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How to solve Quadratic Equations with an Unknown C and other variables. For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?
And also : If $x^2 +px + q$ is a factor of $x^3 + ax^2 + b$,
prove that:
$b = q(a-p)$
| A: $3x^2-11x +r=0$
The codition is that discreminet $\Delta\geqslant 0$, so we must have:
$\Delta= 11^2-12 r\geqslant0$
Or: $r\leqslant 10$
Now $121$ is odd and $12r$ is even and we have to find odd perfect squares less than $121$ which are $81, 49, 25$ and $9$. Only $49$ and $25$ give integers for $r$:
$r= 121-49=72=12r$ so $r=6$
$r=121-25=96=12r$ so $r=8$
B: $x^2+px+q|x^3+ax+b$
By direct division you find that thr remainder is $(a-p)x^2-qx+b$ which must be equal to zero, so we have:
$x[(a-p)x-q]=-b$
Therefore b must divide both $q$ and $(a-p)$ because $(a-p)$ and $q$ have no common divisors, hence we must have $b|q(a-p)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all natural $n$ numbers such that Find all natural $n$ numbers such that
$15(n!)^2+1$ is divisible by $2n-3$.
My try: First I assumed $2n-3$ is not prime number. Let $a$ be divisor of $2n-3$. It's clear that $a<n-1$, so $15(n!)^2$ is divisible by $a$. Which means $15(n!)^2+1$ is not divisble by $a$. But it is given that $15(n!)^2+1$is divisble by $n$ which means it is divisible by $a$ too. But we have already proved it is not. Contradiction!
So $2n-3$ must be prime number. Now if we change $2n-3$ as $p$. We can say $15*((p+3)/2)!*((p+3)/2)!$ is congruent to $-1$ by p module. By Wilson's theorem $15*((p+3)/2)!*((p+3)/2)!$ is congruent to $(p-1)!$ by p module. From here I don't know how to continue.
| You do need Wilson's Theorem, but observe:
\begin{align}n &\equiv -(n-3) \pmod {2n-3}\\
n-1 &\equiv -(n-2) \pmod {2n-3}\\
&\ \vdots\\
1 &\equiv -(2n-4)\pmod {2n-3}\end{align}
This gives:
\begin{align}(n!)^2&\equiv n!(-1)^n(2n-4)(2n-5)\dots(n-2)(n-3) \pmod {2n-3}\\&\equiv(-1)^n(2n-4)!(n)(n-1)(n-2)(n-3) \pmod {2n-3}\\\text{(Wilson)} &\equiv(-1)^{n+1}(n)(n-1)(n-2)(n-3)\pmod {2n-3}\\
&\equiv(-1)^{n+1}n^2(n-1)^2 \pmod {2n-3}\end{align}
Now we seperate into cases where $n$ is odd or even. For $n$ odd satisfying the division condition,
$$15(n!)^2+1\equiv 15n^2(n-1)^2+1 \equiv 0 \pmod {2n-3}$$
$z = \dfrac {15n^2(n-1)^2+1}{2n-3}$ is an integer iff $16z$ is. Simplifying we have
$$16z=120 n^3 - 60 n^2 + 30 n + 45 + \frac {151} {2 n - 3}$$
So the above is an integer only if $2n-3$ divides $151$, which is a prime, giving $2n-3 = 151$, $n = 77$.
The case for $n$ even should be similar. Technically, you also need to consider the cases $2n-3 = \pm 1$ separately, since $\pm1$ are neither prime nor composite.
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite product of $\sqrt{2}$ I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to evaluate the double sum. Furthermore, I tried to square and reorder (although it should not be possible), but haven't quite got the right track.
Could someone give me a hint for this problem?
| $$\begin{align}\frac{2}{1}\frac{2}{3} &= 4^1\frac{2!^3}{4!1!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7} &= 4^2\frac{4!^3}{8!2!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11} &= 4^3\frac{6!^3}{12!3!^2}\\\ldots\end{align}$$
Use Stirlings approximation $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ to establish the limit (the $e$'s and $\pi$'s will factor out).
| {
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"timestamp": "2023-03-29T00:00:00",
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indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$ I tried integration by parts considering $dx=du$
$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.
My attempt....
$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$
How do we calculate $\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$ and
$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$
How do we do it?
Please help!
| Let $x=\sinh^2t$. Then
$$\int\frac{dx}{\sqrt{x(x+1)(x+2)}}=\int\frac{2\sinh t\cosh t\,dt}{\sqrt{\sinh^2t\cosh^2t(\sinh^2t+2)}}=\sqrt2\int\frac{dt}{\sqrt{1+\frac12\sinh^2t}}
\\=\sqrt2\int\frac{dt}{\sqrt{1-\frac12\sin^2it}}.$$
| {
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Proving or disproving basic facts about sequences in Real Analysis I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.
\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):
(a) sequences $(x_n)$ and $(y_n)$, which both diverge, but whose sum $(x_n + y_n)$ converges;
(b) sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges, and $(x_n + y_n)$ converges;
(c) a convergent sequence $(b_n)$ with $b_n \ne 0$ for all $n$ such that $1/b_n$ diverges;
(d) an unbounded sequence $(a_n)$ and a convergent sequence $(b_n)$ with $(a_n - b_n)$ bounded
(e) two sequences $(a_n)$ and $(b_n)$, where $(a_n b_n)$ and $a_n$ converge but $(b_n)$ does not.
Solution.
(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.
Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.
(b) This request is impossible. If the $(x_n + y_n)$ is to be convergent, it implies we are able to make the distance $\vert{(x_n + y_n) - (x + y)}\vert$ as small as like. However, we cannot make $y_n$ to lie eventually in a set $(y - \epsilon, y + \epsilon)$. Hence, the sum cannot be convergent.
(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.
(d) This request is impossible. The key here is to show that, assuming here $(a_n)$ is bounded leads to the contradiction, leading to their difference also being bounded.
If $(a_n)$ is a bounded sequence, there exists a large number $M > 0$, such that $\vert{a_n}\vert < M$ for all $n \in \mathbf{N}$. If $(b_n)$ is a bounded sequence, there exists a large number $N > 0$, such that $\vert{a_n}\vert < N$ for all $n \in \mathbf{N}$.
Thus,
\begin{align*}
\vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\
&\le \vert{a_n}\vert + \vert{-b_n}\vert\\
&< M + N
\end{align*}
(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not.
| $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.
(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $0$.
Also, consider the sequence $(x_n)$ given by $x_n = n$ and the sequence $(y_n)$ given by $y_n = -\sqrt{n^2 + 2n}$. Both sequences diverge, but the sum $(x_n + y_n)$ converges to $-1$.
Short proof.
Consider $a_n = \sqrt{n + 1} - \sqrt{n}$.
Observe that,
\begin{align*}
\sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
&=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\
&< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}}
\end{align*}
Pick $\epsilon > 0$. We can choose $N > \frac{4}{\epsilon^2}$. To show that this choice of $N$ indeed works, we prove that, that for all $n \ge N$,
\begin{align*}
\absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\
&< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon
\end{align*}
Thus, $(\sqrt{n+1} - \sqrt{n}) \to 0$.
Consider $b_n = n - \sqrt{n^2 + 2n}$
Observe that:
\begin{align*}
n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\
&= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\
&= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\
&< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n}
\end{align*}
Pick an arbitrary $\epsilon > 0$. We choose an $N > \frac{1}{2\epsilon}$. To show that choice of $N$ indeed works, we find that:
\begin{align*}
\absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\
&< \frac{1}{2} \cdot (2\epsilon) = \epsilon
\end{align*}
Thus, $(n - \sqrt{n^2 + 2n}) \to -1$.
(b) This request is impossible. Alternatively, we believe that if $(x_n)$ converges and $(x_n + y_n)$ converges, then $(y_n)$ converges. Let us prove this fact rigorously. Suppose $\lim x_n = a$ and $ \lim x_n + y_n = b$. We shall prove that $\lim y_n = b - a$.
Observe that,
\begin{align*}
\absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\
&= \absval{(x_n + y_n - b) - (x_n - a)}\\
&\le \absval{x_n + y_n - b} + \absval{x_n - a}
\end{align*}
Pick an $\epsilon > 0$. Since $(x_n) \to a$, we can make the distance $\absval{x - a}$ as small as we like. There exists an $N_1$ such that
\begin{align*}
\absval{x_n - a} < \frac{\epsilon}{2}
\end{align*}
for all $n \ge N_1$.
Since $(x_n + y_n) \to b$, we can make the distance $\absval{x_n + y_n - b}$ as small as we like. There exists an $N_2$ such that,
\begin{align*}
\absval{x_n + y_n - b} < \frac{\epsilon}{2}
\end{align*}
Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we find that:
\begin{align*}
\absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\
&< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon
\end{align*}
(c) Consider the sequence $(b_n)$ given by $b_n = \frac{1}{n}$. Then, $(b_n)$ is a convergent sequence but $1/b_n$ is divergent.
Consider $(b_n) = \frac{1}{n}$. Pick an arbitrary $\epsilon > 0$. We can choose $N > \frac{1}{\epsilon}$. To show that this choice of $N$ indeed works, we find that:
\begin{align*}
\absval{\frac{1}{n}} < \epsilon
\end{align*}
for all $n \ge N$. Consequently, $(1/n) \to 0$.
(d) This request is impossible.
(e) Consider the sequence $(a_n)$ given by $a_n = \frac{1}{n}$ and $(a_n b_n)$ given by $a_n b_n= \frac{\sin n}{n}$. Thus, $(a_n b_n)$ and $(a_n)$ converges, but $(b_n)$ does not. Let us prove $\frac{\sin n}{n}$ converges to $0$.
Observe that,
\begin{align*}
\absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\
&\le \frac{1}{n}
\end{align*}
Let $\epsilon > 0$ be an arbitary small but fixed positive real number. We choose an $N > \frac{1}{\epsilon}$. To prove that this choice of $N$ indeed works, we have,
\begin{align*}
\absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\
&< \frac{1}{(1/\epsilon)} = \epsilon
\end{align*}
Consequently, $\frac{\sin n}{n} \to 0$.
| {
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Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $ Solve the ODE with the initial conditions:
$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$
I did the substitution:
$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$
Putting in the ODE:
$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = 1 \Rightarrow \frac{dz}{dy} = \frac{1}{y^2} \Rightarrow dz = \frac{1}{y^2}dy
$$
By integration we get:
$$
z = -\frac{1}{y} + C \Rightarrow y’ = - \frac{1}{y} + C
$$
Using the initial conditions we get:
$$
1 = -\frac{1}{1} + C \Rightarrow C = 2
$$
Therefore:
$$
y’ = -\frac{1}{y} + 2
$$
Now I am not sure how to continue.
| $y^2y''=y'$
Write as
$y''=\frac{y'}{y^2}$
Integrate both sides
$y'=-1/y +C$
$y'(0)=1,y(0)=1\to C=2$
$y'=-1/y+2$
$y=\frac{1}{2} \left(W\left(e^{4 x+1}\right)+1\right)$
Where $W(z)$ is Lambert $W$ function which solves $we^w=z$
The implicit form is
$x=\frac{1}{4} (2 y+\log (2 y-1)-1)$
| {
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Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.
Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.
Assume $y \ge 3$ and $x \ge 1$.
By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.
By $\mod 7$,we get $y \equiv 1 \mod 6$.
I also tried $\mod 2$ but it didn't work.
Please post hints ( not a solution). Thanks in advance!
| examples to study
CW:Catalan Thue Ramanujan Nagell Tijdeman p^x - q^y = C
https://math.stackexchange.com/users/292972/gyumin-roh
Exponential Diophantine equation $7^y + 2 = 3^x$
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. ME! 41, 31, 241, 17
Finding solutions to the diophantine equation $7^a=3^b+100$ 343 - 243 =
100
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
The diophantine equation $5\times 2^{x-4}=3^y-1$
Equation in integers $7^x-3^y=4$
Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. 128 - 125 = 3
Diophantine equation power of 7 and 2
Find natural numbers a,b such that $|3^a-2^b|=1$ did +-1
| {
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$\lim_{x\to 0}\frac{1}{x}−\frac{2}{e^{2x}−1}$ doesn't get solved as expected A question says $f:\Bbb R\setminus\{0\} \to \Bbb R$, $f(x)=\dfrac{1}{x}−\dfrac{2}{e^{2x}−1}$. The limit as the function approaches $x=0$ is logically $0$, as in;
$\lim \limits_{x \to 0}\dfrac{e^{2x}-1}{2x} = 1$ This function then becomes $\dfrac{1}{x}−\dfrac{2}{2x} =0$
But when I graphed it online, the function approached $1$.
| You did an approximation error. When $x\to0$ it is $\frac{2}{e^{2x}-1}=\frac{2}{1+2x+\frac{1}{2}(2x)^2+\text{o}(x^2)-1}=\frac{1}{x+x^2+\text{o}(x^2)}$, so
$$\frac{1}{x}-\frac{2}{e^{2x}-1}=\frac{1}{x}-\frac{1}{x+x^2+\text{o}(x^2)}=\frac{1}{x}\left(1-\frac{1}{1+x+\text{o}(x)}\right)$$
Using the fact that $\frac{1}{1+\varepsilon}=1-\varepsilon+\text{o}(\varepsilon)$ when $\varepsilon \to 0$, you have
$$\frac{1}{x}\left(1-\frac{1}{1+x+\text{o}(x)}\right)=\frac{1}{x}\left(1-1+x+\text{o}(x)\right)=1+\text{o}(1) \to 1 \ \text{when} \ x \to 0$$
| {
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How do you solve the equation $3x^2+2\sin x=2$? This seems like a simple equation to solve, but I can't find a way to do it and wolfram alpha can't give me an exact answer.
| This is a transcendental equation; so, you will need numerical methods and then good starting guesses.
Consider that you look for the zero's of function $f(x)=3x^2+2\sin x-2$.
For example, using Taylor series built around $x=0$
$$f(x)=-2+2 x+3 x^2-\frac{x^3}{3}+O\left(x^5\right)$$ Ignoring the term in $x^3$, solving the quadratic would give as estimates
$$x_1=-\frac{\sqrt{7}+1}{3} \qquad \qquad x_2=\frac{\sqrt{7}-1}{3}$$ We could solve the cubic but this is not very pleasant.
But what we can do is to take quite many terms in the series expansion and use series reversion. Looking at the values of $x_1$ and $x_2$, reasonable base points could be $-\frac \pi 3$ and $\frac \pi 6$. This would give as approximation
$$x_1=-\frac{\pi }{3}+t+\frac{\left(6+\sqrt{3}\right) }{2 (2 \pi
-1)}t^2+\frac{\left(59+18 \sqrt{3}-\pi \right) }{3 (2 \pi
-1)^2}t^3+O\left(t^4\right)$$ where $t=\frac{\pi ^2-3 \left(2+\sqrt{3}\right)}{3(2 \pi -1)}$. Numerically, this gives $x_1\approx -1.12638$ while Newton method would give as solution $x_1= -1.12630$.
Doing the same for the second root
$$x_2=\frac{\pi }{6}+t-\frac{5 }{2 \left(\sqrt{3}+\pi \right)}t^2+\frac{\left(78+\sqrt{3}
\pi \right) }{6 \left(\sqrt{3}+\pi \right)^2}t^3+O\left(t^4\right)$$ where $t=\frac{1-\frac{\pi ^2}{12}}{\sqrt{3}+\pi }$. Numerically, this gives $x_2\approx 0.559374$ while Newton method would give as solution $x_2= 0.559372$.
For sure, at the price of more messy terms, we could be even much closer to the solution.
| {
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.
| From Minkowski inequality: $\sqrt{x^2 + a^2} + \sqrt{y^2 + b^2} \ge \sqrt{(x + y)^2 + (a + b)^2}$
applied to the first two terms, with the notation of @Albus, we get:
$\sqrt{\sqrt{3}^2 +x^2} + \sqrt{\sqrt{3}^2 +y^2} \ge \sqrt{(2\sqrt{3})^2 + (x+ y)^2}$. Let's denote RHS of the inequality by $X$.
Then, apply Minkowski inequality again to $ X + \sqrt{3 + z^2}$, to obtain:
$\sqrt{27 + (x + y + z)^2} \ge 9$, which is equivalent to (1) from @Albus Dumbledore's answer.
| {
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Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:
If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$
I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.
| Notice that:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right).$$
The minimum of the function $f(w) = w + \frac{1}{w}$ is attained for $w=1$. More specifically, $f(1)= 2.$ See addendum section below for further details.
Then:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right) \geq 2 \cdot 2 \cdot 2 = 8.$$
Addendum
This addendum uses calculus tools, not just pre-calculus ones!
$w=1$ is the minimum of $f(w) : \mathbb{R}^+ \to \mathbb{R}^+.$
Indeed $f(w)$ is continuous for $w>0$. Moreover, notice that:
$$f'(w) = \frac{w^2-1}{w^2}.$$
Hence, $f'(w) = 0 \Rightarrow w = 1$.
Moreover:
$$f''(w) = \frac{2}{w^3} > 0$$ for all $w > 0.$
Finally, notice that:
$$\lim_{w=0^+} f(w) = +\infty,$$
and
$$\lim_{w=+\infty} f(w) = +\infty.$$
Then, $w=1$ is the unique minimum for $w > 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong? I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$
First of all I saw that it won't be possible to do that in any traditional way and actually calculate the limit, because of the form of expression. I mean - it's a sum with sqares on $n$ so I can't use Stolz lemma that easy. But, I thoght, that the solution is probably $0$, because probably every element of the sum is $0$ when $n \implies \infty$ and the limitation of sum in $\infty$ = sum of limitations in $\infty$. So I just went with that and decided to prove that using induction.
My base is:
$$\lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1}} = 0$$
My assumption:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +...+ \frac{1}{\sqrt{n^2 +2n}}\right) = 0$$
My induction:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0 + \lim_{n \to \infty} \frac{1}{\sqrt{n^2 +2n + 1}}) = 0$$
So the limit is:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0$$
But then my grader at university said that at first sight it looks totaly wrong but he actualy needs to think about it a bit. So here is my question - is that wrong? How is that wrong?
Ok, thank you for your answers. I thought I can solve that exercise that way because I asked not long ago very similar question on that forum: Find the limit of of such a sequence defined by recurrence
Because of your answers I think that the problem is actually that in this case I am dealing with a SUM of elements, am I right (or it the answer that I got in other case wrong?)?
| Clearly,
$$
\frac{1}{n+1}\le \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{n}
$$
for $k=1,2,\ldots,2n+1$. Hence
$$
\frac{2n+1}{n+1}<\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}<\frac{2n+1}{n}
$$
and
$$
\lim_{n\to\infty}\frac{2n+1}{n+1}=\lim_{n\to\infty}\frac{2n+1}{n}=2.
$$
Therefore
$$
\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\to 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| If $x^2 + 5x+12 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (-8)(-6)-48 = 0.$$
If $x^2 + 5x-2 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = 6\cdot 8-48 = 0.$$
Therefore LHS and RHS have the same zeroes. Since RHS has no double roots and both sides are polynomials of degree $4$, we conclude LHS = RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$
Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$
Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$
Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$
Now I can use the squeeze theorem and get:
*
*$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$
*$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$
And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?
