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Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?
| $\displaystyle \because \int\dfrac { 1 } { 1 + x ^ { 3 } } d x{\displaystyle
=\int\dfrac { 1 - x ^ { 2 } + x ^ { 2 } } { 1 + x ^ { 3 } } d x } $
$ {\displaystyle =\int \dfrac { 1 - x } { 1 - x + x ^ { 2 } } d x + \dfrac { 1 } { 3 } \int \dfrac { d ( 1 + x ^ { 3 } ) } { 1 + x ^ { 3 } } }$
$\displaystyle =-\dfrac { 1 } { 2 } \int \dfrac { d ( 1 - x + x ^ { 2 } ) } { 1 - x + x ^ { 2 } } + \dfrac { 1 } { 2 } \int \dfrac { d x } { ( x - \dfrac { 1 } { 2 } ) ^ { 2 } + ( \dfrac { \sqrt { 3 } } { 2 } ) ^ { 2 } } + \dfrac { 1 } { 3 } \ln | 1 + x ^ { 3 } | $
$ { \displaystyle =\dfrac { 1 } { 6 } \ln \left( \dfrac { ( 1 + x ) ^ { 2 } } { 1 - x + x ^ { 2 } }\right ) + \dfrac { 1 } { \sqrt { 3 } } \tan ^ { - 1 } \left( \dfrac { 2 x - 1 } { \sqrt { 3 } }\right) + C }$
$\therefore \begin{aligned}
\int_{0}^{\infty} \frac{1}{1+x^{3}} d x &=\frac{1}{6} \left[\ln\frac{(1+x)^{2}}{1-x+x^{2}}\right]_{0}^{\infty}+\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^\infty=\frac{2 \pi}{3 \sqrt{3}}
\end{aligned}\quad \blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving $r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} < a^k$ by INDUCTION. Let $a$ be a natural number $>1$. For all integers $r_0, r_1, \dots, r_{n-1}$ with $0\leq r_{j} < a$, then
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{n-1}a^{n-1} < a^n.
\end{eqnarray}
Consider the base case for $n_0=2$. Thus $r_0+r_1a<a^2$, which is not so obviously true, but given that $0\leq r_{j} < a$ and the conditions on both $a$ and the $r_j$'s, then the maximum value that $r_0+r_1a$ can attain is when $r_0=r_1=(a-1)$. Thus, we have
\begin{eqnarray}
r_0+r_1a & = & (a-1)+(a-1)a\\
& = & (a-1)(a+1)\\
& = & a^2-1,
\end{eqnarray}
which clearly suffices to show that the inequality $r_0+r_1a<a^2$ holds, namely $a^2-1<a^2$. Now, let $k$ be some number $>2$. By assumption
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} < a^k,
\end{eqnarray}
with the same conditions as are listed in the theorem above. However, we must show that
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k < a^{k+1}.
\end{eqnarray}
To continue, let us add $r_ka^k$ to both sides of the inequality we made by assumption:
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k & < & a^k+r_ka^k\\
& = & a^k(r_k+1)\cdots \text{Where to now? :-(}
\end{eqnarray}
| Consider the base case for $n_0=2$. Thus $r_0+r_1a<a^2$, which is not so obviously true, but given that $0\leq r_{j} < a$ and the conditions on both $a$ and the $r_j$'s, then the maximum value that $r_0+r_1a$ can attain is when $r_0=r_1=(a-1)$. Thus, we have
\begin{eqnarray}
r_0+r_1a & = & (a-1)+(a-1)a\\
& = & (a-1)(a+1)\\
& = & a^2-1,
\end{eqnarray}
which clearly suffices to show that the inequality $r_0+r_1a<a^2$ holds, namely $a^2-1<a^2$. Now, let $k$ be some number $>2$. By assumption
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} < a^k,
\end{eqnarray}
with the same conditions as are listed in the theorem above. However, we must show that
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k < a^{k+1}.
\end{eqnarray}
To continue, let us subtract $r_ka^k$ from both sides of the inequality above:
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} & < & a^{k+1}-r_ka^k\\
& = & a^k(a-r_k).
\end{eqnarray}
Now, since $r_k<a$, then $(a-r_k)$ is positive and thus $a^k(a-r_k)>a^k$, in particular
\begin{eqnarray}
{\bf{r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}<a^k}}<a^k(a-r_k),
\end{eqnarray}
for which the bold part we made by assumption. Therefore,
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k<a^{k+1}
\end{eqnarray}
holds and suffices to attest to the existence of
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{n-1}a^{n-1}<a^n
\end{eqnarray}
for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical Expressions; School Homework $y + 3$ is always $5$ more than $y – 2$
so $y + 3 – (y – 2) = 5$
$(y + 4 ) – (y – 3) = ~ ?$
$(y - 2) – (y - 3) = ~ ?$
How would you work it out? I know that $y + 3$ is 5 more than $y - 2$, so should I do trial method or is there a method that's easier and faster?
Oh and why type of mathematical equation is this? Please make it easy. Thanks!
| When be subtract a quantity $b$, we can think of it as "adding $-b$. So $a - b = a + -b$.
When we have an expression of the form $a - (b + c) = a + -(b + c)$ we can always distribute $-1$ (i.e., multiply through by $-1$)
$$
\begin{align} a + -1\times(b + c)
& = a + -1\times b + -1\times c \\ \\
& = a + - b + - c \\ \\
& = a - b - c
\end{align}
$$
Using these facts, we can proceed:
$$
\begin{align} (y + 4 ) – (y – 3) & = y + 4 + -1\cdot (y + - 3)) \\ \\
& = y + 4 + -1\cdot y + (-1)(-3) \\ \\
& = y + 4 - y + 3 = 7 \\ \\
& = (y - y) + (4 + 3) \\ \\
& = 7
\end{align}
$$
So $\displaystyle\quad (y + 4) - (y - 3) = 7 \implies \quad (y + 4)$ is 7 more than $(y - 3)$
We can think of is also as just distributing the $-$ sign over the quanity:
$$
\begin{align} (y – 2) – (y – 3) & = (y - 2) + (- y - (-3))\\ \\
& = y - 2 - y + 3 \\ \\
& = -2 + 3 \\ \\
& = 1
\end{align}$$
So $\displaystyle \quad(y - 2) - (y - 3) = 1 \implies \quad (y - 2)$ is 1 more than $(y - 3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find ${{{d^2}y} \over {d{x^2}}}$ when $x = {1 \over {t + 1}}$ and $y = {1 \over {t - 1}}$ The answer in the text book is:
${{{d^2}y} \over {d{x^2}}} = 4{\left( {{{t + 1} \over {t - 1}}} \right)^3}$
I've tried this:
$\eqalign{
& x = {1 \over {t + 1}},y = {1 \over {t - 1}} \cr
& x = {(t + 1)^{ - 1}},y = {(t - 1)^{ - 1}} \cr
& {{dx} \over {dt}} = - {(t + 1)^{ - 2}} = {{ - 1} \over {{{(t + 1)}^2}}} \cr
& {{dy} \over {dt}} = - {(t - 1)^{ - 2}} = {{ - 1} \over {{{(t - 1)}^2}}} \cr
& {{dy} \over {dx}} = {{dy} \over {dt}} \times \left( {{1 \over {{{dx} \over {dt}}}}} \right) \cr
& {{dy} \over {dx}} = {{ - 1} \over {{{(t - 1)}^2}}} \times {{{{(t + 1)}^2}} \over { - 1}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}} \cr} $
My approach to ${{{d^2}y} \over {d{x^2}}}$ is:
$${{dy} \over {dx}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}}$$
So using the quotient rule:
$\eqalign{
& u = {(t + 1)^2} \cr
& v = {(t - 1)^2} \cr
& {{du} \over {dt}} = 2(t + 1) \cr
& {{dv} \over {dt}} = 2(t - 1) \cr
& {{{d^2}y} \over {d{x^2}}} = {{2{{(t - 1)}^2}(t + 1) - 2{{(t + 1)}^2}(t - 1)} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)\left[ {(t - 1) - (t + 1)} \right]} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)( - 2)} \over {{{(t - 1)}^4}}} \cr
& {{{d^2}y} \over {d{x^2}}} = {{ - 4(t + 1)} \over {{{(t - 1)}^3}}} \cr} $
As the previous attempt yielded an incorrect answer I attempted to answer it this way:
$\eqalign{
& {{dx} \over {dt}} = - {(t + 1)^{ - 2}} \cr
& {{dy} \over {dt}} = - {(t - 1)^{ - 2}} \cr
& {{{d^2}x} \over {d{t^2}}} = 2{(t + 1)^{ - 3}} \cr
& {{{d^2}y} \over {d{t^2}}} = 2{(t - 1)^{ - 3}} \cr} $
$\eqalign{
& {{{d^2}y} \over {d{x^2}}} = {{{d^2}y} \over {d{t^2}}} \times \left( {{1 \over {{{{d^2}x} \over {d{t^2}}}}}} \right) \cr
& {{{d^2}y} \over {d{x^2}}} = {2 \over {{{(t - 1)}^3}}} \times {{{{(t + 1)}^3}} \over 2} \cr
& {{{d^2}y} \over {d{x^2}}} = {{{{(t + 1)}^3}} \over {{{(t - 1)}^3}}} \cr} $
I'd appreciate it greatly if someone could point out my misunderstanding(s) and show me the correct approach.
Thank you very much.
| As an alternative to the implicit approach, we could write $y$ in terms of $x,$ then translate back into terms of $t$ at the end. Take $t\neq\pm 1$. Since $$x=\frac1{t+1},$$ then $$t+1=\frac1x,$$ so $$t-1=\frac1x-2=\frac{1-2x}{x},$$ and so $$y=\frac1{t-1}=\frac{x}{1-2x}.$$ Now, $$\frac{dy}{dx}=\frac{(1-2x)-(-2)x}{(1-2x)^2}=\frac{1}{(1-2x)^2}=(1-2x)^{-2},$$ so $$\frac{d^2y}{dx^2}=-2\cdot(1-2x)^{-3}\cdot\frac{d}{dx}[1-2x]=\frac4{(1-2x)^3},$$ and since $y=\frac{x}{1-2x},$ then $$\frac{d^2y}{dx^2}=4\cdot\frac{y^3}{x^3}=4\left(\frac yx\right)^3.$$ Finally, recalling that $x=\frac1{t+1}$ and $y=\frac1{t-1},$ it follows that $$\frac{d^2y}{dx^2}=4\left(\frac{t+1}{t-1}\right)^3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ Prove that $\lvert\frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}\rvert<\frac{1}{n^2}$ and deduce that $1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}$ is convergent.
Using algebra, the absolute value becomes $\lvert\frac{1}{2n}-\frac{1}{2n-2}\rvert$ which is $\lvert\frac{-(n+1)}{2n(n+1)}\rvert$.
Not entirely sure how to proceed with this proof..
Edit: Incorrect third term, changed to $\frac{1}{2n+2}$.
| For the inequality you ask about, use brute force:
$$ \frac{1}{n} - \frac{1}{2n} - \frac{1}{2n-2} = \frac{2(n-1) - (n-1) - n}{2n(n-1)} = \frac{-1}{2n(n-1)}$$
So, your inequality will follow from $2n(n-1) > n^2$, which is equivalent to $n^2 > 2n$, which is true for $n > 2$.
For convergence, you should use the inequality to bound pieces of your series by $\frac{1}{n^2}$ (in absolute value). Once you do this, from convergence of $\sum_n \frac{1}{n^2}$ it follows that the series converges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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finding the remainder of $x^{100}-2x^{51}+1$ I have never been great with polynomials. Here's my problem.
Find the remainder of $f(x)=x^{100}-2x^{51}+1$ when $f$ is divided by $x^2-1$
This sounds easy right? Why can't I figure it out? My thought was to try and create it such that $f(x)=q(x)g(x)+r(x)$. But I can not get past getting $deg[r(x)]<deg[g(x)].$
$$f(x)=x^{100}-2x^{51}+1$$
$$=x^{100}-x^{51}-x^{51}+x^2-x^2+1$$
$$=x^{51}(x^{49}-1)-x^2(x^{49}-1)-x^2+1$$
$$=(x^{51}-x^2)(x^{49}-1)-x^2+1$$
$$=x^2(x^{49}-1)(x^{49}-1)-x^2+1$$
$$=x^2[(x^{49}-1)^2-1]+1=?.......$$
I don't see what I am missing
| This is the standard approach, especially if you know the roots of the divisor.
Let $f(x) = x^{100} - 2x^{51} + 1$, and $f(x) = g(x) (x^2-1) + ax + b$ be the division
Then, $0 = f(1) = g(1) ( 1^2 - 1) + a (1) + b = a + b$,
and $4 = f(-1) =g(-1) ( (-1)^2 -1) + a(-1) + b = -a + b$.
Hence $a= -2, b = 2$.
Thus the remainder is $-2x+2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the limits related to $a_1=1$, $a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}$ Suppose $a_1=1, a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}, k \in \mathbb{N}$. Find the limits
$$i)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n\sqrt{n}}$$
$$ii)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n^2}$$
I'm puzzled with it. What to do?
| For $n\ge1$, we have
$$
\sum_{k=1}^na_k=a_{n+1}^2\tag{1}
$$
$(1)$ leads to $a_{n+1}^2-a_n^2=a_n$ which implies that $a_n$ is increasing and that
$$
a_{n+1}-a_n=\frac{a_n}{a_{n+1}+a_n}\le\frac12\tag{2}
$$
In fact,
$$
\frac12-\frac{a_n}{a_{n+1}+a_n}=\frac12\frac{a_{n+1}-a_n}{a_{n+1}+a_n}\le\frac12\frac{\frac12}{2a_n}=\frac1{8a_n}\tag{3}
$$
Thus, $(2)$ and $(3)$ say that
$$
\frac38\le\frac12-\frac1{8a_n}\le a_{n+1}-a_n\le\frac12\tag{4}
$$
Summing $(4)$ yields
$$
\frac{3n+5}{8}\le a_n\le\frac{n+1}{2}\tag{5}
$$
We can use $(5)$ to improve $(4)$:
$$
\frac12-\frac1{3n+5}\le a_{n+1}-a_n\le\frac12\tag{6}
$$
Since
$$
\sum_{k=1}^{n-1}\frac1{3n+5}\le\int_0^{n-1}\frac{\mathrm{d}x}{3x+5}=\frac13\log\left(\frac{3n+2}{5}\right)\tag{7}
$$
Summing $(6)$ yields
$$
\frac{n+1}{2}-\frac13\log\left(\frac{3n+2}{5}\right)\le a_n\le\frac{n+1}{2}\tag{8}
$$
To handle $i)$,
$$
\begin{align}
\frac1{n^{3/2}}\sum_{k=1}^na_n
&=\frac1{n^{3/2}}a_{n+1}^2\\
&\ge\frac1{n^{3/2}}\left(\frac{n+2}{2}-\frac13\log\left(\frac{3n+5}{5}\right)\right)^2\\[12pt]
&\to\infty\tag{9}
\end{align}
$$
To handle $ii)$,
$$
\begin{align}
\frac1{n^2}\sum_{k=1}^na_n
&=\frac1{n^2}a_{n+1}^2\tag{10}
\end{align}
$$
Applying $(8)$ yields
$$
\frac1{n^2}\left(\frac{n+2}{2}-\frac13\log\left(\frac{3n+5}{5}\right)\right)^2
\le\frac1{n^2}a_{n+1}^2
\le\frac1{n^2}\left(\frac{n+2}{2}\right)^2\tag{11}
$$
So by the Squeeze Theorem, we have
$$
\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^na_n=\frac14\tag{12}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$? Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$?
How can I prove this? Can anybody help me please?
| This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $\delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $\pi$ be a Gaussian prime that divides $p$. Then $\pi$ divides one of $a+bi$ or $a-bi$. So $\pi$ must be an associate of one of these, and the conjugate of $\pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find Maclaurin series for$f(x) = \frac{2x}{1-5x^3}$ I'm trying to find the Maclaurin series for $f(x) = \frac{2x}{1-5x^3}$, but my solution is different from what I know it supposed to be, which is $2x+10x^4+50x^7+250x^{10}+...$
This is my attempt:
$$f(x) = \frac{2x}{1-5x^3}$$
$$\frac{1}{1-x} = 1+x+x^2+x^3+...$$
$$\frac{2x}{1-x}=2x+2x^2+2x^3+2x^4+...\tag{multiply both sides by $x$}$$
$$\frac{2x}{1-5x^3}=10x^3+50x^6+250x^9+625x^{12}+...\tag{substitute $x$ with $5x^3$}$$
Where is my mistake?
| You might get into less complication if you use summation notation:
$$\frac{1}{1-u} = \Sigma_{k=0}^{\infty} u^k \rightarrow \frac{2x}{1-(5x^3)} = 2x \cdot \frac{1}{1-(5x^3)} = 2x \cdot\Sigma_{k=0}^{\infty} (5x^3)^k $$
$$= 2x \cdot\Sigma_{k=0}^{\infty} 5^k \cdot x^{3k} = \Sigma_{k=0}^{\infty} 2 \cdot 5^k \cdot x^{3k+1} . $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
| Here is the way to produce ALL integer solutions, not necessarily coprime, and not just some random infinite families.
For any integer $m$, let $D_2(m)$ be the set of integers $w$ such that $w^2$ is a divisor of $m$. For example, $D_2(18)=\{\pm 1,\pm 3\}$.
Then the set of all integer solutions to $x^3+y^3=z^2$ can be described as
$$
(x,y,z)=\left(\frac{u(u^3+v^3)}{w^2}, \, \frac{v(u^3+v^3)}{w^2}, \frac{(u^3+v^3)^2}{w^3} \, \right), \quad u,v \in {\mathbb Z}, \quad w \in D_2(u^3+v^3).
$$
Condition $w \in D_2(u^3+v^3)$ ensures that $x,y,z$ are integers, and a direct substitution proves that they satisfy the equation. Let us prove that, conversely, all integer solutions are covered by this family. Indeed, solution $(0,0,0)$ is covered. Let $(x,y,z)$ be any other solution, let $d$ be the greatest common square-free divisor of $x$ and $y$, and write $x=du$, $y=dv$ for some integers $u,v$. Then $d(u^3+v^3)=(z/d)^2$ is an integer, hence $z/d$ is an integer, say $z'$. Then $d \cdot (u^3+v^3)=(z')^2$ implies the existence of integers $a,b,w$ such that $d=ab^2$ and $u^3+v^3=aw^2$. Because $d$ is square-free, $b^2=1$, hence $d=a=\frac{u^3+v^3}{w^2}$. So, take $u,v$ arbitrary, $w \in D_2(u^3+v^3)$, and then express $x,y,z$. We obtain the stated formulas.
The method works for all equations of the form $F(x,y)=z^2$, where $F$ is a cubic form with integer coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Prove that solving equation is possible The equation:
$${2 \over x} + {3 \over y} = {5 \over z} ,\,\,x < z$$
$x,y,$ and $z$ must be natural numbers.
How would I go about proving this equation possible to solve?
I can't seem to figure out how to prove it without doing some "random" calculations to see how numbers fit in the equation.
Example:
The variables for the possibly lowest answer is: $$x=1,y=6,z=2$$
Since:
$$
{2 \over 1} + {3 \over 6} = {5 \over 2} ,\,\,1 < 2
$$
How would it be possible to find this answer, or any other correct answer for that matter, without using "random" placeholder variables to see how to equation evolves?
Simpler put: How to is it possible to solve this equation faster and more effectively? Is it possible to determine whether or not the equation is solvable before trying to solve it?
| Firstly if $x<z$ then $y>z$.Because if $y<z$,$$\frac{2}{x}+\frac{3}{y}>\frac{2}{z}+\frac{3}{z}=\frac{5}{z}$$Contradiction!
So let $x=z-a$ and $y=z+b$.Plug these values in the equation and you will get,$$z=\frac{5ab}{3b-2a}=\frac{5}{\frac{3}{a}-\frac{2}{b}}$$
Hence minimum value of z can be 5.
