x\over 2}\right)\right] dx...$ Proposed:
$$\int_{0}^{\pi/2}\left[2\cos\left({x\over 2}\right)-x\sin\left({x\over 2}\right)\right]\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx=\color{green}{(4-\pi)\sqrt{2}}\tag1$$
My try:
Rewrite as
$$\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx-
\int_{0}^{\pi/2}x\sin\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx=I_1-I_2\tag2$$
$u=2\cos(x/2)$, then $I_1$ becomes
$$I_1=\int_{\sqrt{2}}^{2}{u\over \sqrt{1-u^2}}\ln\left({{1\over 2}u^2}\right)\mathrm du\tag3$$
$$I_1=-\ln(2)\int_{\sqrt{2}}^{2}{u\over \sqrt{1-u^2}}\mathrm du-\int_{\sqrt{2}}^{2}{u\ln(u)\over \sqrt{1-u^2}}\mathrm du\tag4$$
I am sure $(4)$ is from standard integral table but $I_2$ seems difficult to transform
How may one prove $(1)$?
|
Hint. One may write
$$
\begin{align}
&\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx
\\\\&=\int_{0}^{\pi/2}2\cos\left({x\over 2}\right)\ln\left[2\left(1-\sin^2\left({x\over 2}\right)\right)\right]\mathrm dx
\\\\&=4\int_{0}^{\sqrt{2}/2}\ln\left[2\left(1-u^2\right)\right]\mathrm du
\\\\&=4\ln2\int_{0}^{\sqrt{2}/2}\mathrm du+\int_{0}^{\sqrt{2}/2}\ln\left(1+u\right)\:\mathrm du+4\int_{0}^{\sqrt{2}/2}\ln\left(1-u\right)\:\mathrm du
\\\\&= 8 \ln \left(1+\sqrt{2}\right)-4 \sqrt{2}.
\end{align}
$$
Similarly, integrating by parts,
$$
\begin{align}
&\int_{0}^{\pi/2}x\sin\left({x\over 2}\right)\ln\left[2\cos^2\left({x\over 2}\right)\right]\mathrm dx
\\\\&=\left[x \cdot \left(4 \cos \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\right)\right]_{0}^{\pi/2}-\int_{0}^{\pi/2}\left[4 \cos \left(\frac{x}{2}\right)-2 \cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\right]\mathrm dx
\\\\&=\sqrt{2} \:\pi-\int_{0}^{\pi/2}4 \cos \left(\frac{x}{2}\right)\mathrm dx
+2\int_{0}^{\pi/2}\cos \left(\frac{x}{2}\right) \log \left(2 \cos ^2\left(\frac{x}{2}\right)\right)\mathrm dx
\\\\&=\sqrt{2} \:\pi-8\sqrt{2}+8 \ln\left(1+\sqrt{2}\right).
\end{align}
$$ Considering the preceding results leads to the announced equality.
|
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Generating functions ( recurrence relations ) Find $a_n$ using Generating Functions : $a_n = -a_{n-1} + 2a_{n−2}$, $n\ge2$ and $a_0 = 1$, $a_1 = 2$.
Approach : So I will form a characteristic equation $ r^2 + r - 2 = 0$ whose roots are $r_1 = -2$, $r_2 = 1$.
So my general solution is $a_n = α_1r_1^n + α_2r_2^n$.
$a_n = α_1(-2)^n + α_2(1)^n$
When $a_0 = 1$, then $1 = α_1(-2)^0 + α_2(1)^0$, then $α_2 = 1 - α_1 $.
When $a_1 = 2$, then $2 = α_1(-2)^1 + α_2(1)^1$, then $-2α_1 + 1 - α_1 = 2$.
$α_1 = -1/3$ and $α_2 = 4/3 $
So $a_n = -1/3r_1^n + 4/3r_2^n$.
Can anyone tell me if it is correct or not and any help will be appreciated :) .
Also, if I have $a_{n+2}=a_{n+1}+2a_n$. Can someone tell me if its char. equation should be like $r^2-r-2 = 0$? Just asking because of the addition symbol rather than subtraction.
|
Define the generating function:
$\begin{equation*}
A(z)
= \sum_{k \ge 0} a_k z^k
\end{equation*}$
Write the recurrence with no subtraction in indices, multiply by $z^n$, sum over $n \ge 0$:
$\begin{equation*}
\sum_{n \ge 0} a_{n + 2} z^n
= - \sum_{n \ge 0} a_{n + 1} z^n
+ 2 \sum_{n \ge 0} a_n z^n
\end{equation*}$
Recognize the sums at hand:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1} z^n
&= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n
&= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align*}$
Using the initial values given:
$\begin{equation*}
\frac{A(z) - 1 - 2 z}{z^2}
= - \frac{A(z) - 1}{z} + 2 A(z)
\end{equation*}$
A bit of algebra gives, as partial fractions:
$\begin{align*}
A(z)
&= \frac{1 + 3 z}{1 - z - 2 z^2} \\
&= \frac{5}{3} \cdot \frac{1}{1- 2 z}
- \frac{2}{3} \cdot \frac{1}{1 + z}
\end{align*}$
The terms are geometric series, can read off the coefficients:
$\begin{equation*}
a_n
= \frac{5}{3} \cdot 2^n - \frac{2}{3} \cdot (-1)^n
\end{equation*}$
|
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|
Simplify algebraic expression with radicals: $\frac{1- ax}{1+ ax} \cdot \sqrt\frac{1+bx}{1-bx}$ I got stuck trying to simplify this roots forest:
$\frac{1- ax}{1+ ax}*\sqrt\frac{1+bx}{1-bx}$
where x= $\sqrt{\frac{2a}{b}-1}$
So it is:
$\frac{1- a\sqrt{\frac{2a}{b}-1}}{1+ a\sqrt{\frac{2a}{b}-1}}*\sqrt\frac{1+b\sqrt{\frac{2a}{b}-1}}{1-b\sqrt{\frac{2a}{b}-1}}$
|
HINT: we get
$$1-a\frac{\sqrt{2a-b}}{\sqrt{b}}=\frac{\sqrt{b}-a\sqrt{2a-b}}{\sqrt{b}}$$ for $b>0$
I have got the following result
$$\frac{\sqrt{b}-a\sqrt{2a-b}}{\sqrt{b}+a\sqrt{2a-b}}\cdot \frac{\sqrt{1+\sqrt{b}\sqrt{2a-b}}}{\sqrt{1-\sqrt{b}\sqrt{2a-b}}}$$
if $$2a-b\geq 0$$
|
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|
Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using 3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$
Is everything right here?
But how to check it for convergence?
|
The identity of a series is often ambiguous when only a few example terms are given without a general rule for generating the rest.
I assume that the general rule in this case is that the magnitude of each term is $\newcommand{r}{{\sqrt{10}}} 1/\r$ times the magnitude of the previous term, while the signs repeat every three terms: one positive term followed by two negative terms.
If you try to split the series after the first into two subseries by taking alternating terms (terms $2,4,6,\ldots$ in one subseries and $3,5,7,\ldots$ in the other), you get this:
\begin{multline}
1 - \frac1\r \left(1 - \frac1{10} + \frac1{10^2}
+ \frac1{10^3} - \frac1{10^4} + \frac1{10^5}
+ \frac1{10^6} - \frac1{10^7} + \frac1{10^8} +-\cdots \right)\\
- \frac1{10}\left(1 + \frac1{10} - \frac1{10^2}
+ \frac1{10^3} + \frac1{10^4} - \frac1{10^5}
+ \frac1{10^6} + \frac1{10^7} - \frac1{10^8} +-\cdots \right)\\
\end{multline}
Although the magnitudes of each subseries form a geometric series with ratio $1/10,$ the subseries are not geometric with ratio $1/10,$ because the signs are not all the same.
Just as in the original series, the signs change in patterns that repeat every three terms: $+,-,+$ for the first subseries, $+,+,-$ for the second.
So the method proposed in the question is not correct.
In a geometric series with a positive ratio, all terms have the same sign. In a geometric series with a negative ratio, the terms have alternating signs, for example $+,-,+,-,+,-,\ldots.$
There is no ratio that will produce both positive an negative signs with a pattern that repeats every three terms.
There are various ways to deal with this by rearranging the series,
just not in exactly the way attempted in the question.
Of course, before we start rearranging the series we should make sure that the rearrangement would not affect the convergence or sum of the series. According to the ratio test, the series is absolutely convergent,
and therefore it is OK to split the series into subseries or
to rearrange the terms in other ways.
One rearrangement that works is to separate the series into three subseries as shown in another answer.
Another possible rearrangement is to combine the terms in groups of three:
$$
\left(1-\frac1\r-\frac1{10}\right)
+ \left(\frac1{10\r}-\frac1{10^2}-\frac1{10^2\r}\right)
+ \left(\frac1{10^3}-\frac1{10^3\r}-\frac1{10^4}\right) + \cdots.
$$
This is a geometric series with ratio $\frac1{10\r}$
and first term $\left(1-\frac1\r-\frac1{10}\right),$
which you can evaluate using the usual methods for a geometric series.
|
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|
How to calculate this surface area? (portion of a cylinder inside a sphere ) The surface area of the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$
I'm stuck, so any tip will be helpful
Thanks in advance!
|
$A = \iint dS$
$S: x^2 + y^2 = 8y$
Convert to cylindrical.
$x = r\cos\theta\\
y = r\sin\theta\\
z = z$
Plug these into the equation of the cylinder.
$r = 8\sin\theta$
And substitute back for parameterization of the surface
$x = 8\sin\theta\cos\theta = 4\sin 2\theta\\
y = 8\sin^2\theta = 4 - 4\cos 2\theta\\
z = z$
$dS = $$\|(\frac {\partial x}{\partial \theta}, \frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta})\times (\frac {\partial x}{\partial z}, \frac {\partial y}{\partial z},\frac {\partial z}{\partial z})\|\\
\|(8\cos2\theta, 8\sin 2\theta, 0) \times (0,0,1)\| = \|(8\sin2\theta, -8\cos 2\theta, 0)\| = 8\ dz\ d\theta$
$\iint 8\ dz\ d\theta$
The sphere will establish the limits for z.
$16\sin^2 2\theta + 16 -32\cos 2\theta + 16\cos^2 2\theta + z^2 = 64\\
z^2 = 32 + 32\cos 2\theta = 64\cos^2\theta$
$2\int_0^{\pi}\int_0^{8\cos\theta} 8 \ dz\ d\theta$
|
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|
Combinatorics: How many distinct binary strings of length 16 are possible if the zeroes must appear in groups of even number? I am trying to solve two combinatorics problems. May you help me?
First problem,
How many distinct binary strings of length 16 are possible if the zeroes must appear in groups of even number?
I think that the key is to see that there can be even number of zeroes or odd number of zeroes. The total number of possible lists is 2^16. Half of them has even number of zeroes. Thus the answer is 2^15.
Second problem,
What if the zeroes need to be in pairs?
Then lists that start with 01 are not allowed. Also strings that have 101, or 10001, or 1000001, etc. are not allowed. I have been thinking that if the number of zeroes need to be in pairs... The number of ones need to be in pairs, as well!
Then we just need to use the same strategy. We divide the original 16 in squares of size 2. Then we have 8 squares. The total number of possible lists is 2^8. All are even, of course, because 0 is even too.
Is my reasoning correct? Thank you for everything!
|
In order to count the number of strings of length $16$ having even runs of $0$s resp. pairs of $0$s we consider words
with no consecutive equal characters at all.
These words are called Smirnov words or Carlitz words. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}\tag{1}
\end{align*}
First problem: Even runs of $0$s
We replace occurrences of $0$ in a Smirnov word by an even number of zeros. This corresponds to a substitution of
\begin{align*}
z&\longrightarrow z^2+z^4+z^6+\cdots=\frac{z^2}{1-z^2}\tag{2}\\
\end{align*}
and we replace occurrences of $1$ in a Smirnov word by any run of $1$'s with length $\geq 1$.
\begin{align*}
z&\longrightarrow z+z^2+z^3+\cdots=\frac{z}{1-z}\tag{3}\\
\end{align*}
We obtain by substituting (2) and (3) in (1) a generating function A(z)
\begin{align*}
A(z)&=\left(1-\frac{\frac{z^2}{1-z^2}}{1+\frac{z^2}{1-z^2}}-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}\right)^{-1}\\
&=\frac{1}{1-z-z^2}\\
&=1+z+2z^2+3z^3+5z^4+8z^5+13z^6\cdots+\color{blue}{1597}z^{16}+\cdots
\end{align*}
with $A(z)$ a generating function of the Fibonacci numbers.
Second problem: All runs of $0$ having length $2$
We can do a similar approach as above. Here we use the substitution
\begin{align*}
z&\longrightarrow z^2\tag{4}\\
z&\longrightarrow z+z^2+z^3+\cdots=\frac{z}{1-z}\tag{5}\\
\end{align*}
We obtain by substituting (4) and (5) in (1) a generating function B(z)
\begin{align*}
B(z)&=\left(1-\frac{z^2}{1+z^2}-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}\right)^{-1}\\
&=\frac{1+z^2}{1-z-z^3}\tag{6}\\
&=1+z+2z^2+3z^3+4z^4+6z^5+9z^6\cdots+\color{blue}{406}z^{16}+\cdots
\end{align*}
The series expansion of $A(z)$ and $B(z)$ was done with the help of Wolfram Alpha.
We can derive a recurrence relation from the generating function easily and we obtain from (6) (similarly as it is shown in @MarkoRiedel's answer)
\begin{align*}
[z^n]B(z)(1-z-z^3)=[z^n](1+z^2)=0
\end{align*}
and conclude
\begin{align*}
g_n-g_{n-1}-g_{n-3}&=0\\
g_0&=g_1=1\\
g_2&=2
\end{align*}
Finally we look at an example and consider the number of valid words of length $6$.
We obtain nine solutions
$$\begin{array}{cccccc}
1&1&1&1&1&1\\
\color{blue}{0}&\color{blue}{0}&1&1&1&1\\
1&\color{blue}{0}&\color{blue}{0}&1&1&1\\
1&1&\color{blue}{0}&\color{blue}{0}&1&1\\
1&1&1&\color{blue}{0}&\color{blue}{0}&1\\
1&1&1&1&\color{blue}{0}&\color{blue}{0}\\
\color{blue}{0}&\color{blue}{0}&1&\color{blue}{0}&\color{blue}{0}&1\\
\color{blue}{0}&\color{blue}{0}&1&1&\color{blue}{0}&\color{blue}{0}\\
1&\color{blue}{0}&\color{blue}{0}&1&\color{blue}{0}&\color{blue}{0}\\
\end{array}
$$
in accordance with $[z^6]B(z)=9$.
|
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|
Solving $|x^2-2x|+|x-4|>|x^2-3x+4|$
How do I solve $|x^2-2x|+|x-4|>|x^2-3x+4|?$
I can see the difference of the two terms on the left gives the term on the right. Now,what should I do? Is there any general method for solving $$|a|+|b|>|a-b|?$$
Thanks for any help!!
|
Note that $|x^2-2x|+|x-4|=|x^2-2x|+|-x+4|\ge|x^2-3x+4|$ with the equality holds if and only if $x^2-2x$ and $-x+4$ are of the same sign.
Therefore, $|x^2-2x|+|x-4|>|x^2-3x+4|$ if and only if $x^2-2x$ and $x-4$ are of the same sign. So we have
\begin{align*}
(x^2-2x)(x-4)&>0\\
x(x-2)(x-4)&>0\\
0<x<2 \quad\textrm{or}\quad x&>4
\end{align*}
|
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|
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$$
Note that:
$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$$
Now we have $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$, i.e.$\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ and $\tan h \sim_{0}h.$
I'm stuck.
Update: here is another way :
\begin{aligned}
\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\
&=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
\end{aligned}
Thus
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left(\dfrac{\pi x}{2a}\right)} =e^{\dfrac{2}{\pi}} $$
*
*Am i right beside i'm intersted in way which use equivalents
|
We can also apply L'Hopitals directly to $\log\left(2-\frac{x}{a}\right)\tan \frac{\pi x}{2a}$ and then appeal to continuity of $\exp$. In particular we have $$\lim_{x\to a} \frac{\sin \left(\frac{\pi x}{2a}\right) \log\left(2 - \frac{x}{a}\right) }{\cos \left(\frac{\pi x}{2a}\right)} =\lim_{x\rightarrow a} \frac{\frac{\pi}{2a}\cos(\frac{\pi x}{2a})\log(2-\frac{x}{a})-\frac{1}{a}\left(2-\frac{x}{a}\right)^{-1}\sin(\frac{\pi x}{2a})}{-\frac{\pi}{2a}\sin(\frac{\pi x}{2a})} = \frac{-1/a}{-\pi/2a} = \frac{2}{\pi}.$$
Hence the required limit is $\exp(2/ \pi).$
|
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|
Show that $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ How do I show that: $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ using contours and residues
My attempt:
I know that the singular points are $i,-i,-1,1,0$
consider $f(z)= \frac{\ln z}{(z^2+1)(z^2-1)}$
and the branch $|z|>0$, $0<\theta<2\pi$
$u: z=r, \rho\le r \le R$ (u is the upper edge)
$-l: z=r, \rho\le r \le R$ (lower edge)
$\int_ufdz-\int_{-l}fdz=\int_\rho^R \frac{\ln r + i0}{(z^2+1)(z^2-1)}-\int_\rho^R \frac{\ln r + i2\pi}{(z^2+1)(z^2-1)}$
How do I continue from here?
|
Observe that
\begin{eqnarray*}
\frac{1}{(x^2+1)(x^2-1)}=\frac{1}{2(x^2-1)} + \frac{-1}{2(x^2+1)}
\end{eqnarray*}
So the original integral splits into the following two integrals
\begin{eqnarray*}
I_1=\int_{0}^{\infty} \frac{ln(x)}{2(x^2+1)} \\
I_2=\int_{0}^{\infty} \frac{ln(x)}{2(x^2-1)} .\\
\end{eqnarray*}
Both of these integrals can be evaluated by spliting the inteval into $[0,1]$ and $[1, \infty)$ and then doing the substitution $x=1/y$. The first integral is zero and the second integral will be
\begin{eqnarray*}
I_1=0 \\
I_2=\int_{0}^{1} \frac{ln(x)}{(x^2-1)} dx \\
\end{eqnarray*}
Now geometrically expand $\frac{1}{(x^2-1)}$ and use
\begin{eqnarray*}
\int_{0}^{1} x^{n} \ln(x) dx =\frac{-1}{(n+1)^2} \\
I_2=-\int_{0}^{1} \sum_{n=0}^{\infty} x^{2n} ln(x) dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\
\end{eqnarray*}
It well known that $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\color{red}{\frac{\pi^2}{8}}$.
|
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|
$\alpha + \beta + \gamma = \pi$ , show that $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$
I really didn't know how to solve this problem and I am very unused to the utilization of trigonometric identities, I was wondering if I may have some assistance in this problem with detailed explanations
I was thinking more along the lines of making all of the angles in terms of one angle
|
I think you mean the following problem.
Let $\alpha+\beta+\gamma=\pi$. Prove that:
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1$$
We need to prove that
$$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta\cos(\alpha+\beta)=\sin^2(\alpha+\beta)$$ or
$$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)^2$$ or
$$\cos^2\alpha+\cos^2\beta-2\cos^2\alpha\cos^2\beta=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta$$ or
$$\cos^2\alpha(1-\sin^2\beta)+\cos^2\beta(1-\sin^2\alpha)-2\cos^2\alpha\cos^2\beta=0,$$
which is obvious.
