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Solving cauchy riemann equations, finding all analytic functions I need someone to check my work! I tried doing this as properly as possible, but I have no way to check whether this is correct.
Find $\textit{all}$ analytic functions $f = p(x,y) +iq(x,y) $ such that $p+q = xy$
(I'm using p's and q's since u's and v's look too similar. It gets messy quickly).
We have $$p_x + q_x = y \qquad p_y + q_y = x $$
$f$ is to be analytic so by Cauchy-Reimann we have $p_x = q_y$ and $p_y = -q_x$, thus we have:
$$y- q_x = x - p_y \qquad y+p_y = x- p_y$$
so
$$ p_y = \frac{x-y}2$$
Thus $$p = \frac{xy}{2} - \frac{y^2}{4} + h(x)$$
Since $q = xy - p$ we have
$$ q = xy - \frac{xy}{2} + \frac{y^2}{4} - h(x) = \frac{xy}{2}+ \frac{y^2}{4} - h(x) $$
Again, since $p_x = q_y$ we have
$$\frac{y}2 + h'(x) = x - \frac{x}2 + \frac{y}2 = \frac{x}2 + \frac{y}2$$
so we have:
$$h(x) = \frac{x^2}4 + C $$
Putting it all together:
$$ f = p+iq = \frac{xy}{2} - \frac{y^2}{4} + \frac{x^2}4 + C + i\left( \frac{xy}{2}+ \frac{y^2}{4} - \frac{x^2}4 - C \right)$$
If we would like to rewrite this, we have:
$$f(z) = \frac{z^2}4 + C - i\left( \frac{z^2}4 + C \right) = (1-i)\left( \frac{z^2}4 + C \right) $$
Now I would like to claim that for $C \in \mathbb{R}$, these are all solutions. Can I truly be sure of this? Did I make some mistake somewhere?
Thanks in advance.
|
It is of course possible that I overlooked a mistake, but since the result is correct, and I didn't see any error, I'm reasonably convinced that your work is correct.
Here, as in many situations, we can solve the task in an easier way if we apply a certain transformation to the problem. It is advisable to try to look for such simplifications. Not only does it save work when one spots them, it is also less likely to make mistakes in simpler situations.
With $f = p + iq$, note that $(1+i)f = (p-q) + i(p+q)$, so $xy$ is the imaginary part of $g = (1+i)f = u + iv$. Now the Cauchy-Riemann equations make it easy to find $u$, since
$$u_x = v_y = x \quad \text{and}\quad u_y = -v_x = -y,$$
so $u(x,y) = \frac{1}{2}(x^2 - y^2) + c$ for some $c\in \mathbb{R}$. Writing $g$ as a function of $z = x+iy$, we find
$$g(z) = \frac{z^2}{2} + c$$
for some (arbitrary) $c\in \mathbb{R}$. Since $(1-i)(1+i) = 2$, we then obtain
$$f(z) = \frac{1-i}{2} g(z) = \frac{1-i}{4} z^2 + \frac{1-i}{2}c,$$
with an arbitrary real $c$. (My $\frac{c}{2}$ corresponds to your $C$.)
|
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|
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds:
$$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$
I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).
My effort: I understood how the author defines the variables $M, N$ and $S$.
$$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$
$$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$
$$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$
$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.
Then the author, without any expansion/explanation, says
$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$
Which is also true, since the AM-GM inequality says
$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$
The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.
The next part is the crux of my question.
The author claims,
$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$
This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??
As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.
|
I think, the best way it's C-S:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{\left(\sum\limits_{cyc}a\right)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a-c)^2+(b-d)^2}{\sum\limits_{cyc}(ab+ac)}\geq2.$$
Another way:
Let $(a-c)(b-d)\geq0.$
Thus,
\begin{align}
\sum_{cyc}\frac{a}{b+c}&=\frac{a+d}{b+c}+\frac{b+c}{a+d}+\frac{d}{a+b}-\frac{d}{b+c}+\frac{b}{c+d}-\frac{b}{a+d} \\
&\geq 2+\frac{d(c-a)}{(a+b)(b+c)}+\frac{b(a-c)}{(c+d)(a+d)} \\
&=2+\frac{(a-c)(b-d)(b^2+d^2+ab+ac+ad+bc+bd+cd)}{(a+b)(b+c)(c+d)(d+a)} \\
&\geq2.
\end{align}
|
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|
Binomial expansion in descending power For example, find, in ascending powers of $x$, the first three terms in the expansion of $(2+5x)^7$.
So, $(2+5x)^7=2^7+\binom{7}{1}(2^6)(5x)+\binom{7}{2}(2^5)(5x)^2$.
I've no problem to solve this kind of problem.
But now another question wants me to find, in descending powers of x, the first four terms in the expansion of $(2x+\frac{1}{3x})^6$.
How?
|
Exactly the same way, you can just do this with the binomial theorem.
\begin{align*}
\left(2x+\frac{1}{3x}\right)^6 &= \sum_{k=0}^6 {6 \choose k}(2x)^{6-k}\left(\frac{1}{3x}\right)^k \\ &= 2^6x^6+2^5\frac{1}{3}{6 \choose 1} x^4 + 2^4\frac{1}{3^2}{6 \choose 2} x^2+ 2^3\frac{1}{3^3}{6 \choose 3} + \cdots \\ &= 64x^6+ 64x^4+\frac{80}{3} x^2 + \frac{160}{27}+\cdots \end{align*}
The terms $x^5$, $x^3$ and $x$ have coefficient 0.
|
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|
Find the sum of first $20$ terms of a sequence Define a sequence $$a_n=\sqrt{1+\left(1-\frac{1}{n}\right)^2}+\sqrt{1+\left(1+\frac{1}{n}\right)^2}$$ for $n \geq 1$. Find $$\sum_{i=1}^{20}\frac{1}{a_i}$$ Some insight on the approach is highly appreciated. What is the general way of solving such problems? Thanks.
|
$$\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{x-y},$$
so by choosing $x=1+\left(1+\frac{1}{n}\right)^2$, $y=1+\left(1-\frac{1}{n}\right)^2$ we have:
$$ \frac{1}{a_n} = \frac{n}{4}\left(\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}\right) $$
or:
$$ \frac{1}{a_n} = \frac{1}{4}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right) $$
and our sum turns out to be a telescopic one:
$$ \sum_{n=1}^{20}\frac{1}{a_n}=\frac{1}{4}\left(\sqrt{20^2+21^2}-1\right)=\bbox[5px,border:2px solid #C0A000]{\color{red}{7}}$$
|
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|
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.
The first couple of steps are the same: finding $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}}
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65} \\
\sin x &=\frac {8} {\sqrt{65}}
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$ OR $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant.
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $ OR $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
|
Andre already gave the answer, but let me explain the "generalities".
The main problem is the following:
After you reached the point $\sin(x+y) = \sqrt{2}/2$, you concluded that $x+y$ can take both values $\pi/4$ or $3\pi/4$.
You interpreted this to mean
BOTH $x+y = \pi/4$ and $x+y = 3\pi/4$ are solutions. (That is, the set of solutions is the set $\{\pi/4, 3\pi/4\}$.)
This interpretation is false! The correct interpretation is that
We have ruled out all numbers other than $\pi/4$ and $3\pi/4$ from being possible solutions. (Or, the set of solutions is a subset of the set $\{\pi/4, 3\pi/4\}$.)
There may be other constraints that rule out one (or both) of those values. In your case, knowing that $\cos x \approx 1/8$ should tell you already that $x$ is bigger than $\pi/4$.
Another way to look at this clearly is to look at the directions of implication. From the values of $\sin x$, $\sin y$ etc you can derive the value of $\sin(x+y)$. But just knowing the value of $\sin(x+y)$ you cannot derive the value of $\sin x$, $\sin y$, etc. So the implication going from the "necessary information" step to the "computing $\sin(x+y)$ and $\cos(x+y)$" step is one that loses information.
So that solutions to $\sin(x+y) = \sqrt{2}/2$ are not necessarily solutions to $\cos(x) = 1/\sqrt{65}$ etc.
|
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|
Roots addition.
$$\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}+\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}=Q $$
One is expected to find $Q$ respecting $a>0$, $x>\sqrt{a}$ .
I'd like to have my solution checked; namely the correct answer is $2\sqrt{x}$ but I simply fail to see where I made a mistake.
And it'd be nice to hear if there is any smoother solution, or any other way to solve this.
I introduced two subsequent substitutions.
*
*$\frac{a+x^2}{x}=A, \;\;\;2\sqrt{a}=B \; \Rightarrow \; \sqrt{A-B}+\sqrt{A+B}=Q$
*$\sqrt{A-B}=k, \;\;\; \sqrt{A+B}=n$
Now I have: $ \;\;\;$
$k+n=Q, \;\;\; k^2+n^2=2A, \;\;\; kn=\sqrt{A^2-B^2}$
$$k^2+n^2+2kn=\left ( k+n \right )^2=Q^2$$
$$Q^2=2A+2\sqrt{A^2-B^2}=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left (a+x^2 \right )^2}{x^2}-\left ( 2\sqrt{a} \right )^2}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2+2ax^2+x^4-4ax^2}{x^2}}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2-2ax^2+x^4}{x^2}}=$$
$$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left ( a-x^2 \right )^2}{x^2}}=\frac{2(a+x^2)}{x}+\frac{2\left (a-x^2 \right)}{x}=\frac{4a}{x}$$
And as a result: $$Q=2\sqrt{\frac{a}{x}}$$
|
Since we have $x\gt \sqrt a\gt 0$, we have
$$x^2\gt a\Rightarrow a-x^2\lt 0.$$
Hence, note that we have
$$\sqrt{\frac{(a-x^2)^2}{x^2}}=\frac{\sqrt{(a-x^2)^2}}{\sqrt{x^2}}=\frac{|a-x^2|}{|x|}=\frac{\color{red}{-}(a-x^2)}{x}.$$
|
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|
anyone can help me with solving this $x^{x^{3}}=3$? Find the value of $x$ : $x^{x^{3}}=3$
I tied with "log" but I couldn't. any help?
|
Assume the domain is $(0,\infty)$. You show the equation $x^{x^3} = 3$ has a unique real solution $x = \sqrt[3]{3}$. Look at the function: $f(x) = x^3\ln x , 0 < x < \infty$. If $0 < x < 1 \Rightarrow x^3\ln x < 0$, so $x^3\ln x < \ln 3$, since $\ln 3 > 0$. Thus: $e^{x^3\ln x} < e^{\ln 3}\Rightarrow x^{x^3} < 3$. And for $ 1\leq x <\infty$, the function $f(x) = x^3\ln x$ has $f'(x) = 3x^2\ln x+ x^2 = x^2(3\ln x + 1) > 0$. This means the equation $x^{x^3} = 3$ can only have atmost $1$ solution on $[1,\infty)$. Observe that $x = \sqrt[3]{3}$ is a solution, and so it is the only solution.
|
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|
Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{4}{a^2+b^2}$$
So to prove the required result, we just need to show that
$$\dfrac{4}{a^2+b^2} \geq \dfrac{8}{(a+b)^2}$$
or equivalently
$$(a+b)^2 \geq 2(a^2+b^2)$$
But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work. If anything, the reverse is always true!
What have I done wrong here?
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$\dfrac{1}{a^2}+\dfrac{1}{b^2} \ge \dfrac{2}{ab} \ge \dfrac{8}{(a+b)^2} \iff (a+b)^2 \ge 4ab$
|
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|
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
|
We certainly can use contour integration here. First, through a sub of $x \mapsto x^2$ and a little algebra, we can express the integral as
$$2 \int_0^1 dx \, \sqrt{1-x^2} - 4 \int_0^1 dx \frac{\sqrt{1-x^2}}{2+x^2} = \frac{\pi}{2} - 2 \int_{-1}^1 dx \frac{\sqrt{1-x^2}}{2+x^2}$$
Now consider
$$\oint_C dz \frac{\sqrt{z^2-1}}{z^2+2} $$
where $C$ is (1) the circle $z=R e^{i \theta}$, $\theta \in [-\pi,\pi)$, (2) a line extending from the circle at $\theta=\pi$ to the dogbone contour, (3) the dogbone, and (4) a line back to the circle at $\theta=-\pi$. Note that the integral vanishes along the lines to the dogbone and along the small circles of the dogbone.
Thus, the contour integral is equal to
$$i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{\sqrt{R^2 e^{i 2 \theta}-1}}{R^2 e^{i 2 \theta}+2} + i 2 \int_{-1}^1 dx \frac{\sqrt{1-x^2}}{2+x^2}$$
Note that I have ignored the small semicircular and circular contours, as the integrals around those vanish as their radii vanish. Also, I have skipped over the step of assigning a phase to each branch of the square root along $[-1,1]$.
The contour integral is also equal to $i 2 \pi$ times the residue of the poles inside $C$, which are at $z=\sqrt{2} e^{\pm i \pi/2}$. Thus, taking the limit as $R \to \infty$, we get
$$i 2 \pi + i 2 \int_{-1}^1 dx \frac{\sqrt{1-x^2}}{2+x^2} = i 2 \pi \left (\frac{i \sqrt{3}}{i \sqrt{2}} + \frac{-i \sqrt{3}}{-i \sqrt{2}} \right ) = i \sqrt{6} \pi $$
Putting this altogether, we get
$$\int_{-1}^1 dx \frac{\sqrt{1-x^2}}{2+x^2} = \left (\sqrt{6}-2 \right ) \frac{\pi}{2} $$
and therefore, the original integral is equal to
$$\int_0^1 dx \frac{\sqrt{x-x^2}}{x+2} = \left (\frac{5}{2}-\sqrt{6} \right ) \pi $$
|
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|
Prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$? I have to prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$.
I have proven using congruencies that $2^{15}-1$ is divided by $31$. However we have
$$2^5\equiv 10 \mod{11}$$
$$2^{15}\equiv 10^3=1000\equiv 10 \pmod {11}$$
Therefore
$$2^{15}-1\equiv 9 \pmod{11}.$$
So it is impossible to prove!!
|
You are correct, there is a mistake in the problem. Here are two useful prime factorizations:
$$
\begin{align*}
2^{15}-1&=32767=7\cdot 31\cdot 151\\
2^{15}+1&=32769=3^2\cdot 11\cdot 331
\end{align*}
$$
|
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|
Evaluate the indefinite integral $\int \frac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \!\theta}} \mathrm{d}\theta$ I want to evaluate $$\int \dfrac{\cos \theta \, \mathrm{d}\theta}{ \sqrt{2 - 9 \sin^2 \theta}}$$
but I can't seem to get the answer, my working is as below:
|
Using the substitution $u = \dfrac{\sqrt{2}}{3} \sin \theta \implies \dfrac{\mathrm{d}u}{\mathrm{d}\theta} = \dfrac{\sqrt{2}}{3} \, \cos \theta$. Our integral can be written as $$\int \dfrac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \theta}} \, \frac{\mathrm{d}\theta}{\mathrm{d}u} \mathrm{d}u = \int \dfrac{\cos \theta}{ \sqrt{2 - 9 \sin^2 \theta}} \, \cdot \frac{3 \sec \theta}{\sqrt{2}} \mathrm{d}u$$
So $$\dfrac{3}{\sqrt{2}} \int \dfrac{1}{\sqrt{2 - 9 \sin^2 \theta}} \, \mathrm{d}u = \dfrac{3}{\sqrt{2}} \int \dfrac{1}{\sqrt{2 - (81u^2/2)}} \, \mathrm{d}u$$
And hence we get
$$\int \dfrac{3}{\sqrt{2 - (81u^2/2)}} \, \mathrm{d}u = \frac{1}{3} \arcsin \left(\frac{9u}{2}\right) + \mathrm{C}$$.
Back substituting yields
$$\frac{1}{3} \arcsin \left(\frac{3 \sin \theta}{\sqrt{2}}\right) + \mathrm{C}$$
|
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|
Find remainder when $777^{777}$ is divided by $16$
Find remainder when $777^{777}$ is divided by $16$.
$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.
Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.
Also $777=51\times 15+4$. Therefore,
$777^{777}=777^{51\times 15+4}={(777^{15})}^{51}\cdot777^4\equiv 1^{15}\cdot 9^4 \pmod{16}$ leading to $ 81\cdot81 \pmod{16} \equiv 1 \pmod{16}$.
But answer given for this question is $9$. Please suggest.
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Fermat's theorem will only work with primes.($16$ is not a prime)
But, it can be solved by the general formula, using Euler's theorem.
Since $gcd(9,16)=1, 9^{\phi(16)} \equiv 1 (mod 16)$. Since $16=2^4, \phi(16)=8$.
Therefore, $9^8 \equiv 1 (mod 16)$, and we have $777^{777} \equiv 9^{777} \equiv 9^9 \equiv 9*9^8 \equiv 9 (mod 16)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$? What I know is that for equations of type $x+y=8$, $xy$ attains its maximum value when $x=y$ and this can be proved by either solving the quadratic equation with completing the squares or finding the first and second derivatives of $xy$ w.r.t. $x$ to perform maxima/minima test. But both of these methods are applicable only when there is not more than one independent variable -- here $x$ is independent variable and $y=8-x$ is the dependent one.
I don't understand why $(a+1)(b+1)(c+1)(d+1)$ attains its maximum value at $a=b=c=d$ within the condition of $a+b+c+d=1$. How can we prove this fact? In some other similar posts I saw others using the Lagrange multiplier method. This method is too advanced for me to understand. I'm looking for some kind of algebraic proof similar to comleting with sqaure proof as in case of finding the maximum value of $xy$.
|
In light of all the AM-GM answers, here is a variational approach.
Suppose we know that $a+b+c+d=1$, then we also know that for any direction moved $(\delta a,\delta b,\delta c,\delta d)$ that maintains this condition
$$
\begin{align}
0
&=\delta a+\delta b+\delta c+\delta d\\
&=(\delta a,\delta b,\delta c,\delta d)\cdot(1,1,1,1)\tag{1}
\end{align}
$$
In that same direction, the rate of change of $(a+1)(b+1)(c+1)(d+1)$ satisfies
$$
\begin{align}
\small\frac{\delta\left[(a+1)(b+1)(c+1)(d+1)\right]}{(a+1)(b+1)(c+1)(d+1)}
&=\small\frac{\delta a}{a+1}+\frac{\delta b}{b+1}+\frac{\delta c}{c+1}+\frac{\delta d}{d+1}\\
&\small=(\delta a,\delta b,\delta c,\delta d)\cdot\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)\tag{2}
\end{align}
$$
Case 1: $\boldsymbol{a,b,c,d\gt0}$ (interior maximum)
Suppose we are at the point where $(a+1)(b+1)(c+1)(d+1)$ is maximum given that $a+b+c+d=1$. Suppose further that we can find a direction that is perpendicular to $(1,1,1,1)$, but is not perpendicular to $\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)$. Then, since $a,b,c,d\gt0$, by $(1)$, we can travel in that direction, and its opposite, and $a+b+c+d$ will stay constant. However, by $(2)$, moving in that direction, or its opposite, will cause $(a+1)(b+1)(c+1)(d+1)$ to increase. This contradicts the maximality we assumed.
