Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$
For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$
Can any one help me with this problem?
| Multiplying both LHS and RHS by $a+b+c = s > 0$, you have:
$$\sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot (a+b) \right)+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3(a+b+c) $$
$$\iff 6+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3s $$
Now note that $a^2+b^2 \ge \frac12(a+b)^2$ using AM-GM, so it is enough to show that
$$6 + \frac12\sum_{cyc}(s-c)c \ge 3s \iff 6 + \frac{s^2}2-\frac32 \ge 3s \iff (s-3)^2\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? I'm trying solve: $~a^3 + b^3 = c^3~$ has no nonzero integer solutions.
If $~(c−b)=1~$ then $~c^3-b^3=3c^2-3c+1=a^3,~$
from Wolframalpha get:
$$
c = \dfrac{3- \sqrt{3}\sqrt{4a^{3}-1}}{6} \\
c = \dfrac{\sqrt{3}\sqrt{4a^{3}-1}+3}{6}
$$
How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? (eidt: while $~a,\ b,\ c ~$ are nonzero integers)
| You have to prove that
$$ 12 a^3 = b^2+3 \tag{1}$$
has no integer solutions. Since $3\mid$LHS, we must have $b=3c$, hence:
$$ 4 a^3 = 3c^2+1,\tag{2}$$
and $c$ must be odd, hence $c=2d+1$ and:
$$ a^3 = 3 d^2 + 3 d + 1 = (d+1)^3-d^3.\tag{3}$$
So you just have to prove that the difference between two consecutive cubes is never a cube (except for $1^3-0^3=1^3$). This is usually achieved by splitting the RHS of $(3)$ over $\mathbb{Z}[\sqrt{-3}]$ and by using the fact that the ring of Eisenstein integers $\mathbb{Z}[\omega]$ is an euclidean domain, hence a unique factorization domain. We can use Fermat's descent too, since $(3)$ implies $a=3e+1$ and:
$$ 3e(3e^2+3e+1)= d(d+1).\tag{4}$$
| {
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$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| Since $(a_1)^2(b-1)^2(c-1)^2\geq0$, we can assume that $(a-1)(b-1)\geq0$ or
$c(a-1)(b-1)\geq0$ or $abc\geq ac+bc-c$.
Thus, it remains to prove that
$$a^2+b^2+c^2+ac+bc-c+5\geq3(a+b+c)$$ or
$$c^2+(a+b-4)c+a^2+b^2-3a-3b+5\geq0,$$
for which it's enough to prove that
$$(a+b-4)^2-4(a^2+b^2-3a-3b+5)\leq0$$ or
$$(a-b)^2+2(a-1)^2+2(b-1)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How show that $ABC$ is equilateral? Let $D$, $E$ and $F$ be three points on sides $BC$,$AC$ and $AB$ of triangle $ABC$ such that lines $AD$, $BE$ and $CF$ concur at point $M$. If three trianles $MDB$, $MCE$ and $MAF$ have equal areas and equal perimeters. How show that $ABC$ is equilateral?
| This problem gave me a really hard time, lots of steps are involved.
Step 1. If $ABC$ is equilateral and $MAF,MBD,MCE$ have the same area, then $M$ is the center of $ABC$.
Given that $[x,y,z]$ are the trilinear coordinates of $M$, the areas of our triangles are proportional to: $$\frac{xy}{x+z},\frac{yz}{x+y},\frac{xz}{y+z}$$
hence $M=[1,1,1]$ is the only chance.
Step 2. If the areas of $MAF,MBD,MCE$ are equal, then $M$ is the centroid of $ABC$.
Take an affine map $\Phi$ that brings our original triangle into an equilateral one. Since the affine maps preserve the ratio between the areas, $\Phi(M)$ is the centroid of $\Phi(ABC)$ due to Step1, hence $M$ is the centroid of $ABC$ and $D,E,F$ are the midpoints of the sides.
Call now $m_a,m_b,m_c$ the lengths of the medians through $A,B,C$ and $a,b,c$ the lengths of the sides $BC,AC,AB$ respectively.
Step 3. If $a>b$, then $m_a<m_b$.
This just follows from the Stewart's theorem, granting:
$$ 4m_a^2 = 2b^2+2c^2-a^2,\qquad 4m_b^2=2a^2+2c^2-b^2.$$
Step 4. If $a>b$, then $\frac{a}{2}+\frac{m_a}{3}>\frac{b}{2}+\frac{m_b}{3}$.
This is the crucial part. It is straightforward to check that the inequality is equivalent to:
$$ a^2+b^2-ab > 2c^2-4m_a m_b,$$
but since $a^2+b^2\geq 2ab$, it is sufficient to prove the weaker inequality:
$$2c^2-ab< 4m_a m_b$$
that is equivalent to:
$$a^4+b^4 < 2a^2b^2 + a^2c^2+b^2 c^2+2abc^2$$
or to:
$$(a^2-b^2)^2 < (a+b)^2 c^2$$
or to:
$$(a-b)^2< c^2$$
that is just the triangle inequality $|a-b|<c$. Phew.
Suppose now that $ABC$ is not equilateral. We can assume without loss of generality that $a\geq b\geq c$ and at least one of the two inequalities holds tight. But due to Step3 and Step4 we have:
$$\left(\frac{a}{2}+\frac{m_a}{3}\right)+\frac{2m_b}{3}>\left(\frac{c}{2}+\frac{m_c}{3}\right)+\frac{2m_a}{3},$$
hence the triangles $MAF,MBD$ and $CEF$ cannot all have the same perimeter, contradiction.
| {
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Proving that (4-2/1)(4-2/2)...(4-2/n) in an integer. We have to prove that $(4-2/1)(4-2/2)...(4-2/n)$ is an integer for $n\in\mathbb{N}$. Can we do this by induction?
We prove for $n = 1$, which is trivial as $(4-2/1) = 2$ which is clearly an integer.
next we assume that the statement is true for some integer $n$
i.e. that $(4-2/1)(4-2/2)...(4-2/n) = p$ where p is an integer, and we prove for $n+1$.
So we need to show that $(4-2/1)(4-2/2)...(4-2/n)(4-2/(n+1))$ is also an integer. Which is the same as saying $p\times(4-\frac{2}{n+1})$ is an integer.
We can rewrite $p$ as $\dfrac{2^n(1\times3\times5\times...\times(2n-1))}{n!}$ so essentially all we need to prove is that $n+1$ divides the numerator of $p$. So assume that $n$ is even, which implies that $n+1$ is odd and certainly less than $2n-1$ so therefore it will divide one of the odd numbers in the numerator of $p$. If $n$ is odd, then $n+1$ is even and will not have more factors of $2$ than $2^n$. So if the prime factorization of $n+1$ contains $k$ factors of 2, $k<n$ or $k=n$, implying that $2^k$ divides $2^n$. Also the prime factorisation of $n+1$ will not have have an odd factor that is greater than $2n-1$ as it already has a factor of 2. Therefore $n+1$ divides the numerator of p.
Can we do this? It seems like it is not entirely correct! I may have made a big mistake in my reasoning.
Any help is appreciated.
| [Cannot seem to add a comment (re request above on the induction method), so posting a new answer in response]
Assume result is true for $n$ terms, i.e. product is
$$2n\choose n$$
which is an integer.
Then, for $n+1$ terms, product becomes
$$\begin{align}
{2n\choose n}\left( 4-\frac 2{n+1}\right) &=\frac{4n+2}{n+1} {2n\choose n} \\
&=2\cdot \frac{2n+1}{n+1}\cdot {2n\choose n} \\
&=\frac {2n+2}{n+1}\cdot \frac{2n+1}{n+1}\cdot\frac{2n(2n-1)...(n+2)(n+1)}{1\cdot2\cdot ... n} \\
&=\frac {(2n+2)(2n+1)(2n)...(n+2)}{1\cdot 2\cdot ... n(n+1)}\\
&={{2n+2}\choose{n+1}}\\
&={{2(n+1)}\choose{n+1}}\\
\end{align}$$
i.e. also true for $n+1$.
It can easily be shown that the product is true for $n=1$.
Hence, by induction, the proposition is true for all integer $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate the integral $\int e^{2x} \sin^2x\ dx$ How can I evaluate the following integral?
$$\int e^{2x} \sin^2x\ dx$$
| Using operator D and its properties, we have
$$
\int e^{2x}\sin^2xdx = \dfrac{1}{2D}[e^{2x} - e^{2x}\cos(2x)] = \dfrac{1}{4}e^{2x} - \dfrac{e^{2x}}{2}\dfrac{1}{D+2}\cos(2x)
$$
$$
= \dfrac{1}{4}e^{2x} - \dfrac{e^{2x}}{2}\dfrac{D - 2}{D^2 - 4}\cos(2x) = \dfrac{1}{4}e^{2x} + \dfrac{e^{2x}}{16}[-2\sin(2x) - 2\cos(2x)]
$$
$$
= \dfrac{1}{4}e^{2x} - \dfrac{e^{2x}}{8}[\sin(2x) + \cos(2x)] + C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of infinite $ \sum (\frac{n}{n+1})^{n^2} $? I have being trying to solve this convergence but with no success. Using the ratio test I have reached here:
$$
a_n = \left( \frac{n}{n+1} \right)^{n^2}
$$
And
$$
\frac{1}{a_n} = \left( \frac{n+1}{n} \right)^{n^2}
$$
Also
$$
a_{n+1} = \left(\frac{n+1}{n+2}\right)^{{(n+1)}^2} = \left(\frac{n+1}{n+2}\right)^{n^2+2n+1} = \\ a_{n+1} = \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right)
$$
Therefore
$$
\lim \frac{a_{n+1}}{a_n} = \lim \frac{1}{a_n} \cdot a_{n+1} = \\
\lim \left[ \left( \frac{n+1}{n} \right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right) \right]
$$
And I dont know how to go any further, I have tried many possibilities none of them reaching any result.
| Instead of the ratio test, try the root test:
$\displaystyle\lim_{n \to \infty}|a_n|^{1/n} = \lim_{n \to \infty}\left|\left(\dfrac{n}{n+1}\right)^{n^2}\right|^{1/n} = \lim_{n \to \infty}\left(\dfrac{n}{n+1}\right)^{n} = \lim_{n \to \infty}\left(1-\dfrac{1}{n+1}\right)^{n}$
Do you recongnize this limit?
| {
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Solution of equation $\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$
Solution of equation $\displaystyle \frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$
$\bf{My\; Try::}$ Clearly Here $x>0$, Now Using $\bf{A.M\geq G.M}$
So Here $\displaystyle x\cdot 2014^{\frac{1}{x}}>0$ and $\displaystyle \frac{1}{x}\cdot 2014^x>0$ .
So $\displaystyle \left(\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2}\right)\geq \left(x\cdot 2014^{\frac{1}{x}}\cdot \frac{1}{x}\cdot 2014^x\right)^{\frac{1}{2}} = \sqrt{2014^{\left(x+\frac{1}{x}\right)}} = 2014$
And equality Hold , When $\displaystyle x = \frac{1}{x}\Rightarrow x= 1$
Can we solve it without Using $\bf{A.M\geq G.M}$
If Yes, Then please explain me.
Thanks
| Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that $x\ge0$, we have:
$$\begin{align}
&\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014
\\\implies&x2014^{\frac{1}{x}-1}+\frac{1}{x}2014^{x-1}=2
\\\implies&\left(\sqrt{x2014^{\frac{1}{x}-1}}-\sqrt{\frac{1}{x}2014^{x-1}}\right)^2+2\sqrt{2014^{\frac{1}{x}+x-2}}=2
\\\implies&\left(\sqrt{x2014^{\frac{1}{x}-1}}-\sqrt{\frac{1}{x}2014^{x-1}}\right)^2+2\sqrt{2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}}=2
\\\implies&2\sqrt{2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}}\le2
\\\implies&2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}\le1
\\\implies&\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2\le0
\\\implies&\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2=0
\\\implies&\sqrt{\frac{1}{x}}-\sqrt{x}=0
\\\implies&\frac{1}{x}=x
\\\implies&x=1
\end{align}$$
where we have used the fact that $x\ge 0$ several times.
| {
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How to solve $\log _{x^{2}-3}(x^{2}+6x)<\log _{x}(x+2)$? How to solve the following inequality $$\log _{x^{2}-3}\left(x^{2}+6x\right)<\log _{x}(x+2)\ ?$$
| Note first that $x \in A \cup B$ where $A = (\sqrt3, 2), \; B = (2,\infty)$, for the question to make sense.
Now if $x \in A$, note that $0< x^2-3 < 1 < x < x+2 < x^2+6x$, so LHS $< 0$, while $RHS > 0$ so this is a solution.
For $x \in B$, we have $ 1 < x < x+2 $ and $1 < x^2-3 < x^2+6x$, so both LHS and RHS are positive. Here we have the equivalent inequality:
$$\log_x (x+2) > \log_{x^2-3} (x^2+6x) \iff \log (x^2-3) \log(x+2) > \log x \log (x^2+6x)\\ \iff \log_x (x^2-3) > \log_{x+2} (x^2+6x)$$
Now note that $x^2 - 3 < x^2$ while $x^2+6x > (x+2)^2$ for $x \in B$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrix Multiplication Understanding I have to multiply two 3x3 matrices. I don't understand the answer.
$$A=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
$$B=\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$
The answer given online is $$AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
I don't understand how this is so. I have said that:
$$AB=\begin{pmatrix}(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1)\\(0)(1)+(1)(0)+(0)(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\ (0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1) \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & (1)(-1)+(0)(-1)+(1)(1)\\0 & 1 & (0)(-1)+(1)(-1)+(1)(1) \\ 0 & 0 & 1 \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Can anyone tell me where I am going wrong?
| $\text{You have the right answer.}$
$$AB=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}$$
| {
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{2\choose 2}\dfrac{1}{k!}$.
For $n$ odd ($n=2k+1$) there are ${n\choose 2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{3\choose 2}\dfrac{1}{k!}$
But how do I find the sums?
Seems like I have to use induction. But not quite upto there.
Thanks for the help!!
| If you accept a confluent hypergeometric function as a closed form, then a Computer Algebra System like Mathematica will give you
for even $n = 2w$:
$\left(-\frac{1}{2}\right)^{-w} U\left(-w,\frac{1}{2},-\frac{1}{2}\right)-1$
and for odd $n = 2w-1$:
$\left(-\frac{1}{2}\right)^{1-w} U\left(1-w,\frac{3}{2},-\frac{1}{2}\right)-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sphere equation given 4 points Find the equation of the Sphere give the 4 points (3,2,1), (1,-2,-3), (2,1,3) and (-1,1,2).
The *failed* solution I tried is kinda straigh forward:
We need to find the center of the sphere.
Having the points:
$$
p_{1}(3,2,1),\,
p_{2}(1,-2,-3),\,
p_{3}(2,1,3),\,
p_{4}(-1,1,2)
$$
2 Triangles can be created using these points, let's call $A$ our triangle formed by the points $p_{1},\,p_{2}\, and\,\, p_{3}$; And $B$ our triangle formed by the points $p_{1},\, p_{3}\, and \,\,p_{4}$.
Calculate the centroids of each triangle:
$$
CA = (2,1/3,1/3)\\
CB = (4/3,4/3,2)
$$
And also, a normal vector for each triangle:
$$
\overrightarrow{NA} = \overrightarrow{p_{1}p_{2}} \times \overrightarrow{p_{1}p_{3}}\\
\overrightarrow{NB} = \overrightarrow{p_{1}p_{3}} \times \overrightarrow{p_{1}p_{4}}
$$
$$
\overrightarrow{p_{1}p_{2}} = <-2,-4,-4>\\
\overrightarrow{p_{1}p_{3}} = <-1,-1,2>\\
\overrightarrow{p_{1}p_{4}} = <-4,-1,1>\\
\:\\
\overrightarrow{NA} = <-12, 8, -2>\\
\overrightarrow{NB} = <1, -7, -3>\\
$$
With the centroids and normals of triangles $A$ and $B$, we can build two parametric equations for a line, the first one intersects the centroid of $A$ and the other one the centroid $B$.
$$
Line \enspace A\\
x = 2-12t \quad
y = 1/3+8t \quad
z = 1/3-2t\\
\:\\
Line \enspace B\\
x = 4/3 + s \quad
y = 4/3 - 7s \quad
z = 2 - 3s
$$
The point where this lines intersect should be the center of the sphere, unfortunately this system of equations is not linearly dependent, that means that they do not intersect each other. What could be the problem here?
| I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere $(x-a)^2+(y-b^2)+(y-c)^2=R^2$
| {
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| We need to prove that
$$\frac{a^2+b^2+c^2}{ab+ac+bc}-1\geq1-\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{2(ab+ac+bc)}\geq\frac{\sum\limits_{cyc}c(a-b)^2}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq0.$$
Let $a\geq b\geq c$.
Hence, $(a-c)^2\geq (b-c)^2$, $$a^2b+a^2c+b^2a+b^2c-c^2a-c^2b\geq0$$ and $$a^2b+a^2c+c^2a+c^2b-b^2a-b^2c\geq0.$$
Thus,
$$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq$$
$$\geq(b-c)^2(a^2b+a^2c+c^2a+c^2b-b^2a-b^2c+b^2a+b^2a+c^2a+c^2b-a^2b-a^2c)=$$
$$=2(b-c)^2c^2(a+b)\geq0.$$
Done!
| {
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Product rule for simplex numbers The $n$th triangular number is defined as $T_2(n) = n(n+1)/2$, and there is an interesting product rule for triangular numbers:
$$T_2(mn) = T_2(m)\,T_2(n) + T_2(m-1)\,T_2(n-1).$$
The tetrahedral numbers $T_3(n) = n(n+1)(n+2)/6$ satisfy a similar product rule:
$$T_3(mn) = T_3(m)\,T_3(n) + 4 T_3(m-1)\,T_3(n-1) + T_3(m-2)\,T_3(n-2).$$
Can these product rules be generalized to higher dimensions? Are there analogous formulas for $T_k(mn)$, where $$T_k(n) = \frac1{k!} n(n+1)(n+2)\cdots(n+k-1)\ ?$$
| By denoting with $(n)_k$ the falling Pochhammer symbol
$$(n)_k = n\cdot(n-1)\cdot\ldots\cdot(n-k+1)$$
we have:
$$ (mn+1)_2 = \frac{1}{2}\left((m+1)_2(n+1)_2+(m)_2(n)_2\right)\tag{1}$$
that leads to the first identity. Since:
$$ (mn+2) = \alpha(n+2)(m+2)+\beta(n-1)(m-1) $$
is solved by $\alpha=\frac{1}{3}$ and $\beta=\frac{2}{3}$ and
$$ (mn+2) = \gamma(n+1)(m+1)+\delta(n-2)(m-2)$$
is solved by $\gamma=\frac{2}{3}$ and $\delta=\frac{1}{3}$, multiplying both sides of $(1)$ by $(mn+2)$ we get:
$$ (mn+2)_3 = \frac{1}{6}\left((m+2)_3(n+2)_3+4(m+1)_3(n+1)_3+(m)_3(n)_3\right)\tag{2}$$
that leads to the second identity.
Trying to reproduce the same argument, we may look for $\alpha$ and $\beta$ such that:
$$(mn+3)=\alpha(m+3)(n+3)+\beta(m-1)(n-1),$$
so equating the coefficients of $mn$ we get $\alpha+\beta=1$ and equating the coefficients of $(m+n)$ we get $3\alpha-\beta=0$, from which $\alpha=\frac{1}{4},\beta=\frac{3}{4}$. The constant terms match since $\frac{9}{4}+\frac{3}{4}=3$. So we go on and try to find $\gamma,\delta$ such that:
$$(mn+3)=\gamma(m+2)(n+2)+\delta(m-2)(n-2).$$
However, by choosing $\gamma=\delta=\frac{1}{2}$ the constant term does not match. Ouch. This gives that the formula for pentatope numbers depends on tetrahedral numbers, too. For istance, by multiplying both sides of $(2)$ by $(mn+3)$ we get:
$$\begin{array}{rcl}(mn+3)_4 &=& \frac{1}{24}\left((m+3)_4(n+3)_4+11(m+2)_4(n+2)_4\right.
\\&+&\left.11(m+1)_4(n+1)_4+(m)_4(n)_4\color{red}{-16(m+1)_3(n+1)_3}\right).\end{array}\tag{3}$$
If we don't care introducing further terms, we can simply go on as we did for producing $(3)$ from $(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12} $.