Done!
|
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|
From $a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$ to $a_n$ to be proven by induction
Find and Prove by induction an explicit formula for $a_n$ if $a_1=1$ and for $n \geq 1$
$$a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$$
Checking the pattern:
$$a_1=1$$
$$a_2= 2 \cdot 1$$
$$a_3= (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_4= (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_5= (\frac{5}{4})^4 \cdot (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
I can see the general pattern in the index as $(...)^{(n-1)!}$
and I am tempted to state that it is $( \frac{n!}{(n-1)!})^{(n-1)!}
$
Should this be correct, I do not see the factorial manipulation that allow me to write this.
Otherwise, what would the formula be? What would the approach be?
Much appreciated
|
$a_5= (\frac{5}{4})^4 \cdot (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1
$
$\begin{array}\\
a_n
&=\prod_{k=2}^n (\frac{k}{k-1})^{k-1}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=2}^n (k-1)^{k-1}}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} k^k}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} kk^{k-1}}\\
&=\dfrac{\prod_{k=2}^n k^{k-1}}{\prod_{k=1}^{n-1} k\prod_{k=1}^{n-1} k^{k-1}}\\
&=\dfrac{n^{n-1}\prod_{k=2}^{n-1} k^{k-1}}{\prod_{k=1}^{n-1} k\prod_{k=2}^{n-1} k^{k-1}}\\
&=\dfrac{n^{n-1}}{(n-1)!}\\
&=\dfrac{n^{n}}{n!}\\
\end{array}
$
|
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|
Prove that the following are real numbers
*
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z)}{|z|}$$ which is real number as a division of two real numbers
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
$$\overline{z^3\cdot\overline{z}+z\cdot\overline{z}^3}$$
Due to: $\overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$ and $\overline{z_{1}\cdot z_{2}}=\overline{z_{1}}\cdot\overline{z_{2}}$
we get:
$$\overline{z}^3\cdot{z}+\overline{z}\cdot{z}^3$$
Now we can say that we got the same elements and $(z=\overline{z})$ therefore it is real number?
Is the reasoning in both of these correct?
|
If we know $z + \overline z \in \mathbb R$ and $z*\overline z = |z|^2\in \mathbb R$ and $\overline{\overline z} = z$ and $\overline {z*w} = \overline z*\overline w$ we are pretty much done.
1) $\frac 1{\overline z} *\overline z = 1$
$\overline{\frac 1{\overline z}*\overline z} = \overline 1 = 1$
$\overline{\frac 1{\overline z}}*\overline{\overline z} = \overline{\frac 1{\overline z}}*z = 1$
$\overline{\frac 1{\overline z}} = \frac 1z$ so
$\frac {1}{\overline z} = \overline {\frac 1z}$ so
$\frac 1z + \frac 1{\overline z} = \frac 1z + \overline {\frac 1z} \in \mathbb R$.
2) $z^3*\overline z +z*\overline{z^3}= \overline{\overline {z^3}}*\overline z +\overline{z^3}*z= \overline{\overline {z^3}*z}+\overline{z^3}*z\in \mathbb R$
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|
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
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\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sum_{k = 1}^{n}k\pars{k + 1}\pars{k + 2}\pars{k + 3}
&= \sum_{k = 0}^{n}\pars{k + 3}^{\underline{4}} \\
&= \left.{\pars{k + 3}^{\underline{5}} \over 5}\right\vert_{\ 0}^{\ n + 1} \\
&= {{\pars{n + 4}^{\underline{5}} \over 5} - {3^{\underline{5}} \over 5}}
\\[5mm] &=
\bbox[#ffe,20px,border:1px dotted navy]{\pars{n + 4}\pars{n + 3}\pars{n + 2}\pars{n + 1}n \over 5} \\ &
\end{align}
Note that $\ds{3^{\underline{5}} = 3 \times 2 \times 1 \times 0 \times \pars{-1} = {\large 0}}$.
Reference: See $\ds{\mathbf{2.50}\ \pars{~\mbox{page}\ 50~}}$ in Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik, $\ds{\mrm{6^{th}}}$ ed. Addison-Wesley $\ds{1990}$.
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|
Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and I don't know what am I doing wrong. What is the correct way to prove this equality?
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From the last line of your working, and using $\sin^2x+\cos^2x=1,$
$$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}=\frac{\cos x}{\cos^2x-\sin^2x}=\frac{\cos x}{\cos^2x-(1-\cos^2x)}=\frac{\cos x}{2\cos^2x-1}$$
|
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|
Characteristic curves : Why do I get different results? With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$
I have solved it using two ways.
$$$$
First way:
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$.
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}$$
Therefore, we have $\frac{dy}{ds}=1$, $\frac{dx}{ds}=-\frac{y}{x}$ and $\frac{du}{ds}=\frac{u}{x}$.
From $\displaystyle{\frac{dy}{ds}=1}$ we get $\displaystyle{y=s}$.
From $\displaystyle{\frac{dx}{ds}=-\frac{y}{x}}$ with $x(0)=x_0$ we get $\displaystyle{\frac{dx}{dy}=-\frac{y}{x} \Rightarrow x dx=-y dy \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{-y^2+x_0^2}}$.
From $\displaystyle{\frac{du}{ds}=\frac{u}{x}}$ we get \begin{align*}&\frac{du}{dy}=\frac{u}{\pm \sqrt{-y^2+x_0^2}} \Rightarrow \frac{1}{u}du=\frac{1}{\pm \sqrt{x_0^2-y^2}}dy\Rightarrow \frac{1}{u}du=\frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \\ & \Rightarrow \int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy\end{align*}
We set $\displaystyle{\frac{y}{x_0}=\cos k \Rightarrow \frac{1}{x_0}dy=-\sin k dk \Rightarrow dy=-x_0\sin k \ dk}$.
Therefore we get:
\begin{align*}\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy&=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{1-\cos^2k}}dk=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{\sin^2 k}}dk=\int \frac{-x_0\sin k}{\pm |x_0| |\sin k|}dk \\ & =\mp \int \frac{x_0\sin k}{|x_0 \sin k|}dk=\mp \int \pm 1 dk=k+c\end{align*}
So, we get \begin{align*}&\int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \Rightarrow \ln |u|=k+c \Rightarrow e^{\ln |u|}=e^{k+c} \Rightarrow |u|=e^ce^k \Rightarrow u=\pm e^ce^k \Rightarrow u=Ce^k \\ & \Rightarrow u=Ce^{\arccos \left (\frac{y}{x_0}\right )}\end{align*}
For $y=0$ we have $\displaystyle{u=h(x_0) \Rightarrow Ce^{\arccos \left (\frac{y}{x_0}\right )}=h(x_0) \Rightarrow C=\frac{2h(x_0)}{\pi}}$.
Therefore, the solution is $$u=\frac{2h(x_0)}{\pi}e^{\arccos \left (\frac{y}{x_0}\right )} \Rightarrow u=\frac{2h(\pm \sqrt{x^2+y^2
})}{\pi}e^{\arccos \left (\frac{y}{\pm \sqrt{x^2+y^2
}}\right )}$$
$$$$
Second way:
From $\displaystyle{xu_y-yu_x=u}$ and $\displaystyle{\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}}$ we get : $\displaystyle{\frac{dx}{ds}=-y, \ \frac{dy}{ds}=x, \ \text{ and } \ \frac{du}{ds}=u}$.
From $\displaystyle{\frac{dx}{ds}=-y}$ we get that $\displaystyle{ds=-\frac{1}{y}dx}$ and from $\displaystyle{\frac{dy}{ds}=x}$ we get $\displaystyle{ds=\frac{1}{x}dy}$.
Therefore we get (using that $x(0)=x_0$) $\displaystyle{-\frac{1}{y}dx=\frac{1}{x}dy \Rightarrow -xdx=ydy \Rightarrow -\frac{x^2}{2}=\frac{y^2}{2}-\frac{x_0^2}{2} \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{x_0^2-y^2}}$.
From $\displaystyle{\frac{du}{ds}=u}$ using that $ds=\frac{1}{x}dy$ we get that \begin{align*}\frac{xdu}{dy}&=u \Rightarrow \frac{1}{u}du=\frac{1}{x}dy \Rightarrow \frac{1}{u}du=\frac{1}{\left (\pm \sqrt{x_0^2-y^2}\right )}dy \Rightarrow \ln |u|=\pm \arctan \left (\frac{y}{x_0^2-y^2}\right )+C \\ & \Rightarrow u=\pm e^{\pm C}e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )}\end{align*}
Using the condition for $y=0$ then $\displaystyle{u=h(x_0)}$ we get that $\displaystyle{h(x_0)=\pm e^{\pm C}}$.
Therefore the solution is $$u=h(x_0)e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )} \Rightarrow u=h(\pm \sqrt{x^2+y^2})e^{\arctan \left (\frac{y}{x^2}\right )}$$
Why do I get different results? Have I done something wrong?
|
(The two solutions are the same. The notation tends to obscure the identity of the solutions. You have $h(\pm \sqrt{x^2+y^2})$ and $\dfrac{2h(\pm \sqrt{x^2+y^2})}{\pi}$. Nevertheless, they are not the same function. We need to make the following stipulation: $\dfrac{2h(\pm \sqrt{x^2+y^2})}{\pi}=g(\pm \sqrt{x^2+y^2})$, that it's legitimate as both are functions of the same argument, $\pm \sqrt{x^2+y^2}$) True, but irrelevant now.
Further, $\arctan \left (\dfrac{y}{x^2}\right )$ is not correct, it has to be $\arctan \left (\dfrac{y}{x}\right )$ as the integral of $1/\sqrt{x_0^2-y^2}$ is $\arctan\left(y/\sqrt{x_0^2-y^2}\right)$
And finally,
Corrected (added the $\pm\pi/2$ phase)
$\arctan \left (\pm\dfrac{y}{x}\right )=\arccos \left (\dfrac{y}{\pm \sqrt{x^2+y^2}}\right )\pm\pi/2$ being careful with the signs in order to the identity to hold.
Postcriptum: I don't detelete the answer as it's neat now and I prefer it here than in the never empty trash bin.
|
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|
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmatrix} a^2+bc &b(a+d) \\ c(a+d)&d^2+bc \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
So
$$a^2+bc=1 \tag{1}$$
$$b(a+d)=2 \tag{2}$$
$$c(a+d)=3 \tag{3}$$
$$d^2+bc=4 \tag{4}$$ From $(2)$ and $(3)$ we get
$$\frac{b}{c}=\frac{2}{3}$$
Let $b=2k $ and $c=3k$ Then
$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$ So
$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\ 3k&\sqrt{4-6k^2} \end{bmatrix}$$
So
$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\ 3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4 \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
Now we have to solve for $k$ using equation
$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also
$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$
$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$
Subtracting $(6)$ from $(5)$ we get
$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$ Squaring Both sides we get
$$33k^4-10k^2+1=0$$ we get
$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?
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@ Umesh shankar, this is exactly what should not be done (if you want to make progress)!!
Note that $A^2=trace(A)A-\det(A)I=5A+2I$. Since $B^2=A$ we deduce that $A,B$ commute and since $A$ is not a scalar matrix, $B$ is in the form $B=aI+bA$. Then $(aI+BA)^2=A$, that is $a^2I+b^2(5A+2I)+2abA=A$, that implies
$$a^2+2b^2=0\;,\;5b^2+2ab=1.$$
We deduce $33b^4-10b^2+1=0$,....... and, finally, the four solutions.
EDIT.
Recall that the determinant and the trace of a matrix do not change after a change of basis.
Proposition 1. If $A\in M_2$ is not a scalar matrix, then $I,A$ is a basis of the vector space $E=\{X|AX=XA\}$.
Proof. With a change of basis, we may assume that $A=\begin{pmatrix}0&a\\1&b\end{pmatrix}$ (there is $u$ s.t. $u,Au$ is a basis). Let $X=\begin{pmatrix}x&y\\z&t\end{pmatrix}$. Then $AX=XA$ is equivalent to $y=az,x+bz=t$, that is $2$ linear independent linear relations; thus $dim(E)=4-2=2$.
Proposition 2. If $A\in M_2$ is not a scalar matrix, then $A^2=trace(A)A-\det(A)I$.
Proof. As above, we may assume that $A=\begin{pmatrix}0&a\\1&b\end{pmatrix}$.
$A^2=\begin{pmatrix}a&ab\\b&a+b^2\end{pmatrix}=bA+aI=trace(A)A-\det(A)I$.
|
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|
Express the equation as a single fraction in its simplest form Express the equation
$$ 4 - \left( (x + 3) ÷ {\frac {x² + 5x + 6}{x - 2}} \right) $$
as a single fraction in its simplest form.
So far all I managed to do was factorise the numerator to get this $$(x + 3)(x + 2)$$
Where do I go from here?
|
Well here is an answer for if the expression is as I have written it in my comment above. If my comment indeed has the wrong expression, then the steps are similar, but involves a closer look using the proper order of operations.
We first use the fact that $\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$ where $b,c,d\neq 0$. This yields
$$\begin{align}4-\left((x+3)\div\frac{x^2+5x+6}{x-2}\right)&=4-\left((x+3)\times\frac{x-2}{x^2+5x+6}\right)\\&=4-\left(\frac{(x+3)(x-2)}{x^2+5x+6}\right)\end{align}$$
Now use your factorization $x^2+5x+6=(x+2)(x+3)$ and do the appropriate cancellations. Then I'll leave the rest for you to deal with.
Recall: $4-\frac{a}{b}=\frac{4b-a}{b}$ and $\frac{ab}{bc}=\frac{a}{c}$ if $b,c\neq 0$.
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Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers. Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers.
My attempt:
For $n=1$, trivially there are integer solutions.
For $n=2$, we have the solutions
$$a=b=\pm 1, c=0.$$
For $n=3$, we have the solution
$$a=b=1, c = -1.$$
Anyone can complete the proof for $n=4,5,6$?
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$n=4$:
fourth powers are $\equiv 0$ or $1\pmod{8}$. There is no way to have $(0\text{ or }1)+(0\text{ or }1)\equiv (0\text{ or }1)+4\pmod 8$.
$n=5$:
Fifth powers are $\equiv 0$ or $1$ or $-1\pmod{11}$. Three such numbers combined cannot produce $5$, i.e., $a^5+b^5-c^5\equiv -3,-2,-1,0,1,2,3\not\equiv 5\pmod {11}$.
$n=6$:
Sixth powers are $\equiv 0$ or $1$ or $-1\pmod{13}$. Three such numbers combined cannot produce $6$, i.e., $a^6+b^6-c^6\equiv -3,-2,-1,0,1,2,3\not\equiv 6\pmod {13}$.
|
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Solve $\sin(5\theta)=1$, $0<\theta<2\pi$. Show that the roots of $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, $r=0,2,3,4$. This is a very interesting problem that I came across. I know it's got something to do with trigonometry identities, polynomials and complex numbers, but other than that, I'm not too sure how to approach this. Any guidance hints or help would be greatly appreciated. Thanks in advance. Here's the problem:
Solve the equation $\sin(5\theta)=1$ for $0<\theta<2\pi$ and hence show that the roots of the equation $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, where for $r=0,2,3,4$. Determine the exact value of $\sin\frac\pi{10}\sin\frac{3\pi}{10}$.
It might be useful to find $\sin(5\theta)$ in terms of $sin^n \theta$ and so on.
|
Think of the equation
$1=\sin 5\theta$
$=\sin3\theta cos2\theta+\cos3\theta sin2\theta$
$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(4\cos^3\theta-3\cos\theta) 2\sin\theta \cos\theta$
$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(4\cos^2\theta-3) 2\sin\theta (1-\sin^2\theta)$
$=(3\sin\theta-4\sin^3\theta)(1-2\sin^2\theta)+(1-4\sin^2\theta) 2\sin\theta (1-\sin^2\theta)$
Can you take it from here?
What equation do you get when you cancel out the one from both sides?
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|
How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point?
used partial fraction:
$$\frac{n^2}{n^3+1}=\frac{1}{3}\cdot\frac{2n-1}{n^2-n+1}+\frac{1}{3}\cdot \frac{1}{n+1}$$
I'm stuck with the first term in the partial fraction , the second term simply yields $1-\log(2)$
|
From partial fractions,
$$ \frac{n^2}{n^3 + 1} = \frac{1}{3(n+1)} + \frac{1}{3(n-r)} + \frac{1}{3(n-\overline{r})} $$
where $r$ is a root of $z^2 - z + 1$.
Now $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n +1} = 1 - \log(2) $$
while
$$ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n - r} = \frac{1}{2} \Psi\left(1 - \frac{r}{2}\right) - \frac{1}{2} \Psi\left(\frac{1}{2} -\frac{r}{2}\right) $$
Thus your sum becomes
$$\frac{1-\log(2)}{3} + \frac{1}{6}\left(\Psi \left(\frac{3+i \sqrt{3}}{4}\right)+\Psi \left(\frac{3-i \sqrt{3}}{4}\right) - \Psi \left(\frac{1+i \sqrt{3}}{4}\right) - \Psi \left(\frac{1-i \sqrt{3}}{4}\right)\right)
$$
Now using the identity $\Psi(1-x) = \Psi(x)+\pi \cot(\pi x)$, the sum becomes
$$ \frac{1-\log(2)}{3} - \frac{\pi}{6} \cot\left(\frac{\pi}{4} (3 + i \sqrt{3})\right) - \frac{\pi}{6} \cot\left(\frac{\pi}{4} (3 - i \sqrt{3})\right) $$
and this simplifies to Dr. Graubner's answer.
|
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|
Number Theory - Modular arithmetic with complex numbers So I was looking at a problem that asked me to find an $x$ s.t $x^3 = -1 \pmod{199}$ given that $14^2 = -3 \pmod{199}$.
To do this I was given the hint to look at complex numbers so I did the following
$$2(-1)^{\frac{1}{3}} - 1 = \sqrt{-3} $$
so then I can say that
$$(-1)^{\frac{1}{3}} = (14 + 199 + 1)/2 = 107$$
which is indeed a solution to the problem.
I am looking for an explanation as to why I can do this and any additional information as to how to apply this trick to similar problems.
|
$x^3 + 1 = (x + 1)(x^2 - x + 1)$ and note that $199$ is a Prime, so
$x + 1 = 0 \ \pmod{199}$ or $x^2 - x + 1 = 0 \ \pmod{199}$.