Thus, if $a,b,c,d\gt0$, $\left(\frac1{a+1},\frac1{b+1},\frac1{c+1},\frac1{d+1}\right)$ must be parallel to $(1,1,1,1)$. That is, $a=b=c=d=\frac14$.
Case 2: one or more of $\boldsymbol{a,b,c,d=0}$ (boundary maximum)
Using the same arguments as above, we get that the non-zero variables must be equal. So we must not only check $\left(\frac14,\frac14,\frac14,\frac14\right)$, but also $\left(0,\frac13,\frac13,\frac13\right)$, $\left(0,0,\frac12,\frac12\right)$, and $(0,0,0,1)$.
Since
$$
\left(\frac54\right)^4\ge\left(\frac43\right)^3\ge\left(\frac32\right)^2\ge2^1
$$
we get that the maximum occurs at $\left(\frac14,\frac14,\frac14,\frac14\right)$.
|
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|
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke the identity $(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ by adding and subtracting $3abc$ in the question statement. After doing all this when I substituted back the values of $a,b$ and $c$, I ended up with the initial question statement.
Any hints will be appreciated.
|
HINT:
Following your way,
$$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
|
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|
$P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$ "Let $X_1, X_2, ...$ independent random variables where $X_n\sim B(p_n)$ and $p_n = \frac{1}{n}$. Calculate $P\left(\limsup \left(X_n=0, X_ {n+1}=1,X_ {n+2}=0 \right)\right)$"
I suppose that i can use the lemma of Borel-Cantelli, but I don't know how interpret that limsup...
Thank you very much!
|
Is that really $X_n = 0$, $X_{n+1} = \color{red}{1}$ and $X_{n+2} = 0$ and not $X_n = 0$, $X_{n+1} = \color{red}{0}$ and $X_{n+2} = 0$?
If so, since the $X_n$'s are independent, the events $X_n = 0$, $X_{n+1} = 1$ and $X_{n+2} = 0$ are independent. Thus, we have
$$P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = P(X_n = 0)P(X_{n+1} = 1)P(X_{n+2} = 0) = \frac{n-1}{(n)(n+2)}$$
By limit comparison test with $\sum_{n=1}^{\infty} \frac{1}{n^2}$, we have that
$$\sum_{n=1}^{\infty} P(\{X_n = 0, X_{n+1} = 1, X_{n+2} = 0\}) = \sum_{n=1}^{\infty} \frac{n-1}{(n)(n+2)} < \infty$$
Hence, by BCL1, $P(\limsup \{X_n = 0, X_{n+1} = 1, X_{n+2} = 0\}) = 0$
If it is $X_n = 0$, $X_{n+1} = 0$ and $X_{n+2} = 0$:
Since the $X_n$'s are independent, the events $X_n = 0$, $X_{n+1} = 0$ and $X_{n+2} = 0$ are independent. Thus, we have
$$P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = P(X_n = 0)P(X_{n+1} = 0)P(X_{n+2} = 0)$$
Are the $X_n$'s Bernoulli? If $P(X_n = 1) = p_n = 1 - P(X_n = 0)$, then we have
$$= (1-\frac{1}{n})(1-\frac{1}{n+1})(1-\frac{1}{n+2})$$
$$= (\frac{n-1}{n+2})$$
Thus,
$$\sum_{n=1}^{\infty} P(\{X_n = 0, X_{n+1} = 0, X_{n+2} = 0\}) = \sum_{n=1}^{\infty} (\frac{n-1}{n+2})$$
By the test for divergence, the series diverges. Since the summand is nonnegative, the series diverges to infinity.
However, BCL2 requires independence (or pairwise independence) of the events $\{X_1 = 0, X_{2} = 0, X_{3} = 0\}$, $\{X_2 = 0, X_{3} = 0, X_{4} = 0\}$, ...
Consider the first events. We can see that $\{X_1 = 0, \color{red}{X_{2} = 0, X_{3} = 0}\}$ and $\{\color{red}{X_2 = 0, X_{3} = 0}, X_{4} = 0\}$ are NOT independent.
However, consider a subsequence of those events: $\{X_1 = 0, X_{2} = 0, X_{3} = 0\}$, $\{X_4 = 0, X_{5} = 0, X_{6} = 0\}$, $\{X_7 = 0, X_{8} = 0, X_{9} = 0\}$, ...
These events are independent or pairwise independent. So let us consider this series instead:
$$\sum_{n=1}^{\infty} P(\{X_{3n-2} = 0, X_{3n-1} = 0, X_{3n} = 0\})$$
$$ = \sum_{n=1}^{\infty} \frac{3n-3}{3n}$$
BCL2 gives you that $P(\limsup \{X_{3n-2} = 0, X_{3n-1} = 0, X_{3n} = 0\}) = 1$.
Convince yourself that $P(\limsup \{X_{n} = 0, X_{n+1} = 0, X_{n+2} = 0\}) = 1$
|
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|
Prove that $\frac{8}{5}\le 2a+b\le 8$ Let $a,b,c,d,e$ be real numbers such that
$$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$$
Prove that:
$$\dfrac{8}{5}\le 2a+b\le 8$$
|
Let we set $(a,b,c,d,e)=\frac{8}{5}\cdot(1,1,1,1,1)+(x_1,x_2,x_3,x_4,x_5)$.
Then we have:
$$\left\{\begin{array}{rcl}x_1+x_2+x_3+x_4+x_5&=&0,\\ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2&=&\frac{16}{5}\end{array}\right.$$
and we have to find the stationary points of $2x_1+x_2$. Lagrange multipliers give:
$$ (2,1,0,0,0) = \lambda(1,1,1,1,1)+2\mu(x_1,x_2,x_3,x_4,x_5)$$
so in the stationary points we have $x_3=x_4=x_5=t$, $\lambda+2\mu t=0$,
$$ 2\mu x_1 = 2+2\mu t,\qquad 2\mu x_2 = 1+2\mu t $$
then:
$$ (x_1,x_2,x_3,x_4,x_5) = \left(\frac{1}{\mu}+t,\frac{1}{2\mu}+t,t,t,t\right) $$
or:
$$ (x_1,x_2,x_3,x_4,x_5) = \left(-\frac{7}{3},-\frac{2}{3},1,1,1\right)\cdot t = \left(-\frac{7}{3},-\frac{2}{3},1,1,1\right)\cdot \frac{\pm 3}{5}$$
so that:
$$ -\frac{16}{5}\leq 2x_1+x_2 \leq \frac{16}{5}$$
and:
$$ \frac{8}{5}\leq 2a+b \leq 8 $$
as wanted, with equality attained by $\left(a,b,c,d,e\right) = \frac{8}{5}(1,1,1,1,1)\pm\frac{1}{5}\left(-7,-2,3,3,3\right).$
|
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|
A general method for integration of rational function. $\int\frac {x^3}{1+x^5}$
ATTEMPT:
I did the following substitution:
Let $x=\frac{1}{t}.$
$dx=\frac{-1}{t^2}dt.$
substituting back:
$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.
Next i substituted $x=p^\frac{2}{5}.$
$dx=\frac{2}{5}\frac{p^{3/5}}{1+p^2}.$
Now let $p=tan\theta.$
$dp=sec^2\theta$$d\theta.$
substituting back:
$I=\frac{2}{5}\int tan^{3/5}\theta$$d\theta$ which i couldn't integrate further even by trying By parts method.
Is there a general approach for problems of the form (not the algebraic twins):
$\int \frac{x^m}{1+x^n}dx$ where $n-m \ne1 $
|
There is a general closed form for such integrals but it's not elementary or pretty at all.
$$\int \frac{x^m}{1+x^n} \, \mathrm{d}x = \frac{x^{m+1} \, _2F_1\left(1, \frac{m+1}{n};\frac{m+n+1}{n}; -x^n\right)}{m+1} + \mathrm{C}.$$
Where $_2F_1\left(a, b;c; x\right)$ is the hypergeometric function.
With regards to your specific problem, try partial fractions (although I suspect you may have mistyped, given how ugly this is)
$$\frac{x^3}{1+x^5} = \frac{\sqrt{5}x - 5x + \sqrt{5} + 5}{5\sqrt{5}(2x^2 + \sqrt{5}x - x + 2)} + \frac{\sqrt{5}x + 5x + \sqrt{5} - 5}{5\sqrt{5}(2x^2 - \sqrt{5}x - x + 2)} - \frac{1}{5(x+1)}.$$
|
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|
Sum of trigonometric infinite series I am trying to prove that for any $x\geq 1$ we have:
$$ \sum_{m=1}^{\infty} \frac{\sin\frac{(2m-1)\pi}{x}}{\left(\frac{(2m-1)\pi}{x}\right)^3} = \frac{x}{8}(x-1). $$
Could I have some help please? I am thinking that Fourier series could help, but I found nothing until now. Thank you very much!
|
Yes. Fourier series can help.
It is equivalent to finding the limiting function of the Fourier series
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} $ .
Note that$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {nt} \right)}}{n}} $ converges to $(\pi - t)/2$ for $0 < t < 2 \pi $ .
Observe that
${\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = - 2\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} $ .
We can deduce this from
$Log(1 - {z^2}) = - 2\sum\limits_{n = 1}^\infty {\frac{{{z^{2n}}}}{{2n}}} $.
Hence $\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = - \frac{1}{2}{\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = \frac{1}{2}\frac{1}{2}(\pi - 2t) = \frac{1}{4}(\pi - 2t)$ for $ 0 < t < \pi $.
Therefore,
$\sum\limits_{n = 1}^\infty {\frac{{\sin ((2n - 1)t)}}{{2n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{\sin (nt)}}{n}} - \sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = \frac{1}{2}(\pi - t) - \frac{1}{4}(\pi - 2t) = \frac{\pi }{4}$ for $0 < t < \pi$ .
We may integrate the above Fourier series term by term to give the integral of the function on the right:
$ - \sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} = \frac{\pi }{4}t$ I.e.,
$\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} - \frac{\pi }{4}t = \frac{{{\pi ^2}}}{8} - \frac{\pi }{4}t$ .
Integrating again gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^2}}}{8}t - \frac{\pi }{8}{t^2}$
Now for $x \ge 1$ , $ t = \pi/x \le \pi $.
Substituting this value of $t$ in the above equation gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1){\textstyle{\pi \over x}}} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^3}}}{8}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$ ,
which is equivalent to your equation for $x \ge 1$.
|
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|
Integral of $\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}$ So we have to evaluate $\int\frac{x^2+1}{(1-x^2)\sqrt{1+x^4}}dx$.
My work-
We can write the integrand as $\frac{(x+1)^2-2x}{(1-x)(1+x)\sqrt{1+x^4}}dx$.
So we wish to deduce $\int\frac{(x+1)}{(1-x)\sqrt{1+x^4}}dx-\int\frac{2x}{(1-x^2)\sqrt{1+x^4}}dx$
So lets write it as $I_1+I_2$.
I tried and I can't evaluate $I_1$.
For $I_2$, I took $x^2=t;\ 2xdx=dt$.
The integral becomes $\int\frac{dt}{(1-t)\sqrt{1+t^2}}dt$.
Okay, now we do $t=\tan\theta; \ dt=\sec^2\theta d\theta.$
$\int\frac{\sec\theta}{1-\tan\theta} d\theta$
Now what to do? Please guide me.?
|
$\bf{My\; Solution::}$ Given $$\displaystyle I = \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}dx = \int\frac{1+x^2}{x^2\cdot \left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}-x\right)^2+\left(\sqrt{2}\right)^2}}dx$$
So $$\displaystyle I = \int\frac{\left(\frac{1}{x^2}+1\right)}{\left(\frac{1}{x}-x\right)\sqrt{\left(\frac{1}{x}-x\right)^2+\left(\sqrt{2}\right)^2}}dx$$
Now Put $\displaystyle \left(\frac{1}{x}-x\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$
So Integral $$\displaystyle I = \int\frac{1}{t\sqrt{t^2+\left(\sqrt{2}\right)^2}}dt$$
Now Put $t = \sqrt{2}\tan \theta \;,$ Then $dx = \sqrt{2} \sec^2 \theta d\theta$
So $$\displaystyle I = \int\frac{\sqrt{2}\sec^2 \theta }{2\tan \theta \cdot \sec \theta }d\theta =\frac{1}{\sqrt{2}}\int \csc \theta d\theta = -\frac{1}{\sqrt{2}}\ln \left|\csc \theta + \cot \theta \right|+\mathcal{C}$$
So $$\displaystyle I = -\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{t^2+2}}{t}\right|+\mathcal{C}\;,$$ Where $$\displaystyle \left(\frac{1}{x}-x\right) = t.$$
|
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|
Looking for help to understand example of Group I am looking for someone to help me to understand what is going on in the following example, from Hersteins "Topics in Algebra".
It says,
Let $G$ be the set of all $2*2$ matrices $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ where $a,b,c,d$ are integers modulo $p$, $p$ is a prime number such that $ad-bc \neq 0$, with matrix multiplication having its usual definition, then verify that $G$ is a non-abelian finite group.
I think I am just even confused on some definitions. From what I understood previously, to say $a\equiv b \quad mod \quad n$ is to say that $n | a-b$.
So what is it to mean the these entries are integers modulo $p$? Can anyone share some insight about this? Thank you.
|
$G$ is group because if $x,y \in G$ then $x^{-1} y \in G$ and $G$ is nonabelian because:
$\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix} *\begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \neq \begin{pmatrix} 0 & 1 \\ 1 & 1\\ \end{pmatrix} *\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}$
And at last $G$ is finite because $|G|= (p^2-1)(p^2-p)$
Remarks: addition and multiplication in $G$ is like this:
For example if $p=7$ then $4*5 + 3^2=6+2=1$ ! because $4*5 \equiv 6 \mod {7}$ and $3^2=9 \equiv 2 \mod {7}$ and then $6+2=8 \equiv 1 \mod {7}$ hence:
$\begin{pmatrix} 4 & 3 \\ 0 & 1 \\ \end{pmatrix} *\begin{pmatrix} 5 & 1 \\ 3 & 0 \\ \end{pmatrix}= \begin{pmatrix} 1 & 4 \\ 3 & 0 \\ \end{pmatrix}$
|
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|
If det $A = 0$ and $\det B \neq 0$ then show that $abc = -1$ This has been hurting my head for a while now....
If
$$
\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}=0
$$
And
$$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} ≠0
$$
Then show that $abc=-1$.
|
First note that,
$$\begin{vmatrix}
a & a^2 & a^3+1 \\
b & b^2 & b^3+1 \\
c & c^2 & c^3+1 \\
\end{vmatrix}
=
\begin{vmatrix}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3 \\
\end{vmatrix}
+
\begin{vmatrix}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1 \\
\end{vmatrix}=0$$
Then, $$abc\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}
=
\begin{vmatrix}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3 \\
\end{vmatrix}
=
-\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}$$
Therefore, $abc=-1$
|
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|
Difference quotient of $f(x)= 2-6x+4x^2$ I need to find $f(a), f(a + h)$, and the difference quotient
$$\frac {f(a + h) − f(a)}{h},$$
where $h\neq 0$ and $f(x) = 2-6x+4x^2$.
My work:
$$f(a) = 2-6a+4a^2,\ \ f(a+h) = 2-6(a+h)+4(a+h)^2.$$
I need help on the last one:
$$\frac {f(a+h)-f(a)}{h}.$$
My work:
\begin{split} \frac {f(a+h)-f(a)}{h} &= \frac {(a+h)-(a)}{h}\\
&= \frac {2-6a-6h+4a^2+8ah+8h^2-2+6a-4a^2}{h}\\
&=\frac {-6h+8ah+8h^2}{h}\\
&= {-6+8a+8h}
\end{split}
UPDATE: It looks like I accidentally squared the numerator. So instead it would be $\frac{6+8a+8h}{h}$. Can anyone confirm?
|
You need $\frac{f(a+h)-f(a)}{h}$, but you seem to have computed something else not involving $f$. Try this,
\begin{equation}
\frac{f(a+h)-f(a)}{h}=\frac{(2-6(a+h)+4(a+h)^2)-(2-6a+4a^2)}{h} \\
=\frac{2-6a-6h+4a^2+8ah+4h^2-2+6a-4a^2}{h} \\
=\frac{-6h+8ah+4h^2}{h} \\
=-6+8a+4h
\end{equation}
|
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}
|
Finding subgroups of $\mathbb{Z}_{13}^*$
I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)
My attempt:
$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$
Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$
$$\begin{align}
&12^1=12\mod 13\\
&12^2=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{12,1\}}$ is a subgroup of order $2$
$$\begin{align}
&3^1=3\mod 13\\
&3^2=9\mod 13\\
&3^3=1\mod 13\\
\end{align}$$
$$\begin{align}
&9^1=9\mod 13\\
&9^2=3\mod 13\\
&9^3=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{9,3,1\}}$ is a subgroup of order $3$
$$\begin{align}
&5^1=5\mod 13\\
&5^2=12\mod 13\\
&5^3=8\mod 13\\
&5^4=1\mod 13\\
\end{align}$$
$$\begin{align}
&8^1=8\mod 13\\
&8^2=12\mod 13\\
&8^3=5\mod 13\\
&8^4=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{1,5,12,8\}}$ is a subgroup of order $4$
$$\begin{align}
&4^1=4\mod 13\\
&4^2=3\mod 13\\
&4^3=12\mod 13\\
&4^4=9\mod 13\\
&4^5=10\mod 13\\
&4^6=1\mod 13\\
\end{align}$$
$$\begin{align}
&10^1=10\mod 13\\
&10^2=9\mod 13\\
&10^3=12\mod 13\\
&10^4=3\mod 13\\
&10^5=4\mod 13\\
&10^6=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{4,3,12,9,10,1\}}$ is a subgroup of order $6$
Is it correct? is there any easier method?
|
Since $2$ is a generator of $\Bbb{Z}_{13}^*$ you have that the subgroups are exactly
$$\{ \langle 2^n \rangle : n \mbox{ divides } 12\}$$
i.e.
$$\langle 2 \rangle \\ \langle 2^2 \rangle = \langle 4 \rangle \\ \langle 2^3 \rangle = \langle 8 \rangle \\ \langle 2^4 \rangle = \langle 3 \rangle \\ \langle 2^6 \rangle = \langle -1 \rangle \\ \langle 2^{12} \rangle = \langle 1 \rangle \\$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Can someone please check my work?: $\cos^2(x)=1-\sin(x)$ $$\begin{align}\cos^2(x)&=1-\sin(x)\\
1-\sin^2(x)&=1-\sin(x)\\
(1-\sin x)(1+\sin x)&= 1-\sin(x)
\end{align}$$
divide both sides by $1 - \sin(x)$
End up with $1 + \sin(x)$
The answer is supposed to be in radians between $0$ and $2 \pi$.
