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What's the sum of this series? I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?
Thanks!
| The sum of this series is $\frac{\pi^2}{12}$.
Explantion
We already know that:
$$1+ \frac{1}{2^2} + \frac{1}{3^2} + \text{...} = \frac{\pi^2}{6}$$
HINT
Note that:
$$\large \frac{-1}{2^2} = \frac{1}{2^2} - \frac{1}{2^2} - \frac{1}{2^2}$$
Now,
$$1- \frac{1}{2^2} + \frac{1}{3^2} - \text{...} = \left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right) - 2 * \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ...\right)$$
$$ = \frac{\pi^2}{6} - \frac{2}{2^2}\left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right)$$
$$ = \frac{\pi^2}{6} - \frac{1}{2}*\frac{\pi^2}{6}$$
$$= \frac{\pi^2}{12}$$
Comment if you have questions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)$ I am having a tough time with these TYPE of problems
looking forward ideas, All ideas will be appreciated
| Here is another approach:
$\sqrt{x^2+x+1} - \sqrt{x^2-x} = x (\sqrt{1+{1 \over x}+{1 \over x^2}} - \sqrt{1-{1 \over x}} )$.
For $\delta$ sufficiently small, we have $|\sqrt{1 + \delta}-(1+{\delta \over 2}) | \le |\delta|^2 $ (this follows from the Taylor expansion of $\delta \mapsto \sqrt{1+\delta}$, which is $1+{1 \over 2} \delta - {1 \over 8} \delta^2 + \cdots$).
Hence we have
$|\sqrt{1+{1 \over x}+{1 \over x^2}} - (1+ {1 \over 2}({1 \over x}+{1 \over x^2}))| \le |{1 \over x}+{1 \over x^2}|^2 $ and
$|\sqrt{1-{1 \over x}}-(1- {1 \over 2} {1 \over x})| \le | {1 \over x} |^2 $.
Consequently, we have
$|\sqrt{1+{1 \over x}+{1 \over x^2}} -\sqrt{1-{1 \over x}}-({1 \over x}+{1 \over 2} {1 \over x^2})| \le |{1 \over x}+{1 \over x^2}|^2 + | {1 \over x} |^2 $.
Now multiply by $x$ to get
$|x(\sqrt{1+{1 \over x}+{1 \over x^2}} -\sqrt{1-{1 \over x}})-(1+{1 \over 2} {1 \over x})| \le x(|{1 \over x}+{1 \over x^2}|^2 + | {1 \over x} |^2 ) $.
Now let $x \to \infty$, and we see that the limit is $1$.
| {
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Sum of a series problem involving cubes For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$
How would you go about solving such a problem?
Any help would be appreciated.
| Just split the problem up into "odd" and "even" cases and use Faulhaber's formula (this is perhaps the most straightforward way); e.g.:
$$\begin{align}
7^3-6^3+5^3-4^3+3^3-2^3+1^3 \\
&= (7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\
&=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\
&=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2((2^3)(3^3+2^3+1^3)) \\
&=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-16(3^3+2^3+1^3)
\end{align}$$
Alternatively, notice that if we define:
$$A_n = \sum_{k=1}^n{((-1)^{n-k} \cdot k^3)}$$
for the terms in a sequence, then:
$$A_n + A_{n+1} = (n+1)^3$$
It goes kind of like this: Define $f(x)$ to be:
$$f(x) = \sum_{k=0}^\infty{(A_kx^k)}$$
Then, calculate $(x+1)f(x)=xf(x)-f(x)$ (as a rational function) by applying:
$$A_n + A_{n+1} = (n+1)^3$$
This gives the generating function of $f(x)$ (it is a rational function). Expand it using partial fractions; each partial fraction is a geometric series or a power (i.e., derivative) of a geometric series.
Express each partial fraction as a simple series, add them, and equate coefficients to $A_n$. And you get the answer!!!
See the sequence at OEIS, for example.
The answer is:
$$\frac{1}{2}n^3 + \frac{3}{4}n^2-\frac{1}{8}+\frac{1}{8}(-1)^n$$
| {
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Checking a solution to an indefinite integral I recently did some work to try to find $\int{\frac{dx}{Ax^3 - B}}$, but I'm always paranoid that my solution has some minor trivial error in the middle of the process that screwed up the end result entirely, so could someone please help check my solution?
The first step to my solution is to eliminate $A$ and $B$ from the integrand by pushing them out as constants and leaving $\frac{dx}{x^3 - 1}$. We start with
$$
\begin{align}
\int{\frac{dx}{Ax^3 - B}} & = \int{\frac{dx}{A(x^3 - B/A)}} \\
& = \frac{1}{A}\int{\frac{dx}{x^3 - B/A}}.
\end{align}
$$
To get rid of the $B/A$ term, we make the substitution $x = \sqrt[3]{B/A}u, \ dx = \sqrt[3]{B/A}du$.
$$
\begin{align}
\frac{1}{A}\int\frac{dx}{x^3 - B/A} & = \frac{1}{A}\int{\frac{\sqrt[3]{B/A}du}{(B/A)u^3 - B/A}} \\
& = \frac{1}{A}\cdot\frac{\sqrt[3]{B/A}}{B/A}\int{\frac{du}{u^3 - 1}} \\
& = \mathcal{C}\int{\frac{du}{u^3 - 1}}
\end{align}
$$
with $\mathcal{C} = (AB^2)^{-1/3}$. Now we can just worry about solving $\int{\frac{du}{u^3 - 1}}$.
We can decompose $\frac{1}{u^3 - 1}$ using the fact that $u^3 - 1 = (u - 1)(u^2 + u + 1)$. So
$$
\begin{align}
\frac{1}{u^3 - 1} & = \frac{1}{(u - 1)(u^2 + u + 1)} \\
& = \frac{P}{u - 1} + \frac{Qu + R}{u^2 + u + 1}
\end{align}
$$
Here $P = 1/3$, but $Q$ and $R$ aren't as trivial to find. $u^2 + u + 1$ has no real roots, so I chose to sub in $u = -\sqrt[3]{-1}$ (it was the first root that I found). We then have to solve the equation
$$
(-\sqrt[3]{-1}-1)^{-1} = Q(-\sqrt[3]{-1}) + R
$$
which we can solve by setting $Q = R = [(-1)^{2/3} - 1]^{-1}$ (which will also help when integrating $\frac{Qu + R}{u^2 + u + 1}$ since we can factor $Q$ and $R$ out as a common constant from the numerator).
So
\begin{align}
\mathcal{C}\int{\frac{du}{u^3 - 1}} & = \mathcal{C}\left[\frac{1}{3}\int{\frac{du}{u - 1}} + Q\int{\frac{u + 1}{u^2 + u + 1}du}\right] \\
& = \mathcal{C}\left\{\frac{1}{3}\ln{|u - 1|} + Q\left[\frac{1}{2}\ln{|u^2 + u + 1|} + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2}{\sqrt{3}}u + \frac{1}{\sqrt{3}}\right)} \right] \right\} \\
& = (AB^2)^{-1/3}\left\{\frac{1}{3}\ln{\left|\frac{x}{\sqrt[3]{B/A}} - 1\right|} + \frac{1}{(-1)^{2/3} - 1}\left[\frac{1}{2}\ln{\left|\left(\frac{x}{\sqrt[3]{B/A}}\right)^2 + \frac{x}{\sqrt[3]{B/A}} + 1\right|} \right. \right. \\
& \hspace{15mm} \left. \left. + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2x}{\sqrt{3}\sqrt[3]{B/A}} + \frac{1}{\sqrt{3}}\right)}\right]\right\} + \text{constant}
\end{align}
(I know what $\int{\frac{x + 1}{x^2 + x + 1}dx}$ is from previous problems)
How does it look? Did I do anything severely wrong (I don't feel entirely confident about the partial fractions part)? Any suggestions for how I could get to an answer faster or more efficiently?
| I think that we could make this simpler without involving complex numbers.
Just as rogerl did using partial fractions, we have $$\frac{1}{u^3-1} =\frac{1}{3}\frac{1}{u-1}-\frac{1}{3}\frac{u+2}{u^2+u+1}$$ Now $$\frac{u+2}{u^2+u+1}=\frac{1}{2}\frac{2u+4}{u^2+u+1}=\frac{1}{2}\Big(\frac{2u+1}{u^2+u+1}+\frac{3}{u^2+u+1}\Big)$$ that is to say$$\frac{u+2}{u^2+u+1}=\frac{1}{2}\Big(\frac{2u+1}{u^2+u+1}+\frac{3}{\left(u+\frac{1}{2}\right)^2+\frac{3}{4}}\Big)$$ All pieces can be easily integrated and then $$\int\frac{du}{u^3-1} =\frac{1}{3} \log (1-u)-\frac{1}{6} \log \left(u^2+u+1\right)-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 u+1}{\sqrt{3}}\right)+C$$
| {
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Getting the correct answer for a limit with a cube in the denominator Given:
$$
\lim_{x \to 1} f(x) = \frac3{1-x^3} - \frac1{1-x}
$$
I first used the difference of cubes to get
$$
\lim_{x \to 1} f(x) = \frac3{(1-x)(1+x+x^2)} - \frac1{1-x}
$$
Then multiplied each term by $(1-x)$, cancelling $(1-x)$ in the first term, and making the second term $1$
$$
\lim_{x \to 1} f(x) = \frac3{(1+x+x^2)} - 1
$$
After substitution
$$
\lim_{x \to 1} f(x) = \frac33 - 1
$$
Which equates to $0$. But I know the correct answer is $1$. Where did I slip up?
| You can't multiply each term by $(1-x)$- rather you should factor it out to get:
$\frac{1}{1-x}\frac{3}{(1+x+x^2)-1)}=\frac{1}{1-x}\frac{(2-x-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x+1-x^2)}{(1+x+x^2)}=\frac{1}{1-x}\frac{(1-x)(2+x)}{(1+x+x^2)}$.
Now, cancel, and take the limit- the answer shall be $1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$ In the course of working out the Maclaurin expansions of $e^{-x^2}$ and $cos(x^2)$, I ran into the following nested sum:
$$ \underbrace{ \sum_{a=0}^1 \left( a \sum_{b=0}^{a+1} b \left( \sum_{c=0}^{b+1} c \cdots d \sum_{e=0}^{d+1} \left( e \sum_{f=0}^{e+1} f \right) \right) \right) }_{\text{n sums deep}} $$
To illustrate, for $n=4$ the above will expand to
$$ 1(2(3(4+3+2+1)+2(3+2+1)+1(2+1))+1(2(3+2+1)+1(2+1))) $$
Then from looking at a hand calculation of the first few terms and guessing the pattern, it looks like the above should be equal to
$$ \prod_{k=1}^n (2k-1) $$
Indeed, I numerically confirmed equality up to $n=16$, but haven't had luck at finding a proof yet.
Can you find either a proof or a counterexample?
| Returned to the problem and things became clear. Thank you to almagest for the inspiration to consider his recursive function.
To keep everything in one spot, I'll start from the beginning defintions:
Let $S \colon \mathbb{N}^{2} \to \mathbb{N}$ and define as follows:
$$ \begin{align}
S(0,0) & \equiv 0 \\
S(0,k) & \equiv 0 \\
S(a,0) & \equiv 1 \label{eq_Sa0}\tag{1}\\
S(a,k) & \equiv \sum_{n_{1}=0}^{a} n_{1} \sum_{n_{2}=0}^{n_{1}+1} + ... n_{k-1} \sum_{n_k=0}^{n_{k-1}+1}
\end{align} $$
Then we can recursively define $S$ as follows:
$$ S(a,k+1) = \sum_{n=0}^{a} n S(n+1,k) $$
We wish to show the following since the case $a=1$ reduces to my original question (to be shown below):
$$
S(a,k) = \frac{1}{2^{k}k!} \prod_{p=0}^{2k-1} (a+p)
\label{eq_main}\tag{2} $$
Before we continue we will need to make use of the following lemma (pay close attention to the index $k$):
$$ \sum_{n=0}^{a} \prod_{p=0}^{k-1} (a+p) = \frac{1}{k+1} \prod_{p=0}^{k} (a+p) \label{eq_lemma}\tag{3} $$
To prove \ref{eq_lemma} we will proceed by induction. First notice that \ref{eq_lemma} is trivially true for all $k \in \mathbb{N}$ when $a=0$:
$$ \begin{align}
\sum_{n=0}^{0} \prod_{p=0}^{k-1} (a+p) & = \prod_{p=0}^{k-1} p = 0 \\
& = \frac{1}{k+1} \prod_{p=0}^{k} (0+p)
\end{align} $$
Since we know know \ref{eq_lemma} is true for all $k \in \mathbb{N}$ with at least some given $a \in \mathbb{N}$ then
$$ \begin{align}
\sum_{n=0}^{a+1} \prod_{p=0}^{k-1} (a+p) & = \sum_{n=0}^{a} \prod_{p=0}^{k-1} (a+p) + (a+1)(a+2) \cdots (a+k) \\
& = \frac{1}{k+1} \prod_{p=0}^{k} (a+p) + (a+1)(a+2) \cdots (a+k) \\
& = (a+1)(a+2) \cdots (a+k)\left(\frac{a}{k+1} + 1\right) \\
& = \frac{1}{k+1}(a+1)(a+2) \cdots (a+k)[a + (k+1)] \\
& = \frac{1}{k+1} \prod_{p=0}^{k} (a + 1 +p)
\end{align} $$
Thus quod erat demonstrandum for Lemma \ref{eq_lemma}.
Now let's return to proving \ref{eq_main} and proceed inductively from the trivial case $k=1$ which is clearly true for all $k \in \mathbb{N}$ since by definition we know \ref{eq_Sa0}:
$$ \begin{align}
S(a,1) & = \sum_{n=0}^{a} n S(a+1,0) \\
& = \sum_{n=0}^{a} n \\
& = \frac{1}{2} a(a+1)
\end{align} $$
Thus we know that \ref{eq_main} is true for all $a \in \mathbb{N}$ for some particular $k \in \mathbb{N}$. So we proceed by induction on $k$, keeping in mind Lemma \ref{eq_lemma} from above:
$$ \begin{align}
S(a,k+1) & = \sum_{n=0}^{a} n S(n+1,k) \\
& = \sum_{n=0}^{a} n \frac{1}{2^{k}k!} \prod_{p=0}^{2k-1} (n + 1 +p) \\
& = \frac{1}{2^{k}k!} \sum_{n=0}^{a} \prod_{p=0}^{2k} (n + p) \\
& = \frac{1}{2^{k}k!} \left[\frac{1}{2(k+1)} \prod_{p=0}^{k+1} (a+p)\right] \\
& = \frac{1}{2^{k+1}(k+1)!} \prod_{p=0}^{k+1} (a+p)
\end{align} $$
And Q.E.D. for equation \ref{eq_main}. Now all the remains to be shown is that \ref{eq_main} reduces to the question in the original post when $a=1$:
$$ \begin{align}
\sum_{n_{1}=0}^{1} n_{1} \sum_{n_{2}=0}^{n_{1}+1} + ... n_{k-1} \sum_{n_k=0}^{n_{k-1}+1} &= S(1,k) \\
& = \frac{1}{2^{k}k!} \prod_{p=0}^{2k-1} (p+1) \\
& = \frac{1 \cdot 2 \cdot 3 \cdots (2k-1)2k}{2 \cdot 4 \cdot 6 \cdots (2k-2)2k} \\
& = \prod_{p=1}^{k} (2p-1)
\end{align} $$
Q.E.D.
Thanks again, almagest, for the inspiration!
| {
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What is the correct way to compare to algebraic quantities? If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other?
For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} < \frac{2ds}{s^2-c^2}$ for all positive $d$, $s$, and $c$ I know that I can set them "equal" and treat them as an equation like so:
$$\frac{2d}{\sqrt{s^2-c^2}}=\frac{2ds}{s^2-c^2}$$
$$\frac{2d}{\sqrt{s^2-c^2}}\cdot\left(s^2-c^2\right) = \frac{2ds}{s^2-c^2}\cdot\left(s^2-c^2\right)$$
$$2d\sqrt{s^2-c^2}=2ds$$
$$\sqrt{s^2-c^2}=s$$
$$s^2-c^2=s^2$$
$$0<c^2$$ because we are given $c>0$.
Then you can replace all the equals signs all the way up with less thans since I didn't multiply by $-1$. This doesn't strike me as particularly formal, so what is the formal way to accomplish this, and in addition, is this even a valid line of reasoning?
| After messing around by doing scratch work like what you've done, I like to prove what I originally wanted by using a single chain of operations:
\begin{align*}
\frac{2d}{\sqrt{s^2-c^2}}
&= \frac{2d}{\sqrt{s^2-c^2}} \cdot \frac{\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}} \\
&= \frac{2d\sqrt{s^2-c^2}}{s^2-c^2} \\
&< \frac{2d\sqrt{s^2 + 0}}{s^2-c^2} &\text{since } c > 0 \iff c^2 > 0 \iff -c^2 < 0\\
&= \frac{2ds}{s^2-c^2} &\text{since } s > 0\\
\end{align*}
| {
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Find coefficient of $x^3$ in (2+x) ^(3/2)/(1-x) I can expand $\dfrac{(2+x)^{3/2}}{1-x}=(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}(x+1)+{3/2\choose2}(x+1)^2+\ldots\right)$, but that doesn't seem to lead anywhere.
| Your idea is right but it helps to factor out the $2$ in the numerator so the binomial expansion is in $x$ instead of $x+1$:
\begin{eqnarray*}
\dfrac{(2+x)^{3/2}}{1-x} &=& \dfrac{2^{3/2}(1+\frac{x}{2})^{3/2}}{1-x} \\
&=& 2^{3/2}(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}\frac{x}{2}+{3/2\choose2}\left(\frac{x}{2}\right)^2+{3/2\choose3}\left(\frac{x}{2}\right)^3+\ldots\right).
\end{eqnarray*}
Multiplying out and gathering the $x^3$ terms we get the $x^3$ coefficient:
\begin{eqnarray*}
x^3\mbox{ coeff.} &=& 2^{3/2}\left(1+\frac{3}{2}\cdot\frac{1}{2} + \dfrac{\frac{3}{2}\cdot\frac{1}{2}}{2} \cdot\frac{1}{2^2} + \dfrac{\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{-1}{2}}{2\cdot3} \cdot\frac{1}{2^3}\right) \\
&=& \dfrac{2^{3/2}}{128}\left(128+96+12-1\right) \\
&& \\
&=& \dfrac{235\sqrt{2}}{64}
\end{eqnarray*}
| {
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Integrate $\int\frac{x^3}{(x^3+1)^2}dx$ any ideas on how I could continue this integral
$$\int\frac{x^3}{(x^3+1)^2}dx$$
I am half way done, by entirely using fraction decomposition
$\int\frac{x^3}{(x^3+1)^2}dx=\int\frac{1}{x^3+1}dx-\int\frac{1}{(x^3+1)^2}dx$
$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$ from which it follows that $A=\frac{1}{3}, B=-\frac{1}{3}, C=\frac{2}{3}$
$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$
$\int\frac{1}{3(x+1)}dx=\frac{1}{3}\ln|{x+1}|+C$
$\int\frac{x-2}{3(x^2-x+1)}dx=\frac{1}{3}\left(\int\frac{2x-1}{2(x^2-x+1)}dx-\int\frac{3}{2(x^2-x+1)}dx\right)=\frac{1}{3}\left(\frac{1}{2}\ln|{x^2-1+1}|-\frac{3}{2}\frac{2}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})\right)+C$
$\int\frac{dx}{(x^3+1)^2}=?$
This is the part where I am stuck currently, using fraction decomposition seems to take a lot of time and effort, could I be missing anything?
| The polynomial $x^3+1$ is not irreducible (as you have seen),
$$x^3+1=(x+1)(x^2-x+1).$$
You need to continue decomposing in order to compute the integral.
| {
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Calculus - Derivatives and Tangent Lines I am trying to find the equation of the lines (there's 2) tangent to the graph of $f(x) = x^3$ and parallel to the line $12x − y + 1 = 0$.
So far obviously the slope of the line must be $12$.
The equation of the line tangent to $f(x) = x^3$ is the derivative and must be equal to the slope of the intersecting line, $12$. The derivative of $x^3$ is equal to $2x^2$, so $2x^2 = 12$, and $x=$ $\pm\sqrt{6}$, so $y$ must equal to $\pm6\sqrt{6}$ and thus the equation of the line is $y=12x-6\sqrt{6}$ or $y=12x+6\sqrt{6}$. Is this correct? Thanks.
EDIT
I think I made a mistake it should be $y=12x-5\sqrt{6}$ or $y=12x+5\sqrt{6}$
EDIT again
Thanks to everyone who helped out!
