Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why does $\cos(\pi/2-(\pi/2-x)) = \cos(x)$? In proving $\sin(\pi/2 - x) = \cos(x)$, in my book its given that
$$\sin \left(\frac{\pi}{2} - x\right) = \cos \left( \frac{\pi}{2} - \left(\frac{\pi}{2} - x \right)\right) = \cos x$$
So I understand this but im confused about how $\cos x$ is being obtained from $\cos(\pi/2-(\pi/2-x))$. Please explain it to me.
| \begin{align*}
\sin\left(\frac{\pi}{2}-x\right) &= \left(\sin \frac{\pi}{2}\right)\cos x - \cos \frac{\pi}{2} \sin x \\
&= \cos x &\left(\cos \frac{\pi}{2} = 0 \text{ and } \sin \frac{\pi}{2} =1 \right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| Is this method allowed ?
$$\left[\begin{array}{rr|rr}
3 & 2 & 36 \\
5 & 4 & 64
\end{array}\right]
\sim
\left[\begin{array}{rr|rr}
1 & \frac{2}{3} & 12 \\
5 & 4 & 64
\end{array}\right]
\sim \left[\begin{array}{rr|rr}
1 & \frac{2}{3} & 12 \\
0 & \frac{2}{3} & 4
\end{array}\right] \sim \left[\begin{array}{rr|rr}
1 & 0 & 8 \\
0 & \frac{2}{3} & 4
\end{array}\right] \sim \left[\begin{array}{rr|rr}
1 & 0 & 8 \\
0 & 1 & 6
\end{array}\right]
$$
which yields $x=8$ and $y=6$
The first step is $R_1 \to R_1 \times \frac{1}{3}$
The second step is $R_2 \to R_2 - 5R_1$
The third step is $R_1 \to R_1 -R_2$
The fourth step is $R_2 \to R_2\times \frac{3}{2}$
Here $R_i$ denotes the $i$ -th row.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 6
} |
Using the binomial expansion to approximate a square root.
Find the first four terms of the expansion $\sqrt{1-4x}$ in ascending powers of x.
Hence, approximate $\sqrt{6}$ to four decimal places.
I've expanded it to form this: $1-2x-2x^2-4x^3$
When approximating the value of $\sqrt{6}$, bearing in mind that for the expansion to be valid , $\left|x\right|<\frac{1}{4}$, I took out a value of $\sqrt{10}$ to form $\sqrt{10}\sqrt{0.6}$. Then, letting $x = 0.1$, I substituted $x$ into the expansion, then multiplied by $\sqrt{10}$ to find an approximate value for $\sqrt{6}$. Is there anything invalid about this? Why is it better to take out a factor of $2.5$?
| Here we look at some arguments why taking out a factor $2.5$ is a convenient choice.
Problem: We want to (manually) approximate $\sqrt{6}$ by using the first few terms of the binomial series expansion of
\begin{align*}
\sqrt{1-4x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{4}\\
&= 1-2x-2x^2-4x^3+\cdots\tag{1}
\end{align*}
In order to apply (1) we are looking for a number $y$ with
\begin{align*}
\sqrt{1-4x}&=\sqrt{6y^2}=y\sqrt{6}\tag{2}\\
\color{blue}{\sqrt{6}}&\color{blue}{=\frac{1}{y}\sqrt{1-4x}}
\end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root. We obtain from (2)
\begin{align*}
1-4x&=6y^2\\
4x&=1-6y^2\\
\color{blue}{x}&\color{blue}{=\frac{1}{4}-\frac{3}{2}y^2}\tag{3}
\end{align*}
When looking for a nice $y$ which fulfills (3) there are some aspects to consider:
*
*We have to respect the radius of convergence $|x|<\frac{1}{4}$.
*Since we want to calculate an approximation of $\sqrt{6}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.
*Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{6}$.
We will see it's not hard to find values which have these properties.
We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that
\begin{align*}
\frac{3}{2}y^2
\end{align*}
is close to $\frac{1}{4}$. So, $y^2$ is close to $\frac{1}{4}\cdot\frac{2}{3}=\frac{1}{6}$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $a,b$ so that
\begin{align*}
y^2=\frac{a^2}{b^2}\approx \frac{1}{6}
\end{align*}
This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $6$ we might quickly come to $100$ and $16$. These are two small squares and we have $6\cdot 16=96$ close to $100$. That's all.
Now it's time to harvest. We choose $y^2=\frac{16}{100}=\frac{4}{25}$ resp. $y=\frac{2}{5}$. We obtain for $x$ from (3)
\begin{align*}
x=\frac{1}{4}-\frac{3}{2}y^2=\frac{1}{4}-\frac{3}{2}\cdot\frac{4}{25}=\frac{1}{100}
\end{align*}
We have now an appropriate value $x=\frac{1}{100}$ and we finally get from (1) the approximation:
\begin{align*}
\color{blue}{\sqrt{6}} \approx \frac{5}{2}\left(1-2\cdot 10^{-2}-2\cdot 10^{-4}- 4\cdot 10^{-4}\right)\color{blue}{=2.449\,4}9
\end{align*}
We have $\sqrt{6}=\color{blue}{2.449\,4}89\,7\ldots$ with an approximation error $\approx 2.572\times 10^{-7}$. This result is quite impressive when considering that we have used just four terms of the binomial series.
At this point it might be clear that factoring out $2.5=\frac{5}{2}=\frac{1}{y}$ is a good choice.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim{a_{n}}$ for $a_{n}=((1+\frac{1}{n^{2}})^{n^{2}}(1-\frac{1}{n})^{n}(1+\frac{1}{n}))^{n}$ $$a_{n}=((1+\frac{1}{n^{2}})^{n^{2}}(1-\frac{1}{n})^{n}(1+\frac{1}{n}))^{n}=(e^{\ln((1+\frac{1}{n^{2}})^{n^{2}})+\ln((1-\frac{1}{n})^{n})+\ln((1+\frac{1}{n})^{n})})^{n}$$Then I can use Taylor's theorem - because function $f(x)=e^{x}$ is monotonic I can examine $\ln((1+\frac{1}{n^{2}})^{n^{2}})+\ln((1-\frac{1}{n})^{n})+\ln((1+\frac{1}{n})^{n})$. However I think it is too time-consuming and there is probably a faster way.Have you some ideas?
| You know $$\left(1 + \frac{1}{n}\right)^n \to e$$ as $n \to \infty$. As for
$$\left(1 + \frac{1}{n^2}\right)^{n^3} \left(1 - \frac{1}{n}\right)^{n^2}$$
take logarithms and use $\log(1 + x) = x - x^2/2 + \dots$ as $x \to 0$ to get
$$ n^3 \log\left(1+\frac{1}{n^2}\right) + n^2 \log\left(1 - \frac{1}{n}\right)= n - \dots - n - \frac{1}{2} - \dots \to -\frac{1}{2}$$ as $n \to \infty$.
Thus the limit is $$e^{-1/2} e = \sqrt{e}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proving that $a,b$ are even integers I'm trying to prove the following theorem:
Let $a,b\in\mathbb{Z}$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$
I always struggle to prove some number is odd or even. How to prove it?
I thought of using the $(a+b)^2=a^2+2ab+b^2$ formula but not sure how.
| $$a^{2}+b^{2}\equiv0\pmod 4 \iff(a+b)^2-2ab\equiv 0\pmod{4}.$$
If both $a$ and $b$ are odd, then:
$$\begin{align}a+b\equiv 0\pmod{2} &\Rightarrow (a+b)^2\equiv 0\pmod{4}, \text{but}\\
2ab\equiv 0 \pmod{2} &\Rightarrow \qquad 2ab\not\equiv 0\pmod{4}\end{align}$$
If $a$ is odd and $b$ is even, then:
$$\begin{align}2ab&\equiv 0 \pmod{4}, \text{but}\\
a+b\not\equiv 0\pmod{2} \Rightarrow (a+b)^2&\not\equiv 0\pmod{4}\end{align}$$
Finally, if $a$ and $b$ are even, then:
$$\begin{align}a+b\equiv 0\pmod{2} \Rightarrow (a+b)^2&\equiv 0\pmod 4\\
2ab&\equiv 0\pmod 4\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS = $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When
$LHS_{P}$ =
$RHS_{P}$
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(p+1)\cdot2^{p-1}=p\cdot2^{p}$
Case 3: n = p + 1
$LHS_{P+1}$ = $LHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = (p+1)$\cdot2^{p+1}$
So i need to to prove that:
$RHS_{P+1}$ = $RHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$
Am I thinking right here?
$RHS_{P+1}$ =
$(p+1)2^{p+1}=$
$(p+1)\cdot2^{p}\cdot2$ = ?
(Can I get this equal to $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$ ? )
| $LHS_{p+1} = RHS_{p+1}$
$p\cdot2^{p} + (p+2)\cdot2^{p}=(p+1)\cdot2^{p+1}$
$2^{p}(p+(p+2) = (p+1)2^{p}\cdot2$
$2p+2=2p+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x)))=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{4}(x), r_{4}(x)=o(x)$$
That is why fifth expression of Taylor's is $\frac{2}{15}x^5$ for me. However Mathematica say that: $$\sin \tan x=x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^5)$$ Where have I a mistake?
| The organized and detailed solution must be:
$$\sin y=y-\frac{y^3}{6}+\frac{y^5}{120}+O(y^7);\\
\sin \overbrace{\tan x}^{y}=\overbrace{\tan x}^{y}-\frac{(\overbrace{\tan x}^{y})^3}{6}+\frac{(\overbrace{\tan x}^{y})^5}{120}+O((\overbrace{\tan x}^{y})^7)=\\
\tan x=\color{blue}{x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)};\\
(\tan x)^2=(\tan x)(\tan x)=\left[x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)\right]\left[x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)\right]=\\
x^2+\frac23x^4+O(x^6);\\
(\tan x)^3=(\tan x)^2(\tan x)=\left[x^2+\frac23x^4+O(x^6)\right]\left[x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)\right]=\\
\color{red}{x^3+x^5+O(x^7)};\\
(\tan x)^5=(\tan x)^3(\tan x)^2=\left[x^3+x^5+O(x^7)\right]\left[x^2+\frac23x^4+O(x^6)\right]=\\
\color{green}{x^5+O(x^7)};\\
\sin \tan x=\color{blue}{\left[x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)\right]}-\frac{\color{red}{x^3+x^5+O(x^7)}}{6}+\frac{\color{green}{x^5+O(x^7)}}{120}+O(x^7)=\\
x+\left(\frac{x^3}{3}-\frac{x^3}{6}\right)+\left(\frac{2x^5}{15}-\frac{x^5}{6}+\frac{x^5}{120}\right)+O(x^7)=\\
x+\frac{x^3}{6}-\frac{x^5}{40}+O(x^7).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Representing a function as a power series and finding the coefficients of the terms? My prblem asks me to find the coefficients of the first five terms of the power series representation of the function:
$$f(x)=\frac{10}{17+x}$$
$$f(x)=10*\frac{\frac{1}{17}}{1-\frac{-x}{17}}$$
$$\sum_{n=0}^{\infty}10*\frac{1}{17}*(\frac{-x}{17})^n$$
$$\sum_{n=0}^{\infty}\frac{10}{17}*\frac{(-1)^nx^n}{17^n}$$
First five terms are:
$$\frac{10}{17} + \frac{10}{17}*\frac{-x}{17} + \frac{10}{17}*\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}*\frac{x^4}{17^4}$$
I entered $0$ for the first coefficient. Marked me correct.
I put $\frac{10}{17}$ for the second coefficient. Marked me correct.
I put $\frac{10}{17}$ for the third coefficient. Marked me wrong.
I'm not sure what I'm doing wrong. Looking at the series again, the coefficient should be $-\frac{10}{17}$ for the second coefficient, shouldn't it?
I tried a bunch of stuff for the second coefficient too. $0$, $-1$, $1$, $-10$, $10$, $\frac{10}{17}$, $-\frac{10}{17}$, $\frac{100}{17^2}$, $-\frac{100}{17^2}$, $\frac{10}{17^3}$, $-\frac{10}{17^3}$ (those last few are from different ways of writing the power series I tried). Nothing is working. Where am I going wrong?
Edit: to clarify a bit, the question wants me to find the coefficients of $x^0$, $x$, $x^2$, $x^3$ and $x^4$ in the series.
| The coefficients of the terms of a Maclaurin power series for a function $f(x)$ are numbers $a_0,a_1,a_2,\dots$ such that $$f(x)=\sum_{n=0}^\infty a_nx^n$$ for all $x$ within the interval of convergence of the series. You got very close to finding them, but unfortunately, didn't quite get all the way there, because you didn't get the powers of $x$ separated from the rest. Since
\begin{eqnarray}f(x) &=& \sum_{n=0}^\infty\frac{10}{17}\cdot\frac{(-1)^nx^n}{17^n}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^nx^n}{17\cdot17^n}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^nx^n}{17^{n+1}}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^n,\end{eqnarray} then the coefficients in general have the form $\frac{10\cdot(-1)^n}{17^{n+1}}$ for all $n.$ Hence, the coefficients you're looking for are $$\frac{10}{17},-\frac{10}{17^2},\frac{10}{17^3},-\frac{10}{17^4},\frac{10}{17^5}.$$ I have no idea why $0$ would've been accepted as the first coefficient.
Edit: Thanks to your comment below, I see that the actual function in question is $$g(x)=\frac{10x}{17+x}.$$ Letting $f(x)=\frac{10}{17+x}$ be the initially posted function, we see that $g(x)=x\cdot f(x),$ and so $$\begin{eqnarray}g(x) &=& x\cdot\sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^n\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^{n+1}\\ &=& \sum_{n=1}^\infty\frac{10\cdot(-1)^{n-1}}{17^n}x^n\\ &=& 0x^0+\sum_{n=1}^\infty\frac{10\cdot(-1)^{n-1}}{17^n}x^n,\end{eqnarray}$$ whence the first five coefficients are $$0,\frac{10}{17},-\frac{10}{17^2},\frac{10}{17^3},-\frac{10}{17^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the following manner:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}
This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.
How can I extend the proof for (1) where $r$ is any real number?
When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)
Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}
My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.
| Let $P= \left (1 - \frac 1 n \right )^n.$ Then $\ln P = \frac {\ln \left (1 - \frac 1 n \right )} {\frac 1 n}.\ (*)$
Now as $n \rightarrow \infty$ then $(*)$ is a $\frac 0 0$ form. Applying L'Hospital we have $$\begin{align*} \lim\limits_{n \rightarrow \infty} \ln P & = \lim\limits_{n \rightarrow \infty} \frac {\frac {1} {\left (1 - \frac 1 n \right )} \cdot \frac {1} {n^2}} {-\frac {1} {n^2}}. \\ & = - \lim\limits_{n \rightarrow \infty} \frac {1} {\left (1 - \frac 1 n \right )}. \\ & = -1. \end{align*}$$
This shows that $$\lim\limits_{n \rightarrow \infty} P =\lim\limits_{n \rightarrow \infty} \left (1 - \frac 1 n \right )^n = \frac 1 e.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Does digit $6$ always lead to $\ 25921=161^2\ $? Consider prime numbers with the property that the product of the factorials of the digits plus $1$ is a perfect square, for example the prime $$30241$$ leads to the square $$3!\cdot 0!\cdot 2!\cdot 4!\cdot 1!+1=17^2$$
If the prime number starts with digit $6$, the only square seems to be $\ 25921=161^2\ $ which appears if the prime contains of one digit $6$ , two digits $3$ and the rest digits $0$ or $1$.
Is another square possible, when the prime starts with digit $6$ ? To this, we would need factorials below $10!$ , at least one of them $6!$ , such that their product plus $1$ is a perfect square , other than $6!\cdot 3!\cdot 3!+1$
| A comment on strategy too long for the comment section, but not a complete answer:
You want to find $6!\prod a_i! +1=x^2$ where $a_i$ are single digits. Note that the single digits are $0,1,2,3,2^2,5,2\cdot 3,7,2^3,3^2$, so the factorials will all either be $1$ or will have only the prime factors $2,3,5,7$.
Rearranging, $6!\prod a_i!=x^2-1=(x-1)(x+1)$. LHS is even, so $(x-1)\ \text{and}\ (x+1)$ are both even. Successive even numbers have only a single factor of $2$ in common.
$2^2\cdot3^2\cdot5\cdot\prod a_i!=\frac{(x-1)}{2}\frac{(x+1)}{2}$. RHS are two consecutive numbers; hence their gcd is $1$. So the first problem is to find consecutive numbers which only have prime factors of $2,3,5,7$, and none in common.
$161$ leads to a solution because $\frac{161-1}{2}=80=2^4\cdot 5;\ \frac{161+1}{2}=81=3^4$.
$17$ leads to a solution (although not for the $6$ question) because $\frac{17-1}{2}=8=2^3;\ \frac{17+1}{2}=9=3^2$.
Finding candidate consecutive numbers will not necessarily lead to a solution, as the numbers of various prime factors will have to be distributable among the $a_i!$, which might prove difficult if (for example) the number of factors of $2$ is less than the number of factors of $3$ in the consecutive numbers. If consecutive numbers can be found that satisfy that requirement, then identifying suitable $a_i$ will be possible, and as OP points out, it should then be possible to generate a prime starting number by strategic insertion of digits $0,1$.
| {
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"answer_id": 0
} |
Evaluation of a sum by means of Poisson sum formula and digamma function I have the following series:
$$\sum_{n=-\infty}^{\infty}\frac{1}{(2n+1)^2\pi^2+a^2}=\frac{1}{2a}\tanh\left(\frac{a}{2}\right)$$
and on the text it is written that it can be proven by means of either Poisson sum formula or digamma function. However I didn't manage to do it.
Can anybody help?
Thank you in advance!
| To make it short, start with $$(2n+1)^2\pi^2+a^2=( (2 n+1)\pi -i a)\, ( (2 n+1)\pi +i a)$$ Using partial fraction decomposition
$$\frac 1 {(2n+1)^2\pi^2+a^2}=\frac i {2a}\left(\frac 1{(2 n+1)\pi +i a } -\frac 1{(2 n+1)\pi -i a } \right)$$ make
$$S_m=\sum_{n=-m}^{m}\frac{1}{(2n+1)^2\pi^2+a^2}$$ $$S_m=\frac{2 \pi a+\left(a^2+(2m+1)^2 \pi^2 \right) \left(-i \psi
^{(0)}\left(-\frac{i a}{2 \pi }+m+\frac{1}{2}\right)+i \psi ^{(0)}\left(\frac{i
a}{2 \pi }+m+\frac{1}{2}\right)+\pi \tanh \left(\frac{a}{2}\right)\right)}{2
\pi a \left(a^2+(2 \pi m+\pi )^2\right)}$$
Using asymptotics
$$S_m=\frac{\tanh \left(\frac{a}{2}\right)}{2 a}-\frac{1}{2 \pi ^2 m}+\frac{1}{4 \pi ^2
m^2}+O\left(\frac{1}{m^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve for x: $x^3-\lfloor x\rfloor=5$
Solve for x:$$x^3-\lfloor x\rfloor=5$$
My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here
| Consider the three functions $ y = x-1 $, $ y = x ^ 3-5 $ and $ y = x $. Compare the portion of the graph of $ y = x ^ 3-5 $ that is between the graphs of $ y = x-1 $ and $ y = x $. What can you say about that?
The possible solutions of the equation $x^3-\lfloor x\rfloor=5 $ live in the set of solutions of the inequation $x-1<x^3-5\leq x$.
From the graph we can see that all solutions of the equation $x^3-\lfloor x\rfloor=5 $ are larger than the only real root of $x^3-5= x-1$ and less than the only real root of $x^3-5=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Primality test for numbers of the form $N=k \cdot 3^n-1$ Can you provide proof or counterexample for the claim given below?
Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following claim:
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= k \cdot 3^{n}-1 $ where $k$ is a positive even natural number , $ k<2^n$ and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=S_{i-1}^3-3 S_{i-1}$ with $S_0$ equal to the modular $P_{9k/2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ .
You can run this test here .
I was searching for counterexample using the following PARI/GP code:
CEk31(k1,k2,n1,n2)=
{
forstep(k=k1,k2,[2],
for(n=n1,n2,
if(k<2^n,
N=k*3^n-1;
a=3;
while(!(kronecker(a-2,N)==1 && kronecker(a+2,N)==-1),a++);
S=Mod(2*polchebyshev(9*k/2,1,a/2),N);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==N-2 && !ispseudoprime(N),print("k="k,"n="n)))))
}
| This is a partial answer.
This answer proves that if $N$ is prime, then $S_{n-2}\equiv -2\pmod N$.
Proof :
Let us define $s,t$ as $$s:=\frac{a-\sqrt{a^2-4}}{2},\qquad t:=\frac{a+\sqrt{a^2-4}}{2}$$ where $st=1$.