Of course, the solution provided was one that makes use of a generating function. However, it is unbelievably tedious. Is there a way to do this without making use of a generation function?
For reference, here is the solution given in the book.
If $f(n)$ is that number, then
$$ \begin {gather*}
\sum_{n\ge 0} f(n) x^n =
\frac {1}{(1-x)(1-x^2)(1-x^3)}\qquad\qquad\qquad\qquad \\
= \frac{1}{6(1-x)^3} + \frac{1 }{4(1-x)^2} +
\frac {17}{72(1-x)} + \frac {1}{8(1+x)} + \frac {1}{9(1-\omega x)} +
\frac {1}{9(1-\overline{\omega}x)}.
\end {gather*} $$
If we expand each of
the fractions on the right, we find the formula $$ f(n) = \frac {1}{6}
\dbinom {n+2}{2} + \frac {1}{4} (n+1) + \frac {17}{72} + \frac
{(-1)^n}{8} + \frac {2}{9} \cos \left( \frac {2n \pi}{3} \right), $$
which can be rewritten as
$$ f(n) = \frac {(n+3)^2}{12} + \frac {-7 + 9 (-1)^n
+ 16 \cos \left( \frac {2n\pi}{3} \right)}{72}. $$
The second
fraction cannot exceed $ \frac {32}{72} < \frac {1}{2} $ in absolute
value, so $f(n)$ is the unique integer who distance from
$ \frac {(n+3)^2}{12} $ is less than $\frac{1}{2}$, as required. $\Box$
| The following method could be explained to any student that understands equations for lines and knows how to sum an arithmetic sequence. However, you may find it at least as tedious as the method you already have!
The Idea
Think of points $(x,y)$ in the plane as representing "$x$ 2-cent coins and $y$ 3-cent coins".
Now, the grid points in the first quadrant on or below the line $\mathcal L_N$ given by $$2x+3y=N$$ are in one-to-one correspondence with ways of changing $N$ cents.
Since the area of this triangular region is $\frac 12\frac N2 \frac N3=\frac{N^2}{12}$, we can already see that the number of points should be on the order of $\frac{N^2}{12}$. So far, so good.
An Exact Calculation
To find the exact number of points, we will consider the number $k_N$ of first-quadrant points that lie on $\mathcal L_N$, for each value of $N$.
(Yes, we are taking the first quadrant to include its boundary.)
Since every grid point $(x,y)$ lies on some $\mathcal L_N$ (namely $\mathcal L_{2x+3y}$), the total number $A_N$ of first quadrant grid points on or below $\mathcal L_N$ is given by $A_N = \sum_{n=0}^{N} \mathcal k_n$.
We can see that each $\mathcal L_N$ meets grid points at regular intervals, namely with an inter-point spacing of $\overrightarrow{(3,-2)}$.
Since this is exactly the amount by which the first-quadrant portion of $\mathcal L_N$ lengthens when $N$ increases by $6$, and each grid point on $\mathcal L_N$ is both directly below (by $2$ units) and directly to the left of (by $3$ units) a grid point on $\mathcal L_{N+6}$, we see that $k_{n+6}=k_n+1$.
Thus it suffices to find $k_0$ through $k_5$, and by inspection we see that they are all equal to $1$ except for $k_1$ which is $0$.
$$\begin{eqnarray}(k_0,k_1,k_2,\ldots) =
& 1,0,1,1,1,1, \\
& 2,1,2,2,2,2, \\
& 3,2,3,3,3,3, \\
& etc.
\end{eqnarray}$$
This regularity allows us to find exact formulas for $A_N$ depending on $N$ mod $6$.
$$\begin{eqnarray}
(A_0,\ A_6,\ A_{12},\ \ldots) &=& (1,\ 1+6,\ 1+6+12,\ \ldots) & \\
\mbox{So }\ \ \ \ A_{n=0+6m} &=& 1+6m(m+1)/2 &=& (n^2+6n+12)\;/\;12 \\
\mbox{Similarly, }\ \ \ \ A_{n=1+6m} &=& 1+6m(m+1)/2+m &=& (n^2+6n+5)\;/\;12 \\
A_{n=2+6m} &=& 1+6m(m+1)/2+2m+1 &=& (n^2+6n+8)\;/\;12 \\
A_{n=3+6m} &=& 1+6m(m+1)/2+3m+2 &=& (n^2+6n+9)\;/\;12 \\
A_{n=4+6m} &=& 1+6m(m+1)/2+4m+3 &=& (n^2+6n+8)\;/\;12 \\
A_{n=5+6m} &=& 1+6m(m+1)/2+5m+4 &=& (n^2+6n+5)\;/\;12 \\
\end{eqnarray}$$
We can see that this is always within $\left[\,\frac{\!\!-4}{12}, \frac{3}{12}\right]$ of $(n+3)^2\,/\,12$.
This means that $A_n$ is always the integer nearest to $(n+3)^2\,/\,12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906363",
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"answer_count": 4,
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} |
If $ \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$
If $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$.
$\bf{My\; Try::}$ Given $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\Rightarrow 13!\left(1+\frac{1}{2}+\frac{1}{3}+................+\frac{1}{13}\right) = x$
So $\displaystyle x = \left(13!+\frac{13!}{2}+\frac{13!}{3}+.................+12!\right)$
Now How Can I solve after that
Help me
Thanks
| Imagine bringing the left side to the common denominator $13!$. The numerator $x$ is then a sum of terms of the form $\frac{13!}{r}$. All these terms are congruent to $0$ modulo $11$ (divisible by $11$) except $\frac{13!}{11}$.
So the required remainder is the same as the remainder when $(10!)(12)(13)$ is divided by $11$.
Now by Wilson's Theorem, we have $10!\equiv -1\pmod{11}$. And $(12)(13)\equiv 2\pmod{11}$. Thus $x\equiv -2\equiv 9\pmod{11}$.
Without Wilson's Theorem, we could just calculate directly using $(10!)(12)(13)$.
| {
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"url": "https://math.stackexchange.com/questions/906595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where did I go wrong in completing the square? $$2x^{ 2 }+8x+1=0$$
Move 1 to the other side of the equation:
$$2x^{ 2 }+8x\quad =-1$$
Divide both sides by 2 to get 1 as the leading coefficient:
$$x^{ 2 }+4x\quad =-\frac { 1 }{ 2 } $$
$$(\frac { 4 }{ 2 } )^{ 2 }$$
$$x^{ 2 }+4x+4\quad =-\frac { 1 }{ 2 } +4$$
$$(x+2)^{ 2 }=3.5$$
I get to this and it seems to be the wrong answer:
$$x={ -2 }\pm\sqrt { \frac { 7 }{ 2 } } $$
| Your calculations are right..
$$(x+2)^2=\frac{7}{2} \Rightarrow x+2=\pm \sqrt{\frac{7}{2}} \Rightarrow x=-2 \pm \sqrt{\frac{7}{2}}=-2 \pm \frac{\sqrt{14}}{2}$$
So:
$$x=-2 \pm \frac{\sqrt{14}}{2}$$
| {
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"source": "stackexchange",
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If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$ if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
My try :
$2pp' = (b^2-a^2)\sin 2\theta$
$p'^2 + pp'' = (b^2-a^2)\cos 2\theta$
Thats it ! it doesn't simplify no matter what I try. Any help ?
| We need to establish $$p+p''=\frac{a^2b^2}{p^3}$$
Multiplying by $2p'(\ne0),$ $$(p+p'')2p'=a^2b^2\cdot\frac{2p'}{p^3}$$
Integrating we get,
$$p^2+p'^2=-\frac{a^2b^2}{p^2}+K\text{, where }K\text{ is an arbitrary constant} $$
Multiplying by $p^2(\ne0),$
$$p^4+p'^2p^2=-a^2b^2 +p^2\cdot K\ \ \ \ (1)$$
Now, $$p^4+p'^2p^2=(p^2)^2+(pp')^2=(a^2\cos^2\theta+b^2\sin^2\theta)^2+[(b^2-a^2)\sin\theta\cos\theta]^2$$
$$=a^4\cos^2\theta(\cos^2\theta+\sin^2\theta)+b^4\sin^2\theta(\cos^2\theta+\sin^2\theta)$$
$$=a^2(p^2-b^2\sin^2\theta)+b^2(p^2-a^2\cos^2\theta)$$
$$\implies p^4+p'^2p^2=p^2(a^2+b^2)-a^2b^2$$
Comparing with $(1),K=a^2+b^2$ which definitely a constant independent of $\theta$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Integral of $\arcsin$ of a rational function, using integration by parts I'm a class 12 student and this a question from my textbook:
$$I=\int{\arcsin{2x\over 1+x^2}}\mathrm{d}x$$
I did it using integration by parts like this:
$$I=\arcsin{\left(2x\over 1+x^2\right)}\cdot\int1\cdot\mathrm{d}x-\int(\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\int1\cdot\mathrm{d}x)\mathrm{d}x+c$$
Now,
$$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\
= {1\over\sqrt{1-\left(2x\over1+x^2\right)^2}}
{ 2(1+x^2) - (2x)(2x)\over(1+x^2)^2}\
={1+x^2(2(1-x^2)\over(1-x^2)^2(1+x^2)^2}\
={2\over1-x^4}$$
So
$$I=\arcsin{\left(2x\over 1+x^2\right)}(x)-\int{2\over1-x^4}{x}\mathrm{d}x + c$$
let $I_1=\int{(2x)\over(1-x^4)}\mathrm{d}x$
$$I_1=\int{(2x)\over(1-x^2)(1+x^2)}\mathrm{d}x$$
Let $x^2=t$
So $2x\mathrm{d}x = \mathrm{d}t$ and
$$I_1=\int{\mathrm{d}t\over(1-t)(1+t)}\
=$\int{\mathrm{d}t\over1-t^2}\
={1\over2}\log{|1+t|\over|1-t|}+c_2\
={1\over2}\log{1+x^2\over1-x^2}+c_2$$
From all of this, we conclude
$$I=x \arcsin{\left(2x\over 1+x^2\right)}-{1\over2}\log{|1+x^2|\over|1-x^2|}+c$$
But the answer given in the book is :
$$(2x)\arctan x-{\log(1+x^2)} + c$$
I know they have done this using
$$\arcsin{\left(2x\over 1+x^2\right)}=2\arctan x$$
And then applied integration by parts, but I'd very much like to know where I went wrong.
Any help appreciated.
| HINT
Your calculation of $ \frac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\frac{2x}{1+x^2} \right) $ is wrong. It will be:
$$\begin{align}
\dfrac{\mathrm{d}}{\mathrm{d}x} \sin^{-1}\left (\dfrac{2x}{1+x^2} \right) &= {\dfrac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}} \cdot { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2} \\
&= \color{red}{ \dfrac{2(1+x^2)(1-x^2)}{(1-x^2)(1+x^2)^2}} \\
&= \dfrac{2}{1+x^2}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the sum of this series May I know how I should go about finding the sum of this series?
$\displaystyle\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$
I am really stuck. Thanks!
| $$s=\sum_{n=1}^\infty \dfrac{n}{2^{n-1}}=1+2/2+3/2^2+4/2^3+5/2^4+....(1)$$
By dividing whole equation by $2,$
$$\dfrac{s}{2}=\dfrac{1}{2}(\sum_{n=1}^\infty \dfrac{n}{2^{n}})=1/2+2/2^2+3/2^3+4/2^4+5/2^5+....(2)$$
Now$(1)-(2)$ gives us,
$$\dfrac{s}{2}=1+1/2+1/2^2+1/2^3+1/2^4+....=1+1$$
Hence we can obtain $s=4.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| For the base case simply show that the premise is true for the smallest positive integer: $\mathsf P(1)$.
For the iterative case, you have to show that if the premise holds for $n$, then it is implied to hold for $n+1$. $\mathsf P(n)\implies\mathsf P(n+1)$
Do that by substituting $n+1$ for $n$ and demonstrate that the left and right hand side both contain terms of the original premise and the same modifier.
Here we show that modifier is the addition of $3^n$.
$\begin{align}
\text{Prove }\forall n\in \mathbb Z^+: \sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} & \text{premise }
\\[3ex]
3^0 & = \frac{3^1-1}{2} & \text{base case}
\\[3ex]
3^n + \sum_{k=0}^{n-1} 3^k & = \sum_{k=0}^{n} 3^k
\\[1ex] 3^n+\frac {3^n-1}{2} & = \frac{3\cdot 3^n-1}{2}
\\[2ex] \therefore
\sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} \implies \sum_{k=0}^{n} 3^k = \frac{3^{n+1}-1}{2} &\text{iterative step }
\\[2ex]\therefore \forall n\in\mathbb Z^+: \sum_{k=0}^{n-1} 3^k & = \frac {3^n-1}{2} & \text{by induction}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int_0^\pi \sin(x)\,dx$ explicitly A book asks me to prove that:
$$\int_0^{\pi}\sin(x)\,dx = 2$$
Using the identity:
$$\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right) = \frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
And the famous $\lim_{x\to0}\frac{\sin(x)}{x} = 1$
What I tried:
Using the Right Riemann Sum Method:
$$\int_a^{b}\sin(x)\,dx \approx \Delta x\left[f(a + \Delta x) + f(a + 2\,\Delta x) + \cdots + f(b)\right]$$
By taking $\Delta x = \frac{\pi}{n}$, $a = 0$ and $b = \pi$ we have:
$$\int_0^{\pi}\sin(x)\,dx \approx \Delta x\left[f(\Delta x) + f(2\,\Delta x) + \cdots + f\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}\right]$$
So
$$\int_0^\pi \sin(x)\,dx = \lim_{n\to\infty} \frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
I can't see, however, how to prove this limit to be $=2$
| Notice that we can rewrite the limit as:
$$
\left[\lim_{n\to\infty} \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} \right]\left[\cos\left(\lim_{n\to\infty}\frac{\pi}{2n}\right)-\cos\left(\lim_{n\to\infty}\frac{(2n+1)\pi}{2n}\right) \right]
= 1 \cdot [\cos 0 - \cos \pi] = 2
$$
| {
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Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?
| If you call the distances from $P$ to $A, B, C$, say, $a, b, c$, respectively, then by the Law of Cosines the side length $s$ of the equilateral triangle is given by
$s^2 = a^2 + b^2 - 2 a b \widetilde{C} = b^2 + c^2 - 2 b c \widetilde{A} = c^2 + a^2 - 2 c a \widetilde{B}$ $(\ast)$,
where $\widetilde{A} = \cos \angle BPC$, $\widetilde{B} = \cos \angle CPA$, and $\widetilde{C} = \cos \angle APB$. We don't know these angles, or (almost equivalently) their cosines, so we seek to eliminate these quantities and solve for $s$.
Now, the three angles add up to $2\pi$, and so
$\widetilde{C} = \cos \angle APB = \cos(2 \pi - \angle BPC - \angle CPA) = \cos(\angle BPC + \angle CPA)$
Applying the cosine angle sum identity we get
$\widetilde{C} = \widetilde{A} \widetilde{B} - \sin \angle BPC \sin \angle CPA$.
We can solve for the two sines by introducing square roots, but we can avoid this by rearranging and squaring, giving
$(\widetilde{A} \widetilde{B} - \widetilde{C})^2 = \sin^2 \angle BPC \sin^2 \angle CPA = (1-\widetilde{A}^2)(1-\widetilde{B}^2)$.
Solving for $\widetilde{A}, \widetilde{B}, \widetilde{C}$ in $(\ast)$, and rearranging gives a rational equation in $r_1, r_2, r_3, s$ whose roots $s$ given $r_1, r_2, r_3$ are the potential side lengths. By multiplying through by the common denominator $4 a^2 b^2 c$ we get an equivalent polynomial equation:
$s^2 [s^4 - (a^2 + b^2 + c^2)s^2 + (a^4 + b^4 + c^4 - b^2 c^2 - c^2 a^2 - a^2 b^2)] = 0$.
After discarding the coefficient $s^2$, the resulting equation is quadratic in $s^2$, so we can apply the quadratic equation and take the roots of the resulting expression to find $s$, so in general there may be two solutions that involve two nested radicals (looking back at the only irreversible step, squaring the trigonometric equation, we see that we won't introduce any false solutions). Notice that some of the solutions correspond to triangles for which the distance conditions are satisfied but the reference point lies outside the triangle.
In your example, taking side lengths $4, 5, 6$ give possible side lengths of $\frac{1}{2} \sqrt{154 \pm 30 \sqrt{21}}$. The larger one satisfies the condition that the triangle contain the reference point, since the larger length is larger than $\frac{2}{\sqrt{3}} r$ for $r = a, b, c$, and the smaller solution does not.
Remark We can analyze the quartic polynomial to deduce conditions on $(a, b, c)$ that determine the number of coefficients. Again viewing the polynomial as quadratic in $s^2$, the solutions for $s^2$ are
$s^2 = \tfrac{1}{2}[(a^2 + b^2 + c^2) \pm \sqrt{\Delta}]$,
and we find that its discriminant is
$\Delta = -3 (a + b + c) (a - b - c) (b - c - a) (c - a - b)$.
Analyzing this condition gives that $\Delta$ is
*
*positive iff no one of the three lengths $a, b, c$ is greater than or equal to the sum of the other two
*zero iff one of the lengths is equal to the sum of the other two
*negative iff one of the lengths is greater than the sum of the other two.
This immediately gives the following:
*
*If one of the lengths is larger than the sum of the other two, there are no positive solutions (or even real ones for that matter).
*If one of the lengths is equal to the sum of the other two, say $c = a + b$, there is one positive solution, namely, $s = \sqrt{\frac{1}{2} (a^2 + b^2 + c^2)} = \sqrt{a^2 + ab + b^2}$.
So, henceforth assume that no length is at least the sum of the other two, and so $\Delta > 0$, and $s = \sqrt{\frac{1}{2} \left(a^2 + b^2 + c^2 \pm \sqrt{\Delta}\right)}$ is always a solution when $\pm$ is taken to be $+$. It will give a solution when $\pm$ is taken to be $-$ if $a^2 + b^2 + c^2 > \sqrt{\Delta}$, or equivalently, when $a^2 + b^2 + c^2 - \Delta > 0$. But by definition of $\Delta$, the left-hand side is just $4$ times the constant term in the polynomial, which can be written as
$\frac{1}{2}[(a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2]$,
which is always positive (and zero iff $a = b = c$).
Putting these facts together gives us that:
*
*If none of the lengths is greater than the sum of the other two but not all three lengths are the same, there are two solutions, namely, $\sqrt{\frac{1}{2}(a^2 + b^2 + c^2) \pm \Delta}$.
*If all three lengths are the same, say $r$, there is one positive solution, namely $\sqrt{3} r$, which corresponds to an equilateral triangle centered on the reference point; $0$ is also a solution, and corresponds to a degenerate triangle of side length $0$, i.e., a point, a distance $r$ from the reference point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving determinant using properties of determinants $$\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a^3 & b^3 & c^3
\end{vmatrix} =
(a-b)(b-c)(c-a)(a+b+c)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
| Using $C_2'=C_2-C_1, C_3'=C_3-C_1$
$$\begin{vmatrix}
1 & 1 & 1\\
a & b & c\\
a^3 & b^3 & c^3
\end{vmatrix}$$
$$=\begin{vmatrix}
1 & 0 & 0\\
a & b-a & c-a\\
a^3 & b^3-a^3 & c^3-a^3
\end{vmatrix}$$
$$=-(a-b)(c-a)\begin{vmatrix}
1 & 0 & 0\\
a & 1 & 1\\
a^3 & b^2+ab+a^2 & c^2+ca+a^2
\end{vmatrix}$$
Use $C_2'=C_2-C_3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of a class of continued fractions Is there a closed-form way of writing the continued fraction:
$$
1 + \frac{2}{3+ \frac{4}{5 + \frac{6}{7 + ...}}}
$$
EDIT: Since the above has been determined as $\frac{1}{\sqrt{e}-1}$, is there a similar expression for:
$$
2 + \frac{3}{4+ \frac{5}{6 + \frac{7}{8 + ...}}}
$$
More generally, are there general closed-form expressions for all continued fractions of the form:
$$
a_n = n + \frac{n+1}{(n+2) + \frac{n+3}{(n+4) + ...}} \\
f(x) = x + \frac{x+1}{(x+2) + \frac{x+3}{(x+4) + \cdots}} = x + \frac{x+1}{f(x+2)} \\
f(x) f(x+2) = xf(x+2) + x+1
$$
And can said closed form be extended to all real numbers? For example, I experimented with extending the sequence to negative values of n and found that for all negative odd $n, a_n = -1$.
| It's $1/(e^{1/2}-1)$. You should be able to derive this by doing a term-by-term transformation on an appropriate infinite series.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Checking a solution to an indefinite integral I recently did some work to try to find $\int{\frac{dx}{Ax^3 - B}}$, but I'm always paranoid that my solution has some minor trivial error in the middle of the process that screwed up the end result entirely, so could someone please help check my solution?
