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Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$ Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + \frac{1}{6^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$
I am just clueless. I just have some random thoughts. I can find the sum of this series and then put in the value $1000000$ and then find the floor of that. To do this, I would have to telescope this series which seems impossible to me. But is there any other way to directly find the floor without finding the general sum?
I am very new to calculus, so please provide hints and answers that dio not involve calculus.
| Since
$(x^a)' = ax^{a-1}$,
$a\int_n^{n+1} x^{a-1}dx
=(n+1)^a-n^a
$.
Therefore
$\begin{array}\\
v^a-u^a
&=\sum_{n=u}^{v-1}((n+1)^a-n^a)\\
&=\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx\\
\end{array}
$
If $a > 1$,
since $x^{a-1}$
is increasing,
$n^{a-1}
\lt \int_n^{n+1} x^{a-1}dx
\lt (n+1)^{a-1}
$.
If $a < 1$,
since $x^{a-1}$
is decreasing,
$n^{a-1}
\gt \int_n^{n+1} x^{a-1}dx
\gt (n+1)^{a-1}
$.
In this case,
$a-1 = -1/3 < 0$.
Therefore
$v^a-u^a
=\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx
\lt \sum_{n=u}^{v-1}an^{a-1}
$
so
$\sum_{n=u}^{v-1}n^{a-1}
\gt \frac1{a}(v^a-u^a)
$.
Similarly,
$v^a-u^a
=\sum_{n=u}^{v-1}a\int_n^{n+1} x^{a-1}dx
\gt \sum_{n=u}^{v-1}a(n+1)^{a-1}
= a\sum_{n=u+1}^{v}n^{a-1}
= a\sum_{n=u}^{v-1}n^{a-1}-a(u^{a-1}-v^{a-1})
$
so
$\sum_{n=u}^{v-1}n^{a-1}
\lt \frac1{a}((v-1)^a-u^a)+(u^{a-1}-v^{a-1})
$.
The reverse inequalities hold
if $a > 1$.
In this case,
$a-1=-1/3,
a = 2/3,
v=10^6+1,
u=4$,
so,
if
$s = \sum_{n=u}^{v-1}n^{a-1}$,
then
$s
\gt \frac1{2/3}((10^6)^{2/3}-4^{2/3})
\gt \frac32(10^4-2.519...)
\approx 14996.220
$
and
$s
\lt \frac1{2/3}((10^6)^{2/3}-4^{2/3})+(4^{-1/3}-(10^6)^{-1/3})
\approx 14996.220+0.619
= 14996.839
$
so
$\lfloor s \rfloor
=14996
$.
If the lower and upper bounds
surrounded an integer,
I would manually compute
the first few terms
until the bounds
for the remaining terms
are close enough
that the floor is determined.
| {
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"url": "https://math.stackexchange.com/questions/2911108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$
Find the value of
$$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$
My Try: Using Stolz-Cesaro,
Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n}$
and $b_{n} = \ln(n)$
So $\displaystyle \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = \lim_{n\rightarrow \infty}\frac{1}{{(n+1)^3}}\cdot \frac{1}{\ln\bigg(1+\frac{1}{n}\bigg)} = 0$
Please explain if what I have done above is right.
| Let $H_n=\sum_{k=1}^n \frac 1k$. Since $H_n=\log n +O(1)$, $$H_{n^3} = 3\log n +O(1)$$
thus$$\frac{H_{n^3}}{\log n} = 3+O\left(\frac{1}{\log n} \right)=3+o(1)$$
The limit is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum
$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$$
I have tried to make this series in the form $1 + n x + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (n+2)x^3}{3!} + \cdots$. But I failed to do so. Can anyone please help me
| Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
If $\displaystyle S=\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$
$$2S=\frac {2}{6} + \frac {2\cdot5}{6.12} + \frac {2\cdot5. 8}{6.12.18} + \frac {2\cdot5.8.11}{6.12.18.24} + \cdots$$
$$2S+1+\dfrac13$$
$$=1+\dfrac{-\dfrac23}{1!}\left(-\dfrac36\right)+\dfrac{-\dfrac23\left(-\dfrac23-1\right)}{2!}\left(-\dfrac36\right)^2+\dfrac{-\dfrac23\left(-\dfrac23-1\right)\left(-\dfrac23-2\right)}{3!}\left(-\dfrac36\right)^3+ \cdots$$
$$=\left(1-\dfrac12\right)^{-2/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913065",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)^{n}$
So this takes care of the first terms, but then I just need to show
$\sum_{n = 3}^{\infty} (1 - x)^{n} = (1 - x)^{3}/x.$
Or, maybe I'm approaching this completely incorrectly.
| Notice that the infinite geometric sum is given as following $$a+aq+aq^2+aq^3+\cdots=\dfrac{a}{1-q}$$therefore $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=\dfrac{1}{1-(1-x)}={1\over x}$$and $$(1-x)^3+(1-x)^4+(1-x)^5+\cdots=\dfrac{(1-x)^3}{x}$$so we have $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=1+(1-x)+(1-x)^2+(1-x)^3+(1-x)^4+(1-x)^5+\cdots=1+(1-x)+(1-x)^2+\dfrac{(1-x)^3}{x}=\dfrac{1}{x}$$for $0<x<2$
Alternative way: use the following equality $$a^3-b^3=(a^2+ab+b^2)(a-b)$$with $a=1$ and $b=1-x$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| I admire your efforts, another approach is
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\left(\frac12n-\frac14+ \dfrac{\frac34n - \frac{13}{4}}{2n^2+n-1} \right) =\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.
Base case is easy, n=1.
$$
\frac{1}{2^1}<2
$$
Induction case we assume that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2
$$
Then we get to fun old induction. How do I show that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ?
$$
| Hint:
It's a sum of an arithmetico-geometric progression:
$$S_{n}={\frac {a-(a+nd)\,r^{n}}{1-r}}+{\frac {dr\,(1-r^{n})}{(1-r)^{2}}}$$
Here $d$ is the common difference, $r$ is the common ratio, $n$ denotes the number of terms and $a$ is the first term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Closed form solution for logarithmic inequality We have the following inequality that was very hard to be solved in a closed form. Yet, someone solved it this way, and I can't get to fully understand what is done.
eq: $\sqrt{x-2\sqrt{x-1}} + \sqrt{x+2\sqrt{x-1}} +log_2(x-1)=0$
Solution:
$x\in Df \iff x \geq 1$
We have: $|\sqrt{x-1} -1|+|\sqrt{x-1} +1|+log_2(x-1)=0$
$|\sqrt{x-1} -1|= \{\sqrt{x-1} -1 \ if\ \sqrt{x-1} -1 \geq0\ and\ \sqrt{x-1} -1 \geq0 \Rightarrow x\geq2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \{\ 1 - \sqrt{x-1} \ if\ \sqrt{x-1} -1 \leq0\ and\ \sqrt{x-1} -1 \leq0 \Rightarrow 1\leq x\leq2$
$|\sqrt{x-1} +1|= \{\sqrt{x-1} +1 \ if\ \sqrt{x-1} +1 \geq0\ and\ \sqrt{x-1} +1 \geq0,\ \forall x\geq1$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \{\ -\sqrt{x-1}-1 \ if\ \sqrt{x-1} +1 \leq0\ and\ \sqrt{x-1} +1 \leq0 \Rightarrow x\in \emptyset$
Then, the solution moves to have two intervals for study, (1,2) and (2, +oo[, for which we find the solution to be x = 5/4 on the interval (1,2) and no solution in the second interval. My question is what is the logic of the use of absolute value and replacing x with 1 as you can notice?
| The logic of replacing the square root with an absolute value is the formula $$ \sqrt {x^2} = |x|$$ which is correct.
The problem is that there is a mistake in the solution because $ |\sqrt{x-1} -1|$ is substituted for $ \sqrt{x-\sqrt{x-1}}$ which is not correct.
Note that $$|\sqrt{x-1} -1| =\sqrt {(\sqrt{x-1} -1)^2}= \sqrt{x-2\sqrt{x-1}}$$
Similarly $$|\sqrt{x-1} +1| =\sqrt {(\sqrt{x-1} +1)^2}= \sqrt{x+2\sqrt{x-1}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$ How can I prove this equality:
$$
\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}= \frac{-1}{n(n+1)(n+1)!}
$$
| We have that
$$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}=\color{red}1\cdot \frac{1}{(n+1)!} + \color{red}{\frac{1}{n+1}}\cdot \frac{1}{(n+1)!} - \color{red}{\frac{n+1}{n}}\cdot\frac{1}{(n+1)!}=$$
$$=\left( \color{red}{ 1+ \frac{1}{n+1} - \frac{n+1}{n} }\right)\cdot\frac{1}{(n+1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916693",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$
Calculate $$\int\limits_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}}dx$$
First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t-2}+t^2}}dt$$ and use the fact that $$2(\sqrt{e^{t-2}+t^2})'=\frac{1}{\sqrt{e^{t-2}+t^2}} \cdot(e^{t-2}+2t)$$ Using integration by parts we get that we have to calculate (excluding some terms we know) $$\int\limits_{0}^{2} \sqrt{e^{t-2}+t^2}\cdot \frac{6e^{t-2}-2te^{t-2}+8}{(e^{t-2}+2t)^2}dt$$ which is uglier then the initial problem. Do you have any idea how to solve the problem?
| Hint: Let $y = (x+2)e^{-x/2}.$ You should get $I=-2\sinh^{-1}{2}$.
$\displaystyle y = (x+2)e^{-x/2} \implies dy = -\frac{x}{2}e^{-x/2}\,dx$
and $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$
Therefore we have $\displaystyle I = \int_{-2}^{0}\frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = -2\int_{0}^{2}\frac{dy}{\sqrt{1+y^2}} = -2\sinh^{-1}(2)$, as claimed.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
Here's a picture, since I'm not good with MathJax.
| Multiplying by $$x\ne 0$$ we get
$$gy+2y^2=\frac{gx}{y}+\frac{2x^2}{y^2}$$
Now you can multiply by $$y^2$$
We get
$$gy^3+2y^4=gxy+2x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sort those 3 logarithmic values without using calculator I found this problem interesting, namely we are given three values: $$\log_{1/3}{27}, \log_{1/5}{4}, \log_{1/2}{5}$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $\log_{1/3}{27} = \log_{1/3}{3^3} = \log_{1/3}{(\frac{1}{3})^{-3}} = -3\\ \log_{1/5}{4} = \log_{1/5}{2^2} = 2 \cdot \log_{1/5}{2} = 2 \cdot \log_{5^{-1}}{2} = -2 \cdot \log_{5}{2}$
Since $\log_{5}{2} < 1, \text{then: } -2\cdot \log_{5}{2} > -2 $
And the third value: $\log_{1/2}{5} = \log_{2^{-1}}{5} = -\log_{2}{5}$
Since $2 < \log_{2}{5} < 3, \text{then: } -2 > -\log_{2}{5} > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$\log_{1/3}{27}, \log_{1/2}{5}, \log_{1/5}{4}$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
| As an alternative and to check recall that
$$\log_b a=\frac{\log_c a}{\log_c b}$$
that is
$$\log_{\frac13}27=\frac{\log_2 27}{\log_2\frac13}=-\frac{\log_2 3^3}{\log_2 3}=-3$$
$$\log_{\frac15}4=\frac{\log_2 4}{\log_2 \frac15}=-\frac{\log_2 2^2}{\log_2 5}=-\frac{2}{\log_2 5}$$
$$\log_{\frac12}5=\frac{\log_2 5}{\log_2 \frac12}=-\frac{\log 5}{\log_2 2}=-\log_2 5$$
therefore as you stated
$$\log_{\frac13}27<\log_{\frac12}5<\log_{\frac15}4$$
| {
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"url": "https://math.stackexchange.com/questions/2921146",
"timestamp": "2023-03-29T00:00:00",
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Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
| Just to be different (and to rip off Mark Bennett's answer):
$ab = k> 0$ with $0< a \le b$ means $a \le \sqrt {k}$ and $b \ge \sqrt {k}$. (and if $0< a < b$ then $a < \sqrt k$ and $b > \sqrt{k}$)
So $(n+3)(n+4) = 90$ means $n+3 \le \sqrt{90} \approx 9.smallchange \le n+4$
As $n+3$ and $n+4$ are consecutive we must have:
$(n+3) \le \sqrt{90} < (n+3) + 1$.
Or in other words $n + 3 = \lfloor \sqrt{90} \rfloor = 9$.
So $n = 6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Induction proof of a series Suppose we have the series
$$f(n) = \sum_{i=1}^n \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}.$$
As a hint it was said that this series "telescopes". I observed the pattern to be $f(n)=\frac{n^2}{n^2 +1}$. I wish to prove this via induction. The base case holds, since $f(1)=\frac{1}{2}$ which comes out of both the definition of $f(n)$ and my proposed formula. Now I wish to prove that for some arbitrary $k$ we assume $f(k)=\frac{k^2}{k^2+1}$, then I wish to prove that $f(k+1)=\frac{(k+1)^2}{(k+1)^2+1} (=\frac{k^2 +2k +1}{k^2 +2k +2})$.
I wanted to start out by noticing:
$$f(k+1) = \sum_{i=1}^{k+1} \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}= \\f (k)+ \frac{2(k+1)-1}{(k+1)^4 - 2 (k+1)^3 + 3 (k+1)^2 - 2 (k+1) + 2}$$
This eventually simplifies to
$$ f(n+1)= \frac{k^2}{k^2 +1} + \frac{2k+1}{k^4 +2k^3+3k^2+2k+2}$$
| Note that$$\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}=\frac{2i-1}{i_4-2i^3+3i^2-2i+2}$$and that therefore\begin{align}\sum_{i=1}^n\frac{2i-1}{i^4-2i^3+3i^2-2i+2}&=\sum_{i=1}^n\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}\\&=\frac{n^2}{n^2+1}-\frac{0^2}{0^2+1}\\&=\frac{n^2}{n^2+1}.\end{align}
| {
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Determine the domain of the inverse of $f(x) = 2x^2 + 8x - 7$, which is a function. Determine the domain where the inverse of $f(x) = 2x^2 + 8x - 7$ is a function.
*
*So, I started off switching the $y$- (the $f(x)$) and $x$-value, like so:
$x = 2y^2 + 8y - 7$
*Then, I plugged these values into the quadratic formula, getting:
$f^{-1}(x) = (-8 \pm \sqrt{120})/4$
First off, is the inverse written correctly (this problem is different from the others I worked on)? And second, since I have a ($\pm$) sign, does that mean there will be two different domains?
| Let $a>0$ and $f(x)=ax^2+bx+c$. There is
$$
\varphi: \left[\frac{-\Delta}{4a},+\infty)\right) \to\left(-\infty, -\frac{b}{2a}\right]
\qquad \mbox{and} \qquad
\psi:\left[ \frac{-\Delta}{4a},+\infty)\right) \to \left[ -\frac{b}{2a}, +\infty,\right)
$$
such that
$$
\begin{matrix}
\varphi ( f(x))=x \qquad f(\varphi(y))=y\\
\psi(f(x))=x \qquad f(\psi(y))=y
\end{matrix}
$$
Here $\Delta=b^2-4ac$. Let $y=ax^2+bx+c$. Check for yourself
$$
y=\left[a\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a} \right]
$$
Then
$$
\left( x+\frac{b}{2a}\right)^2=\frac{1}{a}\left( y+\frac{\Delta}{4a}\right)
$$
and
$$
x=-\frac{b}{2a} \pm\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
Finally we have
$$
\varphi(y)=-\frac{b}{2a}-\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
and
$$
\psi(y)=-\frac{b}{2a}+\sqrt{\frac{1}{a}\left(y+\frac{\Delta}{4a}\right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove that all $2 \times 2$ orthogonal matrices can be expressed as rotation or reflection Let A be some $2 \times 2$ matrix with real entries. Prove that $A^T$$A$ = $I$ if and only if $A$ is the rotation matrix or the reflection matrix.
My Progress: It can be shown that if $A$ is either the rotation or reflection matrix, then $A^T A = I$ holds by matrix multiplication. Where I get suck is showing that if $A$ is a $2 \times 2$ orthogonal matrix, then $A$ must either be the Rotation or Reflection Matrix. I suppose that since $A$ is orthogonal, it is distance preserving - and the only $2 \times 2$ matrices that preserve distance are the Rotation and Reflection Matrices, but this isn't really a proof.
| With not so many variables running around, we may verify this claim algebraically. Write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $A^TA = \begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}$. So $\begin{pmatrix} a \\ c\end{pmatrix}$ and $\begin{pmatrix} b \\ d\end{pmatrix}$ are unit vectors so that $\begin{pmatrix} a \\ c\end{pmatrix}\cdot \begin{pmatrix} b \\ d\end{pmatrix} = 0$. We can check that this implies $b = \pm c$ while $d = \mp a$. So $A = \begin{pmatrix} a & c \\ c & -a \end{pmatrix}$ or $A = \begin{pmatrix} a & -c \\ c & a \end{pmatrix}$, which are the forms of a reflection composed with a rotation and a rotation, respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by inspection)?
I am currently using the addition of cosine formula, but it is too slow. I'm thinking perhaps a graphic method can be used?
| Use the fact that when angles correspond to equally spaced points around a circle, their sines add up to zero and their cosines do the same. Thus
$\cos(\frac{\pi}{4})+\cos(\frac{\pi}{4}+\frac{2\pi}{3})+\cos(\frac{\pi}{4}+\frac{4\pi}{3})=0$
Put $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, then:
$\cos(\frac{\pi}{4}+\frac{2\pi}{3})=a-\frac{\sqrt{2}}{4}$
$\cos(\frac{\pi}{4}+\frac{4\pi}{3})=-a-\frac{\sqrt{2}}{4}$
To find $a$, simply subtract and use the appropriate sum-product relation:
$2a=\cos(\frac{\pi}{4}+\frac{2\pi}{3})-\cos(\frac{\pi}{4}+\frac{4\pi}{3})=2\sin(\frac{(\frac{\pi}{4}+\frac{2\pi}{3})+(\frac{\pi}{4}+\frac{4\pi}{3})}{2})\sin(\frac{(\frac{\pi}{4}+\frac{4\pi}{3})-(\frac{\pi}{4}+\frac{2\pi}{3})}{2})=2\sin(\frac{5\pi}{4})\sin(\frac{\pi}{3})$
Then $\sin(\frac{5\pi}{4})=(-\sqrt{2})/2, \sin(\frac{\pi}{3})=(\sqrt{3})/2$ and we get $a=(-\sqrt{6})/4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Volume of $x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1$ Find the volume of the solid defined by $$x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1.$$
I found the intersection of surfaces:
$$S\equiv\begin{cases}
x^2+y^2&=4\\
2+x^2+y^2&=z\\
z&=-1
\end{cases}
\equiv
\begin{cases}
x^2+y^2&=4\\
2+4&\neq-1\\
z&=-1.