Unfortunately, I can not use integrals while doing that exercise.
| $\sum_{k=1}^n k^{1/2}
\approx \int_0^n x^{1/2} dx
=\dfrac{n^{3/2}}{3/2}
=\dfrac23 n^{3/2}
$
and
$\sum_{k=1}^n k^{-1/2}
\approx \int_0^n x^{-1/2} dx
=\dfrac{n^{1/2}}{1/2}
=2n^{1/2}
$
so
$\begin{array}\\
s(n)
&=1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}})\\
&=\sum_{k=1}^n \dfrac{k}{\sum_{j=1}^k j^{1/2}}\\
&\approx\sum_{k=1}^n \dfrac{k}{\dfrac23 k^{3/2}}\\
&=\dfrac32\sum_{k=1}^n k^{-1/2}\\
&\approx\dfrac32(2n^{1/2})\\
&=3n^{1/2}\\
\end{array}
$
so the limit is $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90^\circ$, we have $AD$ the diameter of the circle, so $AD = 2r$.
From here, using Pythagorean Theorem :- $$2x^2 = 4r^2$$
$$\rightarrow x = r\sqrt{2}$$
We have the green angles to be $45^\circ$, now as $ADCB$ is cyclic, $\angle ABC = 135^\circ$ and each of the brown angles are $22.5^\circ$. So we can use :-
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\rightarrow a \div \frac{\sqrt{2 - \sqrt{2}}}{2} = r\sqrt{2} * \frac{2}{\sqrt{2}}$$
$$\rightarrow \frac{2a}{\sqrt{2 - \sqrt{2}}} = 2r$$
$$BC = a = r\sqrt{2 - \sqrt{2}}$$
However, the answer to my question is given as $r\sqrt{\sqrt{2}}$ , so where did I go wrong?
| Angle $ABC$ is $135°$ because it is half the concave angle $COA=270°$.
Thus angle $BAC=22.5°$ and $$BC=2r \sin 22.5°=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Trying to find the sum of cosines I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}{1-\cos(x)-i\sin(x)}\right)\tag{1}$$
$$=\Re\left(\frac{\left[1-\cos((k+1)x)-i\sin((k+1)x)\right]\left[(1-\cos x)+i\sin x\right]}{(1-\cos x)^2+\sin^2x}\right)=\frac{(1-\cos x)(1-\cos((k+1)x))+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$=\frac{1-\cos((k+1)x)-\cos (x)+\cos((k+1)x)\cos (x)+\sin((k+1)x)\sin(x)}{1-2\cos(x)}$$
$$\sin((k+1)x)\sin(x)=\frac{\cos(kx)-\cos((k+2)x)}{2}$$
$$\cos((k+1)x)\cos(x)=\frac{\cos(kx)+\cos((k+2)x)}{2}$$
$$\therefore\sin((k+1)x)\sin(x)+\cos((k+1)x)\cos(x)=\cos(kx)$$
$$S_k(x)=\frac{1-\cos((k+1)x)-\cos(x)+\cos(kx)}{1-2\cos(x)}\tag{2}$$
$$\cos(kx)-\cos((k+1)x)=-2\sin\left(\frac{(2k+1)x}{2}\right)\sin\left(\frac{-x}{2}\right)=2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)$$
$$S_k(x)=\frac{1-\cos(x)+2\sin\left(\left(k+\frac12\right)x\right)\sin\left(\frac x2\right)}{1-2\cos(x)}$$
| You can use this :
$$S_n(x)= \mathcal{Re}\left( \frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)\\
= \mathcal{Re}\left(e^{ikx/2}\frac{e^{-i(k+1)x/2}-e^{i(k+1)x/2}}{e^{-ix/2}-e^{ix/2}} \right)\\
=\mathcal{Re}\left( e^{ikx/2}\frac{\sin((k+1)x/2)}{\sin{(x/2)}}\right)\\
= \cos\left(\frac{kx}{2}\right)\frac{\sin\left(\frac{(k+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
I'm having trouble solving this equation. I've tried squaring both sides and got this
$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$
after simplifying
$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$
and that's where I got stuck.
| There is a formula to simplify nested radicals like that
$$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
It is effective only if $(a^2-b)$ is a perfect square. In your example both of them have this property
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
$$\sqrt {11 - \sqrt {112}} - \sqrt {8 -\sqrt {28}} = a$$
$$\sqrt {11 - \sqrt {112}} =\sqrt{\frac{11+\sqrt{121-112}}{2}}-\sqrt{\frac{11-\sqrt{121-112}}{2}}=\sqrt{\frac{11+3}{2}}-\sqrt{\frac{11-3}{2}}=$$
$$=\sqrt 7-2$$
$$\sqrt {8 -\sqrt {28}} =\sqrt{\frac{8+6}{2}}-\sqrt{\frac{8-6}{2}}=\sqrt 7-1$$
so finally
$$\sqrt 7-2-(\sqrt 7-1)=-1$$
There is a similar formula for the sum
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a linear transformation matrix for a set of vectors I am wondering if we can find a linear transformation matrix $A$ of size $3\times 3$ over the field of two elements $\mathbb{Z}_2$ i.e. a matrix $A$ of zeros and ones s.t.
$$A \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},
A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},
A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$
As can be seen above, the vectors I am looking at are all the possible 3D vectors with zeros and ones. $A$ is supposed to be an invertible transformation such that $A\neq A^{-1}$ because the effect of $A$ is a bijection from the space of all the possible 3D vectors with zeros and ones to that space itself.
I found an invertible $A$ using three of the equations above but that $A$ does not work for other vectors and so am wondering if it is possible to find such an $A$?
| There is no such $A$, invertible or not. For instance your equations lead to
$$
\begin{bmatrix} 1 \\ 0\\ 0\end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix} =A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} +
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} =
A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$ $\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$
How can I calculate this? Since there are $4^n$ and $\cos^2x$, I tried:
$$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}} = 4\sum_{n=1}^{\infty}{\frac{\sin^2{\frac{\pi}{4 \cdot 2^n}}}{4^{n}\sin^2{\frac{\pi}{4\cdot2^{n-1}}}}}$$
to use $2\sin x \cos x = \sin2x$
| Notice
$$\begin{align}\frac{1}{\cos^2\frac{\theta}{2}}
&= \frac{2}{1+\cos\theta}
= 2\frac{1 - \cos\theta}{1-\cos^2\theta}
= \frac{4 - 2(1+\cos\theta)}{1-\cos^2\theta}\\
&= \frac{4}{\sin^2\theta} - \frac{2}{1-\cos\theta}
= \frac{4}{\sin^2\theta} - \frac{1}{\sin^2\frac{\theta}{2}}
\end{align}
$$
We have
$$\begin{align}
\sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
&= \sum_{n=1}^p \left[
\frac{1}{4^{n-1}\sin^2\frac{\pi}{2^{n+1}}} -
\frac{1}{4^n\sin^2\frac{\pi}{2^{n+2}}}
\right]\\
&=\frac{1}{4^{1-1}\sin^2\frac{\pi}{2^{1+1}}} - \frac{1}{4^p\sin^2\frac{\pi}{2^{p+2}}}\\
&= \frac{1}{\sin^2\frac{\pi}{4}} - \frac{\frac{16}{\pi^2}}{\left(\frac{2^{p+2}}{\pi}\sin\frac{\pi}{2^{p+2}}\right)^2}
\end{align}
$$
Since $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$, the denominator in last term tends to $1$ as $p \to \infty$, As a result,
$$\sum_{n=1}^\infty \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
= \lim_{p\to\infty}
\sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
= 2 - \frac{16}{\pi^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.
Can you help me?
I have not seen determinats yet
| Maybe your teacher want you to solve the equations one by one for
each column of $B$. For example let the first column of $Q$ be $q_{1}=\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}$ Then you have the equations $Aq_{1}=b_{1}$, that is
\begin{align*}
x_{1}+\frac{1}{3}x_{3} & =\frac{4}{3}\\
-\frac{x_{3}}{3} & =-\frac{1}{3}\\
x_{1}+x_{2}+\frac{2}{3}x_{3} & =\frac{5}{3}
\end{align*}
From the second line $x_{3}=1$, from the first line $x_{1}=1$ and
from the third line $x_{2}=0$. Hence the first column of $Q$ is
$$
q_{1}=\begin{pmatrix}1\\
0\\
1
\end{pmatrix}.
$$
You can find the other columns simply, in a similar way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$ Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then
(A) $f$ is not invertible on (0,1)
(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$
(C) $f= f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$
(D) $f^{-1}$ is differentiable on (0,1)
My approach is as follow
$f:\left( {0,1} \right) \to R$, $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$,$0 < b < 1$
$f'\left( x \right) = \frac{{\left( {1 - bx} \right)\frac{d}{{dx}}\left( {b - x} \right) - \left( {b - x} \right)\frac{d}{{dx}}\left( {1 - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}}$
$f'\left( x \right) = \frac{{\left( {1 - bx} \right) \times - 1 - \left( {b - x} \right) \times \left( { - b} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( {1 - bx} \right) \times - 1 + \left( {b - x} \right) \times b}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( { - 1 + bx} \right) + \left( {{b^2} - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{{b^2} - 1}}{{{{\left( {1 - bx} \right)}^2}}}$
$x\in(0,1)$ and $b\in(0,1)$, then $bx\in(0,1)$, hence $f'(x)<0$, hence my assumption is that $f$ is invertible as it is decreasing function when $0<x<1$ but official answer is (A) that is $f$ is not invertible on (0,1)
| Just observe that there is no $x$ such that $f(x) =\frac{1}{b}$. If any such $x$ exists then it would imply that $b^2=1$ which is certainly not possible. This shows that $f$ is not surjective which implies that $f$ is not invertible.
Edit:
One more way to see that $f$ is not surjective is to observe that $-1 < f(x) < b$ by simply manipulating the inequalites $0<x<1$ and $0<b<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$ I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4} }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4} } |<ϵ.$$
I have this:
$\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4}} \cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} $
= $\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac{\sqrt[3]{2}}{6} $
=$ \dfrac{6\cdot \sqrt[3]{2+h}-6\cdot \sqrt [3]{2}-\sqrt[3]{2}\cdot h}{6h} $
=$ \left |\dfrac{1}{(\sqrt[3]{2+h})^2 + \sqrt[3]{2+h}\sqrt[3]{2}+(\sqrt[3]{2})^2 } - \dfrac {\sqrt[3]{2}} {6}\right | $
=$ $
I don't know how continue for use the fact $|h|<\delta$
The original exercise says: $f(x)=\sqrt[3]{x}$. Found $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$
| The limit $3\sqrt[3]4$ is given by the problem and it is easy to calculate. What the problem asks is to use the definition of limit to show that this limit is indeed $3\sqrt[3]4$. This presents some difficulty.
Let $\epsilon\gt0$ be given. We need to find out $\delta \gt0$ such that $|x|\le \delta\Rightarrow |3\sqrt[3]4-f(x)|\lt\epsilon$.
We’ll prove that $\boxed{\delta =f(-\epsilon)-f(\epsilon)}$ fits.
Let us first notice that $f$ is strictly decreasing in the neighborhood of $0$ (derivative negative).
Proposition.- Let us $f(x)=\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}$ then for $x\gt0$ and near zero we have $0\lt f(-x)-f(x)\lt x$.
Proof.- (This is graphically verified with any good calculator, for example Desmos). We have
$$0\lt f(-x)-f(x)=\dfrac{\sqrt[3]{2-x}-\sqrt[3]2}{-x}-\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}=\frac{2\sqrt[3]2-\sqrt[3]{2-x}-\sqrt[3]{2+x}}{x} \\0\lt f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(2-\sqrt[3]{1-\frac x2}-\sqrt[3]{1+\frac x2}\right)$$
Using Taylor series $$\sqrt[3]{1-t}=1-\frac t3-\frac{t^2}{9}-\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)\\\sqrt[3]{1+t}=1+\frac t3-\frac{t^2}{9}+\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)$$ It follows
$$ f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(\frac29(\frac x2)^2+\frac{10}{243}(\frac x2)^4+O((\frac x2)^6)\right)=\frac{\sqrt[3]2}{x}\left(\frac{x^2}{18}+\frac{5x^4}{1944}+O\left(\frac{x^6}{64}\right)\right)$$ $$f(-x)-f(x)= x\left(\sqrt[3]2\left(\frac{1}{18}+\frac{5x^2}{1944}+\frac{1}{x^2}O\left(\frac{x^6}{64}\right)\right)\right)=x\left(\frac{1}{9\sqrt[3]4}+\frac{5x^2}{972\sqrt[3]4}+\frac{\sqrt[3]2}{x^2}O\left(\frac{x^6}{64}\right)\right)\lt x$$
because near zero the factor of $x$ is clearly less than $1$.
Now the rest of the proof is immediate. By construction, being $y_0=3\sqrt[3]4$, if $|x|\lt\delta=f(-\epsilon) -f(\epsilon)$ we have $$|y_0-f(x)|\lt f(-\epsilon) -f(\epsilon)\lt\epsilon$$ See the attached figure for clarity.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from there to apply the Euler's theorem
| well if $2^{6k}4\equiv m \pmod {18}$ then $2^{2^{6k}4}\equiv 2^m \pmod {19}$.
$2^{6k}4 \equiv 0 \pmod 2$ and $\phi(9) = 6$ so $2^6 \equiv 1 \pmod 9$ so $2^{6k}4\equiv 4 \pmod 9$. So by Chinese remainder theorem there is one solution $\pmod {18}$ to $2^{6k}4\equiv 0 \pmod 2$ and $6^{6k}4\equiv 4 \pmod 9$.
And that unique solution is $4$
And that's that.
$2^{6k}4 \equiv 4 \pmod{18}$. So there exists an $M$ so that $2^{6k}4=18M + 4$ so $2^{2^{6k}4}=2^{18M + 4} \equiv 2^4 \pmod{19}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule? I can evaluate the limit with L'Hospital's rule:
$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$
But is there any way to do it without using L'Hospital's rule?
| Here's a trick to prove convergence of the continuous limit$$
\lim\limits_{x\rightarrow\infty} x \left(\sqrt[x]4 - 1\right)
$$
if you know also know how to integrate $2^x$.
Observe:\begin{eqnarray}
x \left(\sqrt[x]4 - 1\right) = x \left(\sqrt[x]2 - 1\right)\left(\sqrt[x]2 + 1\right) = 2x\left(\sqrt[2x]4 - 1\right)\frac{\left(\sqrt[x]2 + 1\right)}{2}
\end{eqnarray}
Thus we get $$
x\left(\sqrt[x]4 - 1\right) = \frac{\sqrt[x]2 + 1}2 2x\left(\sqrt[2x]4-1\right) \frac{\sqrt[x]2 + 1}2 \frac{\sqrt[2x]2 + 1}{2} 4x\left(\sqrt[4x]4-1\right) = \left(\prod_{k=0}^{n-1} \frac{\sqrt[2^kx]2+1}{2}\right) 2^n x \left(\sqrt[2^n x]4 -1\right)
$$
Note that the multiplicands are always bigger than $1$, so this implies that $2^n x \left(\sqrt[2^nx]4 - 1\right)$ is decreasing in $n$, so the limit exists for each $x$. It's also pretty clear that the function $x\to x\left(\sqrt[x]4-1\right)$ doesn't have any oscillations in the limit, hence we have that the continuous limit converges, so we just have to compute it for one value of $x$, say $x=1$. Thus we have $$
\lim\limits_{z\rightarrow\infty} z \left(\sqrt[z]4-1\right) = \lim\limits_{n\rightarrow\infty} 2^n \left(\sqrt[2^n]4-1\right) = \lim\limits_{n\rightarrow\infty} \frac{\sqrt[1]4-1}{\prod\limits_{k=0}^{n} \frac{\sqrt[2^k]2+1}{2}} = \frac2 {\prod\limits_{k=1}^\infty \frac{2^{1/2^k} + 1}{2}}
$$
Expanding the partial products, we can see that this product actually isn't hard to evaluate:\begin{eqnarray}
\prod_{k=1}^1 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac12 + 1}2\\
\prod_{k=1}^2 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac14 + 2^\frac12 + 2^\frac34 + 1}4\\
&\vdots&\\
\prod_{k=1}^N \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac1{2^N} + 2^\frac{2}{2^N} + \cdots + 2^\frac{2^N-1}{2^N} + 2^\frac{2^N}{2^N}}{2^N}
\end{eqnarray}
The RHS is a left Riemann sum for $\int_0^1 2^t dt$ splitting the interval into $2^N$ subintervals. Thus we get $$
\prod_{k=1}^\infty \frac{2^{\frac1{2^k}} + 1}{2} = \int_0^1 2^t dt = \frac{2-1}{\ln 2} = \frac1{\ln 2}
$$
Hence we conclude $$
\lim\limits_{z\rightarrow\infty} z\left(\sqrt[z]4-1\right) = \frac{2}{1/\ln 2} = 2\ln 2 = \ln 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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All possible angles of triangle such that the feet of the angle bisectors determine a right angle
Triangle $ABC$ has angle bisectors $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Given that $\angle EDF = 90^\circ$, enter all possible values of $\angle BAC$ (in degrees), separated by commas.
I plan to use vectors here; that is what the hint says. We can see that $\angle{EDF} = 90^\circ$, so $\overrightarrow{DE}\cdot \overrightarrow{DF} = 0$.
However, here I'm stuck. We are dealing with incenters here (intersection of angle bisectors), but I'm not sure how to incorporate angle bisectors or incenters here. Perhaps I could use the fact that since $\angle{BAD} = \angle{CAD}$, then $$\frac{\overrightarrow{AB}\cdot \overrightarrow{AD}}{|AB||AD|} = \frac{\overrightarrow{AC}\cdot \overrightarrow{AD}}{|AC||AD|}$$
However, I don't really see that being useful. Help is greatly appreciated, thanks!
| I entered $120^\circ$ and it was correct, but without much reason to it. Here is their solution.
As usual, let $a = BC,$ $b = AC,$ and $c = AB.$
By the Angle Bisector Theorem, $BD:DC = c:b,$ so
$\overrightarrow{D} = \frac{b}{b + c} \overrightarrow{B} + \frac{c}{b + c} \overrightarrow{C} = \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c}.$ Similarly,
\begin{align*}
\overrightarrow{E} &= \frac{a \overrightarrow{A} + c \overrightarrow{C}}{a + c}, \\
\overrightarrow{F} &= \frac{a \overrightarrow{A} + b \overrightarrow{B}}{a + b}.
\end{align*} If we let $A$ be the origin, then we get
$\overrightarrow{E} = \frac{c \overrightarrow{C}}{a + c}, \quad \overrightarrow{F} = \frac{b \overrightarrow{B}}{a + b}.$ Therefore,
\begin{align*}
\overrightarrow{DE} &= \overrightarrow{E} - \overrightarrow{D} \\
&= \frac{c \overrightarrow{C}}{a + c} - \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c} \\
&= \frac{- b(a + c) \overrightarrow{B} + c(b - a) \overrightarrow{C}}{(a + c)(b + c)},
\end{align*}and
\begin{align*}
\overrightarrow{DF} &= \overrightarrow{F} - \overrightarrow{D} \\
&= \frac{b \overrightarrow{B}}{a + b} - \frac{b \overrightarrow{B} + c \overrightarrow{C}}{b + c} \\
&= \frac{b(c - a) \overrightarrow{B} - c(a + b) \overrightarrow{C}}{(a + b)(b + c)}.