Here, let us prove by induction on $i$ that
$$S_i\equiv s^{3^{i+2}k/2}+t^{3^{i+2}k/2}\pmod N\tag1$$
We see that $(1)$ holds for $i=0$ since $$S_0\equiv P_{9k/2}(a)=s^{9k/2}+t^{9k/2}\pmod N$$ Supposing that $(1)$ holds for $i$ gives
$$\begin{align}S_{i+1}& \equiv S_i^3-3S_i
\\\\&\equiv (s^{3^{i+2}k/2}+t^{3^{i+2}k/2})^3-3(s^{3^{i+2}k/2}+t^{3^{i+2}k/2})\\\\&\equiv s^{3^{i+3}k/2}+t^{3^{i+3}k/2}\pmod N\quad\square\end{align}$$
Now, using $(1)$ and $k\cdot 3^n=N+1$, we get
$$S_{n-2}\equiv s^{3^nk/2}+t^{3^nk/2}\equiv s^{(N+1)/2}+t^{(N+1)/2}\pmod N$$
Using that
$$\sqrt{\frac{a\pm\sqrt{a^2-4}}{2}}=\frac{\sqrt{a+2}\pm\sqrt{a-2}}{2}$$
we have
$$\begin{align}2^{N+1}S_{n-2}&\equiv (\sqrt{a+2}-\sqrt{a-2})^{N+1}+(\sqrt{a+2}+\sqrt{a-2})^{N+1}
\\\\&\equiv (\sqrt{a+2}-\sqrt{a-2})(\sqrt{a+2}-\sqrt{a-2})^{N}
\\&\qquad\quad +(\sqrt{a+2}+\sqrt{a-2})(\sqrt{a+2}+\sqrt{a-2})^{N}
\\\\&\equiv \sqrt{a+2}\sum_{i=0}^{N}\binom Ni(\sqrt{a+2})^i((-\sqrt{a-2})^{N-i}+(\sqrt{a-2})^{N-i})
\\&\qquad \quad +\sqrt{a-2}\sum_{i=0}^{N}\binom Ni(\sqrt{a+2})^i((\sqrt{a-2})^{N-i}-(-\sqrt{a-2})^{N-i})
\\\\&\equiv \sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(a+2)^{j}\cdot 2(a-2)^{(N-(2j-1))/2}
\\& \qquad \quad +\sum_{j=0}^{(N-1)/2}\binom{N}{2j}(a+2)^{j}\cdot 2(a-2)^{(N-2j+1)/2}\pmod N\end{align}$$
Using that $\binom Nm\equiv 0\pmod N$ for $1\le m\le N-1$, we get
$$4\cdot 2^{N-1}S_{n-2}\equiv 2(a+2)(a+2)^{(N-1)/2}+2(a-2)(a-2)^{(N-1)/2}\pmod N$$
Since
$$2^{N-1}\equiv 1,\qquad (a-2)^{(N-1)/2}\equiv 1,\qquad (a+2)^{(N-1)/2}\equiv -1\pmod N$$
we have
$$4S_{n-2}\equiv -2(a+2)+2(a-2)\equiv -8\pmod N$$
from which
$$S_{n-2}\equiv -2\pmod N$$
follows.$\quad\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
5 digits numbers such that when the sum of digits divided by 4 leaves remainder 2. How many 5 digits numbers such that when the sum of digit divided by 4 leaves remainder 2.
Example:-
Consider a 5 digit number-
$(x1,x2,x3,x4,x5)$
Then $(x1+x2+x3+x4+x5)$ must be of form $(4n+2)$
I tried this
(x+x²+x³...+x^9)(1+x+x²+x³....+x^9)⁴
In this sum of coefficient of x^(2,6,10,14....42)
But this involve lot of calculation.!
Please some one provide me something different and smarter solution.
| Call digits $0$ and $1$ small, and digits $2-9$ large.
Given $n\ge 1$, we divide the $n$-digit numbers into two sets: $S$, which contains all numbers consisting entirely of the small digits $0$ and $1$; and $L$, which contains all other numbers. Note that:
*
*$|S|=2^{n-1}$, because each digit apart from the initial $1$ is either $0$ or $1$;
*$|L|=9\cdot 10^{n-1}-|S|$, because there are $9\cdot 10^{n-1}$ $n$-digit numbers in total.
Now, if $c$ and $d$ are large digits, then the number of numbers whose first large digit is $c$ is equal to the number of numbers whose first large digit is $d$. And because the large digits are evenly distributed modulo $4$, this means that the digit sums of the numbers in $L$ are also evenly distributed modulo $4$. So the number of numbers in $L$ with a given digit sum modulo $4$ is $|L|/4$.
This just leaves $S$. But this is easy: the number of numbers in $S$ with digit sum $k$ is the number of ways of choosing $k-1$ positions for the $1$'s (given that the first digit has to be $1$). This is the binomial coefficient $\binom{n-1}{k-1}$.
Thus we see that the number of $n$-digit numbers with a digit sum equal to $m$ mod $4$ is equal to $|L|/4+\sum_k\binom{n-1}{k-1}$, where the sum is taken over all $k$ with $1\le k\le n$ and $k\equiv m$ mod $4$.
Here is a table of the number of $n$-digit numbers with a given digit sum modulo $4$, for $n=1$ to $6$:
$$
\begin{array}{c|lcr}
n & |L|/4 & 0\bmod 4 & 1\bmod 4 & 2\bmod 4 & 3\bmod 4 \\
\hline
1 & 2 & 2 & 3 & 2 & 2\\
2 & 22 & 22 & 23 & 23 & 22\\
3 & 224 & 224 & 225 &226 &225\\
4 & 2248 & 2249 & 2249 & 2251 & 2251\\
5 & 22496 & 22500 & 22498 & \color{red}{22500} & 22502\\
6 & 224992 & 225002 & 224998 & 224998 & 225002\\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $5^n\mid2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$. Show that $5^n\mid 2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$.
I tried using Euler's Theorem: $2^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ and $3^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ but I can't find a way to use it.
I also tried showing that the sum of the remainders of $2^{5^{n-1}}+3^{5^{n-1}}$ is a multiple of $5^n$ with induction but got stuck again.
How do I prove this kind of statement?
| Hint:
If $a+b=c5^r$ where $a,b,c$ are integers
$$(a+b)^5=a^5+b^5+\binom51ab(a^3+b^3)+\binom52a^2b^2(a+b)$$
$$\implies a^5+b^5=(a+b)^5-5ab(a^3+b^3)-10a^2b^2(a+b)$$ is clearly divisible by $5^{r+1}$
Set $a=2,b=5d-2,$ where integer $d$ is not divisible by $5$ then $r=1$
By induction, we can establish the proposition
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$
Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$
My Attempt
$$
2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5
$$
$$
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-\log_{(2x+3)}(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-1=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)=2\\
3\log_{(5x+4)}(2x+3)-\frac{1}{\log_{(5x+4)}(2x+3)}=2
$$
Set $y=\log_{(5x+4)}(2x+3)$
$$
3y^2-2y-1=0\implies y=\log_{(5x+4)}(2x+3)=1,\frac{-1}{3}
$$
Case 1:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(5x+4)}{\log(2x+3)}=1\implies\log(5x+4)=\log(2x+3) \implies \boxed{x=\frac{-1}{3}}
$$
Case 2:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(2x+3)}{\log(5x+4)}=\frac{-1}{3}\implies \color{red} ?
$$
My reference says $x=\frac{-1}{3}$ is the only solution. How do I prove it ?
Possible Solution
$$
y=\log_{(5x+4)}(2x+3)=\frac{-1}{3}<0\:,\quad x>-4/5=-0.8
$$
Case 1: $5x+4>1\implies x>-3/5=-0.6$
$$
0<2x+3<1\implies -3/2<x<-1\implies-1.5<x<-1\\
\text{Not Possible !}
$$
Case 2: $0<5x+4<1\implies -4/5<x<-3/5\implies-0.8<x<-0.6$
$$
2x+3>1\implies x>-1\\
\implies \boxed{x\in(-0.8,-0.6)}
$$
It seems like there could be a solution for "case 2" in my attempt ?
| One can show why this case yields exactly one solution, as follows.
First, the expression on $\text{LHS}$ of the equation $$\log_{(5x+4)}{(2x+3)}=-\frac 13$$ uniquely defines a real number provided that $1\ne 5x+4>0$ and $2x+3>0.$ Thus, any solution of the equation must also satisfy the conditions $$x>-0.8, x\ne -0.6.$$
Having done that, I now show that this equation has exactly one root satisfying these conditions, thereby confirming the claim.
This equation is, by definition, equivalent to the polynomial equation $$40x^4+212x^3+414x^2+351x+107=p(x)=0.$$ First, note that this has no real roots for $x\ge 0$ since then $p(x)>0.$ Now since $p(-1)<0,$ it follows that there is at least one root in $(-1,0).$ We show that there is in fact a unique root in this interval. To see this, take the first derivative of $p(x),$ which gives $$p'(x)=160x^3+636x^2+828x+351.$$ The second derivative is then given by $$p''(x)=40x^2+106x+69,$$ whose discriminant is $14,$ whence its roots are $-1.5,-1.15,$ both outside of the interval $(-1,0).$ It follows that $p''>0$ in this interval. Therefore, the first derivative $p'(x)$ increases for all $x\in(-1,0).$ This means $p$ is convex in that interval; together with the negativity of $p$ at $-1,$ it implies that $p$ has at most one root in this interval. Finally, I show that this root must lie in $(-0.8,-0.6),$ finishing off the problem.
Now, we have that $p(-0.6)>0.$ Furthermore, we have that $p(-0.8)<0.$ Thus, we have confirmed the claim that the unique root in $(-1,0)$ falls within the subinterval $(-0.8,-0.6),$ which lies within the permitted range $-0.6\ne x>-0.8.$ Since there are no roots for $x\ge -0.6,$ it follows that no other root of $p(x)=0$ falls in the permitted range. This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)$
Denote $a_n=\frac{1}{3}\cdot\frac{2}{5}\cdots\frac{n}{2n+1}\cdot\frac{1}{n+1}$.
Please prove
$$\sum_{n=1}^{\infty}a_n=\frac{\pi^2}{8}-1$$
This is the answer given by my friend. He also used the integral $\int_{-\pi}^{\pi}\cos^kx\,dx$ along with much calculation in which the answer came out almost accidentally.
I want to see other ways to deal with it. Feel free to use the integral above. Thanks in advance.
Edit:This may be a little long
(I)$\int_{0}^{\pi}\sin^{2n+1}xdx=\frac{2^n\dot n!}{(2n+1)!!}$
(II)$\sum_{n=1}^\infty \frac{x^{n+1}}{n+1}=-x-ln(1-x)$
(III)$-\int_{-1}^1\frac{ln(\frac{1+x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-\frac{1-x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-t(1-x^2))}{1-x^2}\big|_{t=0}^{1/2}dx=\int_{-1}^1\int_{0}^{1/2}\frac{dtdx}{1-t(1-x^2)}=\int_{0}^{1/2}\int_{-1}^1\frac{dxdt}{1-t(1-x^2)}=2\int_0^{1/2}\frac{\arctan \sqrt{\frac{t}{1-t}}}{\sqrt{t(1-t)}}dt=4\int_0^{1/2}\arctan \sqrt{\frac{t}{1-t}}d\arctan\sqrt{\frac{t}{1-t}}=4\cdot\frac{1}{2}\cdot(\frac{\pi}{4})^2=\frac{\pi^2}{8}$
(IV)$\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{\int_{0}^{\pi}\sin^{2n+1}xdx}{2^{n+1}}\frac{1}{n+1}=\int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin^{2n+1}x}{2^{n+1}}\frac{dx}{n+1}=\int_{0}^{\pi}\frac{1}{\sin x}\sum_{n=1}^{\infty}(\frac{\sin^{2}x}{2})^{n+1}\frac{dx}{n+1}=\int_0^{\pi}\frac{-\frac{\sin^{2}x}{2}-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{ln(1-\frac{\sin^{2}x}{2})}{\sin^2 x}d\cos x=-1+\int_0^{\pi}\frac{ln(\frac{1+\cos^{2}x}{2})}{1-\cos^2 x}d\cos{x}=-1+\frac{\pi^2}{8}$
(Sorry for ugly typing style.)
| Notice we can decompose $a_n$ as a product of two factors
$$a_n = \frac{1}{n+1}\prod_{k=1}^n \frac{k}{2k+1} = \frac{1}{n+1}\frac{n!}{(2n+1)!!}$$
and both factors can be represented as integrals.
$$\begin{align}
\frac{1}{n+1} &= \int_0^1 x^{n} dx\\
\frac{n!}{(2n+1)!!} &= \frac{2^n n!^2}{(2n+1)!} = \frac{2^n\Gamma(n+1)^2}{\Gamma(2n+2)} = \int_0^1 (2t(1-t))^n dt
\end{align}$$
Using these integral representations, we obtain
$$\mathcal{I} \stackrel{def}{=} \sum_{n=0}^\infty a_n = \int_0^1\int_0^1 \sum_{n=0}^\infty (2xt(1-t))^n dt dx
= \int_0^1\int_0^1 \frac{1}{1-2xt(1-t)} dt dx$$
Change variable to $t = \frac{1+u}{2}$, the integral becomes
$$\begin{align}\mathcal{I}
&= \int_0^1\int_{-1}^1 \frac{1}{2 - x(1-u^2)} du dx
= \int_0^1 \int_0^1 \frac{2}{(2-x) + xu^2} du dx\\
&= \int_0^1 \frac{2}{\sqrt{x(2-x)}}\tan^{-1}\sqrt{\frac{x}{2-x}} dx\end{align}$$
Change variable to $y = 1-x$ and then to $\theta = \cos^{-1}y$, we get
$$\begin{align}\mathcal{I}
&= \int_0^1 \frac{2}{\sqrt{1-y^2}}\tan^{-1}\sqrt{\frac{1-y}{1+y}} dy
= 2\int_0^{\pi/2}\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} d\theta\\
&= 2\int_0^{\pi/2}\tan^{-1}\left(\tan\frac{\theta}{2}\right) d\theta
= \int_0^{\pi/2}\theta d\theta
= \frac{\pi^2}{8}\end{align}$$
As a result, $$\sum_{n=1}^\infty a_n = \sum_{n=0}^\infty a_n - a_0 = \mathcal{I} - 1 = \frac{\pi^2}{8} - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a linear system of reciprocals. Solve for $\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$
I turn the equations into $\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$
Not sure if I am doing fine
| Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=\begin{pmatrix}1&1&1&0\\
4&3&2&5\\
3&2&4&-4\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-3R_1}\longrightarrow\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&-1&1&-4\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow$$
$$\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&0&3&-9\end{pmatrix}$$
Try to finish the exercise now, taking into account that the third column represents $\;\cfrac 1z\;$ , the second $\;\cfrac1y\;$ and the first one $\;\cfrac1x\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$-7\cos(x) + 24\sin(x) = -25\sin(x - 16.26..) Why?$ Im confused as to why the $x$ angle is $-16.26..$ and not $360-16.26$ as well as why its $-25$ and not $25$?
I dont understand why its $-25sin(x-16.26..)$ and not $25sin(x+343.74..)$
Can someone please help me! thank you.
| Here. I found this paper that explains why this method works, but I will try and summarize.
This formula is derived from the trig identity $$\cos(x-\alpha) = \cos(x)\cos(\alpha) + \sin(x)\sin(\alpha)$$
Basically,
$$R\cos(x-\alpha) = R\cos(x)\cos(\alpha) + R\sin(x)\sin(\alpha)$$
This means for an equation $b\sin(x)+a\cos(x)$,
$$a = R\cos(\alpha)$$
and
$$b = R\sin(\alpha)$$
Additionally, $a^2+b^2 = R^2(\cos^2(\alpha) + \sin^2(\alpha)) = R^2$, which means $R = \pm\sqrt{a^2+b^2}$. Moreover, $$\frac{b}{a} = \frac{R\sin(\alpha)}{R\cos(\alpha)} = \tan(\alpha)$$
This means $\alpha = \tan^{-1}(\frac{b}{a})$. Thus, your equation can be simplified.
$$-7\sin(x) + 24\cos(x) = -\sqrt{7^2+24^2}\cos(x-\tan^{-1}(\frac{-24}{7})) = -25\cos(x+1.28700221759\ldots)$$
$\cos(x) = \sin(x + \frac{\pi}{2})$ implies the following.\
$$-7\sin(x) + 24\cos(x) =-25\sin(x+1.28700221759\ldots + \frac{(4n-3)\pi}{2})$$
When $n$ is $-2$ you get the equation you had above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.
The equation $x^3-3x^2-3x-1=0$ has exactly one real solution that can be written in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. What is the value of $a+b+c$?
I've found the value of $x$ to be equal to $3.8473221018630726$ by the method of depressing the cubic. Any hints to proceed are appreciated.
Edit:
Would the fact that $x=\dfrac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\dfrac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$ help somewhere in determining $a, b$ and $c$?
| From the numerical solution (which is pretty "small"), you can try to check a few small values for $a,b,c$... which I did:
$$3.8473221018630726=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$$
EDIT: Actually this had to be solved in the following way (for one real solution):
$$x^3 - 3x^2 - 3x - 1 = 0$$
$$2x^3 - (x+1)^3 = 0$$
$$2x^3 = (x+1)^3$$
$$\sqrt[3]2x = x+1$$
$$x = \frac1{\sqrt[3]2-1}$$
$$x = \frac1{\sqrt[3]2-1}\frac{\sqrt[3]{2^2}+\sqrt[3]{2}+1}{\sqrt[3]{2^2}+\sqrt[3]{2}+1}=\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$a$, $b$ and $c$ are three positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Here's what I did.
We have that
$$\left(\frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2}\right)\left(\frac{1}{b} + \frac{1}{c} + \frac{1}{a}\right) \ge \left(\sqrt{\frac{a}{b^3}} + \sqrt{\frac{b}{c^3}} + \sqrt{\frac{c}{a^3}}\right)^2$$
But because of $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$.
$$\implies \frac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{b}\sqrt{\frac{a}{c^3}} + \frac{1}{c}\sqrt{\frac{b}{a^2}} + \frac{1}{a}\sqrt{\frac{c}{b^2}}\right)$$
And I am stuck, I can't think anymore.
| Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, $u\geq0$, $v\geq0$ and we need to prove that
$$\sum_{cyc}\frac{1}{a}\sum_{cyc}\frac{a}{b^2}\geq3\sum_{cyc}\frac{1}{a^2}$$ or
$$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)\geq0,$$ which is true because
$$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)=4(u^2-uv+v^2)a^5+(5u^3-2u^2v+7uv^2+5v^3)a^4+$$
$$+2(u^4+3u^2v^2+6uv^3+v^4)a^3+uv(u^3+10uv^2+5v^3)a^2+2u^2v^3(u+2v)a+u^3v^3\geq0.$$
| {
"language": "en",
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Determine the nature of the singularity of $\frac{z^{3}}{\cos{(z^{5})} - 1}$ at $0$. How can I find if the singularity of $f(z) = \frac{z^{3}}{\cos{z^{5}} - 1}$ at $0$ is essential, removable, or a pole?
I have considered the the Laurent series, and it appears that there are an infinte number of negative powers, so I was inclined to say that this is an essential singularity. However, the solutions say that $f(z)$ has a pole of order $7$ at $0$, by considering zeros of the numerator and denominator.
I am unsure how zeros of the numerator and denominator are related to poles?
Many thanks
| $$cos (z^5) = 1 - \frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + ....$$
So, $$cos(z^5)-1 = (1 -\frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + .... - 1) = -\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...$$
Now, $$\frac{z^3}{cos(z^5)-1} = \frac{z^3}{-\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...} = \frac{1}{-\frac{z^7}{2!} +\frac{(z^{17})}{4!} - ...} = \frac{1}{(z^7)(-\frac{1}{2!} +\frac{(z^{10})}{4!} - ...)}$$
By having a look at the denominator, we can say that the pole is of order 7 at z = 0
.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ when resolved into partial fractions is equal to? $\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ when resolved into partial fractions is equal to?
I solved it using $\frac{23x - 11x^2}{(2x-1)(9-x^2)} = \frac{A}{2x-1} + \frac{B}{3-x} + \frac{C}{3+x}$, but this is very long method to solve, as after comparing I will get three variables equations. Please tell me any shorter method to do it.
| $$\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{23x-11x^2}{(2x-1)(3-x)(3+x)}$$
The denominator of $A$ vanishes when $x=\frac12$, omit the vanishing term, we have
$$A=\frac{23\left( \frac12\right)-11\left( \frac12\right)^2}{(3-\frac12)(3+\frac12)}$$
The denominator of $B$ vanishes when $x=3$, again, we have
$$B = \frac{23(3)-11(3)^2}{(2(3)-1)(3+3)}$$
Similarly for $C$.
The main idea is
$$23x-11x^2 = A(3-x)(3+x)+B(2x-1)(3+x)+C(2x-1)(3-x)$$
By using the terms that the other two terms vanishes, we can solve a variable easily.