The derivative is equal to $3x^2$ not $2x^2$, so $x =\pm 2$ and so $y =\pm 8$ so plugging back into the equation gives $y=12x-16$ or $y=12x+16$, right?
| Yes, the slope of our line(s) needs to be $12$ to ensure they are parallel to the line
$\,12x − y + 1 = 0.$
Next, finding the derivative of $y = x^3$ gives us $\,y' = 3x^2$.
To find the points on the curve at which the tangent lines at those points have slope $12$, we solve $$3x^2 = 12 \iff x^2 = 4 \iff x =\pm 2.$$
Following your edits: Yes, you are correct.
There are two lines tangent to the curve $y = x^3$ with slope $12$. One point at which a line with slope $12$ is tangent is given by $(-2, -8)$ and the other point at which a line with slope $12$ is tangent to the curve is given by $(2, 8)$.
So you have slope $m = 12$, and a point $(x_0, y_0)$ for each of two such tangent lines.
$$(y - y_0) = m(x - x_0)$$
Hence, one line is given by $$y + 8 = 12(x + 2) \iff y = 12 x + 16$$
The other line is given by $$y - 8 = 12(x -2) \iff y = 12x - 16$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
can you show the solution for this in finding the value of a? $$\frac{a^2+4^2-5^2}{8a} = \frac{\sqrt{3}}{2}\frac{a^2+4^2-6^2}{8a}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+4^2-6^2}{8a}\Big)^2}$$
show the solution in finding the value of A
| Let
$$\sin(\theta) = \frac{a^2 + 4^2 - 6^2}{8a}.$$
Then we have:
$$\sin(\theta) + \frac{11}{8a} = \frac{\sqrt{3}}{2}\sin(\theta) + \frac{1}{2}\cos(\theta).$$
Simplifying, we have:
$$\sin(\theta) + \frac{11}{8a} = \sin\left(\theta + \frac{\pi}{6}\right).$$
Consequently:
$$\sin\left(\theta + \frac{\pi}{6}\right) - \sin(\theta) = \frac{11}{8a}.$$
Simplifying the left-hand side of the last equation, we get:
$$\sqrt{2 - \sqrt{3}}\cos\left(\theta + \frac{\pi}{12}\right) = \frac{11}{8a}.$$
Therefore:
$$\cos\left(\theta + \frac{\pi}{12}\right) = \frac{11\left(\sqrt{2 + \sqrt{3}}\right)}{8a}.$$
Hence:
$$\theta = \arccos\left(\frac{11\left(\sqrt{2 + \sqrt{3}}\right)}{8a}\right) - \frac{\pi}{12}.$$
Now plug $\theta$ into:
$$\sin(\theta) = \frac{a^2 - 20}{8a},$$
and then solve for $a$.
Using WolframAlpha,
$$a \approx -0.798038 + 0.540401i,$$
or
$$a \approx -0.798038 - 0.540401i.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\frac{w^2}{2}\text{ and if }x \to \sqrt{2} \text{ then }w\to 2.$$
Is that correct?
| Your work is correct and helps in removing square roots. But you can dispense with the change of variables.
Let $f(x)=x^2+x\sqrt{2}-4$; then $f(\sqrt{2})=0$, so you know $\sqrt{2}$ is a root of the polynomial. Then the usual scheme (you may be used to a different one)
$$
\begin{array}{r|rr|r}
& 1 & \sqrt{2} & -4 \\
\sqrt{2} & & \sqrt{2} & 4\\
\hline
& 1 & 2\sqrt{2} & 0
\end{array}
$$
tells you $x^2+x\sqrt{2}-4=(x-\sqrt{2})(x+2\sqrt{2})$. Similarly, $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$, so you can factor out $x-\sqrt{2}$ from the numerator and denominator.
Alternatively, use l'Hôpital's rule (maybe for checking if you're not officially allowed to use it):
$$
\lim_{x \to\sqrt{2}}
\dfrac{x^{2}-2}{x^{2}+\sqrt{2}x-4}
\overset{\text{(H)}}{=}
\lim_{x\to\sqrt{2}}
\dfrac{2x}{2x+\sqrt{2}}=\frac{2}{3}
$$
| {
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"url": "https://math.stackexchange.com/questions/926507",
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"question_score": "2",
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"answer_id": 2
} |
Limit of a product I need to find the value of $$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$
Is doing this OK?--
$$\begin{align} L &=\lim_{n \rightarrow \infty}\left(1+\dfrac{1^2}{n^2}\right)^{1/n}\left(1+\dfrac{2^2}{n^2}\right)^{1/n} \ ... \ \left(1+\dfrac{n^2}{n^2}\right)^{1/n} \\ &=\lim_{n \rightarrow \infty} e^{1^2/n^3} e^{2^2/n^3} \ ... \ e^{n^2/n^3} \\ &= \lim_{n \rightarrow \infty} e^{n (n+1)(2n+1)/6n^3}\\ &=e^{1/3}
\end{align}$$
I get $L=e^{1/3}$ but the answer is $2e^{\pi/2-2}$.
Where have I gone wrong? How to arrive at the correct answer $2e^{\pi/2-2}$?
Thank you.
| $$L=\lim_{n \rightarrow \infty}\displaystyle\prod_{r=1}^{n} \left(1+\dfrac{r^2}{n^2}\right)^{1/n}$$
$$\ln L=\lim_{n \rightarrow \infty}\frac1n\sum_{r=1}^n\ln\left(1+\dfrac{r^2}{n^2}\right)=\int_0^1\ln(1+x^2)dx$$
$$\int\ln(1+x^2)dx=\ln(1+x^2)\int dx-\int\left(\frac{d[\ln(1+x^2)]}{dx}\int dx\right)dx$$
$$=x\ln(1+x^2)-\int\frac{2x^2}{1+x^2}dx$$
$$=x\ln(1+x^2)-\int\frac{2(x^2+1)-2}{1+x^2}dx$$
$$=x\ln(1+x^2)-2\int dx+2\int\frac1{1+x^2}dx$$
| {
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} |
Rational Expression Simplification The function:
$$f(x) = \frac{3x - 4}{x^2 - 2x}$$
is simplified to:
$$f(x) = \frac{2}{x} + \frac{1}{x - 2}$$
How? And in what way?
| $$f(x)=\frac {3x-4}{x^2-2x}\\
=\frac {3x-4}{x(x-2)}\\
=\frac Ax +\frac B{x-2}\\
=\frac {A(x-2)+Bx}{x(x-2)}$$
Equation numerators:
$$3x-4=A(x-2)+Bx\\
=(A+B)x-2A$$
gives $A=2,B=1$, i.e.
$$f(x)=\frac {3x-4}{x^2-2x}=\frac 2x +\frac 1{x-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$ 2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$
I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems with taking the integration with respect to z because my change of variables is causing me to have a wicked exponential.
Anyway, here is my attempt
The slope is $\frac{2}{1}$ which is is 2, so my characteristic lines must be positive.
The characteristic lines are in the form of $bx-ay=d$ which is for the pde $au_x+bu_y+cu=f(x,y)$
Using my change of variables, I get the following lines
$ 2x - y = w \rightarrow x=\frac{w+z}{2}$
$z = y \rightarrow y = z$
Now I have to take the partial derivatives and the chain rule.
$W_x = 2, W_y = -1, Z_x = 0, Z_y = 1$
$V_wW_x+V_zZ_x+2(V_wW_y+V_zZ_y) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
$V_z -4(v) = e^{\frac{w+z}{2}+z}$
Let $p(a) = -4$, and $q(a) = e^{\frac{w+z}{2}+z}$. Then my integrating factor would be. $e^{\int -4} \rightarrow e^{-4z}$
Multiplying the integrating factor, I get
$e^{-4z}V_z -e^{-4z}4(v) = e^{-4z}e^{\frac{w+z}{2}+z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z}e^{-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}-3z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z-6z}{2}}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w-5z}{2}}$
By reverse product rule
$e^{-4z}(v) = \int e^{\frac{w-5z}{2}}$
Integrating the right hand side, I get
$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$
$ v = \frac{\frac{-2}{5}e^{\frac{w-5z}{2}}}{e^{-4z}}$
$ v = \frac{-2}{5}e^{\frac{w-5z+4z}{2}}$
$ v = \frac{-2}{5}e^{\frac{w-z}{2}} + C(w)$
Substituting back I get
$ u = \frac{-2}{5}e^{\frac{2x-y-y}{2}} +C(2x-y)$
$ u = \frac{-2}{5}e^{\frac{2x-2y}{2}}+C(2x-y)$
from side condition of $u(x,-x) = x$, we have
$x = \frac{-2}{5}e^{\frac{2x-2(-x)}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{2x+2x}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{4x}{2}}+C(x)$
$x = \frac{-2}{5}e^{2x}+C(x)$
The problem is... that's not right because the answer is $u(x,y)=-e^{x+y}+(\frac{2}{3}(x-\frac{1}{2}y+1)e^{\frac{4}{3}(x-\frac{1}{2}y)}e^{2y}$
What happened?
Edit: even if I did rewrite it correctly, the $\frac{-2}{5}$ is still there!!!
$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$
Applying change of variables
$u = \frac{-2}{5}e^{\frac{2x-y-5y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{\frac{2x-6y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x-3y+4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x+y}+C(2x-y)e^{4y}$
| I spotted the following mistake, which invalidates all steps after it: from
$$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$$
you should have gotten
$$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$$
Instead, you omitted $C$, then multiplied by $e^{4z}$, then brought $C$ back in, un-multiplied.
Why don't you do this in a reasonable way?
*
*Particular solution: try $u=ce^{x+y}$, find $ce^{x+y}+2ce^{x+y}-4ce^{x+y}=e^{x+y}$, so $c=-1$.
*Homogeneous equation: $u_x+2u_y-4u=0$. Write $u=Ce^v$, so that $v_x+2v_y-4=0$.
*Solve the equation for $v$ as usual: $v(x,y) = 4x+f(2x-y)$ with arbitrary $f$.
*Collect and clean up: $u(x,y)=-e^{x+y} + e^{4x}h(2x-y)$ is the general solution of original equation.
*Since $u(x,-x)=x$, we have $x=-1+e^{4x}h(3x)$, hence $h(3x)=e^{-4x}(x+1)$. The function $h$ is $h(t) = e^{-4t/3}(t/3+1)$.
*Plug known $h$ into $u$ to get the final answer: $$u(x,y)=-e^{x+y} + e^{4x}e^{-4(2x-y)/3}((2x-y)/3+1)$$
If you want to verify the result of 6, simplify it before taking derivatives.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.
| I think that, in the same spirit as you show in the post, we could do a little faster considering the development of $$1+\sin (x)+a \sin ^2(x)=1+x+a x^2+O\left(x^3\right)$$ Then going to logarithms, series, followed by exponentiation of series and series again $$f(a)=\Big(1+\sin (x)+a \sin ^2(x)\Big)^{1/x}=e+\left(e a-\frac{e}{2}\right) x+O\left(x^2\right)$$ Then, the numerator is just $$f(1)-f(0) \approx e x-ex^2$$ from which the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
What is $\frac{2x}{1-x^2}$ when $x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$? If
$$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
Find
$$\frac{2x}{1-x^2}$$
I got till here by simplification by taking the previous value of x, ie,
$$x={\frac{\sqrt{1-\cos\theta}}{1+\cos\theta}}$$
$$\frac{2\tan\theta\sqrt{1+\cos\theta}}{\cos\theta+3}$$
| $$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
But $\cos2a=2\cos^2a-1=1-2\sin^2a$, so
$$1-\cos\theta=2\sin^2 \frac{\theta}{2}$$
$$1+\cos\theta=2\cos^2 \frac{\theta}{2}$$
Hence
$$x=\left|\tan \frac{\theta}2\right|$$
Therefore
$$\frac{2x}{1-x^2}=\frac{2\left|\tan\frac{\theta}2\right|}{1-\tan^2\frac{\theta}2}=\frac{\left|2\sin\frac{\theta}2\cos\frac{\theta}2\right|}{\cos^2\frac{\theta}2-\sin^2\frac{\theta}2}=\frac{|\sin\theta|}{\cos\theta}$$
Notice that the answer is not $\tan\theta$, when $\sin\theta<0$, that is for
$$\theta \in \bigcup_{k\in\Bbb Z} ](2k-1)\pi,2k\pi[$$
Or, if we remove also the values of $\theta$ for which $\cos\theta=0$,
$$\theta \in \bigcup_{k\in\Bbb Z} ]-\pi+2k\pi,-\pi/2+2k\pi[\;\cup\;]-\pi/2+2k\pi,2k\pi[$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygamma function of order $3$.
Here is my attempt. Using equation (11) from Mathworld Wolfram:
\begin{equation}
\Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right)
\end{equation}
I got
\begin{equation}
\Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right)
\end{equation}
then
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\
&=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
&=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
\end{align}
From the answers of this OP, the integral representation of the latter Euler sum is
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)}
\end{align}
or another simpler form
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\
&=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\
&=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\
\end{align}
I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.
Edit :
Using the integral representation of polygamma function
\begin{equation}
\Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx
\end{equation}
then we have
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\
&=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\
&=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\
\end{align}
I am looking for an approach to evaluate the above integral without using residue method or double summation.
| \begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n}
&=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3)
\end{align}
Let $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_3(-z)}{z}$. Then at the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
&=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{6(-1)^n}{z(z-n)^5}+\frac{6(-1)^n\zeta(2)}{z(z-n)^3}+(-1)^n\frac{(33/2)\zeta(4)+6H_n^{(4)}}{z(z-n)}\right]\\
&=6\sum^\infty_{n=1}\frac{(-1)^n}{n^5}+6\zeta(2)\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{33}{2}\zeta(4)\sum^\infty_{n=1}\frac{(-1)^n}{n}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}\\
&=-\frac{45}{8}\zeta(5)-\frac{9}{2}\zeta(2)\zeta(3)-\frac{33}{2}\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}
\end{align}
At zero,
$${\rm Res}(f,0)=24\zeta(5)$$
At the negative integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,-n)
&=\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n)}{n}\\
&=6\zeta(4)\ln{2}-6\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{n-1}^{(4)}}{n}\\
&=\frac{45}{8}\zeta(5)+6\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}\\
\end{align}
Since the sum of residues is zero,
\begin{align}
12\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}=-24\zeta(5)+\frac{21}{2}\zeta(4)\ln{2}+\frac{9}{2}\zeta(2)\zeta(3)\\
\end{align}
This implies that
\begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n}
&=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3)
\end{align}
Refer to this paper if you have any doubts.
| {
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"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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"answer_id": 2
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Solve the following integration $$\int \sqrt{cot\theta} d\theta$$
I tried to set $t=\sqrt{cot\theta},t^2=cot\theta$ and substitute into the original integration and get$$-\int\frac{2t^2}{1+t^4}dt$$, but then what can I do?
| Consider that $$t^4+1 = (t^4+2t^2+1)-2t^2 = (t^2+1)^2-(\sqrt{2}\,t)^2 = (t^2+\sqrt{2}\,t+1)(t^2-\sqrt{2}\, t+1)$$
hence:
$$\frac{2t^2}{t^4+1}=\frac{t}{\sqrt{2}\, t^2 -2t+\sqrt{2}}-\frac{t}{\sqrt{2}\, t^2 +2t+\sqrt{2}}\tag{1}$$
and the integral of both terms in the RHS is given by:
$$\frac{1}{2\sqrt{2}}\left(\log(\sqrt{2}\, t^2 \pm 2t+\sqrt{2})-2\arctan(1\pm\sqrt{2}\,t)\right).\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
| Another approach is
$\int \frac{y^2-x^2}{(x^2+y^2)^2} dy = \int\frac{y^2+x^2}{(x^2+y^2)^2} dy - \int \frac{2x^2}{(x^2+y^2)^2} dy = \int\frac{1}{x^2+y^2} dy - 2x^2\int \frac{1}{(x^2+y^2)^2} dy$
Once in this form, simple variable substitution for the first one and integration by parts or by $y = tan(u)$ substitution should be very straight forward.
| {
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"url": "https://math.stackexchange.com/questions/936705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$P(X^2+Y^2<1)$ of two independent n(0,1) random variables Suppose that X and Y are independent n(0,1) random variables.
a) Find $P(X^2+Y^2<1)$
Attempt:
a) Let $U = X^2 + Y^2$, $V = Y$.
Then $X = \sqrt{V^2 -U}$, $Y = V$.
$J = \left| \begin{array}{ccc}
\frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\
0 & 1\\
\end{array} \right| $
Then the joint distribution of $f_{u,v}(u,v)$ is:
$$f_{u,v}(u,v)= \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}$$
Then $P(X^2 +Y^2 <1)$ is:
$$\int_0^\infty \int_0^{v^2-u} \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}dudv$$
However, at this point I simply do not know how any tricks to complete this integration.
| Although the following method is more convoluted than necessary, I think it provides a fun way to see the convolution method for computing the density of a sum of independent random variables in action. It also requires a few fun integration tricks.
First we will find the distribution for $X^2$ (and, by symmetry, the distribution of $Y^2$). We consider the cdf $F_{X^2}(t)$, and we see that $$F_{X^2}(t) = P(X^2 \leq t) = P(|X| \leq \sqrt{t}) = P(X \in [-\sqrt{t},\sqrt{t}] = P(X \leq \sqrt{t}) - P(X < \sqrt{t}) $$ Continuing this logic, and noting that $X$ has density $f_X = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, we see that $$ F_{X^2}(t) = F_X(\sqrt{t}) - F_X(-\sqrt{t}) = \displaystyle\int_{-\infty}^{\sqrt{t}} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx - \displaystyle\int_{-\infty} ^{-\sqrt{t}} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx$$ We can then find the density $f_{X^2}$ by differentiating our equation for $F_{X^2}(t)$ with respect to $t$, which gives us $$f_{X^2}(t) = \frac{d}{dt}\left(F_{X^2}(t) \right) = \frac{1}{\sqrt{2 \pi}} e^{- (\sqrt{t})^2 / 2} \left( \frac{1}{2 \sqrt{t}} \right) - \frac{1}{\sqrt{2 \pi}} e^{- (-\sqrt{t})^2 / 2} \left(- \frac{1}{2 \sqrt{t}} \right) $$ and this simplifies to tell us that $$f_{X^2}(t) = f_{Y^2}(t) = \frac{1}{\sqrt{t} \sqrt{2 \pi}} e^{- t/2} $$
Because we know that $X$ and $Y$ are independent, it follows that $X^2$ and $Y^2$ are independent as well. Therefore we can apply the convolution method to find the probability density for $Z = X^2 + Y^2$. We can write the probability distribution as the convolution $$f_Z(z) = \displaystyle\int_{0}^z f_{X^2}(z-y) f_{Y^2}(y) dy $$ where the lower-limit of the integral is 0 because $Y^2$ is a non-negative random variable, and the upper limit is $z$ because it cannot be the case that both $Y^2 = y > z$ and $X^2 < z - y$, as $X^2 $ is a non-negative random variable. Plugging our value for the densities $f_{X^2}(z-y) = \frac{1}{\sqrt{z-y} \sqrt{2 \pi}} e^{-(z-y)/2} $ and $f_{Y^2} = \frac{1}{\sqrt{y} \sqrt{2 \pi}} e^{-y/2}$, we have that $$f_Z(z) = \displaystyle\int_0^z \left( \frac{1}{\sqrt{z-y} \sqrt{2 \pi}} e^{-(z-y)/2} \right) \left( \frac{1}{\sqrt{y} \sqrt{2 \pi}} e^{-y/2} \right) dy = \frac{1}{2 \pi} e^{-z/2} \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}}$$
Now, what we have left to do is to compute the integral on the right. Completing the square under the square root sign we have $$\displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{\frac{z^2}{4} - (\frac{z^2}{4} - zy + y^2) }} = \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{\frac{z^2}{4} - (y - \frac{z}{2})^2 }} $$ Making the u-substitution $u = y - \frac{z}{2}$ (with $du = dy$ and with transformed limits of integration $y = 0 \mapsto u = -z/2$ and $y = z \mapsto z/2$), we obtain the integral $$ \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \displaystyle\int_{-z/2}^{z/2} \displaystyle\frac{du}{\sqrt{\frac{z^2}{4} - u^2}} = \frac{2}{z} \displaystyle\int_{-z/2}^{z/2} \displaystyle\frac{du}{\sqrt{1 - (\frac{2u}{z})^2}} = \frac{2}{z} \left( \left(\frac{1}{\frac{2}{z}} \right) \sin^{-1}(\frac{2u}{z}) \bigg|_{-z/2}^{z/2} \right) $$ This simplifies to $$ \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \sin^{-1}(1) - \sin^{-1}(-1) = \frac{\pi}{2} - (- \frac{\pi}{2}) = \pi $$ and therefore we have that $$f_{Z}(z) = \frac{1}{2 \pi} e^{-z/2} \left( \pi \right) \Longrightarrow \boxed{f_{Z}(z) = \frac{1}{2} e^{-z/2}} $$
And we can compute that $$P(X^2 + Y^2 <1 ) = P(Z < 1) = \displaystyle\int_0^1 f_{Z}(z)dz = \displaystyle\int_0^1 \frac{1}{2} e^{-z/2} dz = - e^{-z/2} \bigg|_0^1 $$ and we obtain our final answer $$ \boxed{P(X^2 + Y^2 < 1) = 1 - e^{-1/2}} $$
| {
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"url": "https://math.stackexchange.com/questions/937972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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In $\triangle ABC$, $D$ is a point on side $BC$ that $\angle BAD = \angle CAD =\angle ABC$. If $BD=1$ and $DC=2$, what would be the length of $AB$?