\end{align*} Since $A$ is the origin, $|\overrightarrow{B}| = c$, $|\overrightarrow{C}| = b$, and by the Law of Cosines,
$\overrightarrow{B} \cdot \overrightarrow{C} = |\overrightarrow{B}| |\overrightarrow{C}| \cos A = bc \cdot \frac{b^2 + c^2 - a^2}{2bc} = \frac{b^2 + c^2 - a^2}{2}.$ We have that $\angle EDF = 90^\circ$ if and only if $\overrightarrow{DE} \cdot \overrightarrow{DF} = 0$, or equivalently,
\begin{align*}
&[-b(a + c) \overrightarrow{B} + c(b - a) \overrightarrow{C}] \cdot [b(c - a) \overrightarrow{B} - c(a + b) \overrightarrow{C}] \\
&= -b^2 (a + c)(c - a) |\overrightarrow{B}|^2 + bc(a + c)(a + b) \overrightarrow{B} \cdot \overrightarrow{C} \\
&\quad + bc(b - a)(c - a) \overrightarrow{B} \cdot \overrightarrow{C} - c^2 (b - a)(a + b) |\overrightarrow{C}|^2 \\
&= -b^2 c^2 (c^2 - a^2) + 2bc(a^2 + bc) \cdot \frac{b^2 + c^2 - a^2}{2} - b^2 c^2 (b^2 - a^2) \\
&= a^2 bc(b^2 + bc + c^2 - a^2) \\
&= 0,
\end{align*}so $a^2 = b^2 + bc + c^2$. Then by the Law of Cosines,
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{-bc}{2bc} = -\frac{1}{2}.$ Therefore, $$A = \boxed{120^\circ}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Can Trigonometric Substitution be used to derive the anti-derivative $\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$? My textbook (Calculus by James Stewart 8th edition) uses partial fraction decomposition to derive the anti-derivative
$$\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$$
But I'm wondering if it's possible to use trigonometric substitution as well.
I tried the following trig. substitution by letting
$$x = a\sec(\theta)$$
$$dx = a\sec(\theta)\tan(\theta)d\theta$$
$$\int \frac{1}{x^2 -a^2}dx = \int \frac{1}{a^2\sec^2(\theta)-a^2} a\sec(\theta)\tan(\theta)d\theta$$
$$= \int \frac{1}{a^2\tan^2(\theta)} a\sec(\theta)\tan(\theta)d\theta$$
$$= \frac{1}{a}\int \frac{1}{\tan(\theta)} \sec(\theta)d\theta$$
$$= \frac{1}{a}\int \frac{1}{\frac{\sin(\theta)}{\cos(\theta)}} \frac{1}{\cos(\theta)}d\theta$$
$$= \frac{1}{a}\int \frac{1}{\sin(\theta)}d\theta $$
$$= \frac{1}{a}\int \csc(\theta) $$
$$= \frac{1}{a} \ln |\csc(\theta)-\cot(\theta)|+ C$$
Then using the relation established by the trig. substitution cosecant and cotangent can be written in terms of $x$.
$$\csc(\theta) = \frac{x}{\sqrt{x^2-a^2}}$$
$$\cot(\theta) = \frac{a}{\sqrt{x^2-a^2}}$$
Now plugging these in results in
$$= \frac{1}{a} \ln|\frac{x-a}{\sqrt{x^2-a^2}}| + C$$
I'm not sure how to proceed from here to get the same results as the book, or if I did something wrong along the way. Any help would be appreciated.
| Use $\frac{x-a}{x+a}=\frac{1-\cos\theta}{1+\cos\theta}=\frac{(1-\cos\theta)^2}{\sin^2\theta}=(\csc\theta-\cot\theta)^2$ or $\left|\frac{x-a}{\sqrt{x^2-a^2}}\right|=\sqrt{\frac{x-a}{x+a}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$
Attempt. Using ratio test for $a_n:=\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}>0$, we have:
$$\frac{a_{n+1}}{a_n}=\frac{\frac{2}{3}+n+1}{\frac{3}{2}+n+1}\to 1,$$
so we do not conclude. Also $\sqrt[n]{(1+a)\ldots(n+a)}\to +\infty$ for $a>0$, so the root test also is inconclusive.
Thanks for the help.
| In terms of the Gamma and Beta functions we are discussing
$$ \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{5}{3}\right)/\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(n+\frac{5}{2}\right)/\Gamma\left(\frac{5}{2}\right)}=\underbrace{\frac{\Gamma(5/2)}{\Gamma(2/3)\Gamma(5/6)}}_{K}\sum_{n\geq 1}B(n+5/3,5/6) $$
or
$$ K \sum_{n\geq 1}\int_{0}^{1} x^{n+2/3}(1-x)^{-1/6}\,dx=K\int_{0}^{1}x^{5/3}(1-x)^{-7/6}\,dx $$
which is clearly divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$? I did
\begin{align}
& \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t^3k+4tk}{(t^2+2)^2} \\[8pt]
= {} & \frac{-4tj+4tk}{(t^2+2)^2}
\end{align}
Is this true?
| \begin{align}
i\times j & = k = -j\times i \\
j\times k & = i = -k\times j \\
k\times i & = j = - i\times k \\
i \times i & = j \times j = k\times k = 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$ I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.
$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$
$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$
| The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq i <j<k}}\frac{1}{5^i3^j2^k}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq i <j<k
\end{align*}
which has $1$ as lower limit and $k-1$ as upper limit. We have two indices $i$ and $j$, which means we can write it as double sum as shown in the evaluation below.
We obtain
\begin{align*}
\color{blue}{\sum_{1\leq i<j<k}\frac{1}{5^{i}3^j2^k}}
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\sum_{j=i+1}^{k-1}\frac{1}{3^j}\tag{2}\\
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\left(\frac{\left(\frac{1}{3}\right)^{i+1}-\left(\frac{1}{3}\right)^k}{1-\frac{1}{3}}\right)\tag{3}\\
&=\frac{1}{2^k}\sum_{i = 1}^{k-2}\frac{1}{5^i}\,\frac{1}{2}\left(\frac{1}{3^i}-\frac{1}{3^{k-1}}\right)\tag{4}\\
&= \frac{1}{2^{k+1}}\sum_{i=1}^{k-2}\frac{1}{15^i}- \frac{1}{2^{k+1}\,3^{k-1}}\sum_{i=1}^{k-2}\frac{1}{5^i}\\
&=\frac{1}{2^{k+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{k-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{k+1}\,3^{k-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{k-1}}{1-\frac{1}{5}}\right)\tag{5}\\
&=\frac{1}{2^{k+2}\cdot7}\left(1-\frac{1}{15^{k-2}}\right)-\frac{1}{2^{k+3 }\,3^{k-1}}\left(1-\frac{1}{5^{k-2}}\right)\\
&\,\,\color{blue}{=\frac{1}{2^{k+2}\cdot7}-\frac{1}{2^{k+3}\,3^{k-1}}+\frac{1}{2^{k+3}\,3^{k-1}\,5^{k-2}\cdot7}}\tag{6}
\end{align*}
Comment:
*
*In (2) we factor out $\frac{1}{2^k}$ and reorder the double sum using another common style.
*In (3) we evaluate the inner sum using the finite geometric summation formula.
*In (4) we do a simplification and multiply out in the next line.
*In (5) we apply the finite geometric summation formula twice and do a simplification in the following lines.
Notes:
*
*Varying $k$ is not admissible since $k-1$ is an upper limit. Here $k$ is a free variable whereas the indices $i$ and $j$ are bound variables. This is different to the situation
\begin{align*}
\sum_{1\leq i <j<k\color{blue}{<\infty}}\frac{1}{5^{i}3^j2^k}
\end{align*}
where $k$ is an index bound by the upper limit $\infty$ and where $k$ varies between $j$ and $\infty$.
*The result (6) is also given by Wolfram Alpha with input
sum(sum 1/(5^i*3^j*2^k), j=i+1..k-1),i=1..k-2
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
| {
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"url": "https://math.stackexchange.com/questions/3960676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very good-looking and nice to work with.
What are the steps to follow? We have $D_x:\begin{cases}x^2-1\ge0\\\dfrac{x+1}{x-1}\ge0\\x-1\ne0\end{cases} \iff x\in(-\infty;-1]\cup(1;+\infty).$ What next?
| $$\sqrt{x^{2}-1}=\left(x+5\right)\frac{\sqrt{x+1}}{\sqrt{x-1}}$$
Both sides are only defined for $|x|>1$ or $x=-1$
Squaring both sides,
$$x^{2}-1=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
We can factor the LHS using difference of squares,
$$\left(x+1\right)\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
Assuming $|x|>1$, we can divide both sides by $x+1$ which is nonzero (We can check $x=-1$ as a sperate case).
$$\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{1}{x-1}\right)$$
Multiplying both sides by $x-1$,
$$\left(x-1\right)^{2}=\left(x+5\right)^{2}$$
$$\Rightarrow \pm\left(x-1\right)=\pm\left(x+5\right)$$
There won't be any solutions for $(x-1)$=$(x+5)$ or $-(x-1)$=$-(x+5)$, so $(x-1)$ and $(x+5)$ must be additive inverses.
$$(x-1)=-(x+5)$$
$$\Rightarrow x=-2$$
Both sides of the original equation are defined for $x=-2$, so it's a solution.
Earlier we only considered values of x that satisfied $|x|>1$. If $x=-1$, both sides of the equation are $0$, so $x=-1$ is also a solution.
| {
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"url": "https://math.stackexchange.com/questions/3962503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
| Note that $x>1$. To avoid squaring equations, factorize directly as follows
\begin{align}
0 &=x+\frac{x}{\sqrt{x^2-1}}-\frac{35}{12}\\
&= \left(x-\frac{35}{12} + \frac{25}{12x}\right)
+ \left( \frac{x}{\sqrt{x^2-1}} -\frac{25}{12x}\right)\\
&=\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
+ \frac{x^2-\frac{25}{12}\sqrt{x^2-1}}{x\sqrt{x^2-1}}\\
& =\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
+ \frac{x^4-\frac{25^2}{12^2}({x^2-1})}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
+ \frac{(x^2-\frac{25}{16})(x^2-\frac{25}{9} )}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
\left( 1+ \frac{(x+\frac{5}{4})(x+\frac{5}{3} )}{\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\right)\\
\end{align}
which leads to the solutions $x= \frac54,\>\frac53$. (The last factor is always positive.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.
Where have I made a mistake. Thank you
| We can use
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$
Since both
$$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}$$
and
$$\int_{-1}^1 2\sqrt{1-x^2}dx$$
converge to $\pi$, you can either use
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty {2n \choose n} \left(\frac{x}{2}\right)^{2n}$$
or
$$2\sqrt{1-x^2} = \frac{2(1-x^2)}{\sqrt{1-x^2}} = 2+2\sum_{n=1}^\infty \left[{2n \choose n} - {2n-2 \choose n-1}\right]\left(\frac{x}{2}\right)^{2n}$$
$$= 2+2\sum_{n=1}^\infty {2n-2 \choose n-1}\left[3-\frac{2}{n}\right]\left(\frac{x}{2}\right)^{2n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$
This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$
I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.
| \begin{aligned}
& \int x^{3} \sqrt{x^{2}-9} d x \\
=& \int \frac{x^{3}\left(x^{2}-9\right)}{\sqrt{x^{2}-9}} d x \\
=& \int x^{2}\left(x^{2}-9\right) d\left(\sqrt{x^{2}-9}\right) \\
=& \int\left(y^{2}+9\right) y^{2} d y,\text { where } y=\sqrt{x^{2}-9} \\
=& \frac{y^{5}}{5}+3 y^{3}+c \\
=& \frac{y^{3}}{5} \cdot\left(y^{2}+15\right)+c \\
=& \frac{\left(x^{2}-9\right)^{\frac{3}{2}}}{5}\left(x^{2}+6\right)+c
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$ Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.
I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$
I think that a geometric solution might be possible, please help me to proceed.
Thanks.
| Remark: @VIVID gave a nice solution. I rewrote it in elementary way (without using derivative, extrema). Note that while we used calculus to motivate our solution, we do not need to include any
calculus in the solution!
Letting $x = 0$, we have $|1 - q| \le \frac{\sqrt{2}-1}{2}$ or
$$1 - \frac{\sqrt{2}-1}{2} \le q \le 1 + \frac{\sqrt{2}-1}{2}. \tag{1}$$
Letting $x = 1$, we have $|-p - q| \le \frac{\sqrt{2}-1}{2}$ or
$$-p - \frac{\sqrt{2}-1}{2} \le q \le -p + \frac{\sqrt{2}-1}{2}. \tag{2}$$
From (1) and (2), we know that $p < 0$ (reason: if $p\ge 0$ then $-p + \frac{\sqrt{2}-1}{2} < 1 - \frac{\sqrt{2}-1}{2}$).
Since $-\frac{p}{\sqrt{1+p^2}} \in [0, 1]$, by letting $x = -\frac{p}{\sqrt{1+p^2}}$,
we have $|\sqrt{p^2+1} - q| \le \frac{\sqrt{2}-1}{2}$ or
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le q \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}. \tag{3}$$
From (2) and (3), since $\sqrt{p^2 + 1} > -p$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le -p + \frac{\sqrt{2}-1}{2}$$
that is
$\sqrt{p^2 + 1} + p - \sqrt{2} + 1 \le 0$ or
$\frac{(p^2+1) - 2}{\sqrt{p^2+1} + \sqrt{2}} + (p+1) \le 0$ or
$\frac{(p - 1 + \sqrt{p^2+1} + \sqrt{2})(p+1)}{\sqrt{p^2+1} + \sqrt{2}} \le 0$ which results in
$p \le -1$.
From (1) and (3), since $1 + \frac{\sqrt{2}-1}{2} \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le 1 + \frac{\sqrt{2}-1}{2}$$
that is $p^2 \le 1$. Thus, we have $p = -1$.
From (1) and (3), we have $q = 1 + \frac{\sqrt{2}-1}{2}$.
Finally, when $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$, it is easy to prove that $|\sqrt{1-x^2} + x - 1 - \frac{\sqrt{2}-1}{2}| \le \frac{\sqrt{2}-1}{2}, \forall x \in [0, 1]$
that is $1\le \sqrt{1-x^2} + x \le \sqrt{2}, \forall x\in [0, 1]$.
Thus, $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$ is the only one pair satisfies the statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Deriving Generating function for centered trinomial coefficients Let $c_n$ denote the $n$-th center trinomial coefficient (OEIS sequence here).
It appears they cannot be generated by a linear recurrence relation, so how should I go about to find the generating function $G(x)$ for the sequence?
$$G(x)=\sum_{n=0}^{∞}c_nx^n=1+x+3x^2+7x^3+19x^4+51x^5+141x^6+...$$
The geometric ratio appears to have a limit close to $$\lim_{n\to ∞}\frac{c_{n+1}}{c_n}=2.95...$$ (these are the successive ratios of the last two listed terms in the OEIS sequence).
Also, what is the interval of convergence (and divergence)? Based on the geometric limit, it seems that $G(1/3)$ will converge.
Edit: The generating function is $$G(x)=\frac{1}{\sqrt{1-2x-3x^2}}$$
Any idea as to how this answer is derived?
| Here is a variation based upon G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. We start with the central trinomial coefficients:
\begin{align*}
[x^n](1+x+x^2)^n\qquad\qquad n\geq 0
\end{align*}
We consider the function
\begin{align*}
f(x)=1+x+x^2\tag{1}
\end{align*}
and derive a function $y=y(x)$:
\begin{align*}
y(x)=\frac{x}{f(x)}=\frac{x}{1+x+x^2}\qquad\qquad y^{\prime}(x)=\frac{1-x^2}{(1+x+x^2)^2 }\tag{2}
\end{align*}
With $f(x)$ and $y(x)=\frac{x}{f(x)}$ we can now apply the substitution rule (Rule 5, one-dimensional case) from section 1.2.2 in G. P. Egorychev's book as follows:
\begin{align*}
\color{blue}{[x^n](f(x))^n=[y^n]\left.\left(\frac{1}{f(x)y^{\prime}(x)}\right)\right|_{x=g(y)}}\tag{3}
\end{align*}
with $g(y)$ the inverted function given by $y=y(x)$ in (2).
We obtain from (1) - (3):
\begin{align*}
\color{blue}{[x^n]}&\color{blue}{\left(1+x+x^2\right)^n}\\
&=[y^n]\left.\left(\frac{1}{\left(1+x+x^2\right)\frac{d}{dx}\left(\frac{x}{1+x+x^2}\right)}\right)\right|_{x=g(y)}\\
&=[y^n]\left.\frac{1+x+x^2}{1-x^2}\right|_{x=g(y)}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-2y-3y^2}}}\tag{4}
\end{align*}
and the claim follows.
In (4) we use the identity
\begin{align*}
2y=\frac{2x}{1+x+x^2}&=1-3\left(\frac{x}{1+x+x^2}\right)^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
&=1-3y^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
\frac{1+x+x^2}{1-x^2}&=\left(1-2y-3y^2\right)^{-\frac{1}{2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book. I'm studying on how to find the vertex of a Parabola and got stuck with this question.
Determine the vertex of $y = x^2 - 6x + 1$
I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it does not have a section where it shows how it got to that answer so I'm doubting if my understanding is incorrect. If I use online algebra calculators, it matches with mine but I just want to make sure that I understand how signs and the formula work.
Formula I used to look for the Vertex:
$$V=\left(-\frac{b}{2a} , \frac{4ac-b^2}{4a}\right)$$
My solution:
$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{-32}{4} = \frac{-8}{1}=-8
\end{align}$$
$$V=(3,-8)$$
What I think was done on the book:
$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{40}{4} = \frac{10}{1}=10
\end{align}$$
$$V=(3,10)$$
Is it correct that I should've added $4-36$ since they are both positive numbers or subtract it?
Any explanation is appreciated. Thanks!
| $4-36$ is indeed $-32$. Whenever you have $a - b$ and $b > a$, you should interpret it as $a - b = -1 \cdot (b - a)$. So your calculation seems to be right. And yes, the vertex of the parabola is at $(3, -8)$, this can be verified by differentiating and setting $y' = 0$.
$y' = 2x - 6 = 0 \implies x = 3$ and then $y(x=3) = 3^2 - 6 \cdot 3 + 1 = 9 - 18 + 1 = -8$. Your method is correct, I just mentioned the differentiation method to confirm the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$
Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?
| $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are both positive in the given domain, so their quotient is also positive, and $|x| = x$ when $x \in \mathbb R^+$. $\sin \alpha + \cos \alpha > 0$ as $\sin \alpha, \cos \alpha$ are positive in the domain. Hence we can focus our attention to just proving $\sin \alpha > \cos \alpha$.
Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign.