The hint $14^2 = -3 \ \pmod{199}$ to find the root of $x^2 - x + 1$ since:
$x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} = 0 \ \pmod{199}$
then
$(2x - 1)^2 = -3 \ \pmod{199}$
p.s.
there are $3$ roots : $\{-1, 107, -106\}$
|
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|
Solving recurrence relations of type $S_n-7S_{n-1}+10S_{n-2}=5\cdot 3^n$ I know how to solve this kind of equasions
$$S_n-7S_{n-1}+10S_{n-2}=5\cdot 3^n$$ $$S_0=0, S_1=1$$ for example...but when there is a constant (example:$(3^n+5))$ or $(5\cdot 3^n)$ i don't know how to solve it. Any tips.
|
Since the characteristic polynomial is $z^2-7z+10 = (z-2)(z-5)$, the solutions of
$$ S_{n}-7 S_{n-1} + 10 S_{n-2} = 0 $$
have the form $S_n = \alpha 2^n+\beta 5^n$. By direct inspection a solution of
$$ S_{n}-7 S_{n-1} + 10 S_{n-2} = 5\cdot 3^n $$
is given by $S_n=-\frac{45}{2}\cdot3^n$, hence the set of solutions of the previous recurrence is given by
$$ S_n = \alpha 2^n+\beta 5^n-\frac{45}{2}3^n $$
and by imposing $S_0=0$ and $S_1=1$ we get:
$$\boxed{ S_n = \color{red}{\frac{44}{3}2^n + \frac{47}{6} 5^n -\frac{45}{2} 3^n}.} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2312712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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|
Show that $\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}?$ How may we show that
$$\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}\color{red}?\tag1$$
$u={x\over 2}\implies 2du=dx$
$$4\int_{0}^{\infty}{u\over (1+4u^2)^2}\coth(\pi u)\mathrm du\tag2$$
$$4\int_{0}^{\infty}{u\over (1+4u^2)^2}\cdot{e^{2u}+1\over e^{2u}-1}\mathrm du\tag3$$
$v=4u^2\implies du={dv\over 8u}$
$${1\over 2}\int_{0}^{\infty}{e^{\sqrt{v}}+1\over e^{\sqrt{v}}-1}\cdot{\mathrm dv\over (1+v)^2}\tag4$$
$${1\over 2}\sum_{k=0}^{\infty}{(2)_k\over k!}(-1)^k\int_{0}^{\infty}{e^{\sqrt{v}}+1\over e^{\sqrt{v}}-1}\cdot v^k \mathrm dv\tag5$$
$e^{\sqrt{v}}=t\implies dv=2\sqrt{v}e^{\sqrt{v}}dt$
$$\sum_{k=0}^{\infty}{(2)_k\over k!}(-1)^k\int_{1}^{\infty}t\cdot{t+1\over t-1}\cdot \ln^{2k+1}(t) \mathrm dt\tag6$$
I don't what to do next ...
|
Another approach is to use the partial fraction expansion of $\coth(z)$.
(See THIS QUESTION for derivations.)
$$\begin{align} & \int_{0}^{\infty} \frac{x}{(1+x^{2})^{2}} \coth \left(\frac{\pi x}{2} \right) \, dx \\ &= \int_{0}^{\infty} \frac{x}{(1+x^{2})^{2}} \left(\frac{2}{\pi x} + \pi x \sum_{n=1}^{\infty} \frac{1}{\frac{\pi^{2}x^{2}}{4}+ n^{2}\pi^{2}} \right) \, dx \\ &= \frac{2}{\pi} \int_{0}^{\infty} \frac{1}{(1+x^{2})^{2}}\, dx+ \frac{4}{\pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{x^{2}}{(1+x^{2})^{2}(x^{2}+ 4n^{2})} \, dx \\ &= \frac{2}{\pi} \int_{0}^{\pi/2} \cos^{2}(u) \, du \\&+ \small \frac{4}{\pi} \sum_{n=0}^{\infty}\left(\frac{4n^{2}}{(4n^{2}-1)^{2}}\int_{0}^{\infty} \frac{1}{1+x^{2}} \, dx- \frac{1}{4n^{2}-1} \int_{0}^{\infty} \frac{1}{(1+x^{2})^{2}} \, dx - \frac{4n^{2}}{(4n^{2}-1)^{2}} \int_{0}^{\infty} \frac{1}{x^{2}+4n^{2}} \, dx\right) \\ &= \frac{2}{\pi} \left(\frac{\pi}{4}\right) + \frac{4}{\pi} \sum_{n=1}^{\infty} \left(\frac{4n^{2}}{(4n^{2}-1)^{2}}\frac{\pi}{2} - \frac{1}{4n^{2}-1}\frac{\pi}{4}- \frac{4n^{2}}{(4n^{2}-1)^{2}} \frac{\pi}{4n} \right) \\ &= \frac{1}{2} + \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\pi}{4} \frac{1}{(2n+1)^{2}} \\ &= \frac{1}{2} + \left(\sum_{n=1}^{\infty}\frac{1}{n^{2}} - \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}-1 \right) \\ &= \frac{1}{2} + \left(\frac{\pi^{2}}{6} - \frac{1}{4} \frac{\pi^{2}}{6}-1 \right) \\ &=\frac{\pi^{2}}{8}-\frac{1}{2} \end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2315485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
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|
If corr(A,B) = x and corr(B,C) = y, what is corr(A,C)? If I know the correlation between two pairs of random variables (A,B) and (B,C), can I determine the correlation of the pair (A,C)? If not, can I at least constrain it to some range?
I'm interesting in generating a covariance matrix where certain pairwise correlation values are determined, and the rest are as small (i.e. close to 0) as possible.
|
Unfortunately no, we cannot determine $\rho_{AC}$ given just $\rho_{AB}$ and $\rho_{BC}$. You can derive the theoretical bounds
\begin{align*}
\rho_{AC} \ge \max\{2(\rho_{AB} + \rho_{BC}) - 3, 2\rho_{AB}\rho_{BC} - 1\}
\end{align*}
Proof. Some notation. I let $\sigma_{AB} = \text{Cov}(A,B)$ and $\sigma_A^2 = \text{Var}(A)$.
Let's first prove $\rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3$. Recall the identity
\begin{align*}
2 E[X^2] + 2E[Y^2] = E[(X+Y)^2] + E[(X-Y)^2]
\end{align*}
hence $2E[Y^2] \le E[(X+Y)^2] + E[(X-Y)^2]$. Set
\begin{align*}
X = \widetilde{B} - (\widetilde{A} + \widetilde{C})/2 \quad \text{and} \quad Y =
(\widetilde{A} - \widetilde{C})/2
\end{align*}
where $\widetilde{C} = (C - E[C])/\sigma_C$, the normalized random variable, and similarly for $\widetilde{A}, \widetilde{B}$. Upon substitution and simplification, we get
\begin{align*}
\frac{1}{2}(2 - 2\rho_{AC}) \le (2 - 2\rho_{AB}) + (2 - 2\rho_{BC}) \iff \rho_{AC} \ge 2(\rho_{AB} + \rho_{BC}) - 3
\end{align*}
To prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$, consider the random variable
\begin{align*}
W = 2 \frac{\sigma_{AB}}{\sigma_B^2}B - A
\end{align*}
We can verify $\sigma_W^2 = \sigma_A^2$, and hence $\sigma_{WC} \le \sigma_{W}\sigma_{C} = \sigma_ A \sigma_C$ by the Cauchy-Schwarz inequality. On the other hand, you may compute
\begin{align*}
\sigma_{WC} = 2 \frac{\sigma_{AB}}{\sigma_B^2}\sigma_{BC} - \sigma_{AC}
\end{align*}
Reorganizing all this, we prove $\rho_{AC} \ge 2\rho_{AB}\rho_{BC} - 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2316314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Why does my parametric trigonometric function appears to be a polynomial? I was fooling around with $(\cos^2(t),\sin^2(at))$ with varying values of $a$, and found that if $a=3$ then $(\cos^2(t),\sin^2(3t))$ gives the graph of $y=-16x^3+24x^2-9x+1$ on the domain $[0,1]$
The calculator won't do parametrics but it just looks like the graph from 0 to 1.
Eliminating the parameter gave me $\sin ^2\left(3\cos ^{-1}\left(\sqrt{x}\right)\right)$
Why is this true? And, how can I solve $\sin ^2\left(3\cos ^{-1}\left(\sqrt{x}\right)\right)=-16x^3+24x^2-9x+1$ to prove that the Cartesian form is equal to the polynomial?
|
We have $$\begin{align*}y=\sin^2 3t &= (3\sin t - 4\sin^3 t)^2 \\ &= \sin^2 t(3-4\sin^2 t)^2 \\& = (1-\cos^2 t)(3-4(1-\cos^2 t))^2 \\ & = (1-x)(4x-1)^2 \\ & =-16x^3+24x^2-9x+1\end{align*}$$
where we derived $\sin 3t = 3\sin t - 4\sin^3 t$ either by expanding $e^{3it}$ and taking imaginary parts or by expanding $\sin 3t = \sin (2t + t)$ and then expanding $\sin 2t$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solve $\sqrt3 \cot^2 x - 2 \cot x - \sqrt3 = 0$ I thought it'd be nice to factor this as follows: $\sqrt3 \cot^2 x - 2 \cot x - \sqrt3 = 0 $, so: $(\sqrt3 \cot x + 1)(\cot x - \sqrt3) = 0$.
So there are 2 equations to solve:
*
*$\sqrt3 \cot x + 1 = 0$. I thought to multiply both sides by $\tan
x$, yielding: $\sqrt3 + \tan x = 0$. Hence: $\tan x = -\sqrt3$ and
$x = \{ -\frac{\pi}{3} + k\pi \}$
*$\cot x - \sqrt3 = 0$, so $\cot x - \sqrt3 = 0$, $\cot x = \sqrt3$.
Multiplying by $\tan x$ yields: $1 = \sqrt3 \tan x$, dividing both
sides by $\sqrt3$: $\frac{\sqrt3}{3} = \tan x$, so $x = \{
\frac{\pi}{6} + k\pi\}$
Is this a correct approach?
|
Assuming $\tan x\neq 0$, multiplying throughout by $\tan^2x$, moving terms to RHS, and putting $t=\tan x$ gives
$$\sqrt3\ t^2+2t-\sqrt3=0\\
(\sqrt3\ t-1)(t+\sqrt3)=0\\
t=\tan x=\frac 1{\sqrt3}, -\sqrt3\\
x=n\pi+\frac {\pi}6, n\pi-\frac {\pi}3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2318653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evalaute $\int_{-\alpha}^{\alpha}{1\over x}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=(-\pi)^{n+1}F(n)$ Proposed:
$$\int_{-\alpha}^{\alpha}{x^k}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=F(k,n)\tag1$$
Let $k=-1$ and $F(-1,n)=F(n)$
$$\int_{-\alpha}^{\alpha}{1\over x}\sqrt{\alpha+x\over \alpha-x}\ln^n\left({\alpha+x\over \alpha-x}\right)\mathrm dx=(-\pi)^{n+1}F(n)\tag2$$
Where $n,\alpha\ge1$
We have the following
$n=1\implies$ $F(1)=1$
$n=2\implies$ $F(2)=-1$
$n=3\implies$ $F(3)=2$
$n=4\implies$ $F(4)=-5$
$n=5\implies$ $F(5)=16$
$n=6\implies$ $F(6)=-61$
How can we find the closed form of $(2)?$
$u={\alpha+x\over \alpha-x}\implies dx={(\alpha-x)^2\over 2\alpha}$
$x=\alpha\cdot{u-1\over u+1}$
$(\alpha-x)^2={4\alpha^2\over (u+1)^2}$
$(2)\implies$
$$2\int_{0}^{\infty}{\sqrt{u}\over u^2-1}\cdot\ln^n(u)\mathrm du\tag3$$
$$....$$
|
The comment of Simply Beautiful Art gives a big hint on how to proceed. I cannot give a general formula to (1), although it is possible albeit complicated.
Let $$I(k,t)=\int_{-1}^{+1}x^k\left(\frac{1+x}{1-x}\right)^t dx \quad\quad J(k,n)=\int_{-1}^{+1}x^k\sqrt{\frac{1+x}{1-x}}\ln^n\left(\frac{1+x}{1-x}\right)dx$$
making $u = \frac{1+x}{1-x}$ gives
$$I(k,t) = 2\int_{0}^{\infty} \frac{u^t(u-1)^k}{(u+1)^{2+k}} dx$$
When $k$ is a positive integer, we can expand and note that $$\int_{0}^{\infty} \frac{u^a}{(u+1)^b} du = B(a+1,b-a-1)$$ with the beta function.
Since the derivative of gamma function at half-integer value can be calculated, the closed form for each $J(k,n)$ can also be found.
For instances
$$J(1,1) = 2\pi \quad J(2,1) = \frac{5\pi}{3} \quad J(2,1) = \frac{5\pi}{3} \quad \\ J(4,1) = \frac{89\pi}{60} \quad J(5,1) = \frac{89\pi}{60} \quad J(6,1) = \frac{381\pi}{280}$$
$$\begin{aligned}
J(1,2) = J(2,2) = 4\pi + \frac{\pi^3}{2} \\
J(3,2) = J(4,2) = \frac{14\pi}{3} + \frac{3\pi^3}{8} \\
J(5,2) = J(6,2) = \frac{439\pi}{90} + \frac{5\pi^3}{16} \\
\end{aligned}$$
$$\begin{aligned}
J(1,3) & = 6\pi^3 \\
J(2,3) &= J(3,3) = 8\pi + 5\pi^3 \\
J(4,3) &= J(5,3) = 12\pi + \frac{89\pi^3}{20} \\
\end{aligned}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2319587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Lack of understanding of steps for second principle of finite induction for: $a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$ The problem i'm trying to solve is:
Use the second principle of finite induction to establish that:
$$a^n -1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$$
for all $n \ge 1$
I have found a solution online that reads:
$$ \mbox{For }k=1: \ a^1-1 =a-1=(a-1)(a^0)=a-1 $$
\begin{align}
k \Rightarrow k+1: \ a^{k+1} -1 &= a^{k+1}-a^k-a+a^k+a-1\\
&=a(a^k-1)+a^k-1-a(a^{k-1} -1)\\
&=(a+1)(a^k-1)-a(a^{k-1}-1)
\end{align}
Use second principle of induction for $k,\ k-1$:
\begin{align}
&=(a+1)[(a-1)(a^{k-1}+a^{k-2}+\cdots+a+1)] \\
&\qquad \; \; -a[(a-1)(a^{k-2}+a^{k-3}+\cdots+a+1)]\\
&=(a-1)[(a)(a^{k-1}+a^{k-2}+\cdots+a+1)\\
&\qquad \quad \ \, +(1)(a^{k-1}+a^{k-2}+\cdots+a+1)\\
&\qquad \quad \ \,-(a)(a^{k-2}+a^{k-3}+\cdots+a+1)]\\
&=(a-1)[(a^k+a^{k-1}+a^{k-2}+\cdots+a^2+a)+1]
\end{align}
$\therefore$ works for $k+1$
My problem:
I understand all the steps used for the principle of second induction however i do not understand how you are supposed to deduce that you need:
$$a^{k+1} -1 = (a+1)(a^k-1)-a(a^{k-1}-1)$$
To clarify i understand the logic that leads to the equation above, but do not understand how you know to try to make it. Is there something specific about the format $ (a+1)(a^k-1)-a(a^{k-1}-1)$ that allows you to know it is needed, or has the format just been found through any possible means that leads to $(a^k-1)$ and $(a^{k-1}-1)$ being used?
|
I'm just going to do the induction step here, for $k \to k + 1$, to show that it can be done in a simpler and more transparent way.
So assume you already know that
$$a^k - 1 = a^{k-1} + a^{k-2} + \dots + a + 1.$$
Then, using the induction hypothesis on the second line, we have
\begin{align}
(a-1)(a^k + a^{k-1} + \dots + a + 1) &= (a-1)a^k + (a-1)(a^{k-1} + a^{k-2} + \dots + a + 1) \\
&= a^{k+1} - a^k + a^k - 1 \\
&= a^{k+1} - 1.
\end{align}
This proves the result for $k + 1$.
Edit Just in case you haven't seen it, here is the proof without induction. (Technically, there's probably some kind of induction hidden somewhere inside it, but it's not formalized.)
$$
\begin{align}
(a-1)(a^{n-1} + a^{n-2} + \dots + a + 1) &= a(a^{n-1} + a^{n-2} + \dots + a + 1) - (a^{n-1} + a^{n-2} + \dots + a + 1)\\
&= a^n + a^{n-1} + \dots + a^2 + a - (a^{n-1} + a^{n-2} + \dots + a + 1) \\
&= a^n - 1.
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2322833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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|
How to solve $\log_{4}(\sqrt{x^{4/3}})+3\log_{x}(16x)=7?$
How to solve $\log_{4}(\sqrt{x^{4/3}})+3\log_{x}(16x)=7?$
I've tried everything from brute force to doing base change but nothing works. I was wondering what was the best way in solving this?
|
\begin{align}
\log_{4}(\sqrt{x^{4/3}})+3\log_{x}(16x)&=7\\
\left(\frac{1}{2}\right)\left(\frac{4}{3}\right)\log_{4}(x)+3\log_{x}(4^2)+3\log_{x}(x)&=7\\
\frac{2}{3}\log_{4}(x)+\frac{6}{\log_4x}+3&=7\\
(\log_4x)^2-6\log_4x+9&=0\\
\log_4x&=3\\
x&=64
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2323086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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|
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
|
The hard way:
You can use the fact that
$$(e^xf(x))'=e^x(f(x)+f'(x)).$$
Then
$$(e^xf(x))'=e^x(x^3+5x^2+x+2).$$
Now, by integration (which can be performed by parts, integrating $e^x$),
$$e^xf(x)=\int e^x(x^3+5x^2+x+2)dx=e^x(x^3+2x^2-3x+5)+C$$ and
$$f(x)=x^3+2x^2-3x+5+Ce^{-x}.$$
[I doubt this is the method you are expected to use...]
Empirically:
Let us assume that the solution is close to $f_0(x)=x^3$ and let us plug in the given equation.
$$f_0(x)+f'_0(x)=x^3+3x^2\ne x^3+5x^2+x+2.$$ But we are off by the polynomial $$2x^2+x+2$$ which is of a lower degree.
So let us try $f_1(x)=x^3+2x^2$, giving
$$f_1(x)+f'_1(x)=x^3+5x^3+4x\ne x^3+5x^2+x+2.$$
This time, we are only off by $-3x+2$, one degree less.
$$f_2(x)=x^3+2x^2-3x$$ then $$f_2(x)+f'_2(x)=x^3+5x^2+x-3.$$
And finally
$$f_3(x)=x^3+2x^2-3x+5$$ has converged to a solution.
Epilogue:
Let us assume that there exists another solution $f$ which is different from $f_3$.
From
$$f(x)+f'(x)=f_3(x)+f'_3(x)= x^3+5x^2+x+2,$$ we can draw that
$$f(x)-f_3(x)+f'(x)-f'_3(x)=0$$ which is of the form
$$g(x)+g'(x)=0.$$
It is not difficult to show that no polynomial is the opposite of its derivative. Actually, there is a single differentiable function having this property: the negative exponential
$$g(x)=e^{-x}$$ (or multiples).