Therefore,
$$\frac{3}{a}-\frac{2}{b}=1\implies b=\frac{2a}{3-a}=\frac{2}{\frac{3}{a}-1}$$
The minimum value of $\frac{3}{a}-1$ is 2 (because if $\frac{3}{a}-1=1$ a = 1.5(a fraction))
Hence minimum value of b=2/2=1.$$\frac{3}{a}-1=2\implies a=1$$
So,$$x=z-a=5-1=4$$ $$y=z+b=5+1=6$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying $\sin(2\tan^{-1} x)$ I've been working on this for a while. The answer in the book is $\frac{2x}{x^2 + 1}$ Here's my workings:
$\sin(2\tan^{-1} x)$
Let $\alpha = \tan^{-1}x \Rightarrow \tan \alpha = x$
$\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha = 2x\cos^2\alpha$
I'm not sure how to proceed to turn that $cos^2\alpha$ into $\frac{1}{x^2 + 1}$
| Making a right-triangle picture can be helpful too, either for getting to the result quickly or checking up on the analytical result. With $\tan \alpha = x = \frac{x}{1}$, we would have $x$ for the leg opposite $\alpha$ and $1$ for the adjacent leg; the hypotenuse is then $\sqrt{x^2 + 1}$ . We wish to evaluate $\sin 2\alpha$ , which is thus
$$2 \sin \alpha \cos \alpha = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} .$$
| {
"language": "en",
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"source": "stackexchange",
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Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge Does the following improper integral converge?
$$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$
| Notice $x^2 + x = (x+1)x = (x+1)^2 - (x+1)$, we have:
$$\begin{align}\int_0^B \sin(x) \sin(x^2) dx
&= \frac12\int_0^B(\cos(x^2-x)-\cos(x^2+x)) dx\\
&= \frac12\left(\int_0^B - \int_1^{B+1}\right)\cos(x^2-x)dx\\
&= \frac12\left(\int_0^1 - \int_B^{B+1}\right)\cos(x^2-x)dx
\end{align}$$
Integrate by parts. For large $B$, we have:
$$\begin{align}
&\int_B^{B+1} \cos(x^2 - x)dx = \left[ \frac{\sin(x^2 - x)}{2x - 1} \right]_B^{B+1} + \int_B^{B+1}\frac{2\sin(x^2 - x)dx}{(2x - 1)^2}\\
\implies & \left|\int_B^{B+1} \cos(x^2 - x)dx\right| \le \frac{2}{2B-1} + \frac{2}{(2B-1)^2}\\
\implies & \lim_{B\to\infty} \int_B^{B+1} \cos(x^2 - x)dx = 0\\
\implies & \lim_{B\to\infty} \int_0^B \sin(x) \sin(x^2) dx = \frac12 \int_0^1 \cos(x^2 - x) dx.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$1+x+\ldots+x^n$ perfect square Let $p$ be the polynomial $p(x)=1+x+\ldots+x^n$.
For which couples $(a, n)\in\mathbb{N}^2$, $p(a)$ is a perfect square? I'm particularly interested in $p(3)$.
| The Diophantine equation $f_n(x)=1+x+x^2+\cdots+x^{n-1}=y^2$.
Ribenboim's book on Catalan's conjecture has a detailed
analysis of this Diophantine equation. Except the cases noted below, there are no more non-trivial solutions.
Here are some of the easy cases:
*
*There are always the two trivial solutions $x=0$ and $x=-1$.
*If $n$ is a square, then $x=1$ is a third solution.
*For $n=3$, if $x\not=0$, then $-2|x|<x<2|x|$ implies
$(|x|-1)^2<1+x+x^2<(|x|+1)^2$ so $f_3(x)$ can only be the square $x^2$.
This gives the other trivial solution $x=-1$.
*For $n=4$, look at Monthly problem 11203 (Feb. 2006) or Exercise 1.10
in Edward's book Fermat's Last Theorem. The only solutions are $x=-1,0,1,7$.
*For $n=5$, see Ed Burger's book Exploring The Number Jungle (Exercise 12.11). By comparing $4f_5(x)$
with $(2x^2+x)^2$ and $(2x^2+x+1)^2$, we find that the only solutions are $x=-1,0,3$.
*For $n=6$, we factor $f_6(x)=f_2(x^3)f_3(x)$.
By the formula $2=f_2(x^3)-(x-1)f_3(x)$ we see that
$\gcd(f_2(x^3),f_3(x))$ divides 2. But $f_3(x)$ is always odd, so the gcd equals 1.
This forces $f_3(x)$ to be a square so we end up with the trivial solutions $x=0,-1$.
*For $n=7$, if $x>5$ we have
$$(16x^3+8x^2+6x+5)^2< 256 f_7(x)< (16x^3+8x^2+6x+6)^2,$$
while
for $x<-4$ we get $$(16x^3+8x^2+6x+5)^2< 256 f_7(x)< (16x^3+8x^2+6x+4)^2.$$
Check the values between and you find that there are only trivial solutions.
By this stage, elementary methods get tougher to push through.
| {
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Computing large modular numbers How do you compute large modular arithmetic such as $8^{128}$ $mod$ $100$ or $10^{111}$ $mod$ $137$ or $3^{100}$ mod $17$? I know that one way is repeated squaring. For the first one, my book says 16, for the second 96, and third 13. It gives the answers but does not show how to do any of the three, the procedure, or even a example. If anyone could show me how the answers are derived, that would be wonderful!!
| Using Fermat's little theorem, we know $a^b \equiv a^{(b \mod \phi(c))} \mod c$. So we can reduce exponents by subtracting any multiple of $\phi(c)$ until we get a number between $0$ and $\phi(c) - 1$.
*
*First, $\phi(100) = \phi(2^25^2) = \frac{1}{2} \cdot \frac{4}{5} \cdot 100 = 40$ so $8^{128} \equiv 8^{128 - 3 \cdot 40} \equiv 8^8 \mod 100$. Now we do need a few computations, for instance using that $2^{10} = 1024$: $$8^8 \equiv 2^{24} \equiv 2^{10} \cdot 2^{10} \cdot 2^4 \equiv (1024) \cdot (1024) \cdot 16 \equiv 24 \cdot 24 \cdot 16 \equiv 16 \cdot 16 \equiv 16 \mod 100.$$
*Since $111 < \phi(137)$, the trick of subtracting multiples of $\phi(137)$ does not help here. But doing a few computations by hand gives us $10^3 = 1000 \equiv 41 \mod 137$ and $10^4 \equiv 410 \equiv -1 \mod 137$ which is a really nice, small number. So:
$$10^{111} = 10^{108} \cdot 10^3 \equiv (10^4)^{27} \cdot 41 \equiv (-1)^{27} \cdot 41 \equiv -41 \equiv 96 \mod 137.$$
*Finally, $\phi(17) = 16$ and $100 = 6 \cdot 16 + 4$, so $$3^{100} \equiv (3^{16})^6 \cdot 3^4 \equiv 3^4 \equiv 81 \equiv 13 \mod 17.$$
| {
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How to solve these inequations I would appreciate if somebody could help me with the following problem:
Q: show that ( $x\in \left[0,\frac{\pi}{2} \right], a,b>0$)
$$a+b\leq \sqrt{a^2 \sin^2 x+b^2 \cos^2x}+\sqrt{a^2 \cos^2x+b^2 \sin^2x}\leq \sqrt{2} \sqrt{a^2+b^2}$$
| Hint: Using $(x+y)^2 \le 2(x^2+ y^2)$ to show the right inequality.
For the left, squaring and canceling the same terms we get
$$ab\le \sqrt{(a^2\sin^2x+b^2\cos^2x)(a^2\cos^2x + b^2\sin^2x)},$$
the rest is yours.
| {
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How to find all integers $a,b > 1$ satisfying $b \mid a^2+1$ and $a^2 \mid b^3+1$?
Let $a,b\in \mathbb{Z}$ with $a,b>1$, and such that $b \mid a^2+1$ and $a^2 \mid b^3+1$. Find all such $a,b$.
I found $a=3,b=2$. Are there any other solutions? Thank you.
yesterday I have find this
since $a,b>1$ so $(a,b)=(3,2)$ or $(a,b)=(3,5)
| Suppose $(a, b)$ is a solution with $a, b \in \mathbb{Z}, a, b>1$. If $b=a^2+1$, then $a^2 \mid (b^3+1)=(a^2+1)^3+1$ so $a^2 \mid 2$, so $a=1$, a contradiction. Thus $\frac{a^2+1}{b}$ is a positive integer $>1$. Note that $a^2 \mid (\frac{a^2+1}{b})^3(b^3+1)=(a^2+1)^3+(\frac{a^2+1}{b})^3$, so $a^2 \mid 1+(\frac{a^2+1}{b})^3$, so if $(a, b)$ is a solution, so is $(a, \frac{a^2+1}{b})$. Note that at least $1$ of $b, \frac{a^2+1}{b}$ is $\leq a$. (Otherwise $a^2+1=b(\frac{a^2+1}{b}) \geq (a+1)^2$, which is impossible) It thus suffices to consider the case where $b \leq a$, since any solution with $b>a$ can be mapped to a solution with $b \leq a$ by above.
If $b=a$, then $b\mid a^2+1=b^2+1$, so $b \mid 1$, so $b=1$, a contradiction. Thus $a \geq b+1$. Write $b^3+1=ca^2$, then $b \mid c(a^2+1)=(b^3+1)+c$ so $b \mid c+1$. Thus $c+1 \geq b$. Therefore $b^3+1=ca^2 \geq (b-1)(b+1)^2=b^3+b^2-b-1$ so $0 \geq b^2-b-2=(b-2)(b+1)$, so $b \leq 2$, so $b=2$. Thus $a^2 \mid (2^3+1)=9$, so $a=3$.
Finally, all solutions are given by $(a, b)=(3, 2)$ and $(a, b)=(3, \frac{3^2+1}{2})=(3, 5)$. These solutions are easily checked to work.
| {
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Proving the combinatorial identity: $\sum_{k=0}^m (-1)^k 2^{2k-1}\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right]=(-1)^m$ Let $m$ be a positive integer. I have trouble proving that
$$\sum_{k=0}^m (-1)^k 2^{2k-1}\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right]=(-1)^m$$
Anyone?
| Suppose we seek to evaluate
$$\sum_{k=0}^m (-1)^k 2^{2k-1}
\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right].$$
There are two pieces here, the first is
$$\frac{1}{2}\sum_{k=0}^m (-1)^k 2^{2k}
{m+k-1\choose 2k}$$
and the second is
$$\frac{1}{2}\sum_{k=0}^m (-1)^k 2^{2k}
{m+k\choose 2k}.$$
We treat these in turn.
Introduce
$${m+k-1\choose 2k}
= {m+k-1\choose m-1-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m-k}} (1+z)^{m+k-1} \; dz.$$
This is zero when $k\ge m$ so we may let $k$ go to infinity.
We thus get for the first piece
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m}
\sum_{k\ge 0} (-1)^k 2^{2k} z^k (1+z)^k
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m}
\frac{1}{1+4z(1+z)}
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m}
\frac{1}{(2z+1)^2}
\; dz.$$
Similarly the second piece is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m}}{z^{m+1}}
\frac{1}{(2z+1)^2}
\; dz.$$
Adding these we obtain
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m-1}}{z^m}
\left(1+\frac{1+z}{z}\right)
\frac{1}{(2z+1)^2}
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m-1}}{z^{m+1}}
\frac{1}{2z+1}
\; dz.$$
Extracting the residue we get
$$\frac{1}{2}
\sum_{q=0}^{m-1} {m-1\choose q} (-1)^{m-q} 2^{m-q}
= -\sum_{q=0}^{m-1} {m-1\choose q} (-1)^{m-1-q} 2^{m-1-q}
\\ = - (1-2)^{m-1} = (-1)^m.$$
| {
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Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| We want to prove that $n^2 - \frac{n^2}{2} = \frac{n^2}{2}$. In order to better understand what we want to do, let us simplify this equation a little bit by adding $\frac{n^2}{2}$ to both sides. We get $n^2 = \frac{n^2}{2} + \frac{n^2}{2}$. So in order to solve the problem, we just need to show that $\frac{n^2}{2} + \frac{n^2}{2} = n^2$. In order to show that these two things are equal, let's start with the left hand side and show that it equal to the right hand side. Since we know how to add fractions with the same denominator, we get
$$\frac{n^2}{2} + \frac{n^2}{2} = \frac{2n^2}{2} = \frac{2}{2}n^2= n^2$$
Now we're done!
| {
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Interception with $x$-axis - not so trivial? I want to find the interception with the x-axis of the following function:
$f(x) = \frac{1}{4}x^4-x^3+2x$.
So putting $0 = \frac{1}{4}x^4-x^3+2x$ I would get $0 = x(\frac{1}{4}x^3-x^2+2)$ but how to check for the other solutions properly?
| Also $x=2$ is a root, so $f(x)=\frac{1}{4}(x-2)x(x^2-2x-4)$. wolframalpha is your friend.
| {
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Write a polynomial equation with the following roots. A quartic funciton with roots of $-3$,$-1$ and $4$? Write a polynomial equation with the following roots. A quartic funciton with roots of $-3,-1$ and $4$?
use each root as a factor
$$f(x) = a(x + 3)(x + 1)(x - 4)^2$$
$$f(x) = a(x^2 + 4x + 3)(x^2 - 8x + 16)$$
$$f(x) = a(x^4 - 4x^3 - 13x^2 + 40x + 48)$$
$$f(5) = 16$$
$$16 = a(5^4 - 4(5^3) - 13(5^2) + 40(5) + 48 = 48a \rightarrow a = 3$$
$$f(x) = 3x^4 - 12x^3 - 39x^2 + 120x + 144$$
Is that correct? Please show another way of doing it please with your steps.
| You did just fine: you equation:
$$f(x) = a(x + 3)(x + 1)(x - 4)^2;\quad a\neq 0\tag{1}$$ is entirely fine: it is a quartic function with three roots: $x = -3, x = -1, \;x = 4$, where the root $x = 4$ has multiplicity of two. Your expansion of the factors is spot on, but not required (and doing the extra work leaves room for a silly mistake, though not in your expansion).
One observation: When evaluating $f(5) = 16$ to solve for $a$, note that $16 = 48a \implies a = \dfrac 13$, which will alter the final equation at the very end.
Note: you could easily have used equation $(1)$ to solve for $a$ as well:
$$f(x) = a(x + 3)(x + 1)(x - 4)^2$$ $$\implies f(5) = 16 = a(5+3)(5+1)(5-4)^2 = 48a $$ $$\iff a = \frac{16}{48} = \frac 13$$
The following functions are also valid quartic functions with the same required roots, as well:
$$f(x) = a(x + 3)^2(x + 1)(x - 4); \quad a\neq 0$$
$$f(x) = a(x + 3)(x + 1)^2(x - 4);\quad a\neq 0$$
| {
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Solve the following limit? Please only give hints? $$\lim_{x \to \frac{\pi}{6}}\frac{2\sin{(x)}-1}{\sqrt{3}\tan{(x)}-1}$$ I tried this and I was able to simplify it down to $$\lim_{x \to \frac{\pi}{6}}\frac{\sin{(2x)}-\cos{(x)}}{2\sin{(x - \frac{\pi}{6})}}.$$ However, I am stuck here and I don't even know if this is the right approach. Thanks!
| Putting $x=\frac\pi6-y$ as $x\to\frac\pi6,y\to0$
$$2\sin x-1=2\sin(\frac\pi6-y)-1=\cos y-\sqrt3\sin y-1$$
$$\sqrt3\tan x-1=\sqrt3\tan(\frac\pi6-y)-1=\sqrt3\cdot \frac{\frac1{\sqrt3}-\tan y}{1+\tan y\frac1{\sqrt3}}-1$$
$$=\sqrt3\cdot\frac{1-\sqrt3\tan y}{\sqrt3+\tan y}-1=\frac{\sqrt3-3\tan y}{\sqrt3+\tan y}-1=\frac{-4\tan y}{\sqrt3+\tan y}$$
$$\text{So,}\frac{2\sin x-1}{\sqrt3\tan x-1}=\frac{(\cos y-\sqrt3\sin y-1)(\sqrt3+\tan y)}{-4\tan y}$$
$$=\frac14\cdot(\sqrt3+\tan y)\cdot\left(\sqrt3\frac{\sin y}{\tan y}+\frac{1-\cos y}{\tan y} \right)$$
$$\text{So,}\lim_{x\to\frac\pi6}\frac{2\sin x-1}{\sqrt3\tan x-1}$$
$$=\lim_{y\to0}\frac14\cdot(\sqrt3+\tan y)\cdot\left(\sqrt3\frac{\sin y}{\tan y}+\frac{1-\cos y}{\tan y} \right)$$
$$= \frac14\cdot\lim_{y\to0}(\sqrt3+\tan y)\cdot\left(\sqrt3 \cdot\lim_{y\to0}\frac{\sin y}{\tan y}+\lim_{y\to0}\frac{\sin^2y\cdot\cos y}{\sin y(1+\cos y)} \right)$$
$$=\frac14\cdot(\sqrt3+0)\cdot\left(\lim_{y\to0}\sqrt3\cos y+\lim_{y\to0}\frac{\sin y\cdot\cos y}{1+\cos y} \right)$$ as $y\to0,\sin y\to 0\implies \sin y\ne0$
$$=\frac{\sqrt3}4\cdot(\sqrt3+0)$$
| {
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How to construct a matrix $A$
Construct a matrix $A$ such that $A^2\ne 0$ but $A^3=0$.
I need your help to find $A$. Please help. Thanks in advance.
| $$\text{Let} \;\;\;A = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$$
$$\text{Then} \;\;\;\; A^2 = \begin{pmatrix} 0 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \neq \bf 0$$
$$\text{But} \;\;\;\; \;A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \bf 0$$
| {
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Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| Find $p$ and $q$ such that
\begin{cases}
-\frac{q}{2}=2\\[3ex]
\frac{p^3}{27}+\frac{q^2}{4}=\frac{100}{27}
\end{cases}
This of course gives $q=-4$ and
$$
\frac{p^3}{27}=\frac{100}{27}-4=-\frac{8}{27},
$$
so $p^3=-8$ and $p=-2$.
Now find the unique real root of the polynomial $x^3+px+q=x^3-2x-4$, which is of course $2$. Indeed the polynomial factor as
$$
x^3-2x-4=(x-2)(x^2+2x+2)
$$
and the second degree factor has no real roots.
This example clearly shows why Cardan's formulas have a very limited usefulness: they present the roots in a form that makes them unintelligible.
Here's Cardan's formula for third degree equations with only one real root: if $x^3+px+q=0$ and the discriminant
$$\frac{p^3}{27}+\frac{q^2}{4}>0$$
then the only real solution for the equation is
$$
\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}
+
\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 0
} |
Absolute values in $\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$ in my math class we were given a list of indefinite integrals, and one of them was:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$
My working:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int \frac{dx}{(x+2)\sqrt{(x+2)^2-1}}$$
Then I used the substitution $x+2=\sec t$ to get:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int \frac{\tan t}{|\tan t|}dt= t\,\text{sgn}\, (\tan t)+C...$$
Then I checked the answer sheet, and this is what they did:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int dt=t+C=\text{arcsec}(x+2)+C$$
What I don't understand is, why are they allowed to say $\sqrt{\sec^2 t-1}=\tan t?$ I tried to put some values in and I have found that:
$$\int_{\sec \left(\frac{8}{5}\right)-2}^{\sec \left(\frac{9}{5}\right)-2} \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}<0$$ but according to the answer sheet I would get $\dfrac{1}{5}$
My answer looks wrong, I would be happy if someone could explain what the problem is, and also why we are allowed to simplify like they did.
| To avoid absolute value
Letting $y=x+2$ yields
$$
\begin{aligned}
\int \frac{1}{(x+2) \sqrt{(x+1)(x+3)}}&=\int \frac{d y}{y \sqrt{y^2-1}}\\&=\int \frac{1}{y^2} d\left(\sqrt{y^2-1}\right) \\
&=\int \frac{\left.d \sqrt{y^2-1}\right)}{\left(\sqrt{y^2-1}\right)^2+1} \\
&=\arctan \left(\sqrt{y^2-1}\right)+C \\
&=\arctan \left(\sqrt{x^2+4 x+3}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find the formula of the sequence using generating functions I have the following sequence given:
$$\sum_{k=1}^{n} (-1)^{k}k^{2}$$
How can I do this? The sequence goes like this:
$$-1 + 4 - 9 + 16 - 25 + 36 - ...$$
So it doesn't have any variables inside. It looks like a geometric sequence for me, so the simple known formula would do the whole thing, but I actually doubt it's that simple.