For example, by letting $x=\frac12$, the last two terms on the right vanishes and we are left with
$$23\left( \frac12\right)-11\left( \frac12\right)^2=A\left( 3-\frac12\right) \left( 3+\frac12\right)$$
This is known as Heaviside cover-up method. Refer to case $2$ of the wikipedia if the factors of the denominator include powers of one expression
| {
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"timestamp": "2023-03-29T00:00:00",
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How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Hint
For $(x+2)(x+1)>0$
We need $x>-1$ or $x<-2$
$2x<-4$ or $>-2$
$2x+3<-1$ or $>1$
WLOG $2x+3=\sec t,-\dfrac\pi2<t<\dfrac\pi2$
$2dx=\sec t\tan t\ dt$
$x^2+3x+2=\dfrac{(2x+3)^2-1}4=\dfrac{\tan^2t}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to do it regular way and I looked for the maximum/maximas finding partial derivatives and setting it to zero and I got $$
\begin{cases}
0=-4x^3+-2x+3x^2y-y^3 \\ 0=-4y^3-2y-3y^2x-x^3\\
\end{cases}
$$
easy to see (0,0) is a solution however I do not know how to prove that it is the only one
| Multiply by $4$ and use $(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$:
$$4(x^4+y^4+x^2+y^2+x^3y+y^3x)\geq 0 \iff \\
(x+y)^4+4x^2+4y^2+\underbrace{3(x^4+y^4)-6x^2y^2}_{\ge 0 \ by \ AM-GM}\ge 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving roots will be real If the roots of the equation $x^2+bx+c=0$ are real.
Show that the roots of the equation $x^2+bx+c(x+a)(2x+b)=0$
are again, real for every real number $a.$
| Not true.
Take $c=-1$. Then the discriminant of $x^2+bx+c$ is $b^2 + 4 > 0$ and so the roots are real.
The discriminant of $x^2+bx+c(x+a)(2x+b)$ is $4 a (a - b)$, which is negative if $0 < a < b$.
Therefore, the roots of $x^2+bx+c(x+a)(2x+b)$ are not real when $a=b/2$ and $b>0$, for instance. Indeed, in this case, the discriminant is $-b^2 < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}*\frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
| "The common denominator will be the product of the denominators so $4a^3$" -- this is wrong.
The common denominator is $4a^2$, not $4a^3$. You multiply denominators only when they don't have a common factor; here they do.
| {
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"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
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What is the area of a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$? I found a "fun algebra problem" that asks you to find the area of a triangle whose sides are $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$. After some algebra hell trying to work with Heron's formula, I plugged the question into Wolfram and it spit out 3.5.
Is there some elegant way to reach this? My algebra kungfu has so far been too weak.
| For another interesting approach, consider the Law of Cosines, $a^2 = b^2 + c^2 - 2bc\cos(\alpha)$. If we let $a = \sqrt{13}$, $b = \sqrt{5}$, and $c = \sqrt{10}$, then we find that $13 = 15 - 10\sqrt{2}\cos(\alpha)$, and thus that $\cos(\alpha) = \sqrt{2}/10$. Using $\cos(\alpha)$, we can calculate $\sin(\alpha)$ through some basic trigonometric manipulation to find that $\sin(\alpha) = \sqrt{98}/10$. Using the area formula for triangles $A =\frac{1}{2}bc\sin(\alpha)$, we find that
$$A = \frac{1}{2}bc\sin(\alpha) = \frac{1}{2} \cdot\sqrt{50}\cdot\frac{\sqrt{98}}{10}=\frac{7}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214136",
"timestamp": "2023-03-29T00:00:00",
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Show that $\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$
Show that $$\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$$
So I have attempted this, but have got a different answer so I have obviously done something wrong but i can't spot it. My workings are as shown. Any help will be great.
I started by writing $\cos\theta + \sqrt{3}\sin\theta$ as $2\cos(\theta - \frac \pi 3)$. This means that integral becoems,
$$\int_0^{\pi/3} \frac{1}{2\cos^2(\theta - \frac\pi3)}d\theta = \int_0^{\pi/3} \frac12\sec^2(\theta - \frac\pi3)d\theta = \left[\frac12 \tan(\theta-\frac\pi3)\right]_0^{\pi/3} $$
$$ = \frac12\tan(0) - \frac12\tan(-\frac\pi3) = 0 - \frac12\sqrt3$$
So I get a final answer of $-\sqrt3/2$ any help will be great.
| You just missed a factor of two in the beginning and have a sign error in the last step.
\begin{align*}
\int_0^{\pi/3}\frac{1}{(\cos t +\sqrt{3}\sin t)^2} dt=&\frac 14 \int_0^{\pi/3} \frac{1}{\left(\frac 12 \cos t + \frac{\sqrt{3}}{2} \sin t\right)^2} dt = \frac 14 \int_0^{\pi/3}\frac{1}{\cos^2(t -\frac{\pi}{3})}dt\\
=&
\frac 14 \left[\tan(t-\frac{\pi}{3})\right]_0^{\pi/3} =\frac{1}{4}(0-(-\sqrt{3})) = \frac{\sqrt{3}}{4}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I should show that
$$\frac{x}{\sin{x}} \le \frac{3x}{2} $$
but last inequality is not true...
| The maximum value of the integrand in the given range is when $x=\pi/2$ and the integrand is also $\pi/2$ so we just get that
$$\int_{\pi/6}^{\pi/2}\frac{x}{\sin{(x)}}\mathrm{d}x\le\frac\pi2\left(\frac\pi2-\frac\pi6\right)=\frac{\pi^2}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
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Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$ Let $a,b\in R^+$ such that $ab+bc+ca=1$. Find the maximize of $$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$$
By Wolframalpha i can see that if $a=b=2-\sqrt 3;c=\sqrt 3$ we will have $P=\dfrac 1 4$
WLOG $a\le b$. I proved that $$P=f(a,b,c)\le f(a,a,c)\le 0$$
$$\Leftrightarrow -(a^2-4a+1)^2\le 0$$
My proof is based on the value of $P$ so it's inconvenient if i dont know value of $P$. This inequality is not homogeneous and symmetric so i dont any idead to solve it
Can you help me solve it without using the equality and value of $P$?
| Find $c$:
$$ab+bc+ca=1 \Rightarrow c=\frac{1-ab}{a+b}$$
Sub to $P$:
$$\begin{align}P&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}=\\
&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{(a+b)^2}{(a^2+1)(b^2+1)}=\\
&=\frac{(a+b)(a-1)(b-1)}{(a^2+1)(b^2+1)}\end{align}$$
Now it is symmetric and the maximum is achieved at $a=b$:
$$P(a)=\frac{2a}{a^2+1}-\frac{4a^2}{(a^2+1)^2}\stackrel{x=\frac{2a}{a^2+1}}{=}x-x^2\le \frac14 \Rightarrow x=\frac12=\frac{2a}{a^2+1} \Rightarrow a=2-\sqrt{3}.$$
| {
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Prove that the number $3^{3^n} + 1$ has at least $2n + 1$ prime factors. For any natural $n,$ prove that $3^{3^n} + 1$ has at least $2n + 1$ prime factors.
My idea was to use induction:
*
*for $n = 1$:
$$f(1) = 3^3 + 1 = 28 = 7*2^2$$
*let it be true for $n = k$, then for $n = k + 1$:
$$f(k + 1) = 3^{3^{k + 1}} + 1 = 3^{3*3^k} + 1 = (3^{3^k} + 1)(3^{2*3^k} - 3^{3^k} + 1) = f(k)\times(3^{2*3^k} - 3^{3^k} + 1)$$
Now I have a problem: how to prove that $(3^{2*3^k} - 3^{3^k} + 1)$ is not a prime number?
Or, if it is harder than solving the original problem, please give a hint where I turned the wrong way.
| $$\large(3^{2\times3^k} - 3^{3^k} + 1) = (3^{3^k}-3^{(3^k+1)/2}+1)(3^{3^k}+3^ {(3^k+1)/2}+1)$$
| {
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Proving $\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$
Prove
$$\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$$
I chose to manipulate the left hand side of the equation, by firstly replacing $\cot^2A$ with $\csc^2A-1$ according to the identities. After doing the same with the denominator, I'm left with,
$$\text{RHS}=\frac{1+(\csc^2A-1)\sin^2C}{1+(\csc^2B-1)\sin^2C}
=\frac{1+\csc^2A\sin^2C-\sin^2C}{1+\csc^2B\sin^2C-\sin^2C}$$
And, by contracting $1-\sin^2C$ to $\cos^2C$, on both top and bottom, I can't think of applying anything else.
Anyone know how to continue?
| Convert $tan^2(C)$ to $\frac{sin^2(C)}{cos^2(C)}$
$$\frac{1+\frac{\csc^2A\sin^2C}{cos^2C}}{1+\frac{\csc^2B\sin^2C}{cos^2C}}$$
Multiply all by $cos^2(C)$
$$\frac{cos^2C+\csc^2A\sin^2C}{cos^2C+\csc^2B\sin^2C}$$
Convert $cos^2C$ to $1-sin^2C$
$$\frac{1-sin^2C+\csc^2A\sin^2C}{1-sin^2C+\csc^2B\sin^2C}$$
Rearrange
$$\frac{1+\csc^2A\sin^2C-sin^2C}{1+\csc^2B\sin^2C-sin^2C}$$
Factor out $sin^2C$
$$\frac{1+sin^2C(csc^2A-1)}{1+sin^2C(csc^2B-1)}$$
Use $csc^2x=cot^2x+1$
$$\frac{1+sin^2C(cot^2A+1-1)}{1+sin^2C(cot^2B+1-1)}$$
Simplify to RHS
$$\frac{1+sin^2Ccot^2A}{1+sin^2Ccot^2B}$$
| {
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Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\int_{0}^{\pi}\frac{1}{1 + \frac{1}{\sec^2x}}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{\sec^2x + 1}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{2 + \tan^2x}dx
$$
Let $\tan x = u$, then $du = \sec^2x dx$
Then,
$$\int \frac{\sec^2x}{2+\tan^2x}dx = \int \frac{du}{2 + u^2} = \frac{1}{\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}+c = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\tan x}{\sqrt{2}}+c
$$
Therefore,
$$
2I = \frac{\pi}{\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\
\Rightarrow I = \frac{\pi}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\= \frac{π}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan \pi}{\sqrt{2}} - \tan^{-1}\frac{\tan 0}{\sqrt{2}}\right]\\=\frac{\pi}{2\sqrt{2}}[\tan^{-1}0 - \tan^{-1}0] \\= 0\\$$
But the answer given in the book is $\frac{\pi^2}{2\sqrt{2}}$
What am I doing wrong ?
| As you have found ever the dicontinuing of $\tan$ is the problem, so you can solve it with your method after this substitution
$$\int_{0}^{\pi}\dfrac{1}{1+\cos^2x}\ dx=2\int_{0}^{\pi/2}\dfrac{1}{1+\sin^2t}\ dt= \color{blue}{\dfrac{\pi}{\sqrt{2}}}$$
where we use the substitution $x=t+\dfrac{\pi}{2}$. Other way is complex integration, with $2x=t$ we have
$$
\begin{align}
\int_{0}^{\pi}\dfrac{1}{1+\cos^2x}\ dx&=
\int_{0}^{2\pi}\dfrac{1}{3+\cos t}\ dt \\
&=\int_{|z|=1}\dfrac{1}{3+\frac12(z+\frac1z)}\dfrac{dz}{iz} \\
&= -2i\int_{|z|=1}\dfrac{1}{(z+3-2\sqrt{2})(z+3+2\sqrt{2})}dz \\
&= \color{blue}{\dfrac{\pi}{\sqrt{2}}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to determine the base of $\ker\phi$ for polynomial function? Given is a base defined as $$B:=(x\mapsto1,x\mapsto x,x\mapsto x^2,x\mapsto x^3 ,x\mapsto x^4)$$ A set V defined as
$$V:= \{ f: \mathbb{R} \mapsto \mathbb{R}\ |\ \exists\ {a_0},...{a_4} \in \mathbb{R}\ : f(x)=\sum_{i=0}^{4}{a_ix^i} \ \forall \ x \in \mathbb{R}\}$$
a function $\phi$ defined as $$\phi(f)(x)=f''(x)-x \cdot f'(x) + f(x-1)$$
I determined the images of $\phi$ regarding the elements in the base B:
$\phi(1)=1$
$\phi(x)=-1$
$\phi(x^2)=-x^2-2x+3$
$\phi(x^3)=-2x^3-3x^2+9x-1$
$\phi(x^4)=-3x^4-4x^3+18x^2-4x+1$
I also calculated the following transformation matrix:
$M_B^B(\phi)=\begin{pmatrix} 1 & -1 & 3 & -1 & 1& \\ 0 & 0 & -2 & 9 & -4&\\ 0 & 0 & -1 & -3 & 18& \\ 0 & 0 & 0 & -2 & -4& \\ 0 & 0 & 0 & 0 & -3& \end{pmatrix}$
From this point on I don't know how to determine the base of $\ker\phi$. I know the definition of $\ker\phi$ is $\ker\phi:=\{v \in V:\phi(v)=0\}.$ However I do not know how to apply this definition to my problem.
| You can ease the computation of the associated matrix by building the matrices of $f(x)\mapsto f''(x)$, $f\mapsto xf'(x)$ and $f(x)\mapsto f(x-1)$ so
$$
\begin{pmatrix}
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 6 & 0 \\
0 & 0 & 0 & 0 & 12 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
-
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 4
\end{pmatrix}
+
\begin{pmatrix}
1 & -1 & 1 & -1 & 1 \\
0 & 1 & -2 & 3 & -4 \\
0 & 0 & 1 & -3 & 6 \\
0 & 0 & 0 & 1 & -4 \\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$
and find
\begin{pmatrix}
1 & -1 & 3 & -1 & 1\\
0 & 0 & -2 & 9 & -4\\
0 & 0 & -1 & -3 & 18\\
0 & 0 & 0 & -2 & -4\\
0 & 0 & 0 & 0 & -3\\
\end{pmatrix}
The RREF is
\begin{pmatrix}
1 & -1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
and a basis of the null space consists of the single vector
\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
The polynomial that has this vector as its coordinate vector is
$$
f(x)=1+x
$$
So a basis of the null space of $\phi$ consists of $1+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$ .
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$
I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
| We also know a well-known indentify as follow for $\theta= \dfrac{\pi}{7}$
$$1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta= 0$$
On the other hand, we have $7- (\!4\sin 2\theta- \tan \theta\!)^{2}= (\!\sec \theta\!)^{2}(\!1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta\!)= 0$
Furthermore $4\sin 2\theta- \tan \theta= \tan \theta (4\cos 2\theta+ 3)> 0\,\therefore\,4\sin 2\theta- \tan \theta= \sqrt{7}$ for $\theta= \dfrac{\pi}{7}$ (OP) .
q.e.d
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Linear dependence of 3 vectors in $\mathbb{R}^4$ Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are linearly dependent on $\mathbb{R}^4$? I know that the usual method of solving this is to show that there exists scalars $x_1,x_2,x_3$ not all zero such that
\begin{align}
x_1\vec{v_1}+x_2\vec{v_2}+x_3\vec{v_3}=\vec{0}.
\end{align}
Doing this would naturally lead us to the augmented matrix
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
1& b & 1 &0\\
-2 & 2 & c &0\\
\end{pmatrix}
Doing some row reduction would lead us to the matrix
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
I'm not quite sure how to proceed after this. Do I take cases on when whether $b+1$ or $c+2$ are zero and nonzero?
| Just build a matrix with these vectors as rows and perform row reduce. The vectors will be linearly dependent if at least one row is made of zeros. The idea is that the rank of a matrix is the maximum number of linearly independent rows (or columns), hence, the rows will be linearly dependent if and only if $r(A) < 3$.
$$
\begin{pmatrix} 1 & 4 & 1& -2 \\ -1 & a & b & 2\\ 1 & 1 & 1 & c\end{pmatrix}\to
\begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & a+4 & b+1 & 0\\ 0 & -3 & 0 & c+2\end{pmatrix}\to
\begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & -3 & 0 & c+2\\ 0 & a+4 & b+1 & 0\end{pmatrix} \to
$$
$$
\begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & -3 & 0 & c+2\\ 0 & 0 & b+1 & \frac{(c+2)(a+4)}{3}\end{pmatrix}
$$
So the vectors are linearly dependent if and only if the last row is filled with zeros, i.e. $b = -1 \wedge (a=-4 \vee c=-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show power series converges for every $x$.
Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$
Now I need to explain that the power series that define $f(x)$ converges for every $x$.
This problem is introduced by two smaller proofs which I will give (or my attempts for them) so the track of thought is clear.
First I found a simple expression for $\frac{a_{2n}}{a{2n-2}}$ for every $n \in \mathbb{N}$. Since this fraction is $\frac{1}{2}$ when $n=1$, it is $\frac{1}{4}$ when $n=2$, it is $\frac{1}{6}$ when $n=3$ and so on, I let
$$\frac{a_{2n}}{a_{2n-2}}=\frac{1}{2n}.$$
Next I had to show that for every fixed $x \in \mathbb{R}$ there exists a $N \in \mathbb{B}$ such that $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ whenever $n \geq N$. This is my attempt:
Let $x \in \mathbb{R}$ be given and let $N > x^{2}$, then for all $n \geq N : \frac{1}{n} < \frac{1}{x^{2}}.$ Then
$$\frac{|a_{2n}|}{|a_{2n-2}|}=\frac{1}{2} \frac{1}{|n|} < \frac{1}{2} \frac{1}{|x^{2}|}=\frac{1}{2} \left| \frac{x^{2n-2}}{x^{2n}} \right|\\
\Longrightarrow\\
|a_{2n}||x^{2n}| < \frac{1}{2} |x^{2n-2}||a_{2n-2}|.
$$
I expect to have to use $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ to show the convergence, but I have no idea how. Any suggestions?
| With $a_{2n}= \frac{1}{2}a_{2n-2}$ you can write your series as the sum of two geometric series. First take n even. Define $b_n= a_{2n}$. Then $b_{n}= \frac{1}{2}b_{n-1}$ so this is a geometric series: $b_0= a_0$, $b_1= a_2= \frac{1}{2}a_0$, $b_2= a_4= \frac{1}{2}a_2= \left(\frac{1}{2}\right)^2a_0$, etc.
And similarly for n odd. Let $c_n= a_{2n+1}$ so that $c_0= a_1$. $c_1= a_3= \frac{1}{2}a_1$, $c_2= a_5= \frac{1}{2}a_3= \left(\frac{1}{2}\right)^2a_1$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.
| Probably not more efficient than straightforwardly factoring out, but it works as well:
$$\begin{eqnarray}\frac{(x+2)(x−3)(x−1)+4}{x-2} & = & \frac{(x-2+4)(x−3)(x−1)}{x-2}+\frac{4}{x-2} \\
& = & (1+\frac{4}{x-2})(x−3)(x−1)+\frac{4}{x-2} \\
& = & (x−3)(x−1) + \frac{4(x−3)(x−2+1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3)(1+\frac{1}{x-2})+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + \frac{4(x−2-1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + 4\\
& = & (x−3)(x+3) + 4\\
& = & x^2 - 5 \; .
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \{2,4\}, & \{2,5\}, & \{2,6\},\\
& & \{3,3\}, & \{3,4\}, & \{3,5\}, & \{3,6\},\\
& & & \{4,4\}, & \{4,5\}, & \{4,6\},\\
& & & & \{5,5\}, & \{5,6\},\\
& & & & & \{6,6\}
\end{array}
but when I roll two different dice, then the probability of getting sum of 7 is $1/6$
\begin{array}{c c c c c c}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array}
How this is possible?
| The second situation is easiest, so let's start there. In the second situation, you throw two six-sided dice. There are 6*6=36 different outcomes (for example [1,6] and [6,1] are two different outcomes), and six of them have a sum of 7. So 6/36 = 1/6 outcomes have a sum of 7, as you said.
In the first situation, the situation is unusual. Instead of picking out of the 6*6=36 different outcomes, you're supposed to pick from the set of outcomes where you can't tell the difference between the two dice. That is, [1,6] and [6,1] are considered to be the same event.