Grouping these results,
$$f(x)=x^3+5x^2+x+2+Ce^{-x}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1 \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.
|
Problem
Diagonalize the matrix
$$
\mathbf{A} =
\left[
\begin{array}{cc}
2 & 3 \\
3 & 4 \\
\end{array}
\right]
$$
Solution
Compute eigenvalues
The eigenvalues are the roots of the characteristic polynomial
$$
p(\lambda) = \lambda^{2} - \lambda \text{ trace }\mathbf{A} + \det \mathbf{A}
$$
The trace and determinant are
$$
\text{ trace }\mathbf{A} = 6, \qquad \det \mathbf{A} = -1
$$
Therefore
$$
p(\lambda)
= \lambda^{2} - \lambda \text{ trace }\mathbf{A} + \det \mathbf{A}
= \lambda^{2} - 6 \lambda - 1
$$
The roots are the eigenvalue spectrum
$$
\lambda \left( \mathbf{A} \right) = 3 \pm \sqrt{10}
$$
Result:
$$
\mathbf{D} = \left[
\begin{array}{cc}
3+\sqrt{10} & 0 \\
0 & 3-\sqrt{10} \\
\end{array}
\right]
$$
Eigenvectors
First
$$
\begin{align}
\left(\mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) w_{1} &= \mathbf{0} \\
%
\left[
\begin{array}{cc}
-1-\sqrt{10}-1 & 3 \\
3 & 1-\sqrt{10} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
w_{x} \\
w_{y} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
0 \\
0 \\
\end{array}
\right]
%
\end{align}
$$
Solution
$$
w_{1} =
\left[
\begin{array}{c}
\frac{1}{3} \left(-1+\sqrt{10}\right) \\
1 \\
\end{array}
\right]
$$
Second
$$
\begin{align}
\left(\mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) w_{2} &= \mathbf{0} \\
%
\left[
\begin{array}{cc}
-1+\sqrt{10} & 3 \\
3 & 1+\sqrt{10} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
w_{x} \\
w_{y} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
0 \\
0 \\
\end{array}
\right]
%
\end{align}
$$
Solution
$$
w_{2} =
\left[
\begin{array}{c}
-\frac{1}{3} \left(1+\sqrt{10}\right) \\
1 \\
\end{array}
\right]
$$
Diagonalization matrix
$$
\mathbf{P} =
\left[
\begin{array}{cc}
\frac{1}{3} \left(-1+\sqrt{10}\right) & -\frac{1}{3} \left(1+\sqrt{10} \right) \\
1 & 1 \\
\end{array}
\right],
\qquad
\mathbf{P}^{-1} =
\frac{1}{2\sqrt{10}}
\left[
\begin{array}{rr}
3 & 1+\sqrt{10} \\
-3 & -1+\sqrt{10} \\
\end{array}
\right]
$$
Validation
You can check that
$$
\mathbf{P} \mathbf{A} \mathbf{P}^{-1} = \mathbf{D}
$$
and
$$
\mathbf{P}^{-1} \mathbf{D} \mathbf{P} = \mathbf{A}
$$
Gaussian elimination
Solving this problem does not require Gaussian elimination. However, since you specifically asked, here is the process:
Clear column 1
$$
\left[
\begin{array}{rc}
\frac{1}{2} & 0 \\
-\frac{3}{2} & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
2 & 3 & 1 & 0 \\
3 & 4 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cr|rc}
1 & \frac{3}{2} & \frac{1}{2} & 0 \\
0 & -\frac{1}{2} & -\frac{3}{2} & 1 \\
\end{array}
\right]
$$
Clear column 2
$$
\left[
\begin{array}{cr}
1 & 3 \\
0 & -2 \\
\end{array}
\right]
%
\left[
\begin{array}{cr|rc}
1 & \frac{3}{2} & \frac{1}{2} & 0 \\
0 & -\frac{1}{2} & -\frac{3}{2} & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|rr}
1 & 0 & -4 & 3 \\
0 & 1 & 3 & -2 \\
\end{array}
\right]
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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|
Trigonometric Equation : $\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$ Please help solving this equation:
$\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$
I used numerical method to solve it and got $x=6^\circ$ but I am not able to solve it by trigonmetry.
Thank!
Best Regards, Michael.
|
Using the same observation as @dantopa we get
\begin{eqnarray}
\sin(x) &=& \sin(12^\circ - x )\\
\sin(x) &=& \sin(12^\circ)\cos(x)-\cos(12^\circ)\sin(x)\\
\tan(x) &=& \sin(12^\circ)-\cos(12^\circ)\tan(x)\\
\tan(x)(1+\cos(12^\circ))&=&\sin(12^\circ)\\
\tan(x)&=&\frac{\sin(12^\circ)}{1+\cos(12^\circ)}\\
\tan(x)&=&\tan(6^\circ)\\
x &=& 6^\circ+180^\circ n
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2324425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Laurent Expansion - Complex Analysis question Exercise :
Find the Laurent Expansion of the function :
$$f(z) = \frac{1}{(z-2i)(z^2 +4)}$$
around $z_0 = -2i$, in the biggest possible ring that includes the point $z=-2 + 2i$.
Attempt :
We have :
$$f(z) = \frac{1}{(z-2i)(z^2 +4)} = \frac{1}{(z-2i)(z + 2i)(z-2i)} = \frac{1}{(z-2i)^2(z + 2i)} $$
We have the rings :
$$D_1 : 0<|z+2i|<4$$
$$D_2 : |z+2i|>4$$
Because $|(-2+2i) + 2i| = |-2 + 4i| = \sqrt{20} > 4$, the biggest ring with center $z_0=-2i$ that contains the point $z=-2+2i$, is :
$$D_2 : |z+2i|>4$$
Now, such exercises can be handled by creating such a fraction, such as to use the usual geometric series :
$$\frac{1}{1-w} = \sum_{n=0}^{\infty}w^n ,|w|<1$$
$$\frac{1}{1+w} = \sum_{n=0}^{\infty}(-1)^nw^n,|w|<1$$
It is :
$$f(z) = \frac{1}{(z-2i)^2(z + 2i)} = \frac{1}{(z-2i)^2[(z - 2i)+4i]} = \frac{1}{(z-2i)^2(z - 2i)\big(1-\frac{4i}{z-2i}\big)} = \frac{1}{(z-2i)^3\big(1-\frac{4i}{z-2i}\big)} $$
But in order to have $|w|<1 $ it must hold that : $|z-2i| > 4$, but our ring is $|z+2i|>4$. My question is : Is the exercise wrong ? Maybe it means around the center $z_0 = 2i$ ? Or is it something that I am doing wrong ? If nothing is wrong, how do I proceed here ?
Except if I use :
$$\frac{1}{(z-2i)^2(z + 2i)} = \frac{1}{(z+2i)^3\big(1-\frac{4i}{z+2i}\big)^2} = \frac{1}{(z+2i)^3}\frac{1}{ \big(1-\frac{4i}{z+2i}\big)^2} = \frac{1}{(z+2i)^3} \Bigg( \sum_{n=0}^{\infty} \bigg(\frac{4i}{z+2i}\bigg)^n\Bigg)^2 $$
because if so, it is $|z+2i|>4$.
I would really appreciate some thorough help.
|
You are on the right track. Just the last step should be revised a little.
The function $$f(z)=\frac{1}{(z-2i)^2}\cdot\frac{1}{(z + 2i)}$$ is to expand around the center $z_0=-2i$. Since there are poles at $z=2i$ and $z=-2i$ we have to distinguish two regions
\begin{align*}
D_1:&\quad 0<|z+2i|<4\\
D_2:&\quad |z+2i|>4
\end{align*}
*
*The first region $D_1$ is a punctured disc with center $z_0=-2i$, radius $4$ and the pole $2 i$ at the boundary of the disc.
In the interior we have a representation of the fractions with a pole at $z=- 2i$ as principal part of a Laurent series at $z_0=-2i$, while the fraction with pole at $z=2i$ admits a representation as power series.
*The other region $D_2$ containing all points outside $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=-2i$.
We have to put the focus on $D_2$ since it contains the point $-2+2i$.
We obtain
\begin{align*}
\color{blue}{f(z)}&\color{blue}{=\frac{1}{(z + 2i)(z-2i)^2}}\\
&=\frac{1}{(z + 2i)}\cdot\frac{1}{\left((z+2i)-4i\right)^2}\\
&=\frac{1}{(z + 2i)^3}\cdot\frac{1}{\left(1-\frac{4i}{z+2i}\right)^2}\tag{1}\\
&=\frac{1}{(z + 2i)^3}\sum_{n=0}^\infty \binom{-2}{n}\left(-\frac{4i}{z+2i}\right)^n\tag{2}\\
&=\sum_{n=0}^\infty (n+1)(4i)^n\frac{1}{(z+2i)^{n+3}}\tag{3}\\
&\color{blue}{=\sum_{n=3}^\infty (n-2)(4i)^{n-3}\frac{1}{(z+2i)^{n}}}\\
\end{align*}
Comment:
*
*In (1) we apply the binomial series expansion.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we shift the index to start with $n=3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove by induction that for $n$ $ \in \mathbb N $ , $ 2 n - 18 < n^2-8n +8 $? Question:
Prove for $n$ $ \in \mathbb N $ , $ 2 n\ -\ 18\ <\ n^2-8n\ +8 $
My attempt:
$ Base\ Case:\ n\ =\ 1,\ it\ holds. $
$I.H:\ Suppose\ 2k-18\ <\ k^2-8k+8,\ where\ k\ is\ a\ natural\ number.$
$ Then,\ \left(k+1\right)^2-8\left(k+1\right)+8\ =\ k^2+2k+1-8k-8+8\ >2k-18+2k+1-8$
I am stuck here. Any help would be appreciated. Thanks.
|
Notice, $2n-18<n^2-8n+8 \implies n^2-10n+26=(n-5)^2+1>0$.
Let $(n-5)=u$, so that we have $u^2+1>0$.
Moving over the $1$ gives us $u^2>-1$. We know this is true because all squares of real (and natural) numbers are greater than $-1$.
Therefore, we have $(n-5)^2>-1$, and $(n-5)^2+1>0$, and $n^2-10n+26>0$.
Adding $2n-18$ to both sides gives us $n^2-8n+8>2n-18 \implies 2n-18<n^2-8n+8$.
Thus, our proof is complete.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Series expansion of Elliptic integral $F(\sin^{-1}(1-x)|-1)$ where $0I understand the leading term $F(\pi/2\mid-1)$ would be $K(-1)$. I also want the next term contributing from $x$. Actually I could not expand $\sin^{-1}(1-x)$ (when $0<x\ll1$) around 1. Any help would be highly appreciated.
Thanks a lot.
|
For more than the first term.
We can start using $$\frac d {dx}F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\frac{1}{\sqrt{1-(1-x)^2} \sqrt{1+(1-x)^2}}\tag 1$$ and use Taylor expansions around $x=0$ $$\sqrt{1-(1-x)^2}=\sqrt{2} \sqrt{x}-\frac{x^{3/2}}{2 \sqrt{2}}-\frac{x^{5/2}}{16
\sqrt{2}}-\frac{x^{7/2}}{64 \sqrt{2}}-\frac{5 x^{9/2}}{1024
\sqrt{2}}+O\left(x^{11/2}\right)$$ $$\sqrt{1+(1-x)^2}=\sqrt{2}-\frac{x}{\sqrt{2}}+\frac{x^2}{4 \sqrt{2}}+\frac{x^3}{8 \sqrt{2}}+\frac{3
x^4}{64 \sqrt{2}}+\frac{x^5}{128 \sqrt{2}}+O\left(x^6\right)$$ which makes the denominator of $(1)$ to be $$2 \sqrt{x}-\frac{3 x^{3/2}}{2}+\frac{7 x^{5/2}}{16}+\frac{5 x^{7/2}}{64}+\frac{11
x^{9/2}}{1024}+O\left(x^{11/2}\right)$$ So
$$\frac d {dx}F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\frac{1}{2 \sqrt{x}}-\frac{3 \sqrt{x}}{8}-\frac{11 x^{3/2}}{64}-\frac{7
x^{5/2}}{256}+\frac{141 x^{7/2}}{4096}+O\left(x^{9/2}\right)$$ Integrating
$$F\left(\left.\sin ^{-1}(1-x)\right|-1\right)=-\sqrt{x}-\frac{x^{3/2}}{4}-\frac{11 x^{5/2}}{160}-\frac{x^{7/2}}{128}+\frac{47
x^{9/2}}{6144}+O\left(x^{11/2}\right)+C$$ and $$C=F\left(\left.\sin ^{-1}(1)\right|-1\right)=F(\frac\pi 2|-1)=K(-1)=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{2 \pi }}$$.
|
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|
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$ if$\ δ=ε$ we have to prove that:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|<ε$$
But I'm having a hard time trying to prove the last part, I tried:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$
|
Using that for $y \neq 0$:
$$|\frac{x^3y^4}{(x^4+y^2)^2}| \leq |\frac{x^3y^4}{(0+y^2)^2}|=|x^3| \to 0$$
As a smaller denominator means larger number in magnitude. We easily conclude by squeeze theorem that the limit is $0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Trigonometric Integral - Complex Integration Exercise :
If $a \in \mathbb R-\{0\}, |a|<1$, calculate the integral and show that :
$$\int_0^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta = \frac{\pi}{\sqrt{1-a^2}}\Bigg(\frac{\sqrt{1-a^2} -1}{a} \Bigg)^n$$
Attempt :
Since $\cos \theta$ is even, we have :
$$\int_0^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta = \frac{1}{2}\int_{-\pi}^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta$$
We substitute $\cos n\theta = \Re(e^{in\theta})$ and we get :
$$\int_0^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta = \frac{1}{2}\int_{-\pi}^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta = \frac{1}{2}\Re\Bigg(\int_{-\pi}^\pi \frac{e^{in\theta}}{1 + a\cos \theta}d\theta \Bigg) $$
Now, I understand that a contour integral can be made around the unit circle $|z|=1$ and find the singularities to work with residues, but I cannot seem how to move one with this procedure.
I would really appreciate a thorough solution or some hints on how to get to the answer.
|
Let $z=e^{i\theta}$ and then
\begin{eqnarray}
&&\int_0^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta\\
& =& \frac{1}{2}\int_{-\pi}^\pi \frac{\cos n\theta}{1 + a\cos \theta}d\theta\\
&=&\frac12\int_{|z|=1}\frac{\frac{z^n+\frac1{z^n}}{2}}{1+a\frac{z+\frac1z}{2}}\frac1{iz}dz\\
&=&\frac1{2i}\int_{|z|=1}\frac{z^{2n}+1}{z^n(2z+a(z^2+1))}dz\\
&=&\frac1{2ai}\int_{|z|=1}\frac{z^{2n}+1}{z^n(z-z_1)(z-z_2)}dz\\
&=&\frac1{2ai}2\pi i\bigg[\textbf{Res}\bigg(\frac{z^{2n}+1}{z^n(z-z_2)},z=z_1\bigg)+\textbf{Res}\bigg(\frac{z^{2n}+1}{z^n(z-z_1)(z-z_2)},z=0\bigg)\bigg]\\
&=&\frac{\pi}{a}\bigg[\frac{z_1^{2n}+1}{z_1^n(z_1-z_2)}+\frac{z_1^n-z_2^n}{z_1^nz_2^n(z_1-z_2)}\bigg]\\
&=&\frac{\pi}{a}\frac{1+z_1^nz_2^n}{(z_1-z_2)z_2^n}\\
&=&\frac{\pi}{\sqrt{1-a^2}z_2^n}
\end{eqnarray}
where
$$z_1=\frac{-1+\sqrt{1-a^2}}{a},z_2=\frac{-1-\sqrt{1-a^2}}{a}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2327403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find high common factor of all possible $n$ If $n$ is an integer such that $2n + 1$ and $3n + 1$ are square numbers, find the highest common factor of all possible $n$.
Do we need to know exactly what all possible $n$ are?
|
As $n=40$ gives us the squares $81=9^2$ and $121=11^2$, the desired number $d$ is a divisor of $40$, which limit sour search from above.
As $2n+1$ is an odd square, it os $\equiv 1\pmod 8$, hence $n\equiv 0\pmod 4$. But then $3n+1$ is also odd, hence $\equiv 1\pmod 8$, hence $8\mid n$.
Thus the only remaining possibilities are $d=0$ or $d=40$, and we need to check $n\bmod 5$.
Note that $2$ and $3$ ar enot squares $\bmod 5$.
If $n\equiv 1\pmod 5$, then $2n+1\equiv 3$; if $n\equiv 2\pmod 5$, then $3n+1\equiv 2$; if $n\equiv 3\pmod 5$, then $2n+1\equiv 2$; if $n\equiv 4\pmod 5$, then $3n+1\equiv 3$. It follows that $n\equiv 0\pmod 5$.
We conclude that $d=40$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Triple integral using cylindrical coordinates with constraints The prompt is to find an iterated triple integral in cylindrical coordinate system where the surface S is defined as the solid cut out from the ball $x^2+y^2+z^2\leq 1$ by the half cone $z = \sqrt{(x^2 + y^2)}$ which represents the volume of the solid S.
So I made the integral $$V = \iiint r.f(x, y, z) dzdrd\theta$$
we know that $x = r\cos\theta$, $y = r\sin\theta$ and z = z.
using equations, $x^2+y^2+z^2\leq 1$ and $z = \sqrt{(x^2 + y^2)}$,
$$r^2 + z^2 = 1$$
$$z = \sqrt{r}$$
$$r^2 + r = 1$$
$$r(r + 1) = 1$$
this gives us $r = -\frac{1}{2}\pm\frac{\sqrt5}{2}$
$$V = \int_0^{2\pi}\int_{-\frac{1}{2}+\frac{\sqrt5}{2}}^{-\frac{1}{2}-\frac{\sqrt5}{2}}\int (r)dzdrd\theta$$
I'm not surehow to find limits for dz
|
Visualizing the regions in question:
We do not want to integrate by $z$ last, as this would require two separate integrals. Noting that the volume we want has rotational symmetry (since both surfaces do) about the $z$-axis, the region on the $x$-$y$ or $r$-$\theta$ plane we want is a circle. So:
$$ \theta: 0 \text{ to } 2\pi $$
Now $r$ goes from $0$ to the value of $r$ where the cone and sphere touch. The cone is represented as $z = r$ and the sphere is represented as $r^2 + z^2 = 1$. Therefore, the value we want is $r^2 = 1/2$, or equivalently, our limits of integration are:
$$ r: 0 \text{ to } \frac{1}{\sqrt{2}}$$
Finally, we will integrate along the $z$ dimension first, integrating from the bottom of the shape, which is the cone so $z = r$, to the top of shape, which is the sphere so $z = \sqrt{1-r^2}$.
Recall that:
$$ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz $$
The volume of the solid region $R$ we have described is:
\begin{align*}
V &= \iiint_R dV \\
&= \int_0^{2\pi} \int_0^\frac{1}{\sqrt{2}} \int_r^\sqrt{1-r^2} r \, dz \, dr \, d\theta \\
&= 2\pi \int_0^\frac{1}{\sqrt{2}} \int_r^\sqrt{1-r^2} r \, dz \, dr \\
&= 2\pi \int_0^\frac{1}{\sqrt{2}} (r\sqrt{1-r^2} - r^2) \, dr \\
&= \pi \int_0^\frac{1}{\sqrt{2}} (2r\sqrt{1-r^2} - 2r^2) \, dr \\
&= \pi \left[-\frac{2}{3}(1-r^2)^{3/2} - \frac{2}{3}r^3 \right]_0^{\frac{1}{\sqrt{2}}} \\
&= \frac{2\pi}{3} \left[(1-r^2)^{3/2} + r^3 \right]_{\frac{1}{\sqrt{2}}}^0 \\
&= \frac{2\pi}{3} \left(1 - \frac{2}{2\sqrt{2}} \right) \\
&= \frac{\pi}{3}(2-\sqrt{2})
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?
|
Let $a+b=2u$ and $ab=v^2$.
Thus, we need to prove that
$$v^4(4u^2-2v^2-2)\geq(4u^2-2v^2)(v^2-1)$$ or
$$v^4(2u^2-v^2-1)\geq(2u^2-v^2)(v^2-1)$$ or
$$2u^2v^4-v^6-v^4\geq2u^2(v^2-1)-v^4+v^2$$ or
$$2u^2(v^4-v^2+1)\geq v^6+v^2.$$
But $v^4-v^2+1>0$ and $u^2\geq v^2$ because it's just $(a-b)^2\geq0$.
Thus, it's enough to prove that
$$2v^2(v^4-v^2+1)\geq v^6+v^2,$$
which is
$$v^2(v^2-1)^2\geq0.$$
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to Wolfram is $(x,y)=(\frac{1}{3},\frac{2}{3}).$
|
Consider the curve
$3x^2-3xy+3y^2-3y+k=0$, which can be written as
$$x^2-xy+y^2-y+\frac k3=0\tag{1}$$
which is an ellipse according to the discriminant criterion.
As the coefficients of $x^2, y^2$ are equal, this implies a rotation of $\frac \pi 4$.