In $\triangle ABC$, $D$ is a point on side $\overline{BC}$ that $\angle BAD = \angle CAD =\angle ABC$. If $\overline{BD}=1$ and $\overline{DC}=2$, what would be the length of $\overline{AB}$ ?
Things I have done: As $\overline{AD}$ is angle bisector this was the first thing came into my mind $$\frac{\overline{BD}}{\overline{DC}}=\frac{1}{2}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\overline{AC}}{2}=\overline{AB}$$
Applying Law of sines gives $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\sin{ \angle C}}{\sin{\angle B}}\times \overline{AC} = \overline {AB}$$
thus $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{1}{2}$$
taking $\angle B = x$ results $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\sin{(180-3x)}}{\sin{(x)}}=\frac{\sin{(3x)}}{\sin{(x)}}=\frac{1}{2}$$
And i stuck here. I solved this question previously by adding some elements and using symmetry and ... . but now I want to solve it without adding any element and constructions.Thanks.
Answer According to answer key: $\frac{\sqrt 6}{2}$
| You are correct that the Angle Bisector Theorem gives you that $|AC| = 2|AB|$. Let's take that as our jumping-off point, assigning $x$ and $y$ to lengths as shown:
(I'm using "$x$" instead of "$1$", so that we can follow the value through the formulas better. "$1$"s are so easy to lose!)
By the Law of Cosines in $\triangle ABD$:
$$\begin{align}
y^2 &= x^2 + x^2 - 2 \cdot x\cdot x \cos \delta \\
\implies \qquad y^2&= 2 x^2 - 2 x^2 \cos\delta \qquad (\star)
\end{align}$$
and in $\triangle ACD$:
$$\begin{align}
(2y)^2 &= x^2 + (2x)^2 - 2\cdot x \cdot (2x) \cos(180^\circ-\delta) \\
\implies \qquad 4y^2 &= 5 x^2- 4x^2\cos(180^\circ-\delta) \\
\implies \qquad 4 y^2 &= 5 x^2 + 4 x^2\cos\delta \qquad (\star\star)
\end{align}$$
Then $2(\star) + (\star\star)$ eliminates the $\cos\delta$ terms, so that:
$$ 6 y^2 = 9 x^2 \quad\implies\quad 2 y^2 = 3 x^2 \quad\implies\quad y = \sqrt{\frac32}\;x = \frac{\sqrt{6}}{2}\;x$$
(where we ignore "$\pm$" issues, as we presume $x$ and $y$ to be non-negative).
The fact that the final relation can be written as
$$\frac{\sqrt{3}}{2} x = \frac{\sqrt{2}}{2} y \qquad\text{that is,}\qquad x\sin 60^\circ = y \sin 45^\circ$$
makes me think that there's a more-clever solution, but I don't see it at the moment.
| {
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"url": "https://math.stackexchange.com/questions/938801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this proof by induction for a sum of odd squares correct? Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$
Proof by induction
-Base case: when $n = 1$
$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$
$1=1$ hence statement holds for $n = 1$
-Inductive step
assume $n = k$ is true
then $1^2 + 3^2 + 5^2 + .. + (2k-1)^2 = (k/3) * (2k - 1) * (2k + 1)$
let $n = k+1$
then $1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$
it follows that $1^2 + 3^2 +5^2 .. + (2k-1)^2 = (k/3) * (2k-1) * (2k+1)$
such that $(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2 = ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$
hence: $(4k^3+12k^2 + 11k + 3)/3 = (4k^3+12k^2 + 11k + 3)/3$
thus the statement holds for $n+1$ and the statement is therefore true
Is this proof correct? Am I missing anything or detail that I can add?
| Your line "hence something complicated is equal to itself" is unconvincing and suggests that you might do better with a different order, perhaps like $$1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$$ $$=(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2$$ $$=(4k^3+12k^2 + 11k + 3)/3$$ $$= ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$$ $$= (n/3)*(2n-1)*(2n+1) \text{ when } n=k+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939004",
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"source": "stackexchange",
"question_score": "2",
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Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now:
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:
$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$
| Since $\sin(x-a)\sin(x+a)=\cos^2(a)-\cos^2(x)=\sin^2(x)-\sin^2(a)$, if we let
$$
u=\frac{\cos(x)}{\cos(a)}\quad\text{and}\quad v=\frac{\sin(x)}{\sin(a)}\tag{1}
$$
then
$$
\begin{align}
&\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\,\mathrm{d}x\\
&=\int\frac{\sin(x-a)}{\sqrt{\sin(x+a)\sin(x-a)}}\,\mathrm{d}x\\
&=\int\left[\frac{\sin(x)\cos(a)}{\sqrt{\cos^2(a)-\cos^2(x)}}-\frac{\cos(x)\sin(a)}{\sqrt{\sin^2(x)-\sin^2(a)}}\right]\,\mathrm{d}x\\
&=-\cos(a)\int\frac{\mathrm{d}u}{\sqrt{1-u^2}}-\sin(a)\int\frac{\mathrm{d}v}{\sqrt{v^2-1}}\\[6pt]
&=\cos(a)\cos^{-1}\left(\frac{\cos(x)}{\cos(a)}\right)-\sin(a)\cosh^{-1}\left(\frac{\sin(x)}{\sin(a)}\right)+C\tag{2}
\end{align}
$$
Since $\cosh^{-1}(x)=\log\left(x+\sqrt{x^2-1}\right)$, $(2)$ is equal to
$$
\cos(a)\cos^{-1}\left(\frac{\cos(x)}{\cos(a)}\right)-\sin(a)\log\left(\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right)+C'\tag{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 2
} |
Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.
|
Hopefully one can see anything in this drawing...
You know that all the green and all the blue lines are equal. Additionally, you know that the top green line is parallel to one of the diagonals and that the diagonals are perpendicular. Therefore, the top green line is perpendicular to the short diagonal as well.
From the Pythagorean theorem follows
$$b^2-\left(\frac{b}{2}\right)^2=\left(n-x\right)^2$$
where $n$ is the short diagonal, $a$ is the length we are looking for and $x$ is the orange bit. This equation still has an $x$ in it; we need it in terms of $n$ and $m$.
Using the Intercept theorem, one can infer
$$\frac{x}{\frac{b}{2}}=\frac{\frac{n}{2}}{\frac{m}{2}} \implies x=\frac{nb}{2m}$$
where $m$ denotes the long diagonal.
Algebra:
\begin{align*}\frac{3}{4}b^2=\left(n-x\right)^2&=\left(n-\frac{nb}{2m}\right)^2\\
\frac{\sqrt{3}}{2}b&=n-\frac{nb}{2m}\\
\frac{\sqrt{3}}{2}b+\frac{nb}{2m}&=n\\
b\left(\frac{\sqrt{3}}{2}+\frac{n}{2m}\right)&=n\\
b=\frac{n}{\frac{\sqrt{3}}{2}+\frac{n}{2m}}&=\frac{2mn}{\sqrt{3}m+n}\end{align*}
Substituting the given values:
$$b=\frac{96}{8\sqrt{3}+6}=\frac{48}{4\sqrt{3}+3}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}+3\right)\left(4\sqrt{3}-3\right)}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}\right)^2-3^2}=\frac{16}{13}\left(4\sqrt{3}-3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$ Integrate:
$$
\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}
$$
| Hint:
We can find the anti-derivative,
$$\begin{align}
\int\frac{\mathrm{d}x}{a^2\cos^2{x}+b^2\sin^2{x}}
&=\int\frac{\sec^2{x}\,\mathrm{d}x}{a^2+b^2\tan^2{x}}\\
&=\int\frac{\mathrm{d}u}{a^2+b^2u^2}\\
&=\frac{\arctan{\frac{bu}{a}}}{ab}+\color{grey}{constant}\\
&=\frac{\arctan{\left(\frac{b\tan{x}}{a}\right)}}{ab}+\color{grey}{constant}.
\end{align}$$
Now integrate by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/941172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Why is $\frac 25$ the real part of $\frac{1}{2+i}$? According to Wolfram Alpha, Re(1/(2+i))=2/5.
How did it calculate that?
| Multiply numerator and denominator by $2-i$:
$$\frac 1{2+i}\cdot \frac{2-i}{2-i} = \frac{2-i}{4-(-1)} = \frac{2-i}{5} = \frac 25 - \frac 15\cdot i$$
Edit for more explanation as to why this strategy works.
We multiply numerator and denominator by the conjugate of $2+ i$, which is $2+i$, to remove the imaginary number from the denominator. Doing so gives us a real-valued denominator: $$(2 + i)(2-i) = 2^2 - (i^2) = 4 - (-1) = 5.$$ Having multiplied by $\frac{2-i}{2-i}=1$, that transforms the numerator to $1\cdot (2-i) = 2- i$. Then all that remained was to transform the fraction $\dfrac{2-i}{5}$ into its real and imaginary components.
The same strategy works for any fraction of the form $\dfrac{z_1}{z_2},\,$ where $z_1, z_2 \in \mathbb C$ and $z_2 = a + bi$ where $a, b \in \mathbb R, b\neq 0$. We simply multiply numerator and denominator by $\overline{z_2}$. The denominator is then $z_2\cdot \overline{z_2} = |z| = a^2 + b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/941617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with Equilateral Triangle Randomly choose a point $P$ in the interior of an equilateral triangle $ABC$, which has a side of length $a$. Let $D,E,F$ be the feet of the perpendiculars from $P$ to $BC,CA,AB$ respectively. Show that $(PDC)+(PFB)+(PEA)=\frac{a^2\sqrt3}{8}$.
Here are my thoughts so far:
We want to show $(PDC)+(PFB)+(PEA)=\frac{ABC}{2}$, therefore it suffices to show that $(PDC)+(PFB)+(PEA)=(PDB)+(PFA)+(PEC)$
I believe that this can be shown using the following two relations:
1)$PD+PE+PF=$height of $ABC$ (Viviani's Theorem)
2)$BD+CE+AF=DC+EA+FB=3a/2$
Here is a proof of 2): By the Pythagorean theorem we get
$(a-BD)^2+PD^2=CE^2+PE^2$
$(a-CE)^2+PE^2=AF^2+PF^2$
$(a-AF)^2+PF^2=BD^2+PD^2$
Addition of the previous three equation yields the result.
The problem now is if and how 1) and 2) can lead to a solution.
| Unfortunately, getting a proof from the two equations reported in the OP seems difficult, since the heights of the three triangles are all different.
The proof can be achieved using an alternative method. Orient an equilateral triangle with side $s$ in a way that $A$ is the vertex at the top, and the base $BC$ is at the bottom. As described in the OP, let us choose an internal point $P$, and let us call $D,E,F$ the feet of the perpendiculars from $P$ to $BC$, $AC$, and $AB$, respectively. Consider the height $AD$ normal to $BC$. It is trivial to show that, if the point $P$ is randomly chosen on $AN$, then the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is equal to half of the total area of the triangle $ABC$, since this is divided in three pairs of symmetric triangles.
Keeping the point $P$ on $AD$, let us set $PD=d$, $BF=CE=b$, and $PF=PE=h$. The triangle $PDC$ has base equal to $\frac{s}{2}$ and heigth equal to $d$. The triangle $PFB$ has base $b$ and height $h$. The triangle $PEA$ has base $s-b$ and height $h$. We can note that $h=\frac{(s \frac{\sqrt{3}}{2}-d)}{2}$ (in fact, looking at the right triangle $PAF$, we have $PA=s \frac{\sqrt{3}}{2}-d$ and angle $PAF=30^o$). By similar considerations, it can also be shown that $b= \frac{2 \sqrt{3}}{3}d + \frac{(s \frac{\sqrt{3}}{2}-d)\sqrt{3}}{6}$.
Now let us move the point $P$ laterally in a direction normal to $AD$ (e.g., leftward) by a distance $x$. The area of the triangle $PDC$ increases by $\frac{dx}{2}$. The area of the triangle $PFB$ changes by $\frac{1}{2}[(b-x/2)(h-x \sqrt{3}/2)-bh]$ (note that the change is negative). The area of the triangle $PEA$ changes by $\frac{1}{2}[(s-b-x/2)(h+x \sqrt{3}/2)-(s-b)h]$ (here the change can be positive or negative, because the base is reduced and the height increases). Summing all changes we have that the overall change $C$ is
$$C=\frac{dx}{2}+\frac{1}{2}[(b-x/2)(h-x \sqrt{3}/2)-bh]+\frac{1}{2}[(s-b-x/2)(h+x \sqrt{3}/2)-(s-b)h]= \\ \frac{dx}{2} +\frac{sx \sqrt{3}}{4}-\frac{bx\sqrt{3}}{2}-\frac{hx}{2}$$
Substituting $b$ and $h$ with the equations reported above, we get
$$C=\frac{dx}{2} +\frac{sx \sqrt{3}}{4}- dx -\frac{sx \sqrt{3}}{8}+\frac{dx}{4}-\frac{sx \sqrt{3}}{8}+\frac{dx}{4}=0$$
which shows that, moving the point $P$ leftward by any distance $x$, the overall change in the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is zero. Because of the symmetry of the problem, the same conclusion can be drawn if we consider a rightward movement.
Since any point inside the triangle $ABC$ can be reached starting from a given point of the height $AD$ (defined by a value of $d$) and moving laterally by a distance $x$, we can conclude that the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is constant whatever the point chosen, and then is equal to $s^2 \frac{\sqrt{3}}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
| Whether $f(x) = -6\sin^4 x = -6(\sin x)^4$ or $f(x) = -6 \sin (x^4)$, the chain rule is the way to go.
If
$f(x) = -6(\sin x)^4, \tag{1}$
set
$u = \sin x \tag{2}$
so that (1) becomes
$f(u(x)) = -6(u(x))^4; \tag{3}$
now we apply the chain rule, which says that $df(u(x))/dx$ is given by
$\dfrac{df(u(x))}{dx} = \dfrac{df}{du}(u(x)) \dfrac{du}{dx}(x). \tag{4}$
We see that
$\dfrac{df(u(x))}{du}(u(x)) = -24(u(x))^3 = -24 (\sin x)^3 \tag{5}$
and
$\dfrac{du}{dx} = \cos x, \tag{6}$
whence, by (4),
$\dfrac{df(u(x))}{dx} = -24 (\sin x)^3 \cos x. \tag{7}$
On the other hand if
$f(x) = -6\sin(x^4), \tag{8}$
then we take
$u(x) = x^4 \tag{9}$
so that (8) yields
$f(u(x)) = -6 \sin u(x), \tag{10}$
from which we have
$\dfrac{df(u(x))}{du}(u(x)) = -6 \cos u(x) = -6 \cos x^4; \tag{11}$
also,
$\dfrac{du}{dx} = 4x^3; \tag{12}$
pulling (11) and (12) together with the aid of (4) gives the result
$\dfrac{df((x))}{dx} = -24x^3 \cos x^4. \tag{13}$
Either way, The Chain Rule rules! Long Live The Chain Rule!
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite product: $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$ Find a closed form for the value of the infinite product $(1-0.5^2)(1-0.5^3)(1-0.5^4)...$
I know it converges. At first I thought it was the Euler–Mascheroni constant, but it's only accurate to about 3 sig fig. I haven't been able to to solve it yet.
| $$A_n=\prod_{i=2}^n(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_n$$ where appears the Pochhammer symbol. For $$A_{\infty}=\prod_{i=2}^{\infty}(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_{\infty }\approx 0.5775761901732048425577994$$ while $$\gamma \approx 0.5772156649015328606065121$$ If now, you use what fixedp suggested and consider the sum of the logarithms and expand each of them as a Taylor series, you should find for the first terms $$\log(A_{\infty})=-x^2-x^3-\frac{3 x^4}{2}-x^5-\frac{11 x^6}{6}-x^7-\frac{7 x^8}{4}-\frac{4
x^9}{3}-\frac{17 x^{10}}{10}-x^{11}-\frac{9 x^{12}}{4}-x^{13}-\frac{23
x^{14}}{14}-\frac{23 x^{15}}{15}-\frac{15 x^{16}}{8}-x^{17}-\frac{19
x^{18}}{9}-x^{19}-\frac{41 x^{20}}{20}+O\left(x^{21}\right)$$ in which you need to replace $x$ by $\frac{1}{2}$; this gives you $$\log(A_{\infty}) \approx -\frac{145045531}{264241152}$$ $$A_{\infty}\approx 0.5775770170415330689637466$$
Adding more terms leads quite quickly to the value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complete the square to change into standard form Here is the equation:
$x^2 + y^2 + 4x - 6y - 3 = 0$
Here are the instructions:
Complete the square to change the equation info standard form. Then graph the equation.
Because both $y^2$ and $x^2$ are present, I do not know what to do. Thank you!
| $$\begin{align}x^2+y^2+4x-6y-3=0&\iff x^2+2\cdot 2x+2^2-2^2+y^2-2\cdot 3y+3^2-3^2-3=0\\&\iff (x^2+4x+4)+(y^2-6y+9)=3+4+9\\&\iff (x+2)^2+(y-3)^2=4^2.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}$ using the residue theorem. He was doing this integral using the formula
$$\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}= \frac{2\pi i}{1-e^{-2\pi i\alpha}}(\sum(Res(\frac{F(z)}{z^{\alpha}};z_{k})))$$ where $F(z)=\frac{1}{(x^{2}+1)}$, $\alpha=\frac{1}{2}$ and $z_{k}$ is a pole of $F(z)$.
The poles of $F (z)$ are $i$ and $-i$ which are of order $1$. But it has not given me the answer which is $\frac{\pi}{\sqrt{2}}$.
I think my problem is calculating the residues or operate for to give me the answer. Any hint is appreciated.
| Here are some relevant hints including basic complex number arithmetic.
Put $$\sqrt{z} = \exp(1/2\log z)$$
where the logarithm has the branch cut on the positive real axis
(argument from zero to $2\pi$.)
Use a keyhole contour with the slot of the key on the positive real
axis. On the circular part of the contour we have
$1/\sqrt{z}/z^2\in\Theta(R^{-5/2})$ and since $\lim_{R\to\infty} 2\pi
R \times R^{-5/2} = 0$ the contribution from the circular arc is
zero.
This leaves the contribution above and below the positive real axis.
Above the real axis we get
$$\int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx$$
with $\sqrt{x}$ being the real square root function.
Below the real axis we have
$$\int_\infty^0
\frac{1}{\exp(1/2\log x + 1/2\times 2\pi i)(x^2+1)} dx$$
which is
$$- e^{-\pi i} \int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx$$
with the square root again being the real square root function.
Collecting the two contributions we have
$$(1-e^{-\pi i})
\int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx
\\= 2\pi i
\left(\mathrm{Res}\left(\frac{1}{\sqrt{z} (z^2+1)}; z=i\right)
+\mathrm{Res}\left(\frac{1}{\sqrt{z} (z^2+1)}; z=-i\right)\right).$$
This implies that
$$\int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx
= \pi i
\left(\frac{1}{\sqrt{i} (+2i)}
+ \frac{1}{\sqrt{-i} (-2i)}\right)
\\ = \pi i
\left(\frac{1}{e^{i\pi/4} (+2i)}
+ \frac{1}{e^{3i\pi/4} (-2i)}\right)
= \pi i e^{-i\pi /4}
\left(\frac{1}{2i} - \frac{1}{2i e^{i\pi/2}}\right)
\\= \pi i e^{-i\pi /4}
\left(- i\frac{1}{2} + \frac{1}{2}\right)
= \frac{\pi}{\sqrt{2}} i e^{-i\pi /4}
\left(- i\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right)
\\ = \frac{\pi}{\sqrt{2}} i e^{-i\pi /4} e^{-i\pi /4}
= \frac{\pi}{\sqrt{2}} i e^{-i\pi /2}
= \frac{\pi}{\sqrt{2}} i (-i) \\
= \frac{\pi}{\sqrt{2}}.$$
Addendum. If we want to be extra careful about it we also have to
check the integral along the miniature circle of radius $\epsilon$
enclosing the origin, which is on the order of
$2\pi\epsilon/\sqrt{\epsilon}= 2\pi\sqrt{\epsilon}\to 0$ as
$\epsilon\to 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of ways the letters of two k-letter words can match $M(k) :=$ the number of ways the letters of two (ordered) k-letter words can match.