So I get $1+\sin(x)=0$
$$\sin(x)=-1 = -90\text{ degrees } = -\pi/2 \text{ or }3\pi/2$$
|
There are two correct ways to proceed from
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
first way
This will be true if
$1 - \sin x = 0$
$x \in \{90^{\circ}, 270^{\circ}\}$
If not, then we can divide both sides by $1 - \sin x$, getting
$1 + \sin x = 1$
$\sin x = 0$
$x \in \{0^{\circ}, 180^{\circ}\}$
second way
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
$$(1-\sin x)(1+\sin x) - (1-\sin x) = 0$$
$$(1-\sin x)(1+\sin x - 1) = 0$$
$$(1-\sin x)\sin x = 0$$
$$1 - \sin x = 0 \; \text {or} \; \sin x = 0\ldots$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the equation $\{x\}+\{\frac{1}{x} \}=1$ Solve the equation $$\large\{x\}+\left\{\dfrac{1}{x}\right\}=1$$
given $x\in\mathbb R\setminus \{0\}$ and $\{x\}$ denotes the fractional part of $x$.
|
$\{x\}+\{\frac{1}{x}\}=1$
$x+\frac{1}{x}=1+[x]+[\frac{1}{x}]$
$x+\frac{1}{x}$ is an integer.
let $n=x+\frac{1}{x}$
$n=1+[x]+[n-x]=1+[x]+n+[-x]$
$[x]+[-x]=-1$
$x\not\in\mathbb Z$
$n=x+\frac{1}{x}$
$x^2-nx+1=0$
since x is real, discriminant $\geq 0$
$n^2\geq 4$
$|n|\geq 2$
$|n|=2 \iff |x|=1 \implies x\in\mathbb Z$ , contradiction
when $n>2$ ,
$0 < \frac{n-\sqrt{n^2 - 4}}{2} < 1$ or $n-1 < \frac{n+\sqrt{n^2 - 4}}{2} < n$ , so $x\not\in\mathbb Z$
when $n<-2$ ,
$n < \frac{n-\sqrt{n^2 - 4}}{2} < n+1$ or $-1 < \frac{n+\sqrt{n^2 - 4}}{2} < 0$ , so $x\not\in\mathbb Z$
the solns are $x=\frac{n\pm\sqrt{n^2 - 4}}{2}$ , where $n\in\mathbb Z\setminus\{-2,-1,0,1,2\}$
|
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|
If $x+y+z=3$, then $\sum_{\text{cyc}}\frac{x^2}{2y^2-y+3}\ge\frac{3}{4}$
Let $x,y,z>0$, be such that $x+y+z=3$. Show that
$$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$
I've tried many things but all have failed.
$$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum_{\text{cyc}}(2y^2-y+3)\right)\ge (x+y+z)^2=9.$$
But
$$\sum_{\text{cyc}}(2y^2-y+3)=2\sum_{\text{cyc}}x^2+6\ge 12.$$
|
I would go into homogeneous coordinates,
$$A=3=x+y+z$$
$$X=2x-y-z \quad x=(A+X)/3$$
and cyclically for $Y$ and $Z$. The coordinates $X$, $Y$, $Z$ are orthogonal to $A$, so whatever their values, they respect the constraint to the plane. Additionally, $X+Y+Z=0$.
With this, the sum is rewritten:
$$\sum \frac{(A+X)^2}{2(A+Y)^2-3(A+Y)+27}=\sum\frac{9+6X+X^2}{36+3Y+2Y^2}$$
The $X=Y=Z=0$ lies on the symmetry axis, and there, the sum is exactly $\frac{3}{4}$. We now just have to prove that the function is increasing off-axis.
The denominator has no poles on the domain, so the Taylor expansion converges. Just write
$$\frac{1}{4}\sum (1+X/2+X^2/9)(1-(Y/12+Y^2/18)+(Y/12+Y^2/18)^2-(Y/12+Y^2/18)^3+\cdots)$$
and you can extract the second derivatives quite easily.
It may be a bit tricker to prove the function does not turn around and reach below the axis value, but at least you know where to start.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determining $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$
$$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$
Attempt:
Simplification of the root factor:
$$\sqrt{x^4+2x^3-x^2+2x+1}=\frac{1}{x}\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}.$$
Arranging the rest of the factors as:
$$\frac{x-1}{x^2(x+1)}=\frac{x^2-1}{x^2(x+1)^2}$$
Now I did the following substitution: let $(x+\frac{1}{x})=t$, so $(1-\frac{1}{x^2})dx=dt$
Arranging the integral: $$I=\int \frac{\sqrt{t^2+2t-3}}{t+2}dt.$$
But what to do next?
I tried integration by parts for this but couldn't simplify my result.
|
Maybe it's not the most rapid way, but it seems work. We have, taking $u=t+1
$ $$\int\frac{\sqrt{t^{2}+2t-3}}{t+2}dt=\int\frac{\sqrt{u^{2}-4}}{u+1}du
$$ and taking $u=2\sec\left(v\right)
$ we have $$=4\int\frac{\tan^{2}\left(v\right)\sec\left(v\right)}{2\sec\left(v\right)+1}dv=4\int\frac{\tan^{2}\left(v\right)}{\cos\left(v\right)+2}dv
$$ using $\tan^{2}\left(v\right)=\sec^{2}\left(v\right)+1
$. Now we can take $w=\tan\left(v/2\right)
$ and get $$=32\int\frac{w^{2}}{w^{6}+w^{4}-5w^{2}+3}dw=32\int\frac{w^{2}}{\left(w^{2}-1\right)^{2}\left(w^{2}+3\right)}dw
$$ and now using a boring partial fractions you can transform the integral in a tractable form.
|
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|
Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent. I know that to prove the existence of a square root of $-1$ in $\mathbb{Z}_5$, I can just plug $x = -5$ and $a = 1/2$ into the Taylor expansion$$(1 + x)^a = \sum_{n=0}^\infty \binom{a}{n} x^n.$$However, I am confused on how I would make sure that the resulting sum is convergent. Could anyone help me with that? Thanks very much in advance.
|
The convergence should be in $\mathbb{Z}_5$, that works differently than in $\mathbb{R}$.
I try first an approximation
$$2^2 = 4 = -1 + 5$$
Say I have $a$ with $a^2 = -1$. Then
$$\left(\frac{a}{2}\right)^2 = \frac{-1}{-1+5}= \frac{1}{1-5}$$
So take
$$\frac{a}{2} = (1-5)^{-1/2}$$
We have the binomial formula
$$(1-t)^{-1/2} = 1+\frac{t}{2}+\frac{3 t^2}{8}+\frac{5 t^3}{16}+\frac{35 t^4}{128}+\frac{63 t^5}{256}+\frac{231 t^6}{1024}+\frac{429 t^7}{2048}+\\
+\frac{6435 t^8}{32768}+\frac{12155 t^9}{65536}+\frac{46189 t^{10}}{262144}+ \cdots$$
convergent in $\mathbb{Z}_5$ if $|t|_5<1$. Now, $t=5$ and $|5|_5=1/5 <1$. Therefore:
$$a = 2\cdot ( 1+\frac{5}{2}+\frac{3\cdot 5^2}{8}+\frac{5 \cdot5^3}{16}+\frac{35\cdot 5^4}{128}+\frac{63\cdot 5^5}{256}+\frac{231\cdot 5^6}{1024}+\frac{429\cdot 5^7}{2048}+\\
+\frac{6435\cdot 5^8}{32768}+\frac{12155\cdot 5^9}{65536}+\frac{46189 \cdot5^{10}}{262144} + \cdots)$$
with the other root $-a$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the intersection point between two lines using a matrix I'm trying to find the intersection point (if any) of two lines. Long story short, here are the parametric equations:
For the first line:
$$x = 3 + 4\lambda_1\\
y = 4 + \lambda_1\\
z = 1$$
For the second line:
$$x = -1 + 12\lambda_2\\
y = 7 + 6\lambda_2\\
z = 5 + 3\lambda_2$$
If you try to solve the equations
$$3 + 4\lambda_1 = -1 + 12\lambda_2\\
4 + \lambda_1 = 7 + 6\lambda_2\\
1 = 5 + 3\lambda_2$$
You can easily see that a solution is
$$\lambda_1 = -5\\
\lambda_2 = -\frac{4}{3}$$
But I wanted to solve the equation with a matrix instead (I like to complicate things):
$$\begin{bmatrix}
4 && -12 && -4\\
1 && -6 && 3\\
0 && 3 && 4
\end{bmatrix}$$
$$\frac{1}{4}r_1$$
$$\begin{bmatrix}
1 && -3 && -1\\
1 && -6 && 3\\
0 && 3 && 4
\end{bmatrix}$$
$$r_1-r_2$$
$$\begin{bmatrix}
1 && -3 && -1\\
0 && -3 && 4\\
0 && 3 && 4
\end{bmatrix}$$
$$r_3+r_2$$
$$\begin{bmatrix}
1 && -3 && -1\\
0 && 0 && 8\\
0 && 3 && 4
\end{bmatrix}$$
... And we got something odd! The second row says that $0\lambda_1 + 0\lambda_2 = 8$, but that can't be true. So, doesn't this mean that there is no solution because the system has an inconsistency? What did I do wrong?
|
There is an error in your initial matrix: the second row should be [0 3 -4] instead of [0 3 4] because the equation is $3\lambda_2=-4$.
I row reduced the matrix with that last row and got the same values: $\lambda_1=-5$ and $\lambda_2=\frac{-4}{3}$
Also note that you only really need two rows because you have two unknowns.
Row reduction:
$$ \begin{bmatrix} 1 & -6 & 3 \\ 0 & 3 & -4 \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -5 \\ 0 & 3 & -4\end{bmatrix} \implies \lambda_1=-5, 3\lambda_2=-4$$
EDIT: using the three equations, you just get a row of zeros
$$ \begin{bmatrix} 1 & -6 & 3 \\ 4 & -12 & -4 \\ 0 & 3 & -4 \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -5 \\ 0 & 12 & -16 \\ 0 & 3 & -4\end{bmatrix}\to \begin{bmatrix} 1 & 0 & -5 \\ 0 & 0 & 0 \\ 0 & 3 & -4\end{bmatrix}$$
|
{
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|
How do I evaluate this integral :$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx$? Is there someone who can show me how to evaluate this integral:
$$\int \frac{\sqrt{-x^2-x+2}}{x^2}dx.$$
I have tried many changes of variables but I haven't succeeded yet. Thank you for any help.
|
$\begin{gathered}
\int {\frac{{\sqrt { - {x^2} - x + 2} }}{{{x^2}}}} \,dx \hfill \\
= - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{\left( { - 2x - 1} \right)}}{{2\sqrt { - {x^2} - x + 2} }}} \frac{{ - 1}}{x}\,dx\quad \quad \left( {{\text{Integrating by parts}}} \right) \hfill \\
= - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{2x + 1}}{{2x\sqrt { - {x^2} - x + 2} }}} \,dx \hfill \\
= - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {\int {\frac{{dx}}{{\sqrt {\left( {2 + x} \right)\left( {1 - x} \right)} }} + \int {\frac{{dx}}{{2x\sqrt { - {x^2} - x + 2} }}} } } \right] \hfill \\
= - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \left[ {2{{\sin }^{ - 1}}\sqrt {\frac{{x + 2}}{3}} - \frac{1}{{2\sqrt 2 }}\log \left| {\frac{1}{{x\sqrt 2 }} - \frac{1}{{4\sqrt 2 }} + \sqrt {\frac{1}{{2{x^2}}} - \frac{1}{{4x}} - \frac{1}{4}} } \right|} \right]+c \hfill \\
\end{gathered} $
|
{
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|
Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$
I tried to integrate that function for $n=1,2,3$ to see whether there's a pattern (recurrence relation), but I just can't figure out how to write the pattern and how to prove that statement. Must I use induction?
|
Here is a way you could do it without induction, using the Beta function and the Gamma function. Let $\sqrt{t}=x$, so that
$$
I_n=\int_0^1(1-t)^n\frac{t^{1/2}}{2}dt.
$$
This now looks like the Beta function, so we have
$$
I_n=\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma ((n+1)+1/2)}.
$$
Using the fact that $\Gamma(1/2)=\sqrt{\pi}$, $\Gamma(n+1)=n!$ for positive integer $n$, and $\Gamma((n+1)+1/2)=\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt{\pi}$ for positive integer $n$ (all of which can be found in linked wikipedia page), we have
$$
I_n=\frac{\sqrt{\pi}n!4^{n+1}(n+1)!}{2(2n+2)!\sqrt{\pi}}=\frac{n!4^{n+1}(n+1)!}{2(2n+2)!}=\frac{n!(n+1)!2^{2n+1}}{(2n+2)!}=\frac{2\cdot4\cdot\dots\cdot 2n}{3\cdot5\cdot\dots 2n+1}
$$
Edit: The last equality is somewhat subtle. Note that $(n+1)!\cdot 2^{n+1}$ cancels the even terms in the denominator. The remaining powers of $2$ distribute over the $n!$.
|
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|
How to solve a system of logarithmic equations? I need to create a function with the following properties:
$$f(1)=1$$
$$f(65)=75$$
$$f(100)=100$$
Additionally, the function needs to grow logarithmically. So that gives three equations:
$$A \cdot \ln(B \cdot 1 + C) = 1$$
$$A \cdot \ln(B \cdot 65 + C) = 75$$
$$A \cdot \ln(B \cdot 100 + C) = 100$$
I am having trouble with using substitution to solve this. Barring there is no analytic way to solve this system, how would I use a numerical approximation for $A, B$ and $C$?
|
This is not an answer, just what I tried using Count Iblis's hint.
Let $$f(x) = A\ln(x+B)+ C$$
You can always write the function with the coefficient of $x$ being $1$, so we assume that it is.
\begin{align*}
f(1)&=A\ln(1+B)+ C=1\\
f(65)&=A\ln(65+B)+ C=75\\
f(100)&=A\ln(100+B)+ C=100
\end{align*}
We use Count Iblis's strategy (dividing the second and third equations by the first) to get these equations:
\begin{align*}
\frac{75-C}{1-C}&=\frac{\ln(65+B)}{\ln(1+B)}\\
\frac{100-C}{1-C}&=\frac{\ln(100+B)}{\ln(1+B)}
\end{align*}
We can rewrite these as
\begin{align*}
(1+B)^{\frac{75-C}{1-C}} &=65+B\\
(1+B)^{\frac{100-C}{1-C}} &=100 + B
\end{align*}
If we let $m = 1+B$ and $n=1-C$ we get
\begin{align*}
m^{1 + (74/n)} &= m + 64\\
m^{1 + (99/n)} &= m + 99
\end{align*}
Not sure what to do from here ...
|
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|
Examining the convergence of $\int_{0}^{1}\left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx$ Okay so I'm trying to determine whether $\int_{0}^{1}\left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx$ converges and if so, to what value? So the function $\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor = 0$ when $x=\frac{1}{k}$ where $k$ is a natural number, and everywhere else $\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor = 1$ From there I split the integral:
\begin{align*}
\int_0^1 \left(\left\lceil \frac{1}{x} \right\rceil-\left\lfloor \frac{1}{x} \right\rfloor\right) \, dx &= \int_{1/2}^1 1 \, dx +\int_{1/3}^{1/2} 1 \, dx + \int_{1/3}^{1/4} 1 \, dx + \cdots \\[10pt]
&= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \cdots \\[10pt]
&= \lim_{m \to \infty} 1-1/m = 1\;.
\end{align*}
Am I correct?
|
Yes, you are correct. But it is much faster if you say that $\lceil \frac{1}{x} \rceil-\lfloor \frac{1}{x} \rfloor$ is $1$ except in a set of measure zero. If you don't know what the meausre of a set is, just say that the set is countable. Then
$$\int_0^1\left(\left\lceil \frac{1}{x}\right \rceil-\left\lfloor \frac{1}{x} \right\rfloor\right)dx=\int_0^11dx=1 $$
Your solution is still very nice.
|
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|
Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$
Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$
I tried to simplify it ,
$2^{a+3}=4^{a+2}-48\\
2^{a+3}=2^{2(a+2)}-2^4\cdot 3\\
2^{2a}-2^{a-1}- 3=0\\
$
I don't know how to go from here.
This question is from chapter quadratic equations, so i think there must be hidden quadratic idea in it.
I look for a short and simple way.
I have studied maths up to $12$th grade.
|
$$2^{a+3}=4^{a+2}-48\Longleftrightarrow$$
$$48+2^{a+3}-4^{a+2}-48=0\Longleftrightarrow$$
$$-8\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)=0 \vee \left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$2^a=2 \vee 3+2^{a+1}=0\Longleftrightarrow$$
$$a=\frac{\log_{10}(2)}{\log_{10}(2)} \vee 2^{a+1}=-3\Longleftrightarrow$$
$$a=1 \vee a+1=\frac{\log_{10}(-3)}{\log_{10}(2)}\Longleftrightarrow$$
$$a_1=1 \vee a_2=\frac{\log_{10}(-3)}{\log_{10}(2)}-1\Longleftrightarrow$$
$$a_1=1 \vee a_2=\frac{\ln(3)+\pi i}{\ln(2)}-1$$
And we see that $a_2$ isn't an real solution so the only real solution is $a=1$.
|
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|
Infinite number of ways to write $1=\frac{1}{n}+\frac{1}{a_1}+\cdots+\frac{1}{a_k}$ How can I show that there is an infinite number of ways in which $1$ can be written in the form $$1=\frac{1}{n}+\frac{1}{a_1}+\cdots+\frac{1}{a_k},$$ where $n>1$ is an integer (this number is fixed), each $a_i$ is an integer and $n<a_1<\cdots<a_k.$
I don't know what to do..
Thank you for your help.
|
Hint: For any integer $a_k > 1$, we have $\dfrac{1}{a_k} = \dfrac{1}{a_k+1} + \dfrac{1}{a_k^2+a_k}$, where $a_k+1 < a_k^2+a_k$.
So, if $1 = \dfrac{1}{n}+\dfrac{1}{a_1}+\cdots+\dfrac{1}{a_{k-1}}+\dfrac{1}{a_k}$ for some integers $n < a_1 < \cdots < a_{k-1} < a_k$, then we also have $1 = \dfrac{1}{n}+\dfrac{1}{a_1}+\cdots+\dfrac{1}{a_{k-1}}+\dfrac{1}{a_k+1} + \dfrac{1}{a_k^2+a_k}$, where $n < a_1 < \cdots < a_{k-1} < a_k+1 < a_k^2+a_k$.
You should be able to finish the proof from this.
|
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|
Is the substitution of standard angles while proving the equality of trigonometric formulas allowed? Here is a problem that my class 10 maths teacher gave me:
Prove that $\sec^4\theta$ - $\sec^2\theta$ = $\tan^4\theta$ + $\tan^2\theta$
She expected me to use trigonometric identities to prove such equality, but I instead substituted $\theta$ with standard angles, and substituted that with their values.