After some manipulation we find that equation $(1)$ can be written as
$$\frac{ \left( \frac {x+y}{\sqrt2}-\frac 1{\sqrt2} \right) ^2 }{\frac 23(1-k)}
+\frac{ \left( \frac {x-y}{\sqrt2}+\frac 1{3\sqrt2} \right) ^2 }{\frac 29(1-k)}=1$$
or, putting $u,v=\frac {x\pm y}{\sqrt2}$,
$$\frac{ \left( u-\frac 1{\sqrt2} \right) ^2 }{\frac 23(1-k)}
+\frac{ \left( v+\frac 1{3\sqrt2} \right) ^2 }{\frac 29(1-k)}=1$$
which is an ellipse rotated by $\frac{\pi}4$ with centre offset $\big(\frac 1{\sqrt2}, \frac 1{3\sqrt2}\big)$ in the $u,v$ directions respectively (which is equivalent to $\big(\frac 13, \frac 23\big)$ in the $(x,y)$ directions respectively - see workings below), and semi-major and semi-minor axis of $\frac 1{\sqrt3}\sqrt{2(1-k)}$ and $\frac 13\sqrt{2(1-k)}$ respectively, where $k<1$.
As $k$ approaches $1$, the ellipse shrinks to a dot at $\left(
\frac 13, \frac 23\right)$, which is the solution for equation $(1)$ with $k=1$.
See desmos implementation here.
$\hspace{3cm}$
From analysis above, $\tan\alpha=1$.
From diagram below, $\tan\beta=\frac 13$.
Hence $\tan(\alpha+\beta)=2$.
This gives
$$\small h=\frac{\sqrt5}3\cos(\alpha+\beta)=\frac {\sqrt5}3\cdot \frac 1{\sqrt5}=\frac 13\\
\small k=\frac{\sqrt5}3\sin(\alpha+\beta)=\frac {\sqrt5}3\cdot \frac 2{\sqrt5}=\frac 23$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range in an Interval I want to find the range of the function, $$f(x) = {{x + 1} \over {{x^2} + 1}}\,\,\;{\rm{when }}\;x \in \left[ { - 1,1} \right].\;\;\;$$
I have isolated $x$ as - $$x = {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}}$$
But I am unable to solve this inequality, $$ - 1 \le {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}} \le 1$$
as it involved $\, \pm \sqrt {} \,$ part. Please tell me how to proceed from here!
|
Here is a method that avoids calculus completely.
Write $y = x+1$, then $\frac{x+1}{x^2+1} = \frac{y}{y^2-2y+2}$. Then note that $$\frac{y^2-2y+2}{y} = y-2+\frac{2}{y}.$$ Keeping in mind that the range of $y$ is $[0, 2]$, $y+\frac{2}{y}$ approaches $\infty$ as $y \to 0$, and the minimum value over positive $y$ is $2\sqrt{2}$ when $y = \sqrt{2}$ by AM-GM ($y+\frac{2}{y} \geq 2\sqrt{y\cdot\frac{2}{y}}$.)
So $y-2+\frac{2}{y}$ ranges from $[2\sqrt{2}-2, \infty]$ when $y$ is in $[0, 2]$. Hence the original fraction which is the reciprocal ranges from $[0, \frac{1}{2\sqrt{2}-2}]$, hitting the bounds at $x = -1$ and $x = \sqrt{2}-1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$ or $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!
|
There also exists a (seemingly) different approach. Recall that $$\sin x-\sin y =2\sin\frac{x-y}{2}\cos\frac{x+y}{2},$$
hence reforming your equation we obtain \begin{align}
0&=\sin x-\sin\left(x-\frac{\pi}{3}\right)\\&=2\sin\frac{x-\left(x-\frac{\pi}{3}\right)}{2}\cos\frac{x+\left(x-\frac{\pi}{3}\right)}{2}\\
&=2\sin\frac{\pi}{6}\cos\left(x-\frac{\pi}{6}\right)
\end{align} That is we have to find the roots of $\cos\left(x-\frac{\pi}{6}\right)$, which are where $$x-\frac{\pi}{6}=k\pi+\frac{\pi}{2}$$ or $$x=k\pi+\frac{2\pi}{3}$$
obtaining $x=-\frac43\pi,-\frac13\pi,\frac23\pi$ and $\frac53\pi$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
LU decomposition- Suppose we have to find LU decomposition of
$A = \begin{bmatrix}
3 & 1 & -1 \\
1 & 0 & 1 \\
4 & 2 & 2
\end{bmatrix}$
I correctly found $M_{1}=\begin{bmatrix}
1 & 0 & 0 \\
-4/3 & 1 & 0 \\
-1/3 & 0 & 1
\end{bmatrix} $ and $M_{2}=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1/2 & 1
\end{bmatrix} $
The problem happens when I try to find $L=(M_{2}M_{1})^{-1}$
In my notes, its done as $\Bigg(\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1/2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 \\
-4/3 & 1 & 0 \\
-1/3 & 0 & 1
\end{bmatrix}\Bigg)^{-1}$
Which equals $\begin{bmatrix}
1 & 0 & 0 \\
-4/3 & 1 & 0 \\
-1 & 1/2 & 1
\end{bmatrix}^{-1}$ which is $\begin{bmatrix}
1 & 0 & 0 \\
4/3 & 1 & 0 \\
1 & -1/2 & 1
\end{bmatrix}$
I used the fact that $(AB)^{-1}=B^{-1}A^{-1}$ which then works to the following:
$\begin{bmatrix}
1 & 0 & 0 \\
4/3 & 1 & 0 \\
1/3 & 0 & 1
\end{bmatrix}$ $\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & -1/2 & 1
\end{bmatrix}$ which equals $\begin{bmatrix}
1 & 0 & 0 \\
4/3 & 1 & 0 \\
1/3 & -1/2 & 1
\end{bmatrix}$
Its almost the same answer as my notes, but the value at $a_{31}$ is different. For me I am getting $a_{31}=\frac{1}{3}$, but in my notes $a_{31}=1$. Every other value matches.
I am really not sure where I am going wrong, can someone please help?
|
$\begin{bmatrix}
1 & 0 & 0 \\
-4/3 & 1 & 0 \\
-1 & 1/2 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
4/3 & 1 & 0 \\
1 & -1/2 & 1
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
2/3 & 0 & 1
\end{bmatrix}.$
Thereore
$\begin{bmatrix}
1 & 0 & 0 \\
-4/3 & 1 & 0 \\
-1 & 1/2 & 1
\end{bmatrix}^{-1}\ne
\begin{bmatrix}
1 & 0 & 0 \\
4/3 & 1 & 0 \\
1 & -1/2 & 1
\end{bmatrix}
$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What number do you remove from $1!2!\cdots 99!100!$ to get a perfect square?
The title says it all, there is a product, as shown above, one of the factorials must be removed and the product will make a perfect square. Which one?
For example, you could remove $54!$?
|
First of all, you can write every $(n!)$ as $(n-1)!\cdot n$.
However, for now, just do this for all the odd values of $n$ and you get
$$(2!)(2!)\cdot 3 \cdot(4!)(4!)\cdot 5\cdots (98!)(98!)\cdot 99\cdot 100!$$
Now call $(2!)(4!)\cdots (98!)=C$ to simplify and your expression is equal to
$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot (100!)$$
Write out $100!$ and that's equal to
$$C^2\cdot 3\cdot 5\cdot 7\cdots 99\cdot 2\cdot 3\cdot 4\cdots 100$$
Rearrange order to get
$$C\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdot 7\cdots 99\cdot 2\cdot 4\cdot 6\cdots 100$$
Now call $3\cdot 5\cdot 7\cdots 99=D$ and that's equal to
$$C^2D^2 2\cdot 4\cdot 6\cdots 100 = C^2D^2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 2\cdot 4\cdots 2\cdot 50$$
Group all the twos together and you see that's equal to $$C^2D^22^{50}\cdot 50!$$
|
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|
Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+11$ . Where did I go wrong ? Thanks !!
|
There are a few issues here - some stray parentheses and an incorrectly written derivative which became corrected on the next step. I'd proceed like this:
\begin{align*}
\frac d{dx} (2x-1)(x+3)^{1/2} &= \left[\frac d{dx} (2x-1)\right] \cdot (x+3)^{1/2} + (2x-1) \cdot \frac d{dx} (x+3)^{1/2}\\[0.3cm]
&= 2 (x+3)^{1/2} + (2x-1) \cdot \frac12(x+3)^{-1/2} \cdot \frac d{dx}(x+3)\\[0.3cm]
&= 2 (x+3)^{1/2} + \frac{2x-1}{2(x+3)^{1/2}}\\[0.3cm]
&= 2 (x+3)^{1/2}\cdot \color{red}{\frac{2(x+3)^{1/2}}{2(x+3)^{1/2}}} + \frac{2x-1}{2(x+3)^{1/2}}\\[0.3cm]
&= \frac{4(x+3) + 2x-1}{2\sqrt{x+3}}\\[0.3cm]
&= \frac{6x+11}{2\sqrt{x+3}}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
The sum of consecutive integers is $50$. How many integers are there? I started off by calling the number of numbers in my list "$n$". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were "$n$" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.
|
If your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$
If $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.
If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$
If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$
So there are 4 answers: $n=1, 4, 5, $ and $20$.
Edit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.
Note that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.
|
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|
Series $\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right]$ How can we find the sum of this hard series any hint please
My trial is to express zeta(8) as a series but I don't know how and what to do
$$\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right]$$
|
We shall begin by calculating the multivariate zeta value $\zeta(7,1)$ so we have it to hand later. Using "well known formula" and Euler reflection formula, we have
\begin{eqnarray*}
\zeta(8)&=&\zeta(7,1)+\zeta(6,2)+\zeta(5,3)+\zeta(4,4)+\zeta(3,5)+\zeta(2,6) \\
\zeta(2)\zeta(6)&=&\zeta(8)+\zeta(6,2)+\zeta(2,6) \\
\zeta(3)\zeta(5)&=&\zeta(8)+\zeta(5,3)+\zeta(3,5) \\
\zeta(4)^2&=&\zeta(8)+2\zeta(4,4). \\
\end{eqnarray*}
Linear algebra gives $\zeta(7,1)=\frac{1}{2}(7\zeta_8-2\zeta_6\zeta_2 -2\zeta_5\zeta_3 -\zeta_4^2)$ & recall that $\zeta_2=\frac{\pi^2}{6},\zeta_4=\frac{\pi^4}{90},\zeta_6=\frac{\pi^6}{945},\zeta_8=\frac{\pi^8}{9450}$.
OK ... the question
\begin{eqnarray*}
\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right]
= \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{H_n}{m^8} \\
= \sum_{n=1}^{\infty} \sum_{i=1}^{\infty} \frac{H_n}{(i+n)^8} = \sum_{n=1}^{\infty} \sum_{i=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(i+n)^8}
\end{eqnarray*}
Invert the order of the $j$ and $n$ plums, substitute $n=j+k-1$,
\begin{eqnarray*}
\sum_{i=1}^{\infty} \sum_{n=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(i+n)^8} =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=j}^{\infty}\frac{1}{j(i+n)^8} =\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{j(i+j+k-1)^8}
\end{eqnarray*}
substitute $p=i+k-1$ & partial fractions $\frac{p}{j(j+p)}=\frac{1}{j}-\frac{1}{p+j}$
\begin{eqnarray*}
\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{j(i+j+k-1)^8} =\sum_{p=1}^{\infty} \sum_{j=1}^{\infty}\frac{p}{j(p+j)^8} =\sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{j(p+j)^7}-\sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{(p+j)^8}
\end{eqnarray*}
note that the penultimate sum is $\zeta(7,1)$ & substitute $q=p+j$ in the last sum
\begin{eqnarray*}
\sum_{p=1}^{\infty} \sum_{j=1}^{n}\frac{1}{(p+j)^8} = \sum_{q=1}^{\infty} \frac{q-1}{q^8} = \zeta(7)- \zeta(8).
\end{eqnarray*}
And so we have
\begin{eqnarray*}
\sum_{n=1}^{\infty} H_n\left[\zeta(8)-\frac{1}{1^8}-\frac{1}{2^8}-\frac{1}{3^8}\cdots\frac{1}{n^8}\right] = \color{red}{\frac{\pi^8}{4200}-\zeta(3)\zeta(5)-\zeta(7)}.
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration to solve this problem? and if I should use spherical or double polar or cylindrical coordinate system?
Also how to find the limits of integration in this particular problem as an example?
This is what the cross section looks like
It is clearly visible that $\theta$ goes from $0$ to $\pi$
|
$\textbf{Third method}$: find the volume of the cylinder, where $0\leq z\leq 4$, and then subtract from it the volume of the cylinder whose lower bound is the paraboloid and upper bound is the plane $z=4$.
So rather than working with the equations:
$$
\color{blue}{z=0, \hspace{4mm} z=x^2+y^2, \hspace{4mm} \mbox{ and } \hspace{4mm} x^2+(y-1)^2=1,}
$$
imagine picking up the solid and then moving it so that the axis of rotation for the cylinder $\underline{\textbf{is}}$ the $z$-axis. This solid is now bounded by the equations:
$$
\color{green}{z=0, \hspace{4mm} z=x^2+(y+1)^2, \hspace{4mm} \mbox{ and } \hspace{4mm} x^2+y^2=1,}
$$
which has been plotted below:
The volume of the cylinder is given by
$$
\text{Volume}(\text{cylinder}) = \pi R^2 h = \pi 1^2(4) = 4\pi.
$$
The volume of the cylinder $E'$ whose lower bound is the paraboloid and upper bound is the plane $z=4$ is
$$
\begin{align*}
\text{Volume}(E')
&= \int_0^{2\pi} \int_{0}^{1}\int_{(r\cos\theta)^2+(r\sin\theta+1)^2}^{4}r\: dz dr d\theta \\
&= \int_0^{2\pi} \int_{0}^{1}
4r-\left( r^3\cos^2\theta + r(r\sin\theta+1)^2\right) \: dr d\theta \\
&= \int_0^{2\pi} \int_{0}^{1}
\left( 3 r - r^3 (\cos^2\theta + \sin^2\theta )- 2 r^2 \sin\theta \right) \: dr d\theta \\
&= \int_0^{2\pi} \int_{0}^{1}
\left( 3 r - r^3 - 2 r^2 \sin\theta \right) \: dr d\theta \\
&= \int_0^{2\pi}
\left( \frac{3}{2} r^2 - \frac{1}{4}r^4 - \frac{2}{3} r^3 \sin\theta \right) \Bigg|_0^1 \: d\theta \\
&= \int_0^{2\pi}
\left( \frac{5}{4} - \frac{2}{3} \sin\theta \right) \: d\theta \\
&= \frac{5}{4}\theta +\frac{2}{3} \cos\theta\Bigg|_0^{2\pi} \\
&= \frac{5}{2}\pi.
\end{align*}
$$
Thus
$$
\text{Volume}(\text{solid}) = 4\pi - \frac{5}{2}\pi = \boxed{\frac{3}{2}\pi}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Suppose $f(x)$ is a real valued polynomial function of degree $6$ satisfying the following conditions
Suppose $f(x)$ is a real-valued polynomial function of degree $6$ satisfying the following conditions
(a) $f$ has minimum value at $x=0$ and $2$
(b) $f$ has maximum value at $x=1$
(c) $\displaystyle\lim_{x\to 0}\frac{1}{x}\ln\,$$\det\pmatrix{\frac{f(x)}{x}&1&0\\0&\frac{1}{x}&1\\1&0&\frac{1}{x}} = 2$
Determine $f(x)$.
I let
$$f(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$$
and I took $f'(0)=0$, $f'(2)=0$, $f'(1)=0$ and $$\lim_{x\to 0}\frac{1}{x}\ln(\frac{f(x)}{x^3}+1)=2$$ But I am not able to solve it further.
|
Finding such a polynomial is not terribly difficult. Starting with
$$p(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g,$$
$p'(0)=0$ implies $f=0$, and in order for $\log(1+p(x)/x^3)/x$ to tend to $2$ when $x\to 0$, we need $\log(1+p(x)/x^3)$ to behave like $2x$ when $x\to 0$, so we could ask $p(x)/x^3$ to behave like $2x$ when $x\to 0$, because we know that $\log(1+\xi)\approx \xi$ for small $\xi$. So if we set $c=2,d=e=f=0$, we get:
$$p(x)=ax^6+bx^5+2x^4$$
and so
$$\lim_{x\to 0}\frac{\log(1+\frac{ax^6+bx^5+2x^4}{x^3})}{x}=\lim_{x\to 0}\frac{\log(1+2x)}{x}=2$$
Now, the rest of the given information helps to determine the coefficients $a,b$. Namely:
$$p'(1)=0\Longrightarrow 8+6a+5b=0,\enspace p'(2)=0\Longrightarrow 64+192a+80b=0$$
solving these two equations gives $a=\frac{2}{3}, b=-\frac{12}{5}$, hence a polynominal satisfying the requirements is given by:
$$p(x)=\frac{2x^6}{3}-\frac{12x^5}{5}+2x^4$$
To prove that only this polynomial satisfies the requirements takes a little more effort.
|
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|
For how many integers $(3n)^4-(n-10)^4$ is a perfect square?
For how many integers $n$ the expression
$$(3n)^4-(n-10)^4$$
becomes a perfect square?
$$$$One way is: Let $x=3n$ and $y=n-10$ to get $$x^4-y^4=z^2.$$ This equation does not have non-trivial solutions in integers (I do not want to discuss about this).
I was trying to solve this problem with simpler elementary methods. For example we can write the equation as
$$40(n+5)(2n-5)(n^2-2n+10)=m^2.$$
Observe that $40 \mid m^2$. Thus $20 \mid m$. Write $m=20k$ to get
$$(n+5)(2n-5)(n^2-2n+10)=10k^2,$$
Where $k$ is a non-negative integer. Look mod $5$. We get $n^3(n-2) \equiv 0 \pmod{5}$. So we have either $n \equiv 0 \pmod{5}$ or $n \equiv 2 \pmod{5}$.
If $n \equiv 0 \pmod{5}$ then let $n=5N$ to get
$$125(N+1)(2N-1)(5N^2-2N+2)=10k^2$$
It follows that $25 \mid k^2$ and $k=5K$. Now we have the following equation $$(N+1)(2N-1)(5N^2-2N+2)=2K^2$$
Is this getting simpler or harder this way? Do you have a suggestion?
|
Using
$$\gcd(N+1,2N-1)=\gcd(N+1,-3)\mid 3$$
$$\gcd(N+1,5N^2-2N+2)=\gcd(N+1,-7N+2)=\gcd(N+1,9)\mid 9$$
$$\gcd(2N-1,5N^2-2N+2)=\gcd(2N-1,5)\mid 5 $$
we see that the factors are almost coprime, hence must almost (i.e., up to small factors involving at most a $2$, a $3$, and/or a $5$) be square.
Then from $\frac{2N-1}{N+1}\approx 2$ you might arrive at a contradiction.
|
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|
What is the limit of the following function? $\lim_{n\rightarrow \infty }\left ( n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right )$
|
Note that
$$
{{\Gamma (n/2)} \over {\Gamma (n/2 - 1/2)}} = \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} }
$$
where $x^{\,\overline {\,a\,}} $ denotes the Rising Factorial (rising Pochammer).
It is an increasing function for $1<n$.