There are 21 different pairings in this case (not 23?): {1,1} ... {1,6}, {2,2}, ..., {2,6}, {3,3},...{3,6}, ..., ..., {5,6}, {6,6}. In the physically realistic case, the probability of each pairing is not the same. For example, when rolling dice it's twice as likely to get {2,4} than to get {3,3} because there are two of the 36 outcomes yield {2,4} and only one yields {3,3}. In a physically unrealistic case, you can imagine a model where you choose one of these pairings uniformly at random. There are only three of those {1,6}, {2,5}, {3,4}, so the chances of getting a sum of 7 are 3/21 = 1/7 in this physically unrealistic case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$ if $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$
My try
Since this is a factor I tired finding A, B, C & D such that,
$(x^2+kx+1)(Ax^3+Bx^2+cx+D)$ =$px^5+qx^2+r$
if I can find k in terms of p, q & r I can try to get the given condition, But when I tried this equating coefficients, I ended up with quadratic expression for k? Is there a better way to approach this question? Please Help! thanks
| $\mathrm{px^5+qx^2+r\\if~x^2+kx+1~is ~a ~factor\\ [x^2+kx+1][Ax^3+Bx^2+Cx+D]\equiv px^5+qx^2+r\\
by~expansion:\\x^5:Ax^5=px^5 ~\underbrace{A=p} \cdots(i) \\x^4: Bx^4+kAx^4=0,~\underbrace{B=-kA} \cdots(ii) \\x^3: Cx^3+kBx^3+Ax^3=0; \underbrace{C+kB+A=0} \cdots(iii) \\x^2:~Dx^2+kCx^2+Bx^2=qx^2;~\underbrace{D+kC+B=q} \cdots(iv)\\x:kDx+Cx=0;\underbrace{kD+C=0} \cdots(v)\\\underbrace{D=r} \cdots(vi) \\A=p, B=-kp, C=-kD, D=r~~\Biggr |C=-kr ~considering ~eqn(iii)~and~eqn(iv)~respectively \\kB=-A-C\cdots*;~kC=q+D-B\cdots**\\divide~ eqn(*)~ by~(**)\\\frac{kB} {kC} =\frac{-A-C} {q-D-B};~~\frac{B} {C} =\frac{A+C} {D+B-q};\frac{-kp} {-kr} =\frac{p-kr} {r-kp-q} ; \frac{p} {r}=\frac{p-kr} {r-kp-q}\\p(r-kp-q) =r(p-kr) \\pr-kp^2-pq=rp-kr^2\Biggr|k=\frac{pr-rp-pq}{p^2-r^2}\\(x^2+kx+1)(Ax^3+Bx^2+Cx+D)=\Biggr[x^2+\frac{pr-rp-rq}{p^2-r^2}x+1\Biggr] \cdots\\
} $
$substituting~the~value~of~k~back$
$\\\Biggr\{x^2+\frac{pq}{r^2-p^2}x+1\Biggr\}\Biggr\{px^3-\frac{p^2q}{r^2-p^2}x^2-\frac{pqr}{r^2-p^2}x+r\Biggr\} \equiv{px^5+qx^2+r} \\consider~x^3:\\\frac{-pqr}{r^2-p^2}+\frac{pq}{r^2-p^2}\biggr(\frac{-p^2q}{r^2-p^2}\biggr)+p=0\\-pqr+\frac{pq(-p^2q)}{r^2-p^2}
+p(r^2-p^2)=0$
$-p^2q^2+(r^2-p^2)^2=qr(r^2-p^2)\\(r^2-p^2)^2 -qr (r^2-p^2)=p^2 q^2 \\ (r^2-p^2)(r^2-p^2-qr)= p^2q^2 \\\underbrace{p^2q^2=(p^2-r^2)(p^2-r^2+qr)} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \sqrt{b^2 - 1})$$
$$e^a = b + \sqrt{b^2 - 1}$$
$${1 \over e^a} = {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = b + \sqrt{b^2 - 1} + {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b(b + \sqrt{b^2 - 1}) + \sqrt{b^2 - 1}(b + \sqrt{b^2 - 1}) + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b^2 + b\sqrt{b^2 - 1} + b\sqrt{b^2 - 1} + b^2 - 1 + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {2b^2 + 2b\sqrt{b^2 - 1} \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = 2b$$
This is all well and good, but for the life of me I can't see: how would someone ever know in advance that adding $b + \sqrt{b^2 - 1}$ to its own reciprocal would clean things up so nicely? Is this just a "special case" which needs to be memorized, or is there some general principle behind it which can be used when solving similar equations?
| Let $$\cosh(a)=b$$ and solve for $a$
$$e^a+e^{-a}=2b\implies e^{2a}-2be^a+1=0$$
Solve for $e^a$ with quadratic formula $$ e^a= b\pm \sqrt {b^2-1}$$
Choose the plus sign and take logarithm and you have the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$
The equation looks very simple. But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^2-b^2)-x^2=0\Longrightarrow (x^2-b^2)(x-a^2x+a-1)=0$$
We have $x(1-a^2)=1-a$ and $x≠±b$.
1) $a=1$, $x\in (-\infty;b) ∪(-b;b)∪(b;\infty)$
2) $a=-1$ then $x\in\emptyset$
3) $a≠±1$ and $b≠±\frac{1}{a+1}$ ,then $x=\frac{1}{a+1}$
What I want to know, is there a mistake in the solution, or is there any other point I missed?
Thank you.
| Note that
$$\frac{b^2}{b^2-x^2}=-\frac{b^2}{x^2-b^2},$$
so, the equation becomes
$$x-a^2x+\frac{b^2}{x^2-b^2}+a=\frac{x^2}{x^2-b^2}$$
Passing $\frac{b^2}{x^2-b^2}$ to the RHS you obtain
$$x-a^2x+a=1.$$
This equation becomes $x(1-a^2)+a-1=0$.
The LHS becomes, $x(1-a)(1+a)-(1-a)=(1-a)\Big(x(1+a)-1\Big).$
So, if $a=1$, you have $x\in(-\infty,-b)\cup(-b,b)\cup(b,+\infty)$.
If $a=-1$ the equation has no solutions (it becomes 2=0).
If $a\neq-1$ and $a\neq1$, then $x=\frac{1}{1+a}.$
Now, in those solutions, we need to exclude when $x=b$ or $x=-b$. If $a=1$ we have already done that, so we just need to update it when $a\neq1$. In those cases, $x=\frac{1}{1+a}$, so if $\frac{1}{1+a}=\pm b$ the equation has no solutions.
To sum up,
If $a=-1$ the equation has no solutions.
If $a=1$,
$x\in(-\infty,-b)\cup(-b,b)\cup(b,+\infty)$.
If $a\neq1$ and $\frac{1}{1+a}=\pm b$ the equation has no solutions.
Otherwise, $x=\frac{1}{1+a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$ $f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.
Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^n$, but there is a problem that $\frac{1}{1 - (-x)} = \sum_{n=0}^\infty (-1)^nx^n$ is convergent only for $|x| < 1$, so not for $1$. How can I go about this?
| How about:
$$f(x) = (1+x)^{-1}+2(2-x)^{-1}$$
$$f'(x) = -(1+x)^{-2}+2(2-x)^{-2}$$
$$f^{(2)}(x) = 2(1+x)^{-3}+4(2-x)^{-3}$$
.
.
.
$$f^{(9)}(x) = -9!(1+x)^{-10}+2\cdot 9!(2-x)^{-10}$$
Now, plug in 1:
$$f^{(9)}(1) = 2\cdot 9!-\dfrac{9!}{2^{10}}$$
You may want to check I did not miss a negative sign, but this should be correct and does not require geometric sums that have limited intervals of convergence.
| {
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"url": "https://math.stackexchange.com/questions/3241008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$ a) Give a combinatorial proof that for every $n \geq r \geq 1$ that:
$$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$
And use (a) to concoct a formula for $1^2 + 2^2 + ... + n^2$
| The starting-point is $\binom{i}{j}+\binom{i}{j+1}=\binom{i+1}{j+1}$, of which the combinatorial proof is famous. We can then telescope:$$\sum_{i=0}^n\binom{i}{r-1}=\sum_{i=0}^n\left(\binom{i+1}{r}-\binom{i}{r}\right)=\binom{n+1}{r}.$$The case $r=3$ gives $$\frac{n(n+1)(n-1)}{6}=\binom{n+1}{3}=\sum_{i=0}^n\binom{i}{2}=\sum_{i=0}^n\frac{i(i-1)}{2}=\frac{\sum_{i=1}^ni^2-\sum_{i=1}^ni}{2}.$$Hence$$\sum_{i=1}^ni^2=\sum_{i=1}^ni+\frac{n(n+1)(n-1)}{3}=\frac{n(n+1)}{2}+\frac{n(n+1)(n-1)}{3}=\frac{n(n+1)(2n+1)}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$ Prove
$$0.9999^{\!101}<0.99<0.9999^{\!100}$$
I think its original idea is
$$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$
For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.
Furthermore $0.99^{\!-99}\times0.99^{\!100}=0.99<1.01^{\!100}\times0.99^{\!100}=0.9999^{\!100}$.
For $x=-100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,1.01^{\!101}<0.99^{\!-100}$.
Furthermore $1.01^{\!101}\times0.99^{\!101}=0.9999^{\!101}<0.99^{\!-100}\times0.99^{\!101}=0.99$.
| Here is an outline of a proof that $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1,$
using some of your tags such as derivatives and logarithms.
Let $f(x)=\dfrac{(1+x)^{\frac1x}}{(1-x)^{(1-\frac1x)}}=\dfrac{(1-x^2)^{\frac1x}}{1-x}.$
Note that $f(0)$ is not defined, but $\lim_{x\to0}f(x)=1$.
$f'(x)=-(1-x^2)^{\left(\frac1x-1\right)}\left[1+\dfrac{1+x}{x^2}\ln(1-x^2)\right].$
For $0<x<1$, $\ln(1-x^2)<-x^2<-\dfrac{x^2}{1+x}$, so $f'(x)>0.$
For $-1<x<0, \ln(1-x^2)>-\dfrac{x^2}{1+x},$ so $f'(x)<0.$
It follows that $f(x)>1$ when $0<|x|<1,$
and thus $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1.$
| {
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"source": "stackexchange",
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Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C$ and $R^2\cdot8(1+\cos A\cos B\cos C)= a^2+b^2+c^2$
Applying the $cosine$ rule I got to something like $$\cos (HIO)=\frac{2r^2-2Rr+R^2\cdot 4\cos A\cos B\cos C}{2(2r^2-4R^2\cos A\cos B\cos C)(R^2-2Rr)}$$
I need to show that the LHS is between $\left[-\frac{1}{\sqrt{2}},-1\right]$
How t0 proceed?
Is there any simpler way to prove the question (without applying cosine rule) ?
| It is not always true that $\angle{HIO}\gt 135^\circ$.
A counterexample is $(a,b,c)=(85,13,88)$.
$\qquad\qquad$
For $(a,b,c)=(85,13,88)$, we have
$$\begin{align}R&=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{2431}{4\sqrt{186}}
\\\\r&=\frac{1}{2}\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}}=\frac{40\sqrt{186}}{93}
\\\\OI^2&=R^2-2Rr=\frac{1451307}{992}
\\\\OH^2&=9R^2-(a^2+b^2+c^2)=\frac{2712387}{992}
\\\\HI^2&=2r^2+4R^2-\frac{a^2+b^2+c^2}{2}=\frac{109875}{248}
\\\\\cos(\angle{HIO})&=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}=-\frac{13693}{\sqrt{708721585}}\approx -0.5143524
\\\\\angle{HIO}&\approx 120.95^\circ\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove a summation equals to one How to prove that
$$
S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1,
$$
For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$.
Also, WolframAlpha gave the result
$$
S = 2 - \binom{2K-1}{K-1} \frac{(xy)^K}{(x+y)^{2K}} \left\{ {}_2F_1\left(1,2K;K+1;\frac{a}{a+b} \right) + {}_2F_1\left(1,2K;K+1;\frac{b}{a+b} \right) \right\},
$$
but $S = 1$ only when $0 < \frac{a}{a+b} < 1$ and $0 < \frac{b}{a+b} < 1$.
| Starting from
$$S = \sum_{q=0}^{K-1} {K-1+q\choose K-1}
\frac{a^q b^K + a^K b^q}{(a+b)^{q+K}}$$
we get two pieces
$$\frac{b^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q}{(a+b)^q}
\\ + \frac{a^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{b^q}{(a+b)^q}.$$
This is
$$\frac{b^K}{(a+b)^K}
[z^{K-1}] \frac{1}{1-z} \frac{1}{(1-az/(a+b))^K}
\\ + \frac{a^K}{(a+b)^K}
[z^{K-1}] \frac{1}{1-z} \frac{1}{(1-bz/(a+b))^K}.$$
Call these $S_1$ and $S_2.$ The first sum is
$$S_1 = \frac{b^K}{(a+b)^K}
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{(1-az/(a+b))^K}
\\ = b^K
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{(a+b-az)^K}
\\ = \frac{b^K}{a^K}
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z}
\frac{1}{((a+b)/a-z)^K}
\\ = (-1)^{K+1} \frac{b^K}{a^K}
\mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{z-1}
\frac{1}{(z-(a+b)/a)^K}.$$
Now residues sum to zero so we compute this from the residues
at the poles at $z=1$ and $z=(a+b)/a.$ The residue at infinity
is zero by inspection. The residue at $z=1$ is
$$(-1)^{K+1} \frac{b^K}{a^K} \frac{1}{(1-(a+b)/a)^K}
= (-1)^{K+1} b^K \frac{1}{(a-(a+b))^K}
\\ = (-1)^{K+1} b^K \frac{1}{(-b)^K} = -1.$$
For the residue at $z=(a+b)/a$ we require
$$\frac{1}{(K-1)!}
\left(\frac{1}{z^K} \frac{1}{z-1} \right)^{(K-1)}
\\ = \frac{1}{(K-1)!}
\sum_{q=0}^{K-1} {K-1\choose q}
(-1)^q \frac{(K-1+q)!}{(K-1)!} \frac{1}{z^{K+q}}
(-1)^{K-1-q} \frac{(K-1-q)!}{(z-1)^{K-q}}
\\ = (-1)^{K+1} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{1}{z^{K+q}}
\frac{1}{(z-1)^{K-q}}.$$
Evaluating the residue we find
$$\left. (-1)^{K+1} \frac{b^K}{a^K}
(-1)^{K+1} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{1}{z^{K+q}}
\frac{1}{(z-1)^{K-q}} \right|_{z=(a+b)/a}
\\ = \frac{b^K}{a^K}
\sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{1}{((a+b)/a-1)^{K-q}}
\\ = \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{b^q}{a^q}
\frac{b^{K-q}}{a^{K-q}}\frac{1}{((a+b)/a-1)^{K-q}}
\\ = \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}}
\frac{b^q}{a^q}
\\ = \frac{a^K}{(a+b)^K} \sum_{q=0}^{K-1}
{K-1+q\choose K-1} \frac{b^q}{(a+b)^{q}} = S_2.$$
We recognise $S_2$ and hence we have shown that
$$S_1-1+S_2 = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
\sum_{q=0}^{K-1} {K-1+q\choose K-1}
\frac{a^q b^K + a^K b^q}{(a+b)^{q+K}} = 1}$$
as claimed.
Addendum. This is a special case with $x=a/(a+b)$ of the identity at this MSE link.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Domain of $f(x,y) = {\sqrt{x+y-1 \over x-y+1}}$ How I can get the domain of the function
$$ f(x,y) = \sqrt{x+y-1 \over x-y+1}?$$
I know that is: $x+1 \neq y$ and $x^2 \ge (y-1)^2$
But I don't know how to get the second condition.
| Hint: Where defined, $f(x,y)$ equals
$$f(x,y)=\sqrt{\frac{x+y-1}{x-y+1}\cdot\frac{x-y+1}{x-y+1}}=\frac{\sqrt{(x+(y-1))\cdot(x-(y-1))}}{|x-y+1|}=\frac{\sqrt{x^2-(y-1)^2}}{|x-y+1|}. $$
| {
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"url": "https://math.stackexchange.com/questions/3245587",
"timestamp": "2023-03-29T00:00:00",
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Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime.
Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$
Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equation, but it doesn't work. I have been taught Congruence
| Notice that $\text{gcd}(X^2+2X+2,X^2-2X+2)=\text{gcd}(X^2+2X+2,4X)=2$ or $1$. If $2|X$ then 2 must also divide $Y$ and thus: $4X^4+1=4Y^4$ a contradiction.
Then $\text{gcd}(X^2+2X+2,X^2-2X+2)=1$ and since $(X^2+2X+2)(X^2-2X+2)=pY^4$ this implies:
1) $X^2+2X+2={Y_1}^4$ and $X^2-2X+2=p{Y_2}^4$ or
2) $X^2+2X+2=p{Y_1}^4$ and $X^2-2X+2={Y_2}^4$.
Let us consider the first case: write ${Y_1}^4-(X+1)^2=1$ and observe that $({Y_1}^2-X-1)({Y_1}^2+X+1)=1 \implies {Y_1}^2-X-1={Y_1}^2+X+1=1$. Can you take it from here for both cases?
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1}\right)-\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\sum_{k=1}^{n}\left(\sqrt{k+2}-\sqrt{k+1}\right)-\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)\right]\\
&=\lim_{n \to \infty}\left[\left(\sqrt{n+2}-\sqrt{2}\right)-\left(\sqrt{n+1}-1\right)\right]\\
&=\lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n+2}}+1-\sqrt{2}\right)\\
&=1-\sqrt{2}.
\end{align*}
But how to determine the convergence directly by comparison and without evaluating?
| MVT:
$A:=\sqrt{n+2}-\sqrt{n+1}=$
$ (1/2)\dfrac{1}{√s},$ $n+1 <s <n+2$;
$B:= \sqrt{n+1}-√n= (1/2)\dfrac{1}{√t}$, $n <t <n+1$;
MVT:
$A-B=(1/2)(-1/2)\dfrac{1}{r^{3/2}}(s-t)$, $t <r <s$;
Finally :
$|A-B| < (1/4)\dfrac{1}{n^{3/2}} \cdot 2$, since
$r >n$, and $s-t <2$.
Comparison test.
| {
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"timestamp": "2023-03-29T00:00:00",
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Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\frac{3}{2}}}{8a^3}+\frac{3b^2\sqrt{b^2-4ac}}{8a^3}-\frac{3b\left(b^2-4ac\right)}{8a^3}$$ multiplying by $ 8a^3$ and simplifying we get $$4a^3=-b^3+3abc+\left(b^2-4ac\right)^{\frac{1}{2}}\left(b^2-ac\right)$$ Now $a$ , $b$ and $c$ are integers and the left hand side is an integer and so must the right hand side be and this happens only if the term $b^2-4ac$ is a perfect square. Let $b^2-4ac = m^2$. This implies that $$x=\frac{m-b}{2a}$$ Substituting $ x =2^{\frac{1}{3}}$ and multiplying by $2$ yields $$2^{\frac{4}{3}}=\frac{m-b}{a}$$ so $2=\left(\frac{m-b}{a}\right)^{\frac{3}{4}}$ which implies that all powers of two can be written in this form. choosing $ 2^4=16$ shows that $16^{\frac{1}{3}}=\frac{h}{a}\ $ where $h=m-b$. Now all I have to do is prove that $16^{\frac{1}{3}} = 2\ \left(2^{\frac{1}{3}}\right)$ is an irrational number. Now $h=2a\left(2^{\frac{1}{3}}\right)$. the number $a$ is the product of some primes and this product must contain a $\left(2^{\frac{r}{3}}\right)$ where $ r \equiv 2 mod(3)$ but that would mean that $r$ is not divisible by $3$ which means that the term $\left(2^{\frac{r}{3}}\right)$ can't exist in the prime factorization of any integer which shows that the initial assumption ( that $a$ , $b$ and $c$ are integers) is wrong.
The same argument works for $x=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ but the signs are different.
edit: Guys this is proof verification. please don't suggest alternative methods without mentioning why the proof is wrong.
| The solution seems alright, but it is too complicated. An easier solution would be to notice that $x^3 - 2$ is satisfied by $\sqrt[3]{2}$. Then considering $x^3 - 2$ and $ax^2 + bx + c = 0$, as elements of $\mathbb{Q}[x]$ we have that they have a common divisor. However, $x^3 - 2$ is irreducible (by Eisenstein criterion, for example), which contradicts the fact that the two polynomials have a common divisor. Thus $\sqrt[3]{2}$ isn't a solution of $ax^2 + bx + c = 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$.
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$$
This is (obviously) adapted from a recent competition. There ought to be better solutions that the one I have provided below. So if you could, please post them.
| We have that $$\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy} \le \frac{x^2}{2x^2\sqrt{yz}} + \frac{y^2}{2y^2\sqrt{zx}} + \frac{z^2}{2z^2\sqrt{xy}}$$
$$ = \frac{1}{2}\left(\frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{yz}} + \frac{1}{\sqrt{zx}}\right) = \frac{z\sqrt{xy} + x\sqrt{yz} + y\sqrt{zx}}{2xyz} \le \frac{z^2 + x^2 + y^2}{2xyz} = \frac{3xyz}{2xyz} = \frac{3}{2}$$
The equality sign occurs when $x = y = z = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c^3\equiv3abc$(mod 6). Now I am not sure how to show $3abc\equiv 0$(mod6), and even doing that, we have only half of the proof because then we need to prove that the converse is also true.
| a^3+b^3+c^3-a-b-c = (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1)
(a-1)a is divisible by 2.
(a-1)a(a+1) is divisible by 3.
Hence, (a-1)a(a+1) is divisible by 6.
Similarly, (b-1)b(b+1) and (c-1)c(c+1) are also divisible by 6.
Since a^3+b^3+c^3 is divisible by 6 and (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1) is divisible by 6,
a+b+c is also divisible by 6.