So there are some generating functions that are quite similar, like:
$$\sum (-1)^{n}x^{n} = x - x^{2} + x^{3} - x^{4} + ...$$
But, uhm, well...what's next? The lack of x variable seems a bit strange to me.
| The generating function is
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^kk^2x^n
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^kk^2x^n\\
&=\sum_{k=0}^\infty(-1)^kk^2\frac{x^k}{1-x}\\
&=\frac1{1-x}\sum_{k=0}^\infty(-1)^kk^2x^k\tag{1}
\end{align}
$$
Starting with $\frac1{1+x}$ and taking derivatives, we get
$$
\begin{align}
\frac1{1+x}&=\sum_{k=0}^\infty(-1)^kx^k\\
\frac1{(1+x)^2}&=\sum_{k=0}^\infty(-1)^k(k+1)x^k\\
\frac2{(1+x)^3}&=\sum_{k=0}^\infty(-1)^k(k+2)(k+1)x^k
\end{align}\tag{2}
$$
Since $k^2=(k+2)(k+1)-3(k+1)+1$, we have
$$
\begin{align}
\sum_{k=0}^\infty(-1)^kk^2x^k
&=\frac2{(1+x)^3}-\frac3{(1+x)^2}+\frac1{1+x}\\
&=\frac{(x-1)x}{(1+x)^3}\tag{3}
\end{align}
$$
Thus, the generating function we want is
$$
\begin{align}
\frac1{1-x}\sum_{k=0}^\infty(-1)^kk^2x^k
&=\frac1{1-x}\frac{(x-1)x}{(1+x)^3}\\
&=-\frac{x}{(1+x)^3}\\
&=-\sum_{n=0}^\infty\binom{-3}{n-1}x^n\\
&=\sum_{n=0}^\infty(-1)^n\binom{n+1}{n-1}x^n\\
&=\sum_{n=0}^\infty(-1)^n\binom{n+1}{2}x^n\tag{4}
\end{align}
$$
Therefore,
$$
\sum_{k=0}^n(-1)^kk^2=(-1)^n\binom{n+1}{2}\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$? According to an example in my text book:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$$
I don't see how this works. The closest I can get is:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2\sqrt{x}}{\sqrt{x}(\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1})} = \frac{2}{\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1}}$$
Which is slightly off. But I'm not even sure if I calculated that correctly. What am I missing, what's the way to think?
| $$ \frac{(x + \sqrt{x}) - (x - \sqrt{x} )}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}} = \frac{2 \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}}$$
Divide by $\sqrt{x}$ throughout. It simplifies to
$$\frac{2}{\dfrac{\sqrt{x + \sqrt{x}} }{\sqrt{x}} + \dfrac{\sqrt{x - \sqrt{x}}}{\sqrt{x}}}.$$
Now the denominator simplifies to
$\sqrt{\dfrac{x + \sqrt{x}}{x}} + \sqrt{\dfrac{x - \sqrt{x}}{x}} $ and the answer comes out nicely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A hard 'if and only if' trigonometric identity proof Prove
$$
\frac{-2+2\tan A+2\cos B\cdot\sin B+\cot^2 A\cdot({\sec^4A-\operatorname{cosec}^2A-2)}}{2+\tan^2A-2\sin^2A} =(\sin A+\cos A)^2
$$
if and only if B is the double angle of A, or $B=2A+2k\pi$, $k=0,1,2,3...$
Advice is welcome as to improve notation and format.
i understand the RHS can be simplified to 1+sin(2A), but that doesn't seen to go anywhere .
Also, I have been able to simplify the left hand side of the expression down a little, but it's not taking me anywhere at all. I have a feeling that the difficulty of this proof lies in using the 2's in the LHS.
again, many thanks,
Yun Fei
| Abbreviating your equation as
$$\frac{P + 2\sin B \cos B}{Q} = R$$
we have
$$\sin 2B = Q R - P$$
where
$$\begin{align}
R &= \left(\cos A + \sin A \right)^2 = \cos^2 A + \sin^2 A + 2 \sin A \cos A \\
&= 1 + \sin 2A \\[4pt]
Q &= 2 - 2 \sin^2 A + \tan^2 A = 2 \cos^2 A + \frac{\sin^2 A}{\cos^2 A} \\
&= \frac{1}{\cos^2 A}\left(2 \cos^4 A + \sin^2 A\right)
\end{align}$$
For $P$, we'll conveniently add and subtract $\csc^4 A$:
$$\begin{align}
P &= 2 \tan A - 2 + \cot^2 A \left(- 2 + \sec^4 A - \csc^2 A \right) + \left( \csc^4 A - \csc^4 A \right)\\
&= 2 \tan A - 2 \left(1 +\cot^2 A \right) + \frac{\cos^2 A}{\sin^2 A}\left(\frac{1}{\cos^4 A} - \frac{1}{\sin^2 A} \right) + \frac{1}{\sin^4 A} &- \csc^4 A\\
&= 2 \tan A - 2 \csc^2 A + \frac{1}{\sin^2 A \cos^2 A} + \left(-\frac{\cos^2 A}{\sin^4 A} + \frac{1}{\sin^4 A}\right) &-\csc^4 A\\[4pt]
&=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{2}{\sin^2 A} + \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt]
&=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt]
&=\frac{2\sin A}{\cos A}+ \frac{1-\cos^2 A}{\sin^2 A\cos^2 A} &-\csc^4 A \\[4pt]
&= \frac{2\sin A \cos A}{\cos^2 A} + \frac{1}{\cos^2 A} &-\csc^4 A \\[4pt]
&= \frac{1+\sin 2A}{\cos^2 A} &-\csc^4 A
\end{align}$$
Therefore,
$$\begin{align}
QR-P &= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A \right)\left( 1 + \sin 2A \right) - \frac{1}{\cos^2 A}\left( 1 + \sin 2 A \right) + \csc^4 A \\[4pt]
&= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&= \frac{1}{\cos^2A}\left( 2 \cos^4 A - \cos^2 A\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&=\left( 2 \cos^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&=\cos 2A \left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
\end{align}$$
So, your equation reduces to
$$\sin 2 B = \cos 2A ( 1 + \sin 2 A ) + \csc^4 A$$
which is decidedly not equivalent to $B = 2 A + 2k\pi$. Perhaps your original equation has a typo or something.
Note that, even had the equation become
$$\sin 2B = \sin 4A$$
(which seems closest to what you might be anticipating), we'd have $2B = 4A + 2k\pi$ OR $2B = \pi - 4A + 2k\pi$, whence $B = 2A+k\pi$ OR $B = \frac{\pi}{2}-2A + k\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point.
| $$
p = x + \sqrt{x^2+3} \Rightarrow \sqrt{x^2+3} = p-x \\
p^2 = x^2 + x^2+3 + 2x \sqrt{x^2+3} = 2x^2+3 + 2x(p-x) = 2xp+3 \\
x = \frac {p^2-3}{2p}
$$
Analogously, if $y + \sqrt{y^2+3} = q$, then
$$
y = \frac {q^2-3}{2q}
$$
Initial expressions becomes $pq = 3$. Now, if calculate $x+y$ you'll get
$$
x+y = \frac {p^2-3}{2p} + \frac {q^2-3}{2q} = \frac 12 \left (\frac {p^2q-3q+q^2p-3p}{pq} \right ) = \frac {p+q}2 \left( pq-3\right) = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$ I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$
But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$
I ran out of imagination as what to try now
| Note that
$$\int_1^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{y^2dy}{1+y^4}$$
Hence,
$$I=\int_0^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{1+y^2}{1+y^4}dy$$
We have $y^4+1 = (y^2+1+y\sqrt2)(y^2+1-y\sqrt2)$. Hence,
$$1+y^2= \dfrac{(y^2+1+y\sqrt2) + (y^2+1-y\sqrt2)}2$$
Hence, we get that
\begin{align}
I & = \dfrac12\int_0^1\dfrac{dy}{1+y^2-y\sqrt2} + \dfrac12\int_0^1\dfrac{dy}{1+y^2+y\sqrt2}\\
& = \dfrac12\int_0^1\dfrac{dy}{\left(y-\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2} + \dfrac12\int_0^1\dfrac{dy}{\left(y+\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2}\\
& = \dfrac1{\sqrt2} \left(\arctan \left(y\sqrt2-1\right) + \arctan \left(y\sqrt2+1\right) \right)_{y=0}^1\\
& = \dfrac1{\sqrt2} \left(\arctan(\sqrt2 - 1)+\arctan(\sqrt2+1)\right) = \dfrac1{\sqrt2} \left(\arctan\left(\dfrac1{1+\sqrt2} \right)+\arctan(1+\sqrt2)\right)\\
& = \dfrac1{\sqrt2} \times \dfrac{\pi}2 = \dfrac{\pi}{2 \sqrt2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even I want to know for what valuyes this function is even
I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!
| Having even values is different to being an even function. For example, the function $\operatorname{f} : x \mapsto x$ has many even values, e.g. $2 \mapsto 2$ and $4 \mapsto 4$. However, $\operatorname{f}$ is not an even function since $\operatorname{f}(-x) \neq x$ for all $x \neq 0$.
In the example you give, the function cannot be even if it contains odd powers of $x$. Consider
\begin{array}{ccc}
\operatorname{f}(-x) &\equiv& (-x)^6 + a(-x)^3 - 2(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& (-x)^6 + (a-2)(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& x^6 +(2-a)x^3 -2x^2 + 1
\end{array}
Hence, $\operatorname{f}(x) \equiv \operatorname{f}(-x)$ if and only if $a-2=2-a$, i.e. $a=2$. Notice that $a=2$ if and only if the $x^3$, i.e. the odd powered term, vanishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve equation $\sqrt{4t + 1} = 3 - 3t$ Solve equation $\sqrt{4t + 1} = 3-3t$
→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$
→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$
I'm stuck at that point. Can someone tell me what I am doing wrong?
| Given: $\sqrt{4t+1}=3-3t$
$$\bigg(\sqrt{4t+1}\bigg)^2=\bigg(3-3t\bigg)^2$$
$$4t+1=\bigg(3-3t\bigg)\bigg(3-3t\bigg)$$
$$4t+1=9-9t-9t+9t^2$$
$$4t+1=9-18t+9t^2$$
$$0=9-1-18t-4t+9t^2$$
$$0=9t^2-22t+8$$
Which is a quadratic in the form:$$0 =ax^2+bx+c$$
Use the quadratic equation:
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x = \frac{ -(-22)\pm \sqrt{(-22)^2-4(9)(8)}}{2(9)}$$
$$x = \frac{22\pm\sqrt{484-288}}{18}$$
$$x = \frac{22\pm\sqrt{196}}{18}$$
$$x = \frac{22\pm{14}}{18}$$
$$x = 2 $$ $$or$$ $$x =\frac{4}{9} $$
Understand that it still be checked that these are indeed solutions to your initial equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$? I suspect that
$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
{{\pi}\over{2}}$$
for $C>0$.
I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?
| Let $f : \Bbb{R} \to \Bbb{C}$ be an integrable even function. Then we claim that
$$ \int_{0}^{\infty} f\left( y - \frac{1}{y} \right) \, dy = \int_{0}^{\infty} f(x) \, dx. \tag{1} $$
Indeed, let $I$ denote the LHS of $(1)$. Then by the substitution $y \mapsto y^{-1}$ we have
$$ I = \int_{0}^{\infty} f\left( \frac{1}{y} - y \right) \, \frac{dy}{y^2} = \int_{0}^{\infty} \frac{1}{y^2} f \left( y - \frac{1}{y} \right) \, dy. $$
Thus with the substitution $x = y - y^{-1}$, we have
$$ 2I = \int_{0}^{\infty} \left( 1 + \frac{1}{y^2} \right) f\left( y - \frac{1}{y} \right) \, dy = \int_{-\infty}^{\infty} f(x) \, dx = 2 \int_{0}^{\infty} f(x) \, dx $$
and the conclusion follows.
Now you can plug
$$ f(x) = \frac{1}{C + x^2} $$
to $(1)$ to obtain the conclusion:
$$ \int_{0}^{\infty} \frac{y^2}{Cy^2 + (y^2 - 1)^2} \, dy = \int_{0}^{\infty} \frac{dy}{C + (y - y^{-1})^2} \, dy = \int_{0}^{\infty} \frac{dx}{C + x^2} = \frac{\pi}{2\sqrt{C}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integration $\int \left(x-\frac{1}{2x} \right)^2\,dx $ Evaluate $$\int\!\left(x-\frac{1}{2x} \right)^2\,dx. $$
Using integrating by substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$. In the end, I came up with the answer to the integral as :
$$\left(\frac{1}{3}+\frac{2x^2}{3}\right)\left(x-\frac{1}{2x}\right)^3.$$
Any mistake ?
| Just expand
$$
\left(x-\frac{1}{2x}\right)^2 = x^2 - 1 +\frac{1}{4x^2}
$$
and integrate term by term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For what $n$ and $m$ is this number a perfect square? Find all positive integers $n$ and $m$ such that $$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3$$ is a perfect square.
Of course $n=m=1$ is a trivial solution and if you put $n=m+1$ (assume WLOG that $n \geq m$), the above expression is equal to $4n^4$ which is a perfect square - But I don't know if this is the only solution. I would really appreciate any help.
| (Too long for a comment) I think the question can be made clearer with a rearrangement. The $2(2mn)(m+n)$ smells like a quadratic:
$$((m+n)-(2mn+1))^2=(2mn+1)^2-2(2mn+1)(m+n)+m^2+n^2+2mn$$
$$=(2mn+1)^2-4mn(m+n)-2(m+n)+m^2+n^2+2mn$$
Rearranging,
$$=(2mn+1)^2-4mn(m+n)+m^2+n^2+2mn-2(m+n)$$
$$=[(2mn+1)^2-4mn(m+n)+m^2+n^2]+2mn-2(m+n)$$
And substituting,
$$(2mn+1)^2-4mn(m+n)+m^2+n^2+(n-1)^2+(m-1)^2-3$$
$$=((m+n)-(2mn+1))^2-2mn+2(m+n)+(n-1)^2+(m-1)^2-3$$
$$=((m+n)-(2mn+1))^2-2mn+2m+2n+n^2-2n+m^2-2m-1$$
$$=((m+n)-(2mn+1))^2-2mn+n^2+m^2-1$$
$$=(m+n-2mn-1)^2+(n-m)^2-1$$
It's clear now why $n=m+1$ worked. It's equally clear that $m+n-2mn-1=1 \Rightarrow n=\frac{m-2}{2m-1}$ is a solution (although no positive integer solutions exist).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove divisibility using linear congruences I need to prove that:
$$10|(53^{53} - 33^{33})$$
I can and should only use linear congruences ($a \equiv b \mod n$) - how can I do this?
| $$53^{53}-33^{33}\equiv 3^{53}- 3^{33}\pmod {10}$$
Now, $$3^4=81\equiv 1\pmod 4\implies 3^{53}- 3^{33}\equiv 3^1-3^1\pmod {10}\equiv0$$ as $33\equiv1\pmod 4, 53\equiv1\pmod 4$
Alternatively, $$3^{53}- 3^{33}\equiv 3^{33}(3^{20}-1)\pmod {10}$$
Now, $3^4\equiv1\pmod 4\implies 3^{20}=(3^4)^5\equiv1^5\pmod {10}\equiv1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove that $\sin (10^\circ), \sin(1^\circ), \sin(2^\circ), \sin(3^\circ), \tan(10^\circ)$ are irrational
Prove that $\sin (10^\circ)$, $\sin(1^\circ)$, $\sin(2^\circ)$, $\sin(3^\circ)$, and $\tan(10^\circ)$ are irrational.
My Attempt:
Let $x = 10^\circ$. Then
$$
\begin{align}
x &= 10^\circ\\
3x &= 30^\circ\\
\sin (3x) &= \sin (30^\circ)\\
3\sin (10^\circ)-4\sin^3(0^\circ) &= \frac{1}{2}
\end{align}
$$
Now let $y = \sin (10^\circ)$. Then
$$
\begin{align}
3y-4y^3 &= \frac{1}{2}\\
6y-8y^3 &= 1\\
\tag18y^3-6y+1 &= 0
\end{align}
$$
How can I calculate the roots of $(1)$?
| Here is a different approach for $\sin(1^\circ)=\sin\left(\frac{2\pi}{360}\right)$ (the other cases are similar).
The complex numbers $\zeta_{1,2}=e^{\pm\frac{2\pi}{360}i}$ are algebraic integers (are roots of $x^{360}-1$) and therefore $\zeta_1+\zeta_2=2\sin\left(\frac{2\pi}{360}\right)$ is algebraic integer. If $\sin\left(\frac{2\pi}{360}\right)$ is rational then $2\sin\left(\frac{2\pi}{360}\right)$ is rational and therefore integer (the only rational algebraic integers are the integers).
So $2\sin(1^\circ)=-2,-1,0,1$ or $2 \Rightarrow\Leftarrow$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$ $\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$
*
*$x=2\tan^{-1}t$
*$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$
*$dx=\dfrac{2}{1+t^2}\,dt$
*$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$
Is this the right idea? If so what do I do next?
$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$
$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$
$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$
| If $\;t = \tan\left(\frac 12 x\right)$, i.e., $\,x = 2\tan^{-1}t,\,$ what should $\dfrac 1{2 - \cos x}$ then be?
We need to replace the function (integrand) of $x$ to one expressed as a function of $t$.
What are the new limits for $\,t\,$ if $\;t = \tan\left(\frac 12 x\right)$?
When $x = 0,\;$ $t = \tan\left(\frac 02\right) = 0$. Okay. But, when $x = \pi/2$, the upper limit of integration needs to be $t = \tan\left(\pi/4\right)$
See Weierstrass Substitution for why $\;\cos x = \dfrac{1-t^2}{1+t^2},\;$ and in general, for the logic of using "$t$-substitution": $t = \tan \frac x2$.
ADDED:
After substituting all of the above, we should have the integrand:
$$\frac{1}{2-\dfrac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}=\frac{2}{2(1+t^{2})-(1-t^{2})}=\frac{2}{1+3t^{2}}=\frac{2}{1+(\sqrt{3}t)^{2}}$$
So we have
$$\int_0^1 \frac{2}{1+(\sqrt{3}t)^{2}}\,dt = 2\int_0^1 \frac{1}{1+(\sqrt{3}t)^{2}}\,dt\tag{1}$$
Now, I'm afraid to say, the work isn't done yet. We cannot use $\ln|f(t)|$ where $f(t) =1 + \sqrt{3}t)^{2}$ because we do not have an integrand in the form of $\;\dfrac{f'(t)}{f(t)} \,dx$.
But we're all set up with $(1)$ to use the substitution $$u = \sqrt 3 t.\,\implies du = \sqrt 3 dt \implies dt = \dfrac{1}{\sqrt 3} du$$
Then we have an integrand of the form $$2 \int_0^{\sqrt 3} \dfrac{1}{\sqrt 3} \dfrac{1}{1 + u^2}\,du = \dfrac{2}{\sqrt 3}\int_0^{\sqrt 3} \dfrac{1}{1 + u^2}\,du\tag{2}$$
Now, we recall that $$\int \dfrac {1}{1 + u^2} \,du = \tan^{-1}u + C\tag{$\star$}$$
Can you try and finish it from here? Apply $\star$ to the integral given by $(2)$
$(\star)$ See trigonometric substitution for integrals involving $a^2 + u^2$, where $a$ is a constant. Our integral is of the same form, with $a = 1$:
$$\int \frac{du}{a^2 + u^2} = \frac 1a\tan^{-1}\left(\frac ua\right)\,+ C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.