The first step to my solution is to eliminate $A$ and $B$ from the integrand by pushing them out as constants and leaving $\frac{dx}{x^3 - 1}$. We start with
$$
\begin{align}
\int{\frac{dx}{Ax^3 - B}} & = \int{\frac{dx}{A(x^3 - B/A)}} \\
& = \frac{1}{A}\int{\frac{dx}{x^3 - B/A}}.
\end{align}
$$
To get rid of the $B/A$ term, we make the substitution $x = \sqrt[3]{B/A}u, \ dx = \sqrt[3]{B/A}du$.
$$
\begin{align}
\frac{1}{A}\int\frac{dx}{x^3 - B/A} & = \frac{1}{A}\int{\frac{\sqrt[3]{B/A}du}{(B/A)u^3 - B/A}} \\
& = \frac{1}{A}\cdot\frac{\sqrt[3]{B/A}}{B/A}\int{\frac{du}{u^3 - 1}} \\
& = \mathcal{C}\int{\frac{du}{u^3 - 1}}
\end{align}
$$
with $\mathcal{C} = (AB^2)^{-1/3}$. Now we can just worry about solving $\int{\frac{du}{u^3 - 1}}$.
We can decompose $\frac{1}{u^3 - 1}$ using the fact that $u^3 - 1 = (u - 1)(u^2 + u + 1)$. So
$$
\begin{align}
\frac{1}{u^3 - 1} & = \frac{1}{(u - 1)(u^2 + u + 1)} \\
& = \frac{P}{u - 1} + \frac{Qu + R}{u^2 + u + 1}
\end{align}
$$
Here $P = 1/3$, but $Q$ and $R$ aren't as trivial to find. $u^2 + u + 1$ has no real roots, so I chose to sub in $u = -\sqrt[3]{-1}$ (it was the first root that I found). We then have to solve the equation
$$
(-\sqrt[3]{-1}-1)^{-1} = Q(-\sqrt[3]{-1}) + R
$$
which we can solve by setting $Q = R = [(-1)^{2/3} - 1]^{-1}$ (which will also help when integrating $\frac{Qu + R}{u^2 + u + 1}$ since we can factor $Q$ and $R$ out as a common constant from the numerator).
So
\begin{align}
\mathcal{C}\int{\frac{du}{u^3 - 1}} & = \mathcal{C}\left[\frac{1}{3}\int{\frac{du}{u - 1}} + Q\int{\frac{u + 1}{u^2 + u + 1}du}\right] \\
& = \mathcal{C}\left\{\frac{1}{3}\ln{|u - 1|} + Q\left[\frac{1}{2}\ln{|u^2 + u + 1|} + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2}{\sqrt{3}}u + \frac{1}{\sqrt{3}}\right)} \right] \right\} \\
& = (AB^2)^{-1/3}\left\{\frac{1}{3}\ln{\left|\frac{x}{\sqrt[3]{B/A}} - 1\right|} + \frac{1}{(-1)^{2/3} - 1}\left[\frac{1}{2}\ln{\left|\left(\frac{x}{\sqrt[3]{B/A}}\right)^2 + \frac{x}{\sqrt[3]{B/A}} + 1\right|} \right. \right. \\
& \hspace{15mm} \left. \left. + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2x}{\sqrt{3}\sqrt[3]{B/A}} + \frac{1}{\sqrt{3}}\right)}\right]\right\} + \text{constant}
\end{align}
(I know what $\int{\frac{x + 1}{x^2 + x + 1}dx}$ is from previous problems)
How does it look? Did I do anything severely wrong (I don't feel entirely confident about the partial fractions part)? Any suggestions for how I could get to an answer faster or more efficiently?
| Everything looks fine until your partial fraction decomposition. Indeed $P=\frac{1}{3}$, but to find $Q$ and $R$, start with
$$1 = \frac{1}{3}(u^2+u+1) + (Qu+R)(u-1)$$
and choose $u=0$ to get
$$1 = \frac{1}{3} - R,$$
so that $R = -\frac{2}{3}$. You then get $Q = -\frac{1}{3}$. So the decomposition is
$$\frac{1}{u^3-1} = \frac{1}{3(u-1)} - \frac{u+2}{3(u^2+u+1)}.$$
(Your solution involves complex numbers --- $(-1)^{2/3}$ is ambiguous, but must be complex since otherwise the denominator vanishes --- I doubt that was what you were looking for.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Get $5$ by doing any operations with four $7$s How can one combine four sevens with elementary operations to get $5$? For example $$\dfrac{(7+7)\times7}{7}$$ (though that does not equal $5$). I am not able to do this. Can you solve it or prove that it's impossible?
| How about:
$$7 - \frac{7+7}{7} = 5$$
$$7 - \log_7 (7·7) = 5$$
$$7 - \frac{\ln (7·7)}{\ln 7} = 5$$
$$\left\lfloor \sqrt{\frac{7^7}{7!}} - 7\right\rfloor = 5$$
$$\lfloor 7\sin 777^\circ\rfloor = 5$$
$$\lfloor 7\cos 7^\circ\rfloor - \frac{7}{7} = 5$$
$$\lfloor 7\cos 7\rfloor = 5 \text{ using radians}$$
You can also use base $174$ and write:
$$\sqrt{\frac{77}{7·7}} = 5$$
That can also reduce the amount of sevens by one if you write:
$$\frac{\sqrt{77}}{7} = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/917164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 0
} |
Prove $a^2+b^2+c^2=\frac{6}{5}$ if $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$
if $a,b,c$ are real numbers that $a\neq0,b\neq0,c\neq0$ and $a+b+c=0$ and $$a^3+b^3+c^3=a^5+b^5+c^5$$
Prove that $a^2+b^2+c^2=\frac{6}{5}$.
Things I have done: $a+b+c=0$ So $$a^3+b^3+c^3=a^5+b^5+c^5=3abc$$
also $$a^2+b^2+c^2=-2ab-2bc-2ca$$
I tried multiplying $(a^3+b^3+c^3)$ and $a^2+b^2+c^2$ but I was not able to reach a useful result from that.
| Note that $(a^3+b^3+c^3)-(a+b+c)(a^2+b^2+c^2)+(ab+bc+ca)(a+b+c)-3abc = 0$.
Using the fact that $a+b+c = 0$, this reduces to $a^3+b^3+c^3 = 3abc$. Thus, $a^5+b^5+c^5 = 3abc$.
Now, note that $(a^5+b^5+c^5)-(a+b+c)(a^4+b^4+c^4)+(ab+bc+ca)(a^3+b^3+c^3)-abc(a^2+b^2+c^2) = 0$.
Using the facts that $a+b+c = 0$ and $a^5+b^5+c^5 = a^3+b^3+c^3 = 3abc$, this reduces to
$(a^2+b^2+c^2)-3(ab+bc+ca) = 3$ [*].
Since $a+b+c = 0$, we have $0 = (a+b+c)^2 = (a^2+b^2+c^2)+2(ab+bc+ca)$.
Hence, $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2)$. Substituting this into [*] gives:
$(a^2+b^2+c^2)+\frac{3}{2}(a^2+b^2+c^2) = 3$, i.e. $a^2+b^2+c^2 = \frac{6}{5}$, as desired.
Sidenote: Newton's Sums is very convenient for these types of problems.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove the following limit: $\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0$? How to prove the following limit:
$$\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0?$$
| Starting the same as Kelenner's answer, but using the Beta function, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\frac{n\,\mathrm{d}x}{n+kx}\\
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}\int_0^1\int_0^1n\,t^{n-1+x+(k-1)x}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_0^1\int_0^1\frac nxt^{n/x}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_1^\infty\int_0^1\frac nxt^{nx}(1-t)^{n-1}\,\mathrm{d}t\,\mathrm{d}x\\
&=-\int_1^\infty\frac{n}{x+1}\frac{\Gamma(nx)\Gamma(n)}{\Gamma(n(x+1))}\,\mathrm{d}x\\
&=-\int_1^\infty\frac{n}{x+1}\mathrm{B}(nx,n)\,\mathrm{d}x\tag{1}
\end{align}
$$
We have the asymptotic
$$
\mathrm{B}(x,y)\sim\tilde{\mathrm{B}}(x,y)=\sqrt{\frac{2\pi(x+y)}{xy}}\frac{x^xy^y}{(x+y)^{x+y}}\tag{2}
$$
where for $x,y\ge n$, we have
$$
\tilde{\mathrm{B}}(x,y)\le\mathrm{B}(x,y)\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\tilde{\mathrm{B}}(x,y)\tag{3}
$$
Thus, for $x,n\ge1$,
$$
\begin{align}
\frac{n}{x+1}\mathrm{B}(nx,n)
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\frac{n}{x+1}\tilde{\mathrm{B}}(nx,n)\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{2\pi(x+1)n}{x}}\left(1+\frac1x\right)^{-nx}(x+1)^{-n-1}\\[6pt]
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}(x+1)^{-n-1}\tag{4}
\end{align}
$$
Applying $(4)$ to $(1)$ yields the upper bound
$$
\begin{align}
\hspace{-1cm}\left|\,\sum_{k=0}^n(-1)^k\binom{n}{k}\log(n+k)\,\right|
&\le\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\int_1^\infty(x+1)^{-n-1}\,\mathrm{d}x\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{4\pi n}\,2^{-n}\frac1n2^{-n}\\
&=\left(1+\frac{2-\sqrt\pi}{n\sqrt\pi}\right)\sqrt{\frac{4\pi}n}\,4^{-n}\tag{5}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
lebesgue's identity In the Lebesgue's identity $$ a^2+b^2+c^2=d^2$$ where: $$a=m^2+n^2-p^2-q^2$$ $$b=2(mp+nq) $$ $$c=2(mq-np) $$ $$d=m^2+n^2+p^2+q^2 $$how can we write $(m,n,p,q)$ as a function of the integers $(a,b,c,d)$? I have been trying and failing. Please help.
| Lebesgue's Identity is complete only if you add a scaling factor t. Hence,
$$\begin{align}
a &= (m^2+n^2-p^2-q^2)t\\
b &= 2(mp+nq)t\\
c &= 2(mq-np)t\\
d &= (m^2+n^2+p^2+q^2)t
\end{align}\tag{1}$$
As a function of the integers $a,b,c,d$, then the variables $m,n,p,q,t$ are simply,
$$\begin{align}
m &= \frac{-b-c}{a-d}\\
n &= \frac{-b+c}{a-d}\\
p &= 1\\
q &= 1\\
t &= \tfrac{1}{4}(-a+d)
\end{align}$$
If you substitute these into the eqns in $(1)$, for example,
$$(m^2+n^2-p^2-q^2)t-a = 0$$
then factor it, you get this result, hence is true if indeed $a^2+b^2+c^2 = d^2$. (Similarly for the 2nd, 3rd, 4th eqns.)
For the smallest solution,
$$1^2+2^2+2^2 = 3^2$$
we get $m,n,p,q,t = 2, 0, 1, 1,\tfrac{1}{2}$. And so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Basic partial fractions issue I've been starting to get the hang on partial fractions, whilst I've been able to do most of the basic ones, this kept causing some issues so I assumed:
*
*I'm using the wrong method
*I'm converting values when I shouldn't be
Before going on I'll post the question:
(a) Express $\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2))}$ in partial fractions
After seeing that this question only held 3 marks out of 100 I thought that it was relatively simple. However the values that I was receiving were nothing like what I would normally get. I could post my results but honestly I wrote so many it's irrelevant.
What I initially did was turn the $(x^2 +2x +2)$ into $(x+2)(x+1)$ which would leave us with:
$$\frac{x^2 + 6x + 7}{(x - 3) (x+2)(x+1)}$$
From there I used the usual method by placing the equation like so:
$$\frac{A}{(x-3)}+\frac{B}{(x+2)}+\frac{C}{(x+1)}$$
After that I found the LCM and starts cutting off terms by replacing $x$ with a specific value:
$$A(x+2)(x+1) + B(x-3)(x+1) + C(x-3)(x+2)$$
The rest is basically history, I can barely understand what I was even trying to do. Feel free to guide me to the right direction
Mistakes pointed out:
*
*My factorisation is incorrent for $$x^2 + 2x +2 = (x + 2)(x+1)$$
Extra Attempts
My second attempt was done using the method for quadratic factors inside the denominator. So it's now:
$$\frac{A}{(x-3)}+\frac{Bx+C}{(x^2 + 2x + 2)}$$
From here on I think I'm meant to find the LCM by doing the following:
$$(Bx+C)(x-3) + A(x^2 + 2x +2)$$
I then substitute $x$ with 3 in order to find the value of $A$ which would end up like so:
$$16 + 6x = 17A$$
After that I'm not entirely sure if it's correct(highly doubt so)
However not much has changed in terms of getting a viable answer
| $$x^2+2x+2=0 \Rightarrow \Delta=4-4 \cdot 2=4-8=-4<0$$
So it has no real roots.
Therefore, to express $\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}$ in partial fractions we do the following:
$$\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A}{x-3}+\frac{Bx+c}{x^2+2x+2}$$
(The polynomial at the numerator has to be one degree smaller than the degree of the polynomial of the denominator.)
EDIT:
$$\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A}{x-3}+\frac{Bx+c}{x^2+2x+2} \\ \Rightarrow \frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A(x^2+2x+2)+(Bx+C)(x-3)}{(x-3)(x^2+2x+2)} \\ \Rightarrow x^2+6x+7=Ax^2+2Ax+2A+Bx^2-3Bx+Cx-3C \\ \Rightarrow x^2+6x+7=(A+B)x^2+(2A-3B+C)x+(2A-3C) $$ Now you have to solve the following system:
$$A+B=1 \\ 2A-3B+C=6 \\ 2A-3C=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of the function $f(x)=(5-x^2)(\sqrt{x})$ I am stuck on this question, not sure what I am doing wrong:
\begin{align*}
f(x) &=(5-x^2)(\sqrt{x}) \\
\\
f'(x) &=(5-x^2)(x)^{1/2} \\
&=(-2x)(x)^{1/2}+(-x^2+5)(\dfrac{1}{2})(x)^{-1/2}\\
&=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\\
&=\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg)\\
&=\dfrac{1}{2}x^{-1/2}\bigg(-x^2+10\bigg)
\end{align*}
| Note that
$$-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\not =\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg).$$
We have
$$f'(x)=\cdots=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})=\frac 12x^{-\frac 12}(-4x^2-x^2+5)=\frac{5(1-x^2)}{2\sqrt x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Indefinite integral of a rational function: $\int\frac{6x+4}{x^2+4}\,dx$ Find $\displaystyle\int\frac{6x+4}{x^2+4}\,dx$
The question asks to find integral of the expression so I divided them into two parts:
$$
\int\frac{6x}{x^2+4}\,dx
$$
and
$$\int\frac{4}{x^2+4}.
$$
So, for the first integral, I set $u=x^2+4$ and got $\int\frac{3}{u}\,du$ which is $3\ln|x^2+4|+c$. But I don't know how to integrate the second part.
| Since you've nailed the first of the split integral, I'll address only the second:
If you make the substitution $\,x = 2\tan\theta,\,$ then $\,\theta = \arctan\left(\frac x2\right)\,$ and $\,dx = 2\sec^2 \theta.$
We also make use of the identity: $$\tan^2 \theta + 1 = \sec^2\theta\tag{1}$$
That gives you $$\begin{align} \int\frac{4\,dx}{x^2+4}& =\int \frac{4\cdot 2\sec^2\theta\,d\theta}{4\tan^2\theta + 4} \\ \\
& = \int \frac{2\sec^2 \theta\,d\theta}{\tan^2\theta +1}\\ \\
& = \int \frac{2\sec^2 \theta\,d\theta}{\sec^2\theta}\tag{1}\\
& = \int 2d\theta \\ \\
& = 2\theta + C \\ \\
& = 2\arctan\left(\frac x2\right) + C\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$6$ women and $4$ men wait in line. If their order in line is random, find the probability that all of the women are adjacent to one another. My thoughts on the problem are that the number of ways the women can be adjacent to each other is $5!$ and the total number of arrangements for all the people is $10!$. Is this correct?
| Total number of arrangements was correctly determined as $10!$.
The number of ways the $6$ women can all stand next to each other are easily seen through the following:
$W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __ __
__ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __ __
__ __ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __ __
__ __ __ $W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$ __
__ __ __ __$W_{1} W_{2} W_{3} W_{4} W_{5} W_{6}$
So, we have $5$ ways for the women to be next to each other and $4!$ ways to arrange the men. Next, we need to determine how many different ways the women can be arranged, which is $6!$
Thus, the total number of ways the women can be next to each other is: $(5\cdot 4!)(6!)$
Probability all women are adjacent: $\large\frac{5!6!}{10!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
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} |
What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$? What does this number equal if it goes on forever?
$$\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\frac{\frac{17}{18}}{\frac{19}{20}}}{\frac{\frac{21}{22}}{\frac{23}{24}}}}{\frac{\frac{\frac{25}{26}}{\frac{27}{28}}}{\frac{\frac{29}{30}}{\frac{31}{32}}}}}$$
$$.$$$$.$$$$.$$
Treat the fraction in blocks, one divides what is under it, then that pair is divided by the next pair, those four are divided by the next four. ETC
Edit. I'm reformatting as I can't read the original... aged eyes I guess ;-( but in deference to almagest's comment I am leaving the original too.
$$\frac{\frac{\frac{1}{2}\Big/\frac{3}{4}}{\frac{5}{6}\Big/\frac{7}{8}}\Bigg/
\frac{\frac{9}{10}\Big/\frac{11}{12}}{\frac{13}{14}\Big/\frac{15}{16}}}
{\frac{\frac{17}{18}\Big/\frac{19}{20}}{\frac{21}{22}\Big/\frac{23}{24}}\Bigg/
\frac{\frac{25}{26}\Big/\frac{27}{28}}{\frac{29}{30}\Big/\frac{31}{32}}}
\cdots$$
| HINT:
let $u_n=\dfrac{n}{n+1}$ and $v_n=\dfrac{u_n}{u_{n+1}}$ where $n=1,2,3,..$
I think here you are asking about the$ \lim_{ n \to \infty} v_n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 6,
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} |
derive limit to make a function continuous Here's the problem:
Choose the value of k that makes the following
function continuous at $x = 1$:
$f(x)=\begin{cases}
\frac{-8x^2 + 48x - 40}{x - 1} & x < 1\\
-2x + k &x \geq 1
\end{cases}$
My steps:
$\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow 1} f(x)$
$\frac{-8x^2 + 48x - 40}{x - 1} = -2x + k$
$\frac{-8(x^2 - 6x + 5)}{x - 1} = -2x + k$
$\frac{-8(x - 5)(x - 1)}{x - 1} = -2x + k$
$-8(x - 5) = -2x + k$
$-8x - (-8)\cdot5 = -2x + k$
$-8x + 40 = -2x + k$
$-8x + 2x + 40 = k$ (See note at bottom)
$-6x + 40 = k$
$-6\cdot1 + 40 = k$
$-6 + 40 = k$
$36 = k$
The solution steps:
For f to be continuous at $x = 1$ we need $\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow1} f(x) = f(1)$
First lets evaluate lim x→1− f(x). Since x < 1 as x approaches 1 from the left,
f(x) = (−8x^2 + 48x − 40)/(x − 1) as x→1−
So lim x→1− f(x) = lim x→1− (−8x^2 + 48x − 40)/(x − 1)
Start by factoring the numerator.
In this case we find:
lim x→1− (−8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = lim x→1− (−8x + 40)(x − 1)/(x − 1)
Aside: can anyone explain (−3x^2 − 15x + 18) in the denominator?
Since we are taking the limit as x→1−, we may assume that x ≠ 1.
Canceling factors we see:
\lim x→1− (−8x + 40)(x − 1)/(−3x − 18)(x − 1) = \lim x→1− (−8x + 40)
We can evaluate this linear function at x = 1.
Hence,
lim x→1− −8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = −8(1) + 40.
Thus, lim x→1− f(x) = 32.
Thus lim x→1− f(x) = 32
lim x→1+ f(x) = lim x→ (−2x + k), because x > 1 as x approaches 1 from the right.
lim x→ (−2x + k) = −2⋅1 + k = −2 + k = f(1) because linear functions are continuous.