$M(1) = 2$: the single letters of the two words are either matching or not.
$M(2) = 12$:
$$\begin{array}{ c|c c }
& A & A \\\hline
B & \circ & \circ \\
B & \circ & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & A \\\hline
A & \bullet & \bullet \\
A & \bullet & \bullet \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
A & \bullet & \circ \\
B & \circ & \bullet \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
B & \circ & \bullet \\
A & \bullet & \circ \\
\end{array}$$
$$\begin{array}{ c|c c }
& A & B \\\hline
A & \bullet & \circ \\
C & \circ & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
B & \circ & \bullet \\
C & \circ & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
C & \circ & \circ \\
A & \bullet & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
C & \circ & \circ \\
B & \circ & \bullet \\
\end{array}$$
$$\begin{array}{ c|c c }
& A & A \\\hline
A & \bullet & \bullet \\
B & \circ & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & A \\\hline
B & \circ & \circ \\
A & \bullet & \bullet \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
A & \bullet & \circ \\
A & \bullet & \circ \\
\end{array}
\begin{array}{ c|c c }
& A & B \\\hline
B & \circ & \bullet \\
B & \circ & \bullet \\
\end{array}$$
How to calculate $M(k)$?
The upper bound for $M(k)$ is obviously $2^\left(k^2\right)$, because each letter from the first word is either matching with any of the letters of the second word or not, but not all "match patterns" are possible.
Based on computer simulation $M(3) = 128, M(4) = 2100, M(5) = 48032$.
Two example matches for k=5 out of the 48032:
$$\begin{array}{ c|c c }
& A & B & C & C & D\\\hline
A & \bullet & \circ & \circ & \circ & \circ \\
A & \bullet & \circ & \circ & \circ & \circ \\
B & \circ & \bullet & \circ & \circ & \circ \\
B & \circ & \bullet & \circ & \circ & \circ \\
C & \circ & \circ & \bullet & \bullet & \circ \\
\end{array}\begin{array}{ c|c c }
& A & B & C & D & E\\\hline
A & \bullet & \circ & \circ & \circ & \circ \\
A & \bullet & \circ & \circ & \circ & \circ \\
B & \circ & \bullet & \circ & \circ & \circ \\
C & \circ & \circ & \bullet & \circ & \circ \\
D & \circ & \circ & \circ & \bullet & \circ \\
\end{array}$$
| After some time trying to solve definitively the problem @Tamas finally found a paper with the "solution".
Based on the calculated values of $M$ given by a computer program that checked all possible matrices, $M(n)$ gives the https://oeis.org/A014235 series. It's formula is therfore is $\sum_{k=0}^{n} k! * S(n+1, k+1)^2$ where $S$ is the function to get the Stirling numbers of the second kind.
The general case for a matrix of $n\times m$ is $\sum_{k=0}^{\min(n,m)} k!\cdot S(m+1, k+1) S(n+1, k+1)$
The problem discussed in the referenced paper of the series can be directly applied to this problem. (On the link provided the answer is the section 3.1 named "$\Gamma$-avoiding matrices".)
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminate the 2.
| $$\frac{x^2+3}{x^6(x^2+1)}=\frac{2x^4-2x^2+3}{x^6}-\frac{2}{x^2+1}=2x^{-2}-2x^{-4}+3x^{-6}-2\cdot\frac{1}{x^2+1}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{x-2}$$
But the answer is
$$\frac{-4}{(x-2)\sqrt{x-2}}$$
What should I have done?
| $$\frac{d(f(g(x)))}{dx}\ne f'(g(x))\quad\text{instead}\quad \frac{d(f(g(x)))}{dx}= f'(g(x))g'(x)$$
So
$$(f(\sqrt{x-2}))'=f'(\sqrt{x-2})(\sqrt{x-2})'=f'(\sqrt{x-2})\frac1{2(\sqrt{x-2})}$$
When $f(x)=8/x$ then $f'(x)=-8/x^2$
$$f'(\sqrt{x-2})=\frac{-8}{(\sqrt{x-2})^2}\frac1{2(\sqrt{x-2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
| We have
$$\frac{1}{n+k}>\log\left(1+\frac{1}{n+k}\right) $$
hence:
$$\sum_{k=1}^{n}\frac{1}{n+k}>\log\prod_{k=1}^{n}\frac{n+k+1}{n+k}=\log\frac{2n+1}{n+1}$$
while:
$$\frac{1}{n+k}=\frac{\frac{1}{n+k-1}}{1+\frac{1}{n+k-1}}<\log\left(1+\frac{1}{n+k-1}\right)$$
gives:
$$\sum_{k=1}^{n}\frac{1}{n+k}<\log\prod_{k=1}^{n}\frac{n+k}{n+k-1}=\log2$$
as wanted.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can someone explain how to solve linear inequality?
Can someone explain how to solve this linear inequality?
| We have $$\frac{6+(13-x)(x-3)-3(x-3)-2(x+3)}{x^2-9}\le0$$
$$\iff-\frac{(x-6)(x-5)}{(x-3)(x+3)}\le0$$
For the equality $x=6$ or $5$
For the inequality,
$$-\frac{(x-6)(x-5)}{(x-3)(x+3)}<0\iff\frac{(x-6)(x-5)}{(x-3)(x+3)}>0$$
$$\iff[x-(-3)](x-3)(x-5)(x-6)>0 (\text{multiplying either sides by }$(x-3)(x+3)^2>0)$$
Check for $-\infty<x<-3,-3<x<3,3<x<5,5<x<6,6<x<\infty$
As the product needs to positive, we need even number of multiplicand $>0$
which is evidently true if $x<-3$ or $x>6$
For $-3<x<3,$ only one is $>0$
For $3<x<5,$ two are $>0$
For $5<x<6,$ three are $>0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $4^{7 x - 10} = 10^{7 x - 6}$ for $x$ Solve for $x$ of the following equation.
$$4^{7 x - 10} = 10^{7 x - 6}$$
I tried to make $4$ and $10$ have a common base but I could not find one so I don't know where to go from here.
| $$
\begin{align}
& 4^{7 x - 10} = 10^{7 x - 6} \\[6pt]
\implies & \log_4(4^{7 x - 10}) = \log_4(10^{7 x - 6}) \\[6pt]
\implies & 7x-10 = (7x-6)\log_4(10) \\[6pt]
\implies & 7x-7x\log_4(10) = 10-6\log_4(10) \\[6pt]
\implies & 7x[1-\log_4(10)] = 10-6\log_4(10) \\[6pt]
\implies & 7x = \frac{10-6\log_4(10)}{1-\log_4(10)} \\[6pt]
\implies & x = \frac{10-6\log_4(10)}{7-7\log_4(10)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the given condition from given two quadratic equation
Question: If the quadratic equations $x^2+bx+c=0$ and $bx^2+cx+1=0$ have a
common root then prove that either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$
Till yet,
I had figured the common root of the given two quadratic equation. i.e.
Multiplying first equation by $b$ and eliminating the term $bx^2$ from the equation I get the common root ($\alpha$ say),
$$\alpha=\frac{1 - cb}{b^2 - c}$$
Further putting this value in either of the equation didn't benefited me much.What it gave me was an odd, unfriendly equation. Can anyone help me in this?
Thanks in advance.
| Let $y$ be the common root
So, we have $$y^2+by+c=0, by^2+cy+1=0$$
Solving for $y,y^2$ we get $y^2=\dfrac{b-c^2}{c-b^2},y=\dfrac{bc-1}{c-b^2}$
and using the identity $y^2=(y)^2$ we get $$b^3+c^3+1^3-3\cdot b\cdot c\cdot1=0$$
Now use Factorize the polynomial $x^3+y^3+z^3-3xyz$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving congruence equations Solve: $7x^6\equiv 11 \pmod{23}$ and $5^x\equiv 19 \pmod{23}$
I can solve simple congruence equations but how do I go about solving these?
| Fermat's little theorem says that $n^{22} \equiv 1 \pmod{23}$, which means that the last equation can be rewritten to an equation of the form $x \equiv a \pmod{22}$. To figure out $a$, we just need all powers of $5$ that is congruent to $19 \pmod{23}$. A quick search reveals that $15$ is the only one.
Also, the first equation can be divided by $7$, to get $x^6 \equiv 18 \pmod{23}$
We now have the two equations
$$
x^6 \equiv 18\pmod{23} \qquad x \equiv 15 \pmod{22}
$$
There exist cube roots $\pmod{23}$, but not unique square roots. So the first equation has two solutions $\pmod{23}$.
The cube root of $18$ is calculated like this: Fermat's little theorem says that $18^{22n}\cdot 18 \equiv 18\pmod{23}$. Especially, $18^{45}\equiv 18$. That means that $(18^{15})^3 \equiv 18$, and it's the only congruence class for which this holds. $18^{15} \equiv 4 \pmod {23}$, so our equations are now reduced to
$$
x^2 \equiv 4 \pmod{23} \qquad x \equiv 15 \pmod{23}
$$
The first equation here has two solutions. One is easy to guess (it's $2$), and the other is easy to get from there (it's $-2 \equiv 21$). Thus we get two possible linear congruence equations:
$$
x \equiv 2 \pmod{23}\quad \wedge \quad x \equiv 15 \pmod{22} \\ \text{or} \\ x\equiv 21\pmod{23}\quad \wedge\quad x\equiv 15 \pmod{22}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to prove that $k+1 \ge (1+\frac{1}{k})^{k}$? How to prove that
$$k+1\ge \bigg(1+\frac{1}{k}\bigg)^{k}
$$ when $k>2$
| We will prove this with induction.
Consider the base case of $k=2$. Clearly, $3\ge 2.25$.
Now assume that $k+1\ge \left(1+\frac{1}{k}\right)^k$. We want to show that $\left(k+1\right)+1\ge\left(1+\frac{1}{k+1}\right)^{k+1}$. Consider
$$\begin{align}
\left(1+\frac{1}{k+1}\right)^{k+1}&=\left(1+\frac{1}{k+1}\right)^k\left(1+\frac{1}{k+1}\right)\\
&\le\left(1+\frac{1}{k}\right)^k\left(1+\frac{1}{k+1}\right)\\
&\le\left(k+1\right)\left(1+\frac{1}{k+1}\right)\\
&=(k+1)+\frac{k+1}{k+1}\\
&=(k+1)+1
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime. Evaluate $\gcd(ab,p^4)$ and $\gcd(a+b,p^4)$, given that $\gcd(a,p^2)=p$ and $\gcd(b,p^3)=p^2$, where $p$ is a prime. Is it true if $\gcd(a,b)=\gcd(a,c)$, then
$\gcd(a^2,b^2)=\gcd(a^2,c^2)$?
We have $\gcd(a,p^2)=p\implies au_1+p^2v_1=p, u_1,v_1\in \mathbb Z\implies au_1p^2+p^4v_1=p^3\implies \gcd(a,p^4)=p^3 $
$\gcd(b,p^3)=p^2\implies bu_2+p^3v_2=p^2, u_2,v_2\in \mathbb Z\implies bu_2p+p^4v_2=p^3\implies \gcd(b,p^4)=p^3$
So $\gcd(ab, p^4)=p^3$. How to show next part?
| By assumption, we have that there exist $x,y,z,w\in\mathbb{Z}$ so that
$$
ax+p^2y=p\tag{1}
$$
and
$$
bz+p^3w=p^2\tag{2}
$$
Separating $ax$ and $bz$ and multiplying, then moving the multiple of $p^4$ back to the left, we get
$$
abxz+p^4(w+y-pwy)=p^3\tag{3}
$$
Thus, $\gcd(ab,p^4)\mid p^3$. However, since $p\mid a$ and $p^2\mid b$, $p^3\mid ab$. Therefore,
$$
\boxed{\displaystyle\bbox[5px]{\gcd(ab,p^4)=p^3}}\tag{4}
$$
Since $p^2\mid b$, we have $b=p^2u$, thus $(1)$ implies
$$
\begin{align}
\hspace{-1cm}1
&\hspace{-.7cm}=\left(\frac{a+b}px+p(y-u)\right)^3\\
&\hspace{-.7cm}=\frac{a+b}p\left[\left(\frac{a+b}p\right)^2x^3+3x^2(a+b)(y-u)+3xp^2(y-u)^2\right]+p^3(y-u)^3\\
\hspace{-1cm}p
&\hspace{-.7cm}=(a+b)\left[\left(\frac{a+b}p\right)^2x^3+3x^2(a+b)(y-u)+3xp^2(y-u)^2\right]+p^4(y-u)^3\tag{5}
\end{align}
$$
Thus, $\gcd(a+b,p^4)\mid p$, but since $p\mid a+b$, we have
$$
\boxed{\displaystyle\bbox[5px]{\gcd(a+b,p^4)=p}}\tag{6}
$$
Suppose that $\gcd(a,b)=\gcd(a,c)=d$. Then,
$$
\gcd\left(\frac ad,\frac bd\right)=\gcd\left(\frac ad,\frac cd\right)=1\tag{7}
$$
If $\gcd(a,b)=1$, then we have $x,y$ so that $ax+by=1$ and so
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(ax^3+3bxy)+b^2(3axy^2+by^3)\tag{8}
\end{align}
$$
so $\gcd(a^2,b^2)=1$. Apply this to $(7)$ and multiply by $d^2$ to get
$$
\boxed{\displaystyle\bbox[5px]{\gcd(a^2,b^2)=\gcd(a^2,c^2)=d^2}}\tag{9}
$$
| {
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Evaluate $ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos (2x)}\right) \mathrm{d}x $ Evaluate the following definite integral:
$$ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos(2x)}\right) \mathrm{d}x, $$ where $k \in \mathbb{N}_{>0}$.
| I think we can get the answer into a relatively elementary form. Observe that
$$\cos{(k-1) x} + 2 \cos{k x} + \cos{(k+1) x} = 2 \cos{k x} (1+ \cos{x})$$
Thus the integrand simplifies into
$$4 \int_0^{\pi} dx \cos^2{\frac{x}{2}} \frac{\sin^2{k \frac{x}{2}}}{\sin^2{x}} $$
which is equal to
$$\frac12 \int_0^{2 \pi} du \left (\frac{\sin{k u}}{\sin{u}} \right )^2 $$
This integral may be evaluated using the residue theorem by converting it to a complex integral:
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z^{2 k-1}} \left (\frac{z^{2 k}-1}{z^2-1} \right )^2$$
The only (non-removable) pole is at $z=0$, but this pole has multiplicity $2 k-1$. Thus, by the residue theorem, the integral is equal to
$$\frac{\pi}{(2 k-2)!} \left [\frac{d^{2 k-2}}{dz^{2 k-2}} \left (\frac{z^{2 k}-1}{z^2-1} \right )^2 \right ]_{z=0} = \frac{\pi}{(2 k-2)!} \left [\frac{d^{2 k-2}}{dz^{2 k-2}} \left (1+z^2+z^4+\cdots z^{2 k-2} \right )^2 \right ]_{z=0}$$
Now, this may be evaluated as follows. The derivative term is equal to
$$2 \frac{d^{2 k-3}}{dz^{2 k-3}} \left (1+z^2+z^4+\cdots z^{2 k-2} \right )^2 \left (2 z+4 z^3+\cdots+(2 k-2) z^{2 k-3} \right )$$
This is obviously a polynomial, and at $z=0$, the only surviving term will be the $z^{2 k-3}$ term. The coefficient of that term is
$$2+4+6+\cdots+(2 k-2) = k (k-1)$$
Therefore, the integral is
$$\frac{2 \pi}{(2 k-2)!} k (k-1) (2 k-3)! = \pi k$$
ADDENDUM
In response to @Igor's challenge, I can at least show that the value of the integral for half-integer values of $k$ is a rational number. (Note that the above analysis is only valid for integer $k$.) For these values of $k$, the above contour integration does not work because the extension of the integration interval to $[0,2 \pi]$ does not apply. Thus, the recurrence approach of @Nico is better here. If we define
$$J_m = I_{m+\frac12}$$
Then the recurrence relation we seek is
$$J_{m+1}-2 J_m + J_{m-1} = 4 \frac{(-1)^m}{2 m+1}$$
Thus, to show that $J_m \in \mathbb{Q}$, we need only show that $J_0 \in \mathbb{Q}$ and $J_1 \in \mathbb{Q}$. So we evaluate the integrals.
$$\begin{align}J_0 &= 2 \int_0^{\pi/2} du \frac{\sin^2{(u/2)}}{\sin^2{u}} \\ &= \frac12 \int_0^{\pi/2} du \, \sec^2{\frac{u}{2}} \\ &= \left [ \tan{\frac{u}{2}}\right ]_0^{\pi/2} = 1 \end{align}$$
$$\begin{align}J_1 &= 2 \int_0^{\pi/2} du \frac{\sin^2{(3 u/2)}}{\sin^2{u}} \\ &= \frac12 \int_0^{\pi/2} du \, \sec^2{\frac{u}{2}} (1+2 \cos{u})^2 \\ &= \left [ \tan{\frac{u}{2}} + 4 \sin{u}\right ]_0^{\pi/2} = 5 \end{align}$$
Thus, the $J_m \in \mathbb{Q}$ as asserted.
| {
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"source": "stackexchange",
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} |
Prove that $\gcd (n^3-1,n+1)=1$ for all even $n$. Prove that if $n$ is even, then
$$\gcd(n^3-1,n+1)=1.$$
I really don't have a clue with this one. Any help would be appreciated.
| If you polynomial long divide $n+1$ into $n^3-1$ (or use the Euclidean Algorithm), you obtain the expression:
\begin{align*}
n^3-1&=(n+1)(n^2-n+1)-2 \\
\implies 2&=(n+1)(n^2-n+1)-(n^3-1)
\end{align*}
The greatest common divisor of $n^3-1$ and $n+1$ divides both terms on the right, so it must divide their difference, 2, as well. So it must be either 1 or 2.
Note that $n+1$ is odd though, so the $\gcd$ cannot be 2. Therefore $$\gcd(n^3-1,n+1)=1.$$
| {
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differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x^2 + 3}{(x^3 - 4)^3}$
f ' (x)
$= \frac{(4x) (x^3 - 4)^3 - (2x^2 + 3)[3(x^3 -4)^2 (3x)}{[(x^3 - 4)^3]^2}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)(x^3 -4)^2}{(x^3 - 4)^6}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)}{(x^3 - 4)^4}$ ---quotient rule
3) f(x) = sin(x cos x)
$f ' (x) = \cos [x \cos(x)] (\cos x - x \sin x)$
4) f(x) = $x^2$ tan 2x
$f ' (x) = 2x \tan (2x)$ + $2x^2$ $\sec^2 2x$
5) f(x) = $ 3 \ln (\cos 3x) $
$f ' (x) = -9 \tan 3x$
| I haven't been through all of them, but I suggest you redo 1 and 2.
In 1, it should be $x^{-6}$, not $x^{-5}$
In 2, it should be $-9x^{2}(2x^{2}+3)$, not $-3(2x^{2}+3)$
| {
"language": "en",
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"answer_id": 1
} |
Solving for $x$ using exponential log laws For
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
I moved all the $x$ to left side, used got rid of log and got
$x-(x-1)^2 - (x+1) = 4$
Simplyifing I get
$x^2-2x=4$
The answer should be $x = 4$ (I checked on wolfram alpha)
Help please?
| Here are the steps I would use:
$\log_2(x) + 2\log_2(x-1) = 2 + \log_2(2x+1)$
$\log_2(x) + \log_2\left((x-1)^2\right) = \log_2\left(2^2\right) + \log_2(2x+1)$
That step turns each term into a base $2$ logarithm. Now they can be combined using the product rule for logs:
$\log_2\left(x(x-1)^2\right) = \log_2\left(4(2x+1)\right)$
Now you can get rid of the logarithms, and you get:
$x(x-1)^2 = 4(2x+1)$,
or
$x^3 - 2x^2 - 7x - 4 = 0$
This can be factored as $(x-4)(x + 1)^2=0$, which has solutions $x=4$ and $x=-1$. However, the second of these solutions is extraneous, which you can see by trying to plug it into the original equation.
Does that all makes sense?
| {
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For which values of $a$ is the solution for $x^2 - y^2 = a^3$ unique?
For which values of $a$ is the solution unique? $$x^2-y^2=a^3$$
I'm not sure how to do this, so I've been looking at this guy's solution.