Here's what I did:
$\sec^4\theta -\sec^2\theta = \tan^4\theta + \tan^2\theta$ take $\theta$ as $45^o$ $\sec^4 45^o - \sec^2 45^o =\tan^4 45^o + \tan^2 45^o$ $(\sqrt2)^4 - (\sqrt2)^2 = (1)^4 +(1)^2$ $4 - 2 = 1+1$ $2 = 2$ therefore LHS = RHS Reason: Standard angles are universal truths, henceforth they work throughout the universe. Also, the functions used require a parameter (although not required in proving this using identities), so I substituted them all with $45^o$.
So, my question is 'Is it okay if I substitute standard angle values for trigonometric functions while proving their equality?'.
|
LHS:
$$
\sec^4\theta - \sec^2\theta = \sec^2\theta(\sec^2\theta - 1) = \frac{1}{\cos^2\theta}\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^4\theta}
$$
RHS:
$$
\tan^4\theta + \tan^2\theta = \tan^2\theta (1 + \tan^2\theta) = \frac{\sin^2\theta}{\cos^2\theta}\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^4\theta}
$$
It's all.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequence are $2n$ and $4n+2$ bigger than $(n+1)$ in the original i.e. $(3n+1)(5n+3).$
|
You can use the following relation :
$$\sum_{k=n+1}^{2n}k^3=\sum_{k=1}^{\color{red}{2n}}k^3-\sum_{k=1}^{n}k^3$$
(To get the first sum on RHS, you only need to replace the $n$ in the given equation with $\color{red}{2n}$.)
|
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|
Finding $4$ variables using $3$. if I have:
$ x=\dfrac{a-.5b-.5c+.25d}{a+b+c+d}$
$ y=\dfrac{\dfrac{b\sqrt{3}}{2}+\dfrac{c\sqrt{3}}{2}+\dfrac{d\sqrt{3}}{4}}{a+b+c+d}$
$ z=a+b+c+2d $
Then how do I get back to:
$ a= $ , $ b= $ , $ c= $ , and $ d= $ ?
When there is a divide by zero error, $ x=0 $ and $ y=0 $.
When
$(a,b,c,d)=(e,e,0,0)$ and $e<>0$
then
$(x,y,z)=(.25,\dfrac{\sqrt{3}}{4},2e)$
but also when
$(a,b,c,d)=(0,0,0,e)$ and $e<>0$
then
$(x,y,z)=(.25,\dfrac{\sqrt{3}}{4},2e)$
again. So, everytime
($a=b$ and $c=0$)
then
$(a,b,0,0)=(0,0,0,a+b)$.
$x>=-.5$
$x\leq1$
$y\geq\dfrac{\sqrt{3}}{2}$
$y\leq-\dfrac{\sqrt{3}}{2}$
$z\geq0$
$z\leq765$ this could mathematically be up to $1020$ but will be always be $\leq 765$, because $a+b+2d \leq 510$.
$a\geq0$
$b\geq0$
$c\geq0$
$d\geq0$
$a\leq255$
$b\leq255$
$c\leq255$
$d\leq255$
$a$, $b$, $c$, and $d$ are all integers.
I can find $a$, $b$, and $c$, when I don't use $d$. but I would like to find what the highest $d$ value can be; while $a$, $b$, and $c$ are between or equal to $0$ and $255$.
When $d=0$, I get the formulas I want:
$a=\dfrac{z}{3}(2x+1)$
$b=\dfrac{z}{3}*(\sqrt{3}y-x+1)$
$c=-\dfrac{z}{3}(x+\sqrt{3}y-1)$
$d=0$
When $θ$ of $(x,y)$ from $(0,0)$ is $60°$ then only $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is between $-120°$ and $60°$ then only $a$,$c$, and $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is between $60°$ and $240°$ then only $b$,$c$, and $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is $240°$ (aka $-120°$) then only $c$ exists.
But I'd like the highest $d$ value; not lowest.
note: $atan(\dfrac{y}{x})$.
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Usually you can't :
The system can be rewritten as
$$x(z-d) - \frac{d}{4} = a - \frac{b}{2} - \frac{c}{2}$$
$$y(z-d) - \frac{d\sqrt{3}}{4} = \frac{b\sqrt{3} }{2} - \frac{c\sqrt{3}}{2}$$
$$ z-2d = a+b+c$$
This can be seen as a linear system with unknown $a$,$b$ and $c$, and it's not hard to show that that this system has an unique solution for every value of $d$
|
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|
Partial derivative of $f(x,y) = (x/y) \cos (1/y)$ So I'm not really sure whether I'm correct as several people are saying some of my syntax is wrong, where others are saying I have a wrong answer. I have checked my answer using Wolfram Alpha and it appears to be correct; could anyone please confirm/clarify?
Calculate the partial Derivative $\frac{\partial f}{\partial y}$ of
$$f(x,y) = \frac{x}{y} \cos\left(\frac{1}{y}\right).$$
Is this correct? (Slightly reduced working because it's super long to type out)
$x$ is constant
$$\frac{\partial f}{\partial y} \left(\frac{\cos(\frac{1}{y})}{y}\right)$$
Quotient Rule
$$\frac{\dfrac{\partial f}{\partial y} \left(\dfrac{\cos(\frac{1}{y})}{y}\right)y-\dfrac{\partial f}{\partial y}(y) \cos(\frac{1}{y})}{y^2}$$
Chain Rule
$$\frac{\partial f}{\partial y} \left(\cos(\tfrac{1}{y})\right) = \frac{\sin(\frac{1}{y})}{y^2}$$
$$x\frac{\frac{\sin\frac{1}{y}}{y^2}y-1\cos(\frac{1}{y})}{y^2}$$
Simplified Answer
$$\frac{x \left(\sin\frac{1}{y} -y\cos(\frac{1}{y})\right)}{y^3}$$
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$$\frac { \partial f\left( x,y \right) }{ \partial y } =\frac { \partial \left( \frac { x }{ y } \cos { \frac { 1 }{ y } } \right) }{ \partial y } =x\frac { \partial \left( \frac { 1 }{ y } \right) }{ \partial y } \cos { \frac { 1 }{ y } +\frac { x }{ y } \frac { \partial \left( \cos { \frac { 1 }{ y } } \right) }{ \partial y } } =\\ =-\frac { x }{ { y }^{ 2 } } \cos { \frac { 1 }{ y } +\frac { x }{ y^{ 3 } } \sin { \frac { 1 }{ y } } } =\frac { -x\left( y\cos { \frac { 1 }{ y } -\sin { \frac { 1 }{ y } } } \right) }{ { y }^{ 3 } } $$
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|
Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \frac {c_1}{x}$ = x + 1 + $\frac{0}{x}$ = x + 1, which is true.
Assume the case for $n=k$ is true, so $x^{3k} = a_kx + b_k + \frac{c_k}{x}$
So for $n = k+1$ (and using $a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,...), $ we have:
$ \begin{align}
x^{3(k+1)} &= (a_{k+1-1}x + b_{k+1-1}){x} + (a_{k+1-1} + b_{k+1-1} + c_{k+1-1}) +
( \frac{(a_{k+1-1} + c_{k+1-1})}{x}) \\
&= (a_k + b_k){x} + ( a_k + b_k + c_k) + \frac{(a_k + c_k)}{x} \\
&= a_{k+1}{x} + b_{k+1} + \frac{c_{k+1}}{x} \\\end{align} $
This is the same form as $ x^{3k} $ but for n=k+1.
Therefore, if the result is true for k, it is also true for (k+1).
Would you proceed like this or would you add the $(k+1)$th term and show that it is equal to the $k$th term with $n$ (or $k$) replaced by $k+1$?
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The basic idea is that $x^3=x+1$ implies
$$x^2=1+{1\over x}\quad\text{and}\quad x^4=x^2+x=1+{1\over x}+x$$
So if $x^{3k}=a_kx+b_k+{c_k\over x}$ then
$$\begin{align}
x^{3(k+1)}&=x^3\cdot x^{3k}\\
&=x^3(a_kx+b_k+{c_k\over x})\\
&=a_kx^4+b_kx^3+c_kx^2\\
\end{align}$$
Now replace the $x^4$, $x^3$, and $x^2$ in the last line with $1+{1\over x}+x$, $x+1$, and $1+{1\over x}$, respectively, simplify, and voila.
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|
$\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$
$\bf{My\; Solution::}$ We Can Write Given Series as
$$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2+r\cdot n}} = \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\frac{1}{\sqrt{1+\frac{r}{n}}}\cdot \frac{1}{n}$$
Now Convert into Reinman Integral,
Put $\displaystyle \frac{r}{n} = x\;,$ Then $\displaystyle \frac{1}{n} = dx$ and Changing Limits, We get
$$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\frac{1}{\sqrt{1+\frac{r}{n}}}\cdot \frac{1}{n} = \int_{0}^{1}\frac{1}{\sqrt{1+x}}dx\;,$$
Put $(1+x) = t^2$ and $dx = 2tdt$ and Changing Limits, We Get
$$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x}}dx =2 \int_{1}^{\sqrt{2}}\frac{t}{t}dt = 2\left(\sqrt{2}-1\right)$$
My Question is Can We solve It Using any Other Method Like
Using Squeeze Theorem or any Other Method.
Help Required Thanks,
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It can be shown that, for $\alpha\in(-1,0]$ and $N\in\mathbb{N}$, $$\frac{(N+1)^{\alpha+1}-1}{\alpha+1}\leq\sum_{k=1}^N\,k^\alpha\leq\frac{N^{\alpha+1}}{\alpha+1}\,.$$ Define $S_N$ for each $N\in\mathbb{N}$ to be $$\sum_{k=1}^N\,\frac{1}{\sqrt{k}}=\sum_{k=1}^N\,k^{-\frac{1}{2}}\,,$$ we have $$2(\sqrt{N+1}-1)\leq S_N \leq 2\sqrt{N}\,.$$
The required limit is $$\lim_{n\to\infty}\,\frac{S_{2n-1}-S_{n-1}}{\sqrt{n}}\,.$$
Now, for each $n\in\mathbb{N}$,
$$2\left(\sqrt{2n}-1\right)-2\sqrt{n-1}\leq S_{2n-1}-S_{n-1} \leq 2\sqrt{2n-1}-2\left(\sqrt{n}-1\right)\,,$$
which means
$$2\left(\frac{\sqrt{2n}-\sqrt{n-1}}{\sqrt{n}}\right)-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq 2\left(\frac{\sqrt{2n-1}-\sqrt{n}}{\sqrt{n}}\right)+\frac{2}{\sqrt{n}}\,.$$
That is,
$$\frac{2\left(n+1\right)}{\sqrt{n}\left(\sqrt{2n}+\sqrt{n-1}\right)}-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq \frac{2\left(n-1\right)}{\sqrt{n}\left(\sqrt{2n-1}+\sqrt{n}\right)}+\frac{2}{\sqrt{n}}\,,$$
or
$$\frac{2\left(1+\frac{1}{n}\right)}{\sqrt{2}+\sqrt{1-\frac{1}{n}}}-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq \frac{2\left(1-\frac{1}{n}\right)}{\sqrt{2-\frac{1}{n}}+1}+\frac{2}{\sqrt{n}}\,.$$
Both sides of the inequalities above go to $\frac{2}{\sqrt{2}+1}=2\left(\sqrt{2}-1\right)$, as $n\to\infty$. By the Squeeze Theorem, $$\lim_{n\to\infty}\,\frac{S_{2n-1}-S_{n-1}}{\sqrt{n}}=2\left(\sqrt{2}-1\right)\,.$$
|
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|
Sum of the series $\sum\limits_{n=0}^\infty \frac{1}{(3n+1)^3}$ The following result matches very good numerically:
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^3}=\frac{13}{27}\zeta(3)+\frac{2\pi^3}{81\sqrt{3}}.$$
Though I'm not sure how to approach this. How can we prove it? Also, is it possible to find closed form for $$\sum_{n=0}^\infty \frac{1}{(3n+1)^5}$$ or the like?
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Let $$S_1=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3} ,S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^3}.$$
It's easily seen that $$S_1+S_2+\sum_{n=1}^{\infty}\frac{1}{(3n)^3}=\zeta(3).$$
That is, $$S_1+S_2=\zeta(3)-\frac{1}{27}\zeta(3)=\frac{26}{27}\zeta(3)\tag{1}$$
However, $$S_1-S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}=\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^3}=\frac{4\pi^3}{81\sqrt{3}}\tag{2}$$
To calculate $(2)$, consider the Weierstrass form of the sinc function $$\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right)$$
and take $\frac{d^3}{dx^3}\log(\cdot)$ of both sides.
Set $x=\frac{\pi}{3}.$
Update. I found that this question has been already asked here, so I link these two: Sum related to zeta function
Also, I give an alternative way to calculate $(2)$.
Notice that
$$\frac{2}{\sqrt{3}} \,\sin\left( \frac{2 \pi n}{3} \right) =\begin{cases} 0 & n\equiv 0 \\ 1 & n\equiv 1 \\ -1 & n\equiv 2
\end{cases} \pmod{3}.$$
Therefore $$S_1 - S_2 =\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty} \frac{\sin(2\pi n/3)}{n^3}. \tag{3}$$
Finally, $(2)$ follows from the Fourier expansion $$ \sum_{n=1}^{\infty} \frac{\sin(2 \pi x n)}{n^3}=\frac{\pi^3}{3}\left( \{x\}-3\{x\}^2+2\{x\}^3\right), \tag{4}$$
where $\,\{x\}$ is the fractional part of $x$.
|
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|
Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^ 2 + y}$ where $y$ in this case will be $2/3$ ?
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We have, $$3\cos ^2t + 5\sin t = 1$$ $$\implies 3(1-\sin ^2t) + 5\sin t = 1$$ $$\implies 3\sin ^2t+ 5\sin t-2=0$$ Factorizing the expression, we get $$ (3\sin t-1)(\sin t+2)=0$$ $$\text{if}\ 3\sin t-1=0 \implies \sin t=\frac{1}{3}$$$$\implies \color{blue}{t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{Or} \ \color{blue}{t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{if}\ \sin t+2=0 \implies \sin t\neq -2$$ Hence, the general solution is $t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)$ or $t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)$
Where, n is any natural number
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|
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0 $$
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Set $A=x^2,B=x^2+x+1$.
Then,
$$\begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\&=(2A-B)(5A-B)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
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|
creative method to obtain range of newton function ?! I am searching for more proof that the range of $y=\frac{x}{x^2+1}$ is $ \frac{-1}{2}\leq y \leq \frac{+1}{2}$
these are my tries :
domain is $\mathbb{R}$
first : $$\quad{y=\frac{x}{x^2+1}\\yx^2+y=x \rightarrow x^2y-x+y=0 \overset{\Delta \geq 0 }{\rightarrow} 1-4y^2 \geq 0 \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2} }$$
second :
$$ \quad{y=\frac{x}{x^2+1} \overset{x=tan \alpha }{\rightarrow} \space y=\frac{tan \alpha}{tan^2 \alpha+1}=\frac{tan \alpha}{\frac{1}{cos^2\alpha}}=sin \alpha cos \alpha =\\\frac{2}{2} sin \alpha cos \alpha =\frac{1}{2}sin 2\alpha\\ -1\leq sin 2\alpha \leq 1 \rightarrow -\frac{1}{2} \leq \frac{1}{2}sin 2\alpha \leq \frac{1}{2} \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$ 3rd:
$$\quad{\frac{1}{y}=\frac{x^2+1}{x}=x+\frac{1}{x} \overset{|x+\frac{1}{x}|\geq 2}{\rightarrow} |\frac{1}{y}| \geq 2 \rightarrow |y| \leq \frac{1}{2} \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$
4th :$$f(x)=\frac{x}{x^2+1} \\f'=\frac{1-x^2}{1+x^2}=0 \rightarrow x=\pm1 \\ \left\{\begin{matrix}
f(1) &=\frac{1}{2} \\
f(-1) & =-\frac{1}{2}\\
\lim_{x \rightarrow +\infty }\frac{x}{x^2+1} & =0\\
\lim_{x \rightarrow -\infty }\frac{x}{x^2+1} & =0
\end{matrix}\right. \overset{x \in \mathbb{R}}{\rightarrow} \space \frac{-1}{2}\leq y \leq \frac{+1}{2}$$
now : is there (creative ) 5th ,6th ,... method to find range of this function ?
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by $AM-GM$ we have $\frac{x^2+1}{2}\geq |x|$, from here we obtain $\frac{|x|}{x^2+1}\le \frac{1}{2}$
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Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.
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Perform the integration in steps:
$$
\begin{array}{rcl}
I_8
&=& \displaystyle \int_0^1 \frac{x^8}{\sqrt{x^3 + 1}} dx
= \displaystyle \int_0^1 x^6 \frac{x^2}{\sqrt{x^3 + 1}} dx\\
&=& \displaystyle
\left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
- \int_0^1 4 x^3 x^2 \sqrt{x^3+1} dx\\
&=& \displaystyle
\left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
- \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1
+ \int_0^1 \frac{8}{3} x^2 \sqrt{x^3+1}^3 dx\\
&=& \displaystyle
\left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
- \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1
+ \left[ \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\
&=& \displaystyle
\left[ \frac{2}{3} x^6 \sqrt{x^3+1}
- \frac{8}{9} x^3 \sqrt{x^3+1}^3
+ \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\
&=& \displaystyle
\left( \frac{2}{3} - \frac{16}{9} + \frac{64}{45} \right) \sqrt{2} - \frac{16}{45}\\
&=& \displaystyle \bbox[16px,border:2px solid #800000] {\frac{14}{45} \sqrt{2} - \frac{16}{45}}
\end{array}
$$
|
{
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"url": "https://math.stackexchange.com/questions/1375090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
A question about differential function If $f(x)=f'(x^{2})+2x$, then $f(1)=?$ and $f''(1)=?$
Sorry. I am going to check the original problem, and then i will update.
|
Consider the equation $f(x) = f'(x^2) + 2x$. It can be seen that $f(1) = 4$, $f'(1) = 2$ and $f^{(n+2)}(1) = 0$ for $n \geq 0$. The demonstration of how these values are obtained is as follows.
Since $f(x)$ is differentiable, which is evident from the differential equation, then consider the function in a series expansion given by
\begin{align}
f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}
\end{align}
for which the equation given leads to
\begin{align}
a_{0} + a_{1} \, x + a_{2} \, x^2 + a_{3} \, x^3 + \cdots &= 2x + (a_{1} + 2 \, a_{2} \, x^2 + 3 \, a_{3} \, x^4 + 4 \, a_{4} \, x^6 + \cdots ) \\
&= a_{1} + 2 x + 2 \, a_{2} \, x^2 + 3 \, a_{3} \, x^4 + 4 \, a_{4} \, x^6 + \cdots .
\end{align}
Equating coefficients provides $a_{0} = a_{1}$, $a_{1} = 2$, $a_{n+ 2} = 0$ for $n \geq 0$. Hence
\begin{align}
f(x) = 2 (x+1).