\end{cases}$$
The surfaces are not cut, and therefore do not define a volume in space. Is it right?
Thanks!
| FALSE. Not all three have to intersect; notice that the surfaces $x^2+y^2=4$ and $z=x^2+y^2+2$ intersect, and the surfaces $x^2+y^2=4$ and $z= -1$ intersect, forming a cylinder-like volume with a circular base and a concave paraboloid-shaped "lid."
You may calculate the volume by converting to polar coordinates and considering the integral
$$\int_0^{2\pi} \int_{0}^{2} (r^2+3)\cdot r\space dr\space d\theta$$
| {
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"url": "https://math.stackexchange.com/questions/2925946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implication of equations Just an interesting question I saw online. :)
Is the following statement true or false?
$$\large \dfrac{ab}{c^2} + \dfrac{bc}{a^2} + \dfrac{ca}{b^2} = 3 \iff \dfrac{a + b}{c} + \dfrac{b + c}{a} + \dfrac{c + a}{b} \in \{-3; 6\}$$
$a + b + c \ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
| Notice that the equality $\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0\leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 \hspace{0.3cm}(1)$ or $a=b=c \hspace{0.3cm}(2)$.
*
*Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
\begin{align}
&\frac{a+b}{c}=-\frac{ab}{c^2}\\
&\frac{c+b}{a}=-\frac{bc}{a^2}\\
&\frac{a+c}{b}=-\frac{ac}{b^2}
\end{align}
$$
and we finally find that
$$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=-\frac{ab}{c^2}
-\frac{bc}{a^2}
-\frac{ac}{b^2}=-3$$
by the original equation.
*
*Case(2):
when $a=b=c$ it's easy to see that
$$\frac{a+b}{c}+\frac{c+b}{a}+\frac{a+c}{b}=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{\dfrac{-5x+1}{x}}{\dfrac{\sqrt{x^2-5x+1}+x}{x}}$$
$$\lim_{x\to\infty} \dfrac{-5+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{5}{x}+\dfrac{1}{x^2}}+1}$$
From here, I know that $\lim_{x\to\infty} \dfrac{1}{x} = 0$, $\lim_{x\to\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to\infty} \dfrac{1}{x^2} = 0$
$$\lim_{x\to\infty} \dfrac{-5}{\sqrt{1}+1}$$
$$\lim_{x\to\infty} \dfrac{-5}{2}$$
$$\lim_{x\to\infty} f(x) = \dfrac{-5}{2}$$
Everything up to here seems fine. The issue is when I try to find $\lim_{x\to-\infty} f(x)$
I also know that $\lim_{x\to-\infty} \dfrac{1}{x} = 0$, $\lim_{x\to-\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to-\infty} \dfrac{1}{x^2} = 0$
This would make me conclude that $\lim_{x\to-\infty}f(x) = \dfrac{-5}{2}$.
However, this is not the case because $\lim_{x\to-\infty}f(x) = \infty$
Desmos view of $f(x)$
Why am I arriving to the wrong answer and how can I algebraically prove that the answer is $\infty$?
| as far as understanding, you have, in essence,
$$ \left| x - \frac{5}{2} \right| - x . $$
for large positive $x$ you get close to $-5/2,$ but for large negative $x,$ meaning $x$ is negative and $|x|$ is large, you have roughly
$$ 2 |x| + \frac{5}{2} $$
which grows without bound. For example, if $x = -10,$ the original expression becomes
$$ \sqrt{100 - (-50) + 1} - (-10) = \sqrt {151} + 10 \approx 22.2882 $$
If $x = -100,$
$$ \sqrt{10000 - (-500) + 1} - (-100) = \sqrt {10501} + 100 \approx 202.474 $$
If $x = -1000,$
$$ \sqrt{1000000 - (-5000) + 1} - (-1000) = \sqrt {1005001} + 1000 \approx 2002.49738 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Random walk Catalan sum Starting at $0$ on the number line, you go right $1$ unit with probability $p$ and left $1$ unit with probability $1-p$. What's the probability of ever getting to $n>0$, and how many steps are expected?
Let $q_k$ be the probability of ever getting to $k\ge 0$. $q_k=q_1^k$ since you ever get to $1$ with probability $q_1$, and the first time you get to $1$ the probability becomes $q_{k-1}$. $$q_1=pq_0+(1-p)q_2=p+(1-p)q_1^2$$ so $$q_1\in\left\{1,\dfrac{p}{1-p}\right\}$$ If $p\ge\tfrac{1}{2}$, then $\tfrac{p}{1-p}\ge 1$, so $q_1=1$, so $q_n=1$. If $p<\tfrac{1}{2}$, I don't know which root to pick.
Let $x_k$ be the expected number of steps to get to $k\ge 0$. By similar reasoning, $x_k=kx_1$. $$x_1=1+px_0+(1-p)x_2=1+(1-p)2x_1$$ so $x_1=\tfrac{1}{2p-1}$. If $p\le\tfrac{1}{2}$, then $$1=(2p-1)x_1\le 0$$ so the expected number of steps is infinite. If $p>\tfrac{1}{2}$, $x_n=\tfrac{n}{2p-1}$.
We can also sum the probability of getting to $1$ in exactly $2k+1$ steps using Catalan numbers to get $q_1$, and sum the probability times $2k+1$ to get $x_1$. The last step must be right and the remaining $2k$ steps can be rearranged in $C_k=\tfrac{1}{k+1}\tbinom{2k}{k}$ ways, so $$q_1=\sum_{k=0}^\infty\dfrac{1}{k+1}\dbinom{2k}{k}p^{k+1}(1-p)^k$$ and $$x_1=\sum_{k=0}^\infty\dfrac{2k+1}{k+1}\dbinom{2k}{k}p^{k+1}(1-p)^k=\sum_{k=0}^\infty\dbinom{2k+1}{k}p^{k+1}(1-p)^k$$ which I have no ideas to compute.
How do you compute the sums? Can you show that $q_1<1$ if $p<\tfrac{1}{2}$ without computing the sum? Are there other ways to answer the original question?
| We consider the case $p<\frac{1}{2}$.
We obtain
\begin{align*}
\color{blue}{q_1(p)}&=\sum_{k=0}^\infty\frac{1}{k+1}\binom{2k}{2k}p^{k+1}(1-p)^k\\
&=p\sum_{k=0}^\infty\frac{1}{k+1}\binom{2k}{k}\left(p(1-p)\right)^k\\
&=p\cdot\frac{1-\sqrt{1-4p(1-p)}}{2p(1-p)}\tag{1}\\
&=p\cdot\frac{1-(1-2p)}{2(1-p)}\\
&\,\,\color{blue}{=\frac{p}{1-p}}
\end{align*}
In (1) we use the generating function of the Catalan numbers.
We obtain
\begin{align*}
\color{blue}{x_1(p)}&=\sum_{k=0}^\infty\binom{2k+1}{k}p^{k+1}(1-p)^k\\
&=p\sum_{k=0}^\infty\binom{2k+1}{k}\left(p(1-p)\right)^k\\
&=p\cdot\frac{1}{\sqrt{1-4p(1-p)}}\left(\frac{1-\sqrt{1-4p(1-p)}}{2p(1-p)}\right)\tag{2}\\
&=p\cdot\frac{1}{1-2p}\cdot\frac{1-(1-2p)}{2p(1-p)}\\
&\,\,\color{blue}{=\frac{p}{(1-2p)(1-p)}}
\end{align*}
In (2) we use the generating function of the shifted central binomial coefficients. See (2.5.15) in H. Wilf's Generatingfunctionology.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
| You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Generating function: Collecting dollars from children and adults. Here is the problem:
In how many ways can I collect a total of $20$ dollars from $4$ different children and $3$ different adults, if each child can contribute up to $6$ dollars, each adult can give up to $10$ dollars, and each individual gives a nonnegative whole number of dollars?
What I have tried:
So the generating function for the children is $(1+x+x^2+\dots+x^6)$ and for the adults $(1+x+x^2+\dots+x^{10})$. So the total generating function is $(1+x+x^2+\dots+x^6)^4(1+x+x^2+\dots+x^{10})^3$. This can be re-written as $\frac{(1-x^6)^4(1-x^{10})^3}{(1-x)^7}$. This is equivalent to $\frac{1-4x^6-3x^{10}+6x^{12}+12x^{16}-4x^{18}+3x^{20}}{(1-x)^7}$(We ignore all terms in the numerator with denominator greater than 20). This means the final answer is $\binom{26}{6}-4\binom{20}{6}-3\binom{16}{6}+6\binom{14}{6}+12\binom{10}{6}-4\binom{8}{6}+3\binom{6}{6}$, which is equivalent to $71595$.
But sadly, this is not the right answer. Where am I wrong, and how can I fix the part where I messed up? ONLY hints please!
Thanks,
Max0815
| You have made two mistakes in summing the geometric series.
$$1 + x + x^2 + \dots + x^6 = \frac{1-x^\color{red}{7}}{1-x}$$
and
$$1 + x + x^2 + \dots + x^{10} = \frac{1-x^\color{red}{11}}{1-x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Both of $\sin(x),\cos(x)$ must be nonnegative since, if one of them was negative, the equation $\sin^3(x)+\cos^3(x)=1$ would imply that the other one is more than $1$, contradiction.
Thus, we have $0\le \sin(x) \le 1$ and $0\le \cos(x)\le 1$.
If both of $\sin(x),\cos(x)$ are less than $1$, then, since they are both nonnegative, we would have
$$
\begin{cases}
0\le \sin^3(x) < \sin^2(x)\\[4pt]
0\le \cos^3(x) < \cos^2(x)\\
\end{cases}
$$
but that would imply
$$\sin^3(x)+\cos^3(x) < \sin^2(x)+\cos^2(x)=1$$
contradiction.
It follows that one of $\sin(x),\cos(x)$ must be equal to $1$, and the other must be equal to zero.
From that information, I'm sure you can finish the solution.
| {
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Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Since,
$$2^{12} \equiv 1 \pmod {13}\Rightarrow (2^{12})^5\equiv 1 \pmod {13}\Rightarrow2^{60} \equiv 1 \pmod {13}$$Again, $$2^4 \equiv 3 \pmod {13}\Rightarrow2^8.2^2 \equiv 10\pmod {13}$$Using both result: $2^{70} \equiv 10\pmod {13}$I failed again to show that. Any hints or solution will be appreciated.Thanks in advance.
| In (mod 13): $ 3^{70} = 9^{35} = (-4)^{35} = -(4^{35}) = -(2^{70}) $
So that: $ 2^{70} + 3^{70} = 0 (mod 13) $
The main idea is: $ 3^2 = (-1)2^2 $ (mod 13)
For all odd n: $ 13 $ $|$ $2^{2n} + 3^{2n} $
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website
| There are many ways, and a natural temptation is to exploit the fact that $g(z)=\frac{1-z}{1+z}$ is an involution, $g(g(z))=z$. In particular
$$ \int\sqrt{\frac{1+x}{1-x}}\,dx = \int \frac{dx}{\sqrt{g(x)}}\stackrel{x\mapsto g(z)}{=}\int\frac{g'(z)\,dz}{\sqrt{z}}\stackrel{z\mapsto u^2}{=}2\int g'(u^2)\,du =4\int\frac{du}{(1+u^2)^2}$$
and the last integral is elementary. By partial fraction decomposition or IBP
$$ \int\frac{du}{(1+u^2)^2}=C+\frac{1}{2}\left(\frac{u}{1+u^2}+\arctan u\right) $$
and now it is enough to substitute $u=\sqrt{g(x)}$ in the RHS.
| {
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Locus problem for vertex of equilateral triangle Question
Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$
My attempt
*
*Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $\tan \angle Q$ and $\tan\angle R.$
*Similarly for $\angle P$ to equate all 3 angles. Got the equation.
*Also use the distance method and then finally get the required locus.
My question
Is there any other method which is easier?
| I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:
$$\pmatrix{x'\\y'} = \pmatrix{x\\y}-\pmatrix{1\\ 3}$$
Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $\pm\frac {2\pi}6$ about the origin $P$.
For $i \in \{0, 1\}$:
$$\begin{align*}
R_i=\pmatrix{x'=h'_i\\y'=k'_i} &=
\pmatrix{\cos\frac {2\pi}6 & -\sin\left[(-1)^i\frac {2\pi}6\right]\\
\sin\left[(-1)^i\frac {2\pi}6\right] & \cos\frac {2\pi}6}
\pmatrix{2\\p'}\\
&= \pmatrix{\frac12 & -(-1)^i\frac{\sqrt3}2\\
(-1)^i\frac{\sqrt3}2 & \frac12}
\pmatrix{2\\p'}\\
&= \pmatrix{1-(-1)^i\sqrt3\frac{p'}2\\ (-1)^i\sqrt3+\frac{p'}2}
\end{align*}$$
One way to eliminate the $p'$ is to note that
$$\begin{align*}
\frac{p'}2 &= k'_i-(-1)^i\sqrt3\\
h'_i &= 1-(-1)^i\sqrt3\frac{p'}2\\
&= 1-(-1)^i\sqrt3\left[k'_i-(-1)^i\sqrt3\right]\\
&= 4- (-1)^i\sqrt3 k'_i
\end{align*}$$
i.e. the loci are $x'=4-(-1)^i\sqrt3 y'$.
The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.
$$\begin{align*}
(x-1) &= 4-(-1)^i\sqrt3(y-3)\\
(x-1) + (-1)^i\sqrt3(y-3) - 4 &= 0
\end{align*}$$
Lastly, if you prefer having one equation representing two straight lines:
$$\begin{align*}
\left[(x-1) + \sqrt3(y-3) - 4\right]\left[(x-1) - \sqrt3(y-3) - 4\right] &= 0\\
(x-5)^2-3(y-3)^2 &= 0
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate $\int_0^1{x·\lceil1/x\rceil dx}$ I am trying to calculate following integral:
$$\int_0^1{x·\biggl\lceil \frac{1}{x}\biggr\rceil dx}$$
I tried usual change t=1/x but not able to further advance.
Thanks!
| If $\frac 1{n} \le x < \frac 1{n-1}$ then $\lceil \frac 1x \rceil = n$. and $\int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx= \int_{\frac 1{n}}^{\frac 1{n-1}} xn dx = n\frac {x^2}2|_{\frac 1n}^{\frac 1{n-1}}= \frac n2(\frac 1{(n-1})^2 -\frac 1{n^2}=\frac n2(\frac {2n-1}{n^2(n-1)^2})=\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}$
So $\int_{0}^{1} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty \int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty[\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}]=\sum\limits_{n=1}^\infty[\frac 1{n^2} - \frac {1}{2(n+1)n^2}]$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{n \to \infty}\left(\sqrt{n^2 - n+1}-\left\lfloor\sqrt {n^2 - n+1}\right\rfloor\right)$
How do I evaluate
$$\lim_{n\to\infty}\left(\sqrt{n^2-n+1}-\left\lfloor\sqrt{n^2 - n+1}\right\rfloor\right),n\in\Bbb N$$
Attempt:
I thought of using Squeeze theorem but that could not help.
Secondly, we know that $x- \lfloor x\rfloor=\{x\}$ where $\{\}$ denotes the fractional part function. But I am not sure how to actually evaluate limits involving the fractional part function.
| Since $n - 1 < \sqrt{n^2 - n + 1} < n$, then\begin{align*}
&\mathrel{\phantom{=}}{} \sqrt{n^2 - n + 1} - [\sqrt{n^2 - n + 1}] = \sqrt{n^2 - n + 1} - (n - 1)\\
&= \frac{n}{\sqrt{n^2 - n + 1} + (n - 1)} → \frac{1}{2}. \quad (n → ∞)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding series of two functions multiplied with each other Let's say we are given a function defined as ${\frac{\ln(1+t)}{1-t}}$. We want to find the series expansion up to ${t^4}$.
Now we know that we have two function within this larger function which are ${\ln(1+t)}$ and ${\frac{1}{1-t}}$. Now the series and their respective expansions are know and defined as:
$${\ln(1+t)=\sum\limits_{n=0}^{\infty}(-1)^n\frac{x^n}{2}=\frac{1}{2}-\frac{x}{2}+\frac{x^2}{2}-...}$$
$${\frac{1}{1-t}=\sum\limits_{n=0}^{\infty}x^n=1+x+x^2...}$$
Now to find the series expansion as far as ${t^4}$, do we just add or multiply the two series?
| Note that:
$$\ln(1+t)=\sum\limits_{n=1}^{\infty}(-1)^{n\color{red}{+1}}\frac{\color{red}t^n}{\color{red}n}=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+O(t^5)\\
\frac{1}{1-t}=\sum\limits_{n=0}^{\infty}\color{red}t^n=1+t+t^2+t^3+O(t^4)\\
$$
Hence:
$$\frac{\ln(1+t)}{1-t}=\frac{1}{1-t}\cdot \ln(1+t)=\\
(1+t+t^2+t^3+O(t^4))\cdot\left(t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+O(t^5)\right)=\\
t+\left(t^2-\frac{t^2}{2}\right)+\left(t^3-\frac{t^3}{2}+\frac{t^3}{3}\right)+\left(t^4-\frac{t^4}{2}+\frac{t^4}{3}-\frac{t^4}{4}\right)+O(t^5)=\\
t+\frac{t^2}{2}+\frac{5t^3}{6}+\frac{7t^4}{12}+O(t^5).$$
Alternatively, applying Taylor's formula at $t=0$:
$$\begin{align}y=\frac{\ln(1+t)}{1-t} \Rightarrow (1-t)y=\ln (1+t) &\Rightarrow y(0)=0;\\
-y+(1-t)y'=\frac{1}{1+t} &\Rightarrow y'(0)=1;\\
-2y'+(1-t)y''=-\frac{1}{(1+t)^2} &\Rightarrow y''(0)=1;\\
-3y''+(1-t)y'''=\frac{2}{(1+t)^3} &\Rightarrow y'''(0)=5;\\
-4y'''+(1-t)y^{IV}=-\frac{6}{(1+t)^4} &\Rightarrow y^{IV}(0)=14;\\
\frac{\ln(1+t)}{1-t}=0+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{5}{3!}t^3+\frac{14}{4!}t^4+O(t^5).\end{align}$$
| {
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"source": "stackexchange",
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Simplify $\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$
Simplify
$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$$
Found in a book with tag "Moscow 1982", the stated answer is $1+\sqrt[4]{5}$. Used all tricks that I know but without success. The answer appears to be correct, checked in Wolfram Alpha.
Hints and answers welcomed. Sorry if this is a duplicate.
| Define $a=\sqrt[4]{5}$. Then $a^4 = 5$.