For f to be continuous, these limits have to be equal. Thus 32 = −2 + k
When k = 34, f is continuous.
NOTE: This is where I diverged from the solution by moving -2x to the left. My question is, assuming the given solution is correct, why is this nominally valid algebraic step not permitted?
| To address your aside: this is likely a typo.
To address your note: the $-2x$ term contains a variable $x$ whose scope is unrelated to the other variable on the other side of the equation. It's a bit unfortunate that we reuse the same variable, as this might be confusing. To emphasize the fact that the variables on each side of the equation for the different limits are unrelated, we could have written our work by doing something like:
\begin{align*}
\lim_{x \to 1^-} f(x) &= \lim_{y \to 1^+} f(y) \\
\vdots\qquad &= \qquad\vdots \\
\lim_{x \to 1^-} (-8x + 40) &= \lim_{y \to 1^+} (-2y + k) \\
-8(1) + 40 &= -2(1) + k \\
k &= 34
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the trigonometric equation/inequality $(1).\quad\cos^2(2x) + \sin^4(x) = 2$ $($solve the equation$)$
$(2).\quad2\cos^2(3x) + 5\cos(3x) - 3 < 0$.
For this question, I tried letting $t=\cos(3x)$.
Thus, $2t^2 + 5t - 3< 0$, but that doesn't factor properly.
I'm really stuck... It would be nice if someone could show me how to do this step by step... Thanks.
| For the equation $$\cos^2(2x) + \sin^4x = 2$$ you can use one of the trigonometric formulas for $\cos(2x)$. They are
\begin{align*}
\cos^2(2x) & = \cos^2x - \sin^2x\\
& = 2\cos^2x - 1\\
& = 1 - 2\sin^2x
\end{align*}
If we use the last one, then we will have a polynomial in $\sin x$.
\begin{align*}
\cos^2(2x) + \sin^4x & = 2\\
[\cos(2x)]^2 + \sin^4x & = 2\\
(1 - 2\sin^2x)^2 + \sin^4x & = 2\\
1 - 4\sin^2x + 4\sin^4x + \sin^4x & = 2\\
5\sin^4x - 4\sin^2x - 1 & = 0\\
5\sin^4x - 5\sin^2x + \sin^2x - 1 & = 0\\
5\sin^2x(\sin^2x - 1) + 1(\sin^2x - 1) & = 0\\
(5\sin^2x + 1)(\sin^2x - 1) & = 0\\
(5\sin^2x + 1)(\sin x + 1)(\sin x - 1) & = 0
\end{align*}
Set the factors equal to zero. $5\sin^2x + 1 > 0$ for all real numbers $x$.
\begin{align*}
\sin x + 1 & = 0 & \sin x - 1 & = 0\\
\sin x & = -1 & \sin x & = 1\\
x & = -\frac{\pi}{2} + 2n\pi & x & = \frac{\pi}{2} + 2n\pi
\end{align*}
where $n$ is an integer.
Your strategy for the inequality $$2\cos^2(3x) + 5\cos(3x) - 3 < 0$$ is correct. Factoring yields
\begin{align*}
2\cos^2(3x) + 5\cos(3x) - 3 & < 0\\
2\cos^2(3x) + 6\cos(3x) - \cos(3x) - 3 & < 0\\
2\cos(3x)[\cos(3x) + 3] - 1[\cos(3x) + 3)] & < 0\\
[2\cos(3x) - 1][\cos(3x) + 3] & < 0
\end{align*}
Since $-1 \leq \cos(3x) \leq 1$, $1 \leq \cos(3x) + 3 \leq 4$. Thus, the term $\cos(3x) + 3$ is always positive.
Therefore, the inequality is satisfied when $2\cos(3x) - 1 < 0$.
\begin{align*}
2\cos(3x) - 1 & < 0\\
2\cos(3x) & < 1\\
\cos(3x) & < \frac{1}{2}
\end{align*}
In the interval $[0, 2\pi)$, the inequality $$\cos u < \frac{1}{2}$$ is satisfied if $$\frac{\pi}{3} < u < \frac{5\pi}{3}$$ In general, the inequality is satisfied if $$\frac{\pi}{3} + 2n\pi < u < \frac{5\pi}{3} + 2n\pi$$ If $u = 3x$, then
$$\frac{\pi}{9} + \frac{2}{3}n\pi < x < \frac{5\pi}{9} + \frac{2}{3}n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculus Question: Improper integral $\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx$ How to evaluate integral $$\int_{0}^{\infty}\frac{\cos(2x+1)}{\sqrt[3]{x}}\text dx?$$ I tried substitution $x=u^3$ and I got $3\displaystyle\int_{0}^{\infty}u \cos(2u^3+1)\text du$. After that I tried to use integration by parts but I don't know the integral $\displaystyle\int \cos(2u^3+1)\text du$. Any idea? Thanks in advance.
| $$\color{blue}{\mathcal{I}=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}\approx-0.391190966503539\cdots}$$
\begin{align}
\int^\infty_0\frac{\cos(2x+1)}{x^{1/3}}{\rm d}x
&=\int^\infty_0\frac{\cos(2x+1)}{\Gamma(\frac{1}{3})}\int^\infty_0t^{-2/3}e^{-xt} \ {\rm d}t \ {\rm d}x\tag1\\
&=\frac{1}{\Gamma(\frac{1}{3})}\int^\infty_0t^{-2/3}\int^\infty_0e^{-xt}\cos(2x+1) \ {\rm d}x \ {\rm d}t\\
&=\frac{\cos(1)}{\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{1/3}}{t^2+4}{\rm d}t-\frac{2\sin(1)}{\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-2/3}}{t^2+4}{\rm d}t\tag2\\
&=\frac{\cos(1)}{2^{2/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{1/3}}{1+t^2}{\rm d}t-\frac{\sin(1)}{2^{2/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-2/3}}{1+t^2}{\rm d}t\tag3\\
&=\frac{\cos(1)}{2^{5/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-1/3}}{1+t}{\rm d}t-\frac{\sin(1)}{2^{5/3}\Gamma(\frac{1}{3})}\int^\infty_0\frac{t^{-5/6}}{1+t}{\rm d}t\tag4\\
&=\frac{\pi\left(\cos(1)-\sqrt{3}\sin(1)\right)}{2^{2/3}\Gamma(\frac{1}{3})\sqrt{3}}\tag5\\
&=\frac{2\pi\cos(1+\frac{\pi}{3})}{2^{2/3}\frac{2\pi}{\Gamma(\frac{2}{3})\sqrt{3}}\sqrt{3}}\tag6\\
&=\frac{\Gamma(\frac{2}{3})\cos(1+\frac{\pi}{3})}{2^{2/3}}
\end{align}
Explanation:
$(1)$: $\small{\displaystyle\frac{1}{x^n}=\frac{1}{\Gamma(n)}\int^\infty_0t^{n-1}e^{-xt}{\rm d}t}$
$(2)$: $\displaystyle\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\sin(bx){\rm d}x=\frac{b}{a^2+b^2}}$
$(2)$: $\small{\displaystyle\int^\infty_0e^{-ax}\cos(bx){\rm d}x=\frac{a}{a^2+b^2}}$
$(3)$: $\displaystyle t\mapsto 2t$
$(4)$: $\displaystyle t\mapsto \sqrt{t}$
$(5)$: $\small{\displaystyle\int^\infty_0\frac{x^{p-1}}{1+x}{\rm d}x=\pi\csc(p\pi)}$
$(6)$: $\small{\displaystyle \Gamma(z)=\frac{\pi\csc(\pi z)}{\Gamma(1-z)}}$, $\small{\displaystyle a\cos{x}-b\sin{x}=\sqrt{a^2+b^2}\cos(x+\arctan{\frac{b}{a}})}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$ I am doing some math repetition and am a bit stuck on this exercise:
Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$.
Now, this is a geometric sum on both the $LHS$ and $RHS$, which I guess is something that I should use to solve the equation...
Another way is to simply start to eliminate terms:
$$1+2^x+4^x+8^x+16^x+32^x=3+3 \times 2^x+3\times4^x$$
$$-2 -2\times 2^x -2\times4^x+8^x+16^x+32^x = 0$$
$$8^x+16^x+32^x = 2 +2\times 2^x +2\times4^x$$
$$8^x+16^x+32^x = 2(1 + 2^x + 4^x)$$
But I am stuck here...
| Note that the left side of the equation can be written as
$2^0+2^x+2^{2x}+2^{3x}...2^{5x}$
This is a geometric series, with
a=1, n=6, and $r=2^x$
We use the formula:
$S_n = \frac {a(1-r^n)}{1-r}$
Substitute the values, you get
$S=\frac{(1-2^{6x})}{1-2^x}$
We do the same for the right side
$S=3[\frac{1-2^{3x}}{1-2^x}]$
Equate the terms, and rearrange
$\frac{(1-2^{6x})}{1-2^x}=3[\frac{1-2^{3x}}{1-2^x}]$
$1-2^{6x}=3-3\cdot2^{3x}$
$-2^{6x}+3\cdot2^{3x}-2=0$
And this bit is my fave.
It's just a quadratic!
Because $-2^{6x} = -(2^{3x})^2$
Now you just let $2^{3x}$ = u
$-u^2+3u-2=0$
Solve for u
Then you just sub back $2^{3x}$
And there you have it.
If I have made an error (as I am prone to), notify me and I will withdraw my answer.
Thanks
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Finding the limit by factoring the denominator and canceling I have the problem
$$\lim_{x\to10} \frac{x-3}{x^2+7x-30}$$
If I factor it to $\dfrac{x-3}{(x+10)(x-3)}$ then $x-3$ cancels and I'm left with $0$.
I know the real answer is $1/20$, but why is zero wrong?
| Your cancellation method needs an adjustment:
$$\frac{a}{a\cdot b} = \frac{1}{b}, \quad \textbf{not} \frac{0}{b}$$
To see why, note that $a=a\cdot 1$, so
$$\frac{a}{a\cdot b} = \frac{a\cdot 1}{a\cdot b}= \frac{1}{b}$$
You may also want to experiment with $\dfrac{3}{2\cdot 3}$, for a concrete example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/929301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Partition of a Matrix In Linear Algebra, we have been taught that the partition of a matrix $A$ consists of matrices,or blocks. In other words, its elements are matrices. This same, partitioned matrix, however is said to be equal to the original matrix. But their elements are different, as one contains scalars and the other matrices. Please help me understand.
| You have to concatenate the matrices. Here's an example:
$$A= \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B = \begin{pmatrix} 10 \\ 20 \end{pmatrix}, C = (100,200), D = 1000,$$
then
$$\begin{pmatrix}A & B \\ C & D \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} & \begin{pmatrix} 10 \\ 20 \end{pmatrix} \\ \begin{pmatrix} 100 & 200\end{pmatrix} & 1000 \end{pmatrix} \equiv \begin{pmatrix} 1 & 2 & 10 \\ 3 & 4 & 20 \\ 100 & 2000 & 1000 \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/929719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $ x_{n+1} - x_n = 2n + 3$ Solve $$ x_{n+1} - x_n = 2n + 3, x_0 = 1, n \ge 0$$
I would try to find a homogen solution and used $$ r^2 - r = 0$$ and got $$x^h_n = A1^n$$ but this seems wrong and I'm stuck on how to continue.
=== Edit ===
Homogen solution:
$r^2 - r = 0 $ has the solutions $r=0, r=1 $ So we have $$x_n^h = C1^n$$
Particulair solution:
$x_n^p = B(2n+3)$. Inserting this in the original equation gives:
$$ B(2(n+1)+3) - B(2n + 3) = 2n + 3$$
$$ B = \frac{2n + 3}{8}$$
Solution:
$$ x_n = C 1^n + \frac{2n + 3}{8}(2n + 3)$$
Using $x_0 = 1$ to find $C$ gives $C = - \frac{1}{8}$
So the answer is $$ x_n = x_n^h + x_n^p = - \frac{1}{8}1^n + \frac{2n + 3}{8}(2n + 3)$$
However I know that the answer should be $x_n = (n+1)^2$ so my answer seems to be really off.
=== Edit 2 ===
Okay, so I realise that I should guess a polynom of the same size as my right hand part of the original equation. Since this doesn't work for this particulair problem, I'll try with size + 1. So:
$$ x_n^p = an^2 + bn + c $$
Insert this in the original equation:
$$a(n+1)^2 + b(n+1) + c - an^2 -bn -c = 2n +3$$ And this gives $a = 1, b = 2$.
Then I got $$ x_n = 1^n + n^2 +2n$$ Still not the answer I'm looking for. Specially not the $1^n$ part which will still be there no matter what particulair solution I find.
| $$x_{n+1} - x_n = 2n + 3 \tag A$$
$$n = \frac 12 x_{n+1} - \frac 12 x_n - \frac 32 \tag B$$
$$n-1 = \frac 12 x_{n} - \frac 12 x_{n-1} - \frac 32 \tag C$$
Subtract previous two:
$$1 = \frac 12 x_{n+1} + \left(-\frac 12 - \frac 12\right)x_n + \frac 12 x_{n-1} \tag D$$
$$x_{n+1} = x_{n} - x_{n-1} + 2 \tag E$$
Now it is a regular affine equation.
$$\begin{bmatrix} x_{n+1} \\ x_{n} \\ 1 \end{bmatrix}
=
\begin{bmatrix}
1 & -1 & 2 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}^n
\begin{bmatrix} x_1 \\ x_0 \\ 1 \end{bmatrix} \tag F$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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problem solving logarithmic equation and reaching an equivalence ok so i've had a problem trying to simplify the $\ln\left[ \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right]$ and this is supposed to be equal to : $\ln [ \sqrt{a^2+u^2} + u ]$
how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?
| $\ln x$ is injective so, if $\ln x=\ln y\implies x=y$. So, the following should hold true:\begin{align}
\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}=\sqrt{a^2+u^2}+u\end{align}
However, simply plugging in $u=1, a=2$ gives us
\begin{align*}
&\sqrt{1+\frac{1}{4}}+\frac{1}{2}=\sqrt{4+1}+1\\
&\implies \sqrt{\frac{5}{4}}+\frac{1}{2}=\sqrt{5}+1\\
&\implies \sqrt{\frac{5}{4}}=\sqrt{5}-\frac{1}{2}\\
&\implies \frac{\sqrt{5}}{2}=\frac{2\sqrt{5}-1}{2}\\
&\implies \sqrt{5}=2\sqrt{5}-1
\end{align*}
which is false.
Hence they are not the same.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a sum formula by induction I am to prove through induction that $$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
And well, my method seems to be working, but I get stuck when I'm nearly done. First I prove the formula work for $n = 1$.
$$\sum_{k=1}^1 (2k-1)^2 = 1$$
$$f(1)=\frac{(2*1-1)(2*1+1)}{3} = 1$$
I then assume it works for $n = p$ and want to show it works for $n = (p+1)$
$$\frac{n(2n-1)(2n+1)}{3} + (2(n+1)-1)^2 \stackrel{?}{=} \frac{(n+1)(2(n+1)-1)(2(n+1)+1)}{3}$$
$$\frac{n(2n-1)(2n+1)}{3} + (2(n+1)-1)^2 = \frac{(4n^3+12n^2+11n+3)}{3}$$
I didn't show all the steps in the last equation, however, this is where I'm stuck; how can I factor the numerator? I tried using Wolfram | Mathematica, but it just went straight from the current expression into factored form.
Also I suppose I'm wondering if I chose the easiest path to proving the formula through induction.
| $$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
*
*$$\sum_{k=1}^1 (2k-1)^2=1^2 = \frac{1(2*1-1)(2*1+1)}{3}=1$$ True
*Assuming it is true for $k=n$:
$$\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$ Assumption of Truth
*Now finding for $k=n+1$:
$$\begin{align}\sum_{k=1}^{n+1} (2k-1)^2&=\sum_{k=1}^n (2k-1)^2+(2*(n+1)-1)^2\\&=\frac{n(2n-1)(2n+1)}{3}+(2n+1)^2\\&=\frac{(2n+1)}3(2n^2-n+6n+3)\\&=\frac{(2n+1)}3(2n^2+5n+3)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}\\&=\frac{(n+1)(2*(n+1)-1)(2*(n+1)+1)}{3}\end{align}$$ Establishing the Chain
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$ 2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$
I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems with taking the integration with respect to z because my change of variables is causing me to have a wicked exponential.
Anyway, here is my attempt
The slope is $\frac{2}{1}$ which is is 2, so my characteristic lines must be positive.
The characteristic lines are in the form of $bx-ay=d$ which is for the pde $au_x+bu_y+cu=f(x,y)$
Using my change of variables, I get the following lines
$ 2x - y = w \rightarrow x=\frac{w+z}{2}$
$z = y \rightarrow y = z$
Now I have to take the partial derivatives and the chain rule.
$W_x = 2, W_y = -1, Z_x = 0, Z_y = 1$
$V_wW_x+V_zZ_x+2(V_wW_y+V_zZ_y) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
$V_z -4(v) = e^{\frac{w+z}{2}+z}$
Let $p(a) = -4$, and $q(a) = e^{\frac{w+z}{2}+z}$. Then my integrating factor would be. $e^{\int -4} \rightarrow e^{-4z}$
Multiplying the integrating factor, I get
$e^{-4z}V_z -e^{-4z}4(v) = e^{-4z}e^{\frac{w+z}{2}+z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z}e^{-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}-3z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z-6z}{2}}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w-5z}{2}}$
By reverse product rule
$e^{-4z}(v) = \int e^{\frac{w-5z}{2}}$
Integrating the right hand side, I get
$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$
$ v = \frac{\frac{-2}{5}e^{\frac{w-5z}{2}}}{e^{-4z}}$
$ v = \frac{-2}{5}e^{\frac{w-5z+4z}{2}}$
$ v = \frac{-2}{5}e^{\frac{w-z}{2}} + C(w)$
Substituting back I get
$ u = \frac{-2}{5}e^{\frac{2x-y-y}{2}} +C(2x-y)$
$ u = \frac{-2}{5}e^{\frac{2x-2y}{2}}+C(2x-y)$
from side condition of $u(x,-x) = x$, we have
$x = \frac{-2}{5}e^{\frac{2x-2(-x)}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{2x+2x}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{4x}{2}}+C(x)$
$x = \frac{-2}{5}e^{2x}+C(x)$
The problem is... that's not right because the answer is $u(x,y)=-e^{x+y}+(\frac{2}{3}(x-\frac{1}{2}y+1)e^{\frac{4}{3}(x-\frac{1}{2}y)}e^{2y}$
What happened?
Edit: even if I did rewrite it correctly, the $\frac{-2}{5}$ is still there!!!
$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$
Applying change of variables
$u = \frac{-2}{5}e^{\frac{2x-y-5y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{\frac{2x-6y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x-3y+4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x+y}+C(2x-y)e^{4y}$
| I don't know if anyone will read this after 8 years, but I think there is another mistake higher up concerning the following two lines:
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
In the second line it should instead read:
$2V_w-2V_w+2V_z -4(v) = e^{\frac{w+z}{2}+z}$
You forgot to also multiply the $V_z$ by 2, thus invalidating the rest of the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/931921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
symetric inequality for a rational function of three variables $a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
| We have $\frac{ab^2}{a^2 +b^2}\leq \frac{b}{2} ,\frac{bc^2}{c^2 +b^2}\leq \frac{c}{2} ,\frac{ca^2}{c^2 +a^2}\leq \frac{a}{2}$ hence $$\frac{a^3}{a^2 +b^2} +\frac{b^3}{c^2 +b^2} +\frac{c^3}{a^2 +c^2} =a+b+c - \left(\frac{ab^2}{a^2 +b^2}+\frac{bc^2}{c^2 +b^2}+\frac{ca^2}{c^2 +a^2}\right) \geq \frac{a+b+c}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.
| No the most rigorous, but fast. Define $n=\frac 1x$, and assume it integer (keep in mind the dependency on $x$).
$$\frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}=\frac{(1 + \sin x + \sin^2 x)^n - (1 + \sin x)^n}{x}.$$
Factoring the numerator as a binomial of the form $a^n-b^n$, you will get two factors:
*
*the difference of $a-b$, i.e. $\sin^2x$,
*the sum of products of powers $a^{n-1-k}b^k$. All these terms behave like $(1+x+o(x^2))^{n-1}$ and there are $n$ of them. You can recognize the definition of $e$.