\end{align}
Now it is easy to obtain $f'(x) = 2$ and $f^{(n+2)}(x) = 0$ for $n \geq 0$. Letting $x=1$ provides the desired results.
|
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|
For two vectors $a$ and $b$, why does $\cos(θ)$ equal the dot product of $a$ and $b$ divided by the product of the vectors' magnitudes? While watching a video about dot products (https://www.youtube.com/watch?v=WDdR5s0C4cY), the following formula is presented for finding the angle between two vectors:
For vectors $a$, and $b$,
$$\cos( \theta ) = \frac{(a, b)}{ \| a \| \| b \| }$$
where $(a,b)$ is the dot product of $a$ and $b$.
How is this formula derived?
|
There are several derivations of this online. Here's where you can start.
Define two vectors $\textbf{a}$ and $\textbf{b}$. Then $ \textbf{a} - \textbf{b}$ is the vector that connects their endpoints and makes a triangle.
Therefore, we have a triangle with side lengths $|\textbf{a}|$, $|\textbf{b}|$, and $|\textbf{a} - \textbf{b}|$. Let the angle between the two vectors be $\theta$. By the Law of Cosines, we have
$$|\textbf{a} - \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 - 2 |\textbf{a}| |\textbf{b}| \cos (\theta)$$
Now, use the fact that
$$
\begin{align*}
|\textbf{a}- \textbf{b}|^2 &= (\textbf{a}- \textbf{b}) \cdot (\textbf{a}- \textbf{b})\\ &= \textbf{a} \cdot \textbf{a} - 2 (\textbf{a} \cdot \textbf{b}) + \textbf{b} \cdot \textbf{b} \\ &= |\textbf{a}|^2 - 2 (\textbf{a} \cdot \textbf{b}) + |\textbf{b}|^2
\end{align*}
$$
Simplify this equation, and you will get the desired formula.
|
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|
Prove the relation for cos inverse Prove the relation $\cos^{-1}x_0=\dfrac{\sqrt {1-x^2_0}}{x_1\cdot x_2\cdot x_3\cdots \text{ ad inf.}}$ where the successive quantities $x_r$ are connected by the relation $x_{r+1}=\sqrt{\frac{1}{2}(1+x_r)}$
My attempt:
$$x_1=\sqrt{\frac{1}{2}(1+x_0)}$$
$$x_2=\sqrt{\frac{1}{2}(1+x_1)}$$
$$x_3=\sqrt{\frac{1}{2}(1+x_2)}$$
$$x_4=\sqrt{\frac{1}{2}(1+x_3)}$$
Multiplying all these, we get
$$x_1\cdot x_2\cdot x_3\cdots\text{ ad inf.}=\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}$$
Putting in equation,
$$\cos^{-1}x_0=\frac{\sqrt {1-x^2_0}}{\sqrt{\frac{1}{2}(1+x_0)\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf.}}}$$
$$\cos^{-1}x_0=\frac{\sqrt {1-x_0}}{\sqrt{\frac{1}{2}(1+x_1)\frac{1}{2}(1+x_2)\frac{1}{2}(1+x_3)\cdots \text{ ad inf}}}$$
but i could not solve further. Can someone guide me in this question?
|
Let $\cos(\theta)=x_0$, so that $\sqrt{1-x_0^2}=\sin(\theta)$.
From known trigonometry, we have
$$x_{r+1}=\cos\left(\frac{\arccos(x_r)}2\right),$$
and by recurrence
$$x_r=\cos\left(\frac{\arccos(x_0)}{2^r}\right)=\cos\left(\frac{\theta}{2^r}\right).$$
Hence the denominator is the infinite product
$$\prod_1^\infty\cos\left(\frac\theta{2^r}\right)=\text{sinc}(\theta),$$
and the ratio is $\theta$.
|
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|
Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{\frac{x}{2}}\cos{\frac{x}{2}} + asin^2{\frac{x}{2}} + sin^2{\frac{x}{2}} + a\cos^2{\frac{x}{2}} - \cos^2{\frac{x}{2}} = 1$$
$$ a\left(sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 = 1 - \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}$$
$$ a\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 =2\cos^2{\frac{x}{2}}$$
I can't finish...
|
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
$$ a\sin{x}+a\left(\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}\right)-\left(\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}\right)=1 $$ $$ a\sin x+a-\cos x=1 $$
$$ a\sin x-\cos x=(1-a) $$ $$ \frac{a}{\sqrt{1+a^2}}\sin x-\frac{1}{\sqrt{1+a^2}}\cos x=\frac{(1-a)}{\sqrt{1+a^2}} $$ $$\sin x\cos \alpha-\cos x\sin \alpha=\sin \beta $$ Where $\alpha=\cos^{-1}\left(\frac{a}{\sqrt{1+a^2}}\right)$ & $\beta=\sin^{-1}\left(\frac{1-a}{\sqrt{1+a^2}}\right)$ $$\sin(x-\alpha)=\sin \beta$$ $$x-\alpha=2n\pi+\beta\iff x=2n\pi+\alpha+\beta$$ or $$x-\alpha=(2n+1)\pi-\beta\iff x=(2n+1)\pi+\alpha-\beta$$
|
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|
Parallelogram ABCD There's a parallelogram $ABCD$. I'm given point $A(3,12)$ and point $B(-1,5)$. Given the equations of the lines $BC$ and $AC$ are $y=8x+13$ and $y=3x+3$ respectively.
How to find the coordinates of the point of intersection between the diagonals $BD$ and $AC$? And the coordinates of $D$?
I've no idea. Can someone explain it ? Thanks!
|
Let the coordinated of $D$ be $(a, b)$. Since $AD\parallel BC$ hence the lines $AD$ & $BC: y=8x+13$ have equal gradients $$\frac{b-12}{a-3}=8$$ $$b-12=8a-24$$ $$\implies b=8a-12\tag 1$$ Now, solving $BC: y=8x+13$ & $AC: y=3x+3$ , we get coordinates of the vertex $C(-2, -3)$
Since $CD\parallel AB$ hence the lines $CD$ & $AB$ have equal gradients $$\frac{b-(-3)}{a-(-2)}=\frac{5-12}{-1-3}$$ $$4b+12=7a+14$$ setting the value of $b$ from (1), we get $$4(8a-12)+12=7a+14\implies 25a=50 \implies a=2$$ $$b=8(2)-12=4$$ Hence the coordinates of the point $D$ are $(2, 4)$ Since, the diagonals $BD$ & $AC$ are bisecting each other hence their intersection point will the mid-point of $BD$ joining $B(-1, 5)$ & $D(2, 4)$ given as follows $$\left(\frac{-1+2}{2}, \frac{5+4}{2}\right)\equiv \left(\frac{1}{2}, \frac{9}{2}\right)$$
Hence, $$\bbox[5px, border:2px solid #C0A000]{\text{intersection point of diagonals BD & AC}\equiv \color{blue}{\left(\frac{1}{2}, \frac{9}{2}\right)}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{D\equiv(2, 4)}}$$
|
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|
Multiplicating inequalities I have two inequalities: $|x|\leq\sqrt{x^2+y^2}$ and $|y|\leq\sqrt{x^2+y^2}, \forall x,y \in \Bbb R$, can I multiply these inequalities to get $|xy|\leq x^2+y^2$?
If yes, what is the justification? If not, why?
|
You can multiply these inequalities, because both sides of both inequalities are nonegative. Here is an explaination:
$$|x| \leq \sqrt{x^2+y^2}$$
$|y| \geq 0$, so you can multiply both sides by $|y|$:
$$|xy| \leq |y|\sqrt{x^2+y^2}$$
But we know that $|y| \leq \sqrt{x^2+y^2}$ and $0 \leq \sqrt{x^2+y^2}$, so:
$$|y|\sqrt{x^2+y^2} \leq \sqrt{x^2+y^2} \times \sqrt{x^2+y^2}=x^2+y^2$$
Finally:
$$|xy| \leq |y|\sqrt{x^2+y^2} \leq x^2+y^2$$
|
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|
Finding all the triangles $ABC$ satisfying $\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=2(a^2+b^2+c^2)$ $\triangle ABC$ has $BC=a, CA=b, AB=c$ and satisfies
$$\dfrac{a^{2}\cos\dfrac{B-C}{2}}{\sin\dfrac{A}{2}}+\dfrac{b^{2}\cos\dfrac{C-A}{2}}{\sin\dfrac{B}{2}}+\dfrac{c^{2}\cos\dfrac{A-B}{2}}{\sin\dfrac{C}{2}}=2(a^2+b^2+c^2)$$ Find all the triangles $ABC$.
|
Converting all the angles in sides the given condition will look like
$$\sum \frac{a^{2}\cos\frac{B-C}{2}}{\sin\frac{A}{2}}\frac{2sin(\frac{B+C}{2})}{2cos(\frac{A}{2})}=2(a^2+b^2+c^2)$$ $$\sum \frac{a^{2}(sin{B}+sinC)}{\sin{A}}=2(a^2+b^2+c^2)$$$$\sum \frac{a^{2}(\frac{b}{R}+\frac{c}{R})}{\frac{a}{R}}=2(a^2+b^2+c^2)$$ $$\sum(ab+ac)=2(a^2+b^2+c^2)$$ $$2(ab+bc+ac)=2(a^2+b^2+c^2)$$
Now considering rearrangement inequlity we have$$a^2+b^2+c^2\geq ab+bc+ac $$
now for the equality to hold the condition should be$$a=b=c$$
and you have the triangle.
|
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|
How do you solve for θ in the equation $\tan \frac{\theta}{5} + \sqrt{3} = 0$ $$\tan \frac{\theta}{5} + \sqrt{3} = 0$$
Alright so the $\frac{\theta}{5}$ is confusing me.
Would it be wrong to do
\begin{eqnarray}
\tan \frac{\theta}{5}&=&-\sqrt{3}\\
\frac{\theta}{5}&=&\tan^{-1}(-\sqrt{3})\\
\theta&=& 5\tan^{-1}(-\sqrt{3})\\
\theta&=& -\frac{5\pi}{3} + 5\pi n
\end{eqnarray}
|
hint: Add $n\pi$ to your answer !
|
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|
Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$ Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx $ for $y\in[0,1].$
I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but that also is not working.Can someone please help me in solving this question?
|
Let
$$
I(y) = \int_0^y \sqrt{x^4 + (y-y^2)^2}dx
$$
Let's find $y$ such that $dI/dy=0$. By Leibniz integral rule we have
$$
\frac{dI}{dy} = \int_0^y \frac{\partial}{\partial y}\sqrt{x^4 + (y-y^2)^2}dx + \sqrt{y^4 + (y-y^2)^2}
$$
But $dI/dy>0$ for $y>0$ ($dI/dy=0$ at $y=0$). So, maximum is reached for $y=1$:
$$
I(1)=\int_0^1 x^2\,dx = 1/3
$$
Here is plot of $72I(y)$:
EDIT
(Answer to @PhoemueX's comment)
We should to prove that $dI/dy>0$. We have
$$
\frac{dI}{dy} = \int_0^y \frac{(1-2 y) (y-y^2)}{\sqrt{x^4+\left(y-y^2\right)^2}}dx + \sqrt{y^4 + (y-y^2)^2}.
$$
It's trivial for $y\le 1/2$ (both terms are positive). For $1/2 < y \le 1$
$$
\frac{dI}{dy} > -\int_0^y \frac{(2 y-1) (y-y^2)}{\sqrt{(y-y^2)^2}}dx + \sqrt{y^4 + (y-y^2)^2} = \\= \sqrt{y^4 + (y-y^2)^2}-(2y-1)y\sqrt{y-y^2}=\\=
\sqrt{y^4 + (y-y^2)^2}-\sqrt{y^2(2y-1)^2(y-y^2)}
$$
But
$$
y^4 + (y-y^2)^2>y^2(2y-1)^2(y-y^2)
$$
|
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|
help with trigonometric equations How do I solve this?
$$\cos3x=\cos^2x-3\sin^2x$$
|
$$ \cos(3x)=4\cos^3(x)-3\cos(x),\quad 4\cos^2(x)-3=\cos^2 x-3\sin^2 x $$
hence by setting $z=\cos x$ we have to solve:
$$ 4z^3-4z^2-3z+3 = (z-1)(4z^2-3) = 0 $$
so $\cos x=1$ or $\cos x=\frac{\sqrt{3}}{2}$, from which $x=2k\pi$ or $x=\pm\frac{\pi}{6}+2k\pi$.
|
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|
Orthocenter of triangle $DEF$ is same as the circumcenter of triangle $ABC$ $D,E,F$ are mid points of the sides of the triangle $ABC$,then prove that the orthocenter of triangle DEF is same as the circumcenter of triangle ABC.
I cannnot figure out what coordinates to suppose for A,B,C.I tried taking $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ but calculations go messy and clumsy.Can someone help me in proving this question?
|
Let's consider a $\triangle ABC$ with side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will be at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$. The mid-points $D, E$ & $F$ of the sides $BC$, $AC$ & $AB$ respectively of $\triangle ABC$ are calculated as follows $$D\equiv\left(\frac{0+a}{2}, \frac{0+0}{2}\right)\equiv\left(\frac{a}{2}, 0\right)$$
$$E\equiv\left(\frac{\frac{a\tan C}{\tan B+\tan C}+a}{2}, \frac{\frac{a\tan B\tan C}{\tan B+\tan C}+0}{2}\right)\equiv\left(\frac{a(\tan B+2\tan C)}{2(\tan B+\tan C)},\frac{a\tan B\tan C}{2(\tan B+\tan C)}\right)$$
$$F\equiv\left(\frac{\frac{a\tan C}{\tan B+\tan C}+0}{2}, \frac{\frac{a\tan B\tan C}{\tan B+\tan C}+0}{2}\right)\equiv\left(\frac{a\tan C}{2(\tan B+\tan C)},\frac{a\tan B\tan C}{2(\tan B+\tan C)}\right)$$ Hence, the ortho-center $H$ of $\triangle DEF$ is determined as follows
Now, the orthocenter H can be calculated by finding intersection point of any two altitude equations. We can get orthocenter as $$H\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$
& the circumscribed center say $P$ can be calculated as $$P\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$ Hence, we can find that orthocenter (H) of $\triangle ABC$ will coincide with the circumcenter (P) of $\triangle ABC$ i.e. both H & P are same.
|
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|
Explanation of inverse trig euation solution The equation is:
$\arctan 3 + 2\arctan2 = \pi + arccot 3$
They go on and assign $\arctan 3 = \theta$ and $\arctan2 = \phi$. Therefore, $\frac{\pi}{4}< \theta < \frac{\pi}{2}$, same for $\phi$. Which I follow.
Then they use $\tan(A+B)$ formula to show that that $\tan(\theta+2\phi)=\frac{1}{3}$. Which implies that $\arctan \frac{1}{3} = \theta + 2\phi$, but apparently not so. As it implies $\pi + \arctan \frac{1}{3} = \theta + 2\phi$.
And there I am lost. I can see that when you combine the inequalities you get $$\pi+\frac{\pi}{4} < \theta + 2\phi < \pi + \frac{\pi}{2}$$
And then they go on to show that $$\theta + 2\phi = \pi + \arctan \frac{1}{3} = \pi + arccot 3$$
I do not understand where that $\pi$ is coming from there...
|
First of all, $\arctan x+\text{arccot}x=\dfrac\pi2$ (Proof)
So, the problem reduces to $\arctan 2+\arctan3=\dfrac{3\pi}4$
Now from this or Ex$\#5$ of Page $\#276$ of this,
$$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\dfrac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$
Finally, the principal value of $\displaystyle\tan$ lies $\displaystyle\in\left[-\frac\pi2,\frac\pi2\right]$
|
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|
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.
|
HINT: use that $$4+2\sqrt{3}=(1+\sqrt{3})^2$$
|
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|
Evaluate the integral $\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}$ using the residue theorem Can someone show me how to compute this integral using the residue theorem:
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
|
Notice,
$$\int_0^{\frac{\pi}{2}} \frac{1}{a + \sin^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a\sec^2\theta + \sec^2\theta\sin^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a+a\tan^2\theta + \tan^2 \theta}d\theta$$
$$=\int_0^{\frac{\pi}{2}} \frac{\sec^2\theta}{a+(a+1)\tan^2\theta}d\theta$$
Let, $\tan \theta=t\implies \sec^2\theta d \theta=dt$
$$=\frac{1}{a+1}\int_{0}^{\infty}\frac{dt}{\frac{a}{a+1}+t^2}$$
$$=\frac{1}{a+1}\sqrt{\frac{a+1}{a}}\left[\tan^{-1}\left(t\sqrt{\frac{a+1}{a}}\right)\right]_{0}^{\infty}$$ $$=\frac{1}{\sqrt{a(a+1)}}\left(\frac{\pi}{2}-0\right)$$
$$=\frac{\pi}{2\sqrt{a(a+1)}}$$ Hence, we have
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_0^{\frac{\pi}{2}} \frac{d\theta}{a + \sin^2 \theta}=\frac{\pi}{2\sqrt{a(a+1)}}}}$$
|
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|
Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align*}
I want to apply the method of partial fraction to $P(x)$ now. I wanted to write \begin{align*} P(x) = \frac{1+x}{1-2x-x^2} = \frac{A}{(x- \lambda_1)} + \frac{B}{(x- \lambda_2)} \end{align*} but this equality does not hold. Since $(x-\lambda_1) (x- \lambda_2) = x^2 + 2x -1 \neq 1-2x-x^2$. So what denumerators should I choose to go with $A$ and $B$?
|
Notice, we have $$\frac{x+1}{1-2x-x^2}$$ $$=\frac{x+1}{-(x^2+2x-1)}$$ $$=\frac{-(x+1)}{(x^2+2x+1)-2}$$ $$=\frac{-x-1}{(x+1)^2-(\sqrt2)^2}$$ $$=\frac{-x-1}{(x+1-\sqrt2)(x+1+\sqrt2)}$$ Now, let
$$\frac{-x-1}{(x+1-\sqrt2)(x+1+\sqrt2)}=\frac{A}{x+1-\sqrt2}+\frac{B}{x+1+\sqrt2}$$
By comparing the corresponding coefficients of the sides, we get $$\begin{cases}
A+B=-1\\ A(1+\sqrt2)+B(1-\sqrt 2)=-1\end{cases}$$ By solving the above equations we get $$A=-\frac{1}{2}, \ B=-\frac{1}{2}$$
Hence, we get
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\frac{x+1}{1-2x-x^2}=-\frac{1}{2(x+1-\sqrt2)}-\frac{1}{2(x+1+\sqrt2)}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac{3}{2}$ but what to do with remaining $\frac{1}{x}$ and $dx$ in the integration.Please help....
|
The fact that the limits are reciprocals of each other gives you your first clue. Splitting the integral up about the geometric mean $\sqrt{2\cdot\frac 12} = 1$, we have
$$I = \int_{1/2}^1 \frac{1}{x}\sin (x-1/x) \ dx + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx$$
Now change variables in the first of those two integrals by $t = 1/x$ say, and we have
$$\int_2^1 \frac{t}{-t^2} \sin(1/t - t) \ dt = \int_2^1 \frac{1}{t} \sin(t - 1/t) \ dt = - \int_1^2 \frac{1}{t} \sin(t - 1/t) \ dt$$
Thus
$$I = - \int_1^2 \frac{1}{t} \sin(t - 1/t) \ dt + \int_1^2 \frac{1}{x}\sin (x-1/x) \ dx \ = \ \cdots$$
|
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|
Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer.