Consider the expression (under the radical in the denominator): $$
E = 4-3a+2a^2-a^3.\tag{1}
$$
What if we'll multiply it by $1+a$:
$$E(1+a)=(4-3a+2a^2-a^3)(1+a)\\=(4-3a+2a^2-a^3)+(4a-3a^2+2a^3-a^4)\\=4+a-a^2+a^3-a^4 \\= -1+a-a^2+a^3.\tag{2}
$$
Now what if we'll multiply it by $(1+a)$ again:
$$
E(1+a)^2 = (-1+a-a^2+a^3)(1+a)\\=(-1+a-a^2+a^3)+(-a+a^2-a^3+a^4)\\=
-1+a^4=4.\tag{3}$$
So,
$$
E=\dfrac{4}{(1+a)^2}.$$
Since $1+a>0$, we have $$\sqrt{E}=\dfrac{2}{1+a};$$
$$
\dfrac{2}{\sqrt{E}}=1+a=1+\sqrt[4]{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution to the equation of a polynomial raised to the power of a polynomial. The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^{x^2−7x+6}=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.
Factorising the exponent, we have:
$$ x^{2}-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x \in \{1,6\} $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x \in \{2, 5\} $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?
| Take natural logarithm from both sides:
$$\ln (x^2−7x+11)^{x^2−7x+6}=\ln1 \Rightarrow \\
(x^2-7x+6)\cdot \ln |x^2-7x+11|=0 \Rightarrow \\
1) \ x^2-7x+6=0 \Rightarrow x_{1,2}=1,6; \\
2) \ \ln |x^2-7x+11|=0 \Rightarrow |x^2-7x+11|=1 \Rightarrow x^2-7x+11=\pm 1 \Rightarrow \\
x_{3,4,5,6}=2,5,3,4.$$
Note: The found solutions satisfy the domain of the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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Factorization of $(1+x+x^2+x^3)^2 - x^3$
Factorize : $(1+x+x^2+x^3)^2 - x^3$
I've tried to expand it but the equation will be even more complicated, anyone can give me some hints to solve it without expanding it (or it is necessary to expand it)?
| $(1+x+x^2+x^3)^2-x^3 = (1+x+x^2)^2 + 2x^3(1+x+x^2) + x^3(x-1)(1+x+x^2) = (1+x+x^2)(1+x+x^2+x^3+x^4)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime
Let $p>3$ be a prime number. How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$?
When $p=3$, you also have $p+1$ as a divisor of $\left(p^2+1\right)^2$.
| Assume there are factors of $\def\num{(p^2+1)^2}\num$ which are congruent to $1$ modulo $p$. Then both the factor and the quotient are $\equiv1\pmod p$ as $\num\equiv1\pmod p$, so we can write $(np+1)(mp+1)=\num$.
Then
$$
(nmp+n+m)p+1=p^4+2p^2+1,
$$
so $nmp+n+m=p^3+2p$. This shows that $p$ divides $n+m$.
Without loss of generality, write $n=a_np+k$ and $m=a_mp-k$ where $a_m\geq1$ and $0\leq k<p$.
Substituting in the equation $nmp+n+m=p^3+2p$, we obtain
$$a_na_mp^2+(a_m-a_n)kp-k^2+a_n+a_m=p^2+2.$$
Now $(np+1)(mp+1)=(p^2+1)^2$ means one of $n$ and $m$ is $\leq p$, otherwise $(np+1)(mp+1)>\num$. This implies we only have to consider three cases: $a_n=1,\,k=0$, $a_n=0$ or $a_m=1$.
$a_n=1,\,k=0$:
The equation becomes $$a_mp^2+a_m=p^2+1.$$ This means $a_m=1$, and we obtain the solution $(p^2+1)(p^2+1)=(p^2+1)^2$.
$a_n=0$:
The equation becomes $$a_mkp-k^2+a_m=p^2+2.$$
If $a_m>p$, then the left-hand side is larger than the right-hand side of the equation, so $a_m\leq p$.
Since $a_m-k^2\leq p$, we must have $a_m=p$ and $k=1$. It follows that $a_m-k^2=2$, so $p=3$. And we obtain the case $(p+1)((p^2-1)p+1)=(p^2+1)^2$ for $p=3$.
$a_m=1$:
Write the equation as $$((a_np+k)p+1)((p-k)p+1)=p^4+2p^2+1.$$
Hence $$(a_np+k)(p-k)p^2+(a_np+p)p+1=p^4+2p^2+1$$ and $$(a_np+k)(p-k)+a_n=p^2+1.$$
Namely, $$a_n(p^2-kp+1)=p^2+1-kp+k^2.$$
This means that $p(p-k)+1$ divides $k^2$. Note that here $k\ne0$, otherwise it is just the case $(p^2+1)(p^2+1)=(p^2+1)^2$. Substituting $k$ by $p-k$, we obtain $pk+1\mid(p-k)^2$.
By this question, this is impossible.
Therefore the factors of $\num$ which are $\equiv1\pmod p$ are as described.
Hope this helps. Perhaps I shall post this as community wiki, as it involves an answer from other users.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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RMO practice problem inequality Let $a_n$ & $b_n$ be two sequences such that $a_0$ , $b_0$ > 0 and $a_{n+1}$ = $a_n$ + $\frac{1}{2b_n}$ & $b_{n+1}$ = $b_n$ + $\frac{1}{2a_n}$ $\forall$ n $\geq$ 0. Then prove that $$max(a_{2018},b_{2018}) > 44.$$
Anyone Please help me with this question.. How to approach this?
| Without loss of generality, suppose that $a_k>b_k$ for some k. Then $a_{k+1}=a_k+\frac{1}{2b_k}>a_k+\frac{1}{2a_k}$. Therefore it suffices to prove that $c_{2018}>44$, where $c_0$ is arbitrary positive number and $c_{n+1}=c_n+\frac{1}{2c_k}$.
Claim. $c_n\ge\sqrt{n+1}$ for all $n>0$.
We prove the claim by mathematical induction. For $n=1$, the claim is clear since by AM-GM, $c_{1}=c_0+\frac{1}{2c_0}\ge2\sqrt{\frac{c_0}{2c_0}}=\sqrt{2}$. For induction case, note that the function $x+\frac{1}{2x}$ is increasing for $x\ge\frac{1}{\sqrt{2}}$, which means $c_{n+1}=c_n+\frac{1}{2c_k}\ge\sqrt{n}+\frac{1}{2\sqrt{n}}$. Therefore it is enough to show that $\sqrt{n+1}\le\sqrt{n}+\frac{1}{2\sqrt{n}}$, or $\sqrt{n+1}-\sqrt{n}\le\frac{1}{2\sqrt{n}}$, or $\frac{1}{\sqrt{n+1}+\sqrt{n}}\le\frac{1}{2\sqrt{n}}$, or $\sqrt{n+1}+\sqrt{n}\ge2\sqrt{n}$, which is obvious.
By claim, we know that $c_{2018}\ge\sqrt{2019}$ and $\sqrt{2019}>44$ and we already showed that $\max{(a_n,b_n)}>c_n$ for any $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For what values of k does this system of equations have a unique / infinite / no solutions? My system of equations is:
\begin{cases}
x + 5y- 6z = 2 \\
kx + y - z = 3 \\
5x - ky + 3z = 7
\end{cases}
So the augmented matrix is:
$$ \left[
\begin{array}{ccc|c}
1&5&-6&2\\
k&1&-1&3\\
5&-k&3&7
\end{array}
\right] $$
I reduced it to this (with the shown steps), but I'm not sure if i've done so correctly and i'm not sure if i need to do Gauss-Jordan elimination or just Gauss elimination:
\begin{align}
r_2 -> r_2 - 2r_1
\end{align}
\begin{align}
r_3 -> r+3 - 5r_1
\end{align}
\begin{align}
r_2 -> r_2 * \frac{1}{-9}
\end{align}
\begin{align}
r_1 -> r_1 - 5r_2
\end{align}
\begin{align}
r_3 -> r_3 +27r_2
\end{align}
$$ \left[
\begin{array}{ccc|c}
1-5(\frac{k-2}{-9})&0&\frac{1}{9}&\frac{13}{9}\\
\frac{k-2}{-9}&1&\frac{-11}{9}&\frac{1}{9}\\
27(\frac{k-2}{-9})&-k+2&0&0
\end{array}
\right] $$
From this reduced from that i got, i noticed that there is an infinite solution for k=2.
However, I believe that i need to get one of the values in the 4th column of the matrix in terms of k so that i can find all the solutions i need, but i am unsure on how to do this.
Could someone please help me and show full working, including the steps taken to reduce the matrix?
| From my comments above, I always find the determinant first which gives us
$$\det \begin{bmatrix} 1&5&-6 \\ k&1&-1 \\ 5&-k&3 \end{bmatrix} = 2 (k-2) (3 k-2)$$
This tells us we may have to account for $$k = 2, k = \dfrac{2}{3}$$
The RREF is given by the steps
*
*Swap $R_1$ and $R_2$
*Set $R_2 \longleftarrow R_2 - \dfrac{1}{k} R_1$
*Set $R_3 \longleftarrow R_3 - \dfrac{5}{k} R_1$
*Swap $R_2$ and $R_3$
*Set $R_3 \longleftarrow R_3 - \dfrac{5k-1}{-k^2-5} R_2$
*Set $R_3 \longleftarrow - \dfrac{k^2+5}{2(3k^2-8k+4)} R_3$
*Set $R_2 \longleftarrow R_2 - \dfrac{3k+5}{k} R_3$
*Set $R_1 \longleftarrow R_1 + R_3$
*Set $R_2 \longleftarrow \dfrac{k}{-k^2-5} R_2$
*Set $R_1 \longleftarrow R_1 - R_2$
*Set $R_1 \longleftarrow \dfrac{1}{k} R_1$
This results in the RREF of
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & \dfrac{8}{3 k-2} \\
0 & 1 & 0 & -\dfrac{24}{3 k-2} \\
0 & 0 & 1 & \dfrac{-k-18}{3 k-2} \\
\end{array}
\right]$$
We can see that we have an issue for $k = \dfrac{2}{3}$.
Also, from the determinant, when $k = 2$, we have a RREF of
$$\left[
\begin{array}{ccc|c}
1 & 0 & \dfrac{1}{9} & \dfrac{13}{9} \\
0 & 1 & -\dfrac{11}{9} & \dfrac{1}{9} \\
0 & 0 & 0 & 0 \\
\end{array}
\right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partitioning a number into set of coprimes such that their product is maximum and sum of partition is the number itself Example 1)
Let the number be 7. Then we have a set {3,4}, so the product of the numbers is 10 and the numbers are mutually coprime.
Example 2)
for n=12, coprimes set = {3,4,5}, where product is 60 which is maximum.
Example 3)
n = 9, coprimes set = {4,5}, product is 20 which is maximum product.
| The product is given by the Landau function, shown in OEIS A000793 and beginning
$$ 1, 2, 3, 4, 6, 6, 12, 15, 20, 30, 30, 60, 60, 84, 105, 140, 210,\\ 210, 420, 420, 420, 420, 840, 840, 1260, 1260, 1540, 2310$$
To get the set you can just factor these numbers. The set always consists of powers of distinct primes. The powers are chosen to be "roughly equal". One of the higher values is $$120120=2^3\cdot 3 \cdot 5\cdot 7 \cdot 11 \cdot 13$$
with sum $47$. $48$ has the same partition with a $1$ added. Then at $49$ we get
$$180180=2^2\cdot 3^2\cdot 5\cdot 7 \cdot 11 \cdot 13$$
If there are $p$ factors the highest prime is about $p \log p$, so we can get a guess for the number of factors by solving $p^2 \log p =n$. At n=$49$ this says we should have about $5.4$ factors, each about $9$. We can guess we want $2^2, 2^3,$ or $2^4$ and $3^1$ or $3^2$ and first powers of the others and do a little search.
| {
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Image and Kernel of a linear transformation from a matrix to a polynomial Let the following be the linear transformation for $T$ from a $2$ x $2$ matrix to a second degree polynomial.
$T\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\left(a-c\right)x^2+c\left(x-1\right)+b$
Find a basis for $Im(T)$ and $Ker(T)$.
So this is what I did to find the kernel.
I rearranged the equation to get:
$\left(a-c\right)x^2+\left(c\right)x+\left(-c+b\right)$
So I tried to set everything in the brackets to $0$ and then got that:
$a=c=0,\:b=0,\:d=free$
This led me to think that a basis for $Ker(T)$ would be:
$\left\{\begin{pmatrix}0&0\\ 0&1\end{pmatrix}\right\}$
For the $Im(T)$, I rearranged the equation to get:
$a\left(x^2\right)+b+c\left(-x^2+x-1\right)$
This gave me the following basis:
$\left\{x^2,\:1,\:-x^2+x-1\right\}$
Which had the following matrix:
$\begin{pmatrix}1&0&-1\\ 0&0&1\\ 0&1&-1\end{pmatrix}$
But all three columns of this matrix are linearly independent, which doesn't make any sense as the kernel I have calculated has dimension $1$.
Does anyone know where I have gone wrong here?
Thank you!
| Use a systematic approach. Let
$$
E_1=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad
E_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\quad
E_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\quad
E_4=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
$$
and let $p_1=1$, $p_2=x$, $p_3=x^2$. Then
\begin{align}
TE_1&=p_3\\
TE_2&=p_1\\
TE_3&=p_2-p_3\\
TE_4&=0
\end{align}
so the matrix of $T$ with respect to the bases $\{E_1,E_2,E_3,E_4\}$ and $\{p_1,p_2,p_3\}$ is
$$
A=\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & -1 & 0
\end{pmatrix}
$$
The RREF of $A$ is
$$
U=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}
$$
which has rank $3$. Therefore the image of $T$ is the whole codomain.
The null space of $U$ (and of $A$) is generated by
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
which says that the kernel of $T$ is generated by $E_4$.
You can also go the hard way, but it's much simpler to observe that
$$
p_1=TE_2,\quad
p_2=T(E_3+E_2),\quad
p_3=TE_1
$$
so the image is the whole space. Moreover $E_4$ clearly belongs to the kernel. By the rank-nullity theorem, the kernel has dimension $1$, so it is generated by $E_4$.
| {
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Factor $10^6-1$ completely I know kind of a very elementary method to factor this number. Consider the following:
$$10^6-1 = (10^3-1)(10^3+1)=9 \times 11 \times (10^2+10+1)(10^2-10+1) = 9 \times 11 \times 111\times 91$$
I would then factor each number individually.
Is there a faster method? The great hint is that this number is a rep unit number = $9\ \times 111111$.
| \begin{align}
10^6 - 1 &= 1000000-1\\
&= 999999 \\
&= 3^2 \times 111111 \\
&= 3^2 \times 111\times 1001 \\
&= 3^2 \times 111 \times (1100 - 99)\\
&= 3^2 \times 111 \times 11 \times (100-9)\\
&= 3^2 \times 111 \times 11 \times 91 \\
&= 3^2 \times (3 \times 37)\times 11 \times 7 \times 13
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Not sure how to solve $\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$ So I got this problem:
Determine the following limit value:
$$\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}$$
What I tried is:
$\large{\lim_\limits{x\to0}{{\sqrt{x^2+1}-1}\over\sqrt{x^2+16}-4}\cdot{\sqrt{x^2+16}+4\over\sqrt{x^2+16}+4}=\\\lim_\limits{x\to0}{({\sqrt{x^2+1}-1})(\sqrt{x^2+16}+4)\over x^2+16-16}}$
From this point I just get everything messy, and can't get anything out of it, so I believe this is not the way to solve this. Doing it from a table of values gives 4, but I should solve this without the table.
| Now, you can make the following:
$$\frac{(\sqrt{x^2+1}-1)(\sqrt{x^2+16}+4)}{x^2+16-16}=\frac{(x^2+1-1)(\sqrt{x^2+16}+4)}{x^2(\sqrt{x^2+1}+1)}\rightarrow\frac{4+4}{1+1}=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
sharp bounds for $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$ Is there any good bounds or estimation of $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$ and $(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$, $1 < k < n$?
What I actually want is $\sum_{1\le k \le n}(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{k}{n})$, and $\sum_{1\le k \le n}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$.
| Classic products, whose results are $\left({\pm 1 \over n} \right)^k P(1\pm n,k)$ where $P$ is the Pochammer symbol.
And what you "really" want is:
$$-\frac{2^{2 n+1} e^{-n} n^{n+1} \Gamma \left(\frac{1}{2} (2 n+3)\right) \Gamma (-2 n-1,-n)}{\sqrt{\pi }}+e^{-n} (-n)^{n+1} \Gamma
(-n,-n)-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1} {3+x^2+z^2}\leq \frac {3}{5} . $ Let $x, y, z>0$ s.t. $x+y+z=3$.
Show that $$\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1 } {3+x^2+z^2}\leq \frac {3}{5}\ . $$
My idea: $$3 + x^2 + y^2 \geq 1 + 2x+ 2y=7-2z $$
I notice that $f (t)=\frac {1}{7-2t} $ is a convex function but it's uselessness.
Also I have some troubles with the next inequality $$\frac {a}{a^2+bc} +\frac {b}{b^2+ac}+\frac {c}{c^2+ab}\geq \frac {3}{2} \frac{a+b+c}{a^2+b^2+c^2} $$
My idea is to multiply and to apply Muirhead's inequality. But there are too much terms.
| The second inequality.
By C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+bc}=\sum_{cyc}\frac{a^2(b+c)^2}{(a^3+abc)(b+c)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}a(b+c)\right)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}=\frac{4(ab+ac+bc)^2}{\sum\limits_{cyc}(a^3+abc)(b+c)^2}.$$
Thus, it's enough to prove that
$$8(ab+ac+bc)^2(a^2+b^2+c^2)\geq3(a+b+c)\sum\limits_{cyc}(a^3+abc)(b+c)^2$$ or
$$\sum_{cyc}(5a^4b^2+5a^4c^2-6a^3b^3+4a^4bc-5a^3b^2c-5a^3c^2b+2a^2b^2c^2)\geq0,$$ which is obvious by Schur
$$\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ and Muirhead.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed-form expression for infinite series related to a Gaussian Consider the following infinite series, where $x$ is indeterminate and $r$ is held constant:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^2} + \frac{x^3}{r^3} + ...$
It is relatively easy to see that the above, for $\frac{x}{r} < 1$, converges to
$\displaystyle \frac{1}{1-\frac{x}{r}}$
Now suppose we modify the above to this:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^{2^2}} + \frac{x^3}{r^{3^2}} + ...$
which we can rewrite as
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
Does there then exist a well-known, closed-form expression for this series?
If not for general $r$, then perhaps for certain special values of $r$? For example, if we set $r=e$ above, then we get
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...$
which we can rewrite as
$\displaystyle 1 + xe^{-1^2} + x^2e^{-2^2} + x^3e^{-3^2} + ...$
So that we can see that for x=1, this becomes a series of evenly spaced points on a Gaussian function.
Does there exist a closed-form expression for any of these?
| Your original series is
$f(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{(n^n)}}
$.