Thus you have $\tan \alpha > 1$, which is true because $\tan 45º = 1$ and $\tan x$ is a strictly increasing function in the given range $(45º, 90º)$. This is already sufficient as a justification, but on some insight as to why this is, $\tan x = \frac{\sin x}{\cos x}$, and $\sin x$ is increasing while $\cos x$ is decreasing in the given interval, which both increase the value of the function. Anything more rigorous has to involve a geometric argument with the unit circle or calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 & 4 \\ \hline
1 & 6 & 6 & 0 \\ \hline
2 & 26 & 30 & -4 \\ \hline
3 & 94 & 102 & -8 \\ \hline
4 & 306 & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.
| So, denoting $w_n = f_n -g _n$, you can write down the equation
\begin{align}
w_0 = f_0-g_0 = 4, \qquad &w_{n+1} = f_{n+1}-g_{n+1} = 3(f_n-g_n)+8n -12, \quad\text{i.e.}\\
w_0 = 4, \qquad &w_{n+1} = 3 w_n +8n -12.
\end{align}
This a liner difference equation for which you can have an explicit solution, in the same way that you could have computed explicitly $f_n$ and $g_n$ right from the beginning.
The general solution of the homogeneous equation $w_{n+1} = 3 w_n$ is $w_n^h = c 3^n$, and if you search for a particular solution of the form $w_n^* = a n + b$, you'll get $w_n^* = 4 - 4n$. So the general solution is
$$
w_n = c 3^n +4 -4n.
$$
Finally, using the initial condition $w_0=4$, we get $c=0$ and the solution is $w_n = 4 - 4n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Evaluating a triple summation ${\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k$
If $$a_i= \begin{cases} 1 && 1 \le i \le 10 \\ 2 && 11 \le i \le 20 \end{cases} $$
Find the value of $$\displaystyle {\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k $$
I could only evaluate the summation when $1 \le i < j \le k \le 10$, by writing it as $\displaystyle {\sum _{i=1}^9\sum_{j=i+1}^{10} \sum_{k=j}^{10}} 1$ which came out to be $\displaystyle \binom{11}{3}$.
Should I make different cases like $i \le 10, 11 \le j \le k \le 20$ , etc.?
How do I evaluate it for the other cases? Is there any easier way to evaluate it without evaluating each sum one by one? Could there possibly be a combinatorial approach? I'd be delighted to see different approaches to this problem.
| Let
*
*$ A = \sum_{i < j < k } a_ia_ja_k$
*$ B = \sum_{i = j < k } a_ia_ja_k$
*$ C = \sum_{i < j = k } a_ia_ja_k$
*$ D = \sum_{i = j = k } a_ia_ja_k$
We're after $ A + C$.
Clearly,
*
*$ 30^3 = ( \sum a_i)^3 = D + 3C + 3B + 6A$.
*$ D = 10 ( 1^3 + 2^3 ) = 90$
*$C = 2^2 ( 10 \times 10\times 1 + \frac{10 \times 9 }{ 2} \times 2 ) + 1^2 (\frac{10 \times 9 } { 2} \times 1) = 805$
*$ B = 1^2 (10\times 10 \times 2 + \frac{10 \times 9} { 2} \times 1) + 2^2 (\frac{10 \times 9 }{ 2} \times 2) = 605$
*So, $A = \frac{30^3 - 90 - 3 \times 805 - 3 \times 605 }{ 6} = 3780$.
Hence, $ A + C = 4585$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$? The problem is as follows:
Find the value of:
$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$
Where $A$ is given by:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\right)}$
Also assume this follows the relationship seen in the figure from
below:
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\frac{1}{2}\\
2.&1\\
3.&\frac{1}{4}\\
4.&\frac{1}{3}\\
\end{array}$
My main issue with this problem is the half angle which is presented in the right side of this equation:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$
I was able to spot that the figure might intend to say that $A$ and $B$ are complementary thus:
$A+B=90^{\circ}$
Therefore this makes the previous expression into:
$\tan A = \frac{2+\tan \left(\frac{A}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$
This helps a bit but again the half angle would produce a square root. I tried to follow such path. But it didn't ended very well.
After squaring and using other identities I ended up having a 8th degree equation. And certainly this wasn't intended.
So what's the trick here?. So far I'm still stuck and all sorts of manipulations don't give me results. Can someone guide me in the right path?. I'm also confused on how to better use the figure. Those other letters don't seem to help much.
It would really help me alot a method explained step by step so I can understand.
| Let $\displaystyle t = \tan\frac{A}{2} \quad →
\tan\frac{B}{2} = \tan \left(45°\!-\frac{A}{2}\right) = \frac{1-t}{1+t}$
$\displaystyle \tan A = \frac{2+\cot\frac{B}{2}}{4+\cot\frac{A}{2}} \;\; →
\frac{2t}{1-t^2} = \frac{2+\frac{1+t}{1-t}}{4+\frac{1}{t}}\;\; → 1-t^2 = 6t
\;\; →\cot A = \frac{6t}{2t} = 3$
$\displaystyle R = \frac{\csc^2 A - \cot A}{\csc^2 A + 4}
= \frac{(3^2+1)-3}{(3^2+1)+4}
= \frac{1}{2}
$
For comparison, we solve for $\cot(A)$ with cotangent sum of angles formula.
I think it is easier to remember, negated reciprocal of tangent formula.
Trig identity: $\;\displaystyle \cot(x±y) = - \frac{1 ∓ \cot(x)\cot(y)}{\cot(x) ± \cot(y)}$
Let $\displaystyle c = \cot\frac{A}{2} \quad →
\cot\frac{B}{2} = \cot \left(45°\!-\frac{A}{2}\right) = - \frac{1+c}{1-c}$
$\displaystyle \tan A = \frac{2+\cot\frac{B}{2}}{4+\cot\frac{A}{2}} \;\; →
-\frac{2c}{1-c^2} = \frac{2-\frac{1+c}{1-c}}{4+c}\;\; → c^2-1 = 6c
\;\; →\cot A = \frac{6c}{2c} = 3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with the following values for the constants
$$
\left(x-\frac{1}{2}\right)^2
\left(x+\frac{5}{2}\right)^2+\frac{5}{2} \left(x+\frac{1}{5}\right)^2+\frac{107}{80}
$$
I suppose there are simpler ways to prove that the polynomial is positive, perhaps by using some inequalities.
Please, advice.
| Let the polynomial be denoted by $f(x)$.
$f'(x)=4(x^3+3x^2+2x-1)$ has only one real root $\approx 0.3247$. At that point $f''(x)=4(3x^2+6x+2)>0$. Thus $f(x)$ has minimum value at $\approx 0.3247$ and $f(0.3247)\approx 2.271$.
$\therefore f(x)>0 \;\forall x\in \mathbb R$.
| {
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"answer_count": 4,
"answer_id": 3
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theory of equation let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
( c+d)^{-1}+ (c+d)^{-1}$$
my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck
| If you combine your result with that of B. Goddard, you have
$$\frac{1}{a+b}+\frac{1}{c+d} = \frac{3}{(a+b)(c+d)}$$
Now suppose $a=1+w, b=1+w^2, c=1+w^3, d=1+w^4$ where $w$ is one of the imaginary roots of $w^5=1$, then
$$(a+b)(c+d) = (2+w+w^2)(2+w^3+w^4) \\= 4 + 2(w+w^2+w^3+w^4) + (w^4 + 1 + 1 + w)=4+w+w^4$$
Similarly,
$$(b+c)(a+d)=1$$
$$(a+c)(b+d)=4+w^2+w^3$$
That's probably easier since both $w+w^4$ and $w^2+w^3$ are real. You will need $\cos \left(\frac{2\pi}{5}\right)=\frac{\sqrt 5 -1}{4}$
Edit: Actually you don't need the cosine. Just add $1/(4+w+w^4)$ and $1/(4+w^2+w^3)$ like the above it becomes very clean.
$$(4+w^2+w^3)(4+w+w^4)\\=16+4(w+w^2+w^3+w^4)+(w^3+w+w^4+w^2)\\=16-4-1=11$$
Can you end it now?
| {
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kernel and image of a map help The question states:
a) Suppose that $U = V = M_{2,2} (\mathbb R)$ is the vector space of $2 × 2$ matrices over $\mathbb R $ Let $\phi$ be defined by $\phi(A) = A - A^T $ for each $A \in M_{2,2} \mathbb R$. Determine
$\ker \phi $ and state its dimension.
b) Determine $\operatorname{im}\phi$ and its dimension.
So my thoughts are we use $\ker \phi = \left \{ f \in U \mid \phi(f) = 0 \right \}$ and then we follow all the steps to show that $$\ker \phi = \left \langle \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\right \rangle
$$
and we can conclude that $\dim \ker \phi = 3 $ and so by Rank and Nullity Thm, we say $\dim\operatorname{im}\phi = 1$.
My question is that is this correct, this process to get that answer, and I'm just wondering what $\operatorname{im}\phi$ would be?
Thank you in advance.
Edit: How I got to the kernel:
$$f = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} $$
then $$f^T = \begin{pmatrix}
a & c\\
b & d
\end{pmatrix}$$
then $$f-f^T = \begin{pmatrix}
0 & c-b\\
c-b & 0
\end{pmatrix}$$
So putting into a system of eqs.
$$\left \{ b-c = 0 \right. \\
\left \{ c-b = 0 \right.$$
Which implies that b=c
So $$f = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} = a\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix} +b \begin{pmatrix}
0 & 1\\
0& 0
\end{pmatrix} + d \begin{pmatrix}
0 & 0\\
0& 1
\end{pmatrix} $$
Which allows me to come to the conclusion that :
$$ker \phi = \left \langle \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\right \rangle
$$
Thank you for telling me to edit
I just realised that for kernel 2nd part, it should be $$\begin{pmatrix}
0 &1 \\
1 & 0
\end{pmatrix}$$
| The first part is correct. It is easy to see that the kernel is the set of symmetric matrices, which for dimension $n$ has dimension $n(n+1)/2$.
Since the image is one dimensional, it suffices to find one nonzero element of it, and that will form a basis. For instance, we have that $\begin{pmatrix}1&1\\0&1\end{pmatrix}\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, and the latter matrix spans the image. Notice it is antisymmetric.
| {
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} |
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
| $$\left(\sqrt{x^2+12y}+\sqrt{y^2+12x}\right)^2=(x+y+10)^2$$
$$\implies x^2+12y+y^2+12x+2\left(\sqrt{(x^2+12y)(y^2+12x)}\right)=x^2 + y^2 + 100+ 2xy+20x+20 y $$
$$\implies \sqrt{(x^2+12y)(y^2+12x)}=xy+4(x+y)+50 $$
$$\implies(x^2+12y)(y^2+12x)=(xy+142)^2 $$
$$\implies x^3 - \dfrac{35 x y}{3}+ y^3 - \dfrac{5041}{3}=0$$
$$ \implies (x+y)^3-3xy(x+y)-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
$$\implies 23^3-69xy-\dfrac{35 x y}{3}-\dfrac{5041}{3}=0$$
Finally, we have
$$\begin{cases} x+y=23 \\ xy=130 \end {cases}$$
$$\implies x=13, y=10 $$ and
$$\implies x=10, y=13.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 6,
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} |
How to find $B=50-\overline{(n-1)n}$ when $n=\operatorname{Im}(z+w)$ and it requires the simplification of each complex? The problem is as follows:
First simplify the following complex numbers:
$z=3\left(\frac{1+i}{1-i}\right)+\left(\frac{1+i}{1-i}\right)^2+3\left(\frac{1+i}{1-i}\right)^3+\left(\frac{1+i}{1-i}\right)^4+...+3\left(\frac{1+i}{1-i}\right)^{4n+1}+\left(\frac{1+i}{1-i}\right)^{4n+2}$
$w=\left(\frac{1-2i}{2+i}\right)+\left(\frac{2-3i}{3+2i}\right)^2+\left(\frac{3-4i}{4+3i}\right)^3+\left(\frac{4-5i}{5+4i}\right)^4+...+\left(\frac{(4n+2)-(4n+3)i}{(4n+3)+(4n+2)i}\right)^{4n+2}$
After doing this find the value of $B$ from $n$:
$n=\operatorname{Im}(z+w)$
$B=50-\overline{(n-1)n}$
The alternatives given on my book are as follows:
$\begin{array}{ll}
1.&\textrm{5}\\
2.&\textrm{27}\\
3.&\textrm{38}\\
4.&\textrm{16}\\
\end{array}$
I'm very confused with how to approach this problem?. My best effort was to spot this:
$\left(\frac{1+i}{1-i}\right)=i$
This makes the earlier equation simplified a little bit, hence becoming the expression as follows:
$z=3\left(\frac{1+i}{1-i}\right)+\left(\frac{1+i}{1-i}\right)^2+3\left(\frac{1+i}{1-i}\right)^3+\left(\frac{1+i}{1-i}\right)^4+...+3\left(\frac{1+i}{1-i}\right)^{4n+1}+\left(\frac{1+i}{1-i}\right)^{4n+2}$
$z=3\left(i\right)+\left(i\right)^2+3\left(i\right)^3+\left(i\right)^4+...+3\left(i\right)^{4n+1}+\left(i\right)^{4n+2}$
But the thing is what to do from here?.
The rest, I mean for the second complex number seems that is more convoluted:
$w=\left(\frac{1-2i}{2+i}\right)+\left(\frac{2-3i}{3+2i}\right)^2+\left(\frac{3-4i}{4+3i}\right)^3+\left(\frac{4-5i}{5+4i}\right)^4+...+\left(\frac{(4n+2)-(4n+3)i}{(4n+3)+(4n+2)i}\right)^{4n+2}$
This second complex number doesn't seem to be easy to simplify.
I'm not really sure what sort of algebraic manipulation should be done here?. Can someone help me here?.
Since I'm not very savvy with this I really appreciate that someone could help me here with the most details as possible.
| The complicated expressions are only there to confuse you. You've noticed that $\frac{1+i}{1-i}=i$, so
$$z=[3(i)+(-1)+3(-i)+(1)]_n+3(i)+(-1)$$
where the part in square brackets, which is zero, repeats $n$ times so it may be ignored. Similarly, every fraction in brackets in $w$ is of the form $\frac k{ik}=-i$ for some $k$, so
$$w=[(-i)+(-1)+(i)+(1)]_n+(-i)+(-1)$$
and again the part in square brackets is repeated and is zero.
Thus $z=3i-1$, $w=-i-1$, $z+w=2i-2$, $n=2$ and $B=50-1\cdot2=48$.
| {
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"answer_count": 1,
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Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.
| We use the identity $(u+v)^2=u^2+v^2.$ From $$x^4+x^2+x=0$$ we see that
$$x=x^4+x^2=(x^2)^2+x^2=(x^2+x)^2$$
And we can construct the root of an arbitrary polynomial
$$ax^3+bx^2+cx+d=x(ax^2+c)+bx^2+d=(x^2+x)^2(ax+c)^2+(bx+d)^2=((x^2+x)(ax+c)+(bx+d))^2=(ax^3+(a+c)x^2+(b+c)x+d)^2$$
So we see that every polynomial has exactly one square root.
| {
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"answer_count": 4,
"answer_id": 2
} |
Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$
If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.
I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. Obviously $uvw$ method doesn't help. and since we have absolute value, I think The Buffalo doesn't help.
I wrote $ab+bc+ca$ this way: $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
We know that
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3},$$
so
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\Rightarrow ab+bc+ca\ge9abc$$
I don't know if it helps (or is even true).
I think finding the minimum of $\left(1-\left(ab+bc+ca\right)\right)^2+\left(1-abc\right)^2$ has the same procedure. If anyone knows how to find it, it helps a lot (I think).
| Let $p = a + b + c = 1$, $q = ab + bc + ca$ and $r = abc$.
Fact 1: $q^2 \ge 3pr$.
(The proof is given at the end.)
Fact 2: $p^2 \ge 3q$.
(Proof: $p^2 - 3q = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0$.)
We split into four cases:
*
*$q \ge 0$: By Fact 2, we have $q \le \frac{1}{3}$. By Facts 1-2, we have $r\le \frac{1}{27}$.
We have
$|1- q| + |1 - r| = 1 - q + 1 - r = 2 - q - r \ge 2 - \frac{1}{3} - \frac{1}{27} = \frac{44}{27}$ with equality if $a = b = c = \frac{1}{3}$.
*$q < 0$ and $r > 1$: By Fact 1, we have $q^2 \ge 3$. Thus, $q < -\sqrt{3}$. We have
$|1-q| + |1 - r| = 1 - q + r - 1 = -q + r > \sqrt{3} + 1 > \frac{44}{27}$.
*$q < 0$ and $0\le r \le 1$: By Fact 1, we have $q^2 \ge 3r$. Thus, $q \le -\sqrt{3r}$. We have
$|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 + \sqrt{3r} - r \ge 2 > \frac{44}{27}$.
*$q < 0$ and $r < 0$: We have
$|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 > \frac{44}{27}$.
Thus, the minimum is $\frac{44}{27}$.
$\phantom{2}$
Proof of Fact 1: We have
\begin{align}
q^2 &= (ab)^2 + (bc)^2 + (ca)^2 + 2abc(a+b+c)\\
&\ge ab\cdot bc + bc \cdot ca + ca \cdot ab + 2abc(a+b+c) \\
&= 3abc(a+b+c)\\
&= 3pr
\end{align}
where we have used
\begin{align}
&(ab)^2 + (bc)^2 + (ca)^2 - ab\cdot bc - bc \cdot ca - ca \cdot ab\\
=\ & \frac{1}{2}[(ab - bc)^2 + (bc - ca)^2 + (ca - ab)^2]\\
\ge \ & 0.
\end{align}
| {
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Looking for analytical proof that this function given as a power series is constant. Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$
Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:
*
*$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,
*$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.
I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.
The behavior of the function for $t=1,2,3,4$ is demonstrated below:
| Too long for comments.
Assuming that $t$ is a positive integer, there is no problem to get explicit formulae up to $t=3$.
For $t \geq 4$ appear hypergeometric function and the result write
$$P_t(x)=\frac x {1-x} \big[Q_t(x)-1\big]$$
$$Q_4(x)=\,
_3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{4^4}{3^3} (1-x) x^3\right)$$
$$Q_5(x)=\,
_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{2}{4},\frac{3}{4},\frac{5}{4};\frac{5^5}{4^4} (1-x) x^4\right)$$
$$Q_6(x)=\,
_5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{6^6} {5^5}(1-x) x^5\right)$$
$$Q_7(x)=\,
_6F_5\left(\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{7}{6};\frac{7^7}{6^6} (1-x) x^6\right)$$
For $x=x_*=1-\frac 1t$, $Q_t(x_*)=\frac t{t-1}$ and then $P_t(x_*)=1$ as you observed.
| {
"language": "en",
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"answer_count": 4,
"answer_id": 3
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Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.
| Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $a^2$ to find
$$0=a^2-a-1+\dfrac1a+\dfrac1{a^2}=\left(a-\dfrac1a\right)^2-\left(a-\dfrac1a\right)+1=0$$
$\implies a-\dfrac1a=-w$ where $w$ is a complex cube root of unity
$\implies a^2=1-wa$
$\implies a^3=a(1-wa)=a-wa^2=a-w(1-wa)=a(1+w^2)-w=-wa-w$
Compare the values of $wa$
$$1-a^2=a^3-w$$
$$\iff a^2-a^3=1+w=-w^2\ne1$$
$$\implies(a^2-a^3)^2=?\ne1$$
$$\implies(a^2-a^3)^3=?\ne1$$
$$\implies(a^2-a^3)^6=?$$
| {
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"answer_count": 3,
"answer_id": 2
} |
Factorize polynomial of degree 4 given statements If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?
Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$
Can I solve $a,b,c,d$ from there? How can I do so?
| You know that $abcd = -8$ and $cd = -8.$ Therefore $ab= 1.$
From the coefficient of $x$ in the polynomial, you know that $abc + acd + abd + bcd = -26$ (note: not $+26$).
But since $ab=1$ and $cd=-8$, you can see that
$abc + acd + abd + bcd = c - 8a + d - 8b.$
Therefore
$$ -8a - 8b + c + d = -26. \tag1 $$
But you also know that
$$ a + b + c + d = 1. \tag2 $$
Subtract Equation $(1)$ from Equation $(2)$:
$$ 9a + 9b = 27. $$
That is, $a + b = 3.$ But $b = \frac1a,$ so we have
\begin{align}
a + \frac1a &= 3, \\
a^2 + 1 &= 3a, \\
a^2 - 3a + 1 &= 0. \\
\end{align}
Apply the quadratic formula to solve $y^2 - 3y + 1 = 0.$
Note by symmetry that $a$ and $b$ both are solutions of this equation.
But you are given that $a > b$, so you can see how to match $a$ and $b$ with the two solutions of the quadratic formula.
For $c$ and $d,$ multiply Equation $(2)$ by $8$ and add the result to Equation $(1)$.
You get $9c + 9d = -18.$ But also $d = -\frac8c.$ Again you can get a quadratic equation out of this and solve it, then use the information that $c < d$ to know which root is which.
No guesswork is required, although if you do guess cleverly you can shorten the path a little.
| {
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The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.
So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?
| Since $$0\lt \cot^{-1} (-\frac 34 ) \lt \pi$$
, $$0\lt \frac 12 \cot^{-1} (-\frac 34) \lt \frac{\pi}{2} $$ which means its sine must be positive, so Wolfram’s answer is wrong. $\frac{2}{\sqrt 5}$ is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of square roots inequality
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$
The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.
| If you square both sides:
$(\sqrt a + \sqrt b + \sqrt c +\sqrt d)^2 =a + b+c+d + 2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})$
while $(2\sqrt {a+b+c+d})^2= 4(a+b+c+d)$ so it suffices to prove
$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le 3(a+b+c+d)$.
If we apply AM/GM though, we get $\sqrt{ab} \le \frac {a+b}2$ or in other words $2\sqrt{ab} \le a+b$.
And that does it:
$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le$
$(a+b) + (a+c) + (a+d) + (b+c) + (b+d) = 3(a+b+c+d)$.
| {
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Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$ I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$
$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$
$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)} = \frac{1/6}{x}+\frac{-1/2}{x+1}+\frac{1/2}{x+2} + \frac{-1/6}{x+3}
$$
$$
n=4 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)(x+4)} = \frac{1/24}{x}+\frac{-1/6}{x+1}+\frac{1/4}{x+2} + \frac{-1/6}{x+3} + \frac{1/24}{x+4}
$$
But I don't seem to find a pattern. Is there another method to resolve this? Thank you.
| Hint. If we have that
$$\prod_{i=0}^n \frac{1}{(x+i)}=\sum_{i=0}^n \frac{A_i}{(x+i)}.$$
then, for $k=0,\dots,n$,
$$\lim_{x\to -k}(x+k)\prod_{i=0}^n \frac{1}{(x+i)}=A_k.$$
The general formula of $A_k$ is related to a binomial coefficient.
Once that the partial fraction decomposition is ready, the integration is straightforward.
| {
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Solve Diophantine Equation Show that $ x^3 - 6 = 25y^2 + 35y $ doesn't have any non-zero integer solution.
What I have tried
$$ 25y^2 + 35y - x^3 + 6 = 0 $$
solving for y we get
$$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$
so I need to show that
$$ 25 + 4x^3= z^2 $$
doesn't have a solution
or if it does I need to show that
the fraction is not equal to an integer
but I can't proceed.
| Alternative solution
Suppose $x^3 = 25y^2 + 35y + 6 = (5y+6)(5y+1).$
If $(5y+6)$ and $(5y+1)$ have a common factor, then this common factor must divide $5$. If the common factor is $> 1$, then it must be $5$. But this would imply that $5$ is a factor of $x^3$, which would imply that $5 | x^3$ which would imply that $5|(5y+6)(5y+1)$.
This is impossible. Therefore $(5y+6)$ and $(5y+1)$ are relatively prime. Therefore $(5y+6)$ and $(5y+1)$ must each be a perfect cube. This is impossible, because for any positive integer $n$,
$[(n+1)^3 - n^3] = 3n^2 + 3n + 1 > 5.$
Note: It is therefore trivial that for $(5y+6)$ and $(5y+1)$ any perfect cubes, $(a^3)$ and $(b^3),$ where each of $(a^3)$ and $(b^3)$ may be positive or negative, you can't have $|(a^3) - (b^3)| = 5.$
| {
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Equality of triple integrals over unit sphere $ \iiint_{\text{unit ball}} x e^{ax + by + cz} dV$ I have to calculate
$$ \iiint_{\text{unit ball}} x e^{ax + by + cz} \,dV,$$
where by "unit ball" I mean the region $x^2 + y^2 + z^2 \leq 1$.
I know how to calculate this (rotation matrix that takes $(a,b,c)$ to $(0,0,\sqrt{a^2 + b^2 + c^2})$ and then spherical coordinates). The answer gives
$$\frac{4\pi a}{r^5}((3 + r^2)\sinh(r) - 3r \cosh(r)), \,\,\,\,\,r = \sqrt{a^2 + b^2 + c^2}$$
Now this would suggest that we have the following equality:
$$ \iiint_{\text{unit ball}} \frac{x}{a} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{y}{b} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{z}{c} e^{ax + by + cz} \,dV$$
Is it true? How would one quickly prove it if it is? It would be relevant because if one could quickly spot and prove this, then by naming the value of the above integrals $I$, we would have
$$(a^2 + b^2 + c^2)I = \iiint_{\text{unit ball}} (ax+by+cz) e^{ax + by + cz} \,dV$$
This integral would be much easier to calculate than the first since it wouldn't require to calculate the rotation explicitly (just considering an arbitrary rotation that orients $(a,b,c)$ along the $z$-axis suffices.)
Any ideas?
| In the Cartesian coordinates $(u,v,w)$ with the unit vector $\hat w=\frac{1}{\sqrt{a^2+b^2+c^2}}(a,b,c)$,
$$\hat x= \cos\theta_u \hat u + \cos\theta_v \hat v + \cos\theta_w \hat w$$
where $\cos \theta_w =\frac{a}{\sqrt{a^2+b^2+c^2}}$. Then
\begin{align}
\iiint_{{r<1}} \frac xa e^{ax + by + cz} \,dV
& = \frac1a\iiint_{{r<1}}( \cos\theta_u u + \cos\theta_v v + \cos\theta_w w) e^{\sqrt{a^2+b^2+c^2}w }\,dV\\
&= \frac1a\iiint_{r<1} \cos\theta_w w e^{\sqrt{a^2+b^2+c^2}w } \,dV\\
&= \frac{1}{\sqrt{a^2+b^2+c^2}} \iiint_{r<1} w e^{\sqrt{a^2+b^2+c^2}w } \,dV=I
\end{align}
where the integrations over $u$ and $v$ vanish due to symmetry of the unit ball. Likewise,
\begin{align}
\iiint_{r<1} \frac yb e^{ax + by + cz} \,dV
= \iiint_{r<1} \frac zc e^{ax + by + cz} \,dV
= I
\end{align}
| {
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If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$? If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?
I need to solve it using combinatorics.
I've started with calculating the number of possible solutions without restrictions and I am stuck.
| $a + b + c +d = 12 \ $ where $0 \leq a, b, c, d \leq 6$ and $a, b, c, d \ne 3$
Here is how you can solve using Principle of Inclusion-Exclusion, which is straightforward as well but I have taken more space and time to explain the steps.
A) Number of unrestricted solutions for non-negative numbers
$\displaystyle = {15 \choose 3} = 455$
B) Number of solutions where at least one of the numbers is $3$
$ \displaystyle = 4 {11 \choose 2} - {4 \choose 2} {7 \choose 1} + {4 \choose 1} - {4 \choose 4} = 181$
Explanation: First term is choosing one of the numbers to be $3$ and rest $3$ numbers sum to $9$ but these will overcount cases where two numbers are $3$ and hence the second term subtracts cases where two numbers are $3$ each and then other two sum to $6$. Then the last term is where three of the numbers are each $3$. Please note that if three numbers are $3$, the fourth will also be $3$. Hence in the first term we count such cases $4$ times and then subtract $6$ times in the second term. As there is only one way for three or four numbers to be $3$ each, we add $3$ so we count it one time. The other way you can write it as ${4 \choose 1} - {4 \choose 4}$ but I wanted to explain the rationale behind it.
C) Number of solutions where at least one of the numbers is greater than $6$
$ \displaystyle = 4 {8 \choose 3} = 224$
Explanation: Now if one of the numbers is $\gt 6$, none of the other numbers can be greater than $6$. We first set one of the $4$ numbers as $7$ and then count number of ways for $4$ numbers to sum to $5$.
Now B) and C) both count solutions where one of the numbers is greater than $6$ and one of the numbers is $3$.
D) Solutions where one of the numbers is greater than $6$ and one of the numbers is $3$
$\displaystyle = {4 \choose 3} {3 \choose 1} {4 \choose 2} = 72$.
Explanation: There are $4$ ways to choose one of the numbers being $7$ and one of the numbers being $3$ and then we have $3$ numbers sum up to remaining $2$ (including the number $7$ which can also be $8$ or $9$).
So the number of desired solutions is $ = 455 - 181 - 224 + 72 = \fbox {122}$
| {
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Combinatorics That Looks Similar to Vandermonde's Identity How do I simplify:
$$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$
Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$
This seems very similar to Vandermonde's Identity but because we skip two everytime instead of one, I don't see how I can apply it to this problem.
| Note that
$$\sum_{r\ge 0} a_{2r} = \sum_{r\ge 0} \frac{1+(-1)^r}{2}a_r.$$
Taking $$a_r=\binom{k}{r}\binom{n-k}{k-r}$$ yields
\begin{align}
\sum_{r\ge0} \binom{k}{2r}\binom{n-k}{k-2r}
&= \sum_{r\ge0} \frac{1+(-1)^r}{2} \binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\sum_{r\ge0} \binom{k}{r}\binom{n-k}{k-r} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\binom{n}{k} + \frac{1}{2}\sum_{r\ge0} (-1)^r\binom{k}{r}\binom{n-k}{k-r}.
\end{align}
I don't know whether the remaining sum can be simplified.
| {
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Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin
First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 =0 $$
Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:
$$ C(0,0) C(x,y) = (6x+8y-75)^2$$
This leads to:
$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$
$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$
For applying the result in this answer, then the formula
$$ a= 75 - 36, b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$
Or,
$$ a = 39, b= 11 , h=-48$$
$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$
However, the intended answer was:
$$ \tan \theta = \frac{1}{\sqrt{3} } $$
Where have I gone wrong?
| The mistake lies in using $6x+8y-75=0$ further instead of $-12x-16y+150=0$
in the formula. Correct way is to write:
$$C(0,0) C(x,y)=(-12x-16y+150)^2.$$
to get the correct answer.
| {
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If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
| hint: $x^2+y^2+z^2+t^2=(\frac{x^2}{3}+y^2)+(\frac{x^2}{3}+z^2)+(\frac{x^2}{3}+t^2)
=(\frac{x}{\sqrt{3}}-y)^2+(\frac{x}{\sqrt{3}}-z)^2+(\frac{x}{\sqrt{3}}-t)^2+\frac{2}{\sqrt{3}}\cdot x(y+z+t)\geqslant 0+1\cdot x(y+z+t)$
When would the equality hold?
| {
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Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| Other answers have shown how you could continue with what you had. But just for interest, here's another proof, which doesn't use the difference, but the quotient:
Let's call the formula
$$ a_n = 4^{2^n} + 2^{2^n} + 1 $$
You already showed that $a_1 = 21$ is a multiple of $7$. (So is $a_0 = 4^1+2^1+1 = 7$, so we can actually prove it for every non-negative integer $n$.)
With so many $2$'s involved, a substitution $4=2^2$ seems worth trying:
$$ a_n = (2^2)^{2^n} + 2^{2^n} + 1 = 2^{2 \cdot 2^n} + 2^{2^n} + 1 $$
The multiple exponent operations are a bit confusing, but we can abstract them out by noticing $2^{2 \cdot 2^n} = (2^{2^n})^2$, so if we define $x_n = 2^{2^n}$, then
$$ a_n = x_n^2 + x_n + 1 $$
Take $7 | a_n$ as the inductive hypothesis, and look at the next iteration:
$$ x_{n+1} = 2^{2^{n+1}} = 2^{2 \cdot 2^n} = (2^{2^n})^2 = x_n^2 $$
$$ a_{n+1} = x_{n+1}^2 + x_{n+1} + 1 = x_n^4 + x_n^2 + 1 $$
But this factors as
$$ a_{n+1} = (x_n^2 - x_n + 1) (x_n^2 + x_n + 1) = (x_n^2 - x_n + 1) a_n $$
Since $x_n$ is a positive integer, $(x_n^2 - x_n + 1)$ is also a positive integer. Since $7 | a_n$, we also have $7 | a_{n+1}$. Therefore by induction, $7 | a_n$ for every natural $n$.
| {
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System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right.
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?
| Observe that your system is equivalent to $$\begin{cases}f(x)=y\\ f(y)=z\\f(z)=x\end{cases}$$ Where $$f(x):=3\cdot \left[\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x\right]$$ Let $g(x)=f(x)-x$. Adding up the three equations yields $\sum g(x)=0$, so at least one of $g(x), g(y), g(z)$ has to be $\leqslant 0$. It's easy to see that $g(x)$ is strictly decreasing and has its only root at $x=3$.
Case 1. $g(x)\leqslant 0\iff x\geqslant 3$. Since $f(x)$ is also strictly decreasing, this implies that $y=f(x)\leqslant f(3)=3$. Thus $y\leqslant 3\implies z=f(y)\geqslant f(3)=3$. Hence, $z\geqslant 3\implies x=f(z)\leqslant f(3)=3$. So $x\leqslant 3$. But this is absurd unless $x=y=z=3$.
Cases 2 and 3 are analogous.
| {
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How do I solve this limit $\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$? I need to find the limit of $$\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$$
I did rooting like this:
$$\frac{\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)*\left(x+\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}} = \frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}$$
but I got stuck. How can I continue from here? Thanks in advance.
| Putting the terms in the numerator over the common denominator one obtains:
$$\begin{align}\frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}&=\frac{-6x^2}{(4x-3)x\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}\\
&=\frac{-6}{(4-\frac3{x})\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}.
\end{align}$$
Can you get it from here?
| {
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Uniform Convergence of series expansion of $\frac{\log(1+x)}{x}$ We know that the series expansion of $\frac{\log(1+x)}{x}$ is
$$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$$
It is also known that the value of dilogarithm at 1 is
$$\int_{-1}^{1} \frac{\log(1+x)}{x}=2\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}=\frac{3}{2}\zeta(2)=\frac{3}{2}Li_2(1)$$
For this to be true we must show that the RHS of the first equation that is $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$ is termwise integrable. To prove that this series is termwise integral we must show that it's uniformly convergent in $(-1,1)$. Although it is simple to prove that the series converges by the ratio test.
How to show that $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$ is uniformly convergent in $(-1,1)$?
| For $x \in [-1+\delta,1-\delta]$ where $0 < \delta < 1$, we have
$$\left|(-1)^k \frac{x^k}{k+1}\right|\leqslant \frac{(1-\delta)^k}{k+1} \leqslant (1-\delta)^k$$
As the geometric series $\sum_{k \geqslant 0} (1-\delta)^k$ converges, it follows by the Weierstrass M-test that we have uniform convergence on $[-1+\delta,1+\delta]$ of the series
$$\frac{\log (1+x)}{x} = \sum_{k=0}^\infty(-1)^k \frac{x^k}{k+1}$$
Now the series may be integrated termwise over $[-1+\delta,1-\delta]$ to obtain
$$\begin{align} \int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1+\delta)^{k+1}}{(k+1)^2}\\ &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1)^{k+1}(1-\delta)^{k+1}}{(k+1)^2} \\ &= \sum_{k=0}^\infty \frac{(1-\delta)^{k+1} + (-1)^{k}(1-\delta)^{k+1}}{(k+1)^2}\\&= 2\sum_{j=0}^\infty \frac{(1-\delta)^{2j+1}
}{(2j+1)^2}\end{align}$$
The series on the RHS is uniformly convergent for $\delta \in [0,1]$, and, consequently we may interchange the limit as $\delta \to 0+$ with the sum to obtain
$$ \int_{-1}^{1}\frac{\log(1+x)}{x} \, dx = \lim_{\delta \to 0+}\int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx = 2\sum_{k=0}^\infty \frac{1
}{(2j+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$
$$=\int \frac {1}{(25\tan^2\theta + 25)} $$
$$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$
$$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$
$$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$
$$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$
To generalize for $x$, $\theta = \arctan(7x/5)$
$$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$
$$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$
But taking the derivative of this gets me:
$$ \frac{35}{(49x^2 +25)^2}$$
Where is my mistake?
| You could also evaluate the integral in this way:
$$\int \frac {1}{49x^2 + 25}dx=\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx$$
By knowing that $\int\frac{1}{x^2+a^2}dx=\frac1a\tan^{-1}(\frac xa)$ we have:
$$\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx=\frac1{49}\times\frac75\tan^{-1}\left(\frac{7x}{5}\right)+C=\frac1{35}\tan^{-1}\left(\frac{7x}5\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct result:
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$
Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?
EDIT:
Given that:
$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$
because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.
hence, $\frac{\sin \theta}{\theta}$ approaches 1.
So the answer is $\frac{1}{2}$
| Note that $1-\cos\theta=2\sin^2\frac{\theta}2$:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}=\lim_{\theta\to0}\frac{2\sin^2\frac{\theta}2}{\theta \sin\theta}$$
Also we have $\sin u\sim u$ when $u\to0$ so:
$$\lim_{\theta\to0}\frac{2\times(\frac{\theta}2)^2}{\theta\times\theta}=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
Selecting at least 4 questions of 12 and at least one of the first 3 questions must be taken?
There are 12 questions. Students must take at least 4 questions and at least one of the first 3 questions is taken too.
How many configurations are there?