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there are no integer $n$ and Pythagorean
triple $(a, b, c)$ satisfying $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n$$
The foregoing question is from the 2005 Canada National Olympiad.
I need some help on part (b).
Part (a)
Examine the behaviour of the following function [that embodies the constraint $c^2=a^2+b^2$, $x$ can play the role of either $a$ or $b$] on $(0,c)$:
$$f(x) = \dfrac{c}{x} + \dfrac{c}{\sqrt{c^2-x^2}} \tag{1}$$
Now find the derivative: $$f'(x)=cx\left[\dfrac{1}{(c^2-x^2)^{3/2}}-\dfrac{1}{x^3}\right] \tag{2}$$
To find local extrema:
$$\begin{align}
f'(x)=0 &\implies x^3=(c^2-x^2)^{3/2} \\
&\implies x^6=(c^2-x^2)^3 &(\text{squaring}) \\
&\implies u^3=(k-u)^3 &(u=x^2,k=c^2) \\
&\implies 2u^3-3ku^2+3k^2u-k^3=0 \\
&\implies (2u-k)(u^2+-ku+k^2)=0 \tag{3}\\
\end{align}$$
So, solutions are: $$u=\dfrac{k}{2};u=\dfrac{k\pm\sqrt{k^2-4k^2}}{2}\notin\mathbb{R}$$
So take the only real solution as $$u=\dfrac{k}{2} \implies x^2=\dfrac{c^2}{2} \implies x=\dfrac{c}{\sqrt2}\quad(\text{since }x\in(0,c))$$
This is a local minima because $f$ is continuous on $(0,c)$ and $$\lim_{x\to0^+}{f(x)}=\lim_{x\to c^-}{f(x)}=+\infty$$
The minimum value is $$f\left(\dfrac{c}{\sqrt2}\right)=\sqrt2+\sqrt2=2\sqrt2 \tag{4}$$
This value is not achievable, because $\sqrt{2}$ is not a rational number. Hence,
$$[f(a)]^2 = \left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > (2\sqrt2)^2 = 8 \tag{5}$$
Part (b)
$$\begin{align}
\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n \implies c^2(a^2+b^2)=a^2b^2\cdot n \tag{6}
\end{align}$$
in a perhaps friendlier form.
I am not sure how to use this further.
|
For now, I prove you part (a) with a different approach. Dividing by $c^2=a^2+b^2$, taking the square root, and multiplying by $-1/2$, the inequality can be rewritten as
$$
\frac{2}{\frac{1}{a}+\frac{1}{b}}<\sqrt{\frac{a^2+b^2}{2}},
$$
which is simply HM-QM. Equality cannot hold because it cannot be the case that $a=b$ (indeed $2a^2$ cannot be a square).
|
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|
Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
|
$$ 3 \cos x + 2\sin x=1 $$
$y:=\tan\big(\frac x 2\big)$ then $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$
$$-1+\frac{3}{y^2+1}+\frac{4y}{y^2+1}-\frac{3y^2}{y^2+1}=0$$
$$\frac{2y^2-2y-1}{y^2+1}=0$$
$$2y^2-2y-1=0$$
$$y^2-y=\frac 1 2$$
Add $\frac 1 4$ to both sides:
$$y^2-y+\frac 1 4=\frac 3 4$$
$$\bigg(y-\frac 1 2 \bigg)^2=\frac 3 4$$
$y=\frac 1 2+\frac{\sqrt 3}{2}$ or $y=\frac{\sqrt 3}{2}+\frac 1 2$
$y=\tan\big(\frac{\pi}{2}\big)$
$$\boxed{\color{blue}{x=2\arctan\big(\frac 1 2\pm\frac{\sqrt 3}{2}+2 n \pi\big)\;\;\;\;, n\in \mathbb{Z} }}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why I am getting different answer? I have just started learning single variable calculus. I'm confused in a problem from sometime. I didn't get why my answer is different from the book.
$$
\require{cancel}
\begin{align}
&\int\sin x \sin 2x \sin 3x\,dx\\
&=\int\sin x\;\,2\sin x\cos x \left(3\sin x - 4\sin^3 x\right)\,dx\\
&\qquad\text{Let }\sin x = t, \text{ then}\\
&\qquad\quad\cos x\, dx = dt\\
&=\int t\;2t\left(3t - 4 t^3\right)\,dt\\
&=\int 2t^2\left(3t - 4t^3\right)\,dt\\
&=\int\left(6t^3-8t^5\right)\,dt\\
&=6\int t^3\,dt - 8\int t^5\,dt\\
&=\cancel{6}\,3\frac{t^4}{\cancel{4}2}+c_1-\cancel{8}\,4\frac{t^6}{\cancel{6}3}+c_2\\
&=\frac32t^4-\frac43t^6+C\\
&=\frac32\sin^4x-\frac43\sin^6x+C
\end{align}
$$
The answer given in my book is
$$\displaystyle\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right] + C. \ $$
Where did I go wrong?
|
$\bf{My\; Solution::}$ Let $\displaystyle \int \sin x\cdot \sin 2x \cdot \sin 3x dx$
Using the formula
$\bullet 2\sin A \cdot \sin B = \cos (A-B)-\cos (A+B)$
$\bullet 2\cos A \cdot \sin B = \sin (A+B)-\sin (A-B)$
So $$\displaystyle I = \frac{1}{2}\int \left[2\sin 3x \cdot \sin x\right]\cdot \sin xdx = \frac{1}{2}\int \left[\cos 2x-\cos 4x\right]\sin 2xdx$$
So $$\displaystyle I = \frac{1}{4}\int \left[2\cos 2x \cdot \sin 2x-2\cos 4x\cdot \sin 2x\right]$$
$$\displaystyle I = \frac{1}{4}\int \left[\sin 4x-0\right]dx-\frac{1}{4}\int \left[\sin 6x-\sin 2x\right]dx$$
So $$\displaystyle I = \frac{1}{4}\left[-\frac{\cos 4x}{4}+\frac{\cos 6x}{6}-\frac{\cos 2x}{2}\right]+\mathcal{C}$$
Now $$\displaystyle \cos 6x = 1-2\sin^2(3x) = 1-2\left[3\sin x-4\sin^3 x\right]$$
and $$\displaystyle \cos 4x = 1-2\sin^2(2x) = 1-2\left[2\sin x\cdot \cos x\right]^2 = 1-4\sin^2\cdot (1-\sin^2 x)$$
and $$\displaystyle \cos 2x = 1-2\sin^2 x$$
Now put into final solution of Integral
|
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|
Find the least positive integer $n$ so that $\left ( 1-\frac{1}{s_{1}} \right ) \cdots \left ( 1-\frac{1}{s_{n}} \right )=\frac{51}{2010}$
Find the least positive integer $n$ for which there exists a set $\left \{ s_{1}, s_{2},....,s_{n} \right \}$ consisting of $n$ distinct positive integers such that $$\left ( 1-\frac{1}{s_{1}} \right ) \left ( 1-\frac{1}{s_{2}} \right )\cdots\left ( 1-\frac{1}{s_{n}} \right )=\frac{51}{2010}$$
My idea was assume that $s_{1}< s_{2}< \cdots< s_{n}$, So I can't know how to start, any help will be appreciate it.
|
Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}< s_{2}< ...< s_{n}$. Surely $s_{1}> 1$ since otherwise $1-\frac{1}{s_{1}}=0 $. So we have $2\leq s_{1}\leq s_{2}-1\leq s_3 - 2\leq ....\leq s_{n} - (n-1)$, hence $s_{i}\geq i+1$ for each $i=1,...,n$. Therefore $$\frac{51}{2010}=\left ( 1-\frac{1}{s_{1}} \right )\left ( 1-\frac{1}{s_{2}} \right )...\left ( 1-\frac{1}{s_{n}} \right )\geq \left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )....\left ( 1-\frac{1}{n+1} \right )=\frac{1}{2}\cdot \frac{2}{3} ... \frac{n}{n+1}=\frac{1}{n+1}$$
which implies $$n+1\geq \frac{2010}{51}=\frac{670}{17}> 39.$$
So $n\geq 39$
Now we are left to show that $n=39$ fits. Consider the set $\left \{ 2,3,...,33,35,36...,40,67 \right \}$ which contains exactly $39$ numbers. We have $$\frac{1}{2}\cdot \frac{2}{3}....\frac{32}{33}\cdot \frac{34}{35}...\frac{39}{40}\cdot \frac{66}{67}=\frac{17}{670}=\frac{51}{2010}$$
Hence for $n=39$ there exists a desired example.
AND WE'RE DONE
|
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|
Finding equation of straight line that is tangent to $y = 2^x$. Problem: Find an equation of the straight line that is tangent to $y= 2^x$ and that passes through the point $(1,0)$.
Attempt: Let $(a, 2^a)$ be the point of tangency. Now we have that $y' = 2^x \ln(2)$, which evaluated at the tangency point becomes $y' = 2^a \ln(2)$.
So the general equation of the tangent line is $y = 2^a \ln(2) (x-a) + 2^a$. If this line is to go through $(1,0)$, then we must demand that $0 = 2^a \ln(2) (1-a) + 2^a$. I rewrote this as $-2^a = 2^a \ln(2) (1-a)$, which simplifies to $ -1 = \ln(2) (1-a)$. Now I'm having trouble finding $a$ from this. If I distribute I get $ a \ln(2) = \ln(2) + 1$ or \begin{align*} a = \frac{ \ln(2) + 1}{\ln(2)} \end{align*}
Can I simplify this any further? Because my textbook says the tangent line has equation \begin{align*} y = 2e \ln(2) (x-1) \end{align*} How can I find this answer?
|
To show the answers are the same, we need to show that if $a=\frac{\ln 2+1}{\ln 2}$ then $2^a=2e$.
Note that $a=1+\frac{1}{\ln 2}$, so
$$2^a=2\cdot 2^{1/\ln 2}=2\cdot (e^{\ln 2})^{1/\ln 2}=2e.$$
|
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|
Inverse Trigonometric Function: Find the Exact Value of $\sin^{-1}\left(\sin\left(\frac{7\pi}{3}\right)\right)$ $$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$
I cannot use this formula, correct? $f(f^{-1}(x))=x$
The answer in the book is $\frac{\pi}{3}$
How do I approach solving a problem such as this?
The inverse sin function of $\sin\frac{7\pi}{3}$
Am I saying to myself there were 7 revolutions and $\frac{\pi}{3}$ a corresponding angle to $\frac{7\pi}{3}$
|
You have to consider the restricted ranges of the inverse trig functions. Since $\sin\left(\frac{7\pi}{3}\right)=\frac{\sqrt{3}}{2}$, you want $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$, since the range of inverse sine is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the other end of the Diameter For a National Board Exam Review
A circle has it center at $(3,-2)$ and one end of a diameter at $(7,2)$. Find the other end of the diameter.
Answer is $(-1,6)$
$$m=\frac{y^2-y^1}{x^2-x^1}=\frac{2-(-2) }{7-3}$$
$$=(3-7)^2+(-2-2)^2=r^2=32$$
$$r =\sqrt{32}$$
Plugin $(7,2)$ into
$$y = mx+b$$
$$b = -5$$
Solve two linear equations:
$$32 = (x-h)^2 + (y-k)^2$$
$$32 = (x-7)^2 + ((x-5)-2)^2$$
$$y=x-5$$
I get $(3,-2)$. What am I doing wrong? Any hint?
|
Here's another approach:
The center of a circle is the midpoint of any diameter. If the coordinates of the unknown point are $(a,b)$, using the midpoint formula:
$\frac{a+7}{2}=3$, and...(see if you can work out the equation for $b$).
|
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|
Sum of (arithmetic?) infinite series How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
|
$$\frac{5}{9}\sum \frac{1}{4x^2-1}$$
$$=\frac{5}{18}\sum \frac{1}{2x-1}-\frac{1}{2x+1}$$
$$=\frac{5}{18}[ \frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...]$$
$$=\frac{5}{18}[\frac{1}{5}]$$
$$=\frac{1}{18}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $\int\frac{1}{1-\tan^2 x}dx$ Evaluation of $\displaystyle \int\frac{1}{1-\tan^2 x}dx$
$\bf{My\; Try::}$ We can write $$\displaystyle \int\frac{1}{(1-\tan x)\cdot (1+\tan x)}dx = \frac{1}{2}\int\frac{(1+\tan x)+(1-\tan x)}{(1-\tan x)\cdot (1+\tan x)}dx$$
So We get $$\displaystyle = \frac{1}{2}\int\frac{1}{1-\tan x}dx+\frac{1}{2}\int\frac{1}{1+\tan x}dx = \frac{1}{2}I + \frac{1}{2}J$$
Now Let $$\displaystyle I = \int\frac{1}{1-\tan x}dx = \frac{1}{2}\int\frac{2\cos x}{\cos x-\sin x}dx = \frac{1}{2}\int\frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x-\sin x}dx$$
So we get $$\displaystyle = \frac{1}{2}\int\frac{(\cos x+\sin x)}{(\cos x-\sin x)}dx+\frac{1}{2}x = -\frac{1}{2}\ln\left|\cos x-\sin x\right|+\frac{1}{2}x$$
Now Let $$\displaystyle I = \int\frac{1}{1+\tan x}dx = \frac{1}{2}\int\frac{2\cos x}{\cos x+\sin x}dx = \frac{1}{2}\int\frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x+\sin x}dx$$
So we get $$\displaystyle = \frac{1}{2}\int\frac{(\cos x+\sin x)}{(\cos x-\sin x)}dx-\frac{1}{2}x = +\frac{1}{2}\ln\left|\cos x+\sin x\right|+\frac{1}{2}x$$
So $$\displaystyle \int\frac{1}{1-\tan^2 x}dx = -\frac{1}{2}\ln\left|\cos x-\sin x\right|+\frac{1}{2}\ln\left|\cos x+\sin x\right|+x+\mathcal{C}$$
Can we solve it any short method, If yes then plz explain here, Thanks
|
Perhaps this might be a bit shorter
\begin{align*}
\frac{1}{1-\tan^2 x} & = \frac{\cos^2x}{\cos^2x-\sin^2x}\\
& = \frac{1+\cos2x}{2\cos2x}\\
& = \frac{1}{2}\sec 2x+\frac{1}{2}.
\end{align*}
Now you just have to integrate $\sec 2x$. This is a routine problem (multiply numerator and denominator by $\sec 2x + \tan 2x$).
|
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|
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
|
The hint
$$
\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=1+\sqrt2 \quad \text{for } x\in [0,\pi/2]
$$
is spot on.
We have
$$
\begin{align}
\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}}
&=
\sqrt{10}\left(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}}\right)
\\&=
\sqrt{10}\left(\sqrt{1+\sin(t)}+\sqrt{1+\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sin(t)}+\sqrt{1-\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sqrt{0.01}}+\sqrt{1-\sqrt{0.99}}\right)
\\&=
(1+\sqrt 2)\left(\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{99}}\right)
\end{align}
$$
for some $t$.
Therefore, we can rearrange the numerator to be $1+\sqrt 2$ times the denominator. (You need to handle the middle terms separately but it works out the same.)
|
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|
Prove that every positive integer less than or equal to the square root of a is a near factor of a In many computer languages, the division operation ignores remainders. Let's denote this by the operation $//$, so for instance $13//3 = 4$. If for some $b$, $a//b = c$ then we say that $c$ is a near factor of $a$. Thus, the near factors of $13$ are $1,2,3,4$ and $6$. Let $a$ be a positive integer. Prove that every positive integer less than or equal to $\sqrt{a}$ is a near factor of $a$.
|
Disclaimer: This is not quite a complete solution. It proves that every $\boldsymbol{k}$ with $\boldsymbol{k+1 \le \sqrt{a}}$ is a near factor.
Let $m$ be the smallest integer for which $m(m+1) > a$.
Consider the (not necessarily strictly) decreasing sequence
$$
a // m, a // (m+1), a // (m+2), \ldots, a // (a-1), a // a = 1. \tag{1}
$$
Note that for $m \le k \le a-1$,
\begin{align*}
0 \le a // k - a // (k+1)
&< \left(\frac{a}{k}\right) - \left(\frac{a}{k+1} - 1\right) \\
&= 1 + a \left(\frac{1}{k} - \frac{1}{k+1}\right) \\
&= 1 + \frac{a}{k(k+1)} \\
&\le 1 + \frac{a}{m(m+1)} \\
&< 1 + \frac{a}{a} \\
&= 2.
\end{align*}
But since $a // k - a // (k+1)$ is an integer, and it is $\ge 0$ and $< 2$, we have $a // k - a // (k+1) \in \{0,1\}$.
Therefore the sequence in (1) goes down $1$ or $0$ at a time, and as it ends in $1$ it hits every integer between $1$ and $a // m$.
Just how large is $a // m$? Well,
Since $m$ is the smallest integer for which $m(m+1) > a$,
$(m-1)m \le a$.
So $(m-1)^2 < a$, so $m < \sqrt{a} + 1$.
Then
$$
a // m
> \frac{a}{m} - 1
> \frac{a}{\sqrt{a}+1} - 1
= \frac{a + \sqrt{a}}{\sqrt{a}+1} - \frac{\sqrt{a}}{\sqrt{a}+1} - 1
> \sqrt{a} - 2
> \lfloor \sqrt{a} \rfloor - 2
$$
which implies, since $a // m$ is an integer, that
$$
a // m \ge \lfloor \sqrt{a} \rfloor - 1.
$$
Therefore, every integer $1, 2, 3, \ldots, \lfloor \sqrt{a} \rfloor - 1$ is found in the sequence (1).