For the converse,
a+b+c is divisible by 6, (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1) is also divisible by 6,
hence a^3+b^3+c^3 is also divisible by 6.
I hope that this is the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $5^n + 6^n -1$ is divisble by 10 for all positive integers n. I know the base case is:
$$ n=1,
5^1+6^1-1=10,10|10
$$
But I am struggling about how to go about the inductive step.
| $5^n-1$ is even implies that $5^n+6^n-1$ is divisible by $2$,
$6^n-1=(6-1)(6^{n-1}-1+..)$ ($6=1$ mod $5$ implies that $6^n=1$ mod $5$ and $6^n-1=0$ mod $5$) implies that $5^n+6^n-1$ is divisible by $5$.
Recursive proof
$5^{n+1}+6^{n+1}-1=5.5^n+(5+1)6^n-1=5(5^n+6^n-1)+6^n+4$
Suppose that $6^n=6$ mod $10$, $6^{n+1}=36=6 $ mod $10$ implies that $6^n+4=0$ mod $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
And I got stuck at this point.
Second (desperately), I expanded
$$(a_1 x + b_1y)(a_2 x + b_2 y)(a_3 x + b_3 y)(a_4 x + b_4 y)$$
and compared it to $x^4 + y^4 - x^2y^2$, but let's just say it didn't go smoothly.
I think there should be a nice way to do it. Anybody?
| A smart way to approach this is to solve the equation
$$x^2-y^2+ixy=0,$$ find the solutions $x_\pm(y)$ and then write
$$x^2-y^2+ixy=(x-x_+(y))(x_-(y)).$$
In this case we have that the solutions are
$$x_i(y)=\frac{-iy\pm\sqrt{(iy)^2+4y^2}}{2}=\frac{-iy\pm\sqrt{3}y}{2}=\frac{-1\pm\sqrt{3}}{2}\cdot y$$
thus the factorisation is
$$x^2-y^2+ixy=\left(x+\frac{-1+\sqrt{3}}{2}y\right)\left(x+\frac{-1-\sqrt{3}}{2}y\right).$$
The procedure is analogous for the second term
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $
Solve the inequality
$$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$
Attempt:
I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes
$$ y - 4 \sqrt{y + 4} < 1 $$
$$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$
so from here I get that $0 < y < 9 - 2\sqrt{14} \: \:\cup \:\: y > 9 + 2\sqrt{14}$. This is for $y>0$ or $x(2x+1)>0 \implies x > 0 \:\: \cup \:\: x < -1/2$
I may continue to solve for $x$.. but the answer key says that the solution is $ -7/2 < x < 3$. I doubt that the solution set will be that tidy. Are there better approaches to solve the inequality?
| It's $$2x^2+x+4-4\sqrt{2x^2+x+4}+4<9$$ or
$$\left(\sqrt{2x^2+x+4}-2\right)^2<9$$ or since $\sqrt{2x^2+x+4}>0$, it's
$$\sqrt{2x^2+x+4}<5$$ or
$$2x^2+x-21<0$$ or
$$(x-3)(2x+7)<0$$ or
$$-3.5<x<3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^2\alpha\bigg)$$
$$\tan^{-1}\bigg(\frac{\frac{2\csc\alpha}{1-\csc^2\alpha}+2\sin\alpha\sec^2\alpha}{1-\frac{2\csc\alpha}{1-\csc^2\alpha}2\sin\alpha\sec^2\alpha}\bigg)$$
How do i solve it Help me please
| Note that
$$
2\sin\alpha\sec^2\alpha=\frac{2\sin\alpha}{\cos^2\alpha}=\frac{2\sin\alpha}{1-\sin^2\alpha}
$$
If
$$
f(x)=\arctan\frac{2x}{1-x^2}
$$
then
$$
f'(x)=\frac{2}{1+x^2}
$$
so $f(x)$ differs from $2\arctan x$ by a constant over $(-1,1)$. Since $f(0)=0=2\arctan0$, we can say that
$$
\arctan\frac{2\sin\alpha}{1-\sin^2\alpha}=2\arctan\sin\alpha
$$
For $x>0$, $\arctan(1/x)=\pi/2-\arctan x$, so your expression evaluates to
$$
2\left(\frac{\pi}{2}-\arctan\sin\alpha\right)+2\arctan\sin\alpha=\pi
$$
if $\sin\alpha>0$.
If $\sin\alpha<0$, the expression evaluates to $-\pi$, because for $x<0$ one has $\arctan(1/x)=-\pi/2-\arctan x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :-
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{2 \sqrt{2}} \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} dx - \frac{1}{2 \sqrt{2}} \int \frac{x - \sqrt{2}}{x^2 - \sqrt{2}x + 1} dx $$
Now, to the first integral, I tried making a u-substitution:-
$$ \text{Let }x^2 + \sqrt{2}x + 1 = u \\
\frac{du}{dx} = 2x + \sqrt{2} \\
\implies du = (2x + \sqrt{2}) dx \\ $$
As you can see, it is not the same as the numerator, which is $$ (x + \sqrt{2}) dx $$
Any hints on how to proceed ?
| $\displaystyle \frac{x+\sqrt{2}}{x^2+\sqrt2 x+1}=\frac{x+\sqrt{2}}{\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac12}$.
Let $\displaystyle x+\frac{\sqrt{2}}2=\frac{\sqrt2}2\tan\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Geometric consequence of absolute vectors If $| \mathbf{a} + \mathbf{b} | = | \mathbf{a} | + | \mathbf{b} |$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors, what is the geometrical significance of this?
My first thought was that the vectors $\mathbf{a}$ and $\mathbf{b}$ must be in the first quadrant, only having positive components.
| Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$. Using properties of the dot product,
$$|\mathbf a+\mathbf b|=|\mathbf a|+|\mathbf b|$$
$$\implies|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2+2|\mathbf a||\mathbf b|$$
$$\implies(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)=\mathbf a\cdot \mathbf a+\mathbf b\cdot \mathbf b+2|\mathbf a||\mathbf b|.$$
But the left side is $\mathbf a\cdot \mathbf a + \mathbf b\cdot \mathbf b + 2\mathbf a\cdot \mathbf b = \mathbf a\cdot \mathbf a +\mathbf b \cdot \mathbf b+2|\mathbf a||\mathbf b|\cos\theta$, so this means $\cos\theta=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Common volume of three cylinders with unequal radii I would like to solve it please for the case where the radii can be a similar size - so the case where this statement is NOT true: $$ \mathbf r_1^2 \mathbf \geq \mathbf r_2^2 \mathbf + \mathbf r_3^2 $$
How do you solve for the common volume of 3 cylinders with unequal radii? (If you could please include the integral needed - I think it might need a cartesian equation system? Or if you have any good idea of what direction or things I could read up on to please learn to solve this. Thank you so much for your help!
$$x^2 + y^2 = r_1^2$$
$$x^2 + z^2 = r_2^2$$
$$y^2 + z^2 = r_3^2$$ where $r_1 \neq r_2 \neq r_3$
I can find that the common volume for equal radii using triple integration with circular co-ordinates to get the answer below (following this)
$$V_c = 8\cdot(2-\sqrt 2)\cdot r^3$$
Essentially I want to get to a place where I can find the equation for the common volume of three cylinders with different radii and at different angles.
So if anyone has an idea of how to calculate for cylinders at different angles other than 90 that would be great too! Thank you!
| Let $a<b<c$ with $c<\sqrt{a^2+b^2}$ be the radii of the three cylinders
$$x^2+y^2\leq a^2,\qquad x^2+z^2\leq b^2,\qquad y^2+z^2\leq c^2\ ,$$
and denote by $S$ the part of the intersection lying in the positive octant. The set $S$ is standing on the quarter disc
$$S':\qquad x\geq0,\quad y\geq 0, \quad x^2+y^2\leq a^2$$
in the $(x,y)$-plane. The upper boundary of $S$ is determined by the two other cylinders. Therefore the height $z(x,y)$ of $S$ at the point $(x,y)\in S'$ satisfies
$$z^2(x,y)=\min\{b^2-x^2, \ c^2-y^2\}\ .$$
The two entries in the $\min$ are equal when $y^2=c^2-b^2+x^2$. It follows that
$$z(x,y)=\left\{\eqalign{\sqrt{b^2-x^2}\qquad&(y\leq\sqrt{c^2-b^2+x^2})\cr
\sqrt{c^2-y^2}\qquad&(y\geq\sqrt{c^2-b^2+x^2})\cr}\right.\ .$$
We now have to calculate
$${\rm vol}(S)=\int_{S'}z(x,y)\>{\rm d}(x,y)\ .\tag{1}$$
The hyperbolic arc $y=\sqrt{c^2-b^2+x^2}$ intersects the quarter circle $x^2+y^2=a^2$ at the point
$$P=\left(\sqrt{a^2+b^2-c^2\over2}, \ \sqrt{a^2+c^2-b^2\over2}\right)=:(u,v)\ .$$
The hyperbolic arc and the lines $x=u$, $y=v$ through $P$ partition the quarter disc $S'$ into four sections. From $(1)$ we then obtain
$$\eqalign{{\rm vol}(S)&=\int_0^u \sqrt{c^2-b^2+x^2}\>\sqrt{b^2-x^2}\>dx+\int_u^a\sqrt{a^2-x^2}\>\sqrt{b^2-x^2}\>dx \cr&\qquad + \int_{\sqrt{c^2-b^2}}^v \sqrt{y^2-(c^2-b^2)}\>\sqrt{c^2-y^2}\>dy+\int_v^a\sqrt{a^2-y^2}\>\sqrt{c^2-y^2}\>dy\ .\cr}$$
| {
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"url": "https://math.stackexchange.com/questions/3266348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b. The question is:
Establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b.
And the Hint says:
Mimic the proof of theorem 13.2, which is given below:
My trial is:
For the case: $p = 8k + 1$
I used the theorem that says the Legendre symbol $(2/p) = 1$ if $p \equiv 1 \pmod 8$
And I end up after using Thue lemma that $$2x_{0}^2 \equiv y_{0}^2,$$
But this will lead to $p = 2a^2 - b^2$ which is not the required, could anyone help me in fixing this please?
I feel that I need the Legendre symbol $(2/p) = 1$ if $p \equiv 1 \pmod 8$ to be $(- 2/p) = 1$ if $p \equiv 1 \pmod 8$ instead, but How can I do this ?
What about the second case when $p \equiv 3 \pmod 8$?
| Comment:An experimental approach; following conditions cover a set of these types of primes:
For $8k+1$:
Let $k=k_1+2$ ⇒ $p=8(k_1+2)+1=8k_1+8+8+1=2[4(k_1+1)]+3^2$
If $k_1+1=c^2$ then we have:
$p=3^2+2\times (2c)^2$
Examples; $k_1=3$ ⇒ $p=41=3^2+2 (2\times 2)^2=3^2+2\times 4^2$
$k_1=15$ ⇒ $c^2=15+1=4^2$ ⇒ $p=137=3^2+2\times 8^2$
For $p=8k+3$ it can be seen that:
$p=8k+3 ≡ c^2k^2 \mod 3^2$;
Type 1: $p=(3d)^2 +2(ck)^2$
Type 2: $p=(ck)^2 +2(3d)^2$
apart from ck=1 or $(3d)^0=1$ we have following values for ck and 3d:
$ck=1, 5^2, 7^2, 11^2, (2n+1)^2$; $2n+1 ≠ 3 t$
$3d = 3, 9, 15, . . .(2n+1)3$
Examples:
$p=8\times 52+3=419=9^2+2\times 13^2$
$p=8\times 58+3=467=15^2 +2\times 11^2$
In both cases $p≡ c^2k^2 \mod 3^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Extreme values of function with one condition What is the simple way to find extreme values of a function $f(x,y) = 5*x^2 + 3*x*y + y^2$ with conditions $x^2 + y^2 = 1$.
I tried to make partial derivatives and initialize them with $0$, but the result is not good.
| As araomis's comment suggested, you can try using polar co-ordinates with $r = 1$. This is because the unit circle $x^2 + y^2 = 1$ can be parametrized by $x = \cos(\theta)$ and $y = \sin(\theta)$ for $\theta \in [0,2\pi)$. Using this, along with the unit circle, then
$$f(x,y) = 5x^2 + 3xy + y^2 = 4x^2 + 3xy + 1 \tag{1}\label{eq1}$$
becomes
$$g(\theta) = f(\cos(\theta),\sin(\theta)) = 4\cos^2(\theta) + 3\cos(\theta)\sin(\theta) + 1 \tag{2}\label{eq2}$$
Wikipedia's List of trigonometric identities includes several useful identities. These include $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ so
$$3\cos(\theta)\sin(\theta) = \frac{3}{2}\sin(2\theta) \tag{3}\label{eq3}$$
Also, $2\cos^2(\theta) = 1 + \cos(2\theta)$ so
$$4\cos^2(\theta) = 2 + 2\cos(2\theta) \tag{4}\label{eq4}$$
\eqref{eq3} and \eqref{eq4} in \eqref{eq2} gives
$$g(\theta) = 2\cos(2\theta) + \frac{3}{2}\sin(2\theta) + 3 \tag{5}\label{eq5}$$
Finally, the Sine and Cosine sub-section gives that $a\sin x+b\cos x=c\sin(x+\varphi)$ where $c = \sqrt{a^2 + b^2}$ and $\varphi = \text{atan2}(b,a)$. In this case, $x = 2\theta$, $a = \frac{3}{2}$ and $b = 2$, so
$$c = \sqrt{\frac{9}{4} + 4} = \frac{5}{2} \tag{6}\label{eq6}$$
Note there's no need to calculate $\varphi$ as it's just a fixed offset $\in (-\pi,\pi]$. Thus, \eqref{eq5} can be rewritten as
$$g(\theta) = \frac{5}{2}\sin(2\theta + \varphi) + 3 \tag{7}\label{eq7}$$
Since $-1 \le \sin(2\theta + \varphi) \le 1$, the minimum value of $g(\theta) = f(x,y)$ is $3 - \frac{5}{2} = \frac{1}{2}$ (when $2\theta + \varphi = -\frac{\pi}{2}, \frac{3\pi}{2} \; \text{and/or} \; \frac{7\pi}{2}$) and the maximum value is $3 + \frac{5}{2} = \frac{11}{2}$ (when $2\theta + \varphi = \frac{\pi}{2}, \frac{5\pi}{2} \; \text{and/or} \; \frac{9\pi}{2}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding angle in the isosceles triangle There is the following triangle. I infer that the triangle is isosceles. But I cannot go further.
| Without trigonometry
Let $D$ be the symmetric of $C$ with respect to $A$, $H$ and $K$ the orthogonal projections of $A$ and $C$ on $BC$ and $AB$.
$BCD$ is a right triangle. By the midpoint theorem, $CK=2MN=6$ and $MN=MA=4$ so $BK=2$. By Pythagoras theorem, $BC=\sqrt{BK^2+CK^2}=2\sqrt{10}$ and $AH=\sqrt{AB^2-BH^2}=3\sqrt{10}$.
$\dfrac{NC}{ND}=\dfrac{5}{15}=\dfrac{1}{3}$ and $\dfrac{BC}{BD}=\dfrac{BC}{2AH}=\dfrac{1}{3}$ so, by the angle bisector theorem, $BN$ is the interior bisector of the right angle $CBD$. Thus $x=45^\circ$.
With trigonometry
$\cos BAC=4/5$ and $BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos BAC=40$ by the law of cosines.
$NC^2=NB^2+BC^2-2NB\cdot BC\cdot\cos x$ gives $x=45^{\circ}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate squared gaussians I need to compute the integral
$$
\int_{-\infty}^\infty e^{-a^4 - (1-a-b)^4} da
$$
for which it seems there is no analytic solution. Trying with Mathematica also gives no solution.
Do you have any suggestion, or possible method to solve this? (e.g. smart change of variables, ...)
| First let's rewrite the integral a bit. Introducing $1-b=2c$, we have:
\begin{align}
&\int_{-\infty}^\infty e^{-a^4-(1-a-b)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-a^4-(2c-a)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-(c+a-c)^4-(c+c-
a)^4}\mathrm{d}a=\\
&=\int_{-\infty}^\infty e^{-(c+x)^4-(c-x)^4}\mathrm{d}x=
=\int_{-\infty}^{\infty}e^{-2(x^4+6c^2x^2+c^4)}\mathrm{d}x=e^{16c^4}\int_{-\infty}^{\infty}e^{-2(x^2+3c^2)^2}\mathrm{d}x
\end{align}
We now want to find the last integral. Wolfram Alpha gives
$$
\int_{-\infty}^\infty e^{-2(x^2+y^2)^2}\mathrm{d}x=\frac{|y|}{\sqrt{2}}e^{-y^4}K_{\frac{1}{4}}(y^4)
$$
Here $K$ is a modified Bessel function of the second kind. Plug in $y^2=3c^2$ and then $2c=1-b$ to get the original integral expressed as a function of $b$:
$$
\int_{-\infty}^\infty e^{-a^4-(1-a-b)^4}\mathrm{d}a=\sqrt{\frac{3}{8}}\,\,|b-1|\,\,e^{\frac{7}{16}(b-1)^4}\,K_{\frac{1}{4}}\!\!\left(\frac{9}{16}(b-1)^4\right)
$$
If $b=1$ the integral is easily expressible in terms of the Gamma function and this result agrees with the one found from the general result by taking the limit $b\rightarrow 1$.
I invite anyone to try to calculate the last integral by hand and post a derivation. I suspect it has something to do with the substitution $x\rightarrow y\sinh t$ but I didn't get anywhere.
EDIT: I managed to calculate the integral.
\begin{align}
\int_{-\infty}^\infty e^{-2(x^2+y^2)^2}\mathrm{d}x=2\int_0^\infty e^{-8((x/\sqrt{2})^2+y^2/2)^2}\mathrm{d}x=2\sqrt{2}\int_0^\infty e^{-8(x^2+y^2/2)^2}\mathrm{d}x
\end{align}
Now make the substitution $x=|y|\sinh (t/4)$.
\begin{align}
2\sqrt{2}\int_0^\infty e^{-8(x^2+y^2/2)^2}\mathrm{d}x=\frac{2\sqrt{2}}{4}|y|\int_0^\infty e^{-8y^4(\sinh(t/4)^2+1/2)^2}\cosh(t/4)\mathrm{d}t
\end{align}
Using appropriate identities, one can find that $$\left(\sinh^2\left(\frac{t}{4}\right)+\frac{1}{2}\right)^2=\frac{1}{8}(1+\cosh(t))$$
Therefore:
$$
\frac{2\sqrt{2}}{4}|y|\int_0^\infty e^{-8y^4(\sinh(t/4)^2+1/2)^2}\cosh(t/4)\mathrm{d}t=
\frac{|y|}{\sqrt{2}}\int_0^\infty e^{-y^4(\cosh(t)+1)}\cosh(t/4)\mathrm{d}t
$$
One definition of $K$ is:
$$
K_{\alpha}(x)=\int_0^\infty e^{-x\cosh t}\cosh(\alpha t)\mathrm{d}t
$$
So:
$$
\frac{|y|}{\sqrt{2}}\int_0^\infty e^{-y^4(\cosh(t)+1)}\cosh(t/4)\mathrm{d}t=\frac{|y|}{\sqrt{2}}e^{-y^4}K_{\frac{1}{4}}(y^4)
$$
as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solution Of Diophantine Equations
Find all integer solutions for the equation:
$$(x+y)(y+z)(z+x)=txyz$$
such that gcd$(x, y)=1$, gcd$(y, z)=1$ and gcd$(z, x)=1$
Now we can write our equation as
$$(\frac{x+y}{y})(\frac{y+z}{z})(\frac{x+z}{x})=t$$
This gives $$(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=t$$
Now as $t$ is an integer and gcd$(x, y)=1$, gcd$(y, z)=1$ and gcd$(z, x)=1$ so the only possibility is $$x=y=z=1$$ giving a solution as $$(x, y, z, t)=(1, 1, 1, 8)$$
Am I Right?
| Hint:
Say $x,y,z$ are positive. We have $x|y+z$ and $y|x+z$ and $z|x+y$. We can assume $x\geq y\geq z$. Then $y+z= kx$ where $k=1$ or $k=2$.
If k=1we have $(2y+z)(2z+y)=tyz$ and so $y|2z^2$ so $y=1$ or $y=2$.
If k=2 we have $(3y+z)(3z+y) =2tyz$ so $y| 3z^2$ so $y=1$ or $y=3$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$. Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$.
Answer:
Since $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle=-\cos (2 \pi/6)+i \sin (2 \pi/6)$ is a generator of $U_6$ and $U_6$ has order $\phi(6)=2$, the cyclic subgroup generated by $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ has $2$ elements.
Am I right ?