I tried many method, such as AM-GM, but all of them failed.
Thank you.
| use this
$$(x+y+z)^2+3(x+y+z)\ge x^2+y^2+z^2+x+2y+3z$$
and let $x+y+z=t$,
then $$t^2+3t\ge \dfrac{13}{4}$$
| {
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"source": "stackexchange",
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Proving an identity using formal power series
4.
(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.
(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
$$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$
Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?
| As DonAntonio points out, the first part is just a matter of factoring and doing some algebra. For the second part, note first that
$$\begin{align*}
\frac{1-x^2}{1+x^3}&=(1-x^2)\cdot\frac1{1-(-x^3)}\\
&=(1-x^2)\sum_{n\ge 0}(-1)^nx^{3n}\\
&=\sum_{n\ge 0}(-1)^nx^{3n}-\sum_{n\ge 0}(-1)^nx^{3n+2}\;,
\end{align*}$$
in which the coefficient of $x^{3n}$ is $(-1)^n$, the coefficient of $x^{3n+1}$ is $0$, and the coefficient of $x^{3n+2}$ is $(-1)^n$. Clearly, then, the absolute value of the coefficient of $x^N$ is $0$ if $N\equiv 1\pmod 3$ and $1$ otherwise.
Similarly,
$$\begin{align*}
\frac{1}{1+\frac{x^2}{1-x}}&=\frac1{1-\left(-\frac{x^2}{1-x}\right)}\\
&=\sum_{n\ge 0}(-1)^n\left(\frac{x^2}{1-x}\right)^n\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\frac1{(1-x)^n}\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\sum_{k\ge 0}\binom{n+k-1}kx^k\;,\tag{1}
\end{align*}$$
where the last step uses the standard power series expansion of $\dfrac1{(1-x)^n}$.
The coefficient of $x^N$ in $(1)$ is
$$\sum_{n\ge 0}(-1)^n\binom{N-n-1}{N-2n}\;;$$
to get this I simply write $N=2n+k$ and sum over $n$, so that $k=N-2n$. (The binomial coefficient is $0$ when $n$ is too large, so there’s no need to specify an upper bound.)
The result is now immediate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
I got $$s + 13 = 4 + 4\sqrt{7-s}+ 7 – s$$
I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$
I’m stuck at this point. Does anyone have an idea how to solve this equation?
| $$2s+2=4\sqrt{7-s}$$
Divide by $2$ both sides
$$s+1=2\sqrt{7-s}$$
Square both sides
$$s^2+1+2s=28-4s$$
Put everything on the left
$$s^2+6s-27=0$$
Now solve by radicals
$$s_{1,2}=\frac{-6\pm\sqrt{6^2-4(-27)}}{2}$$
giving you the two solutions
$$s_1=3\qquad;\qquad s_2=-9$$
Finally discard $s_2$ since it is not a solution of your title equation
| {
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"timestamp": "2023-03-29T00:00:00",
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Find an interval of convergence and an explicit formula for $f(x)$ Let $f(x) = 1 + 2x + x^2 + 2x^3 +x^4+...$
If $c_{2n} = 1$ and $c_{2n+1} = 2$ $\forall n \ge 0$ find the interval of convergence and an explicit formula for $f(x)$.
The answers are $I = (-1,1)$ and $f(x) = \frac{1 + 2x}{1 - x^2}$
Can anyone please give me an idea how to get it...
Thanks in advance
| The radius of convergence is $\dfrac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}}$ and since $1\le a_n\le2$, the Squeeze Theorem says the radius of convergence is $1$.
$$
\begin{align}
&\hphantom{(}1+2x\hphantom{)}+\hphantom{(}x^2+2x^3\hphantom{)}+\hphantom{(}x^4+2x^5\hphantom{)}+\hphantom{(}x^6+2x^7\hphantom{)}+\dots\\[8pt]
=&(1+2x)+(1+2x)x^2+(1+2x)x^4+(1+2x)x^6+\dots\\[8pt]
=&(1+2x)(1+x^2+x^4+x^6+\dots)\\[4pt]
=&\frac{1+2x}{1-x^2}
\end{align}
$$
$$
\begin{align}
&1+2x+x^2+2x^3+x^4+2x^5+x^6+2x^7+\dots\\[12pt]
=&1+\hphantom{2}x+x^2+\hphantom{2}x^3+x^4+\hphantom{2}x^5+x^6+\hphantom{2}x^7+\dots\\
&\hphantom{1}+\hphantom{2}x\hphantom{\,+\;x^2}+\phantom{2}x^3\hphantom{\,+\;x^4}+\phantom{2}x^5\hphantom{\,+\;x^4}+\phantom{2}x^7\dots\\[4pt]
=&\frac1{1-x}+\frac{x}{1-x^2}\\[9pt]
=&\frac{1+2x}{1-x^2}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$ Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?
Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.
| Let $a = \sqrt{x+4}$ and $b=\sqrt{x+1}$. So that $a^2 = x + 4$ and $b^2 = x + 1$.
From the given equation,
$$\sqrt{x+4} - \sqrt{x+1}=1 \implies a - b = 1 \ \ \ \text{and} \ \ \ a^2-b^2=3.$$
So we have that,
$$a^2-b^2 =(a-b)(a+b)=1 \cdot(a+b)=3 \implies a+b=3.$$
We see that
$$(a+b)+(a-b) = 2a.$$
Also,
$$(a+b)+(a-b)=3+1=4.$$
Therefore,
$$2a=4 \implies a=2 \implies \sqrt{x+4}=2.$$
Solving for $x$,
$$x+4=4 \implies x=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Inverse | Modulo | Power Describe the inverse of $5$ modulo $18$ as a positive power of $5\pmod{18}$.
I've got that the inverse of $5$ is $11$, but is this question asking to find a $t$ such that
$$ 11=5^t\pmod{18}?$$
| $$5^2=25\equiv 7\equiv-11 \pmod {18}, \quad 5^3\equiv 5\cdot7\equiv-1 \pmod {18}$$
$$\implies 11=(-11)(-1)\equiv 5^2\cdot5^3\equiv5^5 \pmod {18}$$
Now, as $5^3\equiv-1 \pmod {18}, 5^6=(5^3)^2\equiv(-1)^2\equiv1$
$\implies 5^{6u}=(5^6)^u\equiv1^u\equiv1\pmod{18}$ for any integer $u$
So, the general solution will be $t=6u+5$ as $5^{5+6u}\equiv1\cdot5^5\equiv1\cdot11 \pmod {18}$
If we need $5+6u>0, u\ge 0$
| {
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Proving the product of two real numbers is maximum when the numbers are equal given their sum is constant Let us consider two real numbers $x$ and $y$. How can we prove value of $xy$ is greatest when $x=y$ given the condition $x+y=$ constant?
I have already found a proof, but I am not entirely happy with it yet.
| Start with this lemma. Given real numbers $a, b, c \geq 0$, we have:
$$a(a + b + c) \leq (a + c)(a + b)$$
The proof is easy - just expand out the products on both sides. The RHS has an extra $bc$ term.
From this, we can prove what the question asks for the case where $x, y \geq 0$. Suppose $x + y = C$. Then WLOG $x \leq y$ and so $y = x + 2b$ for some $b \geq 0$. And so:
$$xy = x(x + 2b) = x(x + b + b) \leq (x + b)(x + b) = \left(\frac{C}{2}\right)^2$$
So the product is clearly maximised when $x = y = \frac{C}{2}$.
Generalisation
Using the above lemma, we can go further, and generalise the result to the product of $n$ real numbers $x_1, x_2, \dots, x_n \geq 0$ where $\sum x_i = C$. We want to show that the product $\prod x_i$ is maximised when $x_i = \frac{C}{n}$ for all $i$.
We can prove this by induction. In the base case we have $n = 1$ and the only choice is $x_1 = C$ which trivially maximises the result.
In the inductive case, we want to show:
$\prod_{i=1}^{n+1} x_i \leq \left(\frac{C}{n+1}\right)^{n+1}$
The product can be rewritten as:
$x_{n+1} \prod_{i=1}^n x_i$
Let $B = \sum_{i=1}^n x_i$. By I.H. we have:
$x_{n+1} \prod_{i=1}^n x_i \leq x_{n+1}\left(\frac{B}{n}\right)^n$
Now, let $\Delta = \frac{B}{n} - \frac{C}{n+1}$. We know that since $x_{n+1} = C - B$, then:
$$x_{n+1} = \frac{C}{n+1} - n\Delta$$
Define a sequence $S_i$ for $0 \leq i \leq n$ as follows:
$$S_i = \left(x_{n+1} + i\Delta\right)\left(\frac{C}{n+1}\right)^i\left(\frac{B}{n}\right)^{n-i}$$
Let's show that $S_i$ is monotonically non-decreasing i.e. $S_i \leq S_{i+1}$ for all $i < n$. Well:
$$S_i = \left(x_{n+1} + i\Delta\right)\left(\frac{B}{n}\right) \left(\frac{C}{n+1}\right)^i\left(\frac{B}{n}\right)^{n-(i+1)} \\= \left(\frac{C}{n+1} - (n - i)\Delta\right)\left(\frac{C}{n+1} + \Delta\right) \left(\frac{C}{n+1}\right)^i\left(\frac{B}{n}\right)^{n-(i+1)}$$
Now consider the leading terms $\left(\frac{C}{n+1} - (n - i)\Delta\right)$ and $\left(\frac{C}{n+1} + \Delta\right)$. Let's consider the cases for $\Delta$:
*
*If $\Delta = 0$, then $\left(\frac{C}{n+1} - (n - i)\Delta\right) = \left(\frac{C}{n+1} + \Delta\right) = \frac{C}{n+1}$ and so $S_i = \left(\frac{C}{n+1}\right)^{n+1}$ for all $i$. So certainly $S_i$ is non-decreasing as it is constant in this case.
*If $\Delta < 0$ then $\frac{C}{n+1} - (n - i)\Delta = \frac{C}{n+1} + (n - i - 1)(-\Delta) - \Delta$ and so by our above lemma, $\left(\frac{C}{n+1} - (n - i)\Delta\right)\left(\frac{C}{n+1} + \Delta\right) \leq \left(\frac{C}{n+1} - (n - i - 1)\Delta\right)\left(\frac{C}{n+1}\right)$ and so $S_i \leq S_{i+1}$.
*If $\Delta > 0$, then $\frac{C}{n+1} + \Delta = \left(\frac{C}{n+1} - (n - i)\Delta\right) + (n - i)\Delta + \Delta$ and so again we can apply the lemma to get that $\left(\frac{C}{n+1} - (n - i)\Delta\right)\left(\frac{C}{n+1} + \Delta\right) \leq \left(\frac{C}{n+1} - (n - i - 1)\Delta\right)\left(\frac{C}{n+1}\right)$ and so $S_i \leq S_{i+1}$.
Putting it all together then:
$$\prod_{i=1}^{n+1} x_i = x_{n+1} \prod_{i=1}^n x_i \leq x_{n+1}\left(\frac{B}{n}\right)^n = S_0 \leq S_1 \leq S_2 \dots \leq S_n = \left(\frac{C}{n+1}\right)^{n+1}$$
Which completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Primality of the numbers in the form of $2n^2-1$ I have a question about primality of integers in the form of $2n^2-1$.
I can prove that for the certain type of n such integers are always composite. For example, if $n=7k+2$ or $n=7k+5$, the whole expression would be always divisible by $7$.
The same is applicable to the whole (probably infinite) set of numbers in the form of $n=ak+b$, where $a$ is $7$, $17$, $23$ etc. and $b$ usually has two values (like $2$ and $5$ for $a=7$).
($a$ is prime here and $b \le a-1$)
I also suspect that the only composite values of $2n^2-1$ are those whose factors are from that set of a (like $7$, $17$, $23$ etc.).
I am trying to see if there anything else can be said about primality or compositness of those numbers $n$. Is there any other forms of n that can guarantee compositness (or primality)?
I would appreciate any ideas.
Thanks!
| Let $n=ak+b$, as you have it. Suppose prime $p|a$ and $2b^2\equiv 1\pmod{p}$. Then $2n^2-1=2(ak+b)^2-1=a(2ak^2+4abk)+2b^2-1\equiv 0\pmod{p}$, so $p|2n^2-1$ and hence $2n^2-1$ is composite.
We can test if $2b^2\equiv 1$ using Legendre symbols. Such a $b$ exists (in fact two exist) if $\left(\frac{1/2}{p}\right)=1$. We multiply both sides by $\left(\frac{2}{p}\right)$ to get $1=\left(\frac{1}{p}\right)=\left(\frac{2}{p}\right)$. This has solutions if $p\equiv 1,7\pmod{8}$.
Example: Let $a=51=3\times 17$. Since $17\equiv 1\pmod{8}$, there are two choices for $b$ such that $2b^2\equiv 1\pmod{17}$, namely $3,14$. Hence $n=51k+3$ and $n=51k+14$ will always be composite for all $k$.
| {
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Logarithmic Equations How does one go about solving:
$(5x+2)^{\frac{4}{3}} = 16$
I'm confused as how to parse through the equation to solve it using logs.
| $$(5x+2)^{\frac{4}{3}} = 16$$
$$\left((5x+2)^{\frac{4}{3}}\right)^{\frac{3}{4}} = 16^{\frac{3}{4}}$$
$$(5x+2)^{\frac{4}{3}\cdot \frac{3}{4}}=2^{4\cdot \frac{3}{4}}$$
$$5x+2=2^3$$
$$5x+2=8$$
$$x=\frac{6}{5}$$
| {
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help with $\lim\limits_{(x,y) \to (0,0)} f(x,y) = {\cos(x) -1 - {x^2/2} \over x^4 + y^4}$ $\lim\limits_{(x,y) \to (0,0)} f(x,y) = \dfrac{\cos(x) -1 - {x^2 \over 2}}{x^4 + y^4}$
Is the following approach correct?
If we approach the origin from $y$ , that is $x = 0$:
$\lim\limits_{(x=0,y) \to (0,0)} f(0,y) = {0 \over y^4} = 0$
Now we approach the origin from $x$ and use $\cos(x) \sim_{0} 1 - {x^2\over 2}$
$\lim\limits_{(x,y=0) \to (0,0)} f(0,y) = { -x^2 \over x^4} =- {1 \over x^2}=-\infty$
Then we can conclude that the limit does not exist.
My teacher took another approach where he uses the $\cos (2x)$ formula and ends up using $\lim\limits_{u\to0}\dfrac{\sin u}{u}=1$ so I'm wondering if my solution is OK or not.
| Use the straight line $y=mx$ for $m \in \mathbb{R}$:
$$
\lim_{x \to 0} \frac{\cos x -1 -\frac{x^2}{2}}{(1+m^2)x^2} = \lim_{x \to 0} \frac{1-\frac{x^2}{2}+\frac{c^4}{4!}-1-\frac{x^2}{2}}{(1+m^2)x^2}=\lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2}
$$
for some $c \in (0,x)$. Therefore
$$
\lim_{x \to 0} \frac{-x^2+\frac{c^4}{4!}}{(1+m^2)x^2} = -\frac{1}{1+m^2},
$$
which clearly depends on $m$. The original limit cannot exist.
| {
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"answer_id": 0
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A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$ Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in \mathbb{N}$.
| The binomial expansion tells you the following $$(1+\frac{1}{n})^n= 1^n + n*1^{n-1}*\frac{1}{n}+\binom{n}{2}*1^{n-2}*(\frac{1}{n})^2 + \cdots +(\frac{1}{n})^n $$ $$= 1+ 1+ \frac{1}{2!}(1-{1\over n}) + \cdots + \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1 - \frac{n-1}{n})$$ $$\lt 1+1+\frac{1}{2!}+\cdots+\frac{1}{n!} $$
I recommend you to do the calculation to see how it does this, it really helps in the future.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Number Of Solutions $X^{2}=X$ The equation $x^{2}=x$ $,x\in \mathbb{R}$ has only the solutions $x=0$ and $x=1$.
For a $2 \times 2$ matrix X, how many solutions does $X^{2}=X$ have?
What if $X$ is symmetric?
| Let $\textbf{X} = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)$, then ${\textbf{X}}^2-\textbf{X} =\textbf{X}(\textbf{X} - \textbf{I})= 0$ is equivalent to
$$ \left( {\matrix{
a & b \cr
c & d \cr
} } \right)\left( {\matrix{
a-1 & b \cr
c & d-1 \cr
} } \right) = 0$$
Solve this equation for $a,b,c,d$, I get that
$\left\{\left\{c=\frac{a-a^2}{b},d=1-a\right\},\{a=0,b=0,d=1\},\{a=1,b=0,d=0\},\{a=0,b=0,c=0,d=0\},\{a=0,b=0,c=0,d=1\},\{a=1,b=0,c=0,d=0\},\{a=1,b=0,c=0,d=1\}\right\}$
Obviously, there are infinite solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solving For Variables In Simultaneous Equations I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$
1)
$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$
2)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$
| I tried to solve 1) in your question.
$$\begin{cases}
a^2+b^2=a ----(1)\\
(a+d)b=b ----(2)\\
b^2+d^2=d ----(3)
\end{cases}$$
when $b=0$, the equations (1) and (3) boil down to $a^2=a$ and $d^2=d$. In this case, we have the solutions below
$$\{a=0,b=0,d=0\},\\
\{a=0,b=0,d=1\},\\
\{a=1,b=0,d=0\},\\
\{a=1,b=0,d=1 \}.$$
If $b\ne0$, then equations (1), (2) and (3) becomes
\begin{cases}
a^2-a +b^2=0 ----(4)\\
a+d =1 -------(5)\\
d^2-d +b^2= 0----(6)
\end{cases}
From (4),(5)and (6), we can see $a$ and $d$ are the solutions of the equation $x^2-x+b^2=0$, where $b$ can be any real number. So in this case, the solutions could be
$$\left\{a=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right),d=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right)\right\},\\
\left\{a=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right),d=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right)\right\}.
$$
Hint: your question 2) can be solved in a similar way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
My son's Sum of Some is beautiful! But what is the proof or explanation? My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$
Why is this so? What is the proof or explanation? Is it true for any $n$?
His own presentation of his finding:
Every one of these numbers is two times the number before it.
$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.
If I then skip three ones and read the fourth, that one equals my sum!
| Here is another formula similar to your son's, using powers of $5$:
$$1000000\cdot5^n+100000\cdot5^{n+1}+10000\cdot5^{n+2}+1000\cdot5^{n+3}+100\cdot5^{n+4}+5^{n+6}=5^{n+9}$$
We can derive identities of a similar form more generally. Consider any $m$th degree polynomial $f(x)\in\mathbb{Z}[x]$ of the following form
$$f(x)=a_0+a_1x+\ldots+a_{m-1}x^{m-1}-x^m$$
Notice that all rational roots of $f(x)$ are integers. Suppose $f$ has a rational root $k$. Then we have the following equality:
$$\begin{align}k^{n+m}&=k^nk^m\\&=k^n(a_0+a_1k+a_2k^2+\ldots+a_{m-1}k^{m-1})\\&=a_0k^n+a_1k^{n+1}+a_2k^{n+2}+\ldots+a_{m-1}k^{n+m-1}\end{align}$$
Your son's formula involves the $7$th degree polynomial $f(x)=100+10x+x^3-x^7$ and the root $k=2$. Above I've used the polynomial $f(x)=10^6+10^5x+10^4x^2+10^3x^3+10^2x^4+x^6-x^9$ and the root $k=5$.