This is not the same as
$g(x, r)
=\sum_{n=0}^{\infty} \dfrac{x^n}{(r^n)^n}
=\sum_{n=0}^{\infty} \dfrac{x^n}{r^{n^2}}
$.
Also,
you went from
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...
$
to
$\displaystyle 1 + x \cdot e^{-1^2} + x \cdot e^{-2^2} + x \cdot e^{-3^2} + ...
$,
somehow losing the
exponent of $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which number can I erase? All positive integers greater than $2$ are written on a board. First we erase number $3$ and $5$.
With 4 positive integers $a,b,c,d$ satisfying $a+b=c+d$, if $ab$ is erased, then $cd$ can be erased, otherwise $cd$ cannot be erased.
For example, $3=3 \times 1$, $3+1=4=2+2$ , then $2 \times 2 = 4$ is erased.
a. What are the conditions of a number that can be erased ?
b. If not only $3$ and $5$, but every prime number is erased at the beginning, can all other numbers be erased as well? If not, what are the conditions of a number to be erased ?
(Sorry for my last question, English is my second language)
| Somewhat a long comment (not a complete answer).
The first thing that I would suggest is that you should work through some examples and see what happens. I'll help with some of the initial steps.
To think about what you're erasing, let $n$ be a number that has been erased. Then, for each pair of factors $ab$ so that $ab=n$, compute $a+b=m$. Next, you find all other ways to write $m$ as a sum $m=c+d$ and then delete each $cd$ found in this way. Then, you continue until there isn't anything more to delete (or you see the pattern).
In your set-up, you start with $3$ and $5$ deleted.
Deleted: $3$ and $5$.
*
*Using the deleted $3$, you know that $3$ factors as $3=1\cdot 3$. Therefore, $a+b=1+3=4$. The only sums equal to $4$ are $1+3$ and $2+2$. Taking $c=2$ and $d=2$ gives $cd=4$, so $4$ is deleted. Since we've considered all pairs of factors of $3$, there is nothing more that $3$ can tell us.
Deleted: $3$, $4$, and $5$.
*
*Using the deleted $4$, you know that $4$ factors as $4=1\cdot 4$ or $2\cdot 2$. In the first case, $1+4=5$, and $5$ can be written as $5=1+4$ or $5=2+3$. The first case leads to $1\cdot 4=4$, which is already deleted, but $2\cdot 3=6$, which can now be deleted. On the other hand, using $2\cdot 2=4$, we have two ways to write $4$ as a sum, $4=1+3$ or $4=2+2$. In either case, the product is $1\cdot 3=3$ or $2\cdot 2=4$, both of which have already been deleted. Therefore, there is nothing more that $4$ can tell us.
Deleted: $3$, $4$, $5$, and $6$.
*
*Using the deleted $5$, you know that $5$ factors as $1\cdot 5=5$. Therefore, the sum of these two is $1+5=6$. There are three ways to write $6$ as a sum, $1+5=6$, $2+4=6$, and $3+3=6$. The first pair has product $1\cdot 5=5$, which has already been deleted. The second pair has product $2\cdot 4=8$, which can now be deleted. The third pair has product $3\cdot 3=9$, which can also be deleted.
Deleted: $3$, $4$, $5$, $6$, $8$, and $9$.
*
*Using the deleted $6$, you know that $6$ factors as $1\cdot 6=6$ and $2\cdot 3=6$. Therefore, the corresponding sums are $1+6=7$ or $2+3=5$. There are three sums to get $7$, $1+6=7$, $2+5=7$, and $3+4=7$. There are two sums to get $5$, $1+4=5$ and $2+3=5$. These, combined allow one to delete $6$, $10$, $12$, $4$, and 6$.
Deleted: $3$, $4$, $5$, $6$, $8$, $9$, $10$, and $12$.
Continue this until you find a pattern.
| {
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"timestamp": "2023-03-29T00:00:00",
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How big does $r$ need to be, to ensure that $x^4+y^4 >r^2$ for all $x^2+y^2 =r^2$? Consider the circle $x^2+y^2 = r^2$ for some fixed $r>0$. How big does $r$ need to be, to ensure that $$x^4+y^4 >r^4$$ for all $x^2+y^2 =r^2$? I know that $x^2<x^4$ whenever $|x| >1$, but I'm not sure how to use that here.
| By C-S $$(1^2+1^2)(x^4+y^4)\geq(x^2+y^2)^2.$$
Thus, $$x^4+y^4\geq\frac{1}{2}r^4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expected value and standard deviation of a pmf function
In a family, the probability mass function of the number of people $x$ who have contracted the flu is given by $$P(x) = Kx \qquad x\in\{0,1,\ldots,N\}$$ where $N$ is the number of people in the family.
(a) If nine people are expected to have flu in the family, calculate $K$ and $N$.
(b) Calculate the probability that the number of people in the family with flu is within one standard deviation of the mean.
My attempt:
This is a binomial distribution with probability parameter $p=0.5$ and mean $\mu=9$; hence $N = \mu/p=9/0.5 = 18$.
Now summing $Kx$ from $x = 0$ to $x=18$ gives $$\begin{align}
K + 2K + 3K+ \cdots + 18K &= 1 \\
K(1+ 2+ 3+ \cdots +18) &= 1 \\
171K &= 1 \end{align}$$
Hence $K = 1/171$.
The standard deviation for this distribution is $$Np(1-p) = 18\times 0.5\times0.5 = 4.5$$
Am I on the right path?
| This is not a binomial function, more like a triangular one. We cannot immediately derive $N$ from $K$ or the other way round, but we can derive two equations first. For the expected value:
$$1\cdot K+2\cdot2K+\dots+N\cdot NK=9$$
$$K(1^2+2^2+\dots+N^2)=9$$
For the pmf summing to 1:
$$K(1+2+\dots+N)=1$$
Thus we have
$$\frac{1^2+2^2+\dots+N^2}{1+2+\dots+N}=9$$
$$\frac{2N(N+1)(2N+1)}{6N(N+1)}=\frac{2N+1}3=9\qquad N=13$$
Then $K(1+2+\dots+N)=91K=1$ and $K=\frac1{91}$.
The standard deviation is $\sqrt{\frac{1(1-9)^2+2(2-9)^2+\dots+13(13-9)^2}{91}}=\sqrt{10}=3.16\dots$ Thus one standard deviation is anywhere between 6 and 12 inclusive, and the probability of this happening is $\frac{6+7+\dots+12}{91}=\frac9{13}$.
| {
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"url": "https://math.stackexchange.com/questions/2960347",
"timestamp": "2023-03-29T00:00:00",
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Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}=2-\text{something}.$$
First, find the "something" and then use that form of the equation to prove the assertion.
I can't seem to figure out the form that this equation can be written as. Also, once I find the form how would I do the proof.
I understand it involves using a Basic Step and an Induction Step
| Let the sum of the $n$ first terms be $S_n$. We have the recurrence
$$S_{n+1}=S_n+\frac{n+1}{2^{n+1}}$$ or
$$2^{n+1}S_{n+1}=2\cdot2^nS_n+n+1.$$
This hints the change of variable that leads to
$$R_{n+1}=2R_n+n+1.$$
This is a linear recurrence which we will solve a usual:
*
*homogeneous part, $R_{n+1}=2R_n$, so that $R_n=R_12^{n-1}$.
*particular solution found by indeterminate coefficients, using a linear ansatz: $-n-2$.
Now, using the initial condition $R_1=1$,
$$S_n=2-\frac{n+2}{2^n}.$$
We can complete the inductive proof:
$$S_1=\frac12=2-\frac{1+2}{2^1}<2,$$
$$S_n=2-\frac{n+2}{2^n}<2\implies S_{n+1}=2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{(n+1)+2}{2^{n+1}}<2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $P(x,y,z)=x^n+y^n+z^n-\prod\limits_{k=0}^{n-1}(x+\omega_n^ky+\omega_n^{-k}z)$, where $\omega_n$ denotes a primitive $n$th root of unity
Find $P(x,y,z)=x^n+y^n+z^n-\prod\limits_{k=0}^{n-1}(x+\omega_n^ky+\omega_n^{-k}z)$, where $\omega_n$ denotes a primitive $n$th root of unity.
I have manually multiplied the terms of the product and then equate the coefficients to get the polynomial but that's too cumbersome.
Here is my method :
$(x + y + z)(x + y\omega_n + z\omega_n^{n-1})(x + y\omega_n^2 + z\omega_n^{n-2})....(x + y\omega_n^{n-1} + z\omega_n) = x^n(1 + [Y + Z])(1 + [Y\omega_n + Z\omega_n^{n-1}])(1 + [Y\omega_n^2 + Z\omega_n^{n-2}])....(1 + [Y\omega_n^{n-1} + Z\omega_n])$ where $Y=\frac {y}{x}$ and $Z=\frac {z}{x}$
Hence, I applied the formula:
$(1+\alpha)(1+\beta)(1+\gamma)...... = 1 + [\alpha + \beta + \gamma + ...] + [\alpha\beta + \beta\gamma + ....] + ....$
to get an expression for odd values of $n$ :
$$P = nxyz(x^{n−3}+x^{n−5}yz+x^{n−7}y^2z^2+....)$$
My question is : How can I get a general expression for $P$ in a way better than what I have mentioned?
$$EDIT$$
I have a more general expression by now, which I think can be derived elementarily (unfortunately I still don't know how),
$$P=\frac {x^n}{t^n}(L_n(t)-t^n),$$
where $L_n(t)$ is the $n^{th}$ Lucas polynomial in $t:=\frac {ix}{\sqrt {yz}}$
| We need to find a closed form expression of
$$
\prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z),
$$ where $\omega=e^{\frac{2\pi i}{n}}$ is the primitive $n$-th root of unity. Now let us consider the polynomial
$$
F(r) = r^n\prod_{k=0}^{n-1}(x+r\omega^ky+r^{-1}\omega^{-k}z)=\prod_{k=0}^{n-1}(xr+r^2\omega^ky+\omega^{-k}z)
$$
of degree $2n$ in $r$ regarding $x,y,z$ as fixed constants. By solving
$$
x+r\omega^ky+r^{-1}\omega^{-k}z = 0
$$ for each $0\le k\le n-1$,
we can see that $F(r)$ has simple roots
$$
r\omega^k = \frac{-x \pm \sqrt{x^2 - 4yz}}{2y}\Longrightarrow r = \omega^{-k}\frac{-x \pm \sqrt{x^2 - 4yz}}{2y}.
$$
If we denote $$\alpha(x,y,z) = \frac{-x + \sqrt{x^2 - 4yz}}{2y},\qquad \beta(x,y,z)=\frac{-x - \sqrt{x^2 - 4yz}}{2y},$$ we see that the set of all roots of $F(r)$ coincides with that of $$G(r)=(r^n-\alpha^n(x,y,z))(r^n-\beta^n(x,y,z)).$$ Since neither $F$ nor $G$ has multiple roots, it follows that $F$ is a constant multiple of $G$, i.e.
$$
F(r) = \left[\prod_{0\leq k\leq n-1} \omega^ky \right]G(r)=(-1)^{n-1}y^nG(r).
$$ by matching the leading coefficient. Plugging $r=1$ into the expression gives
\begin{align*}
\prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z)=&F(1)\\
=& (-1)^{n-1}y^nG(1)\\
=& (-1)^{n-1}y^n(1-\alpha^n-\beta^n +\alpha^n\beta^n).
\end{align*}Using $\alpha +\beta = -\tfrac x y$ and $\alpha\beta = \tfrac z y$, we get
$$
(-1)^{n-1}y^n(1-\alpha^n-\beta^n +\alpha^n\beta^n)=(-1)^{n-1}(y^n-(y\alpha)^n-(y\beta)^n+z^n ).
$$It remains to evaluate
$$
(y\alpha)^n+(y\beta)^n = \left(\frac{-x + \sqrt{x^2 - 4yz}}{2}\right)^n + \left(\frac{-x - \sqrt{x^2 - 4yz}}{2}\right)^n.
$$ If we let $t= \frac{ix}{\sqrt{yz}}$, we have
\begin{align*}
(y\alpha)^n+(y\beta)^n =& (i\sqrt{yz})^n\left(\left(\frac{t + \sqrt{t^2 +4}}{2}\right)^n + \left(\frac{t - \sqrt{t^2 +4}}{2}\right)^n\right) \\
=& (-1)^nx^nt^{-n}L_n(t)
\end{align*} where $L_n(t)$ is $n$-th Lucas polynomial. Finally, this gives
$$
\prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z) = (-1)^{n-1}y^n+ (-1)^{n-1}z^n+x^nt^{-n}L_n(t),
$$ hence
$$
P(x,y,z) = x^n(1-t^{-n}L_n(t)) + (1+(-1)^n)y^n + (1+(-1)^n)z^n.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}$$ The linked proof relies upon this trigamma identity. Now by rewriting the integral as:
$$I=\int_0^1 \frac{\sqrt{x}\ln x}{x^2-2\cos\left(\frac{\pi}{3}\right)x+1}dx$$
And using that: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$I=\frac{1}{\sin \left(\frac{\pi}{3}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{3} (n+1) \right)\int_0^1 x^{n+1/2} \ln x dx$$
$$\text{Since} \ \int_0^1 x^p \ln x dx= -\frac{1}{(p+1)^2}$$
$$I=-\frac{2}{\sqrt 3} \sum_{n=0}^\infty \frac{\sin\left((n+1)\frac{\pi}{3}\right)}{(n+1+1/2)^2}=-\frac{8}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} $$
And well by using the previous link we can deduce that the series equals to $\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$, where $G$ is Catalan's constant.
I thought this might be a coefficient of some Fourier series, or taking the imaginary part of $\left(\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{3}}}{(2n+1)^2}\right)$, but I was not that lucky afterwards.
Is there a way to show the result without relying on that trigamma identity? Another approach to the integral would of course be enough.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} =
{G \over \root{3}} - {\pi^{2} \over 24}} \approx 0.1176:\
{\Large ?}}$. $\ds{G = 0.9159\ldots}$ is the
Catalan Constant.
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}
{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}}} =
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} +
\sum_{n = 6}^{11}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} + \sum_{n = 12}^{17}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} + \cdots
\\[5mm] = &\
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} +
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\bracks{2n + 2\pars{6} + 1}^{\, 2}} + \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\bracks{2n + 4\pars{6} + 1}^{\, 2}} + \cdots
\\[5mm] = &\
\sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}}
\sum_{k = 0}^{\infty}{1 \over \pars{2n + 12k + 1}^{2}} =
{1 \over 144}\sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}}\,
\Psi\, '\pars{2n + 1 \over 12}
\\[5mm] = &\
{G \over \root{3}}\ -\
\underbrace{\bracks{80G + \Psi\, '\pars{11 \over 12} -
\Psi\, '\pars{5 \over 12}}{\root{3} \over 288}}
_{\ds{=\ {\pi^{2} \over 24}}}
\end{align}
I'll $\ds{\underline{\mbox{have}}}$ to prove that
$\ds{\color{red}{\Psi\, '\pars{11 \over 12} -
\Psi\, '\pars{5 \over 12} \color{black}{=} 4\root{3}\pi^{2} - 80G}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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} |
Successive differentiation BSc 2
Question : if $y^3 + x^3 - 3axy = 0$, show that $$y'' = -\frac{(2a^3 xy)}{(y^2-ax)^3}$$
How I did
I differentiated the whole expression twice (as shown below)
*
*$3y^2y'+3x^2-3x^2-3ay-3axy'=0$
*$6yy''+6x-3yy'-3ay'-3axy''=0$
Then I kept $y''$ on the left and transferred everything to right and plugged in the values of $y'$ and solved
I get
$$y''=\frac{(y')(3y+3a)-(6x)}{6y-3ax}$$
On solving this further after putting values of $y'$ I'm not getting the desired thing: $y''=-2a^3xy/(y^2-ax)^3$
| I got $$3y^2y'+3x^2-3ay-3axy'=0$$ or
$$y'=\frac{ay-x^2}{y^2-ax}$$
so
$$y''=\frac{(ay'-2x)(y^2-ax)-(ay-x^2)(2yy'-a)}{(y^2-ax)^2}$$
$$y''=\frac{y'(-a^2x-ay^2+2xy)-2xy^2+ax^2+a^2y}{(y^2-ax)^2}$$
$$y''=\frac{(ay-x^2)(-ax^2-ay^2+2x^2y)-(2xy^2-ax^2-a^2y)(y^2-ax)}{(y^2-ax)^3}$$
Expanding all i got
$$y''=\frac{-2a^3xy+6ax^2y^2-2x^4y-2xy^4}{(y^2-ax)^3}$$
| {
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Proving that if $x^4 + 5x + 1 < 27$ then $x < 2$ I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form $P \Rightarrow Q$
$$
P: x^4 + 5x + 1 < 27
$$
$$
Q: x < 2
$$
I wanted to try and prove this by contrapositive , so this state would become
If $X \geq 2$ then $x^4 + 5x + 1 \geq 27$
Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me $2^4 + 5 \times 2 + 1 = 27$ and since $27 \geq 27$ this statement is True.
Am I allowed to prove it like this? Is there a different way to prove a question like this?
| If $x\ge 2$ is it true that $x^4 \ge 16$?
If $x \ge 2$ is it true that $5x \ge 10$?
So if $x \ge 2$ is it true that $x^4 + 5x + 1 \ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $x\ge 2 > 0$ then $x^4 \ge 2x^3 \ge 4x^2 \ge 8x \ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $x\ge 2 \implies 5x \ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c \iff a > b$ then $x^4 + 5x + 1 \ge x^4 + 10 + 1 \ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x \ge 2$ then let $d = x - 2 \ge 0$. Then $x = 2 + d$
Now because $d \ge 0$ then $d^k \ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 \ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d \ge 5*2$
So $x^4 + 5x + 1\ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
| {
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A basic theorem of prime numbers. Checking My Proof
*
*Let $n$ be a positive integer. Prove that $n^5 -1$ is prime if and only if $n=2$.
$(\Rightarrow)$ Assume $n^5-1$ is prime. We will show $n=2$. If $n=3$, then clear $n^5-1$ is not prime.
Claim. If $n>2$, then $n^5-1$ is not prime.
Proof of the claim. We will do induction on $n$.
Initial Step. If $n=3$, clearly $n^5-1$ is not prime.
Induction Step. Assume for $n=m\in\mathbb{N}^{>2}$, $m^5-1$ is not prime. We will show $n=m+1\in\mathbb{N}^{>2}$, $(m+1)^5-1$ is not prime. So
$(m+1)^5-1=m((m+1)^4 +(m+1)^3 +(m+1)^2 +(m+1)+1)$
Let $u=m$ and $v=(m+1)^4 +(m+1)^3 +(m+1)^2 +(m+1)+1)$. Then since $m>2$, then $u>2$ and $v>2$. Thus, $uv$ cannot be a prime because $uv$ has no positive integer divisors other than $1$ and $uv$.
Therefore the claim is proved.
So by the claim, $n$ must be $2$.