$x^2 - y^2 = a^3$ is factored into $(x-y)(x+y) = a^3$. He says that the parity of both $x-y$ and $x+y$ is the same because when you add them together you get $2x$. Then he says let $a = {p_1}^{k_1} ....{p_r}^{k_r}$ be the prime factorization. Then he says that $x^2 - y^2 = a^3$ has exactly $2(3k_1 + 1) ... (3k_r + 1)$ solutions for $(x,y)$. I don't understand how he knows that or where he got that from. Can someone explain this to me, thank you.
| Lets focus on the case when $a>0$.
Let $a=p^{k_1}_{1}...p^{k_r}_{r}$ be the prime factorisation of $a$ then the number of distinct positive divisors of $a$ is equal to $\tau(a)=(k_1+1)...(k_r+1)$.
Therefore the number of distinct positive divisors of $a^3$ is $\tau(a^3)=(3k_1+1)...(3k_r+1)$. As said above $$x^2-y^2=a^3\Rightarrow (x-y)(x+y)=a^3$$
$x-y$ and $x+y$ have the same parity either they are both even or both odd. The existence of at least a solution is clear as $a$ and $a^2$ have the same parity and one could conjecture $x-y=a$ and $x+y=a^2$ which would yield $a+2y=a^2\Rightarrow y=\frac{a(a-1)}{2}\in \mathbb{N}$ if $a\in\mathbb{N}$. Now your question is why do we have that the total number of possible solutions is $2{\tau(a^3)}$? Lets consider the case when $a$ is an odd positive integer. Notice that if $d$ is a divisor of $a^3$ so is $a^3/d$. Now because $a$ is odd, $d$ and $a^3/d$ must necessarily have the same parity. They are both odd. Therefore the pair $(d,a^3/d)$ or $(a^3/d,d)$ could be a potential solution to $(x-y,x+y)$. How many such pairs do we have? We have exactly $2\cdot\tau{(a^3)}$.
Now consider when $a$ is positive even integer, then $a=2^{k_1}...p^{k_r}_r$. Notice that in this case $(2^{3k_1},a^3/2^{3k_1})$,$(a^3/2^{3k_1},2^{3k_1})$,$(1,a^3)$ and $(a^3,1)$ can not be candidates for $(x-y,x+y)$ because $2^{3k_1}$ and $a^3/2^{3k_1}$ or $1$ and $a^3$have different parities. So in this case we have $2(\tau(a^3)-2)$ possible solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1+x}{2}\right)dx$ , $\int_0^1\frac{\ln^2(1-x)}{x}\text{Li}_2\left(\frac{1+x}{2} \right)dx$ How can we evaluate the following integrals:
$$\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1 + x}{2} \right)\,dx\\
.\\
\int_0^1\frac{\ln^2(1-x)}{x}\text{Li}_2\left(\frac{1 + x}{2} \right)\,dx$$
| $$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\
\\
{\large\int}_0^1\frac{\ln^2(1-x)\,\operatorname{Li}_2\left(\frac{1+x}2\right)}xdx&=\frac{81\,\zeta(5)}{32}+\frac{5\pi^2}{16}\zeta(3)-\frac{\zeta(3)}8\ln^22+\frac1{15}\ln^52\\&-\frac{\pi^2}{18}\ln^32-\frac{\pi^4}{15}\ln2+2\operatorname{Li}_5\left(\tfrac12\right)+2\operatorname{Li}_4\left(\tfrac12\right)\ln2.
\end{align*}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Proof by induction of a sum? Let $n ∈ N$. Prove by induction that there are $n$ ways to write the number $n$ as a sum $n=x_1+x_2+...+x_k$ where the $x_i$ are natural numbers and $x_1 ≤x_2 ≤...≤x_k ≤x_1+1$.
For example, $5 = 5$, $5 = 1 + 1 + 1 + 1 + 1$, $5 = 1 + 1 + 1 + 2$, $5 = 1 + 2 + 2$, and $5 = 2 + 3$.
| Here's a rough outline of a strategy: With $n=x_1+\cdots+x_k$, you can write $n+1=x_2+\cdots+x_k+(x_1+1)$. There is one allowed way to write $n+1$ that you can't get in this way, however.
Edit: To consider an example, here are the five ways to write $5$, with $x_1$ marked in red:
$5=\color{red}{5}$, $5=\color{red}{1}+1+1+1+1$, $\color{red}{1}+1+1+2$, $5=\color{red}{1}+2+2$, and $5=\color{red}{2}+3$
For each of these, take the red number, add $1$ to it and put it last, thus getting five ways to write $6$:
$6=\color{red}{6}$, $6=1+1+1+1+\color{red}{2}$, $6=1+1+2+\color{red}{2}$, $6=2+2+\color{red}{2}$, and $6=3+\color{red}{3}$.
Notice how this gives all the allowed ways to write $6$ as a sum, except for one, namely $6=1+1+1+1+1+1$. The same procedure works to get from $n$ to $n+1$ for all $n$, not just for $n=5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$. If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.
So far, I let
$d=(a^2+b^2,a+b)$
$\implies d|(a^2+b^2-(a+b)^2)$
$\implies d|(a^2+b^2-(a^2+2ab+b^2))$
$\implies d|(-2ab)$
I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Alternatively, is there some other way of showing it?
EDIT(10:41PM):
Just noticed that
$(a^2+b^2\pm (a+b)(a-b),a+b)$
results in
$d|2a^2$ and $d|2b^2$.
$\implies d|(2a^2,2b^2)$.
$\implies d|2(a^2,b^2)$.
Now I just have to fully understand why
$(a,b)=1\implies (a^2,b^2)=1$.
I sort of intuitively understand this.
| Many gcd questions can be answered by finding a way to use the Euclidean algorithm. By the Euclidean algorithm, I mean $(a, aq+r) = (a,r)$.
Here, $a$ is played by $a+b$, $q$ is played by $a-b$, and $r$ is played by $2b^2$. The Euclidean algorithm then implies that
$$(a+b, (a+b)(a-b) + 2b^2) = (a+b, 2b^2).$$
This reads as $(a^2+b^2, a+b) = (a+b, 2b^2)$. The conclusion should now be clear.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
Log integrals II By considering the integral
\begin{align}
I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, d\theta
\end{align}
derivatives can be taken with respect to $\mu$ to obtain integrals involving logarithms. Let these integrals be
\begin{align}
J_{\mu}^{m} = \partial_{\mu}^{m} I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, \ln^{m}(\cos\theta - \sin\theta) \, d\theta.
\end{align}
What are the closed form values of $I_{\mu}$, $J_{0}^{0}$, $J_{0}^{1}$, and $J_{0}^{2}$ ?
Is it possible to also to extend the results to the integral
\begin{align}
T_{\mu}^{k} = \int_{0}^{\pi/4} \sin(2\theta) \left( \cos^{2/k}\theta - \sin^{2/k}\theta \right)^{\mu} \, d\theta \hspace{3mm} ?
\end{align}
| Another possible closed-form
$$I_\mu=\frac{2\cdot{_2F_1}\left(\begin{array}c\tfrac12,2\\\tfrac{\mu+5}{2}\end{array}\middle|\,-1\right)}{(\mu+3)(\mu+1)},$$
for $\Re(\mu)>-1$.
I don't know how to simplify it further, but I think it's a good idea for integer $\mu$ values to separate the even and the odd $\mu$ cases, since
$$\begin{align}
I_0 & = \frac 12\\
I_1 & = \frac 23 - \frac{\sqrt{2}}{3}\\
I_2 & = \frac 12 - \frac{\pi}{8}\\
I_3 & = \frac 65 - \frac{\sqrt{2}}{5}\\
I_4 & = \frac 56 - \frac{\pi}{4}\\
I_5 & = \frac{46}{21} - \frac{32 \sqrt{2}}{21}\\
I_6 & = \frac 32 - \frac{15\pi}{32}\\
I_7 & = \frac{182}{45} - \frac{128\sqrt{2}}{45}\\
I_8 & = \frac{83}{30} - \frac{7\pi}{8}\\
I_9 & = \frac{2902}{385} - \frac{2048\sqrt{2}}{385}.
\end{align}$$
| {
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"source": "stackexchange",
"question_score": "4",
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Solving the differential equation: $f(x)yy'=(y')^2-0.5$ I am trying to solve this equation:
$f(x)yy'=(y')^2-0.5$
I have already tried traditional methods...
Any ideas?
| We have
\begin{align*}
&y y^\prime = (y^\prime)^2 - 0.5 \\
& (y^\prime)^2 - y y^\prime = 0.5 \\
& (y^\prime)^2 - y y^\prime + \frac{1}{4}y^2 = 0.5 + \frac{1}{4} y^2 \\
& \left(y^\prime - \frac{1}{2} y \right)^2 = 0.5 + \frac{1}{4} y^2 \\
& y^\prime = \frac{1}{2} \left(y \pm \sqrt{1 + y^2} \right)
\end{align*}
You can solve this implicity using separation of variables.
| {
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"source": "stackexchange",
"question_score": "3",
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Calc I limit question involing trig functions Find the limit
$$\lim_{x\to2}\frac{\sin^2(x^2-4)\sec^2(3x-6)}{(x^3-8)\tan(2x-4)}$$
i have been having trouble finding this limit, i have tried using having trig identities and making all terms sin and cos but i cant figure it out, keep in mind i am not allowed to use L'hopital's rule
| A little prep work gives us
$${\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}={\sin^2(x^2-4)\over(x-2)\sin(2x-4)}\cdot{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}$$
Now the second term causes no trouble as $x\to2$. It gives
$$\lim_{x\to2}{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}={\cos0\over12\cos^20}={1\over12}$$
If we rewrite the first term as
$$\begin{align}
{\sin^2(x^2-4)\over(x-2)\sin(2x-4)}&={2(x+2)^2(x-2)\sin^2(x^2-4)\over2(x+2)^2(x-2)^2\sin(2x-4)}\\
\\
&={(x+2)^2\over2}\left({\sin(x^2-4)\over(x^2-4)}\right)^2{2x-4\over\sin(2x-4)}\\
\end{align}$$
we can now use the limit
$$\lim_{u\to0}{\sin u\over u}=1$$
with $u=x^2-4$ for one piece and $u=2x-4$ for another to get
$$\lim_{x\to2}{\sin^2(x^2-4)\over(x-2)\sin(2x-4)}={4^2\over2}\cdot1^2\cdot{1\over1}=8$$
Combining all this, we have
$$\lim_{x\to2}{\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}=8\cdot{1\over12}={2\over3}$$
| {
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"url": "https://math.stackexchange.com/questions/974319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:
Solve the following system for $a,b,c\in \mathbb{R}$:
$$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$
I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6)^2=4(2c+6)$$
Then maybe $a=b=c$ is one case.
Thank you.
| Let $$ 3T:=a^2+b^2+c^2-18 = 2\sqrt{2a+6} +
2\sqrt{2b+6} + 2\sqrt{2c+6} $$
If $$ a= {\rm max}\ \{ a,b,c\},\ b
= {\rm min}\ \{ a,b,c\},\ a>b$$ then
$$T>b^2-6=2\sqrt{2a+6} > T$$
Contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\int_0^{\pi/4}\sec(x)^3dx$ How do I find $$\int_0^{\pi/4}\sec(x)^3dx$$
I arrive at $\sec x\tan x + 1/3 \ln| (\cos^3(x))| dx$
where $u = \sec x$ and $v' = \sec^2(x)$
What's my error? Could I get a step by step solution?
| So you're integrating by parts with $u = \sec x$ and $dv = \sec^2 x = (\tan x)'$. Then:
$$
I := \int_{0}^{\frac{\pi}{4}} \sec^3 x dx = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right) - \int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx
$$
Where the last integral is
$$
\int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx = \int_{0}^{\frac{\pi}{4}} \frac{1-\cos^2 x}{\cos^3 x} = I - \int_{0}^{\frac{\pi}{4}} \sec xdx
$$
Plugging this in the first expression yields:
$$
2I = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right) + \int_{0}^{\frac{\pi}{4}} \sec xdx = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right) + \log\left\lvert \sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right) \right\rvert
$$
So $I = \frac{1}{\sqrt{2}} + \log \sqrt{1+\sqrt{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| This goes back to Newton: The binomial theorem doesn't only work for natural number exponents! If $a > 2$, the following formula holds:
$$\sqrt{a^2+ 4} = (a^2 + 4)^{1/2} = \sum_{k=0}^\infty {{1/2}\choose k}(a^2)^{1/2-k}4^k$$ $$= a + (1/2)\frac{4}{a} + \frac{(1/2)(-1/2)}{2!} \frac{4^2}{a^3} + \cdots$$ $$= a + \frac{2}{a} - \frac{2}{a^3} + \cdots$$
Notice that when $a$ is very large, the later terms become less and less significant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.
Could anyone shine some light on this? Thanks.
| $x$ is either odd or even. If $x$ is even, let $x=2r+1$. Then the right hand side is
$$
\lfloor \frac{(x+1)^2}{4} \rfloor = \lfloor 4 \frac{(r+1)^2}{4} \rfloor
= \lfloor (r+1)^2 \rfloor = (r+1)^2
$$
and the left hand side is
$$
\lfloor \frac{(x+1)^2+1}{4} \rfloor = \lfloor 4 \frac{(r+1)^2+1}{4} \rfloor
= \lfloor (r+1)^2 + \frac{1}{4} \rfloor = (r+1)^2
$$
The other answer covers the even case.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine whether or not a binary number is divisible by $3$ Let $K$ be a natural number with $n$ binary digits.
Is there an $O(n)$ method for deciding whether or not $K$ is divisible by $3$?
$3|K \iff d_1-d_2+d_3-d_4\dots\pm d_n=0$ works correctly up to $20$, but fails for $21$.
| Another way to prove the accepted answer without using the congruence relation. First, note that we have:
$$
\frac{1}{3} \text{ = } 0+\frac{1}{3} \\
\frac{2}{3} \text{ = } 1-\frac{1}{3} \\
\frac{4}{3} \text{ = } 1+\frac{1}{3} \\
\frac{8}{3} \text{ = } 3-\frac{1}{3} \\
\\ \vdots
$$ And, generally:
$$
\frac{2^k}{3}= \begin{cases}
p - \frac{1}{3} & \text{if } k \text{ is odd} \\
q + \frac{1}{3} & \text{if } k \text{ is even} .
\end{cases}
$$
Where $p$ and $q$ are some integers. It can be proved using induction. In order to check whether a binary integer $N$ with $n$ digits is divisble by $3$ we can write $\frac{N}{3}$ as follows:
$$
\begin{split}
\frac{N}{3} = \frac{\sum_{i=0}^{n-1}b_{i}2^{i}}{3} = (\frac{2^0}{3}b_0+\frac{2^2}{3}b_2+\frac{2^4}{3}b_4+\ldots+\frac{2^{2r}}{3}b_{2r}) \\
+ (\frac{2^1}{3}b_1+\frac{2^3}{3}b_3+\frac{2^5}{3}b_5+\ldots+\frac{2^{2s-1}}{3}b_{2s-1}).
\end{split}
$$
Where $r= \left \lfloor \frac{n-1}{2} \right \rfloor$, $s=\left \lfloor \frac{n}{2} \right \rfloor$. Now replacing the $\frac{2^k}{3}$ terms yields:
$$
\begin{align}
\frac{N}{3} & = ((q_0+\frac{1}{3})b_0+(q_2+\frac{1}{3})b_2+\ldots+(q_{2r}+\frac{1}{3})b_{2r})
\\ & \quad + ((p_1-\frac{1}{3})b_1+(p_3-\frac{1}{3})b_3+\ldots+(p_{2s-1}-\frac{1}{3})b_{2s-1}) \\
& = l + \frac{b_0-b_1+b_2-b_3+\ldots+(-1)^{n-1}b_{n-1}}{3}.
\end{align}
$$
Where $l$ is the sum of $p_ib_i$s and $q_ib_i$s and is already an integer. So in order for $N$ to be divisible by $3$ (i.e., the quotient to be an integer), $b_0-b_1+b_2-b_3+\ldots+(-1)^{n-1}b_{n-1}$ must be divisible by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Triplets of distinct integers > 1 that return integer values. If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?
UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$
My further work:
Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$
Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$
Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.
Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.
Case $1$: $k<C$
Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$
By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$
This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.
If $k=1$ we have $A+B=C$ which violates the condition $A>C$.
If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$
However, this leads to the result that $A=B$ which violates that stated conditions.
Case $1$, $k<C$, fails.
Case $2$: $k=C$
Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$
This later equation is clearly false and thus Case $2$, $k=C$ fails.
Case $3$: $k>C$
Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.
Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.
Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$
Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$
By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$
Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.
By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$
and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$
From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.
Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$
Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$
Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.
Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.
Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.
| We can write $f(A,B,C)=\frac {(A^2-1)BC+(B^2-1)AC}{AB(C^2-1)}$ then $f(A,B,C)$ is an integer if $AB(C^2-1)|C((A^2-1)B+(B^2-1)A)$ $\Rightarrow$ because $GCD(C,C^2-1)=1$ then $A|C \ or\ B|C$ and $ (C^2-1)|(A^2-1)B+(B^2-1)A $.
| {
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"timestamp": "2023-03-29T00:00:00",
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minimize a function using AM-GM inequality I want to minimize the function
$$ \frac{x}{1-x^2} + \frac{y}{1-y^2} + \frac{z}{1-z^2} $$ subject to the constraint $$x^2 + y^2 + z^2 = 1 \space\text{and} \space x,y,z > 0$$
Wolfram Alpha tells me that the minimum occurs at $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{3\sqrt{3}}{2})$. I tried to solve this without using Lagrange Multipliers by using AM-GM and Cauchy-Schwarz, but I couldn't figure out how to do it and am interested in a solution.
| We can write the inequality to prove as
$$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\ge \frac{3\sqrt3}2 \tag{$\star$}$$
As equality is achieved for $x=y=z=\frac1{\sqrt3}$, if $(\star)$ holds then we have established the minimum.
Consider function $f(t) = \dfrac{t}{1-t^2}-\frac{\sqrt 3}2 -\frac{\sqrt 3}2(3t^2-1)$. To show the inequality $(\star)$ holds, it is enough to show $f(t)\ge 0$, as the inequality is equivalent to $f(x)+f(y)+f(z) \ge 0$.
Now $f(t) = \dfrac{t(3\sqrt3 t^3-3\sqrt3t+2)}{2(1-t)(1+t)}$, so it is enough to show that $3\sqrt3 t^3+2 \ge 3\sqrt3t$ for $t \in (0, 1)$. But this follows from AM-GM as $3\sqrt3t^3+1+1 \ge 3\left(3\sqrt3 t^3 \times 1 \times 1 \right)^{1/3}=3\sqrt3t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980875",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
| Hint: use also that
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6
$$
$$
1 + (1+2) + \dots + (1 +2+\dots +n) =
\frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2
\\=\frac 12 \left[
(1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n)
\right]
\\=\frac 12 \left[
(1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n)
\right]
$$
| {
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"url": "https://math.stackexchange.com/questions/980941",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Which of the following numbers does not divide $2^{1650}-1$? I'm practicing for a math competition that is coming up, and I got stuck on this question:
Which of the following numbers does not divide $2^{1650}-1$?
$3$, $7$, $31$, $127$, $2047$
I've seen a link on Yahoo Answers that says this:
Note that $3 = 2^2 - 1$, $7 = 2^3 - 1$, $31 = 2^5 - 1$, $127 = 2^7 - 1$ and $2047 = 2^{11} - 1$.
Let's consider $2^n - 1$. If $n = ab$ (i.e. $n$ is divisible by $a$) then we have
$2^{ab} - 1$.
Now in the case of $n = 1650$ is a number that is divisible by $2$, but not by $4$, which means that exactly one of $a$ or $b$ must be even (since when you multiply two odd numbers you get an odd number, and when you multiply two even numbers you get a number divisible by $4$). So let's say, without loss of generality, that the even one is $a$. Since it's even, we can say that $a = 2k$ for some integer $k$.
So we have $2^{1650} - 1$ = $2^{2kb} - 1$. We can employ the laws of exponents on the right hand side to get $(2^{kb})^2 - 1$, a difference of two squares which can be written as $(2^{kb} + 1)(2^{kb} - 1)$.
Now $2^{1650} - 1 = (2^{kb} + 1)(2^{kb} - 1)$.
We can now conclude that if $k$ divides $1650$, then $2^k - 1$ divides $2^{1650} - 1$.
But I don't understand the conclusion. I see how $2^{1650} - 1$ is divisible by $(2^{kb} - 1)$ and $(2^{kb} + 1)$, but not $(2^k - 1)$.
| Here is an easier way of understanding that theorem.