Here are other examples for $k=3$. Consider the polynomials $f(x)=3+8x+80x^3-x^7,$ $g(x)=2130+10x+x^3-x^7$, and $h(x)=99+687x+x^3-x^7$, and notice that $f(3)=g(3)=h(3)=0$. These give us the following identities:
$$3\cdot3^n+8\cdot 3^{n+1}+80\cdot3^{n+3}=3^{n+7}\\2130\cdot3^n+10\cdot3^{n+1}+3^{n+3}=3^{n+7}\\99\cdot3^n+687\cdot3^{n+1}+3^{n+3}=3^{n+7}$$
Try using this method to find identities for other values of $k$ (including negative integers!).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "443",
"answer_count": 10,
"answer_id": 7
} |
Solving a algebraic equation I am to solve the following equation third order equation
$x^3+x^2+x+1=0$
What I've tried so far is writing the equation as
$ x \cdot (x^2+x+1)+1=0$
but that didn't lead anywhere. How do I solve this without using a computer?
| Factor by grouping:
$$
x^3 + x^2 + x + 1 = x^2(x+1)+(x+1) = (x+1)(x^2+1) = 0.
$$
Then $x + 1 = 0$ gives $x = -1$ and $x^2 + 1 = 0$ gives $x = \pm i$ (assuming you want to solve over the complex numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Sum of alternating sign squares of integers stuck with proof by induction Note that
$$
A(1):1=1\\A(2):1-4=-(1+2)\\A(3):1-4+9=1+2+3\\A(4):1-4+9-16=-(1+2+3+4)
$$
Let us set up the $A(k)$:
$$
A(k)=1-4+9-…+(-1)^{k+1}k^2=(-1)^{k+1}(1+2+…+k)
$$
Setting up $A(k+1)$:
$$
A(k+1)=1-4+9-…+(-1)^{k+1+1}(k+1)^2=(-1)^{k+1+1}(1+2+…+k+(k+1))
$$
Knowing that:
$$
1+2+…+n=\frac{n(n+1)}{2}
$$
We simplify right hand sides of $A(k)$ and $A(k+1)$:
$$
A(k)=(-1)^{k+1}(1+2+…+k)=(-1)^{k+1}\frac{k(k+1)}{2}\\A(k+1)=(-1)^{k+1+1}(1+2+…+k+(k+1))=(-1)^{k+1+1}\frac{(k+1)(k+2)}{2}
$$
Then I am trying to show that right hand side of $A(k+1)$ is equal to $A(k) + (-1)^{k+1+1}(k+1)^2$, but it does not work for me. That is what I am getting:
$$
A(k+1)=(-1)^{k+1+1}\frac{(k+1)(k+2)}{2}=(-1)^{k+1+1}\frac{k^2+2k+k+2}{2}=(-1)^{k+1+1}(\frac{k(k+1)}{2}+(k+1))=(-1)A(k)+(-1)^{k+1+1}(k+1)=-(A(k)+(-1)^{k+1}(k+1))
$$
What am I doing wrong?
How to prove $A(k)$ by induction?
| To prove it by induction, first show by calculation that $A(1)$ is true. Then we assume $A(n)$ is true and try to show $A(n+1)$ is true. So we assume $\displaystyle \sum_{i=1}^n (-1)^{i+1} i^2=(-1)^{n+1}\frac 12 n(n+1)$. Now we evaluate the left side of $A(n+1)$ $$\begin {align} \sum_{i=1}^{n+1} (-1)^{i+1} i^2&=(-1)^{n+1}\frac 12 n(n+1)+(-1)^{n+2}(n+1)^2 \\&=(-1)^{n+2}\left((n+1)^2-\frac 12n(n+1) \right)\\&=(-1)^{n+2}\left((n+1)(n+1-\frac 12n) \right)\\&=(-1)^{n+2}\left(\frac 12(n+1)(n+2) \right)\end {align}$$ and we have derived $A(n+1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Why the sum of $n$ seems equals the period of the binary expansion of $1/n$? The sum of $n$(suppose $n$ is positive odd,using $n=23$ as an example):
Step 1 : Get the odd part of $23 + ~~1 $, which is $~~3$,$~~3\times2^3=23 + ~~ 1$,get $s_1 = 3$
Step 2 : Get the odd part of $23 + ~~3 $, which is $13$,$13\times2^1=23 + ~~ 3$,get $s_2 = 1$
Step 3 : Get the odd part of $23 + 13 $, which is $~~9$,$~~9\times2^2=23 + 13$,get $s_3 = 2$
Step 4 : Get the odd part of $23 + ~~9 $, which is $~~1$,$~~1\times2^5=23 + ~~9$,get $s_4 = 5$
Continuing this operation (with $23 + 1$) repeats the same steps as above. There are $4$ steps in the cycle, so the cycle length of $23$ is $4$，and the sum of $23$ is $s_1 + s_2 + s_3 + s_4 = 11$.
The period of the binary expansion of $1/23$ is $11$,so why the sum of $23$ seems equals the period of the binary expansion of $1/23$?
| Consider the sequence $a_i = 2^{-i} \pmod{23}$
$$
\begin{array}{l}
a_0 = 1,~a_1=12,~a_2=6 \\
a_3 = 3 \\
a_4 = 13, ~ a_5=18, \\
a_6=9, ~ a_7 = 16, a_8=8, a_9=4, a_{10}=2 \\
a_{11} = 1 = a_0
\end{array}
$$
Your steps of "get the odd part" are precisely finding the next odd number in this sequence, your $s_i$ are counting all of the steps, and the sum $\sum s_i$ is the order of 2 modulo 23.
Let $n=\operatorname{ord}_p 2$ be the order of 2 modulo a prime $p$ and $m$ the period of the binary expansion of $1/p$. Then for some positive integer $k$
$$
2^n = pk+1 \\
2^n\frac{1}{p} = k + \frac{1}{p}
$$
That is, the binary representation of $1/p$ repeats after $n$ bits, so $n\mid m$. Similarly, for some positive integer $j$
$$
2^m \frac{1}{p} = j + \frac{1}{p} \\
2^m = jp + 1
$$
so also $m\mid n$ and hence $n=m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Verify the real solution of a linear system of differential equation I'm trying to solve $Y' = AY$ where $A= \left[
\begin{array}{ c c }
-2 & 6 \\
-3 & 4
\end{array} \right]$
I have found the eigenvalue $1 \pm 3i$ with eigenvector for $1+3i: $
$v = \left[
\begin{array}{ c c }
1-i \\
1
\end{array} \right]$
Which seems to be correct by testing $Av = (1+3i)v$ but when I try to write it as a real solution I don't seem to get the right answer.
$$\left[
\begin{array}{ c c }
1-i \\
1
\end{array} \right](\cos 3t + i\sin 3t) = \left[
\begin{array}{ c c }
\cos 3t + \sin 3t - i\cos 3t + i\sin 3t \\
\cos 3t + i\sin 3t
\end{array} \right]$$
$$v_1 = \left[
\begin{array}{ c c }
\cos 3t + \sin 3t \\
\cos 3t
\end{array} \right],\ v_2 = \left[
\begin{array}{ c c }
\cos 3t + \sin 3t \\
\sin 3t
\end{array} \right]$$
If I now verify by $v_1' = Av_1$
$$\left[
\begin{array}{ c c }
3(\cos 3t - \sin 3t) \\
-3\sin 3t
\end{array} \right] \neq \left[
\begin{array}{ c c }
4\cos 3t - 2\sin 3t \\
\cos 3t - 3\sin 3t
\end{array} \right]$$
| You are still missing an exponential term, we have:
$e^{\lambda t}v_1 = e^{(1+3i)t}\begin{bmatrix}1-i\\1\end{bmatrix} = e^te^{3it}\begin{bmatrix}1-i\\1\end{bmatrix} = e^t(\cos 3t + i \sin 3t)\begin{bmatrix}1-i\\1\end{bmatrix} = \begin{bmatrix}e^t(\sin 3t + \cos 3t+i (\sin 3 t-\cos 3 t))\\ e^t(\cos 3t + i \sin 3t) \end{bmatrix} $
So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):
$$Y(t) = c_1 e^t\begin{bmatrix}\sin 3t + \cos 3t\\ \cos 3t \end{bmatrix}+ c_2e^t\begin{bmatrix}\sin 3t - \cos 3t\\ \sin 3t \end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now,
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$.
| Incidentally, nobody appeared to have resolved the fallacy of the question, so I have provided an answer.
For all $a\in \mathbf N$, it follows $$\cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots - \cfrac 12}}}$$ such that the number of times the reciprocal in the continued fraction appears is $a$.
Proof. Note the identity $$\cfrac{1}{1+a}=1-\cfrac{1}{1+\color{red}{\cfrac 1a}}.$$ By letting $a=b-1$, it follows $$\cfrac 1b = 1-\cfrac{1}{1+\cfrac{1}{b-1}}.$$ From this we can substitute for $\color{red}{\cfrac 1a}$. $$\therefore \cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{1+\cfrac{1}{a-1}}}.$$ Clearly we can now likewise substitute for $1/(a-1)$, and the pattern will continue until for some $k\in\mathbf N$, the denominator of $1/(a-k)$ reaches $a-k=1$ since it cannot pass $0$. In consequence, we deduce as desired. (And, of course, when $a=0$, we have $1/1 = 1-0$.) This completes the proof. $\;\bigcirc$
And now, since $$\lim_{a\to\infty}\frac{1}{1+a}=0$$ then $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 7,
"answer_id": 6
} |
Prove $BA - A^2B^2 = I_n$. I have a problem with this. Actually, still don't have the right way to start :/
Problem :
Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$.
Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.
Thanks for any help.
| First, I try to demonstrate that $A$ and $B$ are invertible.
Let $(\lambda_j, v_j)$ be a couple of eigenvalue-eigenvector of $A$. Then, $(\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $A^3$. Since $A^3 = -B^3$, then we can say that $(-\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $B^3$. Finally, we can state that $(-\lambda_j, v_j)$ is is a couple of eigenvalue-eigenvector of $B$.
Consider now the following equations:
$$ABv_j - B^2A^2v_j = I_n v_j$$
$$A (-\lambda_j)v_j - B^2 (\lambda_j^2)v_j = v_j$$
$$(\lambda)_j (-\lambda_i)v_j - (\lambda_j^2) (\lambda_j^2)v_j = v_j$$
$$(\lambda_j^4 + \lambda_j^2 + 1)v_j = 0$$
Posing $(\lambda_j^4 + \lambda_j^2 + 1) = 0$, we get that eigenvalues are $\frac{1 \pm i \sqrt{3}}{2}$ and $\frac{-1 \pm i \sqrt{3}}{2}$. This means that neither $A$ nor $B$ has a null eigenvalues and hence $A$ and $B$ are both invertible.
At this point, we have that:
$AB - B^2A^2 = I_n \Rightarrow A = (I_n + B^2A^2)B^{-1}$
Then:
$BA - A^2B^2 = B(I_n + B^2A^2)B^{-1} - A^2B^2 = I_n + B^3A^2B^{-1} -A^2B^2$
We know that $A^3 = -B^3$, and then:
$BA - A^2B^2 = I_n - A^3A^2B^{-1} -A^2B^2 = I_n + A^2 B^3 B^{-1} - A^2B^2 = I_n + A^2B^2 - A^2B^2 = I_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
Determinant of a matrix Having some problems with a determinant of a 4x4 matrix M.
$
M =
\left( {\begin{array}{cc}
1 & 2 & 3 &-1 \\
0 & 1 & 2 & 2 \\
1 &1 &0 &0 \\
3&1&2&0
\end{array} } \right)
$
Went along and developed it according to the 4th column. So I end up with two matrixes A and B.
$
A = -1 \cdot det
\left( {\begin{array}{cc}
0 & 1 & 2 \\
1 & 1 & 0 \\
3 &1 &2 \\
\end{array} } \right)
$
$
B = 2 \cdot det
\left( {\begin{array}{cc}
1 & 2 & 3 \\
1 & 1 & 0 \\
3 &1 &2 \\
\end{array} } \right)
$
I get $A= (-1) \cdot((0 \cdot1\cdot2)+(1\cdot0\cdot3)+(2\cdot1\cdot1)-(3\cdot1\cdot2)-(1\cdot2\cdot2)-(1\cdot1\cdot0)) \\$
$A=(-1) \cdot(-6)=6$
$B= 2 \cdot((1\cdot1\cdot2)+(2\cdot0\cdot3)+(3\cdot1\cdot1)-(3\cdot1\cdot3)-(1\cdot2\cdot2)-(1\cdot1\cdot0))
\\$
$B = 2\cdot8=16$
$A+B=22$
which is wrong. Where is my mistake?
The correct answer should be $-22$ but I don't get why my solution keeps being positive.
Edit: im such a moron: A = 1* det and B = -2 * det. Everythings clearing up while in bed. Hehe!
| You are expanding down the fourth column, which means that the cofactor on entry $ \ a_{14} = -1 \ $ is negative and the cofactor on entry $ \ a_{24} = 2 \ $ is positive. You should have
$$ [ (-1) \cdot (-1) \cdot ( 2 - 6 - 2 ) ] \ + \ [ (+1) \cdot 2 \cdot ( 2 + 3 - 9 - 4 ) ] $$
$$ = \ (-6) \ + \ (-16) \ = \ -22 \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Problem involving cross ratio identity Let $z_1, z_2, z_3, z_4$ be distinct complex numbers. Assume that they lie on the same circle, in that order. Prove that $$|z_1 - z_3||z_2 - z_4| = |z_1 - z_2||z_3 - z_4| + |z_2 - z_3||z_4 - z_1|$$
So let F be a fractional linear map which sends $z_2, z_3, z_4$ to $1,0,\infty$ respectively. Note that since any fractional linear map is just a combination of translations, inversions and multiplications, it has to maintain the order of $z_1, z_2, z_3$ and $z_4$ if these points lie on the same circle. So neccesarily, $F(z_1) \in ]1, \infty[$. Is what I have done useful so far, if so, how do I continue?
Thanks in advance
| Hint: There is a theorem in geometry that says that the product of the diagonals of an inscribed quadrilateral is the sum of the products of the opposing sides.
Here is a proof using trigonometry. I will look for a simpler proof, if no one else does first.
$\hspace{3.6cm}$
Using cross-products to find the area of the quadrilateral yields
$$
\begin{align}
\text{Area of quadrilateral}
&=\frac12(ad+bc)\sin(\phi)\\
&=\frac12(e_1f_1+e_1f_2+e_2f_1+e_2f_2)\sin(\theta)\\
&=\frac12ef\sin(\theta)\tag{1}
\end{align}
$$
Using the Law of Cosines to find $f$ yields
$$
\begin{align}
f^2&=a^2+d^2-2ad\cos(\phi)\\
&=b^2+c^2+2bc\cos(\phi)\\
a^2+d^2-b^2-c^2&=2(ad+bc)\cos(\phi)\tag{2}
\end{align}
$$
Using the Law of Cosines to find $a,b,c,d$ yields
$$
\begin{align}
a^2&=e_1^2+f_2^2+2e_1f_2\cos(\theta)\\
b^2&=e_2^2+f_2^2-2e_2f_2\cos(\theta)\\
c^2&=e_2^2+f_1^2+2e_2f_1\cos(\theta)\\
d^2&=e_1^2+f_1^2-2e_1f_1\cos(\theta)\\
a^2-b^2+c^2-d^2&=2ef\cos(\theta)\tag{3}
\end{align}
$$
Putting together $(1)$, $(2)$, and $(3)$ yields
$$
\begin{align}
e^2f^2&=e^2f^2\cos^2(\theta)+e^2f^2\sin^2(\theta)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2\sin^2(\phi)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2-(ad+bc)^2\cos^2(\phi)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2-\frac14(a^2+d^2-b^2-c^2)\\[4pt]
&=(a^2-b^2)(c^2-d^2)+(ad+bc)^2\\[8pt]
&=(ac+bd)^2\\[9pt]
ef&=ac+bd\tag{4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $m+n=5$ and $mn=3$, find $\sqrt{\frac{n+1}{m+1}} + \sqrt{\frac{m+1}{n+1}}$? It is known that $m+n=5$ and $mn=3$. So what is the value of:
$$
\sqrt{\dfrac{n+1}{m+1}} + \sqrt{\dfrac{m+1}{n+1}}
$$
I think we're suppose to solve for the system of equations first, but I'm not getting any results that's useful.
| Danny's solution is short and nice. Here is an alternate, longer and less elegant solution:
Let $\sqrt{\frac{n+1}{m+1}}+\sqrt{\frac{m+1}{n+1}}=t$. Squaring both sides, we have:
$$\frac{n+1}{m+1}+\frac{m+1}{n+1}+2=t^2$$
Simplifying, we have:
$$\frac{(n+1)^2+(m+1)^2}{(m+1)(n+1)}+2=t^2$$
i.e.,
$$\frac{n^2+2n+1+m^2+2m+1}{mn+m+n+1}+2=t^2$$
i.e.,
$$\frac{m^2+n^2+2(m+n)+2}{mn+m+n+1}+2=t^2$$
But, $m^2+n^2=(m+n)^2-2mn$. Therefore, we have:
$$\frac{(m+n)^2-2mn+2(m+n)+2}{mn+m+n+1}+2=t^2$$
Substituting for $m+n=5$ and $mn=3$, we have:
$$\frac{(5)^2-2(3)+2(5)+2}{3+5+1}+2=t^2$$
i.e.,
$$\frac{31}{9}+2=t^2$$
i.e.,
$$\frac{49}{9}=t^2$$
Thus, we get $t=\frac{7}{3}$. Note that we eliminate the spurious negative solution of $-\frac{7}{3}$ as $t$ must be greater than $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaulate $\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$ Question: Find the limit, $$\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$$
The limit should be $\frac{3}{5}$ since when $x$ approaches infinity since $\frac{3*\infty^2}{5*\infty^2}$ and infinity squared cancels out. Am I correct?
| $$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}$$
To solve this you have two ways, one very simple is to apply L'hopitals' rule two times and get
$$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}=
\lim_{x\to\infty}\frac{ 6x- 36}{10x+113}=\lim_{x\to \infty}=6/10=3/5$$
or you ca divide by $x^2$ as follow
$$\lim_{x\to\infty}\frac{ 3x^2 - 36x+12}{5x^2+113x-2}=
\lim_{x\to\infty}\frac{ \frac{3x^2 - 36x+12}{x^2}}{\frac{5x^2+113x-2}{x^2}}
= \frac{\displaystyle\lim_{x\to\infty} 3 - 36/x+12/x^2}{\displaystyle \lim_{x\to\infty}5+113/x-2/x^2}= \frac{3}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Derivative of $ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$ $$ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$$
I know that it should be the chain rule and product rule used together to get the answer
$$ y = \frac{dx}{dy}((2x - 3)^4) \cdot (x^2 + x + 1)^5 + \frac{dx}{dy}(x^2 + x + 1)^5 \cdot (2x - 3)^4 $$
this gives me something ridiculous like this
$$8(2x-3)^3 \cdot (x^2 + x + 1)^5 + (x^2 + x + 1)^4 \cdot (2x+1) (2x-3)^4$$
This is wrong and I keep getting it, I don't know how to simplify it without expanding everything.
The book Houdini's out $(2x -3)^3 (x^2 + x + 1)^4 (28x^2 - 12x - 7)$
| Your result is not quite correct, and also not the form that it is conventionally left in. You can factor out $ \ (2x+3)^3 \ $ and $ \ (x^2+x+1)^4 \ $ from both terms, and then consolidate the remaining factors in both terms algebraically; you will have to simplify $ \ 8 \cdot (x^2 + x + 1) \ + \ 5 \cdot (2x + 1) \cdot (2x-3) \ $ .
(Currently, you have an error in your use of the Chain Rule, and a typo in the book's answer...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots$ For which values of $a,b\in\mathbb{R}$ the series:
$$\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots$$
converges?
Attempt:
$$\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots=\sum_{3k-2,3k-1}\frac{1}{n^b}+\sum_{3k}\frac{a}{n^b}$$
So, maybe, $b>1$ and $a\in\mathbb{R}$?
| We do indeed have convergence if $b\gt 1$. For let $c=\max(1,|a|)$. Then the $n$-th term of our series has absolute value $\lt \frac{c}{n^b}$. So by comparison, the series converges absolutely, and therefore converges.
We have divergence for $b\le 0$, since the terms do not have limit $0$.
So we only need to worry about $0\lt b\le 1$. In that case, if $a\ne -2$, we have divergence. We give a brief sketch of a proof. Call our series $\sum_1^\infty x_k$.