I think $\cos (2 \pi/n)+i \sin (2 \pi/n)$ has order $\phi(n)$ but in the above case it is like the form $-\cos (2 \pi/n)+i \sin (2 \pi/n)$. This confuses me
| Note that
$\cos \left ( \dfrac{2\pi}{3} \right ) = -\dfrac{1}{2}, \tag 1$
and
$\sin \left ( \dfrac{2\pi}{3} \right ) = \dfrac{\sqrt 3}{2}; \tag 2$
thus,
$-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} = \cos \left ( \dfrac{2\pi}{3} \right ) + i \sin \left ( \dfrac{2\pi}{3} \right ) = e^{2 \pi i /3}; \tag 3$
it follows that, for any $m \in \Bbb Z$,
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^m = e^{2 m \pi i/3}; \tag 4$
in particular we have
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^2 = e^{4\pi i/3}, \tag 5$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^3 = e^{6 \pi i/3} = e^{2\pi i} = 1; \tag 6$
at this point the powers if $e^{2\pi i/3}$ commence to cycle around so that
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^4 = e^{8 \pi i/3} = e^{6\pi i / 3} e^{2 \pi i / 3}= e^{2 \pi i} e^{2 \pi i / 3}= e^{2 \pi i / 3}, \tag 7$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^5 = e^{10 \pi i/3} = e^{6\pi i / 3} e^{4 \pi i / 3}= e^{2 \pi i} e^{4 \pi i / 3}= e^{4 \pi i / 3}, \tag 8$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^6 = e^{12 \pi i/3} = e^{4 \pi i}= (e^{2 \pi i})^2 = 1^2 = 1, \tag 9$
and so forth. Indeed, since the division algorithm yields
$m = 3k + j, \; j = 0, 1, 2, \tag{10}$
(4) may be written
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^m = e^{2 m \pi i/3} = e^{2 (3k + j) \pi i/3} = e^{6k\pi i / 3 + 2j \pi i / 3}$
$= e^{6k \pi i / 3}e^{2j\pi i / 3} = (e^{2\pi i})^k e^{2j \pi i /3} = e^{2j \pi i /3}, 0 \le j \le 2; \tag{11}$
which shows that every power of $e^{2 \pi i/ 3}$ lies in the set
$\{1, e^{2 \pi i / 3}, e^{4 \pi i / 3 } \}, \tag {12}$
which clearly is possessed of a group structure in light of the preceding calculations. Since there exists precisely one group of order $3$,
$\Bbb Z_3 = \{0, 1, 2 \} \tag{13},$
and it is isomorphic to (12) via the map taking
$0 \mapsto 1, \; 1 \mapsto e^{2 \pi i / 3}, \; 2 \mapsto e^{4 \pi i / 3}, \tag{14}$
we conclude that the subgroup of $\Bbb C^\ast$ generated by $e^{2\pi i / 3}$ has exactly $3$ elements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Let $A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$
Let $A=\begin{bmatrix}
1 & 2\\
3& 4
\end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$
det$(A^3-6A^2+5A+3I)=$det$((A^2-5A-2I)(A-I)+2A+I)= $det$(2A+I)=3$, Since a matrix satisfies its characteristic polynomial. Is this right?
| Here is another method. Assume that $\lambda_i\in\mathbb{C}$, $i=1,2$ are the two igenvalues of $A$. Let $f(x)\in \mathbb{C}[x]$ be the characteristic polynomial of $A$. Then $$f(x)=\det(A-xI)=x^2-5x-2.$$
The matrix $A$ can be diagonalized as
$$A=P\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}P^{-1},$$
where $P$ is an invertible matrix over $\mathbb{C}$. Let $g(x)=x^3-6x^2+5x+3$. We have $g(x)=f(x)(x-1)+2x+1$. It follows that
$$\det(A^3-6A^2+5A+3I)=\det(g(A))=\det\left(g\left(\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}\right)\right)=\det\left(\begin{bmatrix} g(\lambda_1) & 0\\ 0& g(\lambda_2) \end{bmatrix}\right)=g(\lambda_1)g(\lambda_2).$$
Note that $f(\lambda_i)=0$, $i=1,2$. Thus $g(\lambda_i)=f(\lambda_i)(\lambda_i-1)+2\lambda_i+1=2\lambda_i+1$. So
$$\det(A^3-6A^2+5A+3I)=(2\lambda_1+1)(2\lambda_2+1)=4\lambda_1\lambda_2+2(\lambda_1+\lambda_2)+1.$$
According to Vieta theorem, we have $$\lambda_1\lambda_2=-2, \lambda_1+\lambda_2=5.$$
Consequently, $$\det(A^3-6A^2+5A+3I)=4\times (-2)+2\times 5 +1=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Evaluate using Stokes' Theorem To evaluate $\oint_{C} -y^3dx+x^3dy+z^3dz,$ where $C$ is the intersection of cylinder $x^2 + y^2 =1$ and plane $x+y+z=1$. The orientation of $C$ is counter-clockwise motion in the $xy$ plane.
Now I have computed $\nabla \times\mathbf{F} = \left(0,0,3\left(x^2+y^2\right)\right).$ I am having difficulty finding out the curve $C$ of intersection and also I am confused about projection on the $xy$ plane. Should I take part inside $x+y=1 $ only or part between $x+y=1$ and $x^2+y^2=1?$
| Let $\mathbf{F}=\left(-y^3,x^3,z^3\right).$
So we have
\begin{align*}
\oint_{C} -y^3dx+x^3dy+z^3dz&=\oint_C\mathbf{F}\cdot d\mathbf{r} \\
&=\iint_S(\nabla\times\mathbf{F})\cdot dS \\
&=\iint_S(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dA \\
&=\sqrt{3}\iint_S\left(x^2+y^2\right)\,dA.
\end{align*}
My hunch is that we can consider the projection into the $xy$ plane of the surface $S$ for this problem, because the integrand $\nabla\times\mathbf{F}$ only has a $z$ component. If that is so, we will want to switch to polar coordinates:
\begin{align*}
\oint&=\sqrt{3}\int_0^{2\pi}\int_0^1\left(x^2+y^2\right)\,r\,dr\,d\theta \\
&=2\sqrt{3}\,\pi\int_0^1 r^3\,dr \\
&=\frac{\sqrt{3}\,\pi}{2}.
\end{align*}
As we can see from this wiki, to adjust for the projection, we need
$$A_{\text{proj}}=\cos(\beta) A, $$
since the angle is constant and pulls out of the integral. We have computed the projected area, so we must compensate by dividing by $\cos(\beta),$ we can calculate via the dot product formula:
$$\frac{1}{\sqrt{3}}(1,1,1)\cdot(0,0,1)=\cos(\beta). $$
This means the final result is
$$\frac{\sqrt{3}\,\pi}{2}\div\frac{1}{\sqrt{3}}=\frac{3\pi}{2}. $$
Now the question is, was the projection justified? Can we verify by, say, computing the original line integral? As suggested by DiegoMath in his answer, we can parametrize the curve $C$ as
\begin{align*}
x&=\cos(t) \\
y&=\sin(t) \\
z&=1-\cos(t)-\sin(t),\\
0&\le t\le 2\pi.
\end{align*}
Then we have
\begin{align*}
\mathbf{r}(t)&=(\cos(t), \sin(t), 1-\cos(t)-\sin(t))\\
\dot{\mathbf{r}}(t)&=(-\sin(t), \cos(t), \sin(t)-\cos(t)) \\
\mathbf{F}(t)&=(-\sin^3(t),\cos^3(t),(1-\cos(t)-\sin(t))^3) \\
\mathbf{F}\cdot\dot{\mathbf{r}}&=\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3 \\
\oint_C\mathbf{F}\cdot d\mathbf{r}&=\oint_C\mathbf{F}\cdot \dot{\mathbf{r}}(t)\,dt \\
&=\int_0^{2\pi}\left[\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3\right]dt \\
&=\frac{3\pi}{2},
\end{align*}
which is the answer we had above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate the integral from $0$ to $2\pi$ I get that it's $0$ because $\tan\left(2x\right)$ evaluates to $0$ at both $x=0$ and $x=2\pi$. Is there another way to solve this integral and get the correct answer ($2\pi\sqrt2$)?
| Given integral $$ I = \int^{2 \pi}_0 \cfrac{1}{\sin^4 x + \cos^4 x} dx = 4 \int^{\pi/2}_0 \cfrac{dx}{\sin^4 x + \cos^4 x} $$
By repeatedly using $ \int^{2a}_0 f(x) dx = \int^a_0 f(x) dx + \int^a_0 f(2a-x) dx$
On multiplying numerator and denominator by $ \sec^4(x)$, followed by substitution of $\tan(x) = u$ $$ \begin{align} I &= 4 \int^{\infty}_0 \cfrac{1 + u^2}{1+ u^4} du \\ &= 4 \int^{\infty}_0 \cfrac{1}{\bigg( u - \frac{1}{u} \bigg)^2 + 2} d \bigg( u - \frac{1}{u} \bigg) \\ &= \cfrac{4}{\sqrt{2}} \arctan \bigg(\cfrac{u-\frac{1}{u}}{\sqrt{2}} \bigg) \bigg|^{\infty}_0 = 2 \sqrt{2} \pi\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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Prove that $\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1}= \frac{3}{2}$, which is a beautiful value! Prove that $\prod\limits_{n= 2}^{\infty} \dfrac{n^{3}+ 1}{n^{3}- 1}= \dfrac{3}{2}$, which is a beautiful value ! A solution due to a p u as follow
$$\begin{align}
\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1} & = \prod\limits_{n= 2}^{\infty}\frac{(n+ 1)(n+ \omega )(n+ \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\\
& = \prod\limits_{n= 2}^{\infty} \frac{(n+ 1)(n- 1- \omega)(n- 1- \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\,(\!1+ \omega+ \omega^{2}= 0\,used\!)\\
& = \lim_{n\rightarrow \infty} \frac{n(n+ 1)(1- \omega)(1- \omega^{2})}{1\cdot 2(n- \omega)(n- \omega^{2})}\\
& = \lim_{n\rightarrow \infty} \frac{3(1+ \frac{1}{n})}{2(1- \frac{\omega}{n})(1- \frac{\omega^{2}}{n})}\,\left (\!(1- \omega)(1- \omega^{2})= 1+ 1+ 1= 3\,used\!\right )\\
& = \frac{3}{2}
\end{align}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
| Here is a much simpler way. Observe that the partial product
$$
\prod_{n=2}^{m} \frac{n^3+1}{n^3-1} = \frac{3 (m^2+m)}{2(m^2+m+1)}.
$$
Then when $m\to\infty$, the product converges to $\displaystyle \frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Given a diagonalizable matrix $A$ and polynomial $f$, prove $f(A)$ is diagonalizable I have been given a diagonalizable matrix $A \in K^{n
\times n}$ and a polynomial $f \in K[X]$ for a field $K$. I need to prove that $f(A)$ is diagonalizable. Because the matrix $A$ is a arbitrary matrix, it follows:
$$A = \left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right) \in K^{n
\times n} \, for \, a_1,\dots,a_n \in K.$$
Assuming that $f$ is a normalized polynomial, it follows:
$$f = X^n + a_{n-1}X^{n-1} + \dots + a_1X + a_0$$
My first idea was to simply insert the given the matrix $A$ in the polynomial $f$ which results in the following:
$$f(A) = \left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right)^n + a_{n-1}
\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right)^{n-1} + \dots +
a_1\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right)^1 + a_0
\left(\begin{array}{cccc}
1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1
\end{array}\right)$$
I know that in order to determine the matrix $A^n$ I can use the following:
$$A^n = TB^nT^{-1} \text{$\,$ where B is a diagonal matrix}.$$
I know such matrix $B$ exists because matrix $A$ is diagonalizable, so $A$ is similar to a diagonal matrix. I don't know if this is the right approach because from this point on I am stuck.
| I think you can do it as this. Let
$$f(X) = X^n + a_{n-1}X^{n-1} + \cdots +a_1 X + a_0, \quad \in K[X]$$
be your polynomial and let $A$ diagonalize as
$$A = TBT^{-1}.$$
Then you have
\begin{align*}
f(A) &=A^n + a_{n-1}A^{n-1} + \cdots + a_1A + a_0I \\
&=TB^nT^{-1} + a_{n-1}TB^{n-1}T^{-1} + \cdots + a_1 TBT^{-1} + a_0 I \\
&=T \left(B^nT^{-1} + a_{n-1}B^{n-1}T^{-1} + \cdots + a_1 BT^{-1} + a_0 T^{-1} \right) \\
&=T \left(B^n + a_{n-1}B^{n-1} + \cdots + a_1 B + a_0 I \right)T^{-1}
\end{align*}
where we use the property, that for scalars $\lambda \in K$ and Matrices $A,B$,
$$\lambda (AB) = (\lambda A)B = A(\lambda B)$$
Now we have to show, that the matrix $B^n + a_{n-1}B^{n-1} + \cdots + a_1 B + a_0 I$is a diagonal one. Let $\lambda_1,\dots, \lambda_n$ be the different eigenvalues, then
$$B^i = \begin{pmatrix} \lambda_1^i &0 &0 &\cdots &0 \\ 0 &\lambda_2^i& 0 &\cdots &0 \\ 0 &0&\lambda_3^i &\cdots &0\\\vdots&&&\ddots\\0&0&0&\cdots & \lambda_n^i\end{pmatrix},\quad \text{ for } 1 \le i \le n.$$
Since the sum of diagonal matrices is diagonal, and multiplication by scalar gives also a diagonal matrix, it should be fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve$(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ We need to solve the following equation $y=(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ and it is equal to 2 while I am getting the value in excel I am not able to solve it manually eventhough the values are conjugate
I tried $y=a-b$
$a=(10+6\sqrt3)^{\frac{1}{3}}$ & $b=(-10+6\sqrt3)^{\frac{1}{3}}$
$y^3=(a-b)^3$
$y^3=a^3-b^3-3ab(a-b)$ after this step I am struck
| Substituting the values of $a$ and $b$, and the relation $y=a-b$, into $y^3=a^3-b^3-3ab(a-b)$ yields
$$y^3=10+6\sqrt3+10-6\sqrt3-3((6\sqrt3+10)(6\sqrt3-10))^{1/3}y$$
$$y^3=20-3(3\cdot36-100)^{1/3}y$$
$$y^3=20-6y$$
$$y^3+6y-20=0$$
$$(y-2)(y^2+2y+10)=0$$
Since the quadratic factor has no real roots, $y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\,n\bmod 42\,$ given $\,7\mid 3^n+4n+1$ I have been working on a problem relating to remainders: "The number $3^n+4n+1$ is divisible by $7$. Find the remainder of n when it is divided by $42$." I have been unable to find a strategic way to solve this problem and so far, I have only found the remainders, $4$, $8$, $13$, $21$, $23$, and $24$. Can anyone confirm whether this is correct?
| Strategy:
$4n + 1 \equiv k \pmod 7$ will have a solution $n_k$ determined by $n \mod 7$. And for $k = 0,... 7$ there will be $7$ solutions $n_0, ... n_6$.
By Fermats little Theorem, $3^6\equiv 1 \pmod 7$ for $3^n \equiv j\pmod 7$ will have a solution $m_j$ determined by $n \pmod 6$. If we assume $3$ is a primitive root (an assumption for now but one easily determined by simple calcultions) for $j= 1,2,3,...6$ there will be six solutions: $m_1,...,m_6$
So to have $3^n + 4n + 1\equiv k - k \equiv 0 \pmod 7$ is a matter of finding the six $n_k, m_{7-k}$ pairs and solving $n \equiv n_k \pmod 7$ and $n_{7-k} \pmod 6$.
By the Chinese Remainder Theorem, for each $n_k, m_{7-k}$ pair there will precisely one solution $\mod 6*7$.
======= Full Answer below=====
The strategic way I'd consider would figure $3^n$ has a cyclic $\mod 7$ and figure what that is. Presumable $3$ is a primitive root and the cycle is and $3^n \equiv 1,3,2,6,4,5 \pmod 7$ if $n \equiv 0,1,2,3,4,5,6 \pmod 6$
Meanwhile $4n + 1 \equiv 1,5,2,6,3,0,4 \pmod 7$ when $n \equiv 0,1,2,3,4,5,6,7$
We need $3^n + 4n + 1 \equiv 0 \pmod 7$ so
if $n \equiv 0 \pmod 7$ then $4n+1\equiv 1$ and we $3^n\equiv -1$ so we would need $n \equiv 0 \pmod 7$ and $n\equiv 3\pmod 6$.
Similarly if $n\equiv 1,2,3,4,5,6 \pmod 7$ we have $4n+1 \equiv 5,2,6,3,0,4 \pmod 7$ and we need $3^n \equiv 2,5,1,4,0,3 \pmod 7$ so we need $n\equiv 2,5,0, 4, impossible,1 \pmod 6$.
Now by CRT th each of these pairs of options has a unique solution:
1) If $n\equiv 0 \pmod 7;n\equiv 3 \pmod 6$ then $n \equiv 21 \pmod {42}$.
2) If $n\equiv 1 \pmod 7;n\equiv 2 \pmod 6$ then $n \equiv 8 \pmod {42}$.
3)If $n\equiv 2 \pmod 7;n\equiv 5 \pmod 6$ then $n \equiv 23 \pmod {42}$.
4)If $n\equiv 3 \pmod 7;n\equiv 0 \pmod 6$ then $n \equiv 24 \pmod {42}$.
5) If $n \equiv 4 \pmod 7; n\equiv 4\pmod6$ then $n \equiv 4 \pmod {42}$
6) If $n \equiv 6 \pmod 7; n\equiv 1\pmod 6$ then $n \equiv 13 \pmod {42}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}
Now i think it gets more complicated , any help would be appreciated
| We do this in one period $[-\pi, \pi)$.
According to the equation, $\cos x = \sqrt \cdots \geqslant 0$, then $x \in [-\pi/2, \pi/2]$. Also inside the square root, $\sin x \geqslant \sqrt \cdots \geqslant 0$, then $x\in [0, \pi) \cup \{-\pi\}$. Thus $x \in [0, \pi/2]$ at least.
To be more exact, we consider
$$
\sin x \geqslant \sqrt {\sin x + \cos x},
$$
equivalently,
$$
\sin^2 x \geqslant \sin x + \cos x.
$$
But in $[0, \pi/2]$, $\cos x \geqslant 0$ and $\sin^2 x \leqslant \sin x$, thus
$$
\sin x \geqslant \sin^2 x \geqslant \sin x + \cos x \geqslant \sin x,
$$
which means all $\geqslant $ are $=$. The $=$ holds iff $\cos x = 0$ and $\sin x = 1$. Thus the only possible $x$ is $\boldsymbol {\pi/2}$.
Clearly $x = \pi/2$ is truly a solution to the equation, so the final solutions are
$$
\boxed {2n\pi +\frac \pi 2, n \in \mathbb Z }\ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral in terms of $u$ and simplified it as much as possible.
$$\implies \int\frac{1}{5+3(\frac{1-u^2}{1+u^2})} \cdot \frac{2du}{1+u^2}= \int \frac{1}{5+\frac{3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2} = \int \frac{1}{\frac{5+5u^2+3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}$$
$$=\int \frac{1}{\frac{8+2u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1}{\frac{2(4+u^2)}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1+u^2}{2(4+u^2)} \cdot \frac{2du}{1+u^2}=\int \frac{1}{4+u^2} \cdot du$$
I know that $\int \frac{1}{1+x^2}= \text{arctan(x)}$ but in my case I have a $4$ and not a $1$. My guess is that I have to use $u$-subsitution again but I can't figure out what $v$ is going to be. I've seen a lot of videos from BlackPenRedPen about $u$-substitution and I've noticed that he's able to see what the end result should be and choose his $u$ based on that, however I have no such talent. Therefore I'm confused as to what my new $u$ which will actually be a $v$ since I've already used a $u$, will be.
| Write $$4+u^2=4\left(1+\left(\frac{u}{2}\right)^2\right)$$ and substitute $$t=\frac{u}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find every $a>0$ such that $\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges We know that $\forall x \in [2,+\infty)$,
$$\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$
And it's easy to prove that $\int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx$ converges if and only if $a>3$.
But is it legitimate to conclude that our first integral converges if and only if $a>3$? The Comparison Criterion says that if $0 \leq f\leq g$, then, if $g$ converges, $f$ converges. But the converse isn't valid. Are we losing any possible values for $a$?
For example, the following equivalence holds:
$$\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx = \int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx - \int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$$
Shouldn't we prove that $\int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges to $0$?
I'm confused.
| Let $f(x)$ be the given integrand and $g(x)=x^{2-a}(1+\cos^{2}x)$. You can easily verify that $\frac {f(x)} {g(x)} \to 1$ as $ x \to \infty$. Hence $\int_2^{\infty} f(x)dx$ is finite iff $\int_2^{\infty} g(x)dx$ is finite. Now using the fact that $x^{2-a} \leq g(x) \leq 2x^{2-a}$ we can conclude that $\int_2^{\infty} g(x)dx$ is finite iff $\int_2^{\infty} x^{2-a}dx$ is finite. Clearly, this is true iff $a >3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$
Given postives $a, b, c$. Prove that $$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$$
This is not an overkill. Sure it is definitely not without the Buffalo Way and a computer borrowed from NASA or one used to study quantum computing.