For any positive integer $m$ we have: $x^m-1 = (x-1)(x^{m-1}+x^{m-2}+\cdots+x+1)$.
Now set $m = b$ and $x = 2^a$, to get $(2^a)^b - 1 = (2^a-1)[(2^a)^{b-1}+(2^a)^{b-1}+\cdots+(2^a)+1]$.
This simplifies as $2^{ab}-1 = (2^a-1)[\text{some integer}]$.
Hence, $2^{ab}-1$ is divisible by $2^a-1$ for any choice of positive integers $a$ and $b$.
Now, to solve this question, use the above statement for different values of $a$ and $b$.
For instance, pick $a = 2$ and $b = 825$ to get that $2^{1650}-1$ is divisible by $2^2-1 = 3$.
Also, pick $a = 3$ and $b = 550$ to get that $2^{1650}-1$ is divisible by $2^3-1 = 7$.
Can you continue from here? Knowing which small numbers divide $1650$ will be useful.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$ Prove that in any triangle inequality holds:
$$a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$$
Is gender inequality that occurs right triangle, not an equilateral triangle. For this reason I suspect you have used other ways different from the usual.
| Let $H$ be the orthocenter of $ABC$. Then
$$ d(H,BC) = 2R\cos B\cos C $$
hence:
$$ \sum_{cyc} 2R a\cos B\cos C = 2S. $$
If $ABC$ is an acute triangle, then $\cos A,\cos B,\cos C>0$ and $a,b,c < 2R$, hence:
$$ \sum_{cyc} a^2\cos B\cos C < \sum_{cyc} 2R a\cos B\cos C = 2S. $$
However, if $ABC$ is an obtuse triangle, the given inequality does not hold: consider, for instance, $(a,b,c)=(\sqrt{3},1,1)$, so that $(A,B,C)=(120^\circ,30^\circ,30^\circ)$,
$$\sum_{cyc}a^2\cos B\cos C = \frac{9}{4}-\frac{\sqrt{3}}{2},\qquad 2S=\frac{\sqrt{3}}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ I am trying to evaluate:
$$\int_0^{\pi/12} \ln(\tan x)\,dx$$
I think the integral is quite simple but I am having a hard time evaluating it. I started with the result:
$$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$
where $G$ is the Catalan's constant. With the change of variables $x\rightarrow 3x$ and using the fact that $\tan(3x)=\tan x\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$, the integral is:
$$\int_0^{\pi/12}\ln(\tan x)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}-x\right)\,dx=-\frac{G}{3}$$
$$\Rightarrow \int_0^{\pi/12}\ln(\tan x)\,dx+\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=-\frac{G}{3}$$
But I do not see how to proceed.
Help is appreciated. Thanks!
| Using the Fourier series of $\ln(\tan{x})$,
\begin{align}
&\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\
=&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\
=&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\
=&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\
&\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\
=&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\
=&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G}
\end{align}
Things could be made clearer if we explicitly write out the terms of the sums. For the red sums,
\begin{align}
&-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\
=&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\
=&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 4
} |
Finding inverse of the rational fucntion I am trying to find the inverse of $f(x)=\frac{2x-1}{x(x-1)}$. So far I figured out that the domain should be (0,1) and the range is R itself. I found that by drawing a graph after I turned the function $f(x)=\frac{1}{x}+\frac{1}{x-1}$. since function is 1-1 and onto between this domain and range I know this function must have an inverse. But i could not go further after the following point.
$f(x)=y=\frac{1}{x}+\frac{1}{x}-1$ and I exchange the $y$ and $x$ and get $x=\frac{1}{y}+\frac{1}{y}-1$ and after that point I know I should solve it with respect to $y$ but I could not any help will be appreciated.
thanks
| If
$y = \frac{2x-1}{x(x-1)}
$,
$yx(x-1) = 2x-1$
or
$yx^2-yx-(2x-1) = 0$
or
$yx^2-(y+2)x+1 = 0
$.
Applying the quadratic formula,
the discriminant
$D^2
=(y+2)^2-4y
=y^2+4y+4-4y
=y^2+4
$
so the roots are
$x
= \frac{y+2\pm D}{2y}
= \frac{y+2\pm \sqrt{y^2+4}}{2y}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $(\frac{\cos x}{1-\sin x})^2$ $(\dfrac{\cos x}{1-\sin x})^2$
$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x+\sin^2x-\cos^2x}{(1-\sin x)^2})$
Does $\sin^2x-\cos^2=1$? or $-1$? Then it could factor with the bottom and the answer would be $\dfrac{2\cos x}{(1-\sin x)^2}$. Is this right?
| Using $\sin^2 x + \cos^2 x=1$,
$$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x(1-\sin x)+\cos^2x}{(1-\sin x)^2})= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{1-\sin x}{(1-\sin x)^2})=\frac{2\cos x}{(1-\sin x)^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I calculate $\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$ limit without using L'Hopital's rule? $$\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$$
Can this limit be calculated without using L'Hopital's rule?
| Clearly $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin x - \log(e^{x}\cos x)}{x\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\cdot\frac{x}{\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}} - \frac{\log \cos x}{x^{2}}\\
&= A - B\end{aligned}$$ Next we can see that $$A = 0$$ from here. And we have $$\begin{aligned}B &= \lim_{x \to 0}\frac{\log\cos x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\lim_{x\to 0}\frac{\cos x - 1}{x^{2}}\\
&= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x\to 0}\frac{\cos x - 1}{x^{2}}\text{ (putting } t = \cos x - 1)\\
&= \lim_{x \to 0}\frac{\cos x - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\cos^{2} x - 1}{x^{2}(1 + \cos x)}\\
&= -\frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = -\frac{1}{2}\end{aligned}$$ It follows that $L = 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Simplifying radicals I am stuck in the following puzzle and couldn't find a way to approach this.
$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$
Please help.
| Hint:
$$\eqalign{
& \sqrt{14 + \sqrt {180}} = 3 + \sqrt 5 \cr
& \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr
& \sqrt {5 + \sqrt 5 + \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Fourier series representation of $\sin^4 x$ I tried solving for fourier coefficients of Fourier series for the multiples of fundamental frequency $\omega_0=2$. So $F_n=\int_0^{\pi} \sin^4 x \, e^{-i2nx} dx$. And my calculator says answer should be $3e^{-i2nx}/8$.
But for some reasons I get $i(2+1/(4-2n)-1/(4+2n))/(16\pi)$. Can you show how one gets the correct answer without using trigonometry and just using integration of complex exponentials?
| Simply rewrite $\sin^4 x$ in terms of multiple angle trigonometric functions. You will get a finite series.
In fact
$$
1 - 2 \sin^2 x = \cos(2x) \quad \text{and} \quad \cos(2x) = 2\cos^2(x)-1\\
\Rightarrow \sin^2 x = \frac{1-\cos(2x)}{2} \\
\Rightarrow \sin^4 x = \frac{1+\cos^2(2x)-2\cos(2x)}{4} \\
\Rightarrow \sin^4 x = \frac{1+(\cos(4x)+1)/2-2\cos(2x)}{4} \\
= \frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find derivative of $f(x)=\frac{1}{\sqrt{x+2}}$ by definition Use the definition of a derivative to find the derivative of:
$$f(x)=\frac{1}{\sqrt{x+2}}+2x$$
my work:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2(x+h)-\frac{1}{\sqrt{x+2}}-2x}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2h-\frac{1}{\sqrt{x+2}}}{h}$$
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}}{h}+2$$
$$\lim_{h\to 0}\frac{1}{h}\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}+2$$
I don't know what to do from here.
| Hint: Note that
$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+2}}+2h-\frac{1}{\sqrt{x+2}}}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+\lim_{h\to0}\frac{2h}{h}=\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]+2$$
Now look that:
$$\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}\right]$$
then now multiply numerator and denominator by $$\sqrt{x+2}+\sqrt{x+h+2}$$
then
$$\lim_{h\to 0}\frac{1}{h}\left[\frac{\sqrt{x+2}-\sqrt{x+h+2}}{\sqrt{x+h+2}\sqrt{x+2}}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{(\sqrt{x+2}-\sqrt{x+h+2})(\sqrt{x+2}+\sqrt{x+h+2})}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{(\sqrt{x+2})^2-(\sqrt{x+h+2})^2}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{x+2-(x+h+2)}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\frac{1}{h}\left[\frac{-h}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\lim_{h\to 0}\left[\frac{-1}{\sqrt{x+h+2}\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+h+2})}\right]=\frac{-1}{2(x+2)\sqrt{x+2}}$$
then $$f'(x)=\frac{-1}{2(x+2)\sqrt{x+2}}+2$$
| {
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"url": "https://math.stackexchange.com/questions/986541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.
5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$
6 $k = 0$, $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$
7 $k = 1$, $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$
8 $k = 2$, $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$
So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6}, \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?
Thanks!
| Way easier way;
$$z^3=i \\
\iff z^3-i=0 \\
\stackrel{-i=i^3}{\iff}z^3+i^3=0 \\
\iff (z+i)(z^2-iz-1) = 0 \\
\iff z_1=-i,\; z_2=\frac12 (i-\sqrt 3), \; z_3=\frac12 (i+\sqrt 3)$$
Disregard this answer if your exercises restrict you to trigonometric/polar form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Find the area of triangle APB, where P is a point $(a\cos\theta, b\sin\theta)$ on an ellipse and $A, B$ are its radii points $(a,0) (0,b)$
A point $P(a\cos\theta, b\sin\theta)$ sits on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The points $A$ and $B$ have coordinates $(a,0)$ and $(0,b)$ respectively. Show that the area of triangle $APB$ is:
$$
\frac{1}{2}AB(\cos\theta + \sin\theta -1)
$$
The length of $AB$ is $\sqrt{a^2 + b^2}$; and $C$, its mid-point, has co-ordinates of $(\frac{a}{2}, \frac{b}{2})$. So the area of $APB$ must be $\frac{1}{2}\cdot AB \cdot CP$, where:
$$
CP = \sqrt{\left(a\cos\theta - \frac{a}{2}\right)^2+\left(b\sin\theta - \frac{b}{2}\right)^2}
$$
But this gets rather messy. Am I going wrong somewhere?
| A triangle with two sides given by vectors ${\bf x}, {\bf y}$ has area
$$\frac{1}{2}|{\bf x} \times {\bf y}|.$$ In this case we may take our vectors to be
\begin{align}
{\bf x} &= (a \cos \theta, b \sin \theta) - (a, 0) = (a(\cos \theta - 1), b \sin \theta)\\
{\bf y} &= (0, b) - (a, 0) = (-a, b).
\end{align}
Substituting and simplifying gives that the area is
$$\frac{1}{2}ab|\cos \theta + \sin \theta - 1|,$$
and checking shows that we can drop the absolute value symbols when $0 \leq \theta \leq \frac{\pi}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/991641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a function with certain properties I ran into a problem, and I'm not sure how to continue.
Problem: Let $f$ be a function such that $\sqrt {x - \sqrt { x + f(x) } } = f(x)$, for $x > 1$. In that domain, $f(x)$ has the form $\dfrac{a+\sqrt{cx+d}}{b}$, where $a$, $b$, $c$, $d$ are integers and $a$, $b$ are relatively prime. Find $a+b+c+d$.
So, I tried to cancel out the radicals, and got $(f(x))^4-2x(f(x))^2-f(x)+x^2-x=0$. Setting $y=f(x)$, I tried to apply the quadratic formula to find $x$ in terms of $y$. I got
$$
x=\frac{x+2xy^2 \pm \sqrt{(x+2xy^2)^2-4(y^4-y)}}{2}.
$$
From here, I tried simplifying the radical, but got
$$
x=\frac{x+2xy^2 \pm \sqrt{(x^2)(2y^2+1)^2-y(4y^3+1)}}{2}.
$$
I don't know if I factored it wrong, or if I'm missing something painfully obvious. Can I have a hint as to how to continue? It would be greatly appreciated.
| We have $f(x)=O(\sqrt x)$. Note that $\lim_{x\to\infty}\frac{f(x)}{\sqrt x}=\frac{\sqrt c}{b}$, whereas
$\lim_{x\to\infty}\frac{\sqrt{x-\sqrt{x+f(x)}}}{\sqrt x}=1$. So we conclude $c=b^2$.
Then $$ \sqrt{x+f(x)}=x-f(x)^2=x-\frac{a^2+2a\sqrt{b^2x+d}+b^2x+d}{b^2}=-\frac{(a^2+d)+2a|b|\sqrt{x+d/b^2}}{b^2}$$
Repeat the trick from above, i.e. divide by $\sqrt x$ and take the limit as $x\to\infty$, to conclude that $-\frac{2a|b|}{b^2}=1$. From this with $\gcd(a,b)=1$, conclude $a=-1$, $b=2$, hence $c=4$.
Can you do the last step to find $d$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Use modular arithmetic to show that a number is divisible by 11 iff the sum of its alternating digits is divisible by 11 We can expand the number $n = n_0 + 10n_1 + ... + (10^s)n_s$
Then we have $10^k ≡ (-1)^k \mod11$.
How do we go from here to here:
$n ≡ n_0 - n_1 + ... + (-1)^s n_s \mod 11$
I do not understand how the right hand side is congruent to (≡) n.
Furthermore, how do we jump from here to
"This shows if $11|n$ , then $n ≡ 0\mod 11$ (understand this) and so $n_0 - n_1 + ... + (-1)^s n_s = 0 \mod 11$ (do NOT understand where this jump in logic came from). Please explain.
| We know that $10^k \equiv (-1)^k \pmod{11}$. If $k$ is even, then $(-1)^k \equiv 1 \pmod{11}$. If $k$ is odd, then $(-1)^k \equiv -1 \pmod{11}$. Hence,
\begin{align*}
n & \equiv n_0 + 10n_1 + 10^2n_2 + 10^3n_3 + \cdots + 10^sn_s \pmod{11}\\
& \equiv n_0 + (-1)^1n_1 + (-1)^2n_2 + (-1)^3n_3 + \ldots + (-1)^sn_s \pmod{11}\\
& \equiv n_0 - n_1 + n_2 - n_3 + \cdots + (-1)^sn_s \pmod{11}
\end{align*}
If $11|n$, then $n \equiv 0 \pmod{11}$, so
$$n_0 - n_1 + n_2 - n_3 + \cdots + (-1)^sn_s \equiv 0 \pmod{11}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to find the roots of $x^4-i=0$ I need the manually analysis to calculate the roots without using the numerical methods
| $$x^4-i=0$$
$$x^4=i=e^{i(2k+1)\pi/2}=\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)$$
$$x=i^{1/4}=e^{i(2k+1)\pi/8}=\left[\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)\right]^{1/4}$$
Using De Moivre's formula
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin (n\theta)$$
$$x=e^{i(2k+1)\pi/8}=\cos\left(\frac{(2k+1)\pi}{8}\right)+i\sin\left(\frac{(2k+1)\pi}{8}\right)$$
$$k=0 , 1 , 2 , 3$$
$$x=e^{i\pi/8} ,\ e^{3i\pi/8}, \ e^{5i\pi/8} ,\ e^{7i\pi/8}$$
or
$$x=\cos\left(\frac{\pi}{8}\right)+i\sin\left(\frac{\pi}{8}\right)$$
$$x=\cos\left(\frac{3\pi}{8}\right)+i\sin\left(\frac{3\pi}{8}\right)$$
$$x=\cos\left(\frac{5\pi}{8}\right)+i\sin\left(\frac{5\pi}{8}\right)$$
$$x=\cos\left(\frac{7\pi}{8}\right)+i\sin\left(\frac{7\pi}{8}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Determining unknown coefficients of cubic splines The problem : Find $c$ in the following cubic spline.
$S \scriptstyle{1}$$(x)$ = $\large4 - \large\frac{11}{4}x + \large\frac{3}{4}x^3$, on $[0,1]$
$S \scriptstyle{2}$$(x)$ = $\large2 - \frac{1}{2}(x-1) + c(x-1)^2 -\frac{3}{4}(x-1)^3$, on $[1,2]$
My attempt:
1.) Checking that end points of the interval $[0,1]$;
$S \scriptstyle {1}$$(1)$ = $2 = S \scriptstyle{2} $$(1)$
2.)Checking the first derivative:
$S' \scriptstyle {1} $$(x)$ = $ -\frac{11}{4} + \frac{9}{4}x^2$;
$S' \scriptstyle {2} $$(x)$ = $ -\frac{1}{2} + 2cx - \frac{9}{4}(x-1)^2$
$S' \scriptstyle {1}$ $(1)$ = $ -\frac{1}{2} $ , $S' \scriptstyle {2} $ $(1)$ = $ -\frac{1}{2} + 2c$
At this point, it seems intuitive that $c = 0$, but the correct answer is $\frac{9}{4}$.
My textbook provides only an example on the topic but I get lost in the notation and get $c = 0$ for this problem and all the following problems of the same form. I'd appreciate a generalized algorithm or literature more than an answer, but anything is welcome.
| A cubic spline matches second derivatives as well. So we get, at $x=1$
$$
\frac{3}{4} 6x = 2c
$$
from which $c = \frac{9}{4}$.
The best explanation of cubic splines in my opinion is in Numerical Recipes, pages 120-124 or so.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$ using a trigonometric substitution I have this integral:
$$\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx$$
I tried to solve it with a trigonometric substitituon but I can't get any result.
I would appreciate if somebody could help me.
| Hint:
Put $x = \frac{3}{2}\tan\theta \Rightarrow dx = \frac{3}{2} \sec^2 \theta \ d\theta$, we have
\begin{align}
\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx&=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\cdot\sec^2\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{\sin^3\theta}{\cos^3\theta}\cdot\cos\theta\,\,d\theta\quad\Rightarrow\quad\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\mbox{and}\,\,\sec\theta=\frac{1}{\cos\theta}\\
&=\frac{3}{16}\int\frac{\sin^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{1-\cos^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\quad\Rightarrow\quad\mbox{set}\,\,t=\cos\theta \Rightarrow dt=-\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{t^2-1}{t^2}\,dt
\end{align}
Can you take it from here?
| {
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Find all numbers $c$ that satisfy Mean Value Theorem Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
My function is a simple one: $x^{1/3}$ on the interval $[0,1]$.
Where I'm screwing up (I believe) is setting the derivative $1/3(x^{-2/3})$ equal to $1$.
Sorry if that too many parentheses or too silly of a question. It's my first on here.
I know (via the back of the book) that the answer should be $\frac{\sqrt{3}}{9}$
Thanks.
| $$\dfrac{1}{3}x^{-2/3} = 1 \\ \implies \dfrac{1}{3}=x^{2/3} \\ \implies \left(\dfrac{1}{3}\right)^{3}=x^2 \\ \implies \sqrt{x^2}= \pm \sqrt{ \left(\dfrac{1}{3}\right)^{3}}$$ And for what it's worth, $$\sqrt{ \left(\dfrac{1}{3}\right)^{3}} = \sqrt{ \left(\dfrac{1}{3}\right)^{2}\left(\dfrac{1}{3}\right)} \\ = \left(\dfrac{1}{3}\right)\frac{1}{\sqrt{3}} \\ = \left(\dfrac{1}{3}\right)\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \\ = \left(\dfrac{1}{3}\right)\frac{1}{3} \cdot \sqrt{3} \\ = \frac{\sqrt{3}}{9}$$
| {
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$a^2\equiv 10\pmod b$ and $a^3\equiv 33\pmod b$ Let $a,b$ be positive integers such that $a^2\equiv 10\pmod b$ and $a^3\equiv 33\pmod b$. What are all possible values of $b$?
We have that $10a\equiv 33\pmod b$, but how does that determine the value of $b$?
[Source: Russian competition problem]
| For integers c, d, n, and k we have $c\equiv d \mod n \Rightarrow c^k\equiv d^k \mod n$. Then, using Mark Bennet's hint, we have $a^6\equiv 1000\mod b$ and $a^6\equiv 1089\mod b \Rightarrow 1000+bn=1089+bm\Rightarrow 89=b(n-m)$. But 89 is a prime number, so $b=1$ or $b=89$. $a=30$ and $b=89$ works. Lots of solutions exist for $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the remainder on division of $z^{400} + z^{303} + 1$ by $z^4-1$? I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-1$.
I expressed f as $f(z) = h(z)(z^4-1) + r(z)$ where $r(z)$ is a polynomial
Realising that $z^4-1 = (z^2-1)(z^2+1) = (z-1)(z+1)(z^2+1)$, I expressed f as
$f(z) = h(z)(z-1)(z+1)(z^2+1) + r(z)$
I also wrote f as $f(z) = q(z)(z-1) + 3$, since $f(1)=3$, and q is some polynomial.