So you end up with
$$\frac{\sin^2x}xne$$
which tends to $e$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
What is $\frac{2x}{1-x^2}$ when $x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$? If
$$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
Find
$$\frac{2x}{1-x^2}$$
I got till here by simplification by taking the previous value of x, ie,
$$x={\frac{\sqrt{1-\cos\theta}}{1+\cos\theta}}$$
$$\frac{2\tan\theta\sqrt{1+\cos\theta}}{\cos\theta+3}$$
| $$x=\sqrt\frac{1-cos(\theta)}{1+cos(\theta)}.$$ Suppose to the given equation, let it be defined such that $$y = \frac{2x}{1-x^2}$$ Just substitute x in the equation y will result $$ 1-x^2 = 1 - \frac{1-cos(\theta)}{1+cos(\theta)} \implies \frac{2 cos(\theta)}{1+cos(\theta)}.$$ similarly $$x=\sqrt\frac{1-cos(\theta)}{1+cos(\theta)} * \sqrt\frac{1+cos(\theta)}{1+cos(\theta)} \implies \sqrt\frac{sin^2(\theta)}{(1+cos(\theta))^2} $$ $$ x = \frac{sin(\theta)}{1+cos(\theta)}$$ Equation y can be rewritten as $$y = \frac{2 sin(\theta)/(1+cos(\theta))}{2cos(\theta)/(1+cos(\theta)} $$ Hence, $$ \frac{2x}{1-x^2} = tan(\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving an equation with exponents by using logarithms Solve the equation
$$0.25^5 = 4^{(5x-3)/3} \cdot (0.125)^{6x}$$
So would I just bring down the exponents by taking the log of each constant?
| Logarithms here are actually unnecessary. Notice that each base is a power of $2$, so we get:
\begin{align*}
0.25^5 &= 4^{\frac{1}{3}(5x - 3)} \cdot (0.125)^{6x} \\
(2^{-2})^5 &= (2^2)^{\frac{1}{3}(5x - 3)} \cdot (2^{-3})^{6x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3)} \cdot 2^{-18x} \\
2^{-10} &= 2^{\frac{2}{3}(5x - 3) - 18x} \\
\end{align*}
Since the bases are the same, we may equate exponents:
\begin{align*}
-10 &= \frac{2}{3}(5x - 3) - 18x \\
-30 &= 2(5x - 3) - 54x \\
-30 &= (10x - 6) - 54x \\
-24 &= -44x \\
x &= 6/11
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$ Solve the PDE $$ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$$
where, $ p=\displaystyle \frac{\partial z}{\partial x}$ and $ q=\displaystyle \frac{\partial z}{\partial y}$
My attempt:
I start with Lagrange's auxiliary equation
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{z(x^2-y^2)}$$
Relation 1: $$\frac{x \, dx+y \, dy- dz}{x^2y^2+x^2z-y^2x^2-y^2z-x^2z+y^2z} = \frac{x \, dx+y \, dy-dz}{0} $$
Integrating $x \, dx+y \, dy-dz=0$
$$\frac{x^2}{2}+\frac{y^2}{2}-z=c \implies x^2+y^2-2z=c_1 $$
I can't see/find a second relation for me to solve the problem. Please help.
| since
$$\dfrac{\dfrac{dx}{x}}{y^2+z}=\dfrac{\dfrac{dy}{y}}{-x^2-z}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{\dfrac{dx}{x}+\dfrac{dy}{y}}{y^2-x^2}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{dx}{x}+\dfrac{dy}{y}=-dz$$
$$\Longrightarrow z=-C\ln|xy|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygamma function of order $3$.
Here is my attempt. Using equation (11) from Mathworld Wolfram:
\begin{equation}
\Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right)
\end{equation}
I got
\begin{equation}
\Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right)
\end{equation}
then
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\
&=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
&=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
\end{align}
From the answers of this OP, the integral representation of the latter Euler sum is
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)}
\end{align}
or another simpler form
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\
&=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\
&=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\
\end{align}
I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.
Edit :
Using the integral representation of polygamma function
\begin{equation}
\Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx
\end{equation}
then we have
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\
&=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\
&=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\
\end{align}
I am looking for an approach to evaluate the above integral without using residue method or double summation.
| Computation of $\displaystyle U=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx$
\begin{align*}
U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\
&=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)\,dt\,dx\\
&=-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt\right)\left(\int_0^1\frac{\ln x}{1+x}\,dx\right)+\\
&2\left(\int_0^1\frac{\ln^3 t}{1+t}\,dt\right)\left(\int_0^1\frac{1}{1+x}\,dx\right)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\
&=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\
&\overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\
&=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\
&=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\
U&=\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 5,
"answer_id": 3
} |
Solve the following integration $$\int \sqrt{cot\theta} d\theta$$
I tried to set $t=\sqrt{cot\theta},t^2=cot\theta$ and substitute into the original integration and get$$-\int\frac{2t^2}{1+t^4}dt$$, but then what can I do?
| Consider the integral
\begin{align}
I = \int \sqrt{\cot(\theta)} \, d\theta.
\end{align}
Make the substitution $t = \sqrt{\cot(\theta)}$ to obtain the integral
\begin{align}
I = -2 \int \frac{t^{2} \, dt}{1+t^{4}}.
\end{align}
Now it can be seen that
\begin{align}
\frac{t^{2}}{1+t^{4}} &= \frac{t^{2}}{(1+i t^{2})(1- i t^{2})} \\
&= \frac{1}{2i} \left( \frac{1}{1-i t^{2}} - \frac{1}{1+ i t^{2}} \right)
\end{align}
which leads to
\begin{align}
I = i \int \left( \frac{1}{1-i t^{2}} - \frac{1}{1+ i t^{2}} \right) \, dt.
\end{align}
Since
\begin{align}
\int \frac{dt}{1+a t^{2}} = \frac{1}{\sqrt{a}} \, \tan^{-1}(\sqrt{a} t)
\end{align}
then the integral $I$ becomes
\begin{align}
I = i \left[ e^{\pi i/4} \tan^{-1}(e^{\pi i/4} t) - e^{- \pi i/4} \tan^{-1}(e^{- \pi i/4} t) \right].
\end{align}
By the $\tan^{-1}(x)$ addition and eliminating the complex components the result becomes
\begin{align}
I = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left( \frac{\sqrt{2} \, t}{1- t^{2}} \right) - \tanh^{-1}\left( \frac{\sqrt{2} \, t}{1+ t^{2}} \right) \right].
\end{align}
This leads to the result
\begin{align}
\int \sqrt{\cot(\theta)} \, d\theta = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left( \frac{\sqrt{2 \cot(\theta)}}{1-\cot(\theta)} \right) - \tanh^{-1}\left( \frac{\sqrt{2 \cot(\theta)}}{1+\cot(\theta)} \right) \right].
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the indefinite integral of a root function I'm stuck on a particular integral problem. The problem is stated as:
$$\int x \sqrt{2x+1} dx$$
My working thus far:
$$\int x \sqrt{2x+1} dx = \frac{1}{2}x^2\frac{2}{3}(2x+1)^\frac{3}{2}$$
$$\frac{1}{3}x^2(2x+1)\frac{3}{2}$$
...
I think I am totally wrong here and completely on another course. Will somebody please point out where I'm screwing up, and possibly guide me to the correct answer?
Thanks!
| It looks like to integrate the product, you found the product of the integration of each factor. We can't do that!
If $\int f(x) \,dx = F(x)$ and $\int g(x)\,dx = G(x)$
$$\int f(x)\cdot g(x) \,dx \neq F(x)G(x) + C$$
Let's start over.
Note that since the integrand is defined only for $2x+1\geq0$, we put
$$\underbrace{u^2 = 2x+1}_{u = \sqrt{2x+1}} \implies 2u\,du = 2\,dx\iff u\,du = dx$$
Note also that we also have $2x = u^2 -1 \iff x = \frac 12( u^2 - 1)$.
Then $$I = \int x \sqrt{2x+1} dx = \frac 12\int (u^2 -1)u\cdot u \,du =\frac 12\int(u^4 - u^2)\,du$$
$$\begin{align} I & = \frac 12\left( \frac {u^5}{5} - \frac{u^3}{3}\right) + C \\
& = \frac 12 \left(\frac{(2x + 1)^{5/2}}{5} - \frac{(2x+1)^{3/2}}{3}\right) + C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now:
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:
$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$
| Here is an alternative way.
\begin{align}\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}&=\frac{\sin(x-a)}{\sqrt{\sin(x-a)\sin(x+a)}}=\frac{\sin x\cos a-\cos x\sin a}{\sqrt{\frac12(\cos 2a-\cos 2x)}}\\&=\frac{\cos a\sin x}{\sqrt{\cos^2a-\cos^2x}}-\frac{\sin a\cos x}{\sqrt{\sin^2 x-\sin^2 a}}\end{align}
Now for the first one, substitute $\cos x=\cos t\cos a$.
$$\int \frac{\cos a\sin x}{\sqrt{\cos^2a-\cos^2x}}dx=\int\cos a\,dt=t\cos a=\cos a\arccos\left(\frac{\cos x}{\cos a}\right)+C_1$$
The second one, substitute $\sin x=\cosh s\sin a$.
\begin{align}\int\frac{\sin a\cos x}{\sqrt{\sin^2 x-\sin^2 a}}dx&=\int\sin a\,ds=s\sin a=\sin a\cosh^{-1}\left(\frac{\sin x}{\sin a}\right)\\&=\sin a\ln\left(\frac{\sin x}{\sin a}+\sqrt{\frac{\sin^2x}{\sin^2a}-1}\right)+C_2\\&=\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+C_3\end{align}
Therefore, $$\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}dx=\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2 x-\sin^2 a})+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.
| Here is a coordinate geometry solution:
The diagonals of a rhombus are perpendicular bisectors of each other. Suppose that the diagonals intersect at the origin of the coordinate plane, that the vertices at the ends of the short diagonal are $A(-3, 0)$ and $C(3, 0)$, that the vertices at the ends of the long diagonal are $B(0, 4)$ and $D(0, -4)$, and that one vertex of the inscribed equilateral triangle is the vertex $C$ of the rhombus.
The other vertices of the equilateral triangle must lie on sides $\overline{AB}$ and $\overline{AD}$ of the rhombus. Let $E$ be the vertex of the equilateral triangle on side $\overline{AD}$ and $F$ be the vertex of the equilateral triangle that lies on side $\overline{AB}$. Since the line containing an altitude of an equilateral triangle is the perpendicular bisector of the base, points $E$ and $F$ are equidistant from the $x$-axis. Thus, if the coordinates of point $F$ are $(a, b)$, then the coordinates of point $E$ are $(a, -b)$. If $s$ is the length of a side of the equilateral triangle, then $s = b - (-b) = 2b$.
In an equilateral triangle of side length $s$, the length of an altitude is $$\frac{s\sqrt{3}}{2}$$
Since $a < 0$, the length of the altitude is
$$3 - a = \frac{s\sqrt{3}}{2}$$
Since $s = 2b$, we obtain
$$3 - a = b\sqrt{3}$$
Solving for $a$ yields
$$3 - b\sqrt{3} = a$$
The equation of $\overleftrightarrow{AB}$ is
$$y = \frac{4}{3}x + 4$$
Hence, at point $F(a, b)$,
\begin{align*}
b & = \frac{4}{3}(3 - b\sqrt{3}) + 4\\
3b & = 4(3 - b\sqrt{3}) + 12\\
3b & = 12 - 4b\sqrt{3} + 12\\
3b + 4b\sqrt{3} & = 24\\
b(3 + 4\sqrt{3}) & = 24\\
b & = \frac{24}{3 + 4\sqrt{3}}\\
b & = \frac{24}{3 + 4\sqrt{3}} \cdot \frac{3 - 4\sqrt{3}}{3 - 4\sqrt{3}}\\
b & = \frac{24(3 - 4\sqrt{3})}{9 - 48}\\
b & = -\frac{24}{39}(3 - 4\sqrt{3})\\
b & = -\frac{8}{13}(3 - 4\sqrt{3})\\
b & = \frac{8}{13}(4\sqrt{3} - 3)
\end{align*}
Hence,
$$s = 2b = \frac{16}{13}(4\sqrt{3} - 3)$$
| {
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Clearing the confusion of the area In the example 1:
They are finding the area as $8$ $\sqrt[2]{30}$. But, I am finding $8$ $\sqrt[2]{15}$.
I have done $\sqrt[2]{16\times5\times8\times3}$
Then,
Breaking it up,
$2\times2\times2\times\sqrt[2]{15}$
Please help me clear my confusion.
| $16 = 2^4$
$8 = 2^3$
$\sqrt{(16)(8)(5)(3)} = \sqrt{(2^4)(2^3)(5)(3)} = \sqrt{(2^{3+4})(5)(3)} = \sqrt{(2^6)(2)(5)(3)}$
$= \sqrt{(8^2)(2)(5)(3)} = 8\sqrt{(2)(5)(3)} = 8 \sqrt{30}$
You honestly just took out an extra two. Happens to the best of us.
:)
| {
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What algebraic manipulation is used to express the solution to this integral? According to WolframAlpha, $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2 \sinh^{-1}\left(\frac{x}{2}\right) + c.$$
I'm wondering how this was obtained, as I got the following:
Let $x = 2\tan\theta$. Then $dx = 2\sec^2\theta$, so \begin{align*}\int\frac{2}{\sqrt{x^2 + 4}}dx &= \int\frac{4\sec^2\theta}{\sqrt{4\tan^2\theta + 4}}d\theta = \int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta = 2\int\frac{\sec^2\theta}{\sec\theta}d\theta\\ &= 2\int\sec\theta{}d\theta = 2\ln\mid{\tan\theta + \sec\theta}\mid + c.\end{align*}
The substitution $x = 2\tan\theta$ implies that $\sec\theta = \frac{\sqrt{x^2 + 4}}{2}$ and $\tan\theta = \frac{x}{2}$. Hence $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2\ln\left|{\frac{x}{2} + \frac{\sqrt{x^2 + 4}}{2}}\right| + c.$$
Either I made a mistake or I don't know the right algebraic manipulation.
| Well by definition, $\sinh^{-1}(z) = \ln\left(z + \sqrt{1+z^2} \right)$. Now, let $z = \frac{x}{2}$.
| {
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How to prove this inequality using am gm? Is it true that
$$\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}>\frac{1}{2k+1}+\frac{1}{2k+2}$$
for all natural k? Is bashing going to work?
| You can compare the RHS and the LHS (line 3 in Jack d'Auruzio's answer) this way:
$\frac{1}{6k+2} \ $ (inside LHS), is bigger than: $\frac{1}{6k+3}\ $ (inside RHS)
$\frac{1}{6k+4} > \frac{1}{6k+6}$ aswell.
We will now prove that what's left still "respects" the original (weaker) inequality:
$\frac{1}{6k+2} + \frac{1}{6k+4} > \frac{1}{6k+3} + \frac{1}{6k+3} \ \ \ \ (1)$.
We could do this very easily without using AM > GM, just do the calculations:
$\frac{(6k+4)+(6k+2)}{(6k+2)(6k+4)} > \frac{(6k+3)+(6k+3)}{(6k+3)(6k+3)}$
$ \frac{12k+6}{36k^2+36k+8} > \frac{12k+6}{36k^2+36k+9}$, which is true for any k in N.
If you want to use AM > GM you can skip my last 2 lines and observe that:
If $AM > GM$, and since $GM > HM$, then $AM > HM$ , therefore $\frac{2}{HM} > \frac{2}{AM}$. Consider $6k+2$ and $6k+4$, their $\frac{2}{HM}$ is just the left side of (1), while the $\frac{2}{AM} $ is just on the right side, therefore:
$\frac{2}{HM} > \frac{2}{AM}$ leads to:
$\frac{1}{6k+2} + \frac{1}{6k+4} > \frac{1}{6k+3} + \frac{1}{6k+3}$
| {
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Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
| As others have mentioned, $\sin^4x$ and $\sin x^4$ are completely different.
By convention, $\sin^4x = (\sin x)^4$, the fourth power of $\sin x$.
Also by convention, $\sin x^4 = \sin (x^4)$, the sine of the fourth power of $x$.
But here are derivatives for both $f(x) = -6 \sin^4 x$ and $g(x) = -6 \sin x^4$:
$$f'(x) = -6 \frac{d}{dx} \left(\sin^4 x\right) = -24 \sin^3x \frac{d}{dx} \sin x = -24 \sin^3x \cos x.$$
$$g'(x) = -6 \frac{d}{dx} \left(\sin x^4\right) = - 6 \cos x^4 \frac{d}{dx} x^4 = -24 x^3 \cos x^4.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Epsilon-N method - Proof verification
Prove (using the epsilon-N method) that the sequence of numbers $\dfrac{5n^3-2}{n^3}$ converges. Calculate the limit first.
First we calculate the limit: $\lim_{n \to \infty} \dfrac{5n^3-2}{n^3} = \lim_{n \to \infty} 5- \dfrac{2}{n^3} = 5$. So we are required to show, for a given $\epsilon > 0 $, that there exists a positive integer $N$ such that if $n>N$, then
$|(5-\dfrac{2}{n^3}) -5 | = |\dfrac{-2}{n^3} | = \dfrac{2}{n^3} < \epsilon $
$$ \dfrac{2}{n^3} < \epsilon \to n > \sqrt[3]{\dfrac{2}{\epsilon}}$$
Therefore, if we choose $N = \lceil \sqrt[3]{\dfrac{2}{\epsilon}} \rceil $, and we let $n$ be any integer such that $n>N$, then $n> \sqrt[3]{\dfrac{2}{\epsilon}}$.
Now we can set up the proof:
Let $\epsilon > 0$. Choose $N = \lceil \sqrt[3]{\dfrac{2}{\epsilon}} \rceil $ and let $n$ be any integer such that $n>N$. Thus $n > \sqrt[3]{\dfrac{2}{\epsilon}}$ and $\epsilon > \dfrac{2}{n^3}$. Therefore
$$|(5-\dfrac{2}{n^3}) -5 | = |\dfrac{-2}{n^3} | = \dfrac{2}{n^3} < \epsilon $$
Thus we conclude that the sequence of numbers $\dfrac{5n^3-2}{n^3}$ converges.
| The proof looks correct. The only thing I noticed was your use of the ceiling function $\lceil$ and $\rceil$.
Unless your professor taught limits in this manner, traditionally one might invoke the Archimedean Principle to show that there exists $N \in \mathbb{N}$ such that $N > a$ for some real number $a$. In fact, the proof of the ceiling function is usually shown as a basic corollary to this principle.
The use of the ceiling function is certainly not incorrect, just make sure that is okay to use in your proof.
| {
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finding the $\lim_{x \rightarrow -2 }\frac{x+2}{x^3+8}$ How do you solve $\lim_{x \rightarrow -2} \frac{x+2}{x^3+8}$ to get 1/12? I tried factoring out the denominator into $(x+2)(x+2)(x+2)$ and cancelling it out with the top but when you plug in 0 for x the denominator is still 0.
| Here are the steps
$$ \lim_{x\to -2} \frac{x+2}{x^3+8}=\lim_{x\to -2} \frac{x+2}{(x+2)(x^2-2x+4)} $$
$$ =\lim_{x\to -2} \frac{1}{x^2-2x+4}= \frac{1}{(-2)^2-2(-2)+4}=\frac{1}{4+4+4}=\frac{1}{12} $$
| {
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Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminate the 2.
| If you don't want to go the formal partial fractions route, you can systematically chip away at the denominator as follows:
$$\begin{align}
{x^2+3\over x^6(x^2+1)}&={3(x^2+1)-2x^2\over x^6(x^2+1)}\\
&={3\over x^6}-{2\over x^4(x^2+1)}\\
&={3\over x^6}-{2(x^2+1)-2x^2\over x^4(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2\over x^2(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2(x^2+1)-2x^2\over x^2(x^2+1)}\\
&={3\over x^6}-{2\over x^4}+{2\over x^2}-{2\over x^2+1}\\
\end{align}$$
The final batch of terms are all easy to integrate, provided you recognize $1/(x^2+1)$ as the derivative of $\arctan x$.
Note: A solid understanding of partial fractions lets you write
$${x^2+3\over x^6(x^2+1)}={Ax^4+Bx^2+C\over x^6}+{D\over x^2+1}$$
(because $x$ appears only to even powers in the expression you're trying to decompose), and then you can solve for the coefficients $A$, $B$, $C$, and $D$. You get the same answer.
| {
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Irrational solutions to some equations in two variables The next statement is a conjecture of mine, so I dont know if it's true (though quite sure):
Let $x,y$ be irrational numbers such that $x^4+y^4=1$. Prove (or disprove) that $x^5+y^5$ is irrational or $x^6+y^6$ is irrational (or both of them are irrational).