One particularly out-of-the-ordinary (not unintelligible) way to solve this inequality is to let
$$\begin{cases}
a + b + c = 3\\
ab + bc + ca = 3(1 - m)\\
a^2 + b^2 + c^2 = 3(1 + 2m)
\end{cases} (0 \le m < 1)$$
(, which leads to nothing. This is not my idea at all.)
To prove the inequality between the first and second expression, we have that $$\sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} = \sum_{cyc}\frac{\dfrac{b^2}{c + 4a}}{b^2(3a + b + c)} \ge \frac{\displaystyle \sum_{cyc}\left(\dfrac{b}{\sqrt{c + 4a}}\right)^2}{\displaystyle \sum_{cyc}b^2(3a + b + c)}$$
$$ \ge \frac{\displaystyle \sum_{cyc}\left(\dfrac{b}{\sqrt{c + 4a}}\right)^2 \cdot \sum_{cyc}b(c + 4a)}{\displaystyle \left[(a^2 + b^2 + c^2)(a + b + c) + 2 \cdot \sum_{cyc}c^2a\right] \cdot \sum_{cyc}b(c + 4a)}$$
$$ \ge \frac{(a + b + c)^3}{\dfrac{5}{3}(a^2 + b^2 + c^2)(a + b + c) \cdot 5(ab + bc + ca)} = \frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)}$$
But I can't prove the one between the second and third. Help me.
| For the proof of the right inequality we can use BW.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$3(ab+ac+bc)\geq625a^2b^2c^2\left(\sum_{cyc}\frac{1}{(3a+b+c)(c+4a)}\right)^2,$$ which is something obvious:
https://www.wolframalpha.com/input/?i=3%28xy%2Bxz%2Byz%29-625x%5E2y%5E2z%5E2%281%2F%28%283x%2By%2Bz%29%284x%2Bz%29%29%2B1%2F%28%283y%2Bx%2Bz%29%284y%2Bx%29%29%2B1%2F%28%283z%2Bx%2By%29%284z%2By%29%29%29%5E2%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum $K$ such that $\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$ The question is to find the minimum $K$ such that $$\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$$ holds true for all non-negative $a,b$ and $c$.
My attempt:
I used Cauchy-Schwarz to get
$$\left(\sum_{cyc} \frac{a}{\sqrt{a+b}}\right)^2=\left(\sum_{cyc} \sqrt{a}\sqrt{\frac{a}{{a+b}}}\right)^2\leq(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$$
I now have to find the maximum value of $\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$. Putting $\frac ba=x, \frac cb=y$ and $\frac ac=z$, this reduces to the problem of finding the maximum value of $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$ under the constraint $xyz=1$.
I didn't get any further than this, and I'm pretty sure that my approach is wrong. For the last expression, we find that the case where $x=y=z=1$ is not the maximum as the expression is greater than $\frac32$ for $x=\frac12,y=\frac12,z=4$ (It is equal to $\frac{23}{15}$).
I found this similar question which doesn't have an answer yet.
Any help would be appreciated!
(I found this question here)
In this similar question, the maximum doesn't appear to be attained in the case where $x=y=z$.
| For $a=3$, $b=1$ and $c=0$ we obtain $k\geq\frac{5}{4}.$
We'll prove that $\frac{5}{4}$ is a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{5}{4}\sqrt{a+b+c}.$$
Now, by C-S
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sqrt{\left(\sum_{cyc}\frac{a}{\sqrt{a+b}}\right)^2}\leq\sqrt{\sum_{cyc}\frac{a}{2a+4b+c}\sum_{cyc}\frac{a(2a+4b+c)}{a+b}}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{2a+4b+c}\sum_{cyc}\frac{a(2a+4b+c)}{a+b}\leq\frac{25(a+b+c)}{16}$$ or
$$2\sum_{cyc}ab(4a+b)(a+2b)^2(a-3b)^2+$$
$$+abc\sum_{cyc}(122a^4+217a^3b+143a^3c+564a^2b^2+1338a^2bc)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\end{array}
$$
Continue this pattern until the box is $10 \times 10$.
Then add all the numbers together.
So this was my attempt.
Set the square $1\times1$ as $a_1$ square $2 \times 2$ as $a_2$ and so on. $a_1$ to $a_2$'s d is $(1 \cdot 2)+2 a_2$ to $a_3$'s d is $(1 \cdot 2)+(2 \cdot 2)+3$ and so on. So then I can the individual ds from each square to the next square.
And because
$$
1+(1+2)+(1+2+3)\ldots+(1+2+3+\ldots+10)
= \sum_{k=1}^{1} 1 + \sum_{k=1}^{2} 1+ \ldots +\sum_{k=1}^{10} 1=\sum_{k=1}^{10} \frac{n(n+1)}{2}
$$
so
$$
(1\cdot2)+(1\cdot2)+(2\cdot2)\ldots+(1\cdot2)+(2\cdot2)\ldots+(10\cdot2)
= \sum_{k=1}^{1} 2k + \sum_{k=1}^{2} 2k+\ldots+\sum_{k=1}^{10} 2k
$$
which also means
$$
\sum_{k=1}^{10} 2 \cdot \frac{n(n+1)}{2}
=\sum_{k=1}^{10} {n^2+n}
$$
And these are my calculations
$$\frac{10 \cdot 11 \cdot 21}{6}+\frac{10 \cdot 11}{2}$$
$$385+55=440$$
$$1+2+3+4+\ldots+10=55$$
$$440+55=495$$
It all seems right to me but the answer says its 385 is there any thing I did wrong?
| Consider the L-shaped figure $L_i$. $L_1$ adds $1.19$, $L_2$ adds $2.17$, $L_3$ adds $3.15$ and so on till $L_{10}$ adds $10.1$ for a total of $385$. $1.19+2.17+3.15+...+10.1=385$
Edit: Another way is to look at it as a 3d pyramid where each cube adds a $1$ to the sum. The ground floor has $10^2$ cubes, first floor has $9^2$ cubes and so on giving $10^2+9^2+...+1^2=385$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A circle of radius $r$ is inscribed into a triangle. A circle of radius $r$ is inscribed into a triangle $ABC$. Tangent lines to this circle parallel to the sides of the triangle cut out three smaller triangles, $\triangle A_cB_cC$, $\triangle A_bBC_b$, $\triangle AB_aC_a$. The radii of the circles inscribed in these smaller triangles are equal to $1$, $2$ and $3$, respectively. Find $r$.
I have no idea how to start...
|
Let $|CE|=H_c$ and $|CD|=h_c$ be the altitudes
of similar triangles $\triangle ABC$ and $\triangle A_cB_cC$.
Then
\begin{align}
\frac{|CD|}{r_c}
&=
\frac{|CE|}{r}
\tag{1}\label{1}
,\\
\frac{H_c-2r}{r_c}
&=
\frac{H_c}{r}
\tag{2}\label{2}
,\\
H_c &= \frac{2r^2}{r-r_c}
\tag{3}\label{3}
.
\end{align}
Similarly, two other altitudes of $\triangle ABC$
in terms of $r,r_a,r_b$ are
\begin{align}
H_a &= \frac{2r^2}{r-r_a}
\tag{4}\label{4}
,\\
H_b &= \frac{2r^2}{r-r_b}
\tag{5}\label{5}
,
\end{align}
and we can apply a well-known relation
\begin{align}
\frac1r&=
\frac1{H_a}+\frac1{H_b}+\frac1{H_c}
\tag{6}\label{6}
\end{align}
to find out that $r$ in terms of $r_a,r_b,r_c$ is just
\begin{align}
r&=r_a+r_b+r_c
\tag{7}\label{7}
.
\end{align}
The original question would be solved by now,
but we can do more than that: we can completely solve the $\triangle ABC$.
Using known Heron-like formula for the area, we have
\begin{align}
S&=
\frac1{\sqrt{
{(\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(-\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}-\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}+\tfrac1{H_b}-\tfrac1{H_c})}
}}
\\
&=\frac{r^{7/2}}{\sqrt{r_a r_b r_c}}
\tag{8}\label{8}
.
\end{align}
Next, we can find the semiperimeter $\rho$
and circumradius $R$ of $\triangle ABC$:
\begin{align}
\rho&=\frac Sr
=\frac{r^{5/2}}{\sqrt{r_a r_b r_c}}
\tag{9}\label{9}
,\\
R&=
\frac{2\,S^2}{H_a H_b H_c}
=\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c}
\tag{10}\label{10}
.
\end{align}
Now we are ready to find the three side lengths of $\triangle ABC$
as the roots of cubic equation in terms of $\rho,r,R$:
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,\rho\,r\,R&=0
\tag{11}\label{11}
.
\end{align}
In particular, for $r_a=1,\ r_b=2,\ r_c=3$ we have
\begin{align}
r&=6
,\quad
S=216
,\quad
\rho=36
,\quad
R=15
\tag{12}\label{12}
,
\end{align}
\eqref{11} becomes
\begin{align}
x^3-72\,x^2+1692\,x-12960&=0
\tag{13}\label{13}
\end{align}
with three roots $\{18,\, 24,\, 30\}$,
that is, the sought triangle is
the famous $3-4-5$ right-angled triangle,
scaled by $6$.
Note that the side lengths
are inversely proportional to corresponding radii of incircles.
For another example, the picture illustrates
a solution for $r_a=7,\ r_b=5,\ r_c=3$.
In this case we have $r=15$ and the side lengths are
\begin{align}
a&=\tfrac{120\sqrt7}7
,\quad
b=\tfrac{150\sqrt7}7
,\quad
c=\tfrac{180\sqrt7}7
\tag{14}\label{14}
.
\end{align}
Edit
In fact, solution of the cubic equation \eqref{11}
is unnecessary: since the area and the altitudes are known,
the side lengths can be found explicitly as
\begin{align}
a&=r\,(r-r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{15}\label{15}
,\\
b&=r\,(r-r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{16}\label{16}
,\\
c&=r\,(r-r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{17}\label{17}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial.....but it didnt help either.
| Observe that there is no cubic term, and the independent term is $3$. Hence the factorization must be
$$(x^2+kx+1)(x^2-kx+3)=x^4+(4-k^2)x^2+2kx+3.$$
By identification $k=4$.
The "standard" solution is by equating the remainder of the long division,
$$(8+14k-k^3)x+16-k^2$$ to zero. The only solution is $k=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = ... = \frac{3}{a_{n+1}} + a_n$ Prove that $a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$ If $$4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = \space ... \space = \frac{3}{a_{n+1}} + a_n$$
Prove that $$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$$
I tried to prove this via induction, but i't didn't work. Any hints with this problems?
| You want to prove
$$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1} \tag{1}\label{eq1}$$
From $4 = \frac{3}{a_1}$, you have $a_1 = \frac{3}{4}$, which matches \eqref{eq1} when $n = 1$ as it gives $a_1 = \frac{9 - 3}{9 - 1} = \frac{6}{8} = \frac{3}{4}$. Assume \eqref{eq1} holds for all $n \le k$ for some integer $k \ge 1$. Now, from the final part of your definition,
$$4 = \frac{3}{a_{n+1}} + a_n \; \implies \; 4 - a_n = \frac{3}{a_{n+1}} \; \implies \; a_{n+1} = \frac{3}{4 - a_n} \tag{2}\label{eq2}$$
Using $n = k$, along with the induction hypothesis, gives
\begin{align}
a_{k+1} & = \frac{3}{4 - a_k} \\
& = \frac{3}{4 - \frac{3^{k+1} - 3}{3^{k+1} - 1}} \\
& = \frac{3\left(3^{k+1} - 1\right)}{4\left(3^{k+1} - 1\right) - \left(3^{k+1} - 3\right)} \\
& = \frac{3^{k+2} - 3}{(4-1)3^{k+1} - 4 + 3} \\
& = \frac{3^{k+2} - 3}{3^{k+2} - 1} \tag{3}\label{eq3}
\end{align}
Note that this matches the right side of \eqref{eq1} for $n = k + 1$, so it's definition of $a_{k+1}$ matches. Thus, this completes the induction step, proving \eqref{eq1} is true for all $n \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find Sum of Sum of Partial Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $? How do I find the Sum of the Sum of Finitely Many Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $?
According to maple it is:
$\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} = \left( \left( m+1 \right) ^{2}-m-4 \right) \left( \Psi \left( m+2
\right) +\gamma \right) + \left( - \left( m+1 \right) ^{2}+2\,m+5
\right) \left( \Psi \left( m+3 \right) +\gamma \right) -3
$
| The iterated sum actually evaluates to
$$\sum_{b = 1}^m \sum_{k = 1}^b \frac{1}k = \left(m + 1\right)H_m - m.$$
You can show by regrouping terms that for a general sum
$$\sum_{a = 1}^x \sum_{b = 1}^a f(b) = \sum_{k = 1}^x \left(x - k + 1\right) f(k).$$
So plugging in $f(b) = \frac{1}b$, we obtain
\begin{align}
\sum_{a = 1}^x \sum_{b = 1}^a f(b) &= \sum_{a = 1}^x \sum_{b = 1}^a \frac{1}b \\
&= \sum_{k = 1}^x \left(x - k + 1\right) \frac{1}k \\
&= \sum_{k = 1}^x \left(\frac{x}k - 1 + \frac{1}k\right) \\
&= \left(x + 1\right)H_x - x.
\end{align}
I'm guessing that Maple just used a weird polygamma identity somewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Simplify $\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$ In relation to this question, consider the quantity
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$
when $n \in \mathbb{N}$ and $n \geq 2$.
Under these conditions, it should be a real number.
Is it possible to exploit the presence of $n$ both in the denominator and in the power, to further simplify the expression?
My attempt:
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n = \left( e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}} \right)^n = \left( i^{\frac{1}{n}} + \frac{1}{i^{\frac{1}{n}}} \right)^n = \left( \frac{i^{\frac{2}{n}} + 1}{i^{\frac{1}{n}}} \right)^n$$
But I am stuck here and I am no more sure that this is a real quantity.
Edit: trying to evaluate $2 \cos \left( \frac{\pi}{2n} \right)$ for some values of $n$, I obtain
$$n = 2; \ 2 \frac{\sqrt{2}}{2} = \sqrt{2}\\
n = 3; \ 2 \frac{\sqrt{3}}{2} = \sqrt{3}\\
n = 4; \ 2 \frac{\sqrt{2 + \sqrt{2}}}{2} = \sqrt{2 + \sqrt{2}}\\
n = 5; \ 2 \frac{\sqrt{5 + \sqrt{5}}}{2 \sqrt{2}} = \frac{\sqrt{5 + \sqrt{5}}}{\sqrt{2}}\\
n = 6; \ 2 \frac{\sqrt{2 + \sqrt{3}}}{2} = \sqrt{2 + \sqrt{3}}$$
I'm not sure about it being a regular succession. The $n = 5$ term is confusing.
As regards the binomial theorem applied to $\left( e^{i \frac{\pi}{2n}} + e^{-i \frac{\pi}{2n}} \right)^n$, I don't think to be able to manage those terms in the attempt to identify something useful.
| Don't see a reason to write in terms of imaginary numbers.
Distribute the power of $n$:
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$
$$= 2^n \left[\cos \left( \frac{\pi}{2n} \right) \right]^n$$
Using the half-angle identity Gerry mentioned:
$$\cos(\frac{x}{2})=\sqrt{\frac{1+\cos x}{2}}$$
Where $x=\pi/n$:
$$\cos(\frac{\pi}{2n})=\sqrt{\frac{1+\cos {\frac{\pi}{n}}}{2}}$$
Subbing back in:
$$= 2^n \left[\frac{1+\cos {\frac{\pi}{n}}}{2} \right]^{n/2}$$
Since $1+\cos {\theta}$ is always positive, this expression will result in a real number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expressing $\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k\,2^k}$ as a finite sum involving $\zeta(\cdot)$, $Li_k(\cdot)$, $\pi$, and $\ln 2$ While working on the integral posted here, through a large amount of skulduggery, I managed to arrive at the following intriguing sum
$$\begin{align}\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k 2^k} &= 2\operatorname{Li}_5 \left (\frac{1}{2} \right ) + \ln 2 \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{7}{16} \zeta (3) \ln^2 2 \\
&- \frac{\pi^2}{36} \ln^3 2 + \frac{\pi^4}{90} \ln 2 - \frac{7}{96} \pi^2 \zeta (3) - \frac{27}{32} \zeta (5) + \frac{1}{30} \ln^5 2
\end{align}$$
Of course I do not know how to arrive at this result in a direct manner coming from it at the sum on the left.
So my question is, does anyone have any suggestions on how one would approach the evaluation of the sum on the left?
| We have $$S=\sum_{n\geq1}\frac{1}{n^{4}}\sum_{k=1}^{n}\frac{1}{k2^{k}}=\sum_{n\geq1}\frac{1}{n^{4}}\int_{0}^{1}\frac{1-x^{n}2^{-n}}{2-x}dx$$ $$=\int_{0}^{1}\frac{1}{2-x}\left(\zeta\left(4\right)-\mathrm{Li}_{4}\left(\frac{x}{2}\right)\right)dx.$$
Now, since we can split the integral, we obtain $$
S=\zeta\left(4\right)\log\left(2\right)-\int_{0}^{1}\frac{\mathrm{Li}_{4}\left(\frac{x}{2}\right)}{2-x}dx$$ $$=\zeta\left(4\right)\log\left(2\right)-\sum_{k\geq0}\int_{0}^{1/2}x^{k}\mathrm{Li}_{4}\left(x\right)dx$$ $$=\zeta\left(4\right)\log\left(2\right)-\frac{1}{2}\sum_{k\geq0}\frac{1}{2^{k}}\sum_{m\geq1}\frac{1}{m^{4}\left(k+m+1\right)2^{m}}\tag{1}$$ and now we can expand the inner series in $(1)$ via partial fractions decomposition and evaluate every single term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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} |
Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$
Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.
My attempt:
*
*$x>3 \implies x^2 > 9$
*$ y < 2 \implies 5 + 2y < 9$
It follows that $5+2y < 9 < x^2$. From this we can see that $5+2y < x^2 \implies x^2 -2y > 5$
Therefore, if $x>3$ and $y<2$, then $x^2 − 2y > 5$.
Is it accurate?
| Yes, your proof is correct. Here is an alternative:
$x^2> 9$ and since $-y>-2$ we have $-2y>-4$, so $$x^2-2y>9-4 =5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the minimum value of $(\sin x+ \csc x)^2 + (\sec x + \cos x)^2$ I made it till $5+\tan^2x + \cot^2x$, but I don’t know how to proceed further.
| Now, by AM-GM
$$5+\tan^2x+\cot^2x\geq5+2=7.$$
The equality occurs for $\tan^2=\cot^2x,$ which says that we got a minimal value.
But for your problem we can use C-S:
$$\left(\sin{x}+\frac{1}{\sin{x}}\right)^2+\left(\cos{x}+\frac{1}{\cos{x}}\right)^2=5+\frac{1}{\sin^2x}+\frac{1}{\cos^2x}=$$
$$=5+(\sin^2x+\cos^2x)\left(\frac{1}{\sin^2x}+\frac{1}{\cos^2x}\right)\geq5+2^2=9.$$
The equality occurs for $(\sin{x},\cos{x})||\left(\frac{1}{\sin{x}},\frac{1}{\cos{x}}\right),$ which says that $9$ is a minimal value.