$(\Leftarrow)$ Assume $n=2$. Then, $n^5 -1=2^5 -1=31$ is prime.
May you check my proof? Thanks...
| Your proof is fine but really long and complicated and hard to read.
Simpler to just say:
$n^5 -1 = (n-1)(n^4 + n^3 + n^2 + n + 1)$.
If $n-1$ and $n^4 + n^3 + n^2 + n + 1$ are non trivial factors (not equal to $1$ or $n^5 -1$) then this is not prime.
So the only way for $n^5 -1$ to be prime is if either $n-1 = 1$ or $n^4 + n^3 + n^2 + n +1 = 1$.
If $n -1 = 1$ then $n =2$ and if $n^4 + n^3 + n^2 + n + 1= 1$ then $n(n^3 + n^2 + n + 1) = 0$ which has no natural number solutions.
So $n^5 -1$ is not prime if $n \ne 2$.
And if $n = 2$ then $n^5 -1 = 2^5 - 1 = 31$ which is prime.
| {
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?
| Let $G(x)=(1+x)^{1/x}$
$\ln G(x)=\dfrac{\ln(1+x)}x=1-\dfrac x2+\dfrac{x^2}3-\dfrac{x^3}4+\cdots$
$$G(x)=e\cdot e^{-x/2}\cdot e^{-x^2/3}\cdot e^{-x^3/4}\cdots$$ ignoring terms containing $x^4$
$$=e\left(1-\dfrac{x}2+\dfrac{\left(-\dfrac x2\right)^2}{2!}+\dfrac{\left(-\dfrac x3\right)^3}{3!}+\cdots\right)\left(1-\dfrac{x^2}3+\dfrac{\left(-\dfrac{x^2}3\right)^2}{2!}+\cdots\right)\left(1-\dfrac{x^3}4+\cdots\right)\cdots$$
$$=?$$
| {
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Finding the limit of the sequence $(1, \frac{1}{2+1}, \frac{1}{2+\frac{1}{1+1}},\cdots)$ I am struggling to find the limit of the sequence:
$$1, \cfrac{1}{2+1}, \cfrac{1}{2+\cfrac{1}{1+1}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+1}}}, \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+1}}}},\cdots$$
Normally, for these types of sequences, I would find a recurrence relation, i.e. $a_n=f(a_{n-1})$, and then say as $n\to\infty$, $a=f(a)$ which I would solve for $a$. I cannot find a recurrence relationship between each term in this sequence, however. What is the recurrence relationship between the terms in this sequence? Alternatively, is there another way of finding its limit?
| Based on the first five terms I got the following:
$a_1=1$, $\hspace {0,2cm}$ $a_2=\frac{1}{3}$, $\hspace {0,2cm}$
$a_3=\frac{1}{2+\frac{1}{1+a_1}}$
$a_4=\frac{1}{2+\frac{1}{1+a_2}}$
$a_5=\frac{1}{2+\frac{1}{1+a_3}}$
So the recurrence relationship $\hspace {0,2cm}$ $a_n=\frac{1}{2+\frac{1}{1+a_{n-2}}}=\frac{1+a_{n-2}}{3+2a_{n-2}}$
The limit of the sequence can be easily computed by the following way:
If the limit existing and $\lim\limits_{n\rightarrow\infty}= L>0$
then solving the $\hspace {0,2cm}$ $L=\frac{1+L}{3+2L}$ eqution we get that $L=\frac{\sqrt{3}-1}{2}$
(Sorry for the delayed ansver)
| {
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prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$ I need to prove that
$ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$
If I define $f(x) = -2 + x + (2+x)e^{-x}$ and plot it I can see it's a monotonously growing function, and f(0)=0. Then $f(x)>0$ if $x>0$.
However I can't find the way to prove this.
Ideally I would like to prove it without deriving the function and by using some inequalities, however I don't know if it is possible. Any hint is really appreciated.
| This approach requires an extra condition $x\lt2.$
\begin{align}\ln\left(1+\dfrac{x}{2}\right)-\ln\left(1-\dfrac{x}{2}\right)&=\left[\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2+\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3+\cdots\right]-\left[-\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2-\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3-\cdots\right]\\&=x+\dfrac{2}{3}\left(\dfrac{x}{2}\right)^3+\cdots\\&\gt x\quad\forall x\lt2\cdots(*)\\\end{align}
$\forall x\gt0,\dfrac{2+x}{2-x}\gt1\implies\ln\left(\dfrac{2+x}{2-x}\right)\gt0$
$\therefore0\lt x\lt\ln\left(\dfrac{2+x}{2-x}\right)\implies e^{-x}\gt\dfrac{2-x}{2+x}\implies -2+x+(2+x)e^{-x}\gt0$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{\sqrt[n]{n!}}.\tag1$$
Moreover, we can find that, for $k=0,1,\cdots,n-1.$ $$\frac{n^k}{k!}< \frac{n^n}{n!}.$$
Thus
\begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\
&=(2n+1)\cdot \frac{n^n}{n!},
\end{align*}
which shows that
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.\tag2$$
Combining $(1)$ and $(2)$, we have
$$e>\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}\to e(n \to \infty).$$
By the squeeze theorem,
$$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
Please correct me if I'm wrong. Many thanks.
| Using Riemann sums:
$$ \log\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^{n}\log\left(\tfrac{k}{n}\right)\to \int_{0}^{1}\log(x)\,dx = -1$$
hence $\frac{n}{\sqrt[n]{n!}}$ converges to $e$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim_{k \to \infty}(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}$ I would like to calculate this limit. I tried to look for known limits which can be substituted to parts of this limit but couldn't find information about the double factorial.
$$\lim\limits_{k \to \infty}(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}$$
Thanks ahead for any help
EDIT
Denote $n=2^k-1=2\cdot 2^{k-1}-1$ so we actually need to calculate:
$$\lim\limits_{n \to \infty}(2n+3)^{\frac{n+1}{4}-2}\cdot\frac{n^2}{n!!}$$
And have by a relation-to-factorial formula for odd n:
$$n!! = \frac{2^k!}{2^{2^{k-1}}\cdot 2^{k-1}!} = \frac{(n+1)!}{2^{\frac{n+1}{2}}\cdot (\frac{n+1}{2})!}$$
Now using the Stirling approximation we have (only for odd n):
$$n!! \sim \frac{\sqrt{2\pi(n+1)}(\frac{n+1}{e})^{n+1}}{2^{\frac{n+1}{2}}\cdot \sqrt{2\pi(\frac{n+1}{2})}(\frac{\frac{n+1}{2}}{e})^{\frac{n+1}{2}}} = \frac{\sqrt{2}(\frac{n+1}{e})^{n+1}}{2^{\frac{n+1}{2}}(\frac{n+1}{e})^{\frac{n+1}{2}}(\frac{1}{2})^{\frac{n+1}{2}}} = \sqrt{2}(\frac{n+1}{e})^{\frac{n+1}{2}}$$
From here it is clear to see that for odd n's $$\lim\limits_{n \to \infty}(2n+3)^{\frac{n+1}{4}-2}\cdot\frac{n^2}{n!!} = 0$$
| HINT
Let indicate $n=2^k-1\to \infty$ then we have
$$(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}=(2n+3)^{\frac{n+1}4-2}\cdot\frac{n^2}{n!!}=(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}$$
then use Stirling approximation or by Root test
$$\sqrt[n]{(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}}
\sim \sqrt[n]{\frac{n^\frac{n}4n^2}{2^{\frac{3n}4}n!}}=\frac{\sqrt[n]{n^2}}{2^\frac34}\cdot\frac{\sqrt[4] n}{\sqrt[n]{n!}}$$
| {
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Trigonometric Identities: Given $\tan(2a)=2$ and $\frac{3\pi}{2}Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)}
\implies\frac{5}{2\sqrt 5}\left(1-\frac{\sqrt{5}}5\right)$$
I then simplified that to $\frac{\sqrt{5}}{2}-1$
Then because it is in fourth quadrant I multiplied by negative 1 and got:
$$1-\frac{\sqrt{5}}{2}$$
Am I doing something wrong, or am I correct?
| You made two mistakes, but your method does work.
If $\tan(2a)=2$, then either
Case 1: $0< 2a +2 n \pi<\pi/2$, $\cos(2a)=1/\sqrt{5}$, and $\sin(2a)= 2/\sqrt{5}$ (first quadrant), or
Case 2: $\pi<2a+2 n \pi<3 \pi/2$, $\cos(2a)=-1/\sqrt{5}$, and $\sin(2a)= -2/\sqrt{5}$ (third quadrant)
for some integer $n$.
The problem stated that $3 \pi/2 < a < 2 \pi$, so $3 \pi < 2 a < 4 \pi$ and that puts us in Case 2 with $n=-1$.
Now using your method
$$\tan(a)=\frac{1 - \cos(2 a)}{\sin(2 a)} = \frac{1+1/\sqrt{5}}{-2/\sqrt{5}}
= - \frac{\sqrt{5}+1}{2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove following inequality Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$
What I tried was to use AM-GM for the left side of this inequality, what I got was
$3(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 3$ and $(\frac{8abc}{(a+b)(b+c)(c+a)})^\frac{2}{9}≥ 1$ or just $\frac{8abc}{(a+b)(b+c)(c+a)}≥ 1$, but this isn't true.
| By AM-GM
$$\sum_{cyc}\left(\frac{2a}{b+c}\right)^{\frac{2}{3}}=\sum_{cyc}\frac{1}{\sqrt[3]{\left(\frac{b+c}{2a}\right)^2\cdot1}}\geq\sum_{cyc}\frac{1}{\frac{\frac{b+c}{2a}+\frac{b+c}{2a}+1}{3}}=\sum_{cyc}\frac{3a}{a+b+c}=3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know how to find the equation whose roots are diminished by $3$ and they get it $2x^3+19x^2+53x+36=0.$
Hence $(x+1)$ is a common factor of $2x^3+x^2-7x-6=0$ and $2x^3+19x^2+53x+36=0.$And they showed all the roots are: $-1,2,\frac{-3}{2}$Now my Question is "Is there exist any easier way to solve it?" Because in this process i need a lot of work in order to find the new equation and find GCD/HCF of these two equation.Any hints or solution will be appreciated.
Thanks in advance.
| Here is one more approach. First, we do not compute anything, just make sentences. One of the paired roots is $a$, say, the other one is $a\pm 3$, and there is one more root, such that the sum of all three is (Vieta) a rational number, well $-1/2$, but it is not so important, it is rational, this is important. So all three roots can be expressed $\Bbb Q$-linearly (or rather affine) using the one root $a$. The "quadratic" Vieta relation delivers thus a quadratic equation for $a$ with rational coefficients. We use division with rest (algorithm of Euclid, roots also annihilate the rest if not zero) of the given polynomial
$$
P = 2x^3+x^2-7x-6
$$
with this "quadratic equation". If we have some (non zero) rest, it has degree one, so $a$ is rational. If there is no rest, the quotient gives an other rational root.
In both cases, we know now that $P$, polynomial with integer coefficients, has a rational root. The rational roots are (in similar cases) always of the shape
$$
\pm\frac
{\text{divisor of the free coefficient}}
{\text{divisor of the leading coefficient}}\ ,
$$
so in our cases we have to try the values
$$
\pm\frac
{\text{$1$ or $2$ or $3$ or $6$}}
{\text{$1$ or $2$}}\ .
$$
Many values? Not so many if we also use some further information, the continuity of the function. We "calculate" $P(0)=-6$, and thus there is at least one root between $0$ and $\infty$. (It may be rational or not so far. We speculate there is at most $6$ a rational root, so we try to take "the half", we compute $P(2)$ and/or $P(3)$.) Well, i compute $P(3)$ unluckily, get $P(3)= 36>0$, so there is one root between $0$ and $3$. We try all values from the list that match, $1$, $3/2$ and $2$, using best the Horner scheme. OK, $2$ is a root. We use the Horner scheme (below) to also make polynomial division, get $2x^2 + 5*x + 3$, and can use the formula, or also the same (now speculative) search for rational roots. Yes, $-1$ is a root, and the remained one is $-3/2$ (Vieta, product is $3/2$).
The Horner scheme is simple:
$$
\begin{array}{r|rrrr}
& 2 & 1 & -7 & -6\\\hline
2 & 2 &2\cdot 2+1=5 & 2\cdot 5-7=3 & 2\cdot 3-6=\boxed0\\
-1& 2 & (-1)\cdot 2+5 = 3&(-1)\cdot 3+3 =\boxed0 &\\
-3/2 & 2 & \boxed0 &&
\end{array}
$$
Well, usually one sees only
$$
\begin{array}{r|rrrr}
& 2 & 1 & -7 & -6\\\hline
2 & 2 & 5 & 3 & \boxed0\\
-1& 2 & 3&\boxed0 &\\
-3/2 & 2 & \boxed0 &&
\end{array}
$$
instead.
The roots are on the left hand side, the rests w.r.t. the corresponding divisions are boxed.
| {
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"answer_id": 2
} |
Find bases B3 and B2 for $\Bbb R^3$ and $\Bbb R^2$ given a linear transformation and its matrix A linear transformation T is defined by
T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$
Find bases $\mathscr B_3'$ and $\mathscr B_2'$ for $\Bbb R^3$ and $\Bbb R^2$ respectively such that $Mat_{\mathscr B_3' , \mathscr B_2'}$ (T) = $\begin{pmatrix} 1&0&-1\\0&1&2 \end{pmatrix}$
So far I have attempted to answer this by producing an arbitrary basis for $\Bbb R^3$, and then using the definition of the transformation matrix, express it as a linear combination of the basis vectors that will make up the basis for $\Bbb R^2$. If my arbitrary basis for $\Bbb R^3$ is {$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$,$\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}$,$\begin{pmatrix}a_3\\b_3\\c_3\end{pmatrix}$}
then I can simplify this to $(\alpha_2+2\alpha_3) \begin{pmatrix}2a_2+b_2\\b_2+2c_2\end{pmatrix}$ + $(\alpha_1-\alpha_3) \begin{pmatrix}2a_1+b_1\\b_1+2c_1\end{pmatrix}$
I'm not sure what else to try or how to proceed if this is the correct approach
| I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\varepsilon_1 = \begin{pmatrix}1\\0\\0\end{pmatrix},
\varepsilon_2 = \begin{pmatrix}0\\1\\0\end{pmatrix},
\varepsilon_3 = \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix},
\begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$
and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other.
Now the $i$th column of Mat$_{\mathscr B_3', \mathscr B_2'}(T)$ is $[T\varepsilon_i]_{\mathscr B_2'}$ where $[ \cdot ]_{\mathscr B_2'}$ denotes the coordinate matrix with respect to ordered basis $\mathscr B_2'.$ That gives us the following three systems of equations:
$$\begin{align}
\begin{pmatrix} 1\\ 0\end{pmatrix} & = \left[ T \begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 2\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
0 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 2 \\ 0 \end{pmatrix},\\\\
\end{align}$$
$$\begin{align}
\begin{pmatrix} 0\\ 1\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 1\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 1\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
0 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
1 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 1 \\ 1 \end{pmatrix},\\\\
\end{align}$$
$$\begin{align}
\begin{pmatrix} -1\\ 2\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 0\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\
& = \left[ \begin{pmatrix} 0\\ 2\end{pmatrix} \right]_{\mathscr B_2'}\\\\
-1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
2 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & =
\begin{pmatrix} 0 \\ 2 \end{pmatrix}.\end{align}$$
Solving for the four $\beta$s, we get
$$\mathscr B_2' = \left\{
\begin{pmatrix} 2\\ 0\end{pmatrix},
\begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$
As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
complex number with high powers The question:
$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}$
I really do not even know where to begin. Am I suppose to expand $(-1-\sqrt{3}i)$ 73 times?
| Ok, what I got was:
$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$
$|z| = \sqrt{(-1/2)^2+\sqrt{3}/2} = \sqrt{1} = 1$
tan$^{-1}$ ($-.5/ \sqrt{3}/2) = \frac{-2}{3}2\pi$
$z^{73} = 1($cos$(\frac{-2}{3}2\pi)+i$sin$(\frac{-2}{3}2\pi)$)
$z = 1^{73}($cos$(\frac{-2}{3}2\pi*73)+i$sin$(\frac{-2}{3}2\pi8*73)$)
= $-.5 + \frac{\sqrt{3}}{2}$ which is the same answer had I not multiplied by 73
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refering to has marked the answer as $19\pi/24$.