Edit: just to remove any doubt, my meaning is to prove that at least one of the numbers $x^5+y^5$ and $x^6+y^6$ has to be irrational.
| It turns out that $K(x^4+y^4,x^5+y^5,x^6+y^6) = K(x+y,xy)$.
Letting $a = x^4+y^4, b = x^5+y^5, c = x^6+y^6$, we have the identities
$$x+y = - \frac{4c^{6}b-10c^{4}ba^{3}+80c^{3}b^{3}a^{2}-100c^{2}b^{5}a+14c^{2}ba^{6}+16cb^{7}-68cb^{3}a^{5}+70b^{5}a^{4}-6ba^{9}}{2c^{6}a-4c^{5}b^{2}-5c^{4}a^{4}+70c^{3}b^{2}a^{3}-170c^{2}b^{4}a^{2}+c^{2}a^{7}+128cb^{6}a+28cb^{2}a^{6}-16b^{8}-37b^{4}a^{5}+3a^{10}}
\\
xy = - \frac{-6c^{6}a+12c^{5}b^{2}+5c^{4}a^{4}-50c^{3}b^{2}a^{3}+90c^{2}b^{4}a^{2}-5c^{2}a^{7}-64cb^{6}a+8b^{8}+9b^{4}a^{5}+a^{10}}{-8c^{5}a^{2}+40c^{4}b^{2}a-40c^{3}b^{4}+8c^{3}a^{5}-60c^{2}b^{2}a^{4}+100cb^{4}a^{3}-4ca^{8}-32b^{6}a^{2}-4b^{2}a^{7}}
$$
So $a,b,c$ are rational if and only if $x+y$ and $xy$ are rational.
Then $x,y$ are irrational whenever $(x-y)^2 = (x+y)^2-4xy$ is not a rational square (and $x,y$ will be in a quadratic extension of $\Bbb Q$)
Since $x^4+y^4 = (x+y)^4 - 4(x+y)^2xy + 2x^2y^2$, to get a counter-example we want to find rational points on the curve $1 = s^4-4s^2p+2p^2$, which is an elliptic curve.
We have the obvious points $s= \pm 1,p=0$, which gives rational $x,y$.
We also have the solutions $s = \pm 1, p =2$.
This gives counter examples $x,y = \frac{1\pm \sqrt{-7}}2$ giving $a=1,b=11,c=9$ ;
and $x,y = \frac{-1\pm \sqrt{-7}}2$ giving $a=1,b=-11,c=9$
There is a rational parametrization sending the curve to the curve $E : y^2 = x^3- x$, and it is known that $E(\Bbb Q) = (\Bbb Z/2\Bbb Z)^2$. The $4$ points I've given are thus the only $4$ rational points, so there isn't any real counter-example.
However, if you allow to replace $\Bbb Q$ with some quadratic extension $K$ of $\Bbb Q$, you can easily find some real counter-example to that more general statement, simply by picking some small $s$ close to $\pm 1$ and then solving for $p$ and choosing the solution close to $0$ (actually the region of the curve where $s^2 > 4p$ is quite small). Most of those $(s,p )$ should correspond to couples $(x,y)$ lying in turn in a quadratic extension of $K$.
| {
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Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{x-2}$$
But the answer is
$$\frac{-4}{(x-2)\sqrt{x-2}}$$
What should I have done?
| Since $(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$, then $$\frac{8}{\sqrt{x-2}}=\frac{0-8(\sqrt{x-2})'}{x-2}=\frac{-4\frac{1}{\sqrt{x-2}}}{x-2}=-\frac{4}{(x-2)\sqrt{x-2}}.$$
| {
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For how many $n$, $x^6+n$ factors? $\textbf{Question}.$
i) For how many integers $n$ with |$n$|$<500$, can the polynomial $p_n(x)=x^6+n$ be written as a product of two non-constant polynomials with integer coefficients?
ii) How will number of solutions depend on $k>0$ if we replace the condition |$n$|$<500$ by |$n$|$<k$.
iii) Does there exist a $m>6$ such that $p_n(x)=x^m+n$ can never be factored for any $n>0$
$\textbf{Thoughts.}$ For last part if $n$ is prime, then by Eisenstein's criterion, it will never be factored as it will be irreducible whatever the $m$ be, but for any general integer $n$, it is too hard to say anything?
For first part that is case of |$n$|$<500$ , again for all primes less than $500$, it is false, so if we write $n=ab$, then $p_n(x)=x^6+ab=(x^3+a)(x^3+b) $ if $(a+b=0)$. But there can be many cases, is there a method or check for all?
| First, consider two cases: $(A, B, k > 0, k \in \mathbb{N})$
Case 1: k is odd.
*
*$A^k+B^k=(A+B)(A^{k-1}+A^{k-2}B+\dots+AB^{k-2}+B^{k-1})$
*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{k-1})$
Case 2: k is even.
*
*$A^k+B^k$ cannot be factorized this way
*$A^k-B^k=(A-B)(A^{k-1}-A^{k-2}B+\dots-AB^{k-2}+B^{k-1})$
*$A^k-B^k=(A+B)(A^{k-1}+A^{k-2}B+\dots+AB^{k-2}+B^{k-1})$
Assuming that no other ways of factorizing polynomial applies, the polynomial $(x^6 + n)$ can only be factored into two polynomials with integer coefficients in the following cases:
(a, b, c are integers larger than $0$)
*
*$n =-a^2$, $a^2 \leq 500 \Rightarrow a \in [1,22]$, for the case $x^6-a^2=(x^3-a)(x^3+a)$
*$n = b^3$, $b^3 \leq 500 \Rightarrow b \in [1,7]$, for the case $x^6+b^3=(x^2+b)(x^4-x^2 b + b^2)$
*$n = -b^3$, $b^3 \leq 500 \Rightarrow b \in [1,7]$, for the case $x^6-b^3=(x^2-b)(x^4+x^2 b + b^2)$
*$n = -c^6$, $c^3 \leq 500 \Rightarrow c \in [1,2]$, for the case $x^6-c^6=(x-c)(x^5+x^4 c+x^3 c^2 + x^2 c^3 +x c^4 + c^5)$
*$n = -c^6$, $c^3 \leq 500 \Rightarrow c \in [1,2]$, for the case $x^6-c^6=(x+c)(x^5-x^4 c+x^3 c^2 - x^2 c^3 +x c^4 - c^5)$
In total, there are $22+7+7+2+2 = 40$ values of n satisfying the conditions. Note that certain values are counted more than once, since we are counting way to distinctly factorize the polynomial. For examples, $n = -64$ is counted twice, as $x^6-64=(x^3-8)(x^3+8)=(x^2-4)(x^4+4x^2+16)$
If $n=0$ is allowed, $x^6=x^5x=x^4x^2=x^3x^3$, adding 3 additional values.
Another way of approaching the problem:
Let $(x^6+n)=(A_5 x^5+A_4 x^4+A_3 x^3+A_2 x^2+A_1 x^1+A_0)(B_5 x^5+B_4 x^4+B_3 x^3+B_2 x^2+B_1 x^1+B_0)$
Solve the system of equations:
*
*$A_5B_5=0$
*$A_5B_4+A_4B_5=0$
*$\dots$
*$A_5B_1+A_4B_2+A_3B_3+A_2B_4+A_1B_5=1$
*$A_5B_0+A_4B_1+A_3B_2+A_2B_3+A_1B_4+A_0B_5=0$
*$\dots$
*$A_0B_0=n$
*$A_i, B_i, n \in \mathbb{Z}, -500<n<500$
| {
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Probability interview question Suppose we have three positive integers $A, B, C$. We randomly choose an integer $a$ in the range $[0,A]$ and an integer $b$ in the range $[0,B]$. Find the probability that $a + b\leq C$.
I am unable to proceed in this question. Any hints .
| Let $S_C$ be the number of couples $(a,b)$, $a\in [0,A]$ and $b\in [0,B]$ such that $a+b \leq C$. It is clear that $S_0=1$ because the only couple that can sum up to zero is $(0,0)$. Moreover if $C>A+B$ then $S_C=S_{A+B}$ since all couples that are counted in $S_C$ were already counted in $S_{A+B}$. Without loss in generality we can assume that $A\leq B$. Let us study the function $T_C=S_{C+1}-S_C$. It corresponds to the number of couples $a,b$ where $a+b=C$. This is easier to count , e.g. if A and B are big enough then for $T_3$ we count the couples $(3,0), (2,1), (1,2)$ and $(0,3)$. In fact we can conclude easily then for $C$ small enough we have $T_C=C+1$. But this only works if $C \leq A$. Once $C>A$ we must start off with $(A,C-A),(A-1,C-A+1),\ldots,(0,C)$ (on the condition that $B$ is big enough so that $C-A \in [0,B]$). In this case we will have $T_C=A+1$. But this can not continue forever since in case $C>B$ we will not even arrive to $(0,C)$ but only to $(C-B,B)$. We conclude that for $C>B$ we have $T_C=A+B-C+1$ so globally we have :$$T_C =
\begin{cases}
C+1, & \text{for $C \leq A$} \\
A+1, & \text{for $A \leq C \leq B$}\\
A+B-C+1, & \text{for $B \leq C \leq A+B+1 $}\\
0, & \text{for $A+B+1 \leq C $}\\
\end{cases} $$
To reconstruct $S$ all we have to do is to "integrate" this function over the different domains. A bit of calculation gives:
$$S_C =
\begin{cases}
C(C+1)/2, & \text{for $C \leq A$} \\
A(A+1)/2+(A+1)(C-A), & \text{for $A \leq C \leq B$}\\
f(C), & \text{for $B \leq C \leq A+B+1 $}\\
(A+1)(B+1) & \text{for $A+B+1 \leq C $}\\
\end{cases} $$
Where $f(C)=(A+1)(B+1)-(A+B+1-C)(A+B+1-C+1)/2$.
| {
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Find the function and domain for $ (f\circ f)$ when $ f(x) = x+ \frac {1} {x} \ $ Find the function and domain for $ (f\circ f)$
My answer is $ \frac {x^4+3x^2+1x} {(x^2+1)(x)}?$
However, the program I am using states I am wrong. What have I done incorrectly?
| $$\ (f\circ f)=x+\frac{1}{x}+\frac{1}{x+\frac{1}{x}}=\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{x^4+x^2+x^2+1+x^2}{x(x^2+1)}=\frac{x^4+3x^2+1}{x(x^2+1)}$$
The domain is right, it's $\ x≠0$, but in the numerator you have 1 and not $\ 1x$
| {
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Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
| Riemann sums do this pretty quickly. $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ is a
lower Riemann sum for $\int_n^{2n} {1 \over x}\,dx = \log 2$, and an upper Riemann sum for
$\int_{n+1}^{2n + 1} {1 \over x}\,dx = \log {2n +1 \over n + 1}$, so one has
$$\log {2n +1 \over n + 1} < \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} < \log 2 $$
| {
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polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \equiv n^4+2n^2 \pmod3 $$$$n^4+2n^2 = n^2(n^2+2)=n^2(n^2-1)=n^2(n+1)(n-1)\equiv 0 \pmod3$$
Divisble by 8: $$n^4+ 6n^3 + 11n^2+6n \equiv n(n^3-2n^2-5n-2) \pmod8 $$$$n(n^3-2n^2-5n-2) = n(n+1)(n^2-3n-2) = n(n+1)(n^2+5n+6) = n(n+1)(n+2)(n+3) = ??? \pmod8$$
So it seems that the polynomial has to be divisible by four, but I'm not sure how to show that it has to be divisible by two one more time to show that it's divisible by 8.
Any other approaches to solving the problem are welcome, as I'm preparing for an exam and would appreciate having multiple ways to tackle the problem. I think this problem was meant to be an exercise in proof by induction, but the other approach seemed more approachable.
| You can always do a brute force method. Check n=0,1,2,3,4,5,6,7
$$
n=0,
n(n+1)(n+2)(n+3)=0(1)(2)(3)=0 (mod 8)
$$
$$
n=1,
1(2)(3)(4)=24=0 (mod8)
$$
$$....$$
$$n=7,
7(8)(9)(10)=7(0)(1)(2)=0 (mod8)
$$
This is sufficient since if $$n=8m+r, r=0,1,2,3,4,5,6,7, m\in Z $$ then
$$
n(n+1)(n+2)(n+3)=(8m+r)(8m+r+1)(8m+r+2)(8m+r+3)=r(r+1)(r+2)(r+3)(mod8)$$
| {
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Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI How many ways can a sequence of three letters be chosen from the letters of MISSISSIPPI?
I'm just a little confused how to go about this since so many letters repeat
I=4
S=4
P=2
So ultimately there are 4 different letters.
| For a generating function solution, use exponential generating functions.
Let's take an example that's a bit simpler than yours. How many three-letter words can be made from the letters in AABBB? If you were simply forming a multiset of three letters, but not arranging them in order, you could let $r$ be the number of As and $3-r$ be the number of Bs. Then $r$ could be any of $0,$ $1,$ or $2.$ So the number of such multisets is $3.$ Notice that $3$ is the coefficient in front of $x^3$ in the expansion of $(1+x+x^2)(1+x+x^2+x^3).$ The first factor reflects the fact that you can take $0,$ $1,$ or $2$ As, and the second that you can take $0,$ $1,$ $2,$ or $3$ Bs. Since As are not distinguishable and Bs are not distinguishable, there is only one way to take a particular number of As or Bs. This is why the coefficients in front of the powers of $x$ are all $1.$
Now let's worry about how many ways there are to arrange the letters in a multiset. If our multiset has $r$ As and $n-r$ Bs, there are $\frac{n!}{r!(n-r)!}={}_nC_r$ ways to arrange the letters. We can account for this by using exponential generating functions instead of ordinary generating functions. The way this works is that we expand
$$
\left(\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}\right)\left(\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}\right).
$$
Again we find the coefficient of $x^3$ in the expanded function, but this time we multiply the coefficient by $3!.$ An equivalent way to say this is that we find the coefficient of $\frac{x^3}{3!}$ in the expansion. This coefficient is guaranteed to be a whole number since it is effectively introducing binomial coefficients into the expansion. In the unordered example above, $x^3$ arose as the sum $1\cdot x^3+x\cdot x^2+x^2\cdot x.$ In the ordered case, we get
$$
\frac{1}{0!\,3!}1\cdot x^3+\frac{1}{1!\,2!}x\cdot x^2+\frac{1}{2!\,1!}x^2\cdot x=\frac{1}{3!}\left({}_3C_0\cdot 1\cdot x^3+{}_3C_1\cdot x\cdot x^2+{}_3C_2\cdot x^2\cdot x\right),
$$
which is exactly what we want. We find that the number of words is $1+3+3=7.$ This is correct: BBB, ABB, BAB, BBA, AAB, ABA, BAA.
You should be able to generalize this idea to your problem.
| {
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Prove the given condition from given two quadratic equation
Question: If the quadratic equations $x^2+bx+c=0$ and $bx^2+cx+1=0$ have a
common root then prove that either $b + c + 1 = 0$ or $b^2 + c^2 + 1 =bc + b + c$
Till yet,
I had figured the common root of the given two quadratic equation. i.e.
Multiplying first equation by $b$ and eliminating the term $bx^2$ from the equation I get the common root ($\alpha$ say),
$$\alpha=\frac{1 - cb}{b^2 - c}$$
Further putting this value in either of the equation didn't benefited me much.What it gave me was an odd, unfriendly equation. Can anyone help me in this?
Thanks in advance.
| Let $y$ be the common root then, consider $y^2+by+c=0$ and $by^2+cy+1=0.$ Clearly $y\not=0.$ Therefore $$y^3+by^2+cy=0$$
$$y^3-1=0$$
$$(y-1)(y^2+y+1)=0.$$
If $y=1,$ then clearly $b+c+1=0.$
If $y^2+y+1=0,$ then note that $by+c=y+1$ and $cy+1=b(y+1).$ Solve this two simple linear equations and equate the value of $y.$
| {
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Convergence and estimation of the rests of $\sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}} $ For $x\in ]-1,1[$ is required to evaluate the expression
$$
S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}
$$
with absolute error $\epsilon= 3\times 10^{-10}.$
*
*I need to prove that the series $\displaystyle\sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} $ converges and $\displaystyle \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}$ also.
*How many terms are required to evaluate the first series in order to have an absolute error lower than $\epsilon$?
*How (if possible) you can re-write $S(x)$ in order to be evaluated faster.
Sorry if that question is very simple, but for me it is hard for me.
Thanks.
| Using cancellations yields smaller errors, since $$\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}=\frac{-2x}{\sqrt{k^6-x^2}\left(\sqrt{k^3+x}+\sqrt{k^3-x}\right)}\in\Theta\left(\frac1{k^{9/2}}\right),$$ which shows that, for every $K\geqslant1$, $$\sum_{k=K+1}^\infty\left(\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right)\in\Theta\left(\frac1{K^{7/2}}\right),$$ with some explicit bounds if one desires so.
Edit: Let us first show that there exists some $c$ such that, for every $|x|\lt1$ and $k\geqslant2$, $$\left|\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right|\leqslant\frac{c}{(k-1)^{7/2}}-\frac{c}{(k+1)^{7/2}}.$$
It is enough to ask that, for every $k\geqslant2$, $$\frac{1}{\sqrt{k^3-1}} - \frac{1}{\sqrt{k^3+1}}\leqslant\frac{c}{(k-1)^{7/2}}-\frac{c}{(k+1)^{7/2}}.$$ Using the mean value theorem twice, one sees that there exists some $u$ and $v$ in $(-1,1)$, depending on $k$, such that $$\frac{1}{\sqrt{k^3-1}} - \frac{1}{\sqrt{k^3+1}}=\frac1{k^{3/2}}\left(\frac{1}{\sqrt{1-1/k^3}} - \frac{1}{\sqrt{1+1/k^3}}\right)=\frac1{k^{3/2}}\frac2{k^3}\frac{1/2}{(1+u/k^3)^{3/2}}$$ and $$\frac{1}{(k-1)^{7/2}}-\frac{1}{(k+1)^{7/2}}=\frac1{k^{7/2}}\left(\frac{1}{(1-1/k)^{7/2}}-\frac{1}{(1+1/k)^{7/2}}\right)=\frac1{k^{7/2}}\frac2{k}\frac{7/2}{(1+v/k)^{9/2}}.$$ Using, for every $k\geqslant2$, $$1+u/k^3\geqslant7/8,\qquad 1+v/k\leqslant3/2,$$ one gets that some suitable $c$ in the inequality above is $$c=\frac{3^{9/2}}{7^{5/2}}\lt1.1.$$ Using a telescoping series, one gets, for every $K\geqslant1$, $$\left|\sum_{k=K+1}^\infty\left(\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3-x}}\right)\right|\leqslant\frac{c}{K^{7/2}}+\frac{c}{(K+1)^{7/2}}\leqslant\frac{2.2}{K^{7/2}}.$$
| {
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Solving congruence equations Solve: $7x^6\equiv 11 \pmod{23}$ and $5^x\equiv 19 \pmod{23}$
I can solve simple congruence equations but how do I go about solving these?
| $$2^2=4,2^3=8,2^4=16\equiv-7,2^5\equiv9,2^6\equiv18\equiv-5,2^{11}\equiv9(-5)\equiv1$$
$5^2\equiv2\pmod{23},5^5=5(5^2)^2\equiv5\cdot2^2\equiv20\equiv-3,5^6\equiv2^3\equiv8,$
$5^{11}=5^5\cdot5^6\equiv-3\cdot8\equiv-1$
So, $5$ is a primitive root $\pmod{23}$ unlike $2$
$7\equiv-16\equiv5^{11}(5^2)^4\equiv5^{19}\implies$ind$_57=19$
$11\equiv-12\equiv2^2(-3)\equiv(5^2)^25^5\equiv5^9$
Using Discrete logarithm, $7x^6\equiv11\pmod{23}$
$\implies$ind$_57+6$ind$_5x\equiv$ind$_511\pmod{22}\implies19+6$ind$_5x\equiv9\pmod{22}$
$\implies6$ind$_5x\equiv-10\pmod{22}\equiv12$
As $6a=12+22b\implies3y=6+11b\implies3a\equiv6\pmod{11}\implies a\equiv2\pmod{11}$ as $(3,11)=1$
$\implies$ind$_5x\equiv2\pmod{11}\implies x\equiv2^5\pmod{23}$
| {
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Simplifying a square root fraction Simplify the following
$$\frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})}$$
Apparently the answer is $\frac{1}{2} (2 + \sqrt{2})$ but can't for the life of me see how to get it. Any help is massively appreciated.