So I had $f(z)= q(z)(z-1) + 3 = h(z)(z-1)(z+1)(z^2+1) + r(z)$
Hence, for some constant k, $q(z)=h(z)(z+1)(z^2+1) + k$
We know $f(-1) = 1$
From $f(z) = q(z)(z-1) + 3$, we have $f(-1)=q(-1)(-2) + 3 = 1$, which implies
$q(-1) = 1$.
Then $q(-1)=h(-1)(0) + k = 1$. Thus, $k=1$
We then have $q(z)=h(z)(z+1)(z^2+1) + 1$, and from this, we have
$f(z)= q(z)(z-1) + 3 = (h(z)(z+1)(z^2+1) + 1)(z-1) + 3$
Which simplifies to $f(z) = h(z)(z^4-1) + z+2$
And gives $r(z) =z+2$.
But I am given the answer is $r(z) = z^3+ 2$. What is going wrong here? Can I use modular arithmetic instead? How?
| Note that we can write $f(z)=h(z)(z^4-1)+r(z)$ where $r(z)$ is a polynomial of degree at most three (one less than the degree of $z^4-1$).
An unknown polynomial of degree $3$ has four coefficients to be determined. If we successively use $z=1, z=-1, z=i, z=-i$ - the four roots of $z^4-1$ the term involving $h(z)$ will vanish.
This gives us $f(1)=r(1)=3$, $f(-1)=r(-1)=1$, $f(i)=r(i)=2-i$, $f(-i)=r(-i)=2+i$
If $r(z)=az^3+bz^2+cz+d$ we have $$a+b+c+d=3$$$$-a+b-c+d=1$$
From which we obtain $b+d=2$ and $a+c=1$
Also $$-ai-b+ci+d=2-i$$ From which $-b+d=2$ so $b=0, d=2$ and $-a+c=-1$ so that $a=1, c=0$
$$r(z)=z^3+2$$
Where you seem to have gone wrong is when you get to
$$f(z)=q(z)(z-1)+3=h(z)(z-1)(z+1)(z^2+1)+r(z)$$
From this you can conclude that $(z-1)\left(q(z)-h(z)(z+1)(z^2+1)\right)=r(z)-3$ but that does not mean that the term in large brackets is a constant. It is that assumption which puts you off track.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Second order cone with quadratic interpretation Could you please help me to understand how the second part of the equation (quadratic form) derived form the first one?
The basic definition of the second-order cone is:
$C = \big\{(x,t) \in \mathbb{R}^{n+1} | \|x\|_2 \leq t \big\}= \big\{(x,t) \in \mathbb{R}^{n+1} | x^Tx \leq t^2 \big\}= \Bigg\{
\begin{pmatrix}
x\\
t
\end{pmatrix}
\bigg|
\begin{pmatrix}
x\\
t
\end{pmatrix}^T
\begin{pmatrix}
I & 0\\
0 & -1
\end{pmatrix}\begin{pmatrix}
x\\
t
\end{pmatrix}\leq 0,
t \geq 0\Bigg\}$
| Hint: Why don't you try expanding that quadratic form in terms of $x$ and $t$.
\begin{align}
\begin{pmatrix}
x\\
t
\end{pmatrix}^T
\begin{pmatrix}
I & 0\\
0 & -1
\end{pmatrix}\begin{pmatrix}
x\\
t
\end{pmatrix}
\end{align}
First expand
\begin{align}
\begin{pmatrix}
I_{n\times n} & 0_{n\times 1}\\
0_{1\times n} & -1
\end{pmatrix}
\begin{pmatrix}
x_{n\times 1}\\
t
\end{pmatrix}
\end{align}
using the rules of block matrix multipication and do again the same using the term on the left side, then note that $x^Tx=||x||_2^2$.
| {
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"url": "https://math.stackexchange.com/questions/1002376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple first order differential equation I've got another first order differential that has me stumped. Please see my work and let me know where I'm going wrong:
$$y' = \frac{6x^2}{y(1+x^3)}$$
$$y'y = \frac{6x^2}{1+x^3}$$
$$ y.\frac{dy}{ dx} = \frac{6x^2}{1+x^3} $$
$$\int y dy = \int 6x^2\frac{1}{1+x^3} dx$$
$$\frac{1}{2}y^2 = 2x^3 \ln|1 + x^3| + c$$
$$y^2 = 4x^3 \ln|1 + x^3| + c$$
$$y = \pm \sqrt{4x^3 \ln |1 + x^3| + c}$$
Unfortunately, my answer is wrong and I'm not quite sure why. Please help.
| Unlike the integral of a sum or difference of functions , where $$\int (f(x) \pm g(x))\,dx = \int f(x) \,dx \pm \int g(x)\,dx,$$ the integral of the product of functions is not the product of the integrals of each function. I.e., $$\int f(x)g(x)\,dx \neq \left(\int f(x)\,dx\right)\cdot \left(\int g(x) \,dx\right)$$
And in any case, $\displaystyle \int \dfrac 1{1+x^3}\,dx \neq \ln|1 + x^3|+C$, because $$u = 1+x^3 \iff x = (u-1)^{1/3} \implies dx = \frac 13(u-1)^{-2/3}\,du$$ to give us $$\int \dfrac 1{1+x^3}\,dx = \frac 13 \int \frac{du}{u(u-1)^{2/3}}\neq \ln|u| + C = \ln|1 + x^3| + C$$
Okay, so lets take a different direction. Note that if $u = (1+x^3), \;du = 3x^2\,dx.\;$ We see that $du$ is very close to our numerator, which is $6x^2 = 2\cdot 3x^2$.
So $$\int \dfrac{6x^2}{1+x^3}\,dx =2\int \dfrac{3x^2\,dx}{1+x^3} =2\int \dfrac{du}{u}= 2 \ln|u| + C$$ $$ =2\ln|1+x^3| + C, \text{ or } \ln((1+x^3)^2) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding $\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x$ How do we prove that $$I(m)=\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x=\frac{\log{m}}{m^2-1}$$
I see that $$I(m)=\frac{\partial}{\partial m} \int_{0}^{\pi/2} \arctan({m\tan x}) \ \mathrm{d}x$$
But I don't see how to use this fact. Can we? Please help me out, and if possible please post a solution using differentiation under the integral sign.
| $\def\artanh{{\rm{artanh}}\;}$Denote the evaluated integral as $I$ and rewrite it as follows
\begin{align}
I&=\frac{1}{2}\int_0^{\Large\frac{\pi}{2}}\frac{\sin2x}{\cos^2x+m^2\sin^2x}\,dx\\
&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin2x}{m^2+1-(m^2-1)\cos2x}\,dx\\
&=\frac{1}{2}\int_0^{\large\pi}\frac{\sin x}{m^2+1-(m^2-1)\cos x}\,dx\\
\end{align}
where we use trigonometric identity $\sin^2\theta=\dfrac{1-\cos2\theta}{2}$ and $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$ and map $2x\mapsto x$. Now, using identity (proof can be seen here)
\begin{equation}
1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)=\frac{a^2-b^2}{a^2+b^2-2ab\cos x}\qquad,\qquad\mbox{for}\, |b|<a
\end{equation}
and the correspondence values $a=\dfrac{m+1}{\sqrt{2}}$ and $b=\dfrac{m-1}{\sqrt{2}}$, one may find
\begin{align}
I&=\frac{1}{4m}\int_0^{\large\pi}\left[\sin x+2\sum_{n=1}^\infty \left(\frac{m-1}{m+1}\right)^n\cos(nx)\sin x\right]\,dx\\
&=\frac{1}{2m}+\frac{1}{2m}\sum_{n=2}^\infty \left(\frac{m-1}{m+1}\right)^n\int_0^{\large\pi}\cos(nx)\sin x\,dx\\
&=\frac{1}{2m}-\frac{1}{2m}\sum_{n=2}^\infty \left(\frac{m-1}{m+1}\right)^n\frac{\cos(n\pi)+1}{n^2-1}\\
&=\frac{1}{2m}-\frac{1}{m}\sum_{n=1}^\infty \left(\frac{m-1}{m+1}\right)^{2n}\frac{1}{4n^2-1}\\
&=\frac{1}{2m}-\frac{1}{2m}\left[1+\left(\frac{m-1}{m+1}\right)\artanh\left(\frac{m-1}{m+1}\right)-\left(\frac{m+1}{m-1}\right)\artanh\left(\frac{m-1}{m+1}\right)\right]\\
&=\frac{2}{m^2-1}\artanh\left(\frac{m-1}{m+1}\right)\\
&=\frac{\ln m}{m^2-1}\qquad\qquad\square
\end{align}
For the last step evaluation, we use Taylor series for $\artanh$function
\begin{equation}
\artanh x=\sum_{n=0}\frac{x^{2n+1}}{2n+1}\qquad,\qquad\mbox{for}\,\, |x|<1
\end{equation}
and identity
\begin{equation}
\artanh x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\qquad,\qquad\mbox{for}\,\, |x|<1
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
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How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$ How to solve:
$$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$
What I have done:
I) $x < 0$:
Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$
II) $x > 0$:
$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$
$$\frac{|x+4|}{|ax+2|} > \frac1x$$
because $x > 0$ we can transform it to:
$x^2 + 4x > |ax + 2|$
$(x^2 + 4x)^2 > (ax + 2)^2$
$(x^2 + 4x)^2 - (ax + 2)^2 > 0$
$(x^2 + 4x - ax - 2) (x^2 +4x +ax + 2) > 0$
$(x^2 + (4-a)x - 2)(x^2 + (4+a)x + 2) > 0$
I am kind of stuck here. I also need to discuss the solution for various values of the parameter a. What is the easiest way out of this step? Something with the discriminants, perhaps?
| The answer depends on the discriminants of the two quadratic polynomials:
$D_1= (a-4)^2+8$ and $D_2= (a+4)^2-8$. This shows that you have up to 4 roots.
Can you go on from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + \frac{7}{9}$$
I also see that
$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$
I know that in a geometric series the first term is $a$ and the second term is $ar$.
This allows me to see that
$$\left(\frac{1}{9}+\frac{7}{9}\right)r=\frac{1}{9^2}+\frac{7^2}{9^2}$$
which when solved for $r$ gives the value $\frac{25}{36}$.
Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.
But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?
| This is not a geometric series.
But it is the sum of two geometric series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}
= \sum_{n=1}^{\infty} {\frac{1}{9^n}}+
\sum_{n=1}^{\infty} {\frac{7^n}{9^n}}
=\frac 19 \frac 1{1-\frac 19} +\frac 79\frac 1{1-\frac 79}$$
because
$$
\left|\frac 19\right|<1
\\\left|\frac 79\right|<1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Where is the error in this proof : Prove that:
$$\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
But I obtain this equal zero.
My proof:
From Weierstrass definition of Gamma we have
$$\frac{1}{\Gamma(z)}=z.e^{\gamma z}.\prod_{n=1}^{\infty}[ (1+\frac{z}{n}).e^{\frac{-z}{n}} ] $$
$$ -log( \Gamma(z))=log (z)+{\gamma z}+\sum_{n=1}^{\infty}[ log((1+\frac{z}{n}))-{\frac{z}{n}} ] $$
$$ log( \Gamma(z))=-log (z)-{\gamma z}-\sum_{n=1}^{\infty}[ log((1+\frac{z}{n}))-{\frac{z}{n}} ] $$
$$\frac {\Gamma'(z)}{\Gamma(z)} =-\frac{1}{z} -{\gamma }-\sum_{n=1}^{\infty}[ ((\frac{1}{z+n}))-{\frac{1}{n}} ]$$
This implies >>
$$\frac {2\Gamma'(2z)}{\Gamma(2z)} - \frac {\Gamma'(z)}{\Gamma(z)} - \frac {\Gamma'(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})}$$
$$= \sum_{n=0}^\infty \frac{1}{z+\frac{1}{2}+n} + \sum_{n=1}^\infty \frac{1}{z+n}-\sum_{n=1}^\infty \frac{2}{2z+n} $$
$$= \sum_{n=0}^\infty \frac{1}{z+\frac{1}{2}+n} + \sum_{n=0}^\infty \frac{1}{z+n+1}-\sum_{n=0}^\infty \frac{2}{2z+n+1}$$ $$=
(\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+1})-(\frac{2}{2z+1})+(\frac{1}{z+\frac{3}{2}})+(\frac{1}{z+2})-(\frac{2}{2z+2})+(\frac{1}{z+\frac{5}{2}})+(\frac{1}{z+3})-(\frac{2}{2z+3})+... $$
$$ = (\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+1})-(\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+\frac{3}{2}})+(\frac{1}{z+2})-(\frac{1}{z+1})+(\frac{1}{z+\frac{5}{2}})+(\frac{1}{z+3})-(\frac{1}{z+\frac{3}{2}})+...=0 $$
Where is the error ?
| Let's rewrite your equation :
$$\tag{1}\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
as
$$\left[\log\Gamma(2z)-\log\Gamma(z)-\log \Gamma\left(z+\frac 12\right)\right]'=2 \log 2$$
or
$$\tag{2}\left[\log\frac{\Gamma(2z)}{\Gamma(z)\,\Gamma\left(z+\frac 12\right)}\right]'=2 \log 2$$
which is easily deduced from the "duplication formula" $\;\displaystyle \Gamma(z)\,\Gamma\left(z+\frac 12\right)=2^{1-2z}\sqrt{\pi}\;\Gamma(2z)$.
Now let's see your derivation. I can follow you until :
$$\tag{3}\psi(z):=\frac {\Gamma'(z)}{\Gamma(z)} =-\frac 1{z} -\gamma-\sum_{n=1}^{\infty}\left[ \frac 1{z+n}-\frac 1n\right]$$
After that (as explained by MPW) you can't remove the $\,\dfrac 1n\,$ parts since you would obtain divergent series.
Let's continue the computations at this point (without expanding the $\psi(2z)$ part) :
\begin{align}
&f(z):=2\,\psi(2z)-\psi(z)-\psi\left(z+\frac 12\right)\\
&=2\,\psi(2z)+\frac 1{z}+\frac 1{z+1/2}+2\gamma+\sum_{n=1}^{\infty}\left[\frac 1{z+n}+\frac 1{z+1/2+n}-\frac 2{n}\right]\\
&=2\,\left[\psi(2z)+\frac 1{2z}+\gamma+\frac 1{2z+1}+\sum_{n=1}^{\infty}\left[\frac 1{2z+2n}+\frac 1{2z+2n+1}-\frac 1{2n}-\frac 1{2n+1}+\\\frac 1{2n+1}-\frac 1{2n}\right]\right]\\
&=2\,\left[\psi(2z)+\frac 1{2z}+\gamma+\sum_{m=1}^{\infty}\left[\frac 1{2z+m}-\frac 1m\right]+\frac 11+\sum_{n=1}^{\infty}\left[\frac 1{2n+1}-\frac 1{2n}\right]\right]\\
&=2\,\left[\psi(2z)-\psi(2z)-\sum_{m=1}^{\infty}\frac{(-1)^m}m\right]\\
&=2\,\log(1+1)\\
\end{align}
(with $m$ combining the $2n$ and $2n+1$ terms)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A thinking problem of limit from my teacher. Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}} \right)} \right)} \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right].$$
| Step 1. Since for small $x$ we have $\frac{1}{\sqrt{1+x}}=1-\frac{x}{2}+\frac{3x^2}{8}+O(x^3),$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3k^2}{8n^4}\right)+O\left(\frac{1}{n^3}\right)$$
hence:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=1-\frac{1}{4n}-\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right).$$
Step 2.
$$\sum_{k=1}^{n}\frac{1}{\left(\sqrt{n^2+k}\right)^3}=\frac{1}{n^2}+O\left(\frac{1}{n^3}\right).$$
Step 3. Since for small $x$ we have $\sin x = x-\frac{x^3}{6}+O(x^5)$,
$$\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}=\pi-\frac{\pi}{4n}-\frac{\pi}{8n^2}-\frac{\pi^3}{6n^2}+O\left(\frac{1}{n^3}\right).$$
Step 4. Since for small $x$ we have $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$,
$$ n\log\left(\frac{1}{\pi}\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}\right)=-\frac{1}{4}-\left(\frac{5}{32}+\frac{\pi^2}{6}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right).$$
Step 5. Exponentiating, it follows that the value of the limit is:
$$ L = -\frac{15+16\,\pi^2}{96\, e^{1/4}}.$$
| {
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"url": "https://math.stackexchange.com/questions/1011878",
"timestamp": "2023-03-29T00:00:00",
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prove that $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$ Let a,b,c,d be positive real numbers such that$\ a+b+c+d=1$, than prove that $$\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$$
This question is from the practice material of Indian Math Olympiad. I think that this will incorporate the AM-GM-HM inequality.I am not able to learn the perfect heuristic to solve this inequality.
| By AM-GM:
$ab\le \left(\frac{a+b}{2}\right)^2=\frac{(a+b)^2}{4}$
Thus
$$\sum_{\text{sym}}\frac{ab}{a+b}\le \sum_{\text{sym}}\frac{a+b}{4}=\frac{3}{4}$$
Equality is achieved iff $a=b=c=d$.
| {
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"url": "https://math.stackexchange.com/questions/1012176",
"timestamp": "2023-03-29T00:00:00",
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running time of an algorithm I am trying to prove an algorithm with input size $n$ satisties the recurence relation (for $n>=1$)
$T(n) = T(n-1)+n$ and an initial condition of $T(1)=1$ has running time in $Θ(n^2$).
By using telescope, I've got up to the point where I got
$T(n) = n+(n-1)+(n-2)+......+ 1$
But I can not go any further from this point.
I've been googling and searching for a while and I saw that
$n+(n-1)+(n-2)+......+ 1$ can be written as $\frac{n(n+1)}{2}$
Can someone explain how would I get $\frac{n(n+1)}{2}$ from $n+(n-1)+(n-2)+......+ 1$?
| First note that
$$ n+(n-1)+(n-2)+\cdots +1$$
Can be rewritten as
$$ 1+2+\dots +n$$
Now let's use mathematical induction
Hypothesis
$$ \forall n\in\mathbb{N}:n\ge 1$$
$$ 1+2+\dots +n=\frac{n(n+1)}{2} $$
Basis Step
Let $n=1$,
$$ 1=\frac{1\cdot (1+1)}{2} $$
$$ 1=\frac{1\cdot 2}{2} $$
$$ 1=1 $$
Inductive Step
Let $n=k+1$,
$$ 1+2+\dots +k+(k+1)=\frac{(k+1)(k+1+1)}{2} $$
$$ 1+2+\dots +k+(k+1)=\frac{(k+1)(k+2)}{2} $$
$$ 1+2+\dots +k+(k+1)=\frac{k^2+2k+k+2}{2} $$
$$ 1+2+\dots +k+(k+1)=\frac{k^2+k}{2}+\frac{2k+2}{2} $$
$$ 1+2+\dots +k+(k+1)=\frac{k^2+k}{2}+(k+1)$$
$$ 1+2+\dots +k=\frac{k^2+k}{2} $$
$$ 1+2+\dots +k=\frac{k(k+1)}{2} $$
Therefore
$$ 1+2+\dots +n=\frac{n(n+1)}{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subtracting Quarters of Squares Equals Multiply?! Can anyone explain to me how/why this works (hopefully in mostly layman's terms)?
It seems pretty magical to me at the moment.
$${{(a+b)^2\over4} - {(a-b)^2\over4}} = a b.$$
| In addition to the direct derivations already shown,
your magical equation is closely related to the formula
$$x^2 - y^2 = (x + y)(x - y).$$
Just set $a = x + y$ and $b = x - y.$
Then $\frac{a+b}2 = x$ and $\frac{a-b}2 = y,$
so $x^2 = \frac{(a+b)^2}4$ and $y^2 = \frac{(a-b)^2}4.$
Use these facts to replace $x^2,\ y^2,\ x + y,$ and $x - y$
in the equation above and you will have derived your magical equation in $a$ and $b.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 4
} |
Find minima of the multivariable function $\frac{4}{x^2+y^2+1}+2 xy$ $$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the domain: }1/5\leq x^2+y^2\leq 4$$
I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the lower boundary:
$$ x=\frac{1}{\sqrt5}\cos(t), \ y=\frac{1}{\sqrt5}\sin(t) \\
\frac{4}{\frac{6}{5}}+\frac{2}{\sqrt5}(\cos(t)\sin(t))\\
\frac{df}{dt}=\frac{2}{\sqrt5}\cos(2t)=0\\
\tan(t)=1\\
x=\frac{1}{\sqrt5}\lvert\cos(\frac{\pi}{4})\rvert \ , \ \ y=\frac{1}{\sqrt5}\lvert\sin(\frac{\pi}{4})\rvert \\
x=\pm\frac{1}{\sqrt{10}},y=\pm\frac{1}{\sqrt{10}}$$
When i resubstitute this into the original function I get the wrong answer.
$$f\left(\frac{1}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)=\frac{4}{\frac{12}{10}}+\frac{2}{10}=53/15
$$
| $f_x(x,y) = 0 \to \dfrac{-8x}{(x^2+y^2+1)^2} = 2y$
$f_y(x,y) = 0 \to \dfrac{-8y}{(x^2+y^2+1)^2} = 2x$.