Am I doing something wrong?
| You correctly have
$$
A=B+\frac{\pi}{4}+n\pi
$$
Therefore $A+B=2B+\pi/4+n\pi$. Hence
$$
\cos\left(2B+\frac{\pi}{4}+n\pi\right)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}
$$
Therefore either
$$
2B+\frac{\pi}{4}+n\pi=\frac{\pi}{6}+2m\pi
$$
or
$$
2B+\frac{\pi}{4}+n\pi=-\frac{\pi}{6}+2m\pi
$$
In the first case
$$
2B=-\frac{\pi}{12}+(2m-n)\pi
$$
which is minimal positive for $2m-n=1$, giving
$$
B=\frac{11}{24}\pi
$$
In the second case,
$$
2B=-\frac{5}{12}\pi+(2m-n)\pi
$$
that gives the minimal positive solution
$$
B=\frac{7}{24}\pi
$$
You are right and the book is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} = 1+\frac{1}{2}+...+\frac{1}{n}$
How can I prove that $$\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k} = 1+\frac{1}{2}+...+\frac{1}{n}.$$
I tried an induction but couldn't prove that $$\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n+1}{k} = \frac{1}{n+1}+\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}.$$
Thanks.
| In order to verify the inductive step, recall that $\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$. Therefore
$$\begin{align}\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n+1}{k}&=
\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n}{k-1}+
\sum_{k=1}^{n+1} \frac{(-1)^{k-1}}{k} \binom{n}{k}\\
&=
\frac{1}{n+1}\sum_{k=1}^{n+1} (-1)^{k-1} \binom{n+1}{k}+
\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}\\
&=
\frac{1}{n+1}+
\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}
\end{align}$$
where at the last step we used
$$1=1-(1-1)^{n+1}=\sum_{k=1}^{n+1} (-1)^{k-1} \binom{n+1}{k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit (Squeeze Theorem?) I have to calculate the limit of this formula as $n\to \infty$.
$$a_n = \frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl)$$
I tried the Squeeze Theorem, but I get something like this:
$$\frac{1}{\sqrt{2}}\leftarrow\frac{n}{\sqrt{2n^2}}\le\frac{1}{\sqrt{n}}\bigl(\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}\bigl) \le \frac{n}{\sqrt{n^2+n}}\to1$$
As you can see, the limits of two other sequences aren't the same. Can you give me some hints? Thank you in advance.
| As an alternative by Stolz-Cesaro
$$\frac{b_n}{c_n} = \frac{\frac{1}{\sqrt{n+1}}+\cdots+\frac{1}{\sqrt{2n}}}{\sqrt n}$$
$$\frac{b_{n+1}-b_n}{c_{n+1}-c_n} = \frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt n}$$
and
$$\frac{\frac{1}{\sqrt{2n+2}}+\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt n}\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}=$$
$$\frac{\sqrt{n+1}+\sqrt n}{\sqrt{2n+2}}+\frac{\sqrt{n+1}+\sqrt n}{\sqrt{2n+1}}-\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}}\to\frac4{\sqrt 2}-2=2\sqrt 2-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
| Make the change: $x-2=t^4$. Then:
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}=\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}=\\
\lim_{t\to1}\frac{\sqrt[3]{t^4+7}-2}{t-1}\cdot \frac{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}{\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2}=\\
\lim_{t\to1} \frac{(t^4+7)-2^3}{(t-1)(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\\
\lim_{t\to1} \frac{\require{cancel} \cancel{(t-1)}(t+1)(t^2+1)}{\cancel{(t-1)}(\sqrt[3]{(t^4+7)^2}+2\sqrt[3]{t^4+7}+2^2)}=\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Show that $2 < (1+\frac{1}{n})^{n}< 3$ without using log or binommial coefficient $2 <(1+\frac{1}{n})^{n}< 3$
Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.
| Let $a_n=\left(1+\frac{1}{n}\right)^n$. For any $n>1$ the inequality $a_n>2$ is trivial. We have $a_{n+1}>a_n$ by AM-GM:
$$\sqrt[n+1]{1\cdot a_n} = \text{GM}\left(1,1+\tfrac{1}{n},\ldots,1+\tfrac{1}{n}\right) \stackrel{\text{AM-GM}}{<} \tfrac{1}{n+1}\left[1+n\cdot\left(1+\tfrac{1}{n}\right)\right] = 1+\tfrac{1}{n+1}.$$
Additionally:
$$ \frac{a_{2n}}{a_n} = \left(1+\frac{1}{4n(n+1)}\right)^n \leq \frac{1}{1-\frac{1}{4(n+1)}} = 1+\frac{1}{4n+3}$$
hence for any $N\geq 1$ we have:
$$ a_N \leq a_1 \prod_{k\geq 0}\left(1+\frac{1}{4\cdot 2^k+3}\right)=\frac{16}{7}\prod_{k\geq 1}\left(1+\frac{1}{4\cdot 2^k+3}\right)\leq \frac{16}{7}\prod_{k\geq 1}\frac{1+\frac{1}{2^{k+1}}}{1+\frac{1}{2^{k+2}}}=\frac{20}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & 8 \\
4 & 3 & -9 & 9 \\
2 & 3 & -5 & 7 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17
\end{array}\right]$$
$$\overset{R_4\to R_4-R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & 10 & 8 & -12
\end{array}\right]
\overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -1 & -2 & -4 \\
0 & 0 & 23 & -17
\end{array}\right]
\overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$
$$\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 0 & -7 & -7 \\
0 & 1 & 2 & 4 \\
0 & 0 & 23 & -17 \\
\end{array}\right]
\overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 1 & 2 & 4 \\
0 & 0 & -7 & -7 \\
0 & 0 & 23 & -17 \\
\end{array}\right]$$
However, the answer in the book $(3, 2, 1)$ fits the system.
Was there an arithmetical mistake, or do I misunderstand something fundamentally?
| Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
$$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) \ \ \text{(respectively)}$$
Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
$$\left[
\begin{array}{ccc|c}
1&8&-7&12\\
0&2&-3&1\\
0&-3&1&-5\\
0&-13&9&-17\\
\end{array}
\right] \Rightarrow \left[
\begin{array}{ccc|c}
1&8&-7&12\\
0&-7&0&-14\\
0&-3&1&-5\\
0&14&0&28\\
\end{array}
\right] \stackrel{\frac{-R_2}{7};\\ \frac{-3R_2}{7}+R_3}{=}\Rightarrow \left[
\begin{array}{ccc|c}
1&8&-7&12\\
0&1&0&2\\
0&0&1&1\\
0&14&0&28\\
\end{array}
\right]$$
The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.
Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a$, $b$, $c$. Express
$$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c}$$
as a rational number in lowest terms
I really don't know to start with this question. My solution: I have written equation of solution of triangle in form of $a$, $b$, and $c$; i.e, $s/(s-a)$, where $s = (a+b+c)/2$.
| The main difficulties of this problem is how to handle the triangle inequalities:
$$a+b > c, b+c > a, c+a > b$$
The beauty of Ravi substitution (mentioned by @maveric in comment)
$$a = v + w, b = u + w, c = u + v$$
is under Ravi substitution,
$a, b, c$ satisfies the triangle inequalities if and only if $u,v,w$ are positive numbers.
If one want to run through all integer combinations of $a,b,c$ which give a triangle,
one just run $u,v,w$ through all positive integer combinations
or positive half-integer combinations
$$(u,v,w) = \text{ one of }\begin{cases}
(i+1, j+1, k+1 )\\
\\
(i+\frac12+i,j+\frac12,k+\frac12)
\end{cases}
\quad\text{ for }i,j, k\in \mathbb{N}$$
This transform the horrible sum to one over geometric series.
$$\begin{align}
\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c} =
&\left(\frac{2^2}{15^2} + \frac{2}{15}\right)\sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty \frac{2^{j+k}}{3^{i+k}5^{i+j}}\\
= & \frac{34}{225}\sum_{i=0}^\infty \left(\frac{1}{15}\right)^i \sum_{j=0}^\infty \left(\frac{2}{5}\right)^j \sum_{k=0}^\infty \left(\frac{2}{3}\right)^k\\
= &\frac{34}{225}\left(\frac{1}{1 - \frac{1}{15}}\right)
\left(\frac{1}{1 - \frac{2}{5}}\right)\left(\frac{1}{1 - \frac{2}{3}}\right)\\
= & \frac{17}{21}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have
\begin{align*}
\cos^6x-4\cos^4x+8\cos^2x
&= (1 - y^2)^3 - 4 (1 - y^2)^2 + 8 (1 - y^2) \\
&= (1 - y^2) [(1 - 2y^2 + y^4) + (4y^2 - 4) + 8] \\
&= (1 - y^2) [5 + 2y^2 + y^4] \\
&= (1 - y^2) [5 + 2y^2 + (2y - 1)] \\
&= 2(1 - y^2) [2 + y + y^2] \\
&= 2 [2 + y + y^2 - 2y^2 - y^3 - y^4] \\
&= 2 [2 + y + (-y)(y + y^2 + y^3)] \\
&= 2 [2 + y - y] \\
&= 4.
\end{align*}
Maybe there is some approach that is more straightforward using clever algebraic manipulation. However, this solution is quite clear from a theoretical point of view: you have a polinomial in $y$ that you want to simplify using the relation given. Then, you can divide it by the polynomial given in the relation and the remainder will be a polynomial with degree at most 2. In this specific case, it was the constant polynomial 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
A rank identity Let $A,B$ be $n\times n$ matrix, $A^2=A$, $B^2=B$, how to show $\operatorname{rank}(A-B)=\operatorname{rank}(A-AB)+\operatorname{rank}(B-AB)$.
It seems impossible to reduce $\left(\begin{array}{cc}
A-AB&0\\
0&B-AB
\end{array}\right)$ to be $\left(\begin{array}{cc}
A-B&0\\
0&0
\end{array}\right).$
| The mistake I made is that when doing row operation matrix has to be multiplied on the left, when column operation it's on the right.
It seems like
$$\begin{align*}
\pmatrix{
A-AB & 0\\
0 & B-AB
}&\to\pmatrix{
A-AB & (A-AB)+-(B-AB)\\
0 & B-AB
}\\
&\to\pmatrix{
A-AB & A-B\\
0 & B-AB
}\\
&\to\pmatrix{
A(A-B) & A-B\\
0 & (B-A)B
}\\
&\to\pmatrix{
0 & A-B\\
-A(B-A)B & (B-A)B
}\\
&\to\pmatrix{
0 & A-B\\
-AB+AB & (B-A)B
}\\
&\to\pmatrix{
0 & A-B\\
0 & 0
}
\end{align*}$$
In your question \begin{array} $\mapsto$ \begin{array}{cc} to fix $-AB\mapsto A-AB$, and $-B\mapsto A-B$. I made an edit suggest but it didn't show.
Edit:
As OP pointed out I have made a mistake but let me keep it there. Now let me fix it
$$\begin{align*}
&\to\pmatrix{
A(A-B) & A-B\\
0 & (B-A)B
}\\
&\to\pmatrix{
A(A-B)+(A-B)B & A-B\\
(B-A)B & (B-A)B
}\\
&\to\pmatrix{
A-B & 0\\
(B-A)B & 0\\
}\\
&\to\pmatrix{
A-B & 0\\
-A(A-B)+(B-A)B & 0
}\\
&\to\pmatrix{
A-B & 0\\
B-A & 0
}\\
&\to\pmatrix{
A-B & 0\\
0 & 0
}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the integral $\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}$ diverge Does the integral
$$J:=\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}dx $$
diverge ?
If we integrate by parts we find
$$J=\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)} -\\
\int_{0}^{\infty}\sin{(x^2-x)}
\left( \frac{3x^2}{(1+x^2)(2x-1)}-\\
\frac{2x^4}{(1+x^2)^2(2x-1)}-\frac{2x^3}{(1+x^2)(2x-1)^2}\right)dx$$
If we integrate by parts again, we find that the latter integral reduces to
a sum of absolutely convergent integrals plus
the boundary term
$$ -\lim_{a\rightarrow +\infty}\,
\cos{(a^2-a)}\left(\frac{3a^2}{(1+a^2)(2a-1)^2}
-\\
\frac{2a^4}{(1+a^2)^2(2a-1)^2}-\frac{2a^3}{(1+a^2)(2a-1)^3}
\right)=0.$$
So the question simplifies to the existence of the limit
$\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)}$ which I believe does not exist.
Correct ?
| The integral is divergent because the limit does not exist.
Let $f(a)=\frac{a^3\,\cos{(a^2-a)}}{(1+a^2)(2a-1)}$.
Take the sequence $a_{n}=:\frac{1+\sqrt{1+8\pi n}}{2}$
that diverges to infinity and satisfies $a_{n}^2-a_{n}=2n\pi$.
Then $f(a_{n})\rightarrow \frac{1}{2}$.
Now, take the sequence $b_{n}:=\frac{1+\sqrt{1+4(2n+1)\pi }}{2}$ that also diverges to infinity and satisfies $b_{n}^2-b_{n}=(2n+1)\pi$. Then
$f(b_{n})\rightarrow -\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Recurrence Relation - # of binary strings with given property Let $a_n$ be the number of binary strings of length $n$ with the property that each entry is adjacent to at least one entry of the same type.
ex: $11000111$ is a valid string but $11011000$ is not valid
$\textbf{(a) Find $a_1,a_2,a_3,a_4,a_5,a_6,a_7$}$
If someone can check that my attempt is correct, I would really appreciate it.
$a_1=0$ since we cannot have just $0$ or just $1$ as there will be no adjacent of the same type
$a_2=2$: either $00$ or $11$
$a_3=2$: either $000$ or $111$
$a_4=4$:
Reasoning:
$\textbf{If we start with a $0$}$: For the second entry we have $1$ choice as we are forced to put a $0$ since we started with a $0$. For the third entry we have $2$ choices, and similarly for the fourth entry we have $1$ choice. So there are $2$ such strings.
$\textbf{If we start with a $1$}$: For the second entry we are forced to put a $1$. For the third entry we have $2$ choices, and for the fourth entry we have $1$ choice. So there are $2$ such strings.
So $a_4=2+2=4$ strings.
Following the same method for the remaining:
$a_5=4$
$a_6=8$
$a_7=8$
$\textbf{(b) Find the recurrence relation for $a_n$}$
$$a_n=
\begin{cases}
2a_{n-2}&n \text{ even},\\
a_{n-1}&n \text{ odd}
\end{cases}$$
| Using $z$ for ones and $w$ for zeros we get the generating function
$$F(z, w) = (1+z^2+z^3+\cdots)
\times \sum_{q\ge 0} (w^2+w^3+\cdots)^q (z^2+z^3+\cdots)^q
\\ \times (1+w^2+w^3+\cdots).$$
This is
$$\left(1+\frac{z^2}{1-z}\right)
\times \sum_{q\ge 0} \frac{w^{2q} z^{2q}}{(1-w)^q (1-z)^q}
\\ \times \left(1+\frac{w^2}{1-w}\right).$$
Continuing without the distinction between ones and zeros we get
$$\left(1+\frac{z^2}{1-z}\right)^2
\sum_{q\ge 0} \frac{z^{4q}}{(1-z)^{2q}}
\\ = \left(1+\frac{z^2}{1-z}\right)^2
\frac{1}{1-z^4/(1-z)^2}
\\ = (1-z+z^2)^2
\frac{1}{(1-z)^2-z^4}.$$
The difference of two squares yields
$$(1-z+z^2)^2
\frac{1}{(1-z+z^2)(1-z-z^2)}.$$
which simplifies to
$$\bbox[5px,border:2px solid #00A000]{
G(z) = \frac{1-z+z^2}{1-z-z^2}.}$$
From the coefficients of this OGF we get the sequence
$$1, 0, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754,
\\ 1220, 1974, 3194, 5168, 8362, \ldots$$
which is OEIS A006355 where these data
are confirmed. Now for the coefficients we have
$$[z^0] G(z) (1-z-z^2) = G_0 = [z^0] (1-z+z^2) = 1$$
and hence $G_0 = 1.$ Furthermore
$$[z^1] G(z) (1-z-z^2) = G_1-G_0 = [z^1] (1-z+z^2) = -1$$
so $G_1 = 0.$ Next we find
$$[z^2] G(z) (1-z-z^2) = G_2-G_1-G_0 = [z^2] (1-z+z^2) = 1$$
so $G_2 = 2.$ For $n\ge 3$ we get
$$[z^n] G(z) (1-z-z^2) = G_n - G_{n-1} - G_{n-2}
= [z^n] (1-z+z^2) = 0$$
so that for $n\ge 3$
$$\bbox[5px,border:2px solid #00A000]{
G_n = G_{n-1} + G_{n-2}.}$$
The following Maple code documents the problem definition
that was used.
ENUM :=
proc(n)
option remember;
local ind, d, res, pos;
if n=0 then return 1 fi;
if n=1 then return 0 fi;
if n=2 then return 2 fi;
res := 0;
for ind from 2^n to 2*2^n-1 do
d := convert(ind, base, 2)[1..n];
if d[1] = d[2] and d[n] = d[n-1] then
for pos from 2 to n-1 do
if d[pos-1] <> d[pos] and
d[pos] <> d[pos+1] then
break;
fi;
od;
if pos = n then
res := res + 1;
fi;
fi;
end;
res;
end;
X := n-> coeftayl((1-z+z^2)/(1-z-z^2), z=0, n);
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Additional methods for integral reduction formula So I have successfully found a reduction formula for
$$I_{m,n}=\int\frac{dx}{\sin^m(ax)\cos^n(ax)}$$
Went as follows:
$$\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\int\csc^m(ax)\sec^n(ax)dx=\int\csc^m(ax)\sec^{n-2}(ax)\sec^2(ax)dx\\
\begin{vmatrix}u=\csc^m(ax)\sec^{n-2}(ax)\\du=-am\csc^m(ax)\cot(ax)\sec^{n-2}(ax)+a(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan(ax)dx\end{vmatrix}\\
\begin{vmatrix}dv=\sec^2(ax)\quad v=\frac{1}{a}\tan(ax)\end{vmatrix}\\
\int udv=uv-\int vdu\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan^2(ax)\\
-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^n(ax)\\
-(n-2)\csc^m(ax)\sec^{n-2}(ax)-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)\int\csc^m(ax)\sec^n(ax)dx\\
+(m+n-2)\int\csc^m(ax)\sec^{n-2}(ax)dx\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)I_{m,n}+(m+n-2)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^m(ax)\sec^{n-2}(ax)\tan(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^{m-1}(ax)\sec^{n-1}(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a(n-1)\sin^{m-1}(ax)\cos^{n-1}(ax)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}$$
This is definitely not a check my proof question, but I was wondering if there are more elegant or creative ways of finding this reduction formula, or if IBP, u-sub, and identities are the only ways of going with formulas like these.
| This is a known integral:
$$\frac{\sin ^{1-m}(a x) \sin ^2(a x)^{\frac{m-1}{2}} \cos ^{1-n}(a x) \,
_2F_1\left(\frac{m+1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(a x)\right)}{a (n-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Combinatorics. A box contains 20 balls numbered $1,2,3,...,20$.
A box contains $20$ balls numbered $1,2,3,...,20$. If 3 balls are randomly taken from the box, without replacement, what is probability that one of them is the average of the other two?
Which one of these is the correct answer?