Thanks
| $\dfrac{\sqrt 3}{\sqrt 2 (\sqrt 6 - \sqrt 3)} = \dfrac{\sqrt 3}{\sqrt 2 \sqrt 3(\sqrt2 - 1)} = \dfrac{1}{\sqrt 2 (\sqrt 2 - 1)} = \dfrac{1}{(2 - \sqrt 2)} = \dfrac{(2 + \sqrt 2)}{(2 - \sqrt 2)(2 + \sqrt 2)} = \dfrac{2 + \sqrt 2}{2}$
The reason we write it this way is that it is bad practice to have roots at the denominator. To get rid of $a+b \sqrt c$ at the denominator, simply multiply the numerator and the denominator by $a-b \sqrt c$.
| {
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Evaluate $ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos (2x)}\right) \mathrm{d}x $ Evaluate the following definite integral:
$$ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos(2x)}\right) \mathrm{d}x, $$ where $k \in \mathbb{N}_{>0}$.
| Let
$$I_k = \int_0^\pi \left( \frac{2 + 2\cos x - \cos(k-1)x - 2\cos kx - \cos(k+1)x}{1 - \cos2x}\right) \mathrm dx.$$
Then, we have:
*
*$I_0 = 0$;
*$I_1 = \pi$.
For any $ k \in \mathbb{N}^*$, we have:
$$ \begin{align}
I_{k+1} - 2I_k + I_{k-1}
&= \int_0^\pi \left( \frac{\cos(-2+k)x - 2\cos kx + \cos (2 + k)x}{-1 + \cos 2x}\right) \mathrm dx \\
&= \int_0^\pi \left( \frac{2\cos kx(-1 + \cos^2x)}{-1 + \cos 2x}\right) \mathrm dx =0
\end{align}$$
Since $ I_{k+1} - I_k = I_k - I_{k-1} $, $(I_k)$ is an arithmetic progression. Hence $ I_k = k\pi $.
| {
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How can I prove this inequality by M.I. or otherwise? This question is from the past examination:
$$f(n)=\frac{2}{3}(n^{3/2}-1)+\sqrt{n}$$
$$g(n)=1+\frac{2}{3}((n+1)^{3/2}-2^{3/2})$$
My task to prove $f(n)≥g(n)$ for all $n≥1$.
I have tried M.I here.
But the problem is that I do not know how to expand the term $(n+2)^{3/2}$ in the inductive steps. Can anyone tell me what method should I use to finish the proof.
| Since $f(n)\geq g(n)$ holds for $n=1$, it is sufficient to show that:
$$\begin{eqnarray*} f(n+1)-f(n) &=& \frac{2}{3}((n+1)^{3/2}-n^{3/2})+(n+1)^{1/2}-n^{1/2}\\&\geq& \frac{2}{3}((n+2)^{3/2}-(n+1)^{3/2})=g(n+1)-g(n)\end{eqnarray*}$$
that is equivalent to:
$$\frac{2}{3}\left((n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}\right)\leq (n+1)^{1/2}-n^{1/2}\tag{1}$$
or to:
$$(2n+4)(n+2)^{1/2}+(2n+3)n^{1/2}\leq (4n+7)(n+1)^{1/2}\tag{2}$$
that is equivalent to:
$$\frac{2n+4}{4n+7}\left(1+\frac{1}{n+1}\right)^{1/2}+\frac{2n+3}{4n+7}\left(1-\frac{1}{n+1}\right)^{1/2}\leq 1. \tag{3}$$
In virtue of the Cauchy-Schwarz inequality the LHS of $(3)$ is less or equal than:
$$\frac{\sqrt{2}}{4n+7}\cdot\sqrt{(2n+4)^2+(2n+3)^2}=\sqrt{1+\frac{1}{(4n+7)^2}},$$
that, however, is bigger than one. Using $\sqrt{1+x}+\sqrt{1-x}\leq 2-\frac{x^2}{4}$ we get that the LHS of $(3)$ is less or equal than:
$$1-\frac{1}{8(n+1)^2}+\frac{1}{8n+14}\left(\left(1+\frac{1}{n+1}\right)^{1/2}-\left(1-\frac{1}{n+1}\right)^{1/2}\right)$$
and since over $(0,1]$ we have $\sqrt{1+x}-\sqrt{1-x}\leq\frac{x}{1-x^2/3}$, it follows that the LHS of $(3)$ is less than:
$$ 1-\frac{1}{8(n+1)^2}+\frac{1}{(8n+14)(n+1)\left(1-\frac{1}{3(n+1)^2}\right)}\leq 1.$$
| {
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Ray Sphere Intersection I'm reading a book on raytracing and there's a part where the author is working out the equation for ray-sphere intersection. There's a part I don't understand that I'm hoping I can get help with.
The equation starts with:
$(\mathbf{p} - \mathbf{c}) \bullet (\mathbf{p} - \mathbf{c}) - r^2 = 0$
Bold values are vectors, and the dot indicates the dot product.
He then substitutes the ray in for p, making the equation:
$(\mathbf{o} + t \mathbf{d} - \mathbf{c}) \bullet (\mathbf{o} + t \mathbf{d} - \mathbf{c}) - r^2 = 0$
Next is the part I couldn't follow. He expands the equation to:
$(\mathbf{d} \bullet \mathbf{d})t^2 + [2(\mathbf{o} - \mathbf{c}) \bullet \mathbf{d}]t + (\mathbf{o} - \mathbf{c}) \bullet (\mathbf{o} - \mathbf{c}) - r^2 = 0$
I know it's kind of a stupid question, but can someone break it down how he got from equation 2 to equation 3 by expanding? I can't figure out all the intermediate steps on my own.
| It's a matter of rearranging the terms inside and applying FOIL:
$$(\mathbf{o} + t \mathbf{d} - \mathbf{c}) \cdot (\mathbf{o} + t \mathbf{d} - \mathbf{c}) \\ = [t \mathbf{d} + (\mathbf{o} - \mathbf{c})] \cdot [t \mathbf{d} + (\mathbf{o} - \mathbf{c})] \\ = t^2 \mathbf{d} \cdot \mathbf{d} + 2t (\mathbf{o} - \mathbf{c}) \cdot \mathbf{d} + (\mathbf{o} - \mathbf{c}) \cdot (\mathbf{o} - \mathbf{c}).$$
Going from the first line to the second, I used the fact that addition and subtraction with vectors are each commutative: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}.$
Going from the second line to the third, I used the fact that the dot product of two vectors is commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}.$
Also "FOIL" for the dot product of the vector sums:
$$(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{c} + \mathbf{d}) = \mathbf{a} \cdot \mathbf{c} + \mathbf{a} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{d}.$$
| {
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differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x^2 + 3}{(x^3 - 4)^3}$
f ' (x)
$= \frac{(4x) (x^3 - 4)^3 - (2x^2 + 3)[3(x^3 -4)^2 (3x)}{[(x^3 - 4)^3]^2}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)(x^3 -4)^2}{(x^3 - 4)^6}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)}{(x^3 - 4)^4}$ ---quotient rule
3) f(x) = sin(x cos x)
$f ' (x) = \cos [x \cos(x)] (\cos x - x \sin x)$
4) f(x) = $x^2$ tan 2x
$f ' (x) = 2x \tan (2x)$ + $2x^2$ $\sec^2 2x$
5) f(x) = $ 3 \ln (\cos 3x) $
$f ' (x) = -9 \tan 3x$
| $$(-\frac3{x^{5/2}})'=-3x^{-5/2}=-3(-\frac52)x^{-\frac52-1}=\frac{15}{2x^{7/2}}$$
$$\left(\frac{2x^2+3}{(x^3-4)^3}\right)'=\frac{(4x)(x^3-4)^3-(2x^2+3)\big(3(3x^2)(x^3-4)^2\big)}{((x^3-4)^3)^2}=\frac{4x(x^3-4)-9(2x^2+3)x^2}{(x^3-4)^4}=-x\frac{14x^3+27x+16}{(x^3-4)^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration of $\sqrt{x+\sqrt{x^2+3x}}$ I faced the following indefinite integration problem:
$$\int \sqrt{x+\sqrt{x^2+3x}}dx$$
This result by WolframAlpha suggests that there is an elementary way to compute this integration. But I don't know how to start. Any hints would be appreciated.
| See the other answer. To find $\int \frac{27}{2(2t^2+3)^2}\, dt$, you can avoid using non-real numbers (notice that $2t^2+3=at^2+bt+c$, where $b^2-4ac<0$, has no real roots and no non-trivial factorization over the real numbers. We can't use partial fractions with real numbers).
See this link (link) (Wikipedia (link) also has some formulas), which shows the integration formulas over the real numbers of $\int \frac{1}{(x^2+bx+c)^n}\, dx$, $\int \frac{x}{(x^2+bx+c)^n}\, dx$, and see the examples there, in particular the example $\int \frac{1}{(x^2+1)^2}\, dx$. It generalizes. Notice that
$$\left(\frac{1}{\left(x^2+bx+c\right)^n}\right)'=\frac{-(2x+b)n}{\left(x^2+bx+c\right)^{n+1}}$$
$$\left(\frac{x}{\left(x^2+bx+c\right)^n}\right)'=\frac{x^2+bx+c-nx(2x+b)}{\left(x^2+bx+c\right)^{n+1}}$$
Use integration by parts, partial fractions. $$\int \frac{1}{2t^2+3}\, dt$$
$$\int u\, dv=uv-\int v\, du$$
$$u=\frac{1}{2t^2+3}$$
$$du=\frac{-4t}{(2t^2+3)^2}\, dt$$
$$dv=dt, v=t$$
$$\int \frac{1}{2t^2+3}\, dt=\frac{t}{2t^2+3}-$$
$$-\int \frac{-4t^2}{(2t^2+3)^2}\, dt$$
Use partial fractions. You could use WolframAlpha (link) if you want, but it's not needed.
$$\frac{-4t^2}{(2t^2+3)^2}=\frac{6}{(2t^2+3)^2}-\frac{2}{2t^2+3}$$
Now find $\int \frac{1}{(2t^2+3)^2}\, dt$ in terms of $$\int\frac{1}{2t^2+3}\, dt=\frac{1}{3}\sqrt{\frac{3}{2}}\int\frac{d\left(\sqrt{\frac{2}{3}}t\right)}{\left(\sqrt{\frac{2}{3}}t\right)^2+1}=$$
$$=\frac{1}{\sqrt{6}}\arctan\left(\sqrt{\frac{2}{3}}t\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Absolute convergence of $\sum_{n=1}^{\infty} (-1)^n (1-\cos1/n)$ I would like to see if $$\sum_{n=1}^{\infty} (-1)^n \left(1-\cos \frac{1}{n}\right)$$ converges absolutely.
| Donote your sum by $S$. Taylor series for $\cos$ gives us $\cos x = 1 - \frac{1}{2}x^2 + o(x^3)$, hence $1 - \cos x = \frac{1}{2}x^2 + o(x^3)$. Since $|(-1)^n (1-\cos\frac{1}{n})| = \cos\frac{1}{n}$, it follows that
$$S \leq \sum_{n=1}^{\infty}\left(\frac{1}{n^2} + |o(\frac{1}{n^3})|\right).$$
Since $\frac{o(x^3)}{x^3}\to 0 $, when $x\to 0$, we know that $o(x^3)\leq x^3$ for all $|x|\leq x_0$ for some $x_0\in (0,\infty)$. It follows that
$$S\leq \sum_{n = 1}^{\infty} \frac{1}{n^2} + \sum_{n = 1}^{n_0}o(\frac{1}{n^3}) + \sum_{n = n_0 + 1}^{\infty}\frac{1}{n^3}<\infty$$
where $n_0$ is a integer for which $\frac{1}{n_0}\leq x_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/970700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Outer Product of Two Matrices? How would I go about calculating the outer product of two matrices of 2 dimensions each? From what I can find, outer product seems to be the product of two vectors, $u$ and the transpose of another vector, $v^T$.
As an example, how would I calculate the outer product of $A$ and $B$, where
$$A = \begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix} \qquad B = \begin{pmatrix}5 & 6 & 7 \\ 8 & 9 & 10\end{pmatrix}$$
| To extend davcha's answer, for your specific example, you would get:
$$A\otimes B=
\left(
\begin{array}{cc}
\left(
\begin{array}{ccc}
5 & 6 & 7 \\
8 & 9 & 10 \\
\end{array}
\right) & \left(
\begin{array}{ccc}
10 & 12 & 14 \\
16 & 18 & 20 \\
\end{array}
\right) \\
\left(
\begin{array}{ccc}
15 & 18 & 21 \\
24 & 27 & 30 \\
\end{array}
\right) & \left(
\begin{array}{ccc}
20 & 24 & 28 \\
32 & 36 & 40 \\
\end{array}
\right) \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:
Solve the following system for $a,b,c\in \mathbb{R}$:
$$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$
I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6)^2=4(2c+6)$$
Then maybe $a=b=c$ is one case.
Thank you.
| $$(b^2-6)^2-(c^2-6)^2=8(a-b)\Leftrightarrow (b^2-c^2)(b^2+c^2-12)=8(a-b)$$
$$(b^2-6)^2-(a^2-6)^2=8(a-c)\Leftrightarrow (b^2-a^2)(b^2+a^2-12)=8(a-c)$$
$$(c^2-6)^2-(a^2-6)^2=8(b-c)\Leftrightarrow (c^2-a^2)(c^2+a^2-12)=8(b-c)$$
Assume that $a\geq b\geq c\geq 0$ then
$$b^2+c^2\geq 12$$
on the other hand
$$b^2+a^2\leq12$$
and $$c^2+a^2\leq12$$
which would yield $$b^2+a^2\leq b^2+c^2\Leftrightarrow a^2\leq c^2\Leftrightarrow a\leq c$$ and
$$c^2+a^2\leq c^2+b^2\Leftrightarrow a^2\leq b^2\Leftrightarrow a\leq b$$
Therefore $a=b=c$.
It remains to check when not all of them are positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $X+Y = 10 $ ($X$ and $Y $ both are positive) then what is the maximum value of $(X^3)(Y^2)$? I can get to the result by trying different values of X and Y but that is of course time taking. I want to know if there is a better way to get to the result?
| $Y = 10 - X$,and $X^3Y^2 = X^3(10-X)^2 = X^3(X^2 - 20X + 100) = X^5 - 20X^4 + 100X^3 = f(X)$.
$f'(X) = 5X^4 - 80X^3 + 300X^2 = 0 \iff 5X^2(X^2 - 16X + 60) = 5X^2(X-6)(X-10) = 0 \iff X = 0, 6, 10$. Since $0 < X < 10$, $X = 6$, and $Y = 10 - 6 = 4$. From this the max value of $X^3Y^2$ is $6^34^2 = 3,456$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is
If $\displaystyle \int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}.$ Then $f(x)$ is
$\bf{Options:}$ $(a)\;\displaystyle \frac{1}{a^2\sin^2 x+b^2 \cos^2 x} \;\;\;\;\;\; (b)\; \frac{1}{a^2\sin^2 x-b^2 \cos^2 x}\;\;\;\;\;\;(c)\; \frac{1}{a^2\cos^2 x-b^2 \sin^2 x}$
$\displaystyle (d)\; \frac{1}{a^2\cos^2 x+b^2 \sin^2 x}$
$\bf{My\; Solution:}$ Given $\displaystyle \int f(x)\sin x\cos xdx = \frac{1}{2(a^2-b^2)}\ln \left|f(x)\right|+\mathcal{C}$
Now Diff. both side w. t. to $x\;,$ We Get $\displaystyle f(x)\sin x\cos x = \frac{1}{2(a^2-b^2)}\cdot \frac{f^{'}(x)}{f(x)}.$
Now We get $\displaystyle \frac{f^{'}(x)}{(f(x))^2} = (a^2-b^2)\cdot \sin 2x,$ Now Integrate both side w. r. to $x$
$\displaystyle \int \frac{f^{'}(x)}{(f(x))^2}dx = (a^2-b^2)\int \sin 2xdx\Rightarrow --\frac{1}{f(x)} = -\frac{(a^2-b^2)}{2}\cos 2x$
So We get $\displaystyle f(x) = \frac{2}{(a^2-b^2)\cos 2x}$
But Answer Given is $(a)\;,$ I did not understand where i have make mistake, Help me
Thanks
| $\displaystyle\int\frac{f^{'}(x)}{(f(x))^2}dx=(a^2-b^2)\int\sin 2xdx\\
\Rightarrow \frac{1}{f(x)}=\frac{(a^2-b^2)}{2}\cos 2x+C\\
=\frac{(a^2-b^2)}{2}(\cos^{2}x-\sin^{2}x)+\frac{(a^2+b^2)}{2}(\cos^{2}x+\sin^{2}x)+C\\
=a^2\cos^{2}x+b^2\sin^{2}x+C$
So We get $\displaystyle f(x) = \frac{1}{a^2\cos^{2}x+b^2\sin^{2}x+C}$ and the only possible answer is (d).
Answer (a) can't be correct, as $(a^2\sin^{2}x+b^2\cos^{2}x)'=2(a^2-b^2)\sin x\cos x$, so
$$\int\frac{2(a^2-b^2)\sin x\cos x}{a^2\sin^{2}x+b^2\cos^{2}x}
=\ln(a^2\sin^{2}x+b^2\cos^{2}x)+C
=-\ln\frac{1}{a^2\sin^{2}x+b^2\cos^{2}x}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\sin^n{x}$ How do you integrate:
$$\int(\sin^n{x}) dx$$
The link to WolframAlpha : (Integration Answer)
No definite limits...
What is that hypergeometric function in that answer. Please help!
Thanks
| Using binomial series (link), we have:
$$(1-\cos^2x)^{\frac{n-1}{2}} = 1 - \frac{n-1}{2}\cos^2x + \frac{n-1}{2}\frac{n-3}{4}\cos^4x-\ldots. $$
Then:
$$ \int\sin^nx dx = \int\sin x (1-\cos^2x)^{\frac{n-1}{2}} dx $$
$$ = \int\sin x \left(1 - \frac{n-1}{2}\cos^2x + \frac{n-1}{2}\frac{n-3}{4}\cos^4x-\ldots\right) dx $$
$$ = \int \left(\sin x + \frac{1-n}{2}\cos^2x \sin x+ \frac{1-n}{2}\frac{3-n}{4}\cos^4x\sin x -\ldots\right) dx $$
$$ = -\cos x - \frac{1-n}{2}\cos^3x/3 - \frac{1-n}{2}\frac{3-n}{4}\cos^5x/5 -\ldots$$
$$ = -\cos x \cdot({\rm hypergeometric \; series}).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.
Could anyone shine some light on this? Thanks.
| Let $x^2+2x+1=(x+1)^2$ but $(x+1)^2\equiv0,1\pmod4$
i.e., $x^2+2x+1=4a$ or $4a+1$ for any integer $a\ge0$
So, $x^2+2x+2=4a+1$ or $4a+2$
In either case, the floor by $\pmod4$ remains same even for $x^2+2x+3$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the $n$th term of a repeating pattern What is the nth term of following sequence
$1,2,3,4,5,6,1,2,3,4,5,6,\ldots$
$(n, n+1, n+2, n+3,\ldots,$ $,n+p, n, n+1, n+2, n+3,$ $n+4, n+5,$ $\ldots,$ $n+p,$ $n,$ $n+1,$ $\ldots)$
Actually, I am trying to solve the following puzzle
$6$ girls pick a captain by forming a circle then eliminating every $m$th girl. The 2nd girl in the counting order can choose $m$. If she wants to be captain what's the smallest $m$ she should pick?
| The $n$-th term of sequence $1,2,3,4,5,6,1,2,3,4,5,6,\ldots$ is given by $a_n=1+(n-1 \bmod 6)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Triplets of distinct integers > 1 that return integer values. If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?
UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$
My further work:
Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$
Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$
Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.
Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.
Case $1$: $k<C$
Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$
By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$
This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.
If $k=1$ we have $A+B=C$ which violates the condition $A>C$.
If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$
However, this leads to the result that $A=B$ which violates that stated conditions.
Case $1$, $k<C$, fails.
Case $2$: $k=C$
Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$
This later equation is clearly false and thus Case $2$, $k=C$ fails.
Case $3$: $k>C$
Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.
Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.
Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$
Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$
By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$
Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.
By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$
and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$
From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.
Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$
Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$
Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.
Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.
Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.
| $f(A,B,C)$ has no integral solutions for distinct integers $A,B,C$.
Let $f(A,B,C)=k$ as in the question. $k<C$ and $k=C$ have been shown above to lead to contradictions. $k>C$ has been shown to reduce to solving $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ $$C=\frac{AB}{A+B}$$ for positive integers as restricted above.
Start with: $$C=\frac{AB}{A+B}$$ This equation is identical to $$\frac{1}{A}+\frac{1}{B}=\frac{1}{C}$$ Let $A=C+s$ and $B=C+t$ with $s,t$ positive integers with $s>t$. We now have $$\frac{1}{C+s}+\frac{1}{C+t}=\frac{1}{C} \Rightarrow (C+s)(C+t)=C(C+s+C+t)$$ $$C^2+C(s+t)+st=2C^2+C(s+t)\Rightarrow C^2=st$$
The solution set for this last equation is $s=na^2, t=nb^2, C=nab$ for positive integral $n,a,b; a>b$. From this we have $A=na^2+nab$ and $B=nb^2+nab$. Substituting into our equation for $p$ we have: $$p=\frac{(na^2+2nab+nb^2)^3-(na^2+nab)(nb^2+nab)(na^2+2nab+nb^2)}{((na^2+nab)(nb^2+nab))^2-(na^2+2nab+nb^2)^2}\Rightarrow$$
$$p=\frac{(n(a+b)^2)^3-n^2ab(a+b)^2(n(a+b)^2)}{(n^2ab(a+b)^2)^2-(n(a+b)^2)^2}\Rightarrow$$
$$p=\frac{n^3(a+b)^6-n^3ab(a+b)^4}{n^4a^2b^2(a+b)^4-n^2(a+b)^4}\Rightarrow$$
$$p=\frac{n(a+b)^2-nab}{n^2a^2b^2-1}=\frac{na^2+nab+nb^2}{n^2a^2b^2-1}\Rightarrow$$
$$p=\frac{na^2}{n^2a^2b^2-1}+\frac{nab}{n^2a^2b^2-1}+\frac{nb^2}{n^2a^2b^2-1}\Rightarrow$$
$$p\approx\frac{1}{nb^2}+\frac{1}{nab}+\frac{1}{na^2}<1|a,b,n\geq2$$
Thus $n,a,b$ must be small. In fact, a quick computer search reveals that $p=\frac{na^2+nab+nb^2}{n^2a^2b^2-1}$ is only greater than $1$ if $n=b=1$. Rewriting for $p$ we have $$p=\frac{a^2+a+1}{a^2-1}$$ which is clearly asymptotic to $1$.
Testing a few values, if $a=2$ then $p=2.\overline{3}$, if $a=3$ then $p=1.625$.
Since as $a$ gets larger $p$ gets smaller, and $p$ is asymptotic to $1$, clearly no integer $a\geq2$ results in an integral $p$.
Since there exist no integral $p$'s that can solve $k=pC+q$ (see the question for how to derive this expression) clearly $k$ is not greater than $C$ and Case $3$, $k>C$ fails.
$k=f(A,B,C)$ is neither less than, greater than, nor equal to $C$, which is clearly a contradiction and so,
$f(A,B,C)$ has no integral solutions for distinct integers $A,B,C$.
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Find the missing angle of similar triangle Find the missing angle $\theta$ in the triangle below given that $R>r$, $l\geq R$, $0< \theta < \frac{\pi}{2}$.
Attempted Solution
I attempted to use similar triangles to find the angle $\theta$ but the resulting expression for $\theta$ is pretty ugly.
$\frac{r}{R\cos\theta+\sqrt{l^2-R^2 \sin^2 \theta}}=\frac{R\sin\theta}{\sqrt{l^2-R^2\sin^2\theta}}$
I would then use some numerical solver to find $\theta$. Any other ways to attack this problem?
| Hint
\begin{equation}
\begin{cases}
S = {S_1} + {S_2} = {1 \over 2}Rr\sin {\theta _1} + {1 \over 2}R{l_2}sin\theta = {1 \over 2}r{l_2} \\ \\
{\cos }{\theta _1} = {{{r^2} + {R^2} - l_1^2} \over {2Rr}} \\ \\
{\cos}\theta = {{l_2^2 + {R^2} - {l^2}} \over {2R{l_2}}} \\ \\
{\cos}{\theta _1} = {\sin}\theta , {\sin}{\theta _1} = {\cos}\theta \\ \\
{r^2} + l_2^2 = {({l_1} + l)^2}
\end{cases}
\end{equation}
We can find $l_1 ~ , l_2$ then we can find $\theta$
$S_1 \to $ The area of the triangle on the left
$S_2 \to $ The area of the triangle on the right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that the gradient of the tangent to $xy=d^2$ is $-\frac{c^2}{4d^2}$ I have this question:
Given that $y=mx+c$ is a tangent to $xy=d^2$ prove that $m=-\frac{c^2}{4d^2}$.
I'm not sure what direction to take - I tried differentiating the hyperbola equation, but that gave me a gradient of $-\frac{y}{x}$. How do I go about proving this?
| $xy = d^2 \Leftrightarrow y = \frac{d^2}{x}$.
Tangent line equation to $f(x)$ in point $x=a$ is
$g(x) = f(a) + f'(a)(x-a)$.
$f'(x) = -\frac{d^2}{x^2}$, so $g(x) = \frac{d^2}{a} - \frac{d^2}{a^2}(x-a) = -\frac{d^2}{a^2}x + \frac{2d^2}{a}$, hence $m = -\frac{d^2}{a^2}, c = \frac{2d^2}{a}$ and $m = -\frac{c^2}{4d^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using recursion tree to solve recurrence $T(n) = 3T(n/2)+n$ I am trying to solve the recurrence $T(n) = 3T(n/2)+n$ where $T(1) = 1$ and show its time complexity.
$n$ can be assumed to be a power of $2$.
So basically, I drew out the tree and found that:
the number of levels in the tree will be $h = \log_2 n+1$.
The number of leaves in the tree will then be $3^{h-1} = 3^{\log_2 (n) } = n^{\log_2 3}$
Now we can write the time formula..
$$T(n) = cn + c(3n/2) + c(9n/4) +\dots + cn(3^{h-2}/2(h-2)) + \Theta(n^{\log_2 3})$$
[stuck here, how should I formulate the summation? The summation diverges at infinity..]
$$T(n) = \Theta (n^{\log_2 3})$$
I can tell that the summation of the number of leaves is $n$ times some constant but I am not sure how to show that step.
| Our time formula is:
$$
T(n) = n + \frac{3}{2}n + \frac{3^2}{2^2}n + \cdots + \frac{3^{\log_2 n}}{2^{\log_2 n}}n
$$
[As an aside, note that the last term is just $n^{\log_2 3}$.] Now this is a finite geometric series with initial term $a = n$, common ratio $r = \frac{3}{2}$, and $N = \log_2 n + 1$ terms. By applying the usual formula, we obtain:
\begin{align*}
T(n)
&= \frac{n((\frac{3}{2})^{\log_2 n + 1} - 1)}{\frac{3}{2} - 1} \\
&= \frac{n(3^{\log_2 n + 1} - 2^{\log_2 n + 1})}{3(2^{\log_2 n}) - 2^{\log_2 n + 1}} \\
&= \frac{n(3(3^{\log_2 n}) - 2(2^{\log_2 n}))}{3(2^{\log_2 n}) - 2(2^{\log_2 n})} \\
&= \frac{n(3(n^{\log_2 3}) - 2n)}{3n - 2n} \\
&= \frac{n(3n^{\log_2 3} - 2n)}{n} \\
&= 3n^{\log_2 3} - 2n \\
\end{align*}
which belongs to $\Theta(n^{\log_2 3})$.
| {
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"url": "https://math.stackexchange.com/questions/983439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $(\frac{\cos x}{1-\sin x})^2$ $(\dfrac{\cos x}{1-\sin x})^2$
$f\;'(x)= 2(\dfrac{\cos x}{1-\sin x}) \times (\dfrac{-\sin x+\sin^2x-\cos^2x}{(1-\sin x)^2})$
Does $\sin^2x-\cos^2=1$? or $-1$? Then it could factor with the bottom and the answer would be $\dfrac{2\cos x}{(1-\sin x)^2}$. Is this right?
| $\sin^2(x)-\cos^2(x)$ does not equal $-1$ or $1$.
$$\sin^2(x)+\cos^2(x)=1,$$
So $\sin^2(x)-\cos^2(x)$ is equal to:
$$\sin^2(x)-(1-\sin^2(x))= 2\sin^2(x)-1,$$ or alternatively,
$$1-\cos^2(x)-\cos^2(x)= 1-2\cos^2(x).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the co-ordinates of the point of intersection I have the function $$y=2x^2-3x$$ How do I find the co-ordinates of the point of intersection of the lines tangent to the curve at $y=-1$?
One point where $y = -1$ is when $x = \dfrac 12$.
I took the derivative of the given $y$: $\dfrac{dy}{dx} = 4x- 3$, then found slope at $x = \dfrac 12$. So slope of the tangent line is $-1$.
So the equation of tangent is $y+1=-1\left(x-\dfrac 12\right)$.
How to solve it further?
| Solve for $x$ when $y = -1$: $$2x^2 - 3x = -1 \iff 2x^2 - 3x + 1 = 0 \iff (2x-1)(x-1) = 0 \implies x = \frac 12 \text{ or } x = 1$$
So, there are two points corresponding to $y = -1$: $\left(\frac 12, -1\right), (1, - 1)$.
Perhaps you need to find the point of intersection of the line tangent to the first point, with the line tangent to the second point.
Indeed, you've found the equation of the line tangent to the point $\left(\frac 12, -1\right)$, since $-1 = \dfrac{dy}{dx}\left(\frac 12\right)$: $$y+1= -\left(x -\frac 12\right)\iff y = -x -\frac 12\tag{1}$$
Now we need to find the equation of the line tangent to $(1, -1)$. $$\frac{dy}{dx}(1) = 4(1)-3 = 1$$
So the equation of the second line is given by $$y +1 = x-1 \iff y = x-2\tag{2}$$
Now, put $y$ from $(1)$ equal to $y$ from $(2)$, and solve for $x$, the x coordinate of the point of intersection. Then solve for $y$ using the equation for either line.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$ and $\int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx$ Evaluation of Some Integrals::
$\displaystyle (a)\;\;\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx\;\;\;\;\;\;(b)\;\; \int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx\;\;\;\;\;\;$
$\displaystyle (c)\;\;\int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx\;\;\;\;\;\; (d)\;\; $
$\bf{My\; Try::}$I have tried for third one:: Let $\displaystyle I = \int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx$
Now Put $\displaystyle x^4=t\;,$ Then $\displaystyle 4x^3dx=dt$ and changing Limit, we Get
$\displaystyle I = \int_{0}^{1}\frac{1+t^{2010}}{(1+t)^{2012}}dt$
Now How can I solve after that , Help me
and i did not understand how can i solve $(a)$ and $(b)\;,$ Help me
Thanks
| Hint
For the first integral, use $\sin(3x)=3\sin(x)-4\sin^3(x)$ . So $$\frac{1+\sin 3x}{1+2\sin x}=\frac{-4 \sin(x)^3+3 \sin(x)+1}{2 \sin(x)+1}$$ Remark that equation $-4z^3+3z+1=0$ has an obvious solution $z=1$ and two other roots equal to $z=-\frac{1}{2}$ ... which is also the root of $2z+1=0$ !
Then,
$$\frac{1+\sin 3x}{1+2\sin x}=\frac{-4 \sin(x)^3+3 \sin(x)+1}{2 \sin(x)+1}=-2 \sin^2(x)+\sin(x)+1=\cos(2x)+\sin(x)$$
I am sure that you can take from here.
| {
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"url": "https://math.stackexchange.com/questions/985354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove the determinant? We have to prove the following result without expanding
$\left|\begin{array}{lll}
a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
\end{array}
\right|=(ab+bc+ca)\left|\begin{array}{lll}
a^2 & a &1 \\
b^2 & b &1\\ c^2 & c &1
\end{array}
\right|$
Progress :
$\left|\begin{array}{lll}
a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
\end{array}
\right|$
$
=\frac{1}{abc}\left|\begin{array}{lll}
a^3 & a^2 &abc \\
b^3 & b^2 &abc\\ c^3 & c^2 &abc
\end{array}
\right|$ by $abc\times C_3\rightarrow C_3'$
$=\left|\begin{array}{lll}
a^2 & a &bc \\
b^2 & b & ca\\ c^2 & c &ab
\end{array}
\right|$ by $\frac{1}{a}\times R_1\rightarrow R_1',\frac{1}{b}\times R_2\rightarrow R_2', \frac{1}{c}\times R_3\rightarrow R_3'$
$=\left|\begin{array}{ccc}
a^2+b^2+c^2 & a+b+c &ab+bc+ca\\
b^2 & b & ca\\ c^2 & c &ab
\end{array}
\right|$ by $ R_1+R_2+R_3\rightarrow R_1' $
Any one tell me how can I show the required result?
| $ \begin{align}\left|\begin{array}{lll}
a^2 & a &bc+ab+ac+a^2 \\
b^2 & b & ca+ba+b^2+bc\\ c^2 & c &ab+ca+cb+c^2
\end{array}
\right|\end{align} $ by $(a+b+c) \times C_2 + C_3\rightarrow C_3' $
Then $
=\left|\begin{array}{lll}
a^2 & a &bc+ab+ac \\
b^2 & b & ca+ba+bc\\ c^2 & c &ab+ca+cb
\end{array}
\right| $ by $C_3 - C_1\rightarrow C_3'$
Hence
$\left|\begin{array}{lll}
a^3 & a^2 &1 \\
b^3 & b^2 &1\\ c^3 & c^2 &1
\end{array}
\right|=(ab+bc+ca)\left|\begin{array}{lll}
a^2 & a &1 \\
b^2 & b &1\\ c^2 & c &1
\end{array}
\right|$
| {
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"url": "https://math.stackexchange.com/questions/986008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.
5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$
6 $k = 0$, $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$
7 $k = 1$, $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$
8 $k = 2$, $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$
So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6}, \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?
Thanks!
| Use polar coordinates.
$z^{3}=i=e^{i(\frac{\pi}{2}+2k\pi)}$, $k\in \mathbb{Z}$
And from here it is much simpler
EDIT: what I mean by much simpler...
The OP kind of used polar form, but not really. You should stock with polar form until the very end. The OP introduces $\theta$, $cosinus$ and $sinus$ functions etc. You mix everything up and forget that $(\frac{\pi}{2}+2\pi)$, divided by $3$, gives $\frac{5\pi}{6}$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the distribution function for Y for the following density function I am to find the distribution function for Y given the following density function
$$f(y)=\begin{cases}
y,\quad 0<y<1\\
2-y, \quad 1 \leq y < 2\\
0, \quad\text{elsewhere}\\
\end{cases}$$
So since $f(y)$ is given by $\frac{dF(y)}{dy}$, then $F(y)$ should be given by the integral of $f(y)$. So by integrating I get the following:
$$F(y) = \begin{cases}
0, \quad y\leq0 \\
\frac12 y^2,\quad 0<y<1\\
2y-\frac12 y^2, \quad 1 \leq y < 2\\
1, \quad\text{elsewhere}\\
\end{cases}$$
But to be a distribution function $F(y)$ must be continuous for $-\infty < y < \infty$, but my function is not continuous at $y=2$.
Can someone help me get a continuous distribution function? Thanks.
| Let us compute $F(x) = \int_{-\infty}^x f(y)dy$ the distribution function:
If $x<0$ then clearly $F(x)=0$.
If $0<x<1$ then
$$F(x) = \int_{-\infty}^x f(y)dy = \int_{0}^x f(y)dy = \int_0^x ydy = \frac{x^2}{2}.$$
If $1<x<2$ then
$$F(x) = \int_{-\infty}^{x} f(y) dy = \int_0^1 f(y)dy+ \int_1^x (2-y)dy = F(1)+ \bigg(2y - \frac{y^2}{2}\bigg|_{y=1}^{y=x} = \frac{1}{2} + 2x-\frac{x^2}{2} - 2+\frac{1}{2} = 2x-\frac{x^2}{2}-1$$
Remember to subtract the last value here, because you want to compute the accumulated area.
Finally, if $x>2$ then $F(x)=1$. Altogether
$$
F(x)=
\begin{cases} 0, \mbox{ if } x<0 \\
\frac{x^2}{2} \mbox{ if } 0<x<1\\
2x-\frac{x^2}{2}-1 \mbox{ if } 1<x<2\\
1 \mbox{ if } x>2
\end{cases}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiation of $f(x)=2x(2+3x^2)^3$ The question:
Differentiate $f(x)=2x(2+3x^2)^3$.
How do I approach this problem? Do I only have to use the product rule...? I have the answer but I don't know how to get there.
Here is my attempt to this problem:
$$\frac{df}{dx}=(2)(2+3x^2)^3 + (2x)(6x)(3)(2+3x^2)^2 = \\ =(2+3x^2)^2 [(2)(2+3x^2) + (2x)(6x)(3)] = \\ = (2+3x^2)^2 [(4+6x^2) + 36x^2] = \\ = (2+3x^2)^2 (42x^2+4)$$
| You get the right answer:
$$(2+3x^2)^2 (42x^2+4)=(2+3x^2)^2 \cdot 2 \cdot (21x^2+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that : $a^n+b^n+c^n=x^n+y^n+z^n$; $\forall n\in \mathbb{N}$ $a;b;c;x;y;z \in \mathbb{R}$ such that : \begin{matrix}
a+b+c=x+y+z & \\
a^2+b^2+c^2=x^2+y^2+z^2 & \\
a^3+b^3+c^3=x^3+y^3+z^3 &
\end{matrix}
Prove that : $a^n+b^n+c^n=x^n+y^n+z^n$; $\forall n\in \mathbb{N}$
P/s : I don't have any ideas about this problem..!!
Thanks :)
| Hint:
Let $p_1=a+b+c, \; q_1 = ab+bc + ca,\; r_1 = abc, \; p_2 = x+y+z, \; q_2 = xy+yz+zx, \; r_2 = xyz$. Then it is easy to show that:
$$p_1 = p_2, \quad q_1 = q_2, \quad r_1 = r_2$$
Further, $a^n+b^n + c^n$ and $x^n + y^n + z^n$ are both symmetric functions and hence can be expressed in terms of $p_1, q_1, r_1$ and $p_2, q_2, r_2$ respectively in exactly the same manner... so they are equal.
| {
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Prove that $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$ I want to prove this limit by using $(\epsilon,\delta)$ definition
$$\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$$
Here is what I have done
$$|\frac{1}{x} - \frac{1}{2}|<\epsilon \Leftrightarrow -\epsilon < \frac{1}{x} - \frac{1}{2} < \epsilon$$
$$\Leftrightarrow -\epsilon + \frac{1}{2} < \frac{1}{x} < \epsilon + \frac{1}{2}$$
$$\Leftrightarrow \frac{1}{-\epsilon + \frac{1}{2}} > x > \frac{1}{\epsilon + \frac{1}{2}}$$
I have problem at this point by the definition, $\epsilon$ can be any positive arbitrary number so $\frac{1}{-\epsilon + \frac{1}{2}}$ can be negative number so $x$ can be both smaller than $\frac{1}{-\epsilon + \frac{1}{2}}$ and bigger than $\frac{1}{\epsilon + \frac{1}{2}}$ then the domain of $x$ will be $[-\infty,\frac{1}{-\epsilon + \frac{1}{2}}]\cup [\frac{1}{\epsilon + \frac{1}{2}},\infty]$ But also by the definition, when $f(x)$ approach the limit, it is very small and we can chose $\epsilon$ to be sufficient small to make $\frac{1}{-\epsilon + \frac{1}{2}}>0$. So which way should I do. By the former, the thing will be a mess because the domain of x must be something like $[a,b]$ not $[a,b]\cup[c,d]$
So what should I do to prove this limit.
| Let $\epsilon >0$. Then you have to bound $|\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}$. So, by choosing $ \delta=min\{1,2\epsilon\} $ you have $-1<x-2<1\Longrightarrow 1<x<3\Longrightarrow \frac{1}{x}<1$ and $|x-2|<2\epsilon$ therefore
$ |\frac{1}{x}-\frac{1}{2}|=\frac{1}{2}\frac{|x-2|}{|x|}<\epsilon$
| {
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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.