Note that $\lim_{k\to\infty}\frac{(3k+1)^b}{(3k+3)^b}=\lim_{k\to\infty}\frac{(3k+2)^b}{(3k+3)^b}=1$. So by taking $k$ large enough, we can make both these ratios as close to $1$ as we please.
Suppose, for example, that $b+2\gt 0$. Let $b+2=c$. Then by taking $k$ large enough, we can make the "sum" of the the $3$ consecutive terms $x_{3k+1}+x_{3k+2}+x_{3k}$ greater than $\dfrac{c/2}{(3k)^b}$. Thus we have divergence by a $p$-series comparison.
If $b+2\lt 0$, let $c=-(b+2)$. Essentially the same argument as the one for positive $b+2$ works.
We have now dealt with all cases except $0\lt b\le 1$. In that case, the term $\frac{a}{(3k)^b}=-\frac{2}{(3k)^b}$ provides enough cancellation for convergence. We first deal with the simplest case, $b=1$. In that case, the three-term sum $x_{3k+1}+x_{3k+2}+x_{3k}$ simplifies to
$$\frac{9k+4}{(3k+1)(3k+2)(3k+3)}.$$
The sum of three-term sums converges by Comparison. And since the $x_n$ have limit $0$, this implies that $\sum x_n$ converges.
For $0\lt b\lt 1$, essentially the same idea works. We need to estimate
$x_{3k+1}+x_{3k+2}+x_{3k}$. Bring to a common denominator as in the case $b=1$. Next use the fact that for $i=1,2,3$ we have
$$(3k+i)^b =(3k)^b\left(1+\frac{i}{3k}\right)^b =(3k)^b\left(1+\frac{ib}{3k}+o(1/k)\right).$$
There is cancellation of "main" terms, and we end up concluding that for some positive constant $c$, if $k$ is large enough then $0\lt x_{3k+1}+x_{3k+2}+x_{3k}\lt \frac{1}{(3k)^b}\cdot \frac{c}{k}$. This is enough to prove convergence of $\sum_k (x_{3k+1}+x_{3k+2}+x_{3k})$, and therefore of the original series.
Summary: We have convergence for arbitrary $a$ and $b\gt 1$. We also have convergence if $a=-2$ and $0\lt b\le 1$. In all other cases we have divergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\!\sqrt{\cot x}\,dx $ Find the integral
$$\int\!\sqrt{\cot x}\,dx $$
How can one solve this using substitution?
Can this be solved by complex methods?
| $\displaystyle \int \sqrt{\cot{x}}dx$
$\displaystyle \cot{x}=z^{2}\Rightarrow 2zdz=-(1+\cot^{2}{x})dx$
$\displaystyle \int \sqrt{\cot{x}}dx=-\int\frac{2z^{2}dz}{1+z^{4}}=-\int\frac{2dz}{z^{2}+\frac{1}{z^{2}}
}$
$\displaystyle\frac{1}{z^{2}+\frac{1}{z^{2}}}=\frac{1}{(z+\frac{1}{z})^{2}-2}=\frac{1}{(z-\frac{1}{z})^{2}+2}$
$\displaystyle u=z+\frac{1}{z},v=z-\frac{1}{z}$
$\displaystyle d(u+v)=2dz=du+dv$
$\displaystyle\frac{2dz}{z^{2}+\frac{1}{z^{2}}}=\frac{du}{u^{2}-2}+\frac{dv}{v^{2}+2}$
$=\displaystyle\frac{1}{2\sqrt{2}}(\frac{du}{u-\sqrt{2}}-\frac{du}{u+\sqrt{2}})+\frac{1}{\sqrt{2}}(\frac{d(\frac{v}{\sqrt{2}})}{(\frac{1}{\sqrt{2}}v)^{2}+1})$
$\displaystyle \int \sqrt{\cot{x}}dx=\frac{1}{2\sqrt{2}}ln\left| \frac{u+\sqrt{2}}{u-\sqrt{2}} \right|-\frac{1}{2} \arctan{\frac{1}{\sqrt{2}}v}+c$
$=\displaystyle \frac{1}{2\sqrt{2}}ln\left| \frac{z+\frac{1}{z}+\sqrt{2}}{z+\frac{1}{z}-\sqrt{2}} \right|-\frac{1}{2} \arctan{\frac{1}{\sqrt{2}}(z-\frac{1}{z})}+c$
$=\displaystyle \frac{1}{2\sqrt{2}}ln\left| \frac{\sqrt{\cot{x}}+\frac{1}{\sqrt{\cot{x}}}+\sqrt{2}}{\sqrt{\cot{x}}+\frac{1}{\sqrt{\cot{x}}}-\sqrt{2}} \right|-\frac{1}{2} \arctan{\frac{1}{\sqrt{2}}(\sqrt{\cot{x}}-\frac{1}{\sqrt{\cot{x}}})}+c\\$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Is there an algorithm to find the number of digits in 2^n, where n is a positive integer? Is there an algorithm to find the number of digits in 2^2030 ?
$2^1=2$
$2^2=4$
$2^3=8$
$2^4=16$
$2^5=32$
$2^6=64$
$2^7=128$
...
$2^{10}=1024$
| The number of digits in $2^n$ is $1+floor(n \log_{10} 2)$.
Using $\log_{10} 2 \approx 0.30103$ gives you 612 digits.
Here is an explanation:
A number $x$ has $k$ digits in base 10 iff $10^{k-1} \le x < 10^k$. Taking logs, we get $k-1 \le \log_{10} x < k$, which means that $floor(\log_{10} x)=k-1$. So, $k=1+floor(\log_{10} x)$.
An answer to the question in the title needs high-precision approximations of $\log_{10} 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b} \ge 3$
If $a,b,c$ are positive numbers then show that
$$\dfrac{a^2+1}{b+c}+\dfrac{b^2+1}{a+c}+\dfrac{c^2+1}{a+b} \ge 3$$
I am stuck at the first stage. Please give me some hints so that I can solve the problem. Thanks in advance.
| Hint: Note that $a^2+1 \ge 2a$ and also use NESBIT Inequality.
Note that $(a-1)^2\ge 0 \implies a^2+1 \ge 2a$
Otherwise use $AM \ge GM $ for $a^2$ and $1$, you get $a^2+1 \ge 2a$.
NESBITT Inequality: $$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \ge \dfrac 3 2$$
For proof of NESBITT Inequality you can see here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solving a 5 dimensional function in a neighbourhood Consider a function $f:\mathbb{R}^5 \to \mathbb{R}^2$ defined by
$$f(u,v,w,x,y)=(uy+vx+w+x^2,uvw+x+y+1)$$
such that $f(2,1,0,-1,0)=(0,0)$
(i) Show that we can solve $f(u,v,w,x,y) = (0,0)$ for $(x,y)$ in terms of (u,v,w) in a neighbourhood of $(2,1,0)$.
(ii)If $(x,y) = \phi(u,v,w)$ is the solution for (i) then show that derivative of $\phi$ at $(2,1,0)$ is
$$D\phi(2,1,0)=\frac13
\begin{bmatrix}
0 & -1 & -3 \\
0 & 1 & -3 \\
\end{bmatrix}$$
Here's how I tried:
Let $F=uy+vx+w+x^2=0$
& $G=uvw+x+y+1=0$
$$\frac{\partial(F,G)}{\partial(x,y)}_{(2,1,0,-1,0)} =
\begin{bmatrix}
v+2x & u \\
1 & 1 \\
\end{bmatrix}$$
$$\qquad \qquad \qquad=
\begin{bmatrix}
-1 & 2 \\
1 & 1 \\
\end{bmatrix}$$
which is non singular, so solution exists.
Part (ii):
we can write
$$x=X(u,v,w)$$
$$y=Y(u,v,w)$$
defined in the neighbourhood of $(2,1,0)$
such that:
$$X(2,1,0)=-1$$
$$Y(2,1,0)=0$$
what to do next? How to find $x=X(u,v,w)$ & $y=Y(u,v,w)$?
| The solution need not be nearly as complicated as what izœc does.
To show that $\phi$ exists in a neighborhood of $(2,1,0)$, we do exactly as you did:
$$Df(u,v,w,x,y) = \begin{pmatrix} y & x & 1 & v + 2x & u \\ vw & uw & uv & 1 & 1 \end{pmatrix},$$
$$Df(2,1,0,-1,0) = \begin{pmatrix} 0 & -1 & 1 & -1 & 2 \\ 0 & 0 & 2 & 1 & 1 \end{pmatrix}.$$
Since the matrix
$$\begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix}$$
corresponding to $(x,y)$ is invertible, the implicit function theorem can be applied.
Now if we write $Df = (A | B)$, where
$$A = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix},$$
a little work with the chain rule shows us that
$$D\phi(2,1,0) = -A^{-1} B.$$
This works for all implicit function theorem problems; see Rudin's Principles of Mathematical Analysis for the details. In our case,
$$A^{-1} = -\frac{1}{3} \begin{pmatrix} 1 & -2 \\ -1 & -1 \end{pmatrix},$$
and hence
\begin{align}
D\phi(2,1,0) & = \frac{1}{3} \begin{pmatrix} 1 & -2 \\\ -1 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & 2 \end{pmatrix} \\
& = \frac{1}{3} \begin{pmatrix} 0 & -1 & -3 \\ 0 & 1 & -3 \end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the $\int \frac{x+2}{x^2-4x+8}$ - a doubt I have to find the antiderivative of $f(x) = \dfrac{x+2}{x^2-4x+8}$
I rewrote it to the form $$ \dfrac{x-2}{x^2 -4x + 8} + \dfrac{1}{\frac{1}{4} (x-2)^2 +1}$$
The next step supposedly is $$F(x) = \dfrac{1}{2}\ln|x^2-4x+8| + 2 \arctan\left(\dfrac{1}{2}(x-2)\right) + c$$
But now there is a problem which I've always had: I don't know why the $\dfrac{1}{2}$ in front of the $\ln(x)$ is there, same for the $2$ and the $\dfrac{1}{2}$ for the $\arctan(x)$. For me simplifying the function to standard integrals is easy, however finding the right factors and such is hard.
| $\int$$\frac{1}{\frac{1}{4} (x-2)^2 +1}$=$4$$\int$$\frac{1}{(x-2)^2+4}$=$\frac{4}{2}$$\int$$\frac{2}{(x-2)^2+4}$=$\frac{4}{2}$$\tan^{-1}\frac{x-2}{2}$=...
You can simplify further. I have used the basics for differentiation of inverse trig functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Show sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing (a) Show that sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing
Proof
Let ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$ then
${a_k} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}}$
${a_{k + 1}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
We Know ${\left( {\frac{3}{5}} \right)^k} \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}}$
${\left( {\frac{3}{5}} \right)^k} + 1 \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}} + 1$
${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant $
${\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
$5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant 5{\left
[{{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
${a_k} \geqslant {a_{k + 1}}$
So ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$ is monotone decreasing .
I not sure it True or False ,I hope someone help me thank.
| Hint: Taking the log-derivative of $5\left(1+(3/5)^n\right)^{1/n}$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}n}\left(\log(5)+\frac1n\log\left(1+(3/5)^n\right)\right)
&=\frac1n\frac{\color{#C00000}{\log(3/5)}(3/5)^n}{1+(3/5)^n}\color{#C00000}{-\frac1{n^2}}\log\left(1+(3/5)^n\right)\end{align}
$$
I've just noticed that André's answer uses the quotient rule while mine uses product rule; otherwise, mine is not much different. To give added value to my answer, I feel I should give an answer that does not use calculus.
Bernoulli's Inequality, the rational version of which is proven here, shows
$$
\begin{align}
\left[1+\left(\frac35\right)^n\right]^{\Large\frac{n+1}{n}}
&\ge1+\frac{n+1}{n}\left(\frac35\right)^n\\
&\ge1+\frac35\left(\frac35\right)^n\\
&=1+\left(\frac35\right)^{n+1}\\
\left[1+\left(\frac35\right)^n\right]^{\Large\frac1n}
&\ge\left[1+\left(\frac35\right)^{n+1}\right]^{\Large\frac1{n+1}}\\
\left[5^n+3^n\vphantom{3^{n+1}}\right]^{\Large\frac1n}
&\ge\left[5^{n+1}+3^{n+1}\right]^{\Large\frac1{n+1}}
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/430548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Trigonometry and algebra question Given:
*
*The total length of ad + dc
*The lengths of each ab, bc and ca
*That angle adb = angle cdb
How can I work out lengths ad and cd
If you can't see the image above, click here
| Geometric Solution:
Let $h=\dfrac{|\overline{bc}|}2$ and $k=\dfrac{|\overline{bd}|+|\overline{cd}|}2$.
Place $c$ at $(-h,0)$ and $b$ at $(+h,0)$. $a$ would then be at the intersection of the circle of radius $|\overline{ac}|$ centered at $c$ and the circle of radius $|\overline{ab}|$ centered at $b$.
Since $|\overline{bd}|+|\overline{cd}|=2k$ (i.e. a constant), $d$ lies on the ellipse
$$
\frac{x^2}{k^2}+\frac{y^2}{k^2-h^2}=1\tag{1}
$$
Graphically,
$\hspace{3.2cm}$
Because of the reflection property of ellipses, $\angle adc=\angle adb$ when $\overline{ad}$ is normal to the ellipse (then $\angle adc$ and $\angle adb$ are supplementary to the angle of incidence). That is, $d$ is the point on the ellipse given in $(1)$ closest to (or furthest from) $a$.
The slope of the normal to the ellipse at $d=(x,y)$ is
$$
\frac yx\frac{k^2}{k^2-h^2}\tag{2}
$$
Therefore, we need
$$
\frac{y-a_y}{x-a_x}=\frac yx\frac{k^2}{k^2-h^2}\tag{3}
$$
which becomes the right hyperbola
$$
\left(x-\frac{k^2}{h^2}a_x\right)\left(y+\frac{k^2-h^2}{h^2}a_y\right)
=-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{4}
$$
Thus, we can find $d$ at the intersection of the ellipse in $(1)$ and the right hyperbola in $(4)$.
For computation of $d$, it is probably easiest to use
$$
u=x-\frac{k^2}{h^2}a_x\quad\text{and}\quad v=y+\frac{k^2-h^2}{h^2}a_y\tag{5}
$$
Then, we have from $(1)$ and $(4)$,
$$
\frac{\left(u+\frac{k^2}{h^2}a_x\right)^2}{k^2}+\frac{\left(v-\frac{k^2-h^2}{h^2}a_y\right)^2}{k^2-h^2}=1\tag{6}
$$
and
$$
uv=-\frac{k^2}{h^2}\frac{k^2-h^2}{h^2}a_xa_y\tag{7}
$$
Using $(7)$ in $(6)$ yields a fourth degree equation to solve.
| {
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relation between arithmetic series and `square` arithmetic series For example:
$$1+2+\text{...}+n=\frac{n(n+1)}{2}~~~(1)$$
$$1^2+2^2+\text{...}+n^2=\frac{n(n+1)(2n+1)}{6}~~~(2)$$
In this equality, I sometimes recall by heart
$\frac{n(2n+1)(2n+3)}{6}$ or others.
Why I cannot memorize some formulas exactly over these years?
Question1
How to derive $\frac{n(n+1)(2n+1)}{6}$ from $1^2+2^2+\text{...}+n^2$?
Question2
Now I happen to notice that $n(n+1)$ are the same to the summation in
$1+2+\text{...}+n=\frac{n(n+1)}{2}$
What's the relation between (1) and (2).
I think there must be large materials about more general cases
$$\sum _{i=0}^n i^k\text{ },k=1,2,\text{...},n~~~(3)$$
What's the formal terminology about that?
| As $(r+1)^{k+1}-r^{k+1}=\binom{k+1}1r^k+ \binom{k+1}2r^{k-1}+\cdots+\binom{k+1}{k-1}r^2+\binom{k+1}kr+1$
Putting $r=1,2,3,\cdots,n-1,n$ and adding them we get
$$(n+1)^{k+1}-1=\binom{k+1}1S_k+\binom{k+1}2S_{k-1}+\cdots+\binom{k+1}{k-1}S_2+\binom{k+1}kS_1+S_0 $$ where $S_k=\sum_{1\le r\le n}r^k$
Now, $S_0=\sum_{1\le r\le n}1=n$
Putting $k=1,(n+1)^2-1=\binom21S_1+S_0$ which will give us $S_1$ as we know $S_0$
Putting $k=2,(n+1)^3-1=\binom31S_2+\binom32S_1+S_0$ which will give us $S_2$ as we know $S_0,S_1$
and so on
| {
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Intersection of parabola and circle Is it possible to place circle and a parabola on the plane so that their intersection consist of exactly two points, one point being a point of tangency, and the other point a transversal intersection?
| Let the circle be the unit circle, and take the parabola to have the equation $y-k=a(x-h)^2$, with $a\neq 0$. Note that the slope of the line tangent to the parabola at $(x,y)$ is given by $m = 2 a (x-h)$.
Let $T(\cos\theta,\sin\theta)$ be the point of tangency. (We may assume throughout that $\sin\theta \neq 0$, since our (presumably non-degenerate) parabola has no vertical tangents. Note also that distinct points of intersection must correspond to distinct $x$ coordinates.) Then, as $T$ must satisfy the parabola equation, we have
$$\sin\theta - k = a ( \cos\theta - h )^2 \qquad (1)$$
Now, the line tangent to the circle at $T$ necessarily has slope $-\cot\theta$; equating this with the slope relative to the parabola gives
$$\cos\theta = -2 a \sin\theta (\cos\theta - h ) \qquad (2)$$
We can solve $(2)$ for $h$, and then $(1)$ for $k$:
$$
h = \frac{\cos\theta \left( 1 + 2 a \sin\theta \right)}{2a\sin\theta}
\qquad
k = \frac{ 4 a \sin^3\theta - \cos^2\theta }{4a\sin\theta^2}$$
whence the parabola equation becomes
$$y \sin\theta = a x^2 \sin\theta - x \cos\theta \left( 1 + 2 a \sin\theta \right) + 1 + a \cos^2\theta \sin\theta $$
Using that equation to eliminate $y$ in $x^2+y^2=1$ gives this polynomial equation in $x$:
$$\left( x - \cos\theta \right)^2 \left(
\left( a \left( x - \cos\theta \right)\sin\theta - \cos\theta\right)^2
+ \sin^2\theta \left( 1 + 2 a \sin\theta \right)\right) = 0 \qquad (3) $$
For the second factor of $(3)$ to have any roots requires the quantity $b := 1+2a\sin\theta$ to be non-positive. If $b$ vanishes, we get the double root $x=-\cos\theta$ (the mirror image of $x=\cos\theta$) corresponding to a second point of circle-parabola tangency. For us to have a point of non-tangent intersection, we require $b$ to be strictly negative, and $x=\cos\theta$ to be a root of that second factor. The latter condition implies, by substitution:
$$2a\sin^3\theta=-1$$
giving $b=-\cot^2\theta$, which is strictly negative (and defined) for $\theta \neq n \pi/2$. Moreover, equation $(3)$ reduces nicely to
$$0 = (x-\cos\theta)^3\left(x-4\cos^3\theta+3\cos\theta\right) = (x-\cos\theta)^3(x-\cos 3\theta)$$
providing the root $x=\cos3\theta$. Double-checking with the parabola equation, we see that the corresponding $y$ value should be $-\sin3\theta$; thus, we can write the second point of intersection as $S(\cos(-3\theta), \sin(-3\theta))$.
The final parabola equation itself is
$$ 2 y \sin^3\theta = - \left( x - \cos^3\theta \right)^2 + \sin^4\theta\left( 3 - \sin^2\theta \right)$$
Here's an animation:
(Animation, and interstitial symbol-crunching, courtesy of Mathematica.)
| {
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Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| Following your idea:
$$ (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
= \sqrt{n^2\left(1 + \frac1n\right)}
- \sqrt[3]{n^3\left(1 + \frac1n\right)}=$$
$$=n\left(\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}\right)=\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}$$
Looking at the above as function of the continuous variable $\,n\,$, we can use l'Hospital to try to calculate its limit:
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}\stackrel{\text{l'H}}=\lim_{n\to\infty}\left(\frac1{2\sqrt{1+\frac1n}}-\frac1{3\sqrt[3]{\left(1+\frac1n\right)^2}}\right)=\frac12-\frac13=\frac16$$
| {
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Quadratic equation which has rational roots
If the following quadratic equation
$$qx^2+(p+q)x+bp=0$$
always has rational roots for any non-zero integers $p$ and $q$ what will be the value of $b$?