In both problems for equality occurring we can take $x=45^{\circ}.$
In the first solution I used $$a^2+b^2\geq2ab,$$ which is just $$(a-b)^2\geq0,$$
for $a=\tan{x}$ and $b=\cot{x}.$
In the second solution I used
$$(a^2+b^2)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\geq4,$$ which is
$$2+\frac{a^2}{b^2}+\frac{b^2}{a^2}\geq4$$ or
$$a^4+b^4-2a^2b^2\geq0$$ or
$$(a^2-b^2)^2\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ using the limit definition of a definite integral I can't evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ using the limit definition I need some help here's what I did:
Formula to use:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{k=1}^n f(c_k)\Delta x_k$$
Get the variables:
$$\Delta x_k : \frac{b-a}{n} = \frac{3-(-3)}{n} = \frac{6}{n}$$
$$c_k : a + k(\Delta x_k) = \left(-3+\frac{6k}{n}\right) = \left(\frac{-3n+6k}{n}\right)$$
Plug variables into formula and simplify observing the properties of sigma:
$$\begin{align}
\int_{-3}^3 \sqrt{9-x^2} dx &= \lim_{n\rightarrow \infty} \sum_{k=1}^n \left(\sqrt{9-\left(\frac{-3n+6k}{n}\right)^2}\right)\left(\frac{6}{n}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\left(\frac{-3n+6k}{n}\right)\left(\frac{-3n+6k}{n}\right)\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\frac{9n^2-36nk+36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\frac{9n^2}{n^2}-\frac{36nk}{n^2}+\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(9-\frac{36k}{n}+\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{\left(\frac{36k}{n}-\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{\frac{36nk-36k}{n^2}}\right)\\
\end{align}$$
That's as far as I got. According to the textbook, the answer to $\int_{-3}^3 \sqrt{9-x^2} dx$ is ${}^{9\pi}/_2$ square units. Where did I go wrong?
| Let's simplify the integral
\begin{align*}
\int_{-3}^3 \sqrt{9-x^2} dx &= \int_{-3}^3 \sqrt{9(1-\frac{x^2}{9})} dx = 3 \int_{-3}^3 \sqrt{1-\left(\frac{x}{3}\right)^2} dx = 9\int_{-1}^1 \sqrt{1-x^2} dx
\end{align*}
We have
\begin{align*}
\frac{2}{n} \sum_{k=0}^{n-1} \sqrt{1-\left(-1 + k \times \frac{2}{n}\right)^2} &= \frac{2}{n} \sum_{k=0}^{n-1} \sqrt{\frac{4k}{n} - \frac{4k^2}{n^2}}\\
&=\frac{4}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n} - \frac{k^2}{n^2}}\\
&=\frac{4}{n^2} \sum_{k=0}^{n-1} \sqrt{k(n-k)}\\
\end{align*}
Well, it seems to me that it is not possible to further simplify the expression. But we can still determine the value of the integral with a geometric interpretation: $\displaystyle\int_{-1}^1 \sqrt{1-x^2} dx$ corresponds to half of the area of a circle of radius $1$. Therefore
$$
\int_{-3}^3 \sqrt{9-x^2} dx =\frac{9\pi}{2}
$$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Differentiation of $\sin^{-1}(1/\sqrt{1+x})$ $$\arcsin \frac{1}{\sqrt{1+x}}.$$ I tried differentiating it but every time my answer came wrong. The answer in my book is $$- \frac{1}{2(x+1)\sqrt x}.$$
| The function
$$
f(x) = \arcsin{\frac{1}{\sqrt{1+x}}}
$$
can be seen as a composite function $f(x) = u(s(x))$, where $u(x) = \arcsin{x}$ and $s(x)= \frac{1}{\sqrt{1+x}}$. Their derivatives are (I will assume these as given - I won't prove these)
$$
u'(x) = \frac{1}{\sqrt{1-x^2}}
$$
$$
s'(x) = -\frac{1}{2(x+1)^{2/3}}
$$
Now, the derivative of a composite function is
$$
f'(x) = s'(x) u'(s(x)) = -\frac{1}{2(x+1)^{2/3}} \frac{1}{\sqrt{1-s^2(x)}}
$$
The latter terms is a bit tricky to evaluate, we should take it aside:
$$
\frac{1}{\sqrt{1-s^2(x)}} = \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+x}} \right)^2}}
= \frac{1}{\sqrt{1-\left(\frac{1}{1+x} \right)}}
= \frac{1}{\sqrt{\frac{1+x}{1+x}-\frac{1}{1+x}}}
= \frac{1}{\sqrt{\frac{x}{1+x}}}
= \frac{(1+x)^{1/2} }{ \sqrt{x} }
$$
therefore, the derivative can be simplified to
$$
f'(x) = -\frac{1}{2(x+1)^{2/3}} \frac{(1+x)^{1/2} }{ \sqrt{x} }
= -\frac{1}{2\sqrt{x}(x+1)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$
$ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
$ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator?
is there also alternative proof using jensen inequality?
| For $x_i>0$ with $i=1,\dots,n$, the given inequality is equivalent to
$$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$
Now note that $x\to 1/x$ is convex in $(0,+\infty)$ and use Jensen's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| You just missed one term. Note that by using the binomial theorem we get
$$(x^2-x+2)^{10}=(x(x-1)+2)^{10}=\sum_{k=0}^{10}\binom{10}{k}2^{10-k}x^k(x-1)^k$$
Hence, since $[x^3](x^k(x-1)^k)=0$ for $k\not\in \{2,3\}$, we have that
$$\begin{align}[x^3](x^2-x+2)^{10}&=[x^3]\sum_{k=2}^{3}\binom{10}{k}2^{10-k}x^k(x-1)^k\\&=\binom{10}{2}2^8(-2)+\binom{10}{3}2^7(-1)\\
&=-23040-15360=-38400.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
On the Clausen triple $8\rm{Cl}_2\left(\frac{\pi}2\right)+3\rm{Cl}_2\left(\frac{\pi}3\right)=12\,\rm{Cl}_2\left(\frac{\pi}6\right)$ While doing research on the Clausen function, I came across this nice identity,
$$8\operatorname{Cl}_2\left(\frac{\pi}2\right)+3\operatorname{Cl}_2\left(\frac{\pi}3\right)=12\operatorname{Cl}_2\left(\frac{\pi}6\right)$$
The two addends on the LHS are Catalan's constant and Gieseking's constants. It made me wonder if there were similar relations. Define,
$$\begin{aligned}
\text{Cl}_m\left(\frac{\pi}2\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}=\sum_{n=0}^\infty\left(\frac1{(4n+1)^m}-\frac1{(4n+3)^m}\right)\\
\text{Cl}_m\left(\frac{\pi}3\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m}=\frac{2^{m-1}+1}{2^m}\sum_{n=0}^\infty\left(\frac{\sqrt3}{(3n+1)^m}-\frac{\sqrt3}{(3n+2)^m}\right)\\
\text{Cl}_m\left(\frac{\pi}6\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m}
\end{aligned}$$
and,
$$\begin{aligned}
a &= 2^{m-1}(3^{m-1}+1)\\
b &= 3^{m-1}\\
c &= 2^m\,3^{m-1}
\end{aligned}$$
Q: How do we prove, with $a,b,c$ as defined above, that,
$$a\operatorname{Cl}_m\left(\frac{\pi}2\right)+b\operatorname{Cl}_m\left(\frac{\pi}3\right)=c\operatorname{Cl}_m\left(\frac{\pi}6\right)$$
or equivalently,
$$a\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}+b\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m} = c\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m}$$
for all integer $m>1$?
|
In general, assuming convergence, we have
$$ \sum_{n=1}^\infty \sin\left(\frac{\pi}{6}\,n\right)\, a_{n} = \sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n}
+ \sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\, \left(\frac12 a_{n}+\frac32 a_{3n} \right)$$
To see this, we partition the naturals into evens ($2n$) and odds ($6n+1,6n+3,6n+5$).
The evens give
$$\sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n} $$
while the odds give
$$\begin{align}
&\sum_{n=0}^\infty \sin\left(\frac{\pi}{6}\,(6n+1)\right)\, a_{6n+1} +\sin\left(\frac{\pi}{6}\,(6n+3)\right)\, a_{6n+3}
+\sin\left(\frac{\pi}{6}\,(6n+5)\right)\, a_{6n+5}
\\&=\sum_{n=0}^\infty \frac12 (-1)^n \, a_{6n+1} +(-1)^n \, a_{6n+3} +\frac12 (-1)^n \, a_{6n+5}
\\&=\sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\left(\frac12 a_{3n-2}+a_{3n}+\frac12 a_{3n+2}\right)
\end{align}$$
But, at the same time
$$\sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\left(a_{3n-2}-a_{3n}+ a_{3n+2}\right)
=\sum_{n=1} ^\infty \sin\left(\frac{\pi}{2}\,n\right)\, a_{n},$$
from which the claim follows.
Yours is the case $a_n = n^{-m}$.
| {
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"url": "https://math.stackexchange.com/questions/3304781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
| $ \frac{9^{n+1} + 4^{n+1}}{9^n + 4^n} = 6 $
$ 9^{n+1} + 4^{n+1} = 6\cdot 9^n + 6 \cdot 4^n $
$3 \cdot 9^n = 2 \cdot 4^n $
$ 3^{2n+1} = 2^{2n+1}$
$(\frac{3}{2})^{2n+1} = 1$
So, $2n+1 = 0 $
$ n = -\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
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Why does $\int_{-\infty}^{+\infty }\arctan\left(\frac{1}{1+x^2}\right)dx$ have a real value when the indefinite integral uses $i$? WolframAlpha gives a real closed form for this definite integral:
$$\int_{-\infty}^{+\infty } \arctan\left(\dfrac{1}{1+x^2}\right)dx = \sqrt{2\left(\sqrt{2}-1\right)}\;\pi$$
Yet, the formula it gives for the indefinite integral uses $i$.
$$\int \tan^{-1}\left(\frac{1}{x^2 + 1}\right) dx = x \tan^{-1}\left(\frac{1}{x^2 + 1}\right)
+ 2 \left(
\frac{\tan^{-1}\left(\frac{x}{\sqrt{1 - i}}\right)}{(1 - i)^{3/2}}
+
\frac{\tan^{-1}\left(\frac{x}{\sqrt{1 + i}}\right)}{(1 + i)^{3/2}}
\right) + C$$
Why isn't the definite integral non-real?
| A pretty integral, and I'm not sure I would be able to guess the primitive right away.
Still, substitution is always a good way to simplify things:
$$I=\int_a^b \arctan\left(\dfrac{1}{1+x^2}\right)dx, \\ 0 \leq a<b$$
$$x= \tan t$$
$$I=\int_{\arctan a}^{\arctan b} \arctan\left(\cos^2 t\right)\frac{dt}{\cos^2 t}$$
$$u = \cos t$$
$$I=\int_{\cos \arctan b}^{\cos \arctan a} \arctan\left(u^2 \right)\frac{du}{u^2 \sqrt{1-u^2}}$$
$$I=\int_{1/\sqrt{1+b^2}}^{1/\sqrt{1+a^2}} \arctan\left(u^2 \right)\frac{du}{u^2 \sqrt{1-u^2}}$$
$$v=u^2$$
$$I=\frac{1}{2} \int_{1/(1+b^2)}^{1/(1+a^2)} \arctan v \frac{dv}{v^{3/2} \sqrt{1-v}}$$
Now we can use integration by parts:
$$\int \frac{1}{2} \frac{dv}{v^{3/2} \sqrt{1-v}}=-\frac{\sqrt{1-v}}{\sqrt{v}}$$
$$(\arctan v)'= \frac{1}{1+v^2}$$
$$I=-\sqrt{\frac{1}{v}-1} \arctan v \bigg|_{1/(1+b^2)}^{1/(1+a^2)}+ \int_{1/(1+b^2)}^{1/(1+a^2)} \frac{\sqrt{1-v} ~dv}{\sqrt{v} (1+v^2)}$$
$$I=b \arctan \frac{1}{1+b^2}-a \arctan \frac{1}{1+a^2} + 2 \int_{1/\sqrt{1+b^2}}^{1/\sqrt{1+a^2}} \frac{\sqrt{1-u^2} ~du}{1+u^4}$$
For the second part we get back to trigonometric functions again:
$$\int_{1/\sqrt{1+b^2}}^{1/\sqrt{1+a^2}} \frac{\sqrt{1-u^2} ~du}{1+u^4}=\int_{\arcsin(1/\sqrt{1+b^2})}^{\arcsin(1/\sqrt{1+a^2})} \frac{\cos^2 w}{1+\sin^4 w} dw$$
Now we use the tangent half angle substitution:
$$p = \tan \frac{w}{2} \\ \cos w= \frac{1-p^2}{1+p^2} \\ \sin w= \frac{2p}{1+p^2} \\ dw = \frac{2dp}{1+p^2}$$
Denoting:
$$\alpha= \tan \left(\frac{1}{2} \arcsin \frac{1}{\sqrt{1+a^2}} \right) \\ \beta = \tan \left(\frac{1}{2} \arcsin \frac{1}{\sqrt{1+b^2}} \right)$$
$$\int_{1/\sqrt{1+b^2}}^{1/\sqrt{1+a^2}} \frac{\sqrt{1-u^2} ~du}{1+u^4}=\int_{\beta}^{\alpha} \frac{2 (1-p^2)^2 (1+p^2) }{(1+p^2)^4+16 p^4} dp$$
Now we have a rational function under the integral, which can be integrated using partial fractions.
I will stop here, because the point was, we don't really need CAS to integrate this, and no complex numbers were harmed in the process.
Well, it might be much easier to do the partial fractions in the last integral using complex numbers. But I'm sure we can avoid them.
Note that for the case in the OP, we have:
$$a=0, b=\infty$$
Which means we need to find:
$$\int_0^1 \frac{ (1-p^2)^2 (1+p^2) }{(1+p^2)^4+16 p^4} dp= \frac{1}{4} \sqrt{\frac{\sqrt{2}-1}{2}} ~ \pi$$
I confirmed this value numerically, but Mathematica can't find it from the integral.
Though it does find:
$$\int_0^\infty \frac{ (1-p^2)^2 (1+p^2) }{(1+p^2)^4+16 p^4} dp= \frac{1}{2} \sqrt{\frac{\sqrt{2}-1}{2}} ~ \pi$$
To be honest, complex residues seems like the best method in this case to me.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are There Infinitely Many Square Triangular Numbers? Toying around with triangular numbers, it seems that there are
already 4 squares below 1000. At least I get this one:
?- between(1,1000,N), T is N*(N+1)//2, sqrtrem(T,_,0).
N = 1,
T = 1 ;
N = 8,
T = 36 ;
N = 49,
T = 1225 ;
N = 288,
T = 41616 ;
No
Are there infinitely many square triangular numbers?
| To reproduce work
done many times before.
If
$T_n = m^2$,
since
$T_n = \dfrac{n(n+1)}{2}$,
this becomes
$n(n+1) = 2m^2$.
Completing the square,
$n^2+n+\frac14
=2m^2+\frac14
$
or
$(n+\frac12)^2
=2m^2+\frac14
$.
Clearing fractions,
this is
$(2n+1)^2
=8m^2+1
$
or
$(2n+1)^2-8m^2
=1
$.
This is a case
of a Pell equation
$x^2-dy^2 = 1$,
and the identity
$\begin{array}\\
(x^2-dy^2)(u^2-dv^2)
&=x^2u^2-d(x^2v^2+y^2u^2)+d^2y^2v^2\\
&=x^2u^2\pm 2dxuyv+d^2y^2v^2-d(x^2v^2\pm 2xuyv+y^2u^2)\\
&=(xu\pm dvy)^2-d(xv\pm yu)^2\\
\end{array}
$
shows that if there is one solution to
$x^2-dy^2 = 1$
then there are
an infinite number.
If $(x_0, y_0)$
satisfies
$x_0^2-dy_0^2 = 1$,
the recurrence is
$x_{n+1} = x_0x_n+dy_0y_n,
y_{n+1}=x_0y_n+y_0x_n
$.
If $d=8$,
since
$3^2-8\cdot 1^2 = 1$,
the recurrence is
$x_{n+1} = 3x_n+8y_n,
y_{n+1}=3y_n+x_n
$.
Starting with
$(x_0, y_0) =(3, 1)$,
and remembering that
the $n$ in $T_n$
satisfies $2n+1 = x$,
this gives
$
(x_1, y_1)
=(3\cdot 3+8\cdot 1, 3\cdot 1+3)
=(17, 6)
\quad (\dfrac{8\cdot 9}{2} = 6^2)\\
(x_2, y_2)
=(3\cdot 17+8\cdot 6, 3\cdot 6+17)
=(99, 25)
\quad (\dfrac{49\cdot 50}{2} = 25^2)\\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$ .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.
| Eliminating $x$ and $z$ between the equations we obtain:
$$3y^6-9y^4+6y^2-1=0.$$
By symmetry we know that $x$ and $z$ must also satisfy this equation. Hence $x^2$, $y^2$ and $z^2$ are roots of $$3y^3-9y^2+6y-1=0.$$
Therefore $(xyz)^2=\frac{1}{3}$ and hence $xyz=\pm\frac{1}{\sqrt 3}$.
In addition I think this may be one of the simplest direct solutions:
Adding the equations: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>xy+yz+zx=0.$$ Clearing the denominators of the original equations ($x^2-xy-1=0, \dots$) and adding we obtain $$x^2+y^2+z^2=xy+yz+zx+3=3$$ Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=3=>(x+y+z)=\pm\sqrt{3}$$Rearranging the original equations ($y+\frac{1}{x}=x,\dots$), multiplying them and simplifying: $$(y+\frac{1}{x})(z+\frac{1}{y})(x+\frac{1}{z})=xyz =>(x+y+z)=-\frac{1}{xyz}.$$ Combining we obtain the result $$xyz=\pm\frac{1}{\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
| $$4 \sin(x)\cos(2x)\sin(3x) - 1 = 32 \sin(x)^6 - 40 \sin(x)^4 + 12 \sin(x)^2 - 1$$
Letting $\sin(x) = s$, this is
$$ 32 s^6 - 40 s^4 + 12 s^2 - 1 = (2s+1)(2s-1)(8s^4 - 8s^2+1)$$
which has roots
$$ s = \pm \frac{1}{2},\; \pm \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
The values of $x$ in $(0,\pi)$ are $$\frac{\pi}{8},\; \frac{\pi}{6},\; \frac{3\pi}{8},\; \frac{5\pi}{8}\; \frac{5\pi}{6},\; \frac{7\pi}{8}$$
or in degrees, $$22.5,\; 30,\; 67.5,\; 112.5,\; 150,\; 157.5 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there any set of numbers that, when multiplied with anything from $\mathbb{N}$, results in an uneven number? I noticed the (very easy to see) fact that, whenever you multiply any natural number $n$ with anything having a prime factor of $2$, you get an even number.
For example:
$ 3 \times 2 = 6$, even
$5 \times 6 = 30 $, even
and so on. It's easy to see that this is true because everytime you multiply, you add prime factors to the set of the resulting set of prime factors. E.g.
$ 3 \times (2 \times 3) = 3 \times 6 = 18$, prime factors are $2, 3, 3$. And whenever you have a prime factor of 2, the number is even.
Now I ask myself: is it possible to construct a set $\mathbb{G}$ of numbers in a way that $\forall x \in \mathbb{G}: \forall n \in \mathbb{N}: \lfloor x \times n\rfloor = $ uneven? Maybe even without flooring them? $\mathbb{G}$ may be in $\mathbb{R}$, but it would be even more interesting without rounding and with $\mathbb{G} \subset \mathbb{N}$, but is this even possible?
It seems way easier to create even numbers than to create odd numbers from any set of natural numbers. Is this true?
| Any real number $r$ is equal to the limit of a series of this form:
$$
n + \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} +
\cdots + \frac{a_i}{2^i} + \cdots, \tag1
$$
where $n$ is an integer and each $a_i$ is either $0$ or $1.$
Moreover, if $r$ is any number in your desired set $\mathbb G,$ we can rule out the case where the limit is an integer (because then if we multiplied by $2$ the result would be even).
So we must have at least one $0$ term in the sequence, since the series
$$\frac12 + \frac1{2^2} + \frac1{2^3} + \cdots + \frac1{2^i} + \cdots$$
has limit $1$ and we would then have $r = n+1,$ an integer.
Now let $a_k = 0,$ that is, $k$ is a positive integer and the $k$th term is a zero term. Let $r$ be the limit of the series in $(1),$ and consider $2^k r,$ which is the limit of the series
$$
2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k + \frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots.
$$
Thus $2^k r$ is the sum of an integer
$$
2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k
$$
and the limit of the series
$$
\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots.
$$
Now there are two cases.
In the first case, the limit of the series
$\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots$
is $1,$ and therefore $2^k r$ is an integer,
$2^{k+1}r$ is an even integer, and $r \not\in \mathbb G.$
In the second case, the limit of the series
$\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots$
is less than $1,$ and therefore
$$
\lfloor 2^k r \rfloor = 2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k.
$$
But since $a_k = 0,$ and all the other terms have at least one factor of $2,$
it follows that $\lfloor 2^k r \rfloor$ is even and that $r \not\in \mathbb G.$
So in either case $r \not\in \mathbb G.$
There is then no way for any real number to be in your desired set $\mathbb G.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote
$y = -2x - 3/2$.
How we get this asymptote?
| You can apply simple algebraic transformations, being careful to always have positive expressions under the square root. For $x<0$, and thus $|x|=-x$, the expression is equal to
\begin{align}
=\frac{-x\sqrt{-x}}{\sqrt{3-x}}-x
&=\frac{x}{\sqrt{3-x}}(-\sqrt{-x}-\sqrt{3-x})\\
\\
&=-2x+\frac{x}{\sqrt{3-x}}(-\sqrt{-x}+\sqrt{3-x})
\\
&=-2x-\frac{-x}{\sqrt{3-x}}\frac{3}{\sqrt{-x}+\sqrt{3-x}}
=-2x-\frac{3}{\sqrt{1-\frac3x}+1-\frac3x}
\end{align}
which asymptotically is equal to $-2x-\frac32$ for $x\to-\infty$.
For $x>3$ the expression is equal to
$$
\frac{x}{\sqrt{x-3}}(\sqrt{x}-\sqrt{x-3})
=\frac{x}{\sqrt{x-3}}\frac{x-(x-3)}{\sqrt{x}+\sqrt{x-3}}
=\frac{3}{\sqrt{1-\frac3x}+1-\frac3x}
$$
which asymptotically converges to the constant $\frac32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.