*
*7/45
*3/38
*6/155
*3/154
| Refer to the table:
$$\begin{array}{c|c|c}
\text{Middle (average) number}&\text{Sets}&\text{Number of sets}\\
\hline
2&\{1,2,3\}&1\\
3&\{1,3,5\},\{2,3,4\}&2\\
4&\{1,4,7\},\{2,4,6\},\{3,4,5\}&3\\
\vdots&\vdots&\vdots\\
9&\{1,9,17\},\{2,9,16\},\cdots,\{8,9,10\}&8\\
10&\{1,10,19\},\{2,10,18\},\cdots,\{9,10,11\}&9\\
11&\{2,11,20\},\{3,11,19\},\cdots,\{10,11,12\}&9\\
12&\{4,12,20\},\{5,12,19\},\cdots,\{11,12,13\}&8\\
\vdots&\vdots&\vdots\\
18&\{16,18,20\},\{17,18,19\}&2\\
19&\{18,19,20\}&1\\
\hline
\text{Total}&&2\cdot \frac{(1+9)\cdot 9}{2}=90
\end{array}$$
Hence, the required probability is:
$$\frac{90}{{20\choose 3}}=\frac{90\cdot 1\cdot 2\cdot 3}{20\cdot 19\cdot 18}=\frac{3}{38}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), on the right hand side of the page. Am I incorrect with my use of the modulus operator after I integrated each side then?
$$\begin{align*}x\frac{dy}{dx}&=y-\frac{1}{y}\quad\quad(-1<y<1)\\
\int\frac{1}{y-\frac{1}{y}}\ dy&=\int\frac{1}{x}\ dx\\
\int\frac{y}{y^2-1}\ dy&=\int\frac{1}{x}\ dx\\
\frac{1}{2}\ln{|y^2-1|}&=\ln{|x|}+c\\
\sqrt{|y^2-1|}&=A|x|\ \text{, where $A=e^c$}\\
|y^2-1|&=A^2x^2\\
\implies y^2-1&=A^2x^2&1-y^2&=A^2x^2\\
y^2&=A^2x^2+1&y^2&=1-A^2x^2\\
\\
\therefore y&=\pm\sqrt{Kx^2+1}&y&=\pm\sqrt{1-Kx^2}\ \text{, where $K=A^2$}\\
\end{align*}$$
Thanks :)
| The equation in question is $$xy'(x)=y(x)-\frac1{y(x)},$$ and you are looking for functions $y:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ which satisfy the above equation and the inequality $-1\lt{y(x)}\lt1,$ as well as the implicit $y(x)\neq0.$ Given these restrictions, we have that $$\frac{y(x)}{y(x)^2-1}y'(x)=\frac1{x},$$ which implies $$\ln[1-y(x)^2]=2\ln(-x)+C_-,\,x\lt0$$ $$\ln[1-y(x)^2]=2\ln(x)+C_+,\,x\gt0.$$ This is equivalent to $$1-y(x)^2=\begin{cases}e^{C_-}x^2&x\lt0\\e^{C_+}x^2&x\gt0\end{cases}$$ which is equivalent to $$y(x)^2=\begin{cases}1-A_-x^2&x\lt0\\1-A_+x^2&x\gt0\end{cases}$$ where $A_-=e^{C_-}$ and $A_+=e^{C_+},$ thus $A_-,A_+\gt0.$ However, further restrictions are needed. The latest equation implies that $\forall{x\lt0},$ $0\leq1-A_-x^2$ and $1-A_-x^2\lt1,$ which is equivalent to $A_-x^2\leq1$ and $A_-x^2\gt0,$ but while the latter is trivial, because $A_-\gt0$ and $x^2\geq0,$ we also have that $\forall{x\lt0},\,A_-x^2\leq1,$ which is impossible, becuase this implies $A_-=0,$ due to the Archimedean property of real numbers. A similar case occurs with $A_+.$ Therefore, there is no such a function satisfying the equation everywhere. Instead, we should consider searching for functions $y:(a,0)\cup(0,b)\rightarrow\mathbb{R}$ satisfying the conditions. Thus, $$y(x)^2=\begin{cases}1-A_-x^2&a\lt{x}\lt0\\1-A_+x^2&0\lt{x}\lt{b}\end{cases},$$ so $\forall{a\lt{x}\lt0},$ $0\leq1-A_-x^2$ and $1-A_-x^2\lt1,$ hence $A_-x^2\leq1$ and $A_-x\gt0,$ with the latter being trivial, and the former being equivalent to $x^2\leq\frac1{A_-}.$ Since we know that $a\lt{x}\lt0,$ we have that $a=\frac1{\sqrt{A_-}},$ so $A_-=\frac1{a^2}.$ The analysis for $A_+$ is completely analogous, and results in $A_+=\frac1{b^2}.$
Therefore, if we look for functions $y:(a,0)\cup(0,b)\rightarrow\mathbb{R}$ satisfying the differential equation and the restrictions, then $y$ must also satisfy $$y(x)^2=\begin{cases}1-\frac{x^2}{a^2}&x\in(a,0)\\1-\frac{x^2}{b^2}&x\in(0,b)\end{cases}.$$
This results in two solutions. However, four solutions could result if $y$ was not constrained by the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Locus problem on circle and parabola
Circles are drawn through the vertex of a parabola $y^2 = 4ax$ to cut the parabola orthogonally at the other point. Find the locus of the centers of the circles.
What I did:
The tangent must surely be the diameter of the circle. Also, the line through the vertex perpendicular to the line joining the vertex to the point on parabola must intersect the circle at the other end of the diameter i.e. where the tangent meets the circle again. The center of the circle must be their midpoint
So, I take a parametric point $P(at^2, 2at)$ on the parabola. The equation of the line perpendicular to $VP$ going through $V$ is $2y = -tx$. It meets the tangent $ty = x + at^2$ at $G$ which is the other diametric end. The coordinates of $G$ are $\Large(\frac{-2at^2}{t^2+2}, \frac{at^3}{t^2+2})$.Now the midpoint of $GP$ is the center $L$ which comes out to be,
$$\large x = \frac{at^4}{2(t^2+2)}$$
$$\large y = \frac{3at^3+4at}{2(t^2+2)}$$
Now, I am not able to eliminate $t$ from these equations. How do I proceed?
Note: This problem is from SL Loney Co-ordinate geometry which was first printed a 100 years ago. So, there surely must be a shorter solution.
| We are given
\begin{gather}
\tag{1}\label{eq:1}
x = \frac{at^4}{2(t^2+2)}, \\
\tag{2}\label{eq:2}
y = \frac{3at^3+4at}{2(t^2+2)}.
\end{gather}
Combining \eqref{eq:1} and \eqref{eq:2},
$$
t\frac{y}{a} - \frac{x}{a} = \frac{2t^4 + 4t^2}{2(t^2+2)} = t^2,
$$
whence
\begin{equation}
\tag{3}\label{eq:3}
\left(t^2 + \frac{x}{a}\right)^2 = t^2\frac{y^2}{a^2}.
\end{equation}
Rewriting \eqref{eq:1},
$$
t^4 - 2\frac{x}{a}t^2 - 4\frac{x}{a} = 0,
$$
whence
\begin{equation}
\tag{4}\label{eq:4}
\left(t^2 - \frac{x}{a}\right)^2 = \frac{x^2}{a^2} + 4\frac{x}{a}.
\end{equation}
From \eqref{eq:3} and \eqref{eq:4},
$$
4t^2\frac{x}{a} = t^2\frac{y^2}{a^2} - \frac{x^2}{a^2} - 4\frac{x}{a},
$$
whence
\begin{equation}
\tag{5}\label{eq:5}
t^2\left(\frac{y^2}{a^2} - 4\frac{x}{a}\right) = \frac{x^2}{a^2} + 4\frac{x}{a}.
\end{equation}
With hindsight, this deduction could have been shortened, but I've preferred to keep it in a sequence that occurred fairly naturally. Anyway, we can now substitute \eqref{eq:5} in \eqref{eq:4}. How to do so, while making the least mess, is debatable. I decided it was about time to get rid of the denominators, thus
$$
(at^2 - x)^2 = x(x + 4a) = t^2(y^2 - 4ax),
$$
leading to this manageable mess,
\begin{align*}
4ax(y^2 - 4ax)^2 & = a^2[t^2(y^2 - 4ax)]^2 - 2ax(y^2 - 4ax)[t^2(y^2 - 4ax)] \\
& = a^2[x(x + 4a)]^2 - 2ax(y^2 - 4ax)[x(x + 4a)],
\end{align*}
whence
$$
\boxed{4(y^2 - 4ax)^2 = x(x + 4a)[a(x + 4a) - 2(y^2 - 4ax)]}
$$
Gathering all terms containing $y$ on the left hand side,
$$
4y^2(y^2 - 8ax) + 2x(x + 4a)y^2 = ax(x + 4a)[x + 4a + 8x] - 4(4ax)^2.
$$
Simplifying,
\begin{align*}
2y^2(2y^2 - 16ax + x^2 + 4ax) & = ax[(x + 4a)(9x + 4a) - 64ax] \\
& = ax(9x^2 - 24ax + 16a^2),
\end{align*}
and finally,
$$
\boxed{2y^2(2y^2 + x^2 - 12ax) = ax(3x - 4a)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y=0$$ we get
$$12y+2\lambda+2\mu y=0 \tag{2}$$
For $$L_z=0$$ we get
$$8z+\lambda=0 \tag{3}$$
From $(1)$ and $(2)$ we get
$$x=\frac{4 \lambda}{1-2\mu}$$
$$y=\frac{8 \lambda}{6-\mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 \frac{\lambda^2}{(1-2\mu)^2}+4 \frac{\lambda^2}{(6-\mu)^2}=1 \tag{4}$$
$$\frac{4 \lambda}{1-2\mu}+\frac{16 \lambda}{6-\mu}+\frac{\lambda}{8}=4 \tag{5}$$
But its tedious to solve above equations for $\lambda$ and $\mu$
Any other approach?
| Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$\iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $\ge0$
$$\implies256(y-2)^2\ge20(22y^2-16y+16-u)$$
$$\implies64(y-2)^2\ge5(22y^2-16y+16-u)$$
$$\implies5u\ge46y^2-176y+176=46\left(y-\dfrac{44}{23}\right)^2+176-46\left(\dfrac{44}{23}\right)^2$$
Now $y^2=16-2x^2\le16\implies-4\le y\le4\iff-4-\dfrac{44}{23}\le y-\dfrac{44}{23}\le4-\dfrac{44}{23}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is integration by substitution always a reverse of the chain rule? To integrate $\int x^3\sin(x^2+1)dx$, I took the following approach:
\begin{align*}
\begin{split}
\int x^3\sin(x^2+1)dx&=\int x^3\sin(u)\cdot\frac{1}{2x}du\\
&=\frac{1}{2}\int x^2\sin(u)du\\
&=\frac{1}{2}\int(u-1)\sin(u)du\\
&=\frac{1}{2}\int u\sin(u)-\sin(u)du\\
&=\frac{1}{2}\left(-u\cos(u)-\int -\cos(u)\cdot1du-\int\sin(u)(du)\right)+c\\
&=\frac{1}{2}\left(-u\cos(u)+\sin(u)+\cos(u)\right)+c\\
&=\frac{1}{2}\left((1-u)\cos(u)+\sin(u)\right)+c\\
&=\frac{1}{2}\left((1-(x^2+1))\cos(x^2+1)+\sin(x^2+1)\right)+c\\
&=\frac{1}{2}\left(\sin(x^2+1)-x^2\cos(x^2+1)\right)+c
\end{split}
\begin{split}
u&=x^2+1\\
\frac{du}{dx}&=2x\\
\frac{dx}{du}&=\frac{1}{2x}\\
dx&=\frac{1}{2x}du\\
\\
u&=x^2+1\\
x^2&=u-1
\end{split}
\end{align*}
When integration by substitution was first introduced to me, I was told that it was the inverse of the chain rule, and that it was used in cases where you were integrating a result from the chain rule i.e. a result in the form $g'(x)f'(g(x))$. Clearly, $x^3\sin(x^2+1)$ is not in this form, so it seems that I have used integration by substitution in a case where it is not the reverse of the chain rule.
My question is:
Have I got confused about this, and is there another way of looking at this where it becomes clear that integration by substitution is, in fact, reversing the chain rule in this problem? If I am correct, what is a better explanation of when integration by substitution can be used to solve an integral?
(Note: I suppose one way you could look at it is that $x^3\sin(x^2+1)=x^2\cdot x\sin(x^2+1)$, and that you're partially reversing the chain rule for the $x\sin(x^2+1)$ bit, but you then have to use integration by parts, as the chain rule bit is being multiplied by $x^2$.)
| The integrand is
\begin{align}
x^3 \sin(x^2+1) &= \frac{x^2}{2} \sin(x^2+1)(2x)\\
&= \frac{(x^2 +1 - 1)}{2} \sin(x^2+1) (2x) \\
&= f(x^2+1) (2x)
\end{align}
where $f$ is the function defined by
$$
f(u) = \frac{(u-1)}{2} \sin(u).
$$
If $F$ is an antiderivative of $f$, then $F(x^2+1)$ is an antiderivative of the integrand. (Here we are using the chain rule in reverse.)
So, evaluating this integral has been reduced to the problem of finding an antiderivative of $f$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$
$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$
$\frac{6}{10} + \frac{5}{10}$
$\frac{11}{10}$
| The $\cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\int\frac{2x-3}{(x^2+x+1)^2}dx$ $\int\frac{2x-3}{(x^2+x+1)^2}dx$
$\int\frac{2x-3}{(x^2+x+1)^2}dx=\int\frac{2x+1}{(x^2+x+1)^2}dx-\int\frac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
| Hint:
As $x^2+x+1=\dfrac{(2x+1)^2+3}4,$ set $2x+1=\sqrt3\tan t$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \frac{2}{y^{2}} - \frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$\displaystyle y = -\frac{5}{2} \pm \frac{\sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $\displaystyle y = \frac{58}{4} \pm \frac{10\sqrt{33}}{4}$
But this isn't the correct solution.
Also for $\#10$ I do the same thing:
Let $y = \sqrt n$ then $y^2 = n$
So I have $\displaystyle \frac{3}{y^2} - \frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $\displaystyle y = -\frac{3}{2}, \frac{1}{3}$
Then $n = \frac{9}{4}, \frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1)\ (x-7)^2 -13(x-7) +36=0 \qquad \qquad 4)\ 3(w/6)^2 -8(w/6) +4=0 \\
2)\ (1-3x)^2 -13(1-3x) +36=0 \qquad \qquad 5)\ 3(w^2-2)^2 -8(w^2-2) +4=0 \\
3)\ x^4 -13x^2 +36=0 \qquad \qquad 6)\ \frac{3}{p^2} -\frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
| You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= \frac {58-10 \sqrt {33}}{4}$$ for your first problem and it does work nicely.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$ Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4\implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.
| Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$\begin{cases}\frac12(x_1+x_2)=0\\ \frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0\end{cases} \Rightarrow (x_1,x_2)=(\pm 1,\mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=\sqrt{(-1-1)^2+(-4-4)^2}=\sqrt{68}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and
$$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$
Find the smallest possible value of
$$a^2 + b^2 + c^2+ d^2$$
I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.
Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.
| Let $\lambda = a + b + c + d$. When you sum over the 3 equations.
$$(a+b)(c+d)=143,\quad (a+c)(b+d)=150,\quad (a+d)(b+c)=169\tag{*1}$$
LHS sums to
$$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
while RHS sums to $462$. This leads to
$$a^2 + b^2 + c^2 + d^2 = \lambda^2 - 462$$
To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $\lambda^2$. If you look at LHS
of the 3 equations, all of them is a product of $2$ factors which sum to $\lambda$.
In general, if we have $p + q = \lambda$, then
$$4pq = (p+q)^2 - (p-q)^2 \le (p+q)^2 = \lambda^2$$
The set of 3 equations tell us
$$\begin{align}\lambda^2 &\ge 4\max\{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) \}\\
&= 4\max\{ 143, 150, 169 \}\\&= 676\end{align}$$
As a result,
$$a^2 + b^2 + c^2 + d^2 = \lambda^2 - 462 \ge 214$$
To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $\lambda^2 = 676 = 26^2$.
Flipping all the signs of $a,b,c,d$ if necessary, we can assume $\lambda = 26$.
The third equation $(a+d)(b+c) = 169 = 13^2 = \frac14 (26)^2$ tell us
$a + d = b + c$.
Introduce $u,v$ such that
$$(a,b,c,d) = \left( \frac{13+u}{2}, \frac{13+v}{2}, \frac{13-v}{2}, \frac{13-u}{2}\right)$$
and substitute into the first and second equation and simplify, we obtain
$$\left(\frac{u+v}{2}\right)^2 = 26\quad\text{ and }\quad \left(\frac{u-v}{2}\right)^2 = 19$$
Using this, we find following 4-tuple
$$(a,b,c,d) = {\small \left(
\frac{13+\sqrt{26}+\sqrt{19}}{2},
\frac{13+\sqrt{26}-\sqrt{19}}{2},
\frac{13-\sqrt{26}+\sqrt{19}}{2},
\frac{13-\sqrt{26}-\sqrt{19}}{2}
\right)}$$
is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.
| {
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Modulo and probability How can I prove that 4 modulo 5 is 4?
My though is floor of (4 / 5) is 0 then the remaining is = to the modulo.Am I right?
| You proved it what your proved $\equiv \pmod n$ was an equivalence relationship.
$4 \equiv 4 \pmod 5$ because, being an equivalence relationship, equivalence modulo $n$ is reflexive. i.e. for all $a$, $a \equiv a \pmod n$.
Of course we had to prove equivalence modulo $n$ was an equivalence relationship in the first place.
Definition: for any $n \in \mathbb N$ and $a, b \in \mathbb Z$ we say $a \equiv b \pmod n$ if $n|a-b$ that is if $\frac {a-b}n$ is an integer.
Theorem: $\equiv \pmod n$ is an equivalence relationship. That is to say it is a) reflexive; b) symmetric; c) transitive.
Pf:
a) $\equiv \pmod n$ is reflexive. That is for all $a \in \mathbb Z$, $a \equiv a \pmod n$.
Pf of a) $a-a=0$. And $\frac 05 = 0$ so $5|a-a$ so $a\equiv a \pmod n$.
That's it, we're done. Your question is answered completely.
b) $\equiv \pmod n$ is symmetric. That is if $a \equiv b \pmod n$ then $b \equiv a \pmod n$.
Pf of b) $b-a = -(a-b)$ and $\frac {b-a}n = - \frac {a-b}n$ and if $\frac {a-b}n$ is an intger so is $-\frac {a-b}n$. So if $n|a-b$ then $n|b-a$ and if $a\equiv b\pmod n$ then $b\equiv a \pmod n$.
c) $\equiv \pmod n$ is transitive. That is if $a\equiv b\pmod n$ anc $b\equiv c \pmod n$ then $a \equiv c\pmod n$.
Pf of c) $\frac {a-c}n = \frac {a-b}n + \frac {b-c}n$. If $\frac {a-b}n$ and $\frac {b-c}n$ are integers, so is $\frac {a-c}n$. So if $n|a-b$ and $n|b-c$ then $n|(a-b) + (b-c) = a-c$. So if $a\equiv b\pmod n$ and $b\equiv c \pmod n$ then $a \equiv c \pmod n$.
So $\equiv pmod n$ is an equivalence relation as it is reflexive, symmetric, and transitive.
| {
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Solve differential equation $\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$ I want to solve the following first- order nonlinear ordinary differential equation:
$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
where a,b and c are constants. I rewrote the equation:
$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\
\leftrightarrow \int 1dt=\int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx
$
for some arbitrary number k. How do I solve the last integral? Wolfram-Alpha tells me that it is
$\frac{2tan^{-1}(\frac{-c-b+2cx}{\sqrt{-c^2-b^2-2cb+4ca}})}{\sqrt{-c^2-b^2-2cb+4ca}}$
But I don't know how to calculate that on my own.
| The main step is converting the fraction
$$\frac{1}{cx^2+x(b+c)+a}$$
into the form, expected from the integral tables:
$$\int\frac{1}{t^2+q^2}dt=\frac{1}{q}\arctan \frac{t}{q}$$
You take out the extra $c$, complete the square and change variables:
$$\frac{1}{c}\frac{1}{\color{red}{(x+\frac{b+c}{2c})}^2-(\frac{b+c}{2c})^2+\frac{a}{c}}$$
Now you have $t=x+\frac{b+c}{2c}$ and $q^2=\frac{a}{c}+(\frac{b+c}{2c})^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017166",
"timestamp": "2023-03-29T00:00:00",
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limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6}\sqrt{(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\require{cancel} \cancel{x^3} \sqrt{(9-\frac {5}{x^2})}}{\require{cancel} \cancel{x^3}(1-\frac {2}{x}+\frac{1}{x^3})}} =
\frac {3}{1} = +3$$
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
I'll be glad to get your help!