My book's solution says the value of $b$ will be $0$ or $1$.
If we consider the discriminant of the equation,
$$D=(p+q)^2-4bqp = p^2+2q(1-2b)p+q^2$$
then $D$ should be a perfect square of a rational number for the equation having rational roots. so the value of b should be 0 0r 1 for given conditions (p and q are non-zero integers and the equation always has rational roots)
But I am not sure whether we can conclude the discriminant $D=p^2+2q(1-2b)p+q^2$ is a perfect square of a rational number if only if b is 0 or 1.
I can see if b is 0 or 1, then $D$ will be $(p+q)^2$ or $(p-q)^2$. So $D$ will be a perfect square of a rational number.
But I can't figure out the other case: If $D$ is a perfect square of a rational number for any non-zero integer p and q
, then the value of b will be 0 or 1.
EDIT 1: b is rational number.
| The key is that it must always have rational roots for any $p, q$.
Claim: Let $f(x) = x^2 + bx + c$, where $b$ and $c$ are rational numbers. If $f(x)$ satisfies the condition that for all rational numbers $r$, $f(r)$ is the square of a rational number, then $f(x)$ must be the square of a polynomial.
Proof: By completing the square, we simply need to consider polynomials of the form $x^2 + \frac{n}{m}$, where $m,n$ are integers and $\gcd(n,m) = 1$.
Consider $x = 0$ and $\frac{1}{m}$, which gives us $\frac{n}{m}$ and $\frac{1+mn}{m^2}$ are both squares of rational numbers. Multiplying both terms by $m^2$, we get that $mn$ and $mn+1$ are both integers that are squares of a rational number, hence they are perfect squares, so $mn=0$.
Since $m\neq 0$, hence $n=0$ and thus $f(x) = x^2$ and we are done. $_\square$
Corollary: $b=0$ or $1$.
Proof: Consider the polynomial $f(x) = \left( \frac{p}{q} \right)^2 + 2(1-2b) \frac{p}{q} + 1$. From the conditions, for rational numbers, this always evaluates to a rational square. Hence, we must have $ f(x) = (ax+b)^2$.
Comparing coefficients, we see that $ a ^2 = 1, b^2 = 1$, which implies that $f(x) = (x+1)^2 $ or $(x-1)^2$, which hence correspond to $b=0$ and $b=1$ respectively. $_\square$
Note: Franklin shows that the claim is true for all polynomials, though it uses more machinery.
| {
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Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Obviously, AM-GM seems to be suitable for LHS.
For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to do.
Can someone please teach me? Thank you.
p.s. That $\sqrt [27]{}$ is really terrible...
| since
$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)=a^3+b^3+c^3+24$$
so
$$(a+b+c)^3=a^3+b^3+c^3+3+3+3+3+3+3+3+3\ge 9\sqrt[9]{(a^3+b^3+c^3)\times 3^8}$$
so
$$\dfrac{a+b+c}{3}\ge\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}$$
| {
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Partial fractions $\int \frac{(3x^2 - 4x + 5)\,dx}{(x-1)(x^2+2)}$ $$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)}$$
I am going to use undetermined coefficients since it seems straightforward, no wacky matrices or tables to memorize.
$$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)} \quad = \quad\int \left(\frac {A}{x} + \frac{B}{(x-1)^2}+\frac{C}{(x-1)}+D\right)\,dx$$
I get $$A(x^3 - 3x^2 + 3x - 1) +B(x^3 - 2x^2 + x) + C(x^2 - x) + Dx$$
This gives me the sets of
$$\begin{align}
A + B & = 0\\ \\
-3A + -2B + C & = 0 \\ \\
3A + B - C + D & = 0 \\ \\
-A &= 0
\end{align}$$
This is obviously wrong because A and B are now 0. What do I do?
| Before moving to partial fractions, it's useful to separate the denominator's derivative, which is $(3x^2-2x+2)$:
$$\begin{align}
\int\frac{3x^2-4x+5}{(x-1)(x^2+2)}\,dx
&=\int\frac{(3x^2-2x+2)+(-2x+3)}{(x-1)(x^2+2)}\,dx\\
&=\ln((x-1)(x^2+2))+\int\frac{-2x+3}{(x-1)(x^2+2)}\,dx\\
\end{align}$$
Now the remaining separation into partial fractions is easier because there's an extra zero in the system of equations. In fact the separation must be in the form:
$$\begin{align}
\ln((x-1)(x^2+2))+\int\left(\frac{A}{x-1}+\frac{-Ax+B}{x^2+2}\right)\,dx\\
\end{align}$$
to make the $x^2$ terms disappear from the numerator. Now the system to solve is $$\left\{\begin{array}{rcl}A+B&=&-2\\2A-B&=&3\end{array}\right.$$ "Adding" equations reveals that $A=1/3$. From there, we deduce $B=-7/3$. So we have $$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{-x+7}{x^2+2}\right)\,dx\\
\end{align}$$
Again, I'd separate a derivative:
$$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)+\frac{1}{3}\int\frac{-x+7}{x^2+2}\,dx\\
=\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\int\frac{2x-14}{x^2+2}\,dx\\
=\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{x^2+2}\,dx\\
\end{align}$$ and the last integral can be handled via substituting $2u^2=x^2$:
$$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{x^2+2}\,dx\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{2u^2+2}\,(\sqrt{2}\,du)\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{7}{3\sqrt{2}}\arctan(u)+C\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{7}{3\sqrt{2}}\arctan(x/\sqrt{2})+C\\
\end{align}$$
| {
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prove：$\left|x+\frac{1}{x}\right|\geq 2$ prove：
$\left|x+\frac{1}{x}\right|\geq 2$
Can I just use
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$
and
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?
| Denote $x+\frac{1}{x}=v$.
Obviously $|v|>2$ if either $v>2$ or $-v<-2$. The first case is therefore $x +\frac{1}{x}>2$ and it becomes $(x-1)^2>0$, which is always true. The second case is $-x-\frac{1}{x}<-2$ or $\frac{x^2-2x+1}{x}>0$ and also breaks down into 2 cases: either $(x-1)^2>0 \cap x>0$ or $(x-1)^2<0 \cap x<0$. Obviously the second case is never true and the first case is always true.
Hence $|x+\frac{1}{x}|$ is always larger than $2$.
| {
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Maxima/Minima Problem I am working on this Maxima and Minima Problem :
Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$
I tried it solving using the lagrange's method of multipliers. I get the following equations
$x+3x\theta+2y\theta=0$
$y+4y\theta+2x\theta=0$
$2z=0$
$3x^2+4xy+6y^2=140$
So i get $z=0$
Please suggest how to go about solving for $x$ and $y$ ?
| An approach with a bit of geometry:
$3x^2 + 4xy + 6y^2 = 140$ is an ellipse rotated about the origin.
Rewriting
$3x^2 + 4xy + 6y^2 = 140$;
$2x^2 + (x+2y)^2 + 2y^2 = 140$;
$(x^2 + y^2) = 140/2 - [(x + 2y)^2]/2$.
Let $r$ be the distance from the origin,
$r^2 = (x^2 + y^2) = 70 - $
$[(x + 2y)^2 ]/2$.
$Max:$
Since the bracket on the r.h.s., a square, is $\\ \geq 0$, we get for the maximum:
$r^2 = 70$, at $ 2y + x = 0$.
The major axis of the ellipse, call it $a$, lies along the line $2y + x = 0$, and has squared length:
$a^2 = 70$.
$Minimum:$
The major axis lies along $2y +x = 0$, or $y = - (1/2)x$ , a straight line with slope $- 1/2$.
The line $y = 2x$ is perpendicular to $y = - (1/2) x$, the orientation of the major axis, I.e. a line along the minor axis.
Intersection of $y = 2x$ with the ellipse:
$3x^2 + 4x(2x) + 6(2x)^2 = 140$;
$35x^2 = 140$;
$\\x^2 = 4$, and using $y= 2x$, we find
$y^2 = 4x^2 = 16$.
$Finally$:
$r^2 = min ( x^2 +y^2) = 20$;
For the squared minor axis, call it $b$, we have
$b^2 = 20$.
| {
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Find $m, n$ such that $\frac{n^2 + 1}{m^2 + 1 }$ is an integer multiple of a perfect square I'm trying to find $n,m \in \mathbb{N}$ such that $\sqrt{ \frac{n^2+1}{2(m^2+1)}}$ is rational.
I see that if $a,b$ are relatively prime $\sqrt{ \frac{a}{b}}$ is rational if and only if $a,b$ are perfect squares. $n^2+1$ can be a perfect square only if n = 0 and $2(m^2+1)$ is a square only when m = 1. For any other solution, we must have that WLOG $a = b\cdot r^2$ for some integer $r$. In other words $\frac{n^2+1}{m^2+1} = 2 \cdot r^2$
How could I go about solving that - or, what seems more likely, showing that there are no solutions? Is there a more general way to show $\frac{n^2+1}{m^2+1} = d \cdot r^2$ can have solutions only for specific $d$.
Thank you
| As leshik points out, there are infinitely many solutions to
$$n^2 - 2m^2 = 1$$
Each of these gives $\frac{n^2 + 1 } { 2(m^2 + 1) } = 1$
I'm not certain if requiring an integer perfect square is possible, but your question states that rational numbers are fine.
Consider the Pell's equation $x^2 - 13 y^2 = -1$. It has a solution $(18, 5)$, hence has infinitely many solution.
Observe that for each solution,
$$\frac{x^2+1} { 2 (5^2 + 1) } = \left( \frac{y}{2} \right) ^2 $$
Hence, we have infinitely many pairs of integers of the form $(x, 5)$ which work.
Since the smallest solution of $x^2- 13 y^2 = 1$ is $(649, 180)$, the next solution is quite large, and is $(x,y) = (23382, 6485)$.
This arose from realizing that we have a solution $(18,5)$. To generalize this approach for any $d$ would require finding a specific solution first, before knowing what $D$ (in Pell's Equation) to use.
| {
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Prove that $\frac{3}{2}\le\frac{1}{a+ab}+\frac{a}{1+ab}+\frac{ab}{1+a}\le\frac{19}{10}$ for $a, b \in [1/2, 2]$
Let
$a,b\in [\frac{1}{2},2]$. Prove that
$$\dfrac{3}{2}\le\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}\le\dfrac{19}{10}.$$
my idea:
$$\dfrac{1}{a+ab}+\dfrac{a}{1+ab}+\dfrac{ab}{1+a}-\dfrac{3}{2}\ge 0?$$
and this problem is from《Mathematics Studying》(2012.7).
See: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487281&p=2730329#p2730329
Thank you everyone.
I have see this same problem
| You can also use the Lagrange Function for the optimisation problem
$$\min_{(a,b) \in [\frac{1}{2},2]^2} f(a,b) := \frac{1}{a+ab} + \frac{a}{1+ab} + \frac{ab}{1+a}$$
This will result in an inner minimum at $(a,b) = (1,1), f_{min} = \frac{3}{2}$ and boundary maxima at $(a,b) \in \{ (\frac{1}{2},\frac{1}{2}), (2,2) \}, f_{max} = \frac{19}{10}$ as proposed. The biggest annoyance in this solution is the Hessian Matrix of $f$, wich is very ugly:
$$\left[ \begin {array}{cc} 2\,{\frac { \left( 1+b \right) ^{2}}{
\left( a+ab \right) ^{3}}}-2\,{\frac {b}{ \left( 1+ab \right) ^{2}}}+
2\,{\frac {{b}^{2}a}{ \left( 1+ab \right) ^{3}}}-2\,{\frac {b}{
\left( 1+a \right) ^{2}}}+2\,{\frac {ab}{ \left( 1+a \right) ^{3}}}&2
\,{\frac {a \left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a
+ab \right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac
{{a}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{
\frac {a}{ \left( 1+a \right) ^{2}}}\\ 2\,{\frac {a
\left( 1+b \right) }{ \left( a+ab \right) ^{3}}}- \left( a+ab
\right) ^{-2}-2\,{\frac {a}{ \left( 1+ab \right) ^{2}}}+2\,{\frac {{a
}^{2}b}{ \left( 1+ab \right) ^{3}}}+ \left( 1+a \right) ^{-1}-{\frac {
a}{ \left( 1+a \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( a+ab
\right) ^{3}}}+2\,{\frac {{a}^{3}}{ \left( 1+ab \right) ^{3}}}
\end {array} \right] $$
| {
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How to show the monotonicity of this function? Thanks for your help. I want to show $f(x)=\displaystyle \frac{1-x^{n+1}(n+2)+(n+1)x^{n+2}}{(1-x^{n+1})(1-x)}$ (strictly) increases w.r.t. $x$ for $x>0$. Here, $n=1,2,3,\dots.$. So far, I can only show $f(x)$ is increasing when $x>1$. Can anyone give me a hint to complete the proof? Thanks a lot.
| I really like this question.
Applying division and then partial fractions, you can write your function as
$$n + 1 + \frac{1}{1-x} - \frac{n+1}{1-x^{n+1}}.$$
Since $f$ is continuous except at $x=1$, and its limit exists there (I'll leave it to you to show this), to show it is strictly increasing we need to prove that its derivative
$$\frac{1}{(1-x)^2} - \frac{(n+1)^2x^n}{(1-x^{n+1})^2}$$
is positive almost everywhere.
First, by AM-GM, for $x\in (0,1)\cup (1,\infty)$
$$\frac{1+x+x^2+\ldots+x^n}{1+n} > x^{n/2}.$$
Therefore
\begin{align*}
\frac{1-x^{n+1}}{1-x} &> (1+n)x^{n/2}\\
\frac{(1-x^{n+1})^2}{(1-x)^2} &> (1+n)^2x^n\\
\frac{1}{(1-x)^2} &> \frac{(1+n)^2x^n}{(1-x^{n+1})^2},
\end{align*}
so
$$\frac{1}{(1-x)^2} - \frac{(1+n)^2x^n}{(1-x^{n+1})^2} > 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the lengths of the heights descending from triangle vertices Given a triangle, calculate the lengths of the heights descending from triangle vertices $A, B$ and $C$ respectively.
$$A(-1,-1,5),~~ B(0,3,-2),~~ C(3,-1,3)$$
I don't get it with which forma i should use and how to solve this question
| 1.) Use the distance formula to determine the side lengths:
$$d = \sqrt {(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$
$$AB = \sqrt {(-1-0)^2+(-1-3)^2+(5-(-2))^2}=\sqrt{66}$$
$$AC = \sqrt {(-1-3)^2+(-1-(-1))^2+(5-3)^2}=\sqrt{20}$$
$$BC = \sqrt {(0-3)^2+(3-(-1))^2+(-2-3)^2}=\sqrt{50}$$
2.) Then use Heron's formula to calculate the area:
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
Where $s = \frac {a+b+c} {2}$.
$$s=\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2}$$
$$Area=\sqrt{(\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{66})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{20})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{50})} \approx 15.780$$
3.) Calculate each height using $Area=\frac12bh$:
$$H_A = 2\frac{Area}{BC}\approx2\frac{15.780}{\sqrt{50}}\approx 4.463$$
$$H_B = 2\frac{Area}{AC}\approx2\frac{15.780}{\sqrt{20}}\approx 7.057$$
$$H_C = 2\frac{Area}{AB}\approx2\frac{15.780}{\sqrt{66}}\approx 3.885$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $xyz=8$, try to prove that $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2$ Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always less than 1 for $t>0$.
I'm just wondering if we can have some elegant simple prove. I guess we should use Jensen's inequality. Thanks.
Suppose $x,y,z\in R^+$ and $xyz=8$, try to prove that $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2.$$
Please note that the usual AM-GM inequality may not do its trick here as the equality is rather hold on the boundary.
Thanks.
| $a=\sqrt{x+1}>1,b=\sqrt{y+1}>1, c=\sqrt{z+1}>1$,for $a,b,c$,at least two of them $>2$ or $\le 2$,WOLG, assume $a,b$ either both $>2$ or both $\le2$, $z=\dfrac{8}{xy}=\dfrac{8}{(a^2-1)(b^2-1)}$,
edit:
case I:
if $a>2$ and $b>2$
it is trivial $\dfrac{1}{a}+\dfrac{1}{b} <1 ,\dfrac{1}{c}<1$
so it is true in this case.
case II: when $a \le 2,b \le 2$,
now we need to prove:
$\dfrac{1}{a}+\dfrac{1}{b}+\sqrt{\dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}}<2 \iff \dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}< \left(2-\dfrac{1}{a}-\dfrac{1}{b} \right)^2 \iff \dfrac{(2ab-a-b)^2}{a^2b^2} > \dfrac{(a^2-1)(b^2-1)}{((a^2-1)(b^2-1)+8)}$
note: $(a^2-1)(b^2-1)+8>a^2b^2 \iff 9>a^2+b^2 \iff (a^2\le 4) \cap (b^2 \le 4) $
now we need to prove $(2ab-a-b)^2>(a^2-1)(b^2-1)$
$(2ab-a-b)^2= (a(b-1)+b(a-1))^2 \ge 4ab(a-1)(b-1) \iff 4ab >(a+1)(b+1) \iff (2a>a+1 )\cap (2b>b+1) \iff (a>1) \cap (b>1)$
it is true.
both cases are true. QED
Indeed, when $xyz=2$, this inequality is also true.
| {
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Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ Please help to prove this inequality.
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.
Thanks.
| Write the means
$$
\begin{align}
H &= \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\
G &= \sqrt[3]{xyz} \\
A &=\left(\frac{x+y+z}{3}\right)
\end{align}
$$
then the power mean (AM-GM-HM) inequality gives $H \le G \le A$. Also note
$$
xy+yz+zx = xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3G^3}{H}
$$
Now given
$$
\begin{align}
8^3 &= (x^2-1)(y^2-1)(z^2-1) \\
&= (x-1)(y-1)(z-1)(x+1)(y+1)(z+1) \\
&= (xyz-(xy+yz+zx)+(x+y+z)-1)(xyz+(xy+yz+zx)+(x+y+z)+1) \\
&= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
\left(G^3\left(1+\frac{3}{H}\right)+3A+1\right)
\end{align}
$$
Then we have either
$$
1-\frac{3}{H} < 0 \\
\implies H < 3 \implies
1< \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
or else
$$
1-\frac{3}{H} \ge 0 \\
\implies G^3(1-3/H) \ge H^3(1-3/H) = H^3-3H^2
$$
and hence
$$
\begin{align}
8^3 &= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right)
\left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \\
&\ge (H^3-3H^2+3H-1)(H^3+3H^2+3H+1) \\
&= (H-1)^3(H+1)^3 \\
&= (H^2-1)^3 \\
\implies H &\le 3 \\
\implies 1 &\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{align}
$$
| {
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What should be the method for solving and solution of $\frac{x^2+y^2}{x^2-y^2}$? If $\frac{x+y}{x-y}=3$
then $\frac{x^2+y^2}{x^2-y^2} = ?$
| $\frac{x+y}{x-y}=3 \implies x=2y$
$\frac{x^2+y^2}{x^2-y^2} = \frac{5y^2}{3y^2}=\frac{5}{3}$
| {
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How to integrate : $\sqrt{\frac{a-x}{x-b}}$ Problem :
How to integrate : $\sqrt{\frac{a-x}{x-b}}$
Unable to find the substitution for this :
$\sqrt{\frac{a-x}{x-b}}$
Please help how to proceed ...........thanks..
| If $a=b,$ $$\frac{a-x}{x-b}=\frac{b-x}{x-b}=-1$$
But for real calculus $$\frac{a-x}{x-b}\text{ must be }>0$$
So, $a\ne b$
Putting $x=a\cos^2y+b\sin^2y$ so that $dx=2(b-a)\sin y\cos ydy$
and $$\frac{a-x}{x-b}=\frac{a-(a\cos^2y+b\sin^2y)}{(a\cos^2y+b\sin^2y)-b}=\frac{(a-b)\sin^2y}{(a-b)\cos^2y}=\tan^2y$$
$$\implies y=\arctan \sqrt{\frac{a-x}{x-b}}$$
$$\int\sqrt{\frac{a-x}{x-b}}dx=\int \tan y\cdot 2(b-a)\sin y\cos ydy$$
$$=(b-a)\int2\sin^2ydy=(b-a)\int(1-\cos2y)dy=(b-a)\left(y-\frac{\sin2y}2\right)+C$$
Now, $x=a\cos^2y+b\sin^2y\implies 2x=a(1+\cos2y)+b(1-\cos2y)$
$\implies \cos2y=\frac{2x-a-b}{a-b}$
$\implies \sin^22y=1-\left(\frac{2x-a-b}{a-b}\right)^2=\frac{(a-b)^2-(a+b-2x)^2}{(a-b)^2}=\frac{(a-b)^2-(a+b)^2-(2x)^2+4x(a+b)}{(a-b)^2}=\frac{4(a-x)(x-b)}{(a-b)^2}$
$\implies \sin2y=\frac{2\sqrt{(a-x)(x-b)}}{a-b} $
$x=a\cos^2y+b\sin^2y$ can be employed for $$\int\sqrt{(a-x)(x-b)}dx$$ and $$\int\frac1{\sqrt{(a-x)(x-b)}}dx$$ as well
| {
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"timestamp": "2023-03-29T00:00:00",
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what is$ \sqrt{8i}$ Very simple question with an answer that I cannot understand:
I have $\sqrt{8i}$, which, I suppose, is the same as $\sqrt{\sqrt{-64}}$.