My attempt
\begin{align}
\text{answer} &= \sum_{i=1}^3\left({3\choose i}\sum_{j=4-i}^9{9 \choose j}\right)\\
&= 3415
\end{align}
Is it correct?
| Yes that is correct answer. Another way to approach it -
If there are $12$ questions, we have $2^{12}$ ways to select any number of questions including no selection $\big[{12 \choose 0} + {12 \choose 1} + ...+ {12 \choose 12}\big]$.
Out of that ${12 \choose 0} + {12 \choose 1} + {12 \choose 2} + {12 \choose 3} = 299$ are selections with less than $4$ questions.
Similarly if we do not select any of the first $3$ questions, there are $2^9$ ways to select questions and out of that ${9 \choose 0} + {9 \choose 1} + {9 \choose 2} + {9 \choose 3} = 130$ are selections with less than $4$ questions.
So the solution we are interested in $ = 2^{12} - 299 - (2^9 - 130) = 3415$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Division in inequality I have the problem
$(x-3)*(x+3) \leq x*(3+x)$
and with the following steps
$(x-3)*(x+3) \leq x*(3+x)$
$x^2-9 \leq 3x+x^2$
$-9 \leq 3x$
$-3 \leq x$.
However, my first thought was not to go this way but rather divide by $(3+x)$, which yields
$x-3 \leq x$.
Obviously this does not work. My first thought was that it is not allowed to divide by $(3+x)$ since this contains the solution -3 for x, which would equal to 0 $(-3+3) = 0$.
Question: Why does this not work? Is my thought (it does not work since it would include a division with 0) correct?
Thank you for your help!
| Actually I think you can solve this problem by dividing by $x+3,$ provided that you handle the sign of this term correctly and also take care of the special case
$x + 3 = 0.$
Consider three cases: $x + 3 = 0,$ $x + 3 > 0$, $x + 3 < 0.$
Case $x + 3 = 0$:
In this case $(x-3)(x+3) = 0 = x(3+x),$ so it is certainly the case that
$(x-3)(x+3) \leq x(3+x)$. In this case $x = -3,$ so $x = -3$ is one solution.
Case $x + 3 > 0$:
In this case, multiplying or dividing by $x + 3$ on each side of an inequality leaves the inequality intact, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \leq x$.
But $x-3 \leq x$ for all $x$, so every $x$ such that $x + 3 > 0$ is a solution.
In other words, every $x > -3$ is a solution.
Case $x + 3 < 0$:
In this case, multiplying or dividing by $x + 3$ on each side of an inequality reverses the direction of the inequality, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \geq x$.
But there is no $x$ such that $x-3 \geq x$ for all $x$, so this case does not find any solutions.
Conclusion: The only solutions are the ones found in the first two cases, namely, $x$ must satisfy either $x = -3$ or $x > -3.$ A more succinct way of expressing this is that $x \geq -3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$ I would like to ask how to integrate the following integration.
\begin{gather*}
\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}
I try to use integration by parts and I stuck with the following
\begin{gather*}
\frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\ln(x^2+1)}{x^2} \
\end{gather*}
I am confused with the continue step.
| Let $u=\log(x^2+1)$ and $dv=\frac{dx}{x^2}$.
Then $du = \frac{2xdx}{x^2+1}$ and $v=\frac{-1}{x}$.
Therefore
$$
\int_0^1 \frac{\log(x^2+1)}{x^2}dx = \left[\frac{-\log(x^2+1)}{x}\right]_0^1 - \int_0^1\frac{2dx}{x^2+1}
$$
The value of $\frac{\log(x^2+1)}{x}$ at 0 is undefined, but we can compute it with de l'Hôpital's rule:
$$
\lim_{x\to0}\frac{\log(x^2+1)}{x} = \lim_{x\to0}\frac{-2x}{x^2+1} = 0.
$$
Plugging everything in, we obtain
$$
\cdots = -\frac{\log(2)}{1} + 2[\arctan(x)]_0^1= -\log(2) + 2( \frac{\pi}{4}-0) = \frac{\pi}{2}-\log(2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The way to deduce the following equations. Given the $5$ equations.
$a_1=a_2+a_5 \tag{1}$
$a_4=a_3+a_5 \tag{2}$
$b=a_1+a_2 \tag{3}$
$b=a_3+\frac{a_4}{2} \tag{4}$
$\frac{a_5}{2}+a_1=a_3 \tag{5}$
I want to deduce the following.
$a_1=\frac{9}{16}*b\tag{6}$
$a_2=\frac{7}{16}*b\tag{7}$
$a_3=\frac{5}{8}*b\tag{8}$
$a_4=\frac{3}{4}*b\tag{9}$
$a_5=\frac{1}{8}*b\tag{10}$
Can anyone tell some hint(s)? so that I can deduce it in my own.
| Hint 1: As the comments have stated, one approach is to write the system as a matrix and then find the RREF. The matrix is given by $$A = \left(
\begin{array}{cccccc}
1 & -1 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & -1 & 1 & 0 \\
1 & 1 & 0 & 0 & 0 & b \\
0 & 0 & 1 & \frac{1}{2} & 0 & b \\
1 & 0 & -1 & 0 & \frac{1}{2} & 0 \\
\end{array}
\right)$$
You already have the result in your question, so know what the RREF should produce.
Hint 2: A second approach is to eliminate equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$ If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
The solution goes as follows:
$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:
$ab+bc+ac+2abc=1$
It's enough if: $(a+b+c)^2\ge 4-14abc$
$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$
From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.
Could you please provide a more intuitive and easier approach?
| if $x,y,z > 0 $, say $a = \frac{x}{y+z} , b=\frac{y}{x+z} , c=\frac{z}{x+y} $
then we'll proof that $$(a+b+c)^2 \ge 4 - 14\cdot a \cdot b \cdot c $$
$(a+b+c)^2 = a^2+2.a.b+b^2+2.a.c+2.b.c+c^2 = a^2+b^2+c^2+2(a.b+b.c+a.c) $
as you said $a.b+a.c+b.c+2.a.b.c = 1$
$a^2+b^2+c^2 +2(1-2.a.b.c) \ge 4 - 14.a.b.c $
$ a^2+b^2+c^2 +2 - 4.a.b.c \ge 4 - 14.a.b.c $
$ a^2+b^2+c^2 +10.a.b.c \ge 2 $
since $x,y,z > 0 $ means that $x$ or $y$ or $z$ are positive numbers, their lowest value is $ 0 + 10^{-n} $ and since our inequality speaks also minimum value, we can get the result by assuming the minimum value of $x,y,z$ there
for the simplest case scenario, this happens when $ x= y= z$ at the lowest level
$ x^2/(z+y)^2+y^2/(z+x)^2+z^2/(y+x)^2 + (10.x.y.z)/((y+x)(z+x)(z+y)) \ge 2 $
so if i say $x=y=z$, the result is $2$ this means that even if $x , y , z$ was $ < 0 $ our inequality would still be greater than 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What is a determinant in matrix ? NOT how to get it but WHAT is it in real life We know to get determinant in a matrix we use sarrus, laplace, x methods and so on
But question : WHAT is it, according to some pages is about permutations the products of element of the matrix , but WHAT IS IT ?
| \begin{align}
& \left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} ax+by \\ cx+dy \end{array} \right] \\[8pt]
& \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} a \\ c \end{array} \right] \\[8pt]
& \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] = \left[ \begin{array}{c} b \\ d \end{array} \right]
\end{align}
Draw the arrows representing $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right].$ From the head of the arrow $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right],$ draw another arrow parallel to $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ and from the head of the arrow $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$ draw another parallel to $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right].$ There you have a square whose area is $1.$
Then do the same with $\left[ \begin{array}{c} a \\ c \end{array} \right]$ and $\left[ \begin{array}{c} b \\ d \end{array} \right]$ and get a parallelogram. The area of that parallelogram is $\left|\det\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\right|.$ The determinant is positive if turning from $\left[ \begin{array}{c} a \\ c \end{array} \right]$ to $\left[ \begin{array}{c} b \\ d \end{array} \right]$ amounts to turning counterclockwise, and negative if clockwise. In other words, positive if the parallelogram has the same orientation as the square, and negative if the orientation is reversed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does the following integral converge or diverge and can it be solved? I have been stuck solving the following integral for some time now: $$\int_0^{2R} \frac{\sin^2\left(\frac{x}{4R}\pi\right)R^2}{\left(\sqrt{\left(d+Z\right)^2+\sin^2\left(\frac{x}{4R}\pi\right)}\right)^3}\,\mathrm{d}x$$
If it isn't possible to solve, is there a way to show whether this integral converges or diverges for $\lim_\limits{R\to\infty}$?
Any help or suggestions on how to solve this are appreciated.
| First note that your integral diverges for $d + Z = 0$. For $d + Z \neq 0$ we can write
\begin{align}
I (R,d,Z) &\equiv \int \limits_0^{2R} \frac{R^2 \sin^2 \left(\frac{\pi x}{4R}\right)}{\left[(d+Z)^2 + \sin^2 \left(\frac{\pi x}{4R}\right)\right]^{3/2}} \, \mathrm{d} x \stackrel{x = \frac{4 R t}{\pi}}{=} \frac{4 R^3}{\pi} \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[(d+Z)^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \\
&\equiv \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) \, .
\end{align}
Here we have defined the function
$$ f \colon (0,\infty) \to (0,\infty) \, , \, f(a) = \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[a^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \, .$$
Since $f$ is always positive, we immediately see that
$$ \lim_{R \to \infty} I (R,d,Z) = \lim_{R \to \infty} \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) = \infty \, .$$
In order to compute your integral we need to find $f$. This is somewhat tricky and involves elliptic integrals, but it can be done: For $a > 0$ we have
\begin{align}
f(a) &= \int \limits_0^{\pi/2} \frac{1 + a^2 - \cos^2(t) - a^2}{[1 + a^2 - \cos^2(t)]^{3/2}} \, \mathrm{d} t = \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2 - \cos^2(t)}} - \frac{a^2}{[1 + a^2 - \cos^2(t)]^{3/2}}\right] \mathrm{d} t \\
&= \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2} \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \frac{1}{(1+a^2)^{3/2}} \left(\frac{a^2}{\left(1 - \frac{\cos^2(t)}{1+a^2}\right)^{3/2}} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\sin(t) \cos(t)}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}}\right)\right] \mathrm{d} t \\
&= \frac{1}{\sqrt{1+a^2}} \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}\right] \mathrm{d} t \\
&= \frac{\operatorname{K} \left(\frac{1}{\sqrt{1+a^2}}\right) - \operatorname{E} \left(\frac{1}{\sqrt{1+a^2}}\right)}{\sqrt{1+a^2}} \, ,
\end{align}
where $\operatorname{K}$ and $\operatorname{E}$ are the complete elliptic integrals of the first and second kind, respectively. Note that the integral of the derivative term in the second line vanishes, so we are allowed to add it without changing the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.
I solved it as follows:
$x+2\sqrt{x}-1\ge 0$
$(\sqrt{x}+1)^2-2\ge 0$
$(\sqrt{x}+1)^2\ge 2$
$\sqrt{x}+1\ge 2$ or $\sqrt{x}+1\le -\sqrt{2}$
The second can't hold, from the first $x\ge 3-2\sqrt{2}$.
Similarly $(\sqrt{x}-1)^2\ge 2$, hence $x\ge 3+2\sqrt{2}$.
I find my solution to be very ugly. Is my solution correct and is there a neater approach?
| Let's use Mathematica for an overview plot:
ReImPlot[Sqrt[x + 2 Sqrt[x] - 1] + Sqrt[x - 2 Sqrt[x] - 1],
{x, -1, 7}, PlotPoints -> 10^3,
GridLines -> {{0, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2]},
{0, 2 Sqrt[Sqrt[2] - 1], 2, 2 Sqrt[1 + Sqrt[2]]}},
PlotTheme -> "Scientific"]
*
*For $x<0$ the function is purely real.
*For $0\le x\le 3-2\sqrt{2}$ the function is purely imaginary.
*For $3-2\sqrt{2}<x<3+2\sqrt{2}$ the function has nonvanishing real and imaginary parts.
*For $x\ge3+2\sqrt{2}$ the function is purely real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to factorize $P=(X^2-4X+1)^2+(3X-5)^2$ in $\mathbb R[X]$? First of all, I searched for roots. I knew that $\exists x\in\mathbb R,\, P(x)=0 \iff \exists x\in\mathbb R,\, x^2-4x+1 = 0 \text{ and } 3x-5 = 0$.
However, It is really easy to say that there is no real roots. So there is only complex non real roots.
And we know that $\deg{(P)}=4$ and $\text{dom}(P)=1$.
Which means that $P=(X^2+aX+b)(X^2+cX+d) = X^4+(a+c)X^3 + (ac+b+d) X^2 + (ad+bc) X+ bd $
With $a^2-4b<0$ , $c^2-4d<0$ and $a,b,c,d\in\mathbb R$ .
$P = (X^2-4X+1)^2+(3X-5)^2 = X^4 - 8 X^3 + 27 X^2 - 38 X + 26$. This gives us a system :
$$\left\{\begin{align}
a+c &= -8\\
ac+b+d&=27\\
ad+bc&=-38\\
bd&=26
\end{align}\right .
$$
However, I tried and i don't know how to solve this system.
I tried secondly with an euclidean division.
I found that $P = (X^2+aX+b)[X^2+(-8-a)X+(27-b+8a+a^2)]+(-38+8b+2ab-27a-8a^2-a^3)X+(-1+b^2-8ab-a^2b)$
With $a,b\in\mathbb R,\, a^2-4b <0$ and $(-8-a)^2-4(27-b+8a+a^2)<0$.
We have now this other system :
$$\left\{\begin{align}
-38+8b+2ab-27a-8a^2-a^3 &= 0\\
-1+b^2-8ab-a^2b&=0
\end{align}\right .
$$
However, I don't know how to solve this.
How would you do this ? And if it is the same thing as me, how could you solve those systems ?
| Assume (hope) that $a,b,c,d\in\Bbb Z$. If $bd=26$ then either:
*
*$b=1,d=26$ or $b=26,d=1$ which gives $ac=0$ but this is absurd;
*$b=2,d=13$ which gives $13a+2c=-38$, equating with $a+c=-8$ gives $a=-2,c=-6$.
Indeed $(X^2-4X+1)^2+(3X-5)^2=(X^2-2X+2)(X^2-6X+13)$. This is the unique irreducible factorisation so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Side of a square attached to two triangles Solve for $l$
I tried to solve this for $l$ with trigonometry, but I stopped at $l^2+4-2l\cdot \cos(x)=16, l^2 + 4 +2l\cdot \sin(x)=25$, where $x$ is the angle of the left triangle between 4 and $l$.
So I've tried with Heron's, and got stuck with some square roots...
According to WolframAlpha, the answer must be something around 3.71374
Any ideas?
------SOLVED----------
Okay, so, thanks to Narasimham I've seen that I took cosine formula wrong, and thanks to cosmo5 I've been able to solve it, here's the solution:
Start with cosine formula
\begin{matrix}
5^2 & = & l^2+4-2\cdot2\cdot l\cdot \cos(270^\circ - \alpha))\\
4^2 & = & l^2+4-2\cdot2\cdot l \cdot \cos(\alpha)
\end{matrix}
Rearrange teh two equation
\begin{matrix}
441-42l^2+l^4 & = &16l^2\cdot\sin^2(\alpha))\\
144-24l^2+l^4 & = &16l^2 \cdot \cos^2(\alpha)
\end{matrix}
find this quadratic formula
\begin{matrix}
585-66l^2+2l^4=16l^2
\end{matrix}
and then found this solutions
\begin{matrix}
+\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
+\sqrt{\dfrac{41-\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}
Then, discart the negative ones
\begin{matrix}
\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}
| Rearrange the two equations as
$$l^2-12=2l\cos x \quad , \quad l^2-21=-2l\sin x$$
On squaring both and adding, $x$ is eliminated,
$$(l^2-12)^2+(l^2-21)^2=(2l)^2$$
$$\Rightarrow 2(l^2)^2-70l^2+585=0$$
and a quadratic in $l^2$ is obtained. Its value can be computed using the quadratic formula.
To compute $l$, remember to take the square root of $l^2$. WA confirms $l=3.71374$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?
How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?
I thought that $a_n$ can be constructed in three ways:
*
*The strings that contain both three consecutive $0s$ and four consecutive $1s$ and end up with a $0$. Hence, there are $a_{n-1}$ such strings.
*The strings that contain both three consecutive $0s$ and four consecutive $1s$ and end up with a $1$. Hence , there are $a_{n-1}$ such strings.
*The strings that end up with $\dots1111000$ or $\dots0001111$ . There are $2 \times 2^{n-7} = 2^{n-6}$ such strings.
As a result ,the number of length-$n$ strings that contain both three consecutive $0s$ and four consecutive $1s$ is equal to $$a_n = 2a_{n-1}+2^{n-6}$$ with $a_0, a_1,\dots, a_6 =0$ and $a_7 = 2$. Is this correct? If not, can you write correct closed formula?
| Well , first of all we should think about basic set theory . We know that all situations consist of $4$ subcases such that ( containing $3$ consecutive zeros and $4$ consecutive ones ) $\color{red}{\cup}$ ( containing $3$ consecutive zeros and do not contain $4$ consecutive ones ) $\color{red}{\cup}$ (do not contain $3$ consecutive zeros and contain $4$ consecutive ones ) $\color{red}{\cup}$ (do not contain $3$ consecutive zeros and do not contain $4$ consecutive ones ) $\color{blue}{=}$ All situations ,i.e , $2^n$ where $n$ is lenght of the string.
Then , lets call our situations such that
$x_1=$ The number of strings that containing $3$ consecutive zeros and $4$ consecutive ones
$x_2=$ The number of strings that containing $3$ consecutive zeros and do not contain $4$ consecutive ones
$x_3=$ The number of strins that do not contain $3$ consecutive zeros and contain $4$ consecutive ones
$x_4 =$ The number of strings that do not contain $3$ consecutive zeros and do not contain $4$ consecutive ones
We said that $x_1 + x_2 + x_3 + x_4 = 2^n$ ,and we are looking for $x_1$ . If we think these like set operations , it is clear that $x_1 = 2^n - [x_2 +x_3 -x_4]$.
Moreover , $x_2 + x_4 =$ the number of strings that do not contain $4$ consecutive $1's$ , $x_3+x_4 =$ the number of strings that do not contain $3$ consecutive $0's$.
Result , $x_2 + x_3 -x_4 =$ (the number of strings that do not contain $4$ consecutive $1's$ ) + ( the number of strings that do not contain $3$ consecutive $0's$) - (The number of strings that do not contain $3$ consecutive zeros and do not contain $4$ consecutive ones)
The rest is the process of finding generating functions for them by Goulden -jackson . I am not get in elaborately that process , you can find explanation on the stack -exchange about it.