Because of that and considering the rules for summing the exponents of the Rising Factorial, we have:
$$
\left( {n/2 - 1/2} \right) = \left( {n/2 - 1/2} \right)^{\,\overline {\,1\,} } = \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2} \right)^{\,\overline {\,\,1/2\,\,} } > \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}
$$
and
$$
\left( {n/2 - 1} \right) = \left( {n/2 - 1} \right)^{\,\overline {\,1\,} } = \left( {n/2 - 1} \right)^{\,\overline {\,\,1/2\,\,} } \left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2}
$$
which means
$$
\left( {n/2 - 1} \right) < \left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} < \left( {n/2 - 1/2} \right)
$$
and therefore we know that the given function is bound between $0$ and $1$
$$
0 < n - 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,\,1/2\,\,} } } \right)^{\,2} < 1\quad \left| {\;1 < n} \right.
$$
The Stirling series for the rising factorial is
$$
z^{\,\overline {\,w\,} } \propto z^{\,w} \left( {1 + {{w\left( {w - 1} \right)} \over {2\,z}} + O\left( {{1 \over {z^{\,2} }}} \right)} \right)\quad \left| \matrix{
\;\left| z \right| \to \infty \hfill \cr
\;\left| {\arg (z + w)} \right| < \pi \hfill \cr} \right.
$$
so we get
$$
\eqalign{
& \left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } \propto \left( {n/2 - 1/2} \right)^{\,1/2\,} \left( {1 + {{\,1/2\,\left( {\,1/2\, - 1} \right)} \over {2\,\left( {n/2 - 1/2} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) = \cr
& = \left( {{{n - 1} \over 2}} \right)^{\,1/2\,} \left( {1 - {{\,1} \over {4\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) \cr}
$$
and finally
$$
\eqalign{
& n + 1 - 2\left( {\left( {n/2 - 1/2} \right)^{\,\overline {\,1/2\,} } } \right)^{\,2} \propto n + 1 - \left( {n - 1} \right)\left( {1 - {{\,1} \over {2\,\left( {n - 1} \right)}} + O\left( {{1 \over {n^{\,2} }}} \right)} \right) = \cr
& = {1 \over 2} + O\left( {{1 \over n}} \right) \cr}
$$
|
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|
If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to proceed anymore. Thanks in advance.
|
If $x=-1$ then $\left|\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\right|=\frac{2}{3}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\left|\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\right|\leq\frac{2}{3}$$ or
$$-\frac{2}{3}\leq\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\leq\frac{2}{3}$$ or
$$\frac{2}{3}\leq\frac{x^2+1}{x^2+x+1}\leq2,$$
which is true because the left inequality it's $(x-1)^2\geq0$
and the right inequality it's $(x+1)^2\geq0$.
Done!
|
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"url": "https://math.stackexchange.com/questions/2341689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find coefficient of $x^2$ in a complicated expansion
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$
I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there I am stuck when it comes to expanding $(1+2x+3x^2)^6$ as I can't see an easy way to use the binomial theorem (for example by factorising $1+2x+3x^2$, which I can't do as there are no real roots). Is there a way to find the coefficient of $x^2$ in this expansion (specifically $(3x^2+2x+6)^6$ as I can get the answer from there)?
|
For a question like this cheating is legal :)
$\left(4-x^2\right) \left[\left(4 x^3+1\right)^5+\left(3 x^2+2 x+1\right)^6\right]=-1024 x^{17}+4096 x^{15}-2009 x^{14}-2916 x^{13}+1718 x^{12}+1844 x^{11}+15417 x^{10}+31144 x^9+33976 x^8+29832 x^7+20681 x^6+10488 x^5+4302 x^4+1428 x^3+310 x^2+48 x+8$
The answer is $310$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following inequality is used:
$$(1+x)^n \ge \frac{n(n-1)}{2}x^2$$
For $v_n$:
$${n \choose k}p^k > \frac{n^kp^k}{2^kk!}\quad \quad(n>2k) $$
I have painstakingly come up with proofs for both of those inequalities, however, and this is especially true for the second inequality, my proofs are not very elegant nor do they give me any idea of the inequality came to be.
Surely Rudin doesn't pull these inequalities out of thin air. Where can I find a more comprehensive derivation of these inequalities? Where do they come from exactly (which field)?
Update:
Here are my proofs. They both rely on induction.
For the first one:
Suppose $x\ge0$. Then, for $n = 2$:
$$ (1+x)^2 = 1+2x+x^2 $$
$$\frac{2(2-1)}{2} x^2 = x^2$$
Since $x\ge0$, we have $(1+x)^n \ge \frac{n(n-1)}{2}x^2 $ for n = 2.
Suppose our inequality is true for a certain n. Then:
$$ (1+x)^n + nx^2 \ge \frac{n(n+1)}{2}x^2$$
Let $D = (1+x)^{n+1} - (1+x)^n - nx^2 $. Since $\forall n \in \mathbb{N}, \forall m = 0,1, ... n, {n+1 \choose m} \ge {n \choose m} $, and ${n+1 \choose 2} - {n \choose 2} - n = 0$, we have:
$$D \ge 0 \Rightarrow (1+x)^{n+1} \ge (1+x)^n + nx^2$$
So:
$$(1+x)^{n+1} \ge \frac{n(n+1)}{2}x^2$$ The theorem follows by induction.
For the second inequality:
We want to prove that $n(n-1)...(n-k+1) > \frac{n^k}{2^k}$, with $n > 2k$
Let $n\in\mathbb{N}, k$ and integer greater than 0 such that $2n > k$. Then for $k=1$, we have: $$n > \frac{n}{2} $$
Now suppose our inequality holds for a certain $k$. Then:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k)$$
Since $n > 2k$, $n-k > \frac{n}{2} > 0$ such that:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k) > \frac{n^{k+1}}{2^{k+1}} $$
By induction, our inequality holds for all $k$ such that $n > 2k$
|
Let $f(x)=(1+x)^n-1-\frac{n(n-1)}{2}x^2$, where $x\geq0$.
For $n=1$ we see that $f(x)\geq0$.
But for $n\geq2$ we obtain: $f''(x)=n(n-1)(1+x)^{n-2}-n(n-1)\geq0$, which gives
$f'(x)=n(1+x)^{n-1}-n(n-1)x\geq f'(0)>0$ and
$f(x)\geq f(0)=0$ and we are done!
|
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|
Prove that $S_i<1$ for all $i$ by induction We define $x_1=\frac12$, and $x_n=(1-\frac{3}{2n})\cdot x_{n-1}$
Then we define $S_i=x_1+x_2+\cdots+x_i$
Prove that $S_i<1$ for all $i$
I can see that $x_n=(1-\frac{3}{2n})\cdot (1-\frac{3}{2n-2})\cdots (1-\frac 34)\cdot \frac12$
But what then? I suppose my observation isn't useful though...
|
I think it will help to observe,
\begin{align*}
x_n &= \frac{1}{2} \cdot \frac{2n - 3}{2n} \cdot \frac{2n - 5}{2n - 2} \cdot \ldots \cdot \frac{1}{4} \\
&= \frac{(2n - 3)(2n - 5) \ldots 1}{(2n)(2n - 2)(2n - 4) \ldots 2} \\
&= \frac{1}{2^n n!}(2n - 3)(2n - 5) \ldots 1 \\
&= \frac{1}{2^n n!} \cdot \frac{(2n - 2)(2n - 3)(2n - 4) \ldots 1}{(2n - 2)(2n - 4) \ldots 2} \\
&= \frac{1}{2^n n!} \cdot \frac{(2n - 2)!}{2^{n-1}(n-1)!} \\
&= \frac{(2n - 2)!}{2 \cdot 4^{n-1} n ((n - 1)!)^2} \\
&= \frac{\binom{2n - 2}{n - 1}}{n 2^{2n-1}} \\
&= \frac{C(n - 1)}{2^{2n - 1}},
\end{align*}
where $C(n)$ is the $n$th Catalan number. According to Wikipedia, the Catalan numbers have the following generating function:
$$c(x) = \frac{1-\sqrt{1-4x}}{2x}.$$
With any luck, we should have,
$$\sum_{n=1}^\infty x_n = \frac{1}{2}c\left(\frac{1}{4}\right) = 1,$$
provided the Maclaurin Series converges (removing the removable discontinuity at $0$) where we want. Unfortunately, this value lies right on the boundary of the radius of convergence, but it does mean that this sum equalling $1$ is equivalent to the Maclaurin series of $\sqrt{1 - x}$ converging to $0$ at $x = 1$. It seems like a more elementary problem, but I don't personally have a proof that it's true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
BMO2 1997 - Combinatorics Find the number of polynomials of degree 5 with distinct coefficients from the set {1, 2, 3, 4, 5, 6, 7, 8} that are divisible by $x^2-x+1$.
I tried multiplying $x^2-x+1$ by $ax^3+bx^2+cx+d$ to get $ax^5+(b-a)x^4+(a-b+c)x^3+(b-c+d)x^2+(c-d)x+d$, but I am struggling to count from here.
|
Filling in the details for @Gerry Myerson's answer:
WLOG assume $d>a$ and $d < b < f$. We will count the number of pairs, and multiply by $3! \cdot 2 = 12$ to account for the symmetry. Do casework on the value of $d-a$.
If $d-a = 1$, then there are $10$ possible values for $(a,d,e,b,c,f)$:
$$(1,2,3,4,5,6), (1,2,3,4,6,7),(1,2,3,4,7,8),(1,2,4,5,6,7),(1,2,4,5,7,8)\\
(1,2,5,6,7,8),(2,3,4,5,6,7),(2,3,4,5,7,8),(2,3,5,6,7,8),(3,4,5,6,7,8) $$
If $d-a = 2$, then there are $6$ possible values for $(a,d,e,b,c,f)$:
$$ (1,3,2,4,5,7),(1,3,2,4,6,8),(1,3,4,6,5,7),(1,3,5,7,6,8),(2,4,3,5,6,8),(2,4,5,7,6,8) $$
If $d-a = 3$, then there are $4$ possible values for $(a,d,e,b,c,f)$:
$$ (1,4,2,5,3,6), (1,4,3,6,5,8), (2,5,3,6,4,7), (3,6,4,7,5,8) $$
If $d-a = 4$, then there are $4$ possible values for $(a,d,e,b,c,f)$:
$$ (1,5,2,6,3,7), (1,5,2,6,4,8), (1,5,3,7,4,8), (2,6,3,7,4,8) $$
If $d-a = 5$, there is one possible value for $(a,d,e,b,c,f)$:
$$ (1,6,2,7,3,8)$$
This gives $25$ total tuples. So there are $12 \cdot 25 = 300$ different possibilities.
Confirmation that the answer is $300$ using a Python program:
import itertools
def main():
good = []
for l in itertools.permutations(range(1,9),4):
a = l[0]
b_a = l[1]#b-a
c_d = l[2]#c-d
d = l[3]
ambpc = c_d + d - b_a#a-b+c
bmcpd = b_a + a - c_d#b-c+d
if ambpc > 0 and ambpc < 9 and bmcpd > 0 and bmcpd < 9:
if ambpc != bmcpd:
if ambpc not in l and bmcpd not in l:
good.append([a,b_a,ambpc,bmcpd,c_d,d])
print(good)
print(len(good))
if __name__=="__main__":
main()
It works by settings the values for $a, b-a, c-d, d$ from the set $\{ 1, \cdots, 8\}$, and then seeing if the corresponding values of $a-b+c = (c-d) + d - (b-a)$ and $b-c+d = (b-a) + a - (c-d)$ are distinct and lie in the set $\{ 1, \cdots 8\}$.
|
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|
An unusual identity Prove that $$\left(\frac{\sqrt[3]{81+33\sqrt6}+\sqrt[3]{81-33\sqrt6}}{6}\right)^m=\left(\frac{\sqrt[5]{41+29\sqrt2}+\sqrt[5]{41-29\sqrt2}}{2}\right)^n$$ for all natural integers $m,n$.
|
Hint: for the LHS let $\,a=\sqrt[3]{81+33\sqrt6}\,, \,b=\sqrt[3]{81-33\sqrt6}\,$ then $\,a^3+b^3=162\,$ and $\,ab=3\,$, so: $$\,(a+b)^3=a^3+b^3+3ab(a+b)=162+9(a+b)\,$$
But the equation $x^3-9x-162=0$ has only one real root which is $x=6$, therefore $a+b=6$.
[ EDIT ] As pointed out by @LordSharktheUnknown, the RHS would need to be $5^{th}$ roots in order for the equality to hold, rather than cube roots as posted.
An argument like the above works in that case, too. Let $a=\sqrt[5]{41+29\sqrt2}\,$, $b=\sqrt[5]{41-29\sqrt2}$ then $a^5+b^5=82$ and $ab=-1\,$, so:
$$
\begin{align}
(a+b)^5 &= a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b) \\
&= a^5+b^5 +5ab\big((a+b)^3-3ab(a+b)\big)+10a^2b^2(a+b) \\
&= 82 - 5\big((a+b)^3+3(a+b)\big)+10(a+b) \\
&= 82 - 5(a+b)^3 - 5(a+b)
\end{align}
$$
But the equation $x^5+5x^3+5x-82=0$ has only one real root $x=2$, therefore $a+b=2$.
|
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|
Using cylindrical coordinates to evaluate the volume of a sphere external to an ellipsoid I need to use cylindrical coordinates to evaluate the volume of the sphere $$x^{2}+y^{2}+z^{2} = a^{2}$$ that's external to the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2} = 1$.
So, $x$ and $y$ are equal for both, and i got $\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{a^2} = 1$ for the sphere but I don't know how to proceed.
|
What you need is to compute the following integral
$$\int\int\int_{\Delta V}r dr d\theta dz$$ in cylindrical coordinates, where $\Delta V$ is your volume of integration. The first thing I notice is that $b<a$, otherwise the sphere is inside the ellipsoid. So now all we need is to find the limits of this integration volume. Due to symmetry in the $xy$ plane, the limits for the $\theta$ integration are from $0$ to $2\pi$. In principle $z$ has to be between $-a$ and $a$. Note however that, if you draw a cross section ($xz$ plane for example), you can identify regions along the $z$ axis where only the sphere is contributing, or you have both the sphere and the ellipse at the same coordinate. The ellipse extends from $-b$ to $b$.
Now for the limits of $r$: if $z>b$ or $z<-b$, you get $0\le r\le \sqrt{a^2-z^2}$. For the region between $-b$ and $b$, you need to be outside of the ellipse, so $\sqrt{a^2-z^2\frac{a^2}{b^2}}\le r \le \sqrt{a^2-z^2}$. This way you volume is $$2\pi\int_{-a}^{-b}dz\int_0^{\sqrt{a^2-z^2}}rdr+2\pi\int_{-b}^{b}dz\int_\sqrt{a^2-z^2\frac{a^2}{b^2}}^{\sqrt{a^2-z^2}}rdr+2\pi\int_{a}^{b}dz\int_0^{\sqrt{a^2-z^2}}rdr$$ Just for convenience, you can rewrite this as $$2\pi\int_{-a}^{a}dz\int_0^{\sqrt{a^2-z^2}}rdr-2\pi\int_{-b}^{b}dz\int_0^{\sqrt{a^2-z^2\frac{a^2}{b^2}}}rdr$$
That is to say volume of the sphere minus the volume of the ellipsoid. Your first integral in this form has to be $\frac{4}{3}\pi a^3$. The second integral is $\frac{4}{3}\pi a^2b$
|
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Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality:
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$.
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)=y(let)$$
Next using AM-GM inequality
$$\cos\left(\frac{A}{2}\right)+\cos\left(\frac{B}{2}\right)+\cos\left(\frac{C}{2}\right)\geq 3 \left(\frac{y}{4}\right)^{\frac{1}{3}}.$$
I'm unable to proceed further. Please help me.
|
Let $a$, $b$ and $c$ be sides-lengths of the triangle and $S$ be an area of the triangle.
We need to prove that
$$\sum_{cyc}\frac{2S}{bc}\leq\frac{3\sqrt3}{2}$$ or
$$4S(a+b+c)\leq3\sqrt3abc$$ or
$$(a+b+c)^3(a+b-c)(a+c-b)(b+c-a)\leq27a^2b^2c^2.$$
Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$(x+y+z)^3xyz\leq\frac{27}{64}(x+y)^2(x+z)^2(y+z)^2.$$
Now, $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\Leftrightarrow$$
$$\Leftrightarrow\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$$
which is true by AM-GM.
Indeed, $$\sum_{cyc}(x^2y+x^2z)=\sum_{cyc}(y^2z+x^2z)\geq\sum_{cyc}(2\sqrt{y^2z\cdot x^2z})=2\sum_{cyc}xyz.$$
Thus, it remains to prove that
$$(xy+xz+yz)^2\geq3xyz(x+y+z),$$
which is AM-GM again.
Indeed, $$(xy+xz+yz)^2-3xyz(x+y+z)=\sum_{cyc}(x^2y^2-x^2yz)=\frac{1}{2}\sum_{cyc}(2x^2y^2-2x^2yz)=$$
$$=\frac{1}{2}\sum_{cyc}(x^2y^2+x^2z^2-2x^2yz)\geq\frac{1}{2}\sum_{cyc}\left(2\sqrt{x^2y^2\cdot x^2z^2}-2x^2yz\right)=0.$$
Done!
|
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|
What is equivalent that logical function?
(A+B) => (B XOR C)
=> is Implication
Options:
BC+!A+!B+!C
!B!C+ABC
B!C+A+!BC
!BC+!ABC
BC+A!B!C
! is NOT
Please tell me how it is solved.
I do not think that this requires knowledge of discrete mathematics.
But I did not find any simple options
|
$$(A+B) \Rightarrow (B \ XOR \ C) = \text{ (Rewrite XOR)}$$
$$(A+B) \Rightarrow (B!C + !BC) = \text{ (Rewrite} \Rightarrow )$$
$$!(A + B)+B!C + !BC = \text{ (DeMorgan)}$$
$$!A!B+B!C+!BC$$
.... which is not any of the options you are providing ...
$BC + !A +!B+!C=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$
$!B!C+ABC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
$B!C + A +!BC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
$!BC+!ABC=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$
$BC+A!B!C=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$
Are you sure you wrote this all correctly from your exercise?
EDIT
Aha! So the original is:
$$!((A+B) \Rightarrow (B \ XOR \ C))$$
So then you get:
$$!((A+B) \Rightarrow (B \ XOR \ C)) = \text{(Rewrite XOR)}$$
$$!((A+B) \Rightarrow (B!C + !BC)) = \text{ (Rewrite} \Rightarrow )$$
$$!(!(A + B) + B!C + !BC) = \text{ (DeMorgan)}$$
$$(A + B)!(B!C + !BC) = \text{ (DeMorgan)}$$
$$(A + B)(!(B!C)!(!BC)) = \text{ (DeMorgan)}$$
$$(A + B)(!B+C)(B+!C) = \text{ (Distribution)}$$
$$A!BB+A!B!C+ACB+AC!C+B!BB+B!B!C+BCB+BC!C = \text{ (Complement)}$$
$$A0+A!B!C+ACB+A0+0B+0!C+BCB+B0 = \text{ (Annihilation)}$$
$$0+A!B!C+ACB+0+0+0+BCB+0 = \text{ (Identity)}$$
$$A!B!C+ACB+BCB = \text{ (Commutation)}$$
$$A!B!C+ABC+BBC = \text{ (Idempotence)}$$
$$A!B!C+ABC+BC = \text{ (Absorption)}$$
$$A!B!C+BC = \text{ (Commutation)}$$
$$BC+A!B!C$$
|
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|
Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$
I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
|
Derivation using series the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$.
