Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
Here is what I got so far (using Cauchy's integral formula) :
$$\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n} ~=~ \sum\limits_{n=0}^\infty \left(\frac{1}{2\pi i}\int_\mathbb{T} \frac{(1+z)^{2n}}{z^{n+1}}\; dz\right) \cdot \frac{1}{x^n} \\=~ \frac{1}{2\pi i}\sum\limits_{n=0}^\infty \frac{1}{x^n}\int_{\mathbb{T}} \frac{1}{z}\cdot \left( \frac{(1+z)^2}{z}\right)^n \; dz \\=~ \frac{1}{2\pi i} \int_{\mathbb{T}} \frac{1}{z} \sum\limits_{n=0}^\infty \left( \frac{(1+z)^2}{xz}\right)^n \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left( \frac{1}{1- \left(\frac{(1+z)^2}{xz}\right)}\right) \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{z} \cdot \left(\frac{xz}{xz- (1+z)^2}\right) \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{xz- (1+z)^2} \; dz \\=~ \frac{1}{2\pi i}\int_\mathbb{T} \frac{x}{-1 + (x-2)z- z^2} \; dz$$
According to Wolframalpha I should get $\sqrt{\frac{x}{x-4}}$. Where did I commit a sin ?
I'm expecting to use the Residue theorem or Cauchy's integral formula at some point but I feel something is fishy.
| You just need to keep going: factor the quadratic $z^2-(x-2)z+1=(z-r_1)(z-r_2)$, apply partial fractions to get
$${1\over(z-r_1)(z-r_s)}={1\over r_1-r_2}\left({1\over z-r_1}-{1\over z-r_2} \right)$$
and then use the fact that one of the roots is inside the unit circle and the other is outside when applying the Residue Theorem. Finally, note that
$$\sqrt{x\over x-4}={x\over\sqrt{x^2-4x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Partial Derivative with product & chain rule I cannot for the life of me work out the answer to this partial derivative.
$$\frac{\delta}{\delta x}\left(\frac{x^2}{(x+y)^2(x+z)^2}\right) $$
My first thought was:
*
*Split into two equations:
$$\frac{\delta}{\delta x}\left(\frac{x}{(x+y)^2}\cdot \frac{x}{(x+z)^2}\right)$$
*Apply chain rule to each side which gives me:
$$\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}$$ and $$ \frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3} $$
*I then try to use the product rule, so:
$$ \left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+z)^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+y)^2}\right\rbrack $$
However I don't end up anywhere near an answer, let alone the right answer.
Apparently the answer is:
$$ -\frac{2(x^3-xyz)}{(x+y)^3(x+z)^3} $$
Is there an easier way to do this?
| $$\left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+{\bf{z}})^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+{\bf{y}})^2}\right\rbrack = \\
\frac{x(y-x)}{(x+y)^3(x+z)^2}+\frac{x(z-x)}{(x+z)^3(x+y)^2} = \\
\frac{x(y-x)(x+z)+x(z-x)(x+y)}{(x+y)^3(x+z)^3} = \\
\frac{x^2y -x^3 +xyz-x^2z + x^2z -x^3 + xyz -x^2y}{(x+y)^3(x+z)^3} = \\
\frac{ -2x^3 +2xyz}{(x+y)^3(x+z)^3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
| This answer uses similar ideas to those in Jack D'Aurizio's solution:
First we let $\displaystyle c_n=\frac{n^2-n+2}{2}$ and let $\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}$ for each n,
and we consider the related series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ obtained by grouping terms with the same sign,
so $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n=1-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\cdots$
1) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}\le\sum_{k=0}^{n-1}\frac{1}{c_n}=\frac{n}{c_n}=\frac{2n}{n^2-n+2}\to0,\;\;$ so $a_n\to 0$.
2) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}>\int_{c_n}^{c_n+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2n}{n^2-n+2}\right)$ for each n, and
$\;\;\;\;\displaystyle a_{n+1}=\sum_{k=0}^{n}\frac{1}{c_{n+1}+k}<\int_{c_{n+1}-1}^{c_{n+1}+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2}{n}\right)$ for each n.
$\hspace{.2 in}$Since $\displaystyle\frac{2n}{n^2-n+2}\ge\frac{2}{n}$ for $n\ge2$, $\;\;a_n > a_{n+1}$ for $n\ge 2$.
Therefore $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ converges by the Alternating Series Test.
Let $(T_n)$ be the sequence of partial sums of this series, and let $T_n\to T$.