How come that $2+2i$ is the same as $\sqrt{8i}$?
My understanding is that $\sqrt{8i}$ is the same as:
(a) $\sqrt{2^3i}$
OR
(b) $2\sqrt{2i}$
I'm pretty sure (a) is correct and (b) might also be correct, but how can you get from there to $2+2i$?
Thanks in advance
| $(a+bi)^2=(a+bi)(a+bi)=a^2-b^2+2abi\implies \sqrt{a^2-b^2+2abi}=a+bi$
and for conjugate $(a-bi)^2=(a-bi)(a-bi)=a^2+b^2-2abi\implies \sqrt{a^2+b^2-2abi}=a-bi$
Now if we find $a$ and $b$ satisfy appropriate equotation:
For $(a+bi)^2$:
$a^2-b^2+2abi=8i$. This is valid for any $a=b$ where $2ab=8$, or
$2a^2=8 \implies a=\pm\sqrt{4}=\pm{2} = b$. Here exists two numbers:
First $$\sqrt{8i}=2+2i$$ and second $$\sqrt{8i}=-2-2i$$
For $(a-bi)^2$:
$a^2+b^2-2abi=8i$. Because $a^2 \ge 0$ and $b^2 \ge 0$, there does not exists solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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What this sine function equation means? Apostol's book "Calculus" asks to prove that
$$\sin\frac{\pi }{6}=\frac{1}{2}$$
using the fact that
$$\sin 3x=3\sin x-4\sin^3 x$$
and
$$\sin \frac{\pi}{2}=1$$
So, we take $x=\frac{\pi}{6}$ and we have
$$\sin\frac{\pi}{2}=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
$$1=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
if we take $y=\sin \frac{\pi}{6}$
$$4y^3-3y+1=0$$
$$\left( {y- \frac{1}{2}} \right)^2(y+1)=0$$
and finally $y=\sin\frac{\pi}{6}=\frac{1}{2}$ or $y=\sin\frac{\pi}{6}=-1$. What it means?, it not should only be $\sin\frac{\pi}{6}=\frac{1}{2}$?
| We simply need to throw out the "solution" $y = \sin\left(\frac \pi 6\right) = -1\,$ because for $0 \lt \theta \lt \pi$, $0 \lt \sin\theta \lt 1$.
| {
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Partial fractions- Equating coefficients This is a partial fraction question. Kindly could you assist me with finding the values for A, B and C as I'm not very clear on how to go about it. The question is as follows;
$$\frac{3-x}{(x^2+3)(x+3)}$$
And we want A, B, C so that $$\frac{3-x}{(x^2+3)(x+3)} = \frac{Ax+B}{x^2 + 3} + \frac C{x+3}$$
I know that $A+B= 3$, $3B+C= -1$ and $3A+3C= 0$
The problem I'm facing here is with equating coefficients. I also know that after equating coefficients, the final values you get for A,B and C are as follows;
\begin{align}
A&= -1/2\\
B&= 1/2\\
C&= 1/2
\end{align}
But how do you solve with equating coefficients? Can you provide me step-by-step working for this please. I'm not facing any difficulty with partial fractions but just equating coefficients part in this case. Please help.
Many thanks.
| Step 1: Write $$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+3}$$
Step 2: Expand the RHS to get
$$\frac{Ax+B}{x^2+3}+\frac{C}{x+3}=\frac{(Ax+B)(x+3)+C(x^2+3)}{(x^2+3)(x+3)}=\frac{(A+C)x^2+(3A+B)x+3(B+C)}{(x^2+3)(x+3)}$$
Step 3: Compare the coefficients.
$$x^2:A+C=0,\quad x:(3A+B)=-1,\quad\text{constant}: 3(B+C)=3$$
Step 4: Solve simultaneous equations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all circles given two points and not the center This is probably pretty straight forward for you.
I have two points on a circle, $(-4, 7)$ and $(-5, 0)$. Given these two points and the radius $5$, what are all the possible equations?
My first idea was to solve the system
$\sqrt{(-4-a)^2 + (7-b)^2} = 5\tag 1$
$\sqrt{(-5-a)^2 + b^2} = 5\tag 2$
Where $(a, b)$ is the center point. However, I don't get the correct answers. Maybe this question should be about how to solve the system.
Thanks in advance!
Edit:
This is the formulation of the problem:
Find all circles through the points $(-4, 7)$ and $(-5, 0)$ with a radius of $5$.
| I'd personally start with the following 2 equations.
$(-4-a)^2 + (7-b)^2 = 25$ (1)
$(-5-a)^2 + b^2 = 25$ (2)
distribute out the squares.
$16 + 8a +a^2 + 49 - 14b + b^2 = 25$ (1)
$a^2 + b^2 + 8a - 14b + 40 = 0$ (1)
$25 + 10a + a^2 + b^2 = 25$ (2)
$a^2 + b^2 + 10a = 0$ now we set them equal
$a^2 + b^2 + 10a = a^2 + b^2 + 8a - 14b + 40$
$2a = -14b + 40$
$a = 20 -7b$ now we plug this into an earlier equation. I'm using equation 2
$(20 - 7b)^2 + b^2 + 10(20 - 7b) = 0$
$400 - 280b + 49b^2 + b^2 + 200 - 70b = 0$
$600 -350b + 50b^2 = 0$
Using the quadratic formula, b equals 4 or 3, so a = -8 or -1 respectively, giving us two possible centers of (-8,4) or (-1,3). This can be verified using the original equation.
| {
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General formula of fraction integral Consider the sequences of integrals:
$\int \frac{x}{1+x} dx, \int \frac{x}{1+x^2}dx, \int \frac{x}{1+x^3}dx,\ldots,\int \frac{x}{1+x^n}dx$.
Question: What is the general formula of $\int \frac{x}{1+x^n} dx$ for $n>0$?
| It's not a complete answer, but too long for a comment. It gives the answer for $n=4m$, $m$ is positive integer. Denote $I_n:=\int\frac{x}{1+x^n}\,dx$. If $x<0$ then substituting $x:=-z$ we can reduce the problem for $x>0$. So we calculate $I_{4m}:=\int\frac{x}{1+x^{4m}}\,dx$. Substitute $x:=\sqrt{z},\,z>0$. Then $I_{4m}=\frac{1}{2}\int\frac{1}{1+z^{2m}}\,dz$. The result of the integral is (Demidovich, Mathematical Analysis, problem 1925)
$$
-\frac{1}{2m}\sum_{k=1}^{m}\cos\frac{\pi(2k-1)}{2m}\ln\left(1-2x\cos\frac{2k-1}{2m}\pi+x^2 \right)$$
$$+\frac{1}{m}\sum_{k=1}^m\sin\frac{\pi(2k-1)}{2m}\arctan\frac{x-\cos\frac{2k-1}{2m}\pi}{\sin\frac{2k-1}{2m}\pi}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisors of $2^kp$ Let $p$ be an odd prime number. What is the necessary and sufficient condition (in terms of $p$ and $k$) such that we can partition the divisors of $2^kp$ into two set with equal sum.
| Let $N=2^kp$. The divisors of $N$ are $1,2,4,\dots, 2^k$ and $p,2p,4p,\dots, 2^kp$. Their sum simplifies to $(2^{k+1}-1)(p+1)$.
If we divide the divisors of $N$ into two sets, then $N$ will be in one of the sets. The sum of the rest of the divisors is $(2^{k+1}-1)(p+1)-2^kp$. This simplifies to
$$N +(2^{k+1}-1-p).$$
If the term $2^{k+1}-1-p$ is negative, then the divisors less than $N$ add up to less than $N$, so we cannot do the required splitting.
If $2^{k+1}-1-p$ is $\ge 0$, then we can do the splitting. In the case $2^{k+1}-1-p=0$, the number $N$ is a perfect number. The splitting consists of $N$ as the only member of one of the sets, and all the rest of the divisor of $N$ in the other set.
Now we verify that we can do the splitting also in the case $2^{k+1}-1-p\gt 0$.
If we put just $N$ into set $A$, and the rest of the divisors into set $B$, then set $A$ has sum $N$ and set $B$ has sum $N +(2^{k+1}-1-p)$. So we must "transfer" from $B$ to $A$ a collection of numbers with sum $\frac{2^{k+1}-1-p}{2}$.
This is easy to do. For the number $\frac{2^{k+1}-1-p}{2}$ is less than $2^{k+1}-1$. And any number $\le 2^{k+1}-1$ can be expressed as a sum of distinct powers of $2$ each $\le 2^k$. These powers of $2$ are among the divisors of $N$, so by shifting suitable divisors of $N$ from $B$ to $A$ we get the desired splitting.
There is even a mechanical method of doing it: just write down the binary representation of $\frac{2^{k+1}-1-p}{2}$.
Conclusion: The divisors of $2^kp$ can be split into two sets with equal sum if and only if $2^{k+1}-1-p\ge 0$, or equivalently if and only if $p\le 2^{k+1}-1$.
| {
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Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$
Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$
Doesn't this thing approaches 0 at the end? why does it approaches 1?
| HINT:
Using Exponent Combination Laws,
$$a^m\cdot a^n\cdot a^p\cdots=a^{m+n+p+\cdot},$$
$$\displaystyle 3^9\cdot 3^3\cdot3\cdot 3^\frac13\cdots=3^{\left(3^2+3+1+\frac13+\cdots\right)}$$
Observe that the power of $3$ is an infinite Geometric Series with the first Term $=9$ and common ratio $=\frac13<1$
| {
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How to go from $5\pi/4$ to $\pi + \pi/4$?
Use standard triangles to find exact value of $\cos(5\pi/4)$.
Example states that $5\pi/4$ is equal to $\pi + \pi/4$ but doesn't list the steps to get $\pi + \pi/4$...
| $\pi$ is just like any other number. So you can do $\frac{5 \pi}{4} = \frac{4 \pi}{4} + \frac{\pi}{4} = \pi + \frac{\pi}{4}$ just like you can do $\frac{5}{4} = \frac{4}{4} + \frac{1}{4} = 1 + \frac{1}{4}$.
| {
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Equation that produces negative one if x is less than zero and vice versa I have tried coming up with something but i couldn't in the end...
basically i want something to produce -1 or 1. given value of x will not be equal to zero at any point. But if you want to add that, do so in another example. I'd like to have both, but need first one.
so
if x=2
y=1
if x=-3435
y=-1
what i've tried:
well i thought i am gonna need a way to make any number a one regardless of sign right.
so..
x-((x*-1)) working on it :D
can't use square root, absolute value(unless there is a equation for deriving absolute value).
| If you want $0$ or $1$:
$$ \frac { x + | x | } { 2 x } $$
So if $ x = - 7 $
$$ \frac { - 7 + \sqrt { ( - 7 ) ^ 2 } } { 2 ( - 7 ) } = 0 $$
If $ x = 7 $
$$ \frac { 7 + \sqrt { 7 ^ 2 } } { 2 \times 7 } = 1 $$
If you want $ - 1 $ and $ 1 $:
$$ \frac x { | x | } $$
If $ x = - 7 $
$$ \frac { - 7 } { \sqrt { ( - 7 ) ^ 2 } } = - 1 $$
If x = 7
$$ \frac 7 { \sqrt { 7 ^ 2 } } = 1 $$
For $ x < 0 $ to become $ 0 $ and $ x > 0 $ remain $ x $:
$$ \frac { x + | x | } { 2 } $$
If $ x = - 7 $
$$ \frac { - 7 + \sqrt { ( - 7 ) ^ 2 } } { 2 } = 0 $$
If $ x = 7 $
$$ \frac { 7 + \sqrt { 7 ^ 2 } } { 2 } = 7 $$
For $ x < 0 $ to remain $ x $ and $ x > 0 $ become $ 0 $:
If $ x = - 7 $
$$ \frac { - 7 + \sqrt { ( - 7 ) ^ 2 } } { 2 } - \sqrt { ( - 7 ) ^ 2 } = -7 $$
If $ x = 7 $
$$ \frac { 7 + \sqrt { ( 7 ) ^ 2 } } { 2 } - \sqrt { ( 7 ) ^ 2 } = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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we can prove $b_{1}=b_{3}=0$? for any reanl numbers $t\in[-\sqrt{2},\sqrt{2}]$,then have
$$-\dfrac{1}{2}\le t^4+b_{3}t^3+b_{2}t^2+b_{1}t+b_{0}\le\dfrac{1}{2}$$
prove or disprove
$$b_{1}=b_{3}=0$$
my idea: I can only prove that
$$b_{1}+b_{3}=0$$
can't prove $b_{1}=b_{3}=0$
my prove $b_{1}+b_{3}=0$ methods:
let $$f(t)=t^4+b_{3}t^3+b_{2}t^2+b_{1}t+b_{0}$$
and let $$g(t)=\dfrac{f(t)+f(-t)}{2}=t^4+b_{2}t^2+b_{0}$$
so we easy to have
$$g(\sqrt{2})+g(0)-2g(1)=2$$
and $$|g(t)|\le\dfrac{1}{2}$$
then we must have
$$g(\sqrt{2})=\dfrac{1}{2},g(0)=\dfrac{1}{2},g(1)=-\dfrac{1}{2}$$
then
$$4+2b_{2}+b_{0}=\dfrac{1}{2},b_{0}=\dfrac{1}{2},1+b_{2}+b_{0}=-\dfrac{1}{2}$$
then
$$b_{0}=\dfrac{1}{2},b_{2}=-2$$
and
$$|f(1)|\le\dfrac{1}{2}\Longrightarrow |1+b_{3}+b_{2}+b_{1}+b_{0}|\le\dfrac{1}{2}\Longrightarrow 0\le b_{1}+b_{3}\le 1$$
$$|f(-1)|\le\dfrac{1}{2}\Longrightarrow |1-b_{3}+b_{2}-b_{1}+b_{0}|\le\dfrac{1}{2}\Longrightarrow -1 \le b_{1}+b_{3}\le 0$$
so $$b_{1}+b_{3}=0$$
But I Guess we following can prove $$b_{1}=b_{3}=0?$$
if ture? How prove it
if not true,why not?
Thank you everyone
| your proof seems questionable.
Since $g^{\prime}(t)=t^3+2b_2t$ the monotonicity of $g(t)$ will depend on the value of $b_2$.
Why do you assert that
$$g(\sqrt{2})=\dfrac{1}{2}\qquad g(0)=\dfrac{1}{2} g(1)=-\dfrac{1}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/463412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Questions involving two polynomials Let $$P(x)=x^{6}-x^{5}-x^{3}-x^{2}-x$$
$$Q(x)=x^{4}-x^{3}-x^{2}-1$$
Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the roots of Q(x)
Prove that $$P\left (z_{1} \right )+ P\left (z_{2} \right )+ P\left (z_{3} \right )+ P\left (z_{4} \right )=6$$
| HINT:
So, $$z_1^4-z_1^3-z_1^2-1=0\ \ \ \ \ (1)$$
$$P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1$$
$$=z_1^2(z_1^4-z_1^3-z_1^2-1)+z_1^4-z_1^3-z_1=z_1^2-z_1+1$$
Let $z_1^2-z_1+1=y$
$$z_1^4-z_1^3-z_1^2-1=z_1^2(z_1^2-z_1+1)-2z_1^2-1=z_1^2y-2(y-1+z_1)-1$$
$$=(y+z_1-1)y-2(y-1+z_1)-1$$
$$\implies (y+z_1-1)y-2(y-1+z_1)-1=0$$
Express $z_1$ in terms of $y$ and put the value in $(1)$
Then use Vieta's formulas 1,2,3
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \implies a < \frac{2}{a}\\
\text{Let}\hspace{1cm} B=\frac{a+\frac{2}{a}}{2}=\frac{a^2+2}{2a}
$$
$B$ is greater than $a$ because:
$$ \begin{array} {aa} B>a & \implies \frac{a^2+2}{2a}>a \\
& \implies a^2 + 2 > 2a^2 \\
& \implies 2 > a^2 \\
& \implies a^2 < 2
\end{array}$$
If $B^2$ is less than $2$, then $B^2-2<0$, but:
$$\begin{array} {aa} B^2-2 < 0 & \implies \left( \frac{a^2+2}{2a} \right)^2 - 2 < 0 \\
& \implies \frac{a^4+4a^2+4}{4a} - \frac{8a^2}{4a^2} < 0 \\
& \implies \frac{(a^2-2)^2}{(2a)^2} < 0
\end{array}$$
Which is a contradiction since the left hand side of the inequality will be positive for all values of $a$.
But I think we must be able to find such a $B$ since based on my understanding of this answer, we can find a another rational number whose distance from $a$ is less than the distance between $a$ and $\sqrt{2}$
Therefore, I have 2 questions to ask:
*
*Why does this approach work in the case of $a^2>2$ but not when $a^2<2$?
*How should I approach these kind of questions since it seems that there are a few ways to construct a $B$ that satisfies a given set of restrictions? For example, see here (the proof is immediately before the section "13. The Completeness Axiom".
| The trouble with taking the arithmetic mean of $a$ and $\dfrac 2 a$ is, as you observe, that it gives a value $> \sqrt 2$. Try taking the harmonic mean instead.
Btw, some of your logic chains are reversed: you have right arrows where you need left arrows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find a function $f$ and a number $a$ such that $6+\int_{a}^{x}\frac{f(t)}{t^2}\:\mathrm{d}t=2\sqrt{x}$ For all $x>0$
Find a function $f$ and a number $a$ such that:
$$
6+\int_{a}^{x}\frac{f(t)}{t^2}\:\mathrm{d}t=2\sqrt{x}
$$
For all $x>0$
From Fundamental Theorem of Calculus section. Having some trouble with this. Any help?
| $$ \int_a^x \frac{f(t)}{t^2} \ dt = 2\sqrt{x} - 6 $$
$$\mathrm{ differentiate \ using \ leibniz \ rule } $$
$$\frac{f(x)}{x^2} = \frac{1}{\sqrt{x} } \Rightarrow f(x) = x\sqrt{x}$$
$$\int_a^x \frac{\sqrt{t}}{t} \ dt = 2\sqrt{x} - 6 $$
$$\int_a^x t^{-\frac{1}{2}} \ dt = \left | 2\sqrt{t} \right|_a^x = 2\sqrt{x} - 2\sqrt{a} = 2\sqrt{x} - 6$$
$$ \Rightarrow 2\sqrt{a} = 6 \Rightarrow a = 9 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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