Thank you.
| You are forgetting that $\sqrt{x^6}=|x^3|$ and $\frac{|x^3|}{x^3}=-1$ when $x$ is negative
| {
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About proof: $\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac x2$ I have the following question:
Prove that: $$ \cot^{-1}\Biggl(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\Biggl) = \frac x2, \ x \in \biggl(0, \frac \pi4\biggl) $$
The solution:
My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ \sqrt{1\pm \sin x} = \sqrt{\sin^2\frac x 2 + \cos^2\frac x 2 \pm2\sin\frac x 2\cos\frac x 2} = \pm \biggl( \cos\frac x 2 \pm \sin\frac x 2 \ \biggl) $$
I've included the $\pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ \cos\frac x 2 - \sin\frac x 2 $) for $ \sqrt{1- \sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ \cot^{-1}\biggl(\tan\frac x 2\biggl) = \cot^{-1}\Biggl(\cot\left(\frac \pi 2 - \frac x 2\right)\Biggl) $$
which yield the result: $$ \frac \pi 2 - \frac x 2 $$
Mathematically, this result is different from that provided in the RHS of question.
Is the question statement wrong or I've been hacked up?
| We see that
$$\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}$$
$$= \sqrt{(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}})^2}$$
$$= \sqrt{\frac{1+\sin x +1-\sin x + 2\cdot\sqrt{1-\sin^2 x}}{1+\sin x +1-\sin x -2\cdot\sqrt{1-\sin^2 x}}}$$
$$= \sqrt{\frac{1+\cos x}{1-\cos x}}$$
$$= \sqrt{\frac{2\cdot\cos^2\frac{x}{2}}{2\cdot\sin^2\frac{x}{2}}}$$
$$= \cot \frac{x}{2}$$
And hence,
$$\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}})$$
$$= \cot^{-1} (\cot \frac{x}{2})$$
$$= \frac{x}{2}$$
Note: as $x\in [0,\frac{\pi}{4}]$, $\sqrt{1-\sin x} \le \sqrt{1+\sin x}$.
And hence
$$0\le\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}$$
| {
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How to prove this algebra question? If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1$$
| hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
| {
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Prove that integral of $ 1-\cos\left(\frac{1}{x^2}\right) $ is finite I need to prove that
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right) dx < \infty $$
My attempt:
$$ \forall x\in[0,\infty] \hspace{1cm} 0 < 1 - \cos\left(\frac{1}{x^2}\right) < 2 \tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - \cos\left(\frac{1}{x^2}\right) \underset{x \to \infty}{\sim} \frac{1}{2x^4} \tag{2} $$
Which means that $ 1 - \cos\left(\frac{1}{x^2}\right)$ behaves like $ \frac{1}{2x^4} $ when $ x \to \infty $.
So I think that, there exist $N \in \mathbb{N} $ such that:
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right)dx < \int_0^{N} 2dx + \int_N^{\infty} \frac{1}{x^4}dx < \infty $$
but I'm not sure. I would appreciate any tips or hints.
| $\int_0^\infty(1-\cos (1/x^2))dx= \int_0^1(1-\cos (1/x^2))dx+\int_1^\infty(1-\cos (1/x^2))dx\leq 2+\int_1^\infty2\sin^2 \left(\frac{1}{2x^2}\right)dx\leq 2+\int_1^\infty2\sin \left(\frac{1}{2x^2}\right)dx\leq2+\int_1^\infty2 \frac{1}{2x^2}dx \text{ (as } \sin x\leq x \text{ forall } x>0) =2+\int_1^\infty \frac{1}{x^2}dx .$
| {
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Trying to prove an equation I would like to receive some help about the next problem.
The problem:
I'm trying to prove the next equation:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = 0 \quad, n = 1, 2, ...$$
My work until now:
$$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = \sum_{k = 0}^{n} \frac{1}{(-1)^{k}k!(n - k)!} =$$
$$= \frac{1}{(-1)^0 \cdot n!} + \frac{1}{(-1)^1 \cdot (n - 1)!} + \frac{1}{(-1)^2 \cdot 2!(n - 2)!} + \frac{1}{(-1)^3 \cdot 3!(n - 3)!} + \cdot \cdot \cdot + \frac{1}{(-1)^{n - 3} \cdot (n - 3)!3!} + \frac{1}{(-1)^{n - 2} \cdot (n - 2)!2!} + \frac{1}{(-1)^{n - 1} \cdot (n - 1)!} + \frac{1}{(-1)^n \cdot n!} \quad (1)$$
Because we have $n + 1$ summands i tried to see what happens when $n$ is even or odd number:
1) $n = 2k - 1$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k$ summands
$$(1) \iff
\frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k - 1}}\right) +
\frac{1}{(2k - 2)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 2}}\right) +
\frac{1}{2!(2k - 3)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 3}}\right) +
\cdot \cdot \cdot +
\frac{1}{(k - 1)!k!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^k}\right) = 0$$
This is equal $0$, because the difference in powers of number $-1$ is odd number, so one of them has to be $1$ and another has to be $-1$.
2) $n = 2k$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k + 1$ summands
$$(1) \iff
\frac{1}{(2k)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k}}\right) +
\frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 1}}\right) +
\frac{1}{2!(2k - 2)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 2}}\right) +
\cdot \cdot \cdot +
\frac{1}{(k - 1)!(k + 1)!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^{k + 1}}\right) +
\frac{1}{(-1)^k \cdot k!k!} \quad \Longrightarrow \quad ?$$
Here, i can't use the same princip as in the first case. Also, the last summand is the $k + 1$ -st summand, he is in the middle of this sum and doesn't have a pair.
My question:
Please, could you help me with some advice or a hint that i could use to finnish my proof of the given equation?
Thank you, for your time and your help!
| $\sum_{k=0}^n \frac{(-1)^{-k}}{k!(n-k)!}=1/(n!)\sum_{k=0}^n \frac{n!(-1)^{k}}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n(-1)^{k}{n\choose k}=\frac{1}{n!}(1-1)^n=0.$
| {
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"answer_id": 0
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For $0 \leq a,b \leq 1$, why does $aFor $0 \leq a,b \leq 1$, why does $a<b$ imply $\frac a{1-a }< \frac b{1-b}$?
Do we need to compute derivatives to validate this inequality? Or talk about increasing/decreasing functions?
| Rearrange the inequality as follows:
\begin{align}&\frac{a}{1-a}<\frac{b}{1-b}\\&\frac{a}{a-1}>\frac{b}{b-1}\\&\frac{a-1+1}{a-1}>\frac{b-1+1}{b-1}\\&1+\frac{1}{a-1}>1+\frac{1}{b-1}\\&\frac{1}{a-1}>\frac{1}{b-1}\end{align}
From $a<b$ it follows that $a-1<b-1<0$ and the inequality is thus satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Expression for $\cos^{-1}x\pm\cos^{-1}y$ As mentioned in Proof for the formula of sum of arcsine functions $\arcsin x+\arcsin y$ for $\sin^{-1}x+\sin^{-1}y$
$$
\sin^{-1}x+\sin^{-1}y=
\begin{cases}
\sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \\\
\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, 0< x,y \le 1\\
-\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, -1< x,y \le 0
\end{cases}
$$
Can we have similar expression for $\cos^{-1}x\pm\cos^{-1}y$ ?
| Alternatively, to get rid of all those cases, we can write:
$$\sin^{-1}x\pm\sin^{-1}y=\text{atan2}(x\sqrt{1-y^2}\pm y\sqrt{1-x^2},\ \sqrt{1-x^2}\sqrt{1-y^2}\mp xy)$$
and:
$$\cos^{-1}x\pm\cos^{-1}y=\text{atan2}(y\sqrt{1-x^2}\pm x\sqrt{1-y^2},\ xy\mp\sqrt{1-x^2}\sqrt{1-y^2})$$
See atan2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I could get help on that part. Here is what I did.
$$a^2+b^2=c^2$$ $$b^2=(c-a)(c+a)$$ Since $a, b, c > 0 \therefore (c+a) \ne 0$ $$\therefore c-a={b^2\over c+a}$$ Similarly we get, $$c-b={a^2\over c+b}$$ $$\therefore {1\over2}(c-a)(c-b)={1\over2}({b^2\over c+a})({a^2\over c+b})$$ $$={(ab)^2\over 2c^2+2ab+2bc+2ca}$$ $$={(ab^2)\over a^2+b^2+c^2+2ab+2bc+2ca}$$ $$={(ab)^2\over (a+b+c)^2}$$ $$=({ab\over a+b+c})^2$$ However, I was unable to prove that ${ab\over a+b+c} \in \mathbb{Z}$ Is there a way to prove it? Thank you
| Multiply by conjugate:
$${ab\over a+b+c}={ab\over a+b+c}\cdot \frac{a+b-c}{a+b-c}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}\in \mathbb Z^+,$$
because:
$$a+b>c$$
and there are two cases for $a^2+b^2=c^2$: 1) $a,b,c$ are even; 2) one is even, the other two are odd. And for each case, $a+b-c$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How to solve for unknown matrix? How can I solve this?
$$ \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
X +
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
X
\begin{bmatrix}
1 & 5 \\
1 & 2 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 1\\
\end{bmatrix}
$$
I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C \implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!
| If we can't see the trick suggested in the comments, by $X=\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}$ we obtain
$$ \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}X= \begin{bmatrix}
a+b & a+2b \\
c+d & c+2d \\
\end{bmatrix}$$
$$
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
X
\begin{bmatrix}
1 & 5 \\
1 & 2 \\
\end{bmatrix}
=$$
$$=\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
\begin{bmatrix}
a+b & 5a+2b\\
c+d & 5c+2d\\
\end{bmatrix}
=\begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\\
\end{bmatrix}$$
then the system
$$\begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\\
\end{bmatrix}= \begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$$\displaystyle =\displaystyle \lim\limits _{x\rightarrow \infty }\dfrac{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }} =$
$\displaystyle =\lim\limits _{x\rightarrow \infty }\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}$
What is the next step should be? Please help!
| Another approach is using substitution $x=\cot t$:
\begin{align}
\lim_{x\to\infty}\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^{4}}\right)
&=\lim_{t\to0}\dfrac{\sqrt{\cos^2t+\cos t}-\sqrt{2\cos^2t}}{\sin t} \\
&=\lim_{t\to0}\dfrac{\cos t(1-\cos t)}{\sin t(\sqrt{\cos^2t+\cos t}+\sqrt{2\cos^2t})} \\
&=\lim_{t\to0}\dfrac{\cos t}{\sqrt{\cos^2t+\cos t}+\sqrt{2\cos^2t}}\dfrac{2\sin^2\frac{t}{2}}{\frac{t^2}{2}}\dfrac{t}{\sin t}\dfrac{t}{2} \\
&=\color{blue}{0}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solutions to $a,\ b,\ c,\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}$ I came across a puzzle in a Maths Calendar I own. Most of them I can do fairly easily, but this one has me stumped, and I was hoping for a hint or solution. The question is:
What are the solutions to
$$\left \{ a,\ b,\ c,\ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a},\ \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \right \} \subset \mathbb{Z}$$
I've tried a few things, but don't think I've made any meaningful progress, besides determining that $a = \pm b = \pm c \ $ are the only obvious possible solutions. My hope is to prove that no other solution can exist.
I don't know if it helps, but I also did a brute force search for coprime numbers $a,b,c$ for which $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \in \mathbb{Z}$, with $1 \leq a \leq b \leq c$, and $a \leq 100, b \leq 1000, c \leq 10000$.
The reason for coprimality is that if a solution has a common factor, we can divide through by the common factor and have another solution that satisfies the conditions.
The triplets I found which satisfy this are:
$(a, b, c) = (1, 1, 1), (1,2,4), (2, 36, 81), (3, 126, 196), (4, 9, 162), (9, 14, 588), (12, 63, 98), (18, 28, 147), (98, 108, 5103)$
None of these except the first satisfy $\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \in \mathbb{Z}$.
| Suppose that $\displaystyle a,b,c,\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \in \mathbb Z$.
Consider polynomial
$$P(x)=\left(x-\frac{a}{b}\right)\left(x-\frac{b}{c}\right)\left(x-\frac{c}{a}\right) = x^3-\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)x^2+\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)x-1.$$
Its coefficients are integers. Since the leading coefficient is $1$, all rational roots of $P$ are integers. Since the constant term is $-1$, it follows that all integer roots of $P$ are $1$ or $-1$ (they must divide the constant term). Since $\dfrac ab, \dfrac bc, \dfrac ca$ are rational roots of $P$, it follows that $\dfrac ab, \dfrac bc, \dfrac ca \in \{-1,1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the secondary equation $x^2 + 2y = 0$ to write $L^2$ as a function of a single variable and then minimizing $L^2$, thus minimizing $L$, I am getting conflicting answers.
Let $D = L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
Approach 1: substituting $y = -\frac{x^2}{2}$ into our primary equation,
$$D = (x-0)^2 + \left(-\frac{x^2}{2}-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dx} = 2x + 2\left(\frac{x^2}{2}-\frac{1}{2}\right)(x)$$
$$\frac{dD}{dx} = x(x^2+1)$$
Therefore, $D$ has a critical point at x = 0. Using the first derivative test, $\frac{dD}{dx} < 0$ for all $x<0$ and $\frac{dD}{dx}>0$ for all $x>0$, therefore a minimum exists at $x = 0$. At $x=0$, $y = 0$.
Approach 2: substituting $x^2 = -2y$ into our primary equation,
$$D = -2y + \left(y-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dy} = -2 + 2\left(y+\frac{1}{2}\right)$$
$$\frac{dD}{dy} = 2y-1$$
Therefore, $D$ has a critical point at $y = \frac{1}{2}$. Using the first derivative test, $\frac{dD}{dy} < 0$ for all $y<\frac{1}{2}$ and $\frac{dD}{dy}>0$ for all $y>\frac{1}{2}$, therefore a minimum exists at $y = \frac{1}{2}$.
Why do the results differ? What faulty assumption(s) have I made? Thanks in advance.
| If $x=2t, y=-2t^2$
We need to minimize $(2t-0)^2+(2t^2+1/2)^2=4t^4+6t^2+1/4=\dfrac{(4t^2+3)^2-8}4\ge\dfrac{0+9-8}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$
$$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x} = \sqrt{3x}$$
$$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$
At this step, I can't continue. Please help me!
| Let's generalise a bit, with parameters say $a, b \in (0, \infty)\setminus \{1\}$ subject to $a^2 \neq b$ and let's try to solve the equation:
$$a^{\log_{b}x+\frac{1}{2}}+a^{\log_{b}x-\frac{1}{2}}=\sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{\log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=b^{\log_{b}a\cdot \log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=x^{\log_{b}a}(\sqrt{a}+\frac{1}{\sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$\frac{(a+1)^2}{a}x^{2\log_{b}a}=x$$
which leads to
$$x^{2\log_{b}a-1}=\frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=\left(\frac{a}{(a+1)^2}\right)^{\frac{1}{2\log_{b}a-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Show that point does not belong to a plane Problem
Show that point $\textbf{q}$ does not belong to plane defined by these 3 points:
$$ \textbf{p}_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \textbf{p}_2=\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \textbf{p}_3 = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} $$
Point $\textbf{q}$ is defined as.
$$ \textbf{q}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$
Attempt to solve
Plane in vector form can be defined as:
$$ \textbf{x} = \textbf{p} + s\textbf{u}+ t\textbf{v}, \{s,t\} \in \mathbb{R} $$
Where vectors $\textbf{u}$ and $\textbf{v} $ define the plane and $\textbf{p}$ is point which belongs to the plane. let $\textbf{p} = \textbf{p}_2$ and vectors $\textbf{u}$ and $\textbf{v}$ can be defined as
$$ \textbf{u} = \textbf{p}_2 - \textbf{p}_1, \textbf{v} = \textbf{p}_2 - \textbf{p}_3 $$
$$ \textbf{u} = \begin{bmatrix} 0 - 2 \\ 1 -1 \\ -1 -1 \end{bmatrix}, \textbf{v} = \begin{bmatrix} 0 - 2 \\ 1 - 4 \\ -1 - 0 \end{bmatrix} $$
$$ \textbf{u} = \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix}, \textbf{v} = \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$
Now we can write the equation as:
$$ \textbf{x}= \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} $$
Now i want to show that $$ \nexists \{s,t\} : \textbf{p} + s\text{u} + t \textbf{v} = \begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} $$
I can form matrix equation:
$$\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$
$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}-\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$
$$ \implies s\begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} + t \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$
$$ \begin{bmatrix} -2 & -2 \\ 0 & -3 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} $$
Now i can solve the equation with row redcution
$$ \begin{bmatrix} -2 & -2 & 1 \\ 0 & -3 & 0 \\ -2 & -1 & 2 \end{bmatrix} $$
Solution is $$ s = -\frac{15}{19}, t = \frac{1}{19} $$
Problem is solution should be impossible. i Was trying to prove poin $\textbf{q}$ doe $\textbf{not}$ belong to plane. Is my approach correct or did something went wrong with my solution ?
| Define ${\bf u} = {\bf p}_2 - {\bf p}_1$ and ${\bf v} = {\bf p}_3 - {\bf p}_1$, a normal vector to the plane is
$$
\hat{\bf n} = \frac{{\bf u}\times {\bf v}}{|{\bf u}\times {\bf v}|} = \frac{1}{\sqrt{19}}\pmatrix{3 \\ -1 \\ -3}
$$
A point ${\bf x}$ belongs to the plane if
$$
\hat{\bf n}\cdot({\bf x} - {\bf p}_1) = 0
$$
If you set ${\bf x} = {\bf q}$ you'll find
$$
\hat{\bf n}\cdot({\bf q} - {\bf p}_1) = -\frac{3}{\sqrt{19}} \not=0
$$
So the point ${\bf q}$ does not belong to the plane
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expressing $\int_0^{\pi/2}\frac{\sin 2x}{x+1}\,dx$ using $A = \int_0^\pi\frac{\cos x}{(x+2)^2}\,dx$ If $$A = \int_0^\pi\frac{\cos x}{(x+2)^2}\,dx,$$
then find the value of $$\int_0^{\pi/2}\frac{\sin 2x}{x+1}\,dx$$ in terms of $A$.
| As pointed out in the comments, the substitution $x \mapsto 2x $ gives
$$ \int_0^{\pi/2} \frac{\sin 2x}{2x+2}2dx = \int_0^\pi \frac{\sin x}{x+2}dx $$
Integration by parts gives
$$ \int_0^\pi \frac{\sin x}{x+2}dx = -\frac{\cos x}{x+2}\Bigg\vert_0^\pi - \int_0^\pi \frac{\cos x}{(x+2)^2}dx = \frac{1}{\pi+2} + \frac12 - A $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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