I.e., every $k$ with $k + 1 \le \sqrt{a}$.
|
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|
Calculus Integral from Partial Fractions When you have an irreducible quadratic factor repeated you can get integrals that look like $\int \dfrac{dx}{(x^2+a)^m}$, where $m>1$, integer, and $a>0$. What is the best way to integrate this function? Is there more than one way?
|
The recurrence relation derived by Jack D'Aurizio in his answer above can actually be solved explicitly for $I_m$. First let me quote it below, but changing the recurrence index to $n$:
$$
I_{n+1} = \frac{2n+1}{2n}\,I_{n} + f_n(t)
\qquad (n \ge 1) \qquad (1)
$$
where
$$
\begin{align}
I_0 &= t \\[0.05in]
I_1 &= \arctan(t) \\[0.05in]
f_n(t) &\equiv \frac{1}{2n}\frac{t}{(1+t^2)^n} \qquad (n \ge 1)
\end{align}
$$
Now define the quantity
$$
P_{n} \equiv \prod_{k\,=\,1}^{n} \left(\frac{2k+1}{2k}\right)
= \prod_{k\,=\,1}^{n} \left(1 + \frac{1}{2k}\right)
$$
Then, dividing (1) by $P_n$, we find
$$
\frac{I_{n+1}}{P_{n}} - \left(\frac{2n+1}{2n}\right)\frac{I_{n}}{P_{n}} = \frac{f_n}{P_{n}}
$$
Since
$$
P_{n} = \left(\frac{2n+1}{2n}\right)P_{n-1}
$$
we have
$$
\frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} = \frac{f_n}{P_{n}}
$$
Now sum both sides on $n$:
$$
\sum_{n\,=\,2}^{m-1} \left( \frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} \right)= \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}}
$$
The LHS telescopes, so
$$
\frac{I_{m}}{P_{m-1}} - \frac{I_{2}}{P_{1}} = \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}}
$$
and
$$
I_{m} = P_{m-1} \left(
\frac{I_{2}}{P_{1}} + \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left(
\frac{I_{2}}{P_{1}} - \frac{f_1}{P_{1}} + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left(
I_1 + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right)
$$
where (1) was used in the last equality to remove $I_2$.
Hence,
$$
I_{m} = P_{m-1} \left(
\arctan(t) + \sum_{n\,=\,1}^{m-1} \frac{t\,(1+t^2)^{-n}}{2n\,P_{n}} \right)
\qquad (m \ge 2)
$$
|
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|
Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper rational fraction we have to divide
$$\frac{x^2 + 1}{x^2-5x+6}$$
I'm trying from sometime but couldn't find the right solution.
Please help! Thank you in advance.
|
Given $\displaystyle \int\frac{x^2+1}{x^2-5x+6}dx = \int\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}dx = \int 1dx+5\int\frac{(x-1)}{(x-2)(x-3)}dx$
Now Using Partial fraction Method::
$\displaystyle \frac{x-1}{(x-2)\cdot (x-3)} = \frac{A}{(x-2)}+\frac{B}{(x-3)}$
So we get $x-1 = A(x-3)+B(x-2) = (A+B)x+(-3A-2B)$
After Camparing the coefficient, we get $\displaystyle A+B=1$ and $3A+2B = 1$
So we get $\displaystyle A = -1$ and $\displaystyle B= 2$
So we get $\displaystyle \frac{x^2+1}{x^2-5x+6}dx = x+5\int \left[-\frac{1}{x-2}+\frac{2}{x-3}\right]dx$
|
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|
Confusing question on exponents and algebra. I was going through some sample papers of math, and I found this question which I cant solve:
If $abc=1$, find $1/(1+a+b^{-1})+1/(1+b+c^{-1})+1/(1+c+a^{-1})$.
Please help me with this.... I have spent almost 3 hours on this question...
Thanks for the help.
|
$$\frac 1{1+a+b^{-1}}=\frac b{b+ab+1}=\frac b{1+b+c^{-1}}$$
The last equation follows because, $ab=1/c$.
In a similar manner,
$$\frac 1{1+c+a^{-1}}=\frac a{a+ac+1}=\frac a{a+b^{-1}+1}=\frac{ab}{1+b+c^{1}}$$
So, putting this altogether gives
$$\begin{align}
& \frac 1{1+a+b^{-1}}+\frac 1{1+b+c^{-1}}+\frac 1{1+c+a^{-1}} \\
= & \frac b{1+b+c^{-1}}+\frac 1{1+b+c^{-1}}+\frac{ab}{1+b+c^{-1}}\\
= & \frac{1+b+c^{-1}}{1+b+c^{-1}}=1.
\end{align}$$
|
{
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|
How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16 =0
$$
here $( x \neq 0; x \neq \frac{2}{3}; x \neq - \frac{2}{3}; 27x^3 - 12x + 8 \neq 0 )$.
Is there some simple (and different) way to solve it?
|
the equation $$81 x^5-81 x^4-90 x^3+36 x^2+16 x-16=0$$ can only be solved by a numerical method, e.g. the Newton method
|
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|
What base system satisfies the equation For the equation:
$$5x^2 - 50x +125 = 0$$
$x=5$ and $x=8$ are solutions.
This is in another base, what are the steps required to find out what base it is in?
Thanks
|
Let $b$ be the base of the coefficients in the quadratic. Then, $(50)_b = 5b$ and $(125)_b = b^2+2b+5$.
Hence, the quadratic in base-$10$ is $5x^2-5bx+(b^2+2b+5) = 0$.
Since $x = 5$ is a solution, we have $125 - 25b + (b^2+2b+5) = 0$, i.e. $b^2-23b+130 = 0$.
Since $x = 8$ is a solution, we have $320 - 40b + (b^2+2b+5) = 0$, i.e. $b^2-38b+325 = 0$.
Can you find what value of $b$ satisfies both of these conditions?
|
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|
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2-1=t$ but no benefit. Please guide me.
|
Given $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int_{1}^{2}\frac{x^2-1}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx$$
$$\displaystyle = \int_{1}^{2}\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$
Now Let $2-2x^{-2}+x^{-4} = t^2\;,$ Then $\displaystyle \left(x^{-3}-x^{-5}\right)dx = \frac{2t}{4}dt$
so we Get $$\displaystyle I = \frac{1}{2}\int_{1}^{2}\frac{t}{t}dt = \frac{1}{2}t = \frac{1}{2}\left[\sqrt{2-2x^{-2}+x^{-4}}\right]_{1}^{2} = \frac{1}{2}\left[\frac{\sqrt{2x^4-2x^2+1}}{x^2}\right]_{1}^{2}$$
So we get $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \frac{1}{8}$$
But Here Given $$\displaystyle I = \frac{1}{8} = \frac{u}{v}.$$ So we get $\displaystyle u=1\;, v=8$
So we get $$\displaystyle \frac{1000u}{v} = \frac{1000 \times 1}{8} = 125.$$
|
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|
Prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ If the bisectors of the angles of a triangle $ABC$ meet the opposite sides in $A',B',C'$,prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}:\cos \frac{A-B}{2}\cos \frac{B-C}{2}\cos \frac{C-A}{2}$
Area of $ABC=$Area of $ABA'+$Area of $ACA'$
$=\frac{1}{2}(c+b)AA'\sin \frac{A}{2}=R(\sin C+\sin B)AA'\sin \frac{A}{2}$
$=R\times 2\sin \frac{B+C}{2} \cos \frac{B-C}{2}AA'\sin \frac{A}{2}$
How to find the area of $A'B'C'$ and get the desired result,help me please.Thanks in advance.
|
Let me try. One has $$\frac{B'A}{B'C} = \frac{BA}{BC} \Rightarrow \frac{B'A}{AC} = \frac{BA}{BA+BC},$$
$$\frac{C'A}{C'B} = \frac{CA}{CB} \Rightarrow \frac{C'A}{AB} = \frac{CA}{CA+CB}$$.
So, one has $$\frac{S_{AB'C'}}{S_{ABC}} = \frac{AB'.AC'}{AC.AC} = \frac{BA.CA}{(BA+BC)(CA+CB)}.$$
Similarly, one has
$$\frac{S_{A'BC'}}{S_{ABC}} = \frac{BA'.BC'}{BA.BC} = \frac{BA.BC}{(AB+AC)(BC+AC)}.$$
$$\frac{S_{A'B'C}}{S_{ABC}} = \frac{CA'.CB'}{CA.CB} = \frac{CA.CB}{(BC+AB)(CA+AB)}.$$
Thus, $$\frac{S_{A'B'C'}}{S_{ABC}} = 1 - \frac{S_{AB'C'}}{S_{ABC}} - \frac{S_{A'BC'}}{S_{ABC}} - \frac{S_{A'B'C}}{S_{ABC}} = \frac{(AB+BC)(BC+CA)(CA+AB) - AB.CA(AB+CA)- AB.BC(AB+BC)-CA.BC(BC+CA)}{(AB+BC)(BC+CA)(CA+AB)}$$
$$= \frac{2AB.BC.CA}{(AB+BC)(BC+CA)(CA+AB)}$$
$$ = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin A + \sin C)}$$
$$ = 2\frac{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{\cos \frac{A-B}{2}\cos \frac{B-C}{2}\cos \frac{C-A}{2}} .$$
|
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|
Where does this sequence $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.... converge? The given sequence is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and so on.
the sequence is increasing so to converge must be bounded above.Now looks like they would not exceed 7. The given options are
*
*${1+\sqrt{33}}\over{2}$
*${1+\sqrt{32}}\over{2}$
*${1+\sqrt{30}}\over{2}$
*${1+\sqrt{29}}\over{2}$
How to proceed now.
Thanks for any help.
|
Here's a quick proof that the sequence converges.
The sequence is given by the recursive formula $a_0 = 0$, and $a_{n+1} = \sqrt{7 + a_n}$.
Using the method suggested by the previous poster, let $L$ be the positive solution to the equation $x^2 = x + 7$.
We can prove that $a_n < L$ for all $L$ by induction. It is clear that $a_0 = 0 < L$. Then if $a_n < L$ we have $a_{n+1} = \sqrt{7 + a_n} < \sqrt{7 + L} = L$.
We can also prove that the sequence is increasing. Since $a_1 = \sqrt{7}$, we have $a_1 > a_0$. Now, if $a_n > a_{n-1}$, we have $a_{n+1} = \sqrt{7 + a_{n}} > \sqrt{7 + a_{n-1}} = a_{n}$.
|
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|
Find the values of 'a' in a $4\times 4$ matrix(A) when the determinant is less than 2012 The matrix is $A \ =\begin{pmatrix}
7 & 1 & 3 & -2\\
-2 & 1 & -12 & -1 \\
1 & 16 & -4 & a \\
2 & 4 & 2 & 2 \\
\end{pmatrix}$
Where $\det(A)\lt 2012$
I have tried reducing the first columns to 0's to reduce it to a $3\times3$ determinant. When I do that it gets messy fast. Is there a special method to solve this problem?
|
$A \ =\begin{pmatrix}
7 & 1 & 3 & -2\\
-2 & 1 & -12 & -1 \\
1 & 16 & -4 & a \\
2 & 4 & 2 & 2 \\
\end{pmatrix}\cong \begin{pmatrix}
13 & 1 & 5 & -5\\
-5 & 1 & -25 & -3 \\
-14 & 16 & -24 & 2a-16 \\
0 & 4 & 0 & 0 \\
\end{pmatrix}$
S0, you would get $\det(A)=-4\det\begin{pmatrix}
13 & 5 & -5\\
-5 & -25 & -3 \\
-14 & -24 & 2a-16 \\
\end{pmatrix}$
Can you finish this and fill the gaps?
|
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|
Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$ Let $a,b,c$ be positive numbers. Then we need to prove
$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$
I have an idea to set $x=\frac a{b+c}$, $y=\frac b{c+a},z=\frac c{a+b}$ then
$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}=2$
and we need to prove
$\frac1{\sqrt x}+\frac1{\sqrt y}+\frac1{\sqrt z}\ge2\left(\sqrt x+\sqrt y+\sqrt z\right)$
But I could not go further.
|
$$\sum_{cyc}\sqrt{\frac{a+b}{c}}-2\sum_{cyc}\sqrt{\frac{c}{a+b}}=\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}=$$
$$=\sum_{cyc}\frac{b-c-(c-a)}{\sqrt{c(a+b)}}=\sum_{cyc}(a-b)\left(\frac{1}{\sqrt{b(a+c)}}-\frac{1}{\sqrt{a(b+c)}}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2c}{\sqrt{ab(a+c)(b+c)}\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)}\geq0.$$
Done!
|
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|
Find the min and max distance from origin of the curve $\vert z+\frac{1}{z}\vert=a$ $z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation \vert z + 1/z \vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+\frac{2(x^2-y^2)}{x^2+y^2}+\frac{(x^2-y^2)}{(x^2+y^2)^2}=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
|
Short Solution:
By the Triangle Inequality,
$$a=\left|z+\frac{1}{z}\right|\geq \left||z|-\frac{1}{|z|}\right|\,,$$
giving
$$-a\leq |z|-\frac{1}{|z|}\leq +a\,,$$
which means
$$\frac{\sqrt{a^2+4}-a}{2} \leq |z| \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
Since $z=\pm \frac{\sqrt{a^2+4}+a}{2}$ and $z=\pm\frac{\sqrt{a^2+4}-a}{2}$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=r\exp(\text{i}\theta)$ with $r>0$ and $\theta\in\mathbb{R}$ is a solution to $\left|z+\frac{1}{z}\right|=a$. Then, $$a^2=\left|z+\frac{1}{z}\right|^2=r^2+\frac{1}{r^2}+2\cos(2\theta)\,.$$
Hence,
$$a^2-2\leq r^2+\frac{1}{r^2} \leq a^2+2\,.\tag{*}$$
If $a>2$, then, $r$ must satisfy $\left(r+\frac{1}{r}\right)^2\geq a^2$, or $r+\frac{1}{r}\geq a$, as well as$\left(r-\frac{1}{r}\right)^2\leq a^2$, or $-a\leq r-\frac{1}{r}\leq +a$. This means
$$0\leq r\leq \frac{a-\sqrt{a^2-4}}{2}\text{ or }r\geq \frac{a+\sqrt{a^2-4}}{2}\,,$$ along with $$\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
Observe that $\frac{a-\sqrt{a^2-4}}{2}> \frac{\sqrt{a^2+4}-a}{2}$ for all $a>2$. Thence,
$$\frac{\sqrt{a^2+4}-a}{2}\leq r \leq \frac{a-\sqrt{a^2-4}}{2}\text{ or }\frac{a+\sqrt{a^2-4}}{2}\leq r\leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
If $0 \leq a\leq 2$, then we have by AM-GM that $ r^2+\frac{1}{r^2}\geq 2\geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $\left(r-\frac{1}{r}\right)^2\leq a^2$, or $-a\leq r-\frac{1}{r}\leq +a$. Therefore, $$\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}\,.$$
Consequently, in both cases, we have $\frac{\sqrt{a^2+4}-a}{2} \leq r \leq \frac{\sqrt{a^2+4}+a}{2}$. The equality on the left side is satisfied when $z=\pm\frac{\sqrt{a^2+4}-a}{2}$, and the equality on the right is satisfied iff $z=\pm\frac{\sqrt{a^2+4}+a}{2}$. Ergo, the maximum value of $|z|=r$ is $\frac{\sqrt{a^2+4}+a}{2}$, whereas the minimum value is $\frac{\sqrt{a^2+4}-a}{2}$.
|
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|
Understanding degrees of freedom in relation to rank for $\sum_{i=1}^{n}(y_i-\bar{y})^2$ So I'm looking at this website which states:
One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathematical answer is a single phrase, "The rank of a quadratic form."
And near the end:
Okay, so where's the quadratic form? Let's look at the variance of a single sample. If $y$ is an $n$ by $1$ vector of observations, then
$$\sum(y_i - \bar{y})^2 = y^{\prime}My\text{, where }M = \begin{pmatrix}
1-\frac{1}{n} & -1/n & \cdot & -1/n \\
-1/n & 1-\frac{1}{n} & \cdot & -1/n \\
\cdot & \cdot & \cdot & \cdot \\
-1/n & -1/n & -1/n & 1-\frac{1}{n}
\end{pmatrix}\text{.}$$
The number of degrees of freedom is equal to the rank of the $n$ by $n$ matrix $M$, which is $n-1$.
I will use $r$ to denote the rank and $C(A)$ to denote the column space of a matrix $A$.
From my notes, I have the definition $r(A) = r\left(C(A)\right)$, which is the number of vectors that create a basis for $C(A)$. Sure.
Let's consider column one:
$$\begin{pmatrix}
1-\frac{1}{n}\\
-1/n \\
\cdot \\
-1/n
\end{pmatrix} = \dfrac{-1}{n}
\begin{pmatrix}
1-n \\
1 \\
\cdot \\
1
\end{pmatrix} = \dfrac{-1}{n}\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix} + \dfrac{-1}{n}\begin{pmatrix}
-n \\
0 \\
\cdot \\
0
\end{pmatrix} = \dfrac{-1}{n}\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}+\begin{pmatrix}
1 \\
0 \\
\cdot \\
0
\end{pmatrix}\text{.}
$$
So based on what I'm seeing here, I'm guessing that a basis for $C(A)$ would be something like
$$\mathscr{V} = \left\{\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}, \begin{pmatrix}
1 \\
0 \\
\cdot \\
0
\end{pmatrix}, \begin{pmatrix}
0 \\
1 \\
\cdot \\
0
\end{pmatrix}, \cdots, \begin{pmatrix}
0 \\
0 \\
\cdot \\
1
\end{pmatrix}\right\}$$
which consists of $n+1$ vectors. However, clearly $$\begin{pmatrix}
1 \\
1 \\
\cdot \\
1
\end{pmatrix}$$ needs to be removed, since it is the sum of the other $n$ vectors, so I get that there are $n$ vectors in $\mathscr{V}$, and therefore $r(\mathscr{V}) = r(C(M)) = n$.
Where is the $n-1$ coming from?
|
The sum of all the $n$ columns is $(0,0,0,\ldots,0)^T$ and so surely the column rank cannot be $n$ as you claim it to be?
|
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|
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
|
There's one more solution (it isn't mine). One can even prove that $(a + b + c) \mid (a^n + b^n + c^n)$ for all $n=3k+1$ and $n=3k+2$. It's enough to prove that
$a + b + c \mid a^n + b^n + c^n$ => $a + b + c \mid a^{n+3} + b^{n+3} + c^{n+3}$.
The proof is here:
https://vk.com/doc104505692_416031961?hash=3acf5149ebfb5338b5&dl=47a3df498ea4bf930e
(unfortunately, it's in Russian but it's enough to look at the formulae). One point which may need commenting:
$(ab+bc+ca)(a^{n-2} + b^{n-2} + c^{n-2})$ is always divisible by $(a+b+c)$ (it's necessary to consider 2 cases: $(a+b+c)$ is odd and $(a+b+c)$ is even).
|
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|
How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
|
If $b=0,ac=0\implies c=0\implies ab+cd=0$
Else $ac+bd=0\iff ac=-bd\iff\dfrac ab=\dfrac{-d}c=k$(say)
$\implies a=bk, d=-ck$
If $a^2+b^2=c^2+d^2,$ not necessarily $=1$
$b^2(1+k^2)=c^2(1+k^2)\implies b^2=c^2$ if $1+k^2\ne0$
Now $ab+cd=(bk)b+c(-ck)=k^2(b^2-c^2)=?$
|
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|
How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to apply Vieta formula to find the relationship between its roots. From there, I can get:
$$a+b+c+d = 7$$
$$ab+ac+ad+bc+bd+cd= 3$$
$$abc+abd+acd+bcd= 21$$
$$abcd= 1$$
Then,
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$$
$$= (a²+b²+ab+3)(c²+d²+cd+3)$$
$$= 3(a^2+b^2+c^2+d^2)+abcd+(ac)^2+(ad)^2+(bc)^2+(bd)^2+a^2cd+b^2cd+abc^2+abd^2$$
From there, I don't have any idea how to go further.