Thus $(x,y) = (0,0)$ is a critical point, also $\dfrac{x}{y} = \dfrac{y}{x} \to x = \pm y$
Consider $g(x) = f(x,x) = \dfrac{4}{2x^2+1} + 2x^2$ with $1/5 \leq 2x^2 \leq 4$. let $u = 2x^2$, then $h(u) = g(x,x) = \dfrac{4}{u+1} + u$, $1/5 \leq u \leq 4\to h'(u) = 1-\dfrac{4}{(u+1)^2} = 0 \iff u+1 = \pm 2 \to u =1,-3 \to u = 1$, since $u > 0$.
$h(1) = 3, h(1/5) = 53/15, h(4) = 24/5$.
And $r(x) = f(x,-x) = \dfrac{4}{2x^2+1} - 2x^2 = \dfrac{4}{u+1}-u = q(u)$, for $1/5 \leq u \leq 4$.
$q'(u) = -1-\dfrac{4}{(u+1)^2} < 0 \to q(u) \geq q(4) = 4/5-4$.
Thus the min value of $f(x,y)$ is $4/5-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $\lvert z\rvert^2z-3\overline z=0$ I was trying to solve the equation using the identities $z=x+iy$; $\overline z=x-iy$ and $\lvert z\rvert^2=x^2+y^2$ so as to get
$(x^2+y^2)(x+iy)-3(x-iy)=0$, that is $x^3+x^2iy+xy^2+iy^3-3x+3iy=0$
Now, a complex number is null $\iff Re(z)=Im(z)=0$, therefore
$x^3+xy^2-3x=0\implies x=0\lor x^2+y^2=3$, that is $\lvert z\rvert=\sqrt 3$
$x^2y+y^3+3y=0\implies y=0\lor x^2+y^2=-3$, that is $\lvert z\rvert=-\sqrt 3$
Is it correct to say that the solutions are $z=0$, $z=\sqrt 3$, and $-\sqrt 3$?
| HINT:
As $|z|^2=z\cdot\bar z,$
we have $\bar z(z^2-3)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lesser known derivations of well-known formulas and theorems What are some lesser known derivations of well-known formulas and theorems?
I ask because I recently found a new way to derive the quadratic formula which didn't involve completing the square as is commonly taught. Doing so I was wondering what other proofs and derivations for other formulas that have remained unknown to most people?
Whether it be because the proof is too complex or less pretty, I still find it insightful to see different ways to solve a problem. To me it makes me understand proofs better, and thus also giving a better comprehension of it.
$$
\begin{align}
&\text{Given a quadratic function } f:\\[0.1em]
f &=ax^2+bx+c = a(x-r_1)(x-r_2) = ax^2-a(r_1+r_2)+ar_1r_2\\[0.1em]
a &= a,\enspace \frac{b}{a} = -(r_1+r_2),\enspace \frac{c}{a} = r_1r_2\\[1em]
f' &= 2ax+b, \enspace f'(x) = 0 \Rightarrow x = -\frac{b}{2a}\\[0.2em]
\text{This is an}& \text{ extremum of } f \text{, and is equidistant from each root } r_1, \enspace r_2 \text{ as shown:}\\[0.4em]
\frac{b}{a} &= -(r_1+r_2) \iff -\frac{b}{2a} = \frac{r_1+r_2}{2} \\[1em]
\Rightarrow \enspace &\text{The roots are of the form } r= -\frac{b}{2a}\pm d\\[1em]
\frac{c}{a} = r_1r_2 &= (-\frac{b}{2a}+d)(-\frac{b}{2a}-d) = \frac{b^2}{4a^2}-d^2\\[0.2em]
\Rightarrow\enspace& d^2 = \frac{b^2}{4a^2}-\frac{c}{a} = \frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}\\[0.2em]
\Rightarrow\enspace&d = \pm\frac{\sqrt{b^2-4ac}}{2a}\\[1em]
\text{Which yields }& r = -\frac{b}{2a}\pm d = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\enspace\square
\end{align}
$$
| The proof in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563
that
$(1+1/n)^n
\to e$
which uses the arithmetic-geometric mean
inequality (AGMI),
which we will use in the
form
$((v_1+v_2+...v_n)/n)^n > v_1v_2...v_n$
(all $v_i$ positive)
with equality if and only if all the $v_i$ are
equal.
Consider $n$ values of $1+1/n$
and $1$ value of $1$.
By the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n
$, or
$(1+1/(n+1))^{n+1} > (1+1/n)^n$.
Consider $n$ values of $1-1/n$
and $1$ value of $1$.
By the AGMI,
$(n/(n+1))^{n+1} > (1-1/n)^n$
or
$(1+1/n)^{n+1} < (1+1/(n-1))^n
$.
Therefore
$(1+1/n)^n$
is increasing
and
$(1+1/n)^{n+1}$
is decreasing.
Since
$0
< (1+1/n)^{n+1}-(1+1/n)^n
= (1+1/n)^{n}/n
<4/n
$,
the two sequences converges to
a common limit.
To show that
$(1+1/n)^{n}
< 4
$,
note that
$(1+1/n)^{n}
<(1+1/n)^{n+1}
\le 2^2
$
(with
$n=1)
=4
$,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 5
} |
How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \geq 4\quad\mbox{and}\quad{\rm gcd}\left(\,a,3\,\right) = 1
$$
| Alternatively, let's consider
$$f(a)=\sum_{n=0}^{\infty }\left(\frac{3}{2n+1-\frac{a}{3}}-\frac{3}{2n+1+\frac{a}{3}}\right)=9\sum_{n=0}^{\infty }\left(\frac{1}{6n+3-a}-\frac{1}{6n+3+a}\right)$$
so that our original sum is $f'(a)$.
$$\begin{align}
f(a)&=9\sum_{n=0}^{\infty }\int_0^1 \left(x^{6n+2-a}-x^{6n+2+a}\right)\,dx\\
&=9\int_0^1\sum_{n=0}^{\infty }\left(x^{6n+2-a}-x^{6n+2+a}\right)\,dx\\
&=9\int_0^1\left(\frac{x^{2-a}-x^{2+a}}{1-x^6}\right)\,dx\\
&=\frac{3}{2}\int_0^1\left(\frac{t^{-\large\frac{3+a}{6}}-t^{-\large\frac{3-a}{6}}}{1-t}\right)\,dt\\
&=\frac{3}{2}\left(\psi\left(\frac{3+a}{6}\right)-\psi\left(\frac{3-a}{6}\right)\right)\\
&=\frac{3\pi}{2}\cot\left(\frac{3-a}{6}\pi\right)
\end{align}$$
See the integral representation and the reflection formula of digamma function. Therefore
$$f'(a)=\sum_{n=0}^{\infty }\left[\frac{1}{\left(2n+1-\frac{a}{3}\right)^2}+\frac{1}{\left(2n+1+\frac{a}{3}\right)^2}\right]=\frac{\pi^2}{4}\sec^2\left(\frac{\pi a}{6}\right)$$
then it follows $f'(a)=\pi^2$ for $a\ge4$ and $\text{gcd }(a,3)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Find the interval of convergence of $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$ How to find the interval of convergence of the following series: $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$
I have no idea what to proceed. Any help? Thanks!
| I would recommend splitting the $x^k*k$ and the $\frac{x^k}{k}$ terms and seeing where the two intervals intersect.
$(∑(2k+1)x^{2k+1}) + (∑\frac{x^{2k}}{2k})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why can we square both sides when LHS and RHS is square rooted? When we have something in this form
$$\sqrt{x + a} = \sqrt{y + b},$$
a common technique to solve is to square both side so that:
$$(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b.$$
I'm an engineer and not a mathematician. As I understand it engineers do lots of things that are mathematically frowned upon. However, while those things do make intuitive sense to me this technique makes no sense to me. Specifically what if we consider the alternative road.
\begin{align*}
\sqrt{x + a} &= \sqrt{y + b}\\
\sqrt{x + a} \cdot \sqrt{y + b} &= y + b\\
&\,\, \vdots
\end{align*}
It is unclear to me how we can conclude that $(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b$ is true.
| Travelling down the alternative road, we see that
\begin{align*}
\sqrt{x+a} &= \sqrt{y+b} & \\
\sqrt{x+a}\times\sqrt{y+b} &= \sqrt{y+b}\times\sqrt{y+b} &(\text{multiplying both sides by}\ \sqrt{y+b})\\
\sqrt{x+a}\times\sqrt{y+b} &= y + b &(\text{simplifying})\\
\sqrt{x+a}\times\sqrt{x+a} &= y + b &(\text{using the equation}\ \sqrt{x+a}=\sqrt{y+b})\\
x + a &= y + b &(\text{simplifying}).
\end{align*}
Multiplying by $\sqrt{x+a}$ has exactly the same effect on the equation as multiplying by $\sqrt{y+b}$, but we can replace one expression by the other whenever it is convenient. For example, going from the third equation to the fourth, $\sqrt{y+b}$ on the left hand side was replaced with $\sqrt{x+a}$; this was done to allow for algebraic simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of $\exp[f(x,y)]$ Here is the question i want to solve.
$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy\,dz$$
I know that $\exp$ is $e^{f(x)}$ and i can find $\int e^{f(x)}\,dx$ but with two variables makes me confusing.
| \begin{align}
\text{Let }u & = y+z, \\
\text{and }v & = y-z. \\[15pt]
\text{It follows that }y & = \frac{u+v} 2, \\[6pt]
\text{and }z & = \frac{u-v} 2.
\end{align}
Then
\begin{align}
y^2-yz+z^2 & = \left( \frac{u+v} 2 \right)^2-\frac{u+v} 2 \cdot \frac{u-v} 2 + \left(\frac{u-v} 2\right)^2 \\[6pt]
& = \frac{u^2+3v^2} 4.
\end{align}
And $dy\,dz = \left|\dfrac{\partial(y,z)}{\partial(u,v)}\right|\,du\,dv= \dfrac 1 2 \,du\,dv$. So
\begin{align}
& \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy\,dz \\[8pt]
= {} & \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \exp\left[\frac{-2}3 \left(\frac{u^2+3v^2} 4\right) \right] \frac{du\,dv} 2\\[8pt]
= {} & \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} \exp\left[\frac{-2}3 \frac{u^2}4 \right] \underbrace{\exp\left[\frac{-2}3 \frac{3v^2}4 \right] \frac 1 2}\ \ du\right)\,dv.
\end{align}
The part over the $\underbrace{\text{underbrace}}$ does not change as $u$ goes from $-\infty$ to $+\infty$ and is a factor of the function being integrated (i.e. one multiplies it by something to get that function). Therefore, it can be pulled out of the inside integral, leaving a standard Gaussian integral on the inside. Once you've reduced that to a number, you've got one more standard Gaussian integral.
PS: Here's another way:
\begin{align}
& \int_{-\infty}^{+\infty} \exp\left[{-2\over3}(y^2-yz+z^2)\right]\,dy \\[8pt]
= {} & \int_{-\infty}^{+\infty} \exp\left[{-2\over3}\left(\left(y^2-yz + \frac{z^2}4\right) + \frac{3z^2}4 \right) \right]\,dy \quad (\text{completing the square}) \\[8pt]
= {} & \int_{-\infty}^{+\infty} \exp\left[{-2\over3} \left(y^2- \frac z 2 \right)^2 \right] \exp\left[{-2\over3}\frac{3z^2}4 \right]\,dy \\[8pt]
= {} & \exp\frac{-z^2} 2 \int_{-\infty}^{+\infty} \exp\left(\frac{-2}3 \left(y - \frac z 2\right)^2\right) \, dy \\[8pt]
= {} & \exp\frac{-z^2} 2 \int_{-\infty}^{+\infty} \exp\left(\frac{-2}3 u^2 \right)\,du \\[8pt]
= {} & A\exp\frac{-z^2}2
\end{align}
where $A$ is the value of the integral with respect to $u$. Then do the outside integral in which the variable is $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral with branch cut ( Problem while calculating residue) While calculating this integral $\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}(1+x^2)}$ , I am really struggling to calculate the residue at (-i), I am getting the value of residue as $\frac{-1}{2\sqrt{2}i}$, but for the value of residue is $\frac{1}{2\sqrt{2}i}$ with no minus sign in it. Now, what I might have been doing wrong since I used $\frac{1}{\sqrt{2}(1+(e^{i.3\pi/2})^2}$. I think I might have been doing some wrong in branch cut. Help me to figure it out.
| We will do the residue calculation since it is necessary to conclude on
this problem.
Suppose we seek to compute
$$Q = \int_{-1}^{+1} \frac{dx}{\sqrt{1-x^2}(1+x^2)}.$$
Re-write this as
$$\int_{-1}^{+1}
\exp(-1/2\mathrm{LogA}(1+z))
\exp(-1/2\mathrm{LogB}(1-z))
\frac{1}{1+z^2} dz$$
and call the function $f(z).$
We must choose two branches of the logarithm $\mathrm{LogA}$ and
$\mathrm{LogB}$ so that the cut is on the real axis from $-1$ to $+1.$
This is accomplished when $\mathrm{LogA}$ has the cut on the negative
real axis and $\mathrm{LogB}$ on the positive real axis. The poles are
simple so to compute the residues we merely need to evaluate the
logarithms at these two points.
For the first one at $\rho_0=i$ we put $$1+i = \sqrt{2} e^{1/4\pi i}$$
and $$1-i = \sqrt{2} e^{7/4\pi i}$$ to get the residue
$$\frac{1}{2i}
\exp(-1/2\times \log\sqrt{2} -1/2\times 1/4\pi i)
\exp(-1/2\times \log\sqrt{2} -1/2\times 7/4\pi i)
\\ = \frac{1}{2i}
\exp(-\log\sqrt{2} - \pi i)
= - \frac{1}{2i\sqrt{2}}.$$
For the second one at $\rho_1=-i$ we put $$1+(-i) = \sqrt{2}
e^{-1/4\pi i}$$ and $$1-(-i) = \sqrt{2} e^{1/4\pi i}$$ to get the
residue
$$-\frac{1}{2i}
\exp(-1/2\times \log\sqrt{2} +1/2\times 1/4\pi i)
\exp(-1/2\times \log\sqrt{2} -1/2\times 1/4\pi i)
\\ = - \frac{1}{2i}
\exp(-\log\sqrt{2})
= - \frac{1}{2i\sqrt{2}}.$$
Now using the dogbone contour shown in the accepted answer and
traversed counterclockwise we pick up $2Q.$ This is because for
$x\in(-1,1)$ we get above the cut the value
$$\exp(-1/2\log (1+x))\exp(-1/2\log (1-x) -1/2\times 2\pi i)
= - \frac{1}{\sqrt{1-x^2}}$$
and below the cut
$$\exp(-1/2\log (1+x))\exp(-1/2\log (1-x))
= \frac{1}{\sqrt{1-x^2}}.$$
Therefore
$$ 2Q= -2\pi i
\left(\mathrm{Res}_{z=i} f(z)
+ \mathrm{Res}_{z=-i} f(z)
+ \mathrm{Res}_{z=\infty} f(z)\right).$$
The residue at infinity is zero because $f(z)$ is $O(1/R^3)$ on the circle.
This gives
$$2Q = - 2\pi i
\left(- \frac{1}{2i\sqrt{2}} - \frac{1}{2i\sqrt{2}}\right)
= \pi \frac{2}{\sqrt{2}}
\quad\text{or}\quad
Q = \frac{\pi}{\sqrt{2}}.$$
In order to be rigorous we also need to show continuity across the two overlapping cuts on $(-\infty, -1)$ as shown in this MSE link.
Remark. It really helps to think of the map from $z$ to $-z$ as a $180$ degree rotation when one tries to visualize what is happening here.
| {
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"url": "https://math.stackexchange.com/questions/1027015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?
| I would like to add to the above answer, the symmetry put a spark in my brain and i came up with this.
As in answer stated by peiter.
Assume
$X=\large\frac{1}{a+1}+\large\frac{1}{b+1}+\large\frac{1}{c+1}$ $\space\space\space\space\space\space\space\space$ and
$\large\frac{1}{a+\omega}+\large\frac{1}{b+\omega}+\large\frac{1}{c+\omega}=2\omega^2$
$\large\frac{1}{a+\omega^2}+\large\frac{1}{b+\omega^2}+\large\frac{1}{c+\omega^2}=2\omega$
Adding the three equation and considering first term related to "$a$"
$X+2\omega+2\omega^2=\large\frac{1}{a+1}+\large\frac{1}{a+\omega}+\large\frac{1}{a+\omega^2}+other\space terms$
simplifies to,
$X+2\omega+2\omega^2=\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$
Now as according to solution given by pieter $X=2$
Therefore
$\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$ must equal zero.
If somehow you can prove that , the answer may become simpler.(i.e., to prove $X=2$)
I would suggest trying to multiplying and dividing the express by $\omega$ and $\omega^2$ to arrive at other expressions that may yield this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions? Let $a,b,c$ be positive integers satisfying
\begin{align}
2a^2-1 &= b^2, \\
2a^2+1 &= 3c^2.
\end{align}
The trivial solution is $(a,b,c)=(1,1,1)$. Are there others?
|
NOTE: As pointed out by Erick Wong, there was a flaw in the original proof. It may be fixable, though I haven't found a fix yet; leaving here for history (and just in case anyone else can take this method to the goal line).
Theorem. The Diophantine system of equations
\begin{align}
2a^2-1 &= b^2, \\
2a^2+1 &= 3c^2
\end{align}
has only the trivial solution $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$.
Proof. Evidently, $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$ is a solution. We also see that $\lvert c \rvert = 1$ forces $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$. We now assume, contrary to the theorem, that there is a solution with $bc > 1$ [positivity assumed without loss of generality].
Multiplying the two original equations yields
\begin{align}
(2a^2-1)(2a^2+1) &= (b^2)(3c^2) \\
(2a^2)^2 - 1 &= 3(bc)^2.
\end{align}
Evidently $bc$ is odd, say $bc=2u+1$ for an integer $u \ge 1$ [because $bc > 1$]. There are exactly two cases to consider.
Case 1: $a \equiv 1\!\pmod{3}$. Say $a=3v+1$ for an integer $v \ge 1$ [because $a > 1$]. Substituting, expanding, and simplifying yields
\begin{align}
\bigl(2(3v+1)^2\bigr)^2 - 1 &= 3(2u+1)^2 \\
324v^4 + 432v^3 + 216v^2 + 48v + 3 &= 12u^2+12u+3 \\
v(3v+2)(9v^2+6v+2) &= u(u+1).
\end{align}
Since all the factors are positive [by hypothesis], the Fundamental Theorem of Arithmetic implies there exist positive integers $p,q,r,s$ such that
\begin{align} \tag{1}
v(3v+2) &= pq, &
u &= pr, \\
9v^2+6v+2 &= rs, &
u+1 &= qs.
\end{align}
The left-hand pair of relations implies $rs-3pq=2$, and the right-hand pair implies $qs-pr=1$. Solving these two equations for $q$ and $r$ in terms of $s$ and $p$ yields
\begin{align} \tag{2}
q &= \frac{s+2p}{s^2-3p^2}, &
r &= \frac{2s+3p}{s^2-3p^2}.
\end{align}
Since $q$ and $r$ are both integers [by hypothesis], it must be that $s^2-3p^2$ divides both $s+2p$ and $2s+3p = 2(s+2p)-p$, and so it must divide $p$. But $qs-pr=1$ implies $\gcd(p,s)=1$, and hence $\gcd(s^2-3p^2,p)=1$. Therefore we conclude $s^2-3p^2 = \pm 1$; since $s^2-3p^2=-1$ has no integer solutions, we have $s^2-3p^2=1$. Substituting this into (2) gives $q=s+2p$ and $r=2s+3p$.
[PROOF INCORRECT AFTER THIS POINT]
NOTE: One can quickly deduce that $(s,p)$ and $(r,q)$ are consecutive solutions to $U^2-3V^2=1$, with $r>s$ and $q>p$. Since $(2a^2,bc)$ is also a solution, we can let $(s,p)$ be the $n$th solution, so that $(r,q)$ is the $(n+1)$th, and $(2a^2,bc)$ is the $(2n+1)$th. One can then show [using standard elementary Pell theory] that $(b,c)=(q+p,s+p)$, so that $b-c=2p$. From that point, I can't get any more useful information with this method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
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