The integrand is an even function & break the interval at $1$ and we have
\begin{eqnarray*}
\int_{-\infty}^{\infty} \frac{dx}{1+x^2} =2 \int_{0}^{\infty} \frac{dx}{1+x^2} =2 \int_{0}^{1} \frac{dx}{1+x^2}+2 \underbrace{\int_{1}^{\infty} \frac{dx}{1+x^2}}_{x \rightarrow \frac{1}{x}}=4 \int_{0}^{1} \frac{dx}{1+x^2}
\end{eqnarray*}
Now geometrically expand the integrand
\begin{eqnarray*}
\frac{1}{1+x^2} =\sum_{i=0}^{\infty} (-1)^n x^{2n}
\end{eqnarray*}
interchange the order of the integration & sum ... & then perform the each integration.
|
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|
Finding the domain and the range of this function Problem: Find the domain and range of the function $f(x)=\log \sqrt{4-x^2}$
My attempt:
For domain
$y=f(x)=\log \sqrt{4-x^2}$
$\implies y=\log \sqrt{4-x^2}$
Since
logarithm of only positive numbers is possible,
$4-x^2>0$
$\implies x^2-4<0$
$\implies(x+2)(x-2)<0$
$\implies -2<x<2$
Therefore $D_f=(-2,2)$
For range
$y=f(x)=\log \sqrt{4-x^2}$
$y=\log \sqrt{4-x^2}$
$y=\frac{1}{2}\log (4-x^2)$
$2y=\log (4-x^2)$
$e^{2y}=4-x^2$
$x=\sqrt{4-e^{2y}}$
Since the exponent can take any real value, $R_f=(-\infty,+\infty)$
My problem: The answer given in my book for $D_f$ matches with my $D_f$ but $R_f=(-\infty,\log 2]$ according to my book. Where have i gone wrong?
|
$\sqrt {4-x^2}$ cannot be greater than $\sqrt 4=2$, so $\log \sqrt{4-x^2}$ cannot be any greater than $\log \sqrt 4=\log 2$. It achieves that value when $x=0.$ As $\sqrt{4-x^2}$ can achieve $0$, you can have $\log \sqrt{4-x^2}$ as negative as you want, so the range is $(-\infty, \log 2]$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find all solutions of $9^x-4^y=5^z$ over the naturals Find all solutions of $9^x-4^y=5^z$ over the naturals.
My attempt: We have
$${(3^x)}^2-{(2^y)}^2=5^z \Rightarrow (3^x+2^y)(3^x-2^y)=5^z.$$
It can be easily proved that $\gcd (3^x+2^y,3^x-2^y)=1$ so we have
$$3^x-2^y=1,3^x+2^y=5^z.$$
What do we have to do from here?
|
As @ Xam said,
if $y=0$, then there is no solution,
if $y=1$, then $x=1$.
Now if $y$ is greater than $1$, then $2^y$ is divisible by $4$.
using modular arithmetic module $4$ we have:
$(-1)^x-0$ is congruent to $1$ module $4$, so $x$ must be even, i.e. $x=2k$.
Now $2^y=3^{2k}-1=(3^k-1)(3^k+1)$, and by the same conclusion you have done, as you explained, we can conclude that:
$(3^k-1)=2$ and $(3^k+1)=2^{y-1}$,
so $k=1$ and $x=2$ and $y-1=2$.
But this values for $x$ and $y$ does not give any solution to the main equation.
Another soloution:
We claim that $y$ could not be greater than $1$.
Suppose on contrary, that y is greater than $1$, then $4^y$ is divisible by $8$.
Using modular arithmetic module $8$ we have:
$1-0$ is congruent to $5^z$ module $8$, so $z$ must be even, i.e. $z=2t$.
Again by considering the main equation module $3$, we have :
$0-1$ is congruent to $(-1^2)^t$ module $3$, i.e. $-1=1$ module $3$, which is a contradiction.
So we have two cases:
*
*$y=1$, then $(3^x-2)(3^x+2)=5^z$, and by the same conclusion as you have been used, we can conclude that:
$(3^x-2)=1$ and $(3^x+2)=5^z$, which gives:
$x=1$ and $z=1$.
*$y=0$, then $(3^x-1)(3^x+1)=5^z$, we can conclude that:
$(3^x-1)=2$ and $(3^x+1)=(1/2)5^z$, which does'nt gives any solution.
|
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|
Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?
|
Classic long method:
Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to
\begin{align}
f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\
f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\
f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b
\end{align}
from which $b = -5$, $c = 11$, and $d = -6$ and
$$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$
With $f(x)$ then
\begin{align}
f(4) &= 64 - 80 + 55 -6 = 22 \\
f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}.
\end{align}
It may also be noticed that $f(x)$ can be seen in the form
$$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$
From this it is easy to see that
\begin{align}
f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\
f\left(\frac{5}{6}\right) &= \frac{59}{216}.
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find the distribution of $Y+X$ Consider the following two experiments: the first has outcome $X$ taking on
the values $0, 1$ and $2$ with equal probabilities; the second results in an (in-
dependent) outcome $Y$ taking on the value $3$ with probability $1/4$ and $4$ with
probability $3/4$. Find the distribution of $Y+X$
Attempt: I assigned the probability $1/3$ to $0,1$ and $2$ and the I set $P(S=3)=m(3)m(0)+m(0)m(3)+m(1)m(2)=7/18$, but I felt like I was getting the wrong answer. How would you solve this?
|
In this case you can enumerate all the possibilities and use independence to find their probabilities. Then you can divide up the possibilities according to what the sum $X+Y$ is.
We have:
$$
P(X=0,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=1,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=2,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=0,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}\\
P(X=1,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}\\
P(X=2,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}
$$
The values of $X+Y$ in each of these cases is $3,4,5,4,5,6$ respectively, so
$$ P(X+Y=3) =\frac{1}{12}\\
P(X+Y=4) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=5) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=6) =\frac{1}{4}\\
$$
|
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|
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!
|
WLOG $a\geq b\geq c.$
Let $f(a,b,c)=a+b+c+1/a+/b+1/c.$
For a given value of $a+b+c,$ let $c$ remain constant while $a,b$ vary , subject to the constraint that $a+b$ is constant, so that $a+b+c$ also remains constant. Then $db/da=-1$ and $d(1/b)/da=1/a^2 .$ So with constant $c$ we have $$df(a,b,c)/da=-1/a^2+1/b^2=(a-b)(a+b)/a^2b^2\geq 0$$ (because $a\geq b$). So we cannot have a minimum of $f(a,b,c)$ for a given value of $a+b+c$ unless $a=b.$
Applying this method again, leaving $a$ constant and letting $b,c$ vary, subject to the constraint that $b+c$ is constant, we see also that we cannot have a minimum of $f$ for a given value of $a+b+c$ unless $b=c.$
Therefore for each $S\in (0,3/2]$ we have $$\min \{f(a,b,c): a+b+c=S\}=f(S/3,S/3,S/3)=S+9/S.$$ The least value of $S+9/S$ for $S\in (0,3/2]$, is $15/2,$ which occurs uniquely at $S= 3/2.$ And as we have seen , this only occurs when $a=b=c=S/3=1/2.$
|
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|
$x^7+y^7+z^7$ is divisible by $7^3$, then $x+y+z$ is divisible by $7^2$
Let $x, y, z$ be positive integers, and $7 \nmid xyz$. If $7^3|x^7+y^7+z^7$, show that $7^2|x+y+z$.
by Fermat's little theorem, $x^7 \equiv x \pmod7$, then $x^7+y^7+z^7\equiv x+y+z \equiv 0 $ (mod 7)
so we have $7 | (x+y+z)$.
what should I do next?
|
Since
$$x^7+y^7+x^7=\sum_{cyc}(x^7-x)+x+y+z,$$
we see that $x+y+z$ is divided by $7$.
In another hand,
$$x^7+y^7+z^7=(x+y+z)^7-7(xy+xz+yz)(x+y+z)^5+7xyz(x+y+z)^4+$$
$$+14(xy+xz+yz)^2(x+y+z)^3-21xyz(xy+xz+yz)(x+y+z)^2-$$
$$-7(xy+xz+yz)^3(x+y+z)+7x^2y^2z^2(x+y+z)+7xyz(xy+xz+yz)^2,$$
which says (see the last term) that $(xy+xz+yz)^2$ is divided by $7$ or
$xy+xz+yz$ is divided by $7$.
Thus, $7x^2y^2z^2(x+y+z)$ is divided by $7^3$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int (x^2+1)/(x^4+1)\ dx$ I have Divided the numerator and Denominator by $x^2$
to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:
$$
\int\frac{x^2+1}{x^4+1} \, dx = \int (u^2+2)^{-1} \, du
$$
After this I need help!
|
Standard tables of integrals say:
$$ \int (1+v^2)^{-1} \, dv = (\arctan v) + \text{constant}. $$
Therefore
\begin{align}
\int (2+u^2)^{-1} \, du & = \frac 1 {2\sqrt 2} \int \left(1 + \left(\frac u {\sqrt 2} \right)^2 \right)^{-1} \, \big( \sqrt 2\, du\big) \\[10pt]
& = \frac 1 {2\sqrt 2} \int (1+v^2)^{-1} \, dv \\[10pt]
& = \frac 1 {2\sqrt 2} \arctan v + C \\[15pt]
& = \frac 1 {2\sqrt 2} \arctan \frac u {\sqrt 2} + C.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2361853",
"timestamp": "2023-03-29T00:00:00",
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|
$\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$
Prove that \begin{align}\begin{vmatrix}
1 & a &bc \\
1& b & ac\\
1&c & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\
1& b&b^2 \\
1& b & c^2
\end{vmatrix}\\&=(c-a)(b-a)(c-b)\begin{vmatrix}
1 & a & a^2\\
0& 1 &b+a \\
0& 0 & 1
\end{vmatrix}\\\\
&=(c-a)(b-a)(c-b)\end{align}
I got the last two equalities but I did't get first two. I need help with first two
|
\begin{align}\begin{vmatrix}
1 & a &bc \\
1& b & ac\\
1&c & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\
1& b&b^2 \\
1& c & c^2
\end{vmatrix}\\\\
\begin{vmatrix}
1 & a &bc \\
0& b-a & ac-bc\\
0&c-a & ab-bc
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\
0& b-a&b^2-a^2 \\
0& c-a & c^2-a^2
\end{vmatrix}\\\\
(b-a)(c-a)\begin{vmatrix}
1 & a &bc \\
0& 1& -c\\
0&1 &-b
\end{vmatrix}&=(b-a)(c-a)\begin{vmatrix}
1 & a &a^2 \\
0& 1&b+a \\
0& 1 & c+a
\end{vmatrix}\\\\
(b-a)(c-a)\begin{vmatrix}
1 & a &bc \\
0& 1& -c\\
0&0 &c-b
\end{vmatrix}&=(b-a)(c-a)\begin{vmatrix}
1 & a &a^2 \\
0& 1&b+a \\
0& 0& c-b
\end{vmatrix}\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere?
|
if you want to prove that the axis of symmetry rotates with the function then simply show that $M(\theta)R(\theta)(x,y)=R(\theta)(x',y')$ where $R$ and $M$ are rotation and reflection matrices respectively given that $\theta=0$ is the line of symmetry.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all possible values of $x+y+z$. Let $x, y, z$ be non-zero integers such that ${x\over y}+{y\over z}+{z\over x}$ is an integer. Find all possible values of $x+y+z$. Please provide a proof with all solutions.
|
Note: this assumes that the question is really about non-zero integers, and not about positive integers. I think the answer is more difficult when having to deal with positive integers.
Now, we can get any non-zero integer $t$ by choosing $(x,y,z)=(t,t,-t)$.
This gives $\frac{t}{t} + \frac{t}{-t} + \frac{-t}{t} = -1$ and $x+y+z=t$.
Now suppose $x+y+z=0$, so $z=-x-y$.
This gives $\frac{x}{y} - \frac{y}{x+y} - \frac{x+y}{x} \in \mathbb Z$, hence $\frac{x}{y} - \frac{y}{x+y} - \frac{y}{x} \in \mathbb Z$.
Hence $\dfrac{x^2(x+y)-y^2(x+y)-y^2x}{xy(x+y)} \in \mathbb Z$.
Note that numerator and denominator are both homogenous of degree 3, so we can divide out any common prime factor. So we can assume that $\gcd(x,y)=1$. Suppose there is a prime $p \mid x+y$. We get $p \mid x+y \mid y^2x$. So $p \mid x$ or $p \mid y$, and $p \mid x+y$ gives that $p \mid x$ and $p \mid y$. Contradiction with $\gcd(x,y)=1$. Therefore, we have $x+y=1$ or $x+y=-1$.
The first case gives $\dfrac{x^2-y^2-y^2x}{xy} \in \mathbb Z$, hence $\dfrac{x^2-y^2}{xy} \in \mathbb Z$.
The second case gives $\dfrac{-x^2+y^2-y^2x}{xy} \in \mathbb Z$, hence $\dfrac{-x^2+y^2}{xy} \in \mathbb Z$, so $\dfrac{x^2-y^2}{xy} \in \mathbb Z$.
Now, if $p \mid x$, then $p \mid y$, contradiction with $\gcd(x,y)=1$. So $x=\pm1$, and analogous $y=\pm1$. But then $x+y$ can't be $1$ or $-1$. So there are no solutions with $x+y+z=0$.
|
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|
Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go about this.
Working
This is convergent if
$$
\lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx
$$
exists.
The indefinite integral is
\begin{equation*}
\begin{aligned}
\int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx
& = \int \left(\frac{nx^2}{x^3 + 1}
\right) - \int \left(\frac{1}{13x + 1} \right)dx \\
&= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1)
\end{aligned}
\end{equation*}
Which gives
\begin{equation*}
\begin{aligned}
&\lim_{b \to + \infty}
\int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\
&= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\
&= \lim_{b \to + \infty}
\left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right)
\\ &= \lim_{b \to + \infty}
\left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln(b^3 + 1) - \ln(2)
\right)
- \frac{1}{13}
\left(
\ln(13b + 1)
- \ln(14)
\right)
\right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln \left(\frac{b^3 + 1}{2}\right)
\right)
- \frac{1}{13}
\left(
\ln \left(\frac{13b + 1}{14}\right)
\right)
\right)
\end{aligned}
\end{equation*}
I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( +
\infty)$. But things went pretty south.
So I'm sure there's a better approach.
|
$\int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$
$\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
=\frac{nx^2(13x+1)-(x^3 + 1)}{(x^3 + 1)(13x + 1)}
=\frac{x^3(13n-1)+nx^2-1}{(x^3 + 1)(13x + 1)}
$.
If $13n-1 \ne 0$,
this behaves like
$\frac1{x}$
and the integral diverges.
Therefore the only $n$
for which the integral
might converge is
$n =\frac1{13}$.
For this $n$,
the integrand behaves like
$\frac1{x^2}$
and the integral of this
does converge.
|
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|
Modulus of complex numbers problem draw the loci Given $|z − 3| + |z + 3| = 12$, $z = x + iy$, find the expression in terms of $x$ and $y$ and draw the loci.
I've tried many ways and I cannot isolate $x$ or $y$. I think that is necessary put the equation in this form $(x-a)^2 + (y-b)^2 = r^2$.
May someone help me?
|
If $z=x+iy$ where $x,y$ are real
For
$$\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$$
$$(x+c)^2+y^2=4a^2+(x-c)^2+y^2-4a\sqrt{(x-c)^2+y^2}$$
$$\iff a\sqrt{(x-c)^2+y^2}=a^2-cx$$
Squaring we get $$a^2\{(x-c)^2+y^2\}=a^4-2a^2cx+c^2x^2$$
$$\iff(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$
For $a\ne c,$ $$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2-c^2}=1$$
|
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Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same. Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same.
The first equation is of the form of Newton's Serpentine. When you graph the second equation it appears to overlap the first equation.
I'm not sure whether these two equations are identities or just very close approximations.
I tried to manipulate both equations to get the other but failed.
How does one explain why these two equations are identical?
|
Since $\sin(2A) = 2 \sin A \cos A$,
\begin{align}2\sin(2 \arctan x) &= 4 \sin (\arctan x) \cos(\arctan x) \\
&=4\left( \frac{x}{\sqrt{1+x^2}} \right)\left( \frac{1}{\sqrt{1+x^2}} \right) \\
&=\frac{4x}{1+x^2}\end{align}
|
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$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$
I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^{\circ} \cos 60^{\circ}-\cos 76^{\circ} \, \cos16^{\circ}.$$
I couldn't simplify after that.
|
Let $c= \cos 16^\circ$ and $s=\sin 16^\circ$. Cosine addition gives ($76^\circ=60^\circ+16^\circ$)
\begin{eqnarray*}
\cos(76^\circ)=\frac{c-\sqrt{3}s}{2}
\end{eqnarray*}
So the expression is
\begin{eqnarray*}
\cos^2(76^\circ)+\cos^2(16^\circ)-\cos(76^\circ)\cos(16^\circ) =\left(\frac{c-\sqrt{3}s}{2}\right)^2+c^2-c\left(\frac{c-\sqrt{3}s}{2}\right) \\=\frac{c^2-2\sqrt{3}cs+3s^2+4c^2-2c^2+2\sqrt{3}cs}{4}=\color{red}{\frac{3}{4}}.
\end{eqnarray*}
|
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Numbers of ways to choose $k$ out of the first $n$ natural numbers so the longest string of consecutive numbers is exactly $m$? How many ways can $k$ numbers be chosen from the first $n$ natural numbers so that the longest string of consecutive numbers is exactly $m$ numbers long
For example, if choosing $k = 7$ distinct numbers from the first $n = 14$ natural numbers ($1-14$), how many combinations of numbers are there that have exactly $m = 3$ consecutive numbers?
Some sets that satisfy this would be $[1,2,3,5,7,10,13]$, $[1,2,3,7,8,10,11]$, or $[1,2,3,6,9,10,11]$.
Obviously $m \le k \le n$, and the order of the numbers picked does not matter, but there is no repetition, so $[1,2,3]$ is the same as $[2,1,3]$, but $[1,1,3]$ is not allowed.
For cases when $m \gt \frac k2$ I believe the equation
$$2* \binom {n-m-1}{k-m} + (n-m-1)*\binom{n-m-2}{k-m}$$
will produce the correct answer, but for values of $m \le \frac k2$ I do not know how to account for multiple strings of consecutive digits in the same set (eg. $[\textbf{1,2,3},6,\textbf{8,9,10}]$) not being counted twice.
If anyone could give me help with this example that I could extrapolate from or (ideally) a formula or any references to solve for a general case it would be much appreciated.
|
Another more practical solution uses a recurrence relation. Let the size of the gaps between the $n-k$ numbers not chosen be $x_1,x_2,...,x_{n-k+1}$. Then the problem reduces to how many ways are there such that $\forall i$, $x_i\in [0,1...m]$, $\sum_{i=1}^{n-k+1}x_i = k$ and at least one of $x_i = m$.
To solve this first consider the solutions of another similar problem, $\forall i$, $x_i\in [0,1...m]$, $\sum_{i=1}^{y}x_i = k$ where the constraint at least one of $x_i = m$ doesn't exist and $n-k+1$ has been replaced by $y$. Let the number of solutions of this problem be the function $S(y,k,m)$. Next consider removing the last $x_i$ in a solution of $y,k,m$. Then the remaining $x_i$ will give us a solution of $y-1,k-a,m$ for some $a\in [0,1...m]$. Therefore $S(y,k,m) = \sum_{a=0}^{m}S(y-1,k-a,m)$. $S(1,k,m) = 1$ iff $k\in [0,1...m]$ which allows us to tabulate all other values.