Can somebody help me to explain to solve this equation?
Thanks.
|
Let me try. You have $$(a+b+c)(a+b+d)(a+c+d)(b+c+d) = (7-a)(7-b)(7-c)(7-d) = 7^4 - 7^3(a+b+c+d) + 7^2(ab+ac+ad+bc+bd+cd)-7(abc+abd+bcd+acd) + abcd$$
|
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|
Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$ I've found the following identity while I was going through a quite difficult path.
$$
\Re\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right) = \frac{\pi^2}{24} -\frac{1}{2}\ln^2 2 - \frac{1}{4}\operatorname{Li}_2\left(\tfrac{1}{4}\right),$$
where $\operatorname{Li}_2$ is the dilogarithm function.
I think we could prove it directly from dilogarithm identites.
How could we prove the identity above?
Furthermore
Could we specify $\Im\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$?
There is a similar question here.
|
Here is a solution only using dilogarithm identities:
$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1} $$
$$ \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2} $$
$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{2}\operatorname{Li}_2(z^2). \tag{3} $$
We remark that each identity is true for all $z \in \Bbb{C}$ that avoids the principal branch cuts of the functions which are involved in it. Then
*
*Applying (1), we have
\begin{align*}
\operatorname{Li}_2(1\pm i\sqrt{3})
&= \zeta(2)-\log(1\pm i\sqrt{3})\log(\mp i\sqrt{3}) - \operatorname{Li}_2(\mp i\sqrt{3}) \\
&= \zeta(2)-\left( \log 2 \pm \tfrac{i\pi}{3} \right) \left( \tfrac{1}{2}\log 3 \mp \tfrac{i\pi}{2} \right) - \operatorname{Li}_2(\mp i\sqrt{3}).
\end{align*}
Taking real part of both sides, we get
$$ \Re \operatorname{Li}_2(1\pm i\sqrt{3}) = - \tfrac{1}{2}\log 2 \log 3 - \Re\operatorname{Li}_2(\mp i\sqrt{3}). $$
*Now we apply (3). Using the fact that $\overline{\operatorname{Li}_2(z)} = \operatorname{Li}_2(\bar{z})$, we have
$$ \Re\operatorname{Li}_2(\mp i\sqrt{3})
= \tfrac{1}{2}\left( \operatorname{Li}_2(i\sqrt{3}) + \operatorname{Li}_2(-i\sqrt{3}) \right)
= \tfrac{1}{4}\operatorname{Li}_2(-3). $$
*Finally, we use (2) with $z = 4$ and (1) with $z = 1/4$. Then
$$ \operatorname{Li}_2(-3) = \operatorname{Li}_2(\tfrac{1}{4}) - \zeta(2) + 2\log 2 \log(\tfrac{2}{3}). $$
*Combining these altogether, we have
\begin{align*}
\Re \operatorname{Li}_2(1\pm i\sqrt{3})
&= - \tfrac{1}{2}\log 2 \log 3 - \Re\operatorname{Li}_2(\mp i\sqrt{3}) \\
&= - \tfrac{1}{2}\log 2 \log 3 - \tfrac{1}{4}\operatorname{Li}_2(-3) \\
&= \tfrac{1}{4}\zeta(2) - \tfrac{1}{2}\log^2 2 - \tfrac{1}{4}\operatorname{Li}_2(\tfrac{1}{4}).
\end{align*}
P.s. I read Jack D'Aurizio's answer now and I see that all the essential idea is exactly the same. I should have checked his answer before I write it. ;(
|
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|
For Which Value The Matrix is Diagonalizable?
For which values of $a$ the matrix $\left(\begin{array}{ccc} 2 & 0 & 0 \\ 2 & 2 & a \\ 2 & 2 & 2 \end{array}\right)$ is diagonalizable:
*
*above $\mathbb{R}$
*above $\mathbb{C}$
We need to look at the characteristic polynomial which is $(x-2)^3-2a(x-2)=x^3-6x^2+2x(6-a)+4(a-2)$ , How do I find the eigenvalues one is $2$ and the other? and how so I continue?
|
$x^3-6x^2+2x(6-a)+4(a-2)$. Clearly it shows that $x=2$ is a zero of the polynomial. Now ,
$x^3-6x^2+2x(6-a)+4(a-2)=0$
$\implies (x-2)(x^2-4x+4-2a)=0$
$\implies x=2 , x=\frac{4\pm\sqrt{16-4(4-2a)}}{2}=2\pm\sqrt{2a}.$
If $a=0$ then , all roots are $2$ and then the matrix is NOT diagonalizable(check!)
If $a\not=2$ then the matrix is diagonalizable.
|
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|
How to compute fraction sums? For example, $$\sum\limits_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
Is there an easier way to evaluate fraction sums (without using partial sums)?
|
A general approach which works quite well is partial fraction decomposition:
$$
\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)
$$
After this, you can use the technique of telescoping.
In your example, this yields:
\begin{align}
& \sum_{k=1}^{n}{\frac{1}{(2k-1)(2k+1)}} = \sum_{k=1}^{n}{\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)} \\
&= \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+…+\frac{1}{2n-3}-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\right) \\
&= \frac{1}{2}\left(1-\frac{1}{2n+1}\right) = \frac{n}{2n+1}
\end{align}
|
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|
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4cos^2(x)-2=\sqrt3$$
$$2(2\cos^2(x)-1)=\sqrt3$$
$$2\cos(2x)=\sqrt3$$
$$\cos(2x)=\frac{\sqrt3}{2}$$
$$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$
$$2x=\pm \frac{\pi}{6}+2n\pi$$
$$x=\pm \frac{\pi}{12}+n\pi$$
Considering the condition $\sin(x)\le 0$, we are left with
$$x=- \frac{\pi}{12}+n\pi$$
But the textbook's answer is
$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$
What did I overlook?
P.S. The problem and the answer from the texbook:
|
Take advantage of the fact that $\cos^2{x}=1-\sin^2{x}$. Then solve for $\sin{x}$, then solve for $x$. $\frac{\pi}{12}=15^o$
|
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|
Limits using definite integration $F(k)$ = $$ \lim_{n\to \infty}{\frac{1^k + 2^k +...+n^k}{(1^2 + 2^2 +...+n^2)*(1^3 + 2^3 +...+n^3)}} $$
I need help in finding $F(5)$ and $F(6)$.
I tried converting it into summation form and using the progression formulas of $n^2$ and $n^3$ but it was of no use.
|
Theorem: Define $S_r(n)=1^r+2^2+...+n^r$, $$\frac{n^{r+1}}{r+1}\leq S_r(n)\leq\frac{(n+1)^{r+1}}{r+1}$$ for any positive $r\geq0$
Then $$\frac{12n^{k+1}}{(k+1)(n+1)^3(n+1)^4}\leq{\frac{1^k + 2^k +...+n^k}{(1^2 + 2^2 +...+n^2)\cdot(1^3 + 2^3 +...+n^3)}}\leq \frac{12(n+1)^{k+1}}{(k+1)n^3n^4}$$
So, $$F(6)=\frac{12}{7}$$ and $F(k)=0$, if $0\leq k<6$ and $F(k)=+\infty$, if $k>6$
|
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|
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$.
I have done the following:
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$
Is this correct?
|
You can take also another approach. Since $x^3-y^3=(x-y)(x^2+y^2+xy)$ you need show that $\frac{xy(x-y)}{x^2+y^2}$ goes to $0$ when $(x,y)\to 0$. So, call $x=r\cos \theta$, $y=\sin \theta$ and $r^2=x^2+y^2$. We have $\frac{xy(x-y)}{x^2+y^2}=\frac{r^3(\cos^2 \theta \sin \theta-\cos \theta \sin^2\theta)}{r^2}$ and this goes to zero when $r\to 0$.
|
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|
Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution.
Let $x,y,z$ be positive real numbers such that $xyz =1$. Prove that
$$
\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+z)(1+x)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}
$$
Using AM-GM in the following form:
$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}{8} + \frac{1+z}{8} \ge \frac{3x}{4}$.
Now, this is where I get lost. Where was this expression derived from?
We conclude that
$\sum_{cyc}\frac{x^3}{(1+y)(1+z)} + \frac{1}{4}\sum_{cyc}(1+x) \ge \sum_{cyc}\frac{3x}{4} \rightarrow \sum_{cyc}\frac{x^3}{(1+y)(1+z)} \ge \frac{1}{4}\sum_{cyc}(2x-1) \ge \frac{3}{4}$
The equality holds for $x=y=z=1$
|
I hope one of the $3$ parts below would help you understand the proof above.
Part 1: the AM-GM ineq is: $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{1+y}{8}+\dfrac{1+z}{8} \geq 3\sqrt[3]{\dfrac{x^3(1+y)(1+z)}{(1+y)(1+z)\cdot 8\cdot 8}} = \dfrac{3x}{4}$.
Part 2: $\sum_{cyclic} \left(\dfrac{1+y}{8}+\dfrac{1+z}{8}\right) = 2\sum_{cyclic} \dfrac{1+x}{8}= \dfrac{1}{4}\sum_{cyclic}(1+x)$
Part 3: $\sum_{cyclic} \dfrac{3x}{4} - \sum_{cyclic} \dfrac{1+x}{4}=\sum_{cyclic} \left(\dfrac{3x}{4}-\dfrac{1+x}{4}\right)=\sum_{cyclic} \dfrac{2x-1}{4}=\dfrac{1}{2}\sum_{cyclic} x - \dfrac{3}{4}\geq \dfrac{1}{2}\cdot 3\sqrt[3]{\prod_{cyclic}x}=\dfrac{3}{2}-\dfrac{3}{4} = \dfrac{3}{4}$
|
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|
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
|
\begin{align}
& \lim_{x\to0} \left( \frac 1 {x \arctan x} - \frac 1 {x^2} \right) = \lim_{x\to0} \frac {x - \arctan x} {x^2 \arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac{1 - \frac 1 {1+x^2}}{\frac{x^2}{1+x^2} + 2x \arctan x } \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to0} \frac{x^2}{x^2 + 2x(1+x^2) \arctan x} = \lim_{x\to0} \frac{x}{x + 2(1+x^2)\arctan x} \\[10pt]
= {} & \lim_{x\to0} \frac 1 {1 + 4x\arctan x + 2} \qquad \text{(L'Hopital)} \\[10pt]
= {} & \lim_{x\to 0 } \frac 1 {3+4x\arctan x} = \frac 1 3.
\end{align}
|
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|
trying to solve $\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$ The equation is
$$\sqrt{\cos(x)-2\cos(2x)}+\sqrt{2}\cos(2x)=0$$
The system is
$$
\begin{cases}
\cos(x)-2\cos(2x)=2\cos^2(2x) \\
-\sqrt{2}\cos(2x)\ge 0 \iff \cos(2x)\le 0
\end{cases}
$$
The equation:
$$\cos(x)-2(2\cos^2(x)-1)=2(2\cos^2(x)-1)^2$$
$$\cos(x)-4\cos^2(x)+2=2(4\cos^4(x)-4\cos^2(x)+1)$$
$$\cos(x)-4\cos^2(x)+2=8\cos^4(x)-8\cos^2(x)+2$$
$$8\cos^4(x)-4\cos^2(x)-\cos(x)=0$$
But what to do next?
If I factor out $\cos(x)$, I get a non-factorizable third-degree polynomial in parentheses
$$\cos(x)(8\cos^3(x)-4\cos(x)-1)=0$$
The problem with its solution (upside-down) from the textbook
|
Hint: You can factorise: $$8c^4-4c^2-c=c(2c+1)(4c^2-2c-1)$$
|
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|
Find all integers $m,n$ for which $m^2+n^2$ is a square and $\sqrt{\frac{2m^2+2}{n^2+1}}$ is rational This is a repost of my old question here. The question is as follows:
Find all integers m and n, such that $m^2 + n^2$ is a square and $\sqrt{\frac{2(m^2+1)}{n^2+1}}$ is rational.
I have made no progress on this problem since I last asked it. Tito gave a very nice answer for if $m,n$ were rational numbers, but it would be awesome if some people could provide some insight into this problem and solve it once and for all. It would also be really cool if we replaced that 2 with another number and investigated that as well.
(For your reference, the original question was 2010 USAJMO problem 6. I tried to bash the question out and somehow I arrived at this problem. I suspect I made a mistake somewhere because the problem is no longer homogeneous.)
|
Here is my amateur attempt to your problem.
The number is rational iff there exist integers $\,p, q\in \mathbb N\,$ such that $\displaystyle\,\sqrt{\frac{k\left(m^2+1\right)}{n^2+1}}=\frac{p}{q}.\,$
$\,m^2 + n^2\,$ is square $\iff$ there exist an integer $\,r\in \mathbb N\,$ such that $\,m^2 + n^2 = r^2\,$
Thus we get a system
$$
\left\lbrace\begin{aligned}
\frac{p^2}{q^2} &= \frac{k\left(m^2+1\right)}{n^2+1} \\
r^2 & = m^2 + n^2 \\
\end{aligned}\right.
\implies
\left\lbrace\begin{aligned}
p^2 & = k\left(m^2+1\right) \\
q^2 & = n^2+1 \\
r^2 & = m^2 + n^2
\end{aligned}\right.
\implies
\left\lbrace\begin{aligned}
m^2 & = \frac{p^2}{k} - 1 \\
n^2 & = q^2 - 1 \\
r^2 & = \frac{p^2}{k} + q^2 - 2
\end{aligned}\right.
\implies
%\left\lbrace\begin{aligned}
% p,q,r \in \mathbb N, \\
% \frac{p^2}{k} \in \mathbb N ,\\
%\end{aligned}\right.
\left\lbrace\begin{aligned}
%\frac{p^2}{k}
p^2&/k\in \mathbb N^+ ,\\
p^2 &\ge k>0 \\
q^2 & = n^2 + 1
\end{aligned}\right.
$$
so the problem is only solvable for $\,k\in \mathbb Q^+.\,$
Also, $\, r^2 = m^2 + n^2 \implies m^2 = r^2 - n^2,\,$ therefore
$$
\left\lbrace\begin{aligned}
p^2 & = k\left(m^2+1\right) \\
q^2 & = n^2+1 \\
r^2 & = m^2 + n^2
\end{aligned}\right.
\Rightarrow
\left\lbrace\begin{aligned}
p^2 & = k \left( r^2 - n^2 + 1 \right) \\
q^2 & = n^2+1 \\
m^2 & = r^2 - n^2
\end{aligned}\right.
\Rightarrow
\left\lbrace\begin{aligned}
p^2 & = k\left(m^2+1\right) \\
q^2 & = 1 \\
n^2 & = 0 \\
\end{aligned}\right.
%\Rightarrow
\implies
\left\lbrace\begin{aligned}
m & = \pm\sqrt{\frac{p^2}{k}-1} \\
% q & = \pm 1 \\
n & = 0 \\
k & \ge 0
\end{aligned}\right.
$$
The problem is only well-defined for $\,k \ge 0.\,$
Note that case $\,k=0\,$ is trivial, so we require $\, k \in \mathbb Q^+.\,$
Additionally, $\,r^2 = m^2 + n^2 \ge 0 \,$ is non-trivial iff $\,r^2 >0.\,$
If $\,r^2=0\,$ we would get $\,m = n = 0.\,$
$$
%m^2= \frac{p^2}{k}-1 \implies
%\frac{p^2}{k}-m^2 = 1 \implies
\left\lbrace\begin{aligned}
&p^2 = k\left(m^2+1\right) \\
% &n^2 = 0 \\
&\frac{p^2}{k} + q^2 > 2
\end{aligned}\right.
\stackrel{q^2 =1}{\implies }
\left\lbrace\begin{aligned}
p^2 & = k\left(m^2+1\right) \\
p^2 & > k
\end{aligned}\right.
\implies m^2 >0 \implies m \neq 0
$$
$$
0<k \in \mathbb Q^+ \implies k = \dfrac{a}{b}, \ a,b\in \mathbb N^+
\implies
%p^2 = \frac{a\left(m^2 + 1 \right)}{b}
m^2 = \frac{b}{a}p^2 - 1\\
\bbox[5pt, border:2pt solid #f10000]{b p^2 - am^2 = a}
$$
In this way we obtained a second order Diophantine equation with $\,a,b\in \mathbb N^+\,$ given and $\,m\,$ can possibly be determined from the equation above by choosing appropriate integer values of $\,p.\,$
Note that if we assume integer $\,k \in \mathbb N^+,\,$ we get Pell-like equation which might be easier to solve.
$$
\bbox[5pt, border:2pt solid #f10000]{m^2 - kp^2 = - 1}
$$
According to WolframAlpha, if one of the solutions of such an equation is known, then the rest can be computed using standard techniques for Pell Equation.
In particular, for $\, k = 2\,$ we get $\, a=2, \ b= 1,\,$ and
$$
\bbox[5pt, border:2pt solid #f10000]{m^2 = 2p^2 - 1}
$$
so that in addition to $\,n=0\,$ we get
$$
\begin{aligned}
p &= 1 &\implies& &m &= \pm 1, \\
p &= 5 &\implies& &m &= \pm 7, \\
p &= 29 &\implies& &m &= \pm 41, \\
\end{aligned}
\\
\cdots
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1414144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Nonseparable differential equations How can I solve the equation
$\frac{dy}{dx} =\frac{x^2-y}{x-y^2}$?
I've tried few substitutions such as $y=xv$ and $y=x/v$ but all to no avail!
Please, help.
|
$$\frac { dy }{ dx } =\frac { x^{ 2 }-y }{ x-y^{ 2 } } \\ \quad \quad \quad \quad $$ $$\Downarrow \\$$$$ \left( x^{ 2 }-y \right) dx-\left( x-y^{ 2 } \right) dy=0\\ \quad \quad \quad \quad \quad $$$$\Downarrow \\ \left( { x }^{ 2 }dx+{ y }^{ 2 }dy \right) -\left( ydx+xdy \right) =0\\ \quad \quad \quad \quad \quad \quad $$$$\Downarrow \\ d\left( \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } \right) -d\left( xy \right) =0\\ \quad \quad \quad \quad \quad \quad $$$$\Downarrow \\ d\left( \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } -xy \right) =0\\ \quad \quad \quad \quad \quad \quad \quad $$
$$\Downarrow \\$$
$$ \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } -xy=C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1415892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know how that helps. I'm greatly seeking your help. Thank you very much.
|
Here's an idea. In general you want to show that
$$\frac{a}{a+1} < \frac{a+1}{a+2}.$$
What happens if you multiply $\frac{a}{a+1}$ by $\frac{(a+1)^2}{a+2}$? What do you know about the number $\frac{(a+1)^2}{a+2}$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.