So ,
the generating function for the number of strings that do not contain $4$ consecutive $1's$ : $$\frac{1-x^4}{1-2x+x^5}$$
the generating function for the number of strings that do not contain $3$ consecutive $0's$ : $$\frac{1-x^3}{1-2x+x^4}$$
The generating functions for the number of strings that do not contain $3$ consecutive zeros and do not contain $4$ consecutive ones : $$\frac{1+2x +3x^2 +3x^3 +2x^4 +x^5}{1-x^2 -2x^3 -2x^4 -x^5}$$
The generating function for $2^n$ is $$\frac{1}{1-2x}$$
Then , $$x_1 = \frac{1}{1-2x} - \Biggl[\frac{1-x^4}{1-2x+x^5} + \frac{1-x^3}{1-2x+x^4} - \frac{1+2x +3x^2 +3x^3 +2x^4 +x^5}{1-x^2 -2x^3 -2x^4 -x^5} \Bigg] $$
Hence , $$x_1 = 2x^7 +8x^8 + \color{blue}{26}x^9 +75x^{10}+...$$
$26x^9$ means that there are $26$ string of lenght $9$ that contain $000$ and $1111$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generating Function for Recursive Sequences (generalization to the binomial series) Let $S_k$ be a recursive sequence defined by
$S_k = (x+1)S_{k-1}+S_{k-2}$ with $k>1$
with initial conditions:
$S_0 = 2, S_1=x+1$.
Then the first few terms are
$2, x + 1, x^2 + 2x + 3, x^3 + 3x^2 + 6x + 4, x^4 + 4x^3 + 10x^2 + 12x + 7,\\ x^5 + 5x^4 + 15x^3 + 25x^2 + 25x + 11, x^6 + 6x^5 + 21x^4 + 44x^3 + 60x^2 + 48x + 18$
For simplicity, the first term $S_0$ will be set to $1$ instead of $2$. We can arrange the coefficients of all the terms of $S_k$ in a triangle like format (Similar to Pascals Triangle).
$[1]$
$[1, 1]$
$[1, 2, 3]$
$[1, 3, 6, 4]$
$[1, 4, 10, 12, 7]$
$[1, 5, 15, 25, 25, 11]$
$[1, 6, 21, 44, 60, 48, 18]$
$...$
Let $P_{t,n}$ denote the $t$-th coefficient of $S_k$ (starts at $t=0$). For example, $P_{3,3} = 4$.
If $t$ is fixed, then $P_{t,n}$ is a polynomial in $n$.
$P_{0,n} = 1$, $P_{1,n} = n$, $P_{2,n} = \frac{n^2 + n}{2}, P_{3,n} = \frac{n^3 + 3n^2 - 10n}{6}$
and so on.
If $t > 1, n > 2$, then $P_{t,n} = P_{t,n-1}+P_{t-1,n-1}+P_{t-2,n-2}$. This is similar to Pascal's Identity for binomial coefficients.
What is the generating function for the power series
$$G(x,n) = \displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = 1 + nx + \frac{n^2 + n}{2}x^2 + \frac{n^3 + 3n^2 - 10n}{6}x^3 \\+ \frac{n^4 + 6n^3 - 37n^2 + 30n}{24}x^4 + \frac{n^5 + 10n^4 - 85n^3 + 50n^2 + 264n}{120}x^5 + O(x^6)$$
Is there a general formula for generating functions involving other polynomial recursive functions?
Edit:
$$\displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = (\sum_{j=0}^{\lceil{t/2}\rceil} \frac{n}{n-j}\binom{n-j}{j}\binom{n-2j}{t-2j})x^t $$
| The solutions to $X^2 -(x+1)X-1=0$ are $X_1= \frac{x+1+\sqrt{(x+1)^2+4}}{2}$ and $X_2= \frac{x+1-\sqrt{(x+1)^2+4}}{2}$. Then, with the given initial values, we have
$$S_k(x)= X_1^k+X_2^k.$$
Since by definition $$S_n(x):=\sum_{t\ge 0}x^{n-t}P_{t,n},$$
we have $$S_n(x^{-1})=\sum_{t\ge 0}x^{t-n}P_{t,n},$$
that is $$\sum_{t\ge 0}P_{t,n}x^{t}=x^nS_n(x^{-1}),$$
that is
$$\sum_{t\ge 0}P_{t,n}x^{t}=\big(\frac{1+x+\sqrt{5x^2+2x+1}}{2}\big)^{n}+\big(\frac{1+x-\sqrt{5x^2+2x+1}}{2}\big)^{n},$$
if I did not make a mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the value of $\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$ My question is to find the value of $$\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$$
Mathematica told me the result is 71 and $$\lfloor (\sqrt{21}+\sqrt{29})^{20k+4}\rfloor \mod 100$$ is always 71 for $k\in \mathbb N$, can anyone explain the result?
| Let $\alpha = (\sqrt{21}+\sqrt{29})^{2} = 50 + 2 \sqrt{609}$. Then $\alpha^n = x_n + y_n \sqrt{609}$ with $x_n, y_n \in \mathbb Z$.
Let $\beta = (\sqrt{21}-\sqrt{29})^{2} = 50 - 2 \sqrt{609}$. Then $\beta^n = x_n + y_n \sqrt{609}$ and so $\alpha^n+\beta^n = 2x_n \in \mathbb Z$.
Since $0 < \beta < 1$, we get $\lfloor \alpha^n \rfloor = 2x_n-1$.
Since $\alpha^2 = 100 \alpha -64$, we get $x_{n+2} = 100 x_{n+1} - 64 x_n$, with $x_0=1$ and $x_1=50$.
Let $z_n = x_n \bmod 100$. Then $z_{n+2} \equiv 36 z_n \bmod 100$ and so $z_{2n} \equiv 36^n \bmod 100$.
Now, $(\sqrt{21}+\sqrt{29})^{1984} = \alpha^{992}$ and so $z_{992} \equiv 36^{496} \bmod 100$.
Finally, $36^n \bmod 100$ is a pre-periodic sequence of period $5$ starting at $n=1$:
$$
1,36,96,56,16,76,36,96,56,16,76,\dots
$$
that is, $36^{n+5} \equiv 36^n \bmod 100$ for $n\ge 1$. In particular, $36^{5n+1} \equiv 36 \bmod 100$ for all $n$.
Since $496 =5 \cdot 99 + 1$, we get $36^{496} \equiv 36 \bmod 100$.
Therefore, $\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor = 2x_{992}-1 \equiv 2z_{992}-1 \equiv 2 \cdot 36 -1 = 71\bmod 100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Probability of Sum being $4$ with unfair dice You roll two six-sided dice. Die $1$ is fair. Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same. What is the probability of the sum of both dice adding up to $4$?
My answer:
I wrote out the combinations to sum $4$ between both dies:
(F1, U3), (F2, U2), (F3, U1) [F = Fair die, U = Unfair die].
My thinking is we need to find the OR probability of these pairs:
P((F1 and U3) or (F2 and U2) or (F3 and U1)):
$$(\frac16 \times \frac23) + (\frac16 \times \frac12) + (\frac16 \times \frac23) = 0.30555555555.$$
Is this correct?
| There is a mistake. For the unfair die, the probability of an odd number is $\frac{2}{3}$ and of an even number is $\frac{1}{3}$. It also says that the probability of each odd number is same and of each even number is same.
So for the unfair die, probability of each odd number is $\frac{2}{9}$ and of each even number is $\frac{1}{9}$.
So the desired probability should be $ = \displaystyle \small 2\cdot \frac{1}{6} \cdot \frac{2}{9} + \frac{1}{6} \cdot \frac{1}{9} = \frac{5}{54}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of an interception point as one parameter changes. Consider the functions
\begin{align*}
h_1(z)&=m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)\\
h_2(z)&=\frac{-M}{(a^2+z^2)^{3/2}}
\end{align*} where $M$, $m$ and $e$ are real parameters and $e<0$. I know that these two only have one interception point, $z_0$ (I checked it graphically). I also know that the limit of $z_0$ as $m$ approaches infinity is equals to $2e$ (It was also checked graphically).
BUT, if possible, I'd like to show it analytically. i.e, show that
$$\lim_{m \rightarrow \infty} z_0 = 2e$$
Is it possible to show it by a "non numerical" approach? Any tips/help will be very appreciated.
| It should actually be $-2e$, which assuming that $e<0$ will be positive. We can show this by considering solutions to $h_1(z) = h_2(z)$, since a point $z = z_0$ which satisfies this equation is an intercept. So, we have
\begin{align*}
h_1(z) &= h_{2}(z)\\[5pt]
\implies m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)&=\frac{-M}{(a^2+z^2)^{3/2}}\\[5pt]
\implies \frac{1}{z^3} + \frac{2e}{z^4} &= \frac{-M/m}{(a^2+z^2)^{3/2}}.
\end{align*}
Now, taking the limit of each side as $m\to \infty$:
$$\lim_{m\to\infty}\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg) =\lim_{m\to\infty}\left(\frac{-M/m}{(a^2+z^2)^{3/2}}\right) = 0.
$$
So, as $m\to\infty$, we have that $z_0$ solves
\begin{align*}
\frac{1}{z_0^3} + \frac{2e}{z_0^4} &= 0\\
\implies\frac{1}{z_0^3} &= -\frac{2e}{z_0^4}\\
\implies z_0 &= -2e.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.
| It's $\,(\color{#0a0}{n^4\!+\!1})(\color{#c00}{n^4\!-\!1})^2$ so $\,n\,$ odd $\,\Rightarrow\begin{align}&\ \,2\mid \color{#0a0}{n^4\!+\!1}\\ &2^4\mid \color{#c00}{n^4\!-\!1}\end{align}\,$ since $\,\begin{align} n&\equiv \pm1\ \ \pmod{\!4}\\\Rightarrow \color{#c00}{n^4}&\equiv \color{#c00}{(\pm1)^4}\!\!\!\!\pmod{\!4^2}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
| Using the result $\int_0^{\infty} \frac{\ln x}{x^2+1} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x &=\frac{1}{a} \int_0^{\infty} \frac{\ln a+\ln y}{y^2+1} d y \\
&=\frac{\ln a}{a} \int_0^{\infty} \frac{d y}{y^2+1}+\frac{1}{a} \int_0^{\infty} \frac{\ln y}{y^2+1} d y \\
&=\frac{\pi \ln a}{2 a}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ yields a general integral
$$
\boxed{\int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2}=\frac{\pi}{4 a^3}(\ln a-1)}
$$
Putting $a=1$ gives our integral $$
\int_0^{\infty} \frac{\ln x}{\left(x^2+1\right)^2} d x=-\frac{\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be? If we let $x = 0$.
\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}
Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.
But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $x$.
Where am I going wrong?
| $3 (x + 7) - y (2 x + 9)=21-9y$ for $x=0$ and
$3 (x + 7) - y (2 x + 9)=24 - 11 y$ for $x=1$
They are the same, so must be
$$21-9y=24-11y\to y=\frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?
This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!
| $$L=\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}$$
Lets try using our dear L'Hopital's rule to simplify the expression.
$$L=\lim_{x\to 0} \frac{2xe^{x^2} -2\sin x }{2x\sin (2x^2)}$$
We are very familiar with these functions! So lets expand them since they're easier to deal with now.
$$L=\lim_{x\to 0} \frac{x(1+x^2+\cdots) -(x-\frac{x^3}{6}+\cdots) }{x(2x^2-\frac{8x^6}{6})}=\lim_{x\to 0} \frac{\frac{7x^3}{6}+\cdots}{2x^3+\cdots}=\bbox[10px,#ffb]{\frac{7}{12}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$ I'm trying to solve the following sum for a project. Many thanks in advance for answering the question!
$$\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m+a)}{\Gamma(k+m+b)}(k+m)$$
where $m$ is an integer but $a$ and $b$ can be any real numbers.
| Using this result for hypergeometric functions of unity argument as well as the property of the gamma function $x\Gamma(x)=\Gamma(x+1)$ we find:
$$
\begin{align}
S
&=\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m
+a)}{\Gamma(k+m+b)}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(m
+a)_k}{(m+b)_k}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}(k+m)\\
&=m\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}
+\sum_{k=1}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}k\\
&=mF(1,m+a;m+b;1)
+\sum_{k=0}^{\infty}\frac{(1)_{k+1}(m
+a)_{k+1}}{(m+b)_{k+1}\,(k+1)!}(k+1)\\
&=m\frac{\Gamma(m+b)\Gamma(b-a-1)}{\Gamma(m+b-1)\Gamma(b-a)}
+\frac{m
+a}{m+b}\sum_{k=0}^{\infty}\frac{(2)_k(m
+a+1)_k}{(m+b+1)_k\,k!}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}F(2,m+a+1;m+b+1;1)\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}\frac{\Gamma(m+b+1)\Gamma(b-a-2)}{\Gamma(m+b-1)\Gamma(b-a)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=\frac{m+b-1}{b-a-1}\left(m
+\frac{m+a}{b-a-2}\right),
\end{align}
$$
which holds so long as $\Re(b-a-2)>0$. This result was numerically evaluated in Mathematica and compared with numerical values of the original sum $S$, which showed agreement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$ With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?
$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$
Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count different cases:
first: having all $1,2,3,4$ digits and choose another digit from $1,3,4$:
$${3\choose1}\times\frac{5!}{2!}$$
second: having two pair of same digit and choose another digit:
$${3\choose2}\times2\times\frac{5!}{2!2!}$$
Summing them we have $3\times\dfrac{5!}{2}+3\times\dfrac{5!}{2}=5!\times3$ but I don't have this in the options.
| We have two multiplicity patterns possible for numbers given the digits available:
$(2,1,1,1) \approx wwxyz$ (A) and $(2,2,1)\approx wwxxy$ (B).
Pattern A can be filled in $\binom 31 \binom 33 = 3$ ways and permuted in $\binom{5}{2,1,1,1}$ $ = 5!/2! = 60$ ways
Pattern B can be filled in $\binom 32 \binom 21 = 6$ ways and permuted in $\binom{5}{2,2,1} = 5!/(2!2!) = 30$ ways
So $180+180 = 360$ options altogether which matches your revised result but again doesn't match any given options.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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matrices such that $A^2B+BA^2=2ABA$ Let be $A$, $B$ two matrices $3 \times 3$ with complex entries.
Prove that if $$A^2B+BA^2=2ABA$$
THEN
$$B^2A+AB^2=2BAB$$
I tried it and do not know how to continue.
If $A$ is invertible then
$$AB^2+A^{-1}BA^2B=2BAB$$
so I have to prove that
$$B^2A=A^{-1}BA^2B$$
How to continue and also have to discuss the case when $A$ is not invertible
| The statement in question is equivalent to $[A,[A,B]]=0\rightarrow[[A,B],B]=0$ and it is false. Here is a counterexample taken from the last section of Irving Kaplansky, Jacobson's Lemma Revisited, Journal of Algebra, 62, 473-476 (1980):
\begin{aligned}
A&=\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1},\quad B=\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0},\\
[A,B]&=\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1}\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1}\\
&=\pmatrix{0&0&1\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&1&0}
=\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0},\\
[A,[A,B]]&=A[A,B]-[A,B]A=[A,B]-[A,B]=0,\\
[[A,B],B]&=\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0}\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}-\pmatrix{0&0&0\\ 0&0&1\\ 1&0&0}\pmatrix{0&0&1\\ 0&0&0\\ 0&-1&0}\\
&=\pmatrix{1&0&0\\ 0&0&0\\ 0&0&-1}-\pmatrix{0&0&0\\ 0&-1&0\\ 0&0&1}
=\pmatrix{1&0&0\\ 0&1&0\\ 0&0&-2}\ne0.
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$ The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have 2 points with a distance $d$ such that
$$d \le \sqrt{4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)}$$
This result was proven by Fejes Tóth. What I wish to show is that $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$.
We can see in the limit as $n \to \infty$, $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \to 0^+$. Also using basic numerical simulation for many and large $n$ I can visually verify that this statement seems to be true. The point now would be to show that for any $n$, then there exists some constant $k$ such that
$$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{k}{n}$$
This would be good enough. I'll be working on it here just thought the community may have some ideas :-)
| By Taylor, we have
$$\csc \left(\frac{\pi n}{6 (n-2)}\right)=2-\frac{2 \pi }{\sqrt{3} n}+\frac{\frac{7 \pi ^2}{9}-\frac{4 \pi
}{\sqrt{3}}}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=4-\frac{8 \pi }{\sqrt{3} n}+\frac{8 \pi \left(5 \pi -6 \sqrt{3}\right)}{9
n^2}+O\left(\frac{1}{n^3}\right)$$
$$4-\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}-\frac{8 \left(\pi \left(5 \pi -6
\sqrt{3}\right)\right)}{9 n^2}+O\left(\frac{1}{n^3}\right)$$
$$4-\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}+O\left(\frac{1}{n^2}\right)$$
Edit
Another way to do it using @Dzoooks' hint.
$$\frac{\pi n}{6 (n-2)}=\frac{\pi (n-2)+2\pi}{6 (n-2)}=\frac \pi 6+\frac{\pi }{3 (n-2)}=\frac \pi 6+a$$
$$\csc \left(a+\frac{\pi }{6}\right)=2-2 \sqrt{3} a+7 a^2-\frac{23 a^3}{\sqrt{3}}+O\left(a^4\right)$$
$$\csc^2\left(a+\frac{\pi }{6}\right)=4-8 \sqrt{3} a+40 a^2+\left(-\frac{92}{\sqrt{3}}-28 \sqrt{3}\right)
a^3+O\left(a^4\right)$$
$$4-\csc^2\left(a+\frac{\pi }{6}\right)=8 \sqrt{3} a-40 a^2+\left(\frac{92}{\sqrt{3}}+28 \sqrt{3}\right)
a^3+O\left(a^4\right)$$ Now, let $a=\frac{\pi }{3 (n-2)}$ and, continuing with Taylor series, the same result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Congruency application I want to prove whether the following is true using congruencies: If $n$ is odd and $3\not | n$, then $n^2\equiv 1\pmod{24}$.
I tried a direct proof.
Let $n=2k+1$ for an integer $k$ such that $3\not| n$, or $3c=n+r$ such that $r=1$ or $r=2$ for an integer $c$. Then in the first case, $3c=n+1$ implies $n\equiv 1\pmod 3$. Or in the second case, $3c=n+2$ implies $n\equiv 2 \pmod 3$. Then $$n^2\equiv 1 \pmod 3 \text{ or } n^2\equiv 4 \pmod 3$$
iff $$3|{n^2-1} \text{ or } 3|n^2-4$$
or $$8n^2\equiv8, 32 \pmod{24}$$
This was the farthest I got. I'm having a hard time going from mod 3 to mod 24 just given the conjunction that $n$ is odd and that $3\not| n$
| If $n$ is odd and not divisible by $3$. Then $n \equiv \pm 1 \pmod{6}$.
$$n = 6k \pm 1$$
$$n^2 = 36k^2 \pm 12k + 1=12k(3k \pm 1) +1$$
We consider two cases if $k$ is even, then $12k(3k \pm 1)$ is divisible by $24$.
If $k$ is odd, then $3k\pm 1$ is even, hence $12k(3k\pm 1)$ is again divisible by $24$.
Hence $n^2 \equiv 1 \pmod{24}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to resolve this integral $J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$? $$J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$$
I'm already try to do this step in below:
$$\overset{split}{=} \int{\frac{2}{2+\cos{x}}dx}+\int{\frac{d(\cos{x}+2)}{2+\cos{x}}} = \int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx} + \ln({2+\cos{x}})$$
| We need to find $$J:=\int \frac{2-\sin(x)}{2+\cos(x)}\operatorname{dx}.$$
The answer is $$\boxed{J=\frac{4}{\sqrt{3}}\arctan\left(\frac{\tan(x/2)}{\sqrt{3}} \right)+\ln(\cos(x)+2)+C}$$
For solve $J$, we can use the substitution recommended by @Prime Mover.