For example in the $m=2$ case:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \text{y\k} & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7} & \text{8} & \text{9} & \text{10} \\
\hline \text{1} & \text{1} & \text{1} & \text{1} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} \\
\hline \text{2} & \text{1} & \text{2} & \text{3} & \text{2} & \text{1} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} & \text{0} \\
\hline \text{3} & \text{1} & \text{3} & \text{6} & \text{7} & \text{6} & \text{3} & \text{1} & \text{0} & \text{0} & \text{0} & \text{0} \\
\hline \text{4} & \text{1} & \text{4} & \text{10} & \text{16} & \text{19} & \text{16} & \text{10} & \text{4} & \text{1} & \text{0} & \text{0} \\
\hline \text{5} & \text{1} & \text{5} & \text{15} & \text{30} & \text{45} & \text{51} & \text{45} & \text{30} & \text{15} & \text{5} & \text{1} \\
\hline \end{array}$$
Then let $P(y,k,m)$ be the number of solutions of the problem with the constraint at least one of $x_i = m$ included. $P(y,k,m) = S(y,k,m) - S(y,k,m-1)$.
For example $P(3,3,2) = S(3,3,2) - S(3,3,1) = 7-1 = 6$ (with $S(3,3,1)$ quickly worked out by hand).
Therefore the number of solutions of the original problem with $n=5, k=3, m=2$ is $P(5-3+1,3,2) = P(3,3,1) = 6$ and as the possible solutions of this problem are $[1,2,4], [1,2,5], [2,3,5], [1,3,4], [1,3,5], [2,3,5]$ this is indeed correct.
Although this is a bit time consuming by hand it should be possible to get a computer to use the recurrence relation to generate the solutions.
|
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|
Complex Integration using Cauchy's Theorem The problem is the integration of
$$I=\int_{\left\lvert z-1\right\rvert=1} f(z) dz$$
where
$$f(z)=\frac{1}{z^3-1}$$
and the path goes $1$ loop in positive direction.
I tried to solve the problem using Cauchy's Theorem by finding $z$ that makes $f(z)$ denominator be $0$. That was $z=1$.
And I got struck.
I think the integral is needed to be treat somehow so that
$$\int_{C_a}\frac{1}{(z-\alpha)^n}dz = 2{\pi}i \text{ when } n=1$$
can be used.
My question is how should I continue with the integral?
|
Note: The hint of @rtybase is essentially the answer to OPs question. Let's recall Cauchy's Integralformula:
If $a$ is in the interior of $\gamma=\{z:|z-1|=1\}$ and a function $g$ is holomorphic in a region which contains the closure of the interior of $\gamma$, then
\begin{align*}
g(a)=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-a}\,dz
\end{align*}
We have the following situation:
The function $f(z)=\frac{1}{z^3-1}$ has three simple poles at $z_0=1$ and $z_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$. The pole $z_0=1$ is in the interior of $\gamma$. We can write $f(z)$ as
\begin{align*}
f(z)=\frac{1}{z^3-1}=\frac{1}{(z-1)(z^2+z+1)}
\end{align*}
and observe: Since $g(z)=\frac{1}{z^2+z+1}$ is holomorph in the interior of $\gamma$ we obtain
\begin{align*}
g(1)&=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-1}\,dz\\
&=\frac{1}{2\pi i}\oint_{\gamma} \frac{1}{(z-1)(z^2+z+1)}\,dz\\
&=\frac{1}{2\pi i}\oint_{\gamma} f(z)\,dz\\
\end{align*}
We finally conclude
\begin{align*}
\oint_{\gamma} f(z)\,dz=2\pi i\, g(1)=\frac{2\pi i}{3}
\end{align*}
|
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|
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even. Question:
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.
I am not quite sure how to prove this. My guess is proof by contradiction.
Assume the contrary, that is, $a^{2} + b^{2} = c^{2}$ for $c$ even and at least one of $a,b$ odd.
|
Suppose $a^2+b^2=c^2$ and $c$ is even. Since $a^2=c^2-b^2=(c+b)(c-b)$, $a$ and $b$ have the same parity. Assume $a$ is odd. Then $a^2+b^2\equiv 2(\mod4)$, whence $c^2\equiv2(\mod 4)$ which is a contradiction since $4\mid c^2$ as $2\mid c$. Hence $a$ and $b$ are even.
|
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Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ .
Answer:
The graph is above :
Since the region is symmetrical ,
$ Area =4 \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $
Am I right ? Is there any help ?
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Convert to polar form $r^2=\dfrac{16}{\cos 4 t+3}$
The area is $$A=\frac{1}{2} \int_0^{2 \pi } \frac{16}{\cos 4 t+3} \, dt=32 \int_0^{\frac{\pi }{2}} \frac{1}{\cos 4 t+3} \, dt$$
$\cos 4t=\cos^2 2t-\sin^2 2t=2\cos^2 2t -1$
so we have to integrate
$\int \dfrac{dt}{2(\cos^2 2 t+1)}=\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}$
Remember the identity $\cos 2t =\dfrac{1-\tan^2 t}{1+\tan^2 t}$
with the substitution $\tan t=u;\;t=\tan^{-1}u;\;dt=\dfrac{du}{1+u^2};\;\cos 2t = \dfrac{1-u^2}{1+u^2}$
Limits of integration become $t=0\to u=0;\;t=\pi/2\to u=\infty$
$$\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}=\dfrac{1}{2}\int \dfrac{\dfrac{du}{1+u^2}}{1+\left(\dfrac{1-u^2}{1+u^2}\right)^2}$$
Which simplifies to
$$\dfrac{1}{2}\int \frac{u^2+1}{ u^4+1} \, du=\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}$$
Area is
$$A=32\left[\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}\right]_0^{\infty}=4 \sqrt{2} \pi$$
|
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|
How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed can someone help me to find the way to simplify the expression.
|
Let $\sqrt{a\sqrt{b}} = \sqrt{a}\sqrt{\sqrt{b}} = \sqrt{a}\cdot\sqrt[4]{b}$
$$\tag1
\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
$$\tag2
\frac{5\sqrt{7}}{4\sqrt{3}\sqrt[4]{5}-4\sqrt{2}\sqrt[4]{5}}- \frac{4\sqrt{5}}{\sqrt{3}\sqrt[4]{5}-\sqrt{2}\sqrt[4]{5}}$$
$$\tag3
\left(\frac{5\sqrt{7}}{4(\sqrt{3}-\sqrt{2})}- \frac{4\sqrt{5}}{\sqrt{3}-\sqrt{2}}\right)\frac{1}{\sqrt[4]{5}}$$
$$\tag4
\left(\frac{5 \sqrt{7} - 16 \sqrt{5}}{4(\sqrt{3}-\sqrt{2})}\right)\frac{1}{\sqrt[4]{5}}$$
|
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Infinite series inequality. I have some trouble with the following problem:
Show that $$\sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt[p]{n}}<p$$ is true for every natural number $p$.
I tried using induction but I can't finish the proof. Or there is a simpler way to solve this? Thank you.
|
Solution 1. Let $a = \sqrt[p]{n}$ and $b = \sqrt[p]{n+1}$. If $p > 1$ is an integer, then
\begin{align*}
\frac{1}{(n+1)\sqrt[p]{n}}
&= \frac{p}{ab \cdot pb^{p-1}} \\
&< \frac{p}{ab \cdot (b^{p-1} + ab^{p-2} + \cdots + a^{p-1})} \\
&= p \cdot \frac{b - a}{ab (b^p - a^p)} \\
&= p \left( \frac{1}{\sqrt[p]{n}} - \frac{1}{\sqrt[p]{n+1}} \right).
\end{align*}
Summing over $n = 1, 2, \cdots$ gives the desired inequality.
Solution 2. Let $p \geq 1$ be real and write
$$ p \left( \frac{1}{n^{1/p}} - \frac{1}{(n+1)^{1/p}} \right)
= \frac{p}{(n+1)n^{1/p}} \left[ (n+1) - n^{\frac{1}{p}}(n+1)^{1-\frac{1}{p}} \right]. $$
By the Jensen's inequality (or by the AM-GM inequality if $p$ is integer), we have
$$n^{\frac{1}{p}}(n+1)^{1-\frac{1}{p}} \leq \frac{1}{p}\cdot n + \left(1-\frac{1}{p}\right)\cdot(n+1) = n+1 - \frac{1}{p} $$
with the equality exactly when $p = 1$. So it follows that
$$ p \left( \frac{1}{n^{1/p}} - \frac{1}{(n+1)^{1/p}} \right) \geq \frac{1}{(n+1)n^{1/p}} $$
with equality if and only if $p = 1$. Summing over $n = 1, 2, \cdots $ gives the desired inequality.
|
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How to turn the ellipse $x^2 - xy + y^2 - 3y - 1 = 0$ to the canonical form using an isometric transformation? There is an exam problem I'm having trouble with, it is as follows:
Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.
Comparing to the general equation for conic sections I see that this is a rotated ellipse. I assume that the solution would be to multiply a rotation matrix (to align the ellipse's axes to the $x$ and $y$ axes) and a translation matrix (to bring the ellipse's center to the origin), and then somehow apply the result to the ellipse.
How would I go about doing this? How do I get the coordinates of the ellipse's center and the angle of its rotation to construct the matrices, and then how would I apply the transformation to the ellipse itself?
|
To keep it understandable albeit inelegant I'll do the passage in two steps
First the translation
$x^2 - xy + y^2 - 3y -1 = 0$
we look for a new centre $(h;\;k)$ so we substitute $x=x'+h;\;y=y'+k$
$-(h+x) (k+y)+(h+x)^2+(k+y)^2-3 (k+y)-1=0$
$x'^2-x' y'+y^2+x' (2 h-k)+y' (-h+2 k-3) +h^2-h k+k^2-3 k-1=0$
To have no first degree terms we put $2h-k=0;\;-h+2 k-3=0$ which gives
$h=1;\;k=2$ and we plug these values in the previous equation
$x'^2 - x' y' + y'^2=4$
Now we want to get rid of the $x'y'$ term. To do so we have to rotate the axis using these equations
$\begin{aligned}x'&=X\cos \theta -Y\sin \theta \\y'&=X\sin \theta +Y\cos \theta \end{aligned}$
$(X \sin \theta+Y \cos \theta)^2-(X \cos \theta-Y \sin \theta) (X \sin \theta+Y \cos \theta)+(X \cos \theta-Y \sin \theta)^2=4$
collecting terms
$X^2 \left(\sin ^2\theta+\cos ^2\theta-\sin \theta \cos \theta\right)+X Y \left(\sin ^2\theta-\cos ^2\theta\right)+Y^2 \left(\sin ^2\theta+\cos ^2\theta+\sin \theta \cos \theta\right)=4$
as we want the term $XY$ off we set $\sin ^2\theta-\cos ^2\theta=0$
$\tan^2\theta=1\to \theta=\pm \dfrac{\pi}{4}$
if we want the major axis of the ellipse to be horizontal we choose $\theta=\dfrac{\pi}{4}$ and substitute this value in the last equation
$\dfrac{X^2}{8}+\dfrac{3 Y^2}{8}=1$
the equations of the roto-translation altogether are
$\begin{aligned}x&=X\cos \frac{\pi}{4}-Y\sin \frac{\pi}{4} +1\\y&=X\sin \frac{\pi}{4}+Y\cos \frac{\pi}{4} +2\end{aligned}$
or
$\begin{aligned}x&=\frac{\sqrt 2}{2}(X-Y) +1\\y&=\frac{\sqrt 2}{2}(X+Y) +2\end{aligned}$
hope this helps
|
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What are the roots of this equation? I have a quadratic equation $ ax^2 +bx+c =0 $, where $ a,b,c $ all are positives and are in Arithmetic Progression.
Also, the roots $\alpha$ and $\beta$ are integers.
I need to find out $ \alpha + \beta + \alpha\beta $.
I have tried taking $ a = a' - d , b = a' , c = a' + d $ because I have supposed $ a' $ is the first term and $ d $ is common difference, I used sum of roots and product of roots rule, but nothing helped..
|
If the roots are integers, we can write $x^2 + bx + c = (x-\alpha)(x-\beta)$. Note that it's okay to take $a=1$, since $b-a = c-b \implies \frac{b-a}{a} = \frac{c-b}{a}$, i.e. the arithmetic progression is preserved.
Now write $c = b + (b-1) = 2b-1$.
Use Vieta's: $\alpha + \beta = -b$ and $\alpha\beta = 2b-1$.
$\alpha = \frac{-b\pm \sqrt{b^2 - 8b +4}}{2}$. But $\alpha$ is an integer, so $(b-4)^2 -12$ is a perfect square means that we must have $b = 8$ (can be checked manually by trying squares up to 6, since difference between consecutive squares is $2n+1$). Then $\alpha + \beta + \alpha\beta = -b + (2b-1) = b - 1 = 7$.
|
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Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM and CS inequality but failed to get a result. Please give me some hint in how to solve this question.
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For $\prod\limits_{cyc}(a-b)\neq0$ we obtain:
$$\sum_{cyc}\frac{a^2}{2a^2+bc}=\sum_{cyc}\frac{a^2}{a(a+b+c)+a^2-ab-ac+bc}=$$
$$=\sum_{cyc}\frac{a^2}{(a-b)(a-c)}=\frac{\sum\limits_{cyc}a^2(c-b)}{\prod\limits_{cyc}(a-b)}=1.$$
|
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Find all solutions to $x^3+(x+1)^3+ \dots + (x+15)^3=y^3$
Find all pairs of integers $(x, y)$ such that
$$x^3+(x+1)^3+ \dots + (x+15)^3=y^3$$
What I have tried so far:
The coefficient of $x^3$ is $16$ in the left hand side. It is not useful then to trying bound LHS between, for example, $(ax+b)^3$ and $(ax+c)^3$ and then say that $ax+b<y<ax+c$.
I also tried to use modulo a prime. But it seems unlikely to bound variables this way.
EDIT : Though, it can be factored as $(2x+15)(x^2+15x+120)=(y/2)^3$. LSH factors are almost co-prime and we can say that $x^2+15x+120=3z^3$ or $x^2+15x+120=5z^3$. These are still too difficult to solve!
Any ideas?
|
This isn't a complete solution, but I hope it gives you an approach. (It's too long for a comment)
Since we have $$\sum_{r=1}^{n} r^3=\left(\frac { n(n+1)}{2}\right)^2$$
You can write
\begin{align}
x^3+(x+1)^3+ \dots + (x+15)^3
&=\sum_{r=1}^{x+15} r^3-\sum_{r=1}^{x-1} r^3
\\
&=\left(\frac { (x+15)(x+16)}{2}\right)^2-\left(\frac { x(x-1)}{2}\right)^2\\
&=\left[\left(\frac { (x+15)(x+16)}{2}\right)-\left(\frac { x(x-1)}{2}\right)\right]\left[\left(\frac { (x+15)(x+16)}{2}\right)+\left(\frac { x(x-1)}{2}\right)\right]\\
\end{align}
Simplifying this, we get $$(x^2+15x+120)(2x+15)=\left(\frac y2 \right)^3$$
|
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Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $[x,y,z]$ and applying the property of orthogonal matrix.
In each approach I got a different solution, which is not correct.
Even I suspect there is an error in given problem.
Please help me in finding the third row.
Adding cross product calculation below.
\begin{vmatrix}
\bar{\imath}&\bar{\jmath}&\bar{k}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}
\end{vmatrix}
\begin{align}
&=\bar{\imath}(0)-\bar{\jmath}\left(-\frac{1}{3\sqrt{2}}\right)+\bar{k}\left(\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\frac{1}{\sqrt{\frac{1}{18}+\frac{1}{18}}}\left(0,\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)
\end{align}
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One of the possible solutions:
$$\frac{\begin{pmatrix} \frac{1}{3} \\\frac{2}{3} \\ \frac{2}{3}\end{pmatrix} \times \begin{pmatrix} 0 \\\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{pmatrix}}{\left\Vert\begin{pmatrix} \frac{1}{3} \\\frac{2}{3} \\ \frac{2}{3}\end{pmatrix} \times \begin{pmatrix} 0 \\\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}}\end{pmatrix}\right\Vert}$$
Why this solution works:
An othogonal matrix has an orthonormal basis in its rows, so by taking the cross product of the two given vectors, you assure that your vector is perpendicular to both given vectors. We then devide the vector by its own norm, because then it has length $1$. Therefore, together with the two given vectors this vector will form an orthonormal basis. Note that the inverse (for addition) of this vector will also be a correct solution.
There are also other possibilities to solve this question. You can solve it by inspection, but this requires some insight. Another way would be to use the Gramm-Schmidt algorithm.
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"url": "https://math.stackexchange.com/questions/2385989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
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Following your rationalizing the denominator method, you err when expanding the square. You should get
$$
\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} = \frac{(1+\sin\theta)^2 - (\cos\theta)^2 + 2i\cos\theta(1+\sin\theta)}{(1+\sin\theta)^2 + (\cos\theta)^2}
$$
This can then be simplified by expanding the $(1+\sin\theta)^2$ term and using $(\sin\theta)^2 + (\cos\theta)^2 = 1$.
$$
\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} = \frac{1 - (\cos\theta)^2 + (\sin\theta)^2 + 2\sin\theta + 2i\cos\theta(1+\sin\theta)}{1+2\sin\theta + (\sin\theta)^2 + (\cos\theta)^2} = \frac{2\sin\theta(1+\sin\theta) + 2i\cos\theta(1+\sin\theta)}{2 + 2\sin\theta} = \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta)
$$
It should be clear then that
$$
\left(\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\right)^n = i^n\left[\cos(n\theta)-i\sin(n\theta) \right]
$$
and you should be able to take real and imaginary parts from there.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding $n$th power of a $3\times 3$ matrix Find the $A^n$ if $$A=\begin{bmatrix}1 & a & b \\0 & 1 &a\\0 &0 &1\end{bmatrix}$$
I tried inductive method to show $$A^n=\begin{bmatrix}1 & na & nb+\frac{n(n-1)}{2}a^2 \\0 & 1 &na\\0 &0 &1\end{bmatrix}$$
now : My question is : Is there other method (idea ) to find $A^n$ ?
Thanks in advance.
Can the idea apply for $$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$ when $c \neq a$ ?
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$$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$
Write $A = I + U$, where $U^2 = \begin{bmatrix} a \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & c \end{bmatrix}$ and $U^3=0$.
$$ A^n=(I+U)^n=I+nU+{n\choose 2}U^2 =
\pmatrix{1&na&nb+\frac{n(n-1)}{2}ac\\0&1&nc\\0&0&1}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2388294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
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To avoid errors or pitfalls it can sometimes be helpful to try an initial 'lazy approach', where you still don't drop mathematical circumspection.
When looking at the equation a third degree polynomial can be sought, but that is no fun and we are really hoping that a quadratic equation 'pops-up'. The equation is ugly but perhaps by inverting the LHS and RHS things will simplify. We check that there is no $x$ with both the LHS and RHS equal to $0$ (see next section), so we can safely invert.
For the LHS,
$\frac {1} { \frac {1}{x-1} - \frac {1}{x-2}} = \frac{(x-1)(x-2)}{x-2-(x-1)} = -(x^2-3x+2)$
For the RHS,
$\frac {1} { \frac {1}{x-3} - \frac {1}{x-4}} = \frac{(x-3)(x-4)}{x-4-(x-3)} = -(x^2-7x+12)$
Combining our work,
$\tag 1 x^2 -3x +2 = x^2 -7x +12$
OMG! It is turning into a linear equation!
$\tag 2 4x = 10$
So $x = 2.5$
Examining
$\tag 3 \frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$
the LHS, $\frac{1}{(x-1)} - \frac{1}{(x-2)}$ is zero if and only if $x -
1 = x - 2$, which is silly. So we can forget about looking for any $x$ solutions that would make the LHS and RHS of (3) both equal to 0.
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"timestamp": "2023-03-29T00:00:00",
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