Let, $\color{blue}{u=\tan(x/2)}$ so we have that $$\operatorname{du}=\frac{1}{2}\sec^{2}(x/2)\operatorname{dx}.$$
Then transform the integrand using the substitution recommended by @tommik, $$\sin(x)=\frac{2u}{u^{2}+1},$$ we can find that $$\cos(x)=\frac{1-u^{2}}{u^{2}+1},$$ and $$\operatorname{dx}=\frac{2}{u^{2}+1}\operatorname{du}.$$
Then, \begin{eqnarray}
J&=&\int \frac{2\left(2-\frac{2u}{u^{2}+1}\right)}{(u^{2}+1)\left(\frac{1-u^{2}}{u^{2}+1}+2 \right)}\operatorname{du}\\
&=&4\int \frac{u^{2}-u+1}{u^{4}+4u^{2}+3}\operatorname{du}\\
&=&2\int \frac{u+2}{u^{2}+3}\operatorname{du}-2\int \frac{u}{u^{2}+1}\operatorname{du}.
\end{eqnarray}
In the last part I used partial fractions, I'll let you fill in those details.
Since that, $$\frac{u+2}{u^{2}+3}=\frac{u}{u^{2}+3}+\frac{2}{u^{2}+3},$$so we have that $$J=2\int \frac{u}{u^{2}+3}\operatorname{du}+4\int \frac{1}{u^{2}+3}\operatorname{du}-2\int \frac{u}{u^{2}+1}\operatorname{du}.$$
The first integral is solved for change of variable $t_{1}:=u^{2}+3$, the second integral for change of variable $t_{2}:=\frac{u}{\sqrt{3}}$ and the third integral for change of variable $t_{3}:=u^{2}+1$.
Finally, return the variable changes and you will get $$J=\frac{4}{\sqrt{3}}\arctan\left(\frac{\tan(x/2)}{\sqrt{3}} \right)+\ln(\cos(x)+2)+C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $3^n \gt n^4 \ \forall n \gt 8$ Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:
Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$
Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$
Inductive step ($P(n) \implies P(n+1)$): $\ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$
$$3^{n+1} = 3^n + 3 = 3^n + 3^n + 3^n = 3^n + 3^n + \frac{(n+1)^4}{3} \gt (n+1)^4 \\ \text{I found } \frac{(n+1)^4}{3} \text{ by dividing 3 on both sides of } 3^{n+1} \gt (n+1)^4 \text{ as soon as I suppose this to be true.}$$
|
Inductive step ($P(n) \implies P(n+1):) \ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$
You need to PROVE $ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$. You can NOT start by assumenting that.
But you start by assuming $3^n > n^3$.
So $3^{n+1} = 3\cdot 3^n > 3n^4$
So how does $3n^4$ compare with $(n^4 + 4n^3 + 6n^2 + 4n + 1)$?
Well $3n^4 = n^4 + 2n^4$.
SO how does $2n^4$ compare with $4n^3 + 6n^2 + 4n + 1$?
Well $n \ge 8$ so $2n^4 = (2n)n^3 \ge 8n^3$.
So
$n^4 + 2n^4 \ge n^4 + 8n^3 = n^4 + 4n^3 + 4n^3$.
So how does $4n^3$ compare with $6n^2 + 4n + 1$.
Well, we just keep doing this $n \ge 8$ so $n^k \ge 8n^{k-1}$ to get:
$3^{n+1} = 3\cdot 3^n >$
$3n^4 = n^4 + 2n^4 \ge $
$n^4 + 8n^3 = n^4 + 4n^3 + 4n^3 \ge$
$n^4 + 4n^3 + 32n^2= n^4 + 4n^3 + 6n^2 + 26n^2\ge$
$n^4 + 4n^3 + 6n^2 + 26\cdot 8 n=$
$n^4 +4n^3 + 6n^2 + 4n + (26\cdot 8 - 4)n \ge$
$n^4 + 4n^3 + 6n + 4n + (26\cdot 8 -4)\cdot 8 >$
$n^4 + 4n^3 + 6n + 4n + 1=(n+1)^4$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$ Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$
This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$. Then $\min P=\sqrt{3}$. But I have no idea to solve it. Please help me a hint. Thank you.
| As $a,c \ge 0$ then
$$P \ge S := ab+bc+ca + \frac{3}{a+b+c}$$
We minimize $S$
As $a^2+b^2 +c^2 = 3$ then
$$S =\frac{1}{2} (a+b+c)^2+\frac{3}{a+b+c} -\frac{3}{2}$$
Denote $x = a+b+c$. Because $a^2+b^2 +c^2 = 3$ then $\sqrt{3} \le x \le 3$
$$S = f(x) := \frac{1}{2}x^2+\frac{3}{x}-\frac{3}{2}$$
As $f'(x) = x-\frac{3}{x^2}>0$ in the interval $[\sqrt{3},3]$ then $f(x)$ is increasing. Hence, $S_1 = f(x)$ is minimized as $a+b+c = x_0 = \sqrt{3}$
Conclusion: we can conclude that
$$P \ge S \ge \sqrt{3} $$
The equality occurs when $(a,b,c) = (0,\sqrt{3},0),(0,0,\sqrt{3})$ or $(\sqrt{3},0,0)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $x^x+y^y\ge x^2+y^2$ for $x,y>0$ and $x+y\le 2$. We may prove the inequality for $x,y\in (0,1]$.
Note that, for $0<x\le 1$, it holds that
\begin{align*}
x^x&=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\cdots\\
&\ge1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\\
&\ge x^2.
\end{align*}
Similarily, for $y \in (0,1]$, it holds that $$y^y\ge y^2.$$
Thus$$x^x+y^y\ge x^2+y^2.$$
But how to prove under the condition $x+y\in (0,2]$?
| Here is a sketch. As you have shown, it suffices to show that
$$x^x+y^y\ge x^2+y^2$$
if one of $x,y$ is bigger than 1. Assume $x\in (1,2]$, then $y\le 2-x\in (0,1).$ One can check that the function $x^x-x^2$ is decreasing on $(0,1)$ and thus
$$y^y-y^2\ge (2-x)^{2-x}-(2-x)^2.$$
Thus it suffices to show that
$$x^x-x^2+(2-x)^{2-x}-(2-x)^2\ge 0$$
for $x\in (1,2].$
This should not be very hard. Actually, the last inequality holds for all $x\in [0,2]$ (since the left side is symmetric about the line $x=1$) and the left side function in increasing on $(1,2)$.
Edit: about the proof of the last inequality. Consider the function
$$f(x)=x^x-x^2+(2-x)^{2-x}-(2-x)^2, x\in [0,2].$$
We want to show that $f'(x)\le 0$ on $[0,1]$ (and $f'(x)\ge 0$ on $[1,2]$). We have
$$f'(x)=x^x(\ln(x)+1)-(2-x)^{2-x}(\ln(2-x)+1)-4x+4,$$
$$f''(x)=x^x((\ln(x)+1)^2+1/x)+(2-x)^{2-x}((\ln(2-x)+1)^2+1/(2-x))-4.$$
Denote $g(x)=x^x((\ln(x)+1)^2+1/x)$. We will check that $g(x)$ is concave up. Thus by Jensen's inequality, we have $g(x)+g(2-x)\ge 2g((x+2-x)/2)=2g(1)=4$, which shows $f''(x)\ge 0$ for $x\in (0,2)$ and thus $f'(x)$ is increasing on $(0,2)$. Note that $f'(1)=0$, we get the result.
Edit 2: The proof of the function $h(x)=x^x-x^2$ is decreasing on $(0,1)$. We have $h'(x)=x^x(\ln(x)+1)-2x=x(x^{x-1}(\ln(x)+1)-2)$. Let $u(x)=x^{x-1}(\ln(x)+1)$. We can check that $u'(x)=x^{x-1}((\ln(x)+1)^2-\ln(x)/x)>0$ on $(0,1)$ since $\ln(x)<0$ on $(0,1)$. Thus $u(x)$ is increasing on $(0,1)$ and $u(x)\le u(1)=1$ for $x\in (0,1)$. Thus $h'(x)=x(u(x)-2)<0$ for $x\in (0,1)$, which shows $h(x)$ is decreasing on $(0,1)$.
Edit 3: A sketch about the proof $g(x)=x^x((1+\ln(x))^2+1/x)$ is concave up on $(0,2)$. We need to show that $g''(x)>0$. Denote $A(x)=x^x$ and $B(x)=((1+\ln(x))^2+1/x)$, so that $g(x)=A(x)B(x)$. We use the formula
$$g''(x)=(AB)''=A''B+2A'B'+AB''.$$
Note that $A''=AB$ and thus $A''B=AB^2>0$. Denote $C=1+\ln(x)$. We have $A'=AC$. Thus $2A'B'+AB''=A(2B'C+B'')$. Thus it suffices to show that $2B'C+B''>0$ on $(0,2)$. One can check that
$$2B'C+B''=x^{-3}(4x^2(\ln(x))^2+4x(2x-1)\ln(x)+4x^2-2x+2).$$
It is not hard to show the last expression is positive on $(0,2)$. The details are omitted. Note that $g''(x)>0$ is equivalent to the 4th-derivative of $x^x$ is positive on $(0,2)$. I suspect that there should be a simpler proof.
| {
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Prove the identity $\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}$ Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})^2}=34-2\sqrt{289-120}=34-2\sqrt{169}=34-2.13=8=(2\sqrt2)^2, $$
so $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
In the hints the authors have written that I should use the fact that the LHS is positive and square it. What would be the problem if it wasn't positive? The identity obviously won't hold because LHS<0, RHS>0...
| both of your original real numbers are roots of
$$ x^4 - 34 x^2 + 169 $$
Standard bit for quartic with no cubic term and no linear,
$$ (x^2 - 13)^2 - 8 x^2 = x^4 - 34 x^2 + 169 $$
$$ (x^2 - 13)^2 - (x \sqrt 8)^2 = x^4 - 34 x^2 + 169 $$
This becomes ( because a difference of squares)
$$ (x^2 - x \sqrt 8 - 13)(x^2 + x \sqrt 8 - 13) $$
so that your numbersare two out of four values
$$ \frac{\pm \sqrt 8 \pm \sqrt {60}}{2} $$
or
$$ \pm \sqrt 2 \pm \sqrt {15} $$
This leads to checking
$$ (\sqrt 2 + \sqrt {15} )^2 = 17 + 2 \sqrt{30} $$
along with
$$ (-\sqrt 2 + \sqrt {15} )^2 = 17 - 2 \sqrt{30} $$
Your original expression is equal to
$$ (\sqrt 2 + \sqrt {15} ) - (-\sqrt 2 + \sqrt {15} ) $$
| {
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
| Using the Quadratic Formula is more general but here is an alternative method:
We have $abx^2-(a^2+b^2)x+ab=0 \iff x^2-\frac{a^2+b^2}{ab}x+1=0 \iff x^2-(\frac{a}{b}+\frac{b}{a})x+1=0$.
Let $r_1$ and $r_2$ be the (possibly equal, possibly complex) roots of the quadratic function $x^2-(\frac{a}{b}+\frac{b}{a})x+1$. Then their product is $1$ and their sum is $\frac{a}{b}+\frac{b}{a}$. This shows that the roots are $\frac{b}{a}$ and $\frac{a}{b}$.
Therefore, $x^2-(\frac{a}{b}+\frac{b}{a})x+1=(x-\frac{a}{b})(x-\frac{b}{a})$ so
$$abx^2-(a^2+b^2)x+ab = ab(x-\frac{a}{b})(x-\frac{b}{a}) = (ax-b)(bx-a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| COMMENT.-For $x=y$ one has $2x^3=44x^2$ so $x=0$ and $x=22$. Besides we have
$(x+y)^3-3xy(x+y)=(x+y)^2+40xy\iff S^3-3SP=S^2+40P$ where $S$ is sum and $P$ is product. Note that if $(x,y)$ is solution so is $(y,x)$.
We have $S|40P$, and $P=\dfrac{S^3-S^2}{3S+40}$. We consider first $S|40$ so $S=1,2,4,5,8,10,20,40$.
$$S=1\Rightarrow 43P=0\Rightarrow (x,y)=(1,0)\\S=2\Rightarrow4=46P\\S=4\Rightarrow P=\frac{48}{52}\\S=5\Rightarrow P=\frac{100}{55}\\S=8\Rightarrow P=\frac{448}{64}=7\Rightarrow X^2-8X+7=0\Rightarrow (x,y)=(7,1)$$
$S=10,20,40$ gives $P=\dfrac{900}{70},76,390$ respectively and the two last integers give $x=10\pm\sqrt{24}$ and $x=20\pm\sqrt{10}$.It remains to look at other possibilities
Thus we have found the solutions $(0,0),(22,22),(1,0),(7,1)$
given above by other users. The other solution is given taking $S=-5$ which gives $P=\dfrac{-150}{25}=-6$ from which the equation $X^2+5X-6=0$ so $(x,y)=(-6,1)$.
Here we want just to give another way of calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
MGF of standard normal raised to a power Let $Z \sim \mathcal N(0,1)$ be a standard normal variable. What is the moment generating function $M_{Z^d}(t)$ for the variable $Z^d$ where $d$ is a positive integer?
For $d = 1$, we have $M_Z(t) = e^{t^2/2}$, and for $d = 2$, thinking of the variable as Chi-Squared with $1$ degree of freedom, we have that $M_{Z^2}(t) = \frac{1}{\sqrt{1 - 2t}}$. Is computation of $M_{Z^d}(t)$ tractable for general $d$?
| It is worth noting that the moment-generating function $M_{Z^d}(t)$ of $Z^d$ corresponds to the moment-generating function of the Chi-square distribution for $d=2$. Furthermore, for all other $d>2$ where $d$ is even, it is only finite if $t \leq 0$, whereas it is infinite for $t>0$. This is because
\begin{align}
\mathbb{E} \left[e^{Z^dt} \right] &= \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}} e^{Z^d t -\frac{1}{2}Z^2} dZ \\
&\ge \frac{1}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} e^{Z^d t -\frac{1}{2}Z^2}dZ+ \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{-\frac{1}{2t}^{\frac{1}{d-2}}} e^{Z^d t -\frac{1}{2}Z^2}dZ \\
& \ge \frac{1}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} 1 dZ+ \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{-\frac{1}{2t}^{\frac{1}{d-2}}} 1 dZ \\ &= \frac{2}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} 1 dZ = \infty,
\end{align}
which holds because $Z^d t -\frac{1}{2}Z^2 \geq 0$ for all $t>0$ and $Z$ with $\vert Z\vert \geq \left(\frac{1}{2t}^{\frac{1}{d-2}}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings? 3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings?
Finding the total different ways is easy, just $\binom{9}{3} * \binom{6}{3} * \binom{3}{3}$.
Then I named them, so $ A_1, A_2, A_3, B_1,B_2, B_3, C_1, C_2, C_3 $
I'll fast forward casework, these are the cases we don't want;
$[ABC]-[AAB]-[BCC]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[BBC]-[CAA]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[CCA]-[ABB]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[ABC]-[ABC]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * \binom{2}{1} * \binom{2}{1} = 3^3 * 2^3$
In total, $3 * 3^4 * 2^2 + 3^3 * 2^3 = 3^3 * 2^2 * (3^2 + 2)$
So the probability of not happening is: $\frac{3^3 * 2^2 * (3^2+2)}{\binom{9}{3} * \binom{6}{3}} = \frac{99}{140}$
Finally, $1 - \frac{99}{140} = \frac{41}{140}$,
This was the solution, and the thing is I just hate casework, there is always the feeling of missing something. I was looking for if there was an another elegant solution that uses something/or a technique that are generally applicable, other than casework?
| I'm assuming that the tents are indistinguishable and can contain $\leq3$ children.
First put the children family-wise into the tents, so that we have the three tents
$$(A_1,A_2,A_3),\quad(B_1,B_2,B_3),\quad(C_1,C_2,C_3)\ .$$
No we can make certain permutations, moving $\leq1$ children from each tent:
*
*Leave it as it is. Makes $1$ case.
*Make one transposition. Choose the two involved tents and one child in each of these. These two children will be interchanged. Makes $3\cdot3^2=27$ cases.
*Make a cyclic permutation. Choose one child in each tent and the direction of the circuit. Makes $3^3\cdot2$ cases.
It follows that there are $1+27+54=82$ admissible placements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When we express $\sin x - \cos x$ as $A \sin (x+c)$, how many solutions are there for $c \in [0, 2\pi)$? This is a problem from problem set 1 of MIT OCW 18.01SC:
express $\sin x - \cos x$ in the form $A \sin (x+c)$.
Their solution is $\sqrt{2} \sin (x - \frac{\pi}{4})$
I found two solutions (for $c \in [0, 2\pi]$):
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
Are there two solutions or is there something amiss with the second solution above?
Here is how I got the solutions:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$
$$A^2 \sin^2 c = 1$$
Sum the two equations
$$A^2(\cos^2 x + \sin^2 x) = 2$$
$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$
$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
| As you wrote, $$A\sin(x+c)=A\sin x\cos c+A\cos x\sin c=\sin x-\cos x$$ leads, by identification, to the system
$$\begin{cases}A\cos c=1,\\A\sin c=-1.\end{cases}$$
This is a Cartesian-to-polar transformation, which can be represented graphically. The point is uniquely determined, and if you assume a positive modulus, the azimuth is also uniquely determined in $[0,2\pi)$ or $(0,2\pi]$.
If you admit a negative modulus (which is unusual), you compensate by a phase shift of $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\lim \sqrt{1-\frac{1}{n}} = 1$ by definition Is my attempt correct?
Proof. Let $\varepsilon > 0$. Take $N > \frac{1}{\varepsilon}$ and let $n \geq N$. Then
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lvert \frac{\sqrt{n-1} - \sqrt{n}}{\sqrt{n}} \right\rvert\\\\
&= \left\lvert \frac{(\sqrt{n-1} - \sqrt{n})(\sqrt{n-1} + \sqrt{n})}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{(n-1)-n}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \left\lvert \frac{-1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})} \right\rvert\\\\
&= \frac{1}{\sqrt{n}(\sqrt{n-1} + \sqrt{n})}\\\\
& < \frac{1}{\sqrt{n}(\sqrt{n})}= \frac{1}{n} \leq \frac{1}{N} < \varepsilon\end{align}$
| As the function values are smaller than $1$, we have to solve the inequation
$$1-\sqrt{1-\frac1n}<\epsilon$$ (we can assume $\epsilon\le1$ because for larger $\epsilon$ the inequation is always true), or
$$1-\epsilon<\sqrt{1-\frac1n},$$
$$(1-\epsilon)^2<1-\frac1n,$$
$$\frac1n<(2-\epsilon)\epsilon,$$
$$n>\frac1{(2-\epsilon)\epsilon}\ge\frac1{2\epsilon}$$
and this gives us a lower bound on $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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