If $(S_k)$ is the sequence of partial sums for the original series,
then for any value of $k$ let $n$ be the largest integer satisfying $n^2-n+2\le 2k$.
Then $\big|S_k-T\big|\le\max\{|T_n-T|, |T_{n+1}-T|\}$, so it follows that $S_k\to T$.
| {
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"url": "https://math.stackexchange.com/questions/1786521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
} |
How to diagonalize a matrix with several eigenvalues of zero? I'm looking to diagonalize a matrix A seen below. (Find a $P$ and a $D$ such that $AP = PD$).
$$
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
\end{bmatrix}
$$
It doesn't seem practical to use determinants in this matrix to find the eigenvalues. I do realize that with row reduction I can limit to the matrix:
$$
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
And further realize that the following vector is in the column space of A.
$$
\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1 \\
1 \\
\end{bmatrix}
$$
By solving $Av_1=λv_1$ I get an eigenvalue of 21. But this is the only eigenvalue that I get and I believe it has a multiplicity of 1. Then is there a $λ=0$ with multiplicity 5? Further, how is it possible to find the matrix $P$ (specifically the eigenvectors that form a basis for the eigenspace associated with $λ=0$)?
| We need to find the eigenvectors $\mathbf{x}$ such that:
$A\mathbf{x} =\mathbf{0}.$ Because the system is homogeneous, we can use the row echelon form of $A$, thus we need to solve the sytem:
$$ \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \cdot \begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\x_ 6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix},$$
Thus, we have the equation:
$$x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 + 6x_6 = 0,$$ which implies that we have $5$ free variables.
So, every eigenvector corresponding to the zero eigenvalue can be written in the form:
$$x_2\cdot\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\0 \\0 \end{bmatrix} +x_3 \cdot \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \\0 \\0 \end{bmatrix} +x_4\cdot \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \\0 \\0 \end{bmatrix}+x_5\cdot\begin{bmatrix} -5 \\ 0 \\ 0 \\ 0 \\1 \\0 \end{bmatrix} +x_6\cdot \begin{bmatrix} -6 \\ 0 \\ 0 \\ 0 \\0 \\1 \end{bmatrix}.$$
The column vectors form an eigenbasis for the eigenspace that corresponds to the zero eigenvalue. So, we have the matrix $$P = \begin{bmatrix}
1 & -2 & -3 & -4 & -5 & -6\\
1 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}$$
and $D = \operatorname{diag}(21\quad 0\quad 0\quad 0\quad 0\quad 0 ).$
We can notice that matrix $A$ is a rank one matrix, so there is only one non-zero eigenvalue, which can be found by adding the diagonal elements of $A$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equiv 1 \pmod{25}$ if $\gcd(x,25) = 1$ and thus $x^4 = x \cdot x^3 \equiv x^{21} \cdot x^3 \equiv x^{24} \equiv (x^3)^8 \pmod{25}$. $\square$
I am not sure if the above proves the result, but it does show that if we have an $8$th power of an integer modulo $25$ and it is relatively prime to $25$, then it is equal to a fourth power of an integer modulo $25$ and vice-versa. Does that therefore mean they are equivalent?
| We only need to prove that
$$(5a+1)^8 + (5a+2)^8 + (5a+3)^8 +(5a+4)^8 \equiv_{25} (5a+1)^4+(5a+2)^4+(5a+3)^4+ (5a+4)^4$$
In this equations the elements of the polynomials that only matter are the elements of the form $\dbinom{8}{i}(5a)^ik^{8-i}$ with $i=0,1$, because for $i>1$, the number is $\equiv_{25} 0$. This is the same for the right side, only matters $\dbinom{4}{i}(5a)^ik^{4-i}$ with $i=0,1$. So let's check the rest of the equation and reduce it using this idea:
$$(5a+1)^8 + (5a+2)^8 + (5a+3)^8 +(5a+4)^8 -(5a+1)^4-(5a+2)^4-(5a+3)^4- (5a+4)^4 \equiv_{25} 0$$
$$\Leftrightarrow (8*5a*1+1) + (8*5a*2+2^8) + (8*5a*3+3^8) +(8*5a*4+4^8) -(4*5a*1+1^4)-(4*5a*2+2^4)-(4*5a*3+3^4)- (4*5a*4+4^4) \equiv 0$$
And using the mod 25, we get:
$$\Leftrightarrow (40a+1) + (20a+6) + (5a+11) +(10a+11) -(20a+1)-(10a+16)-(15a+6)- (5a+6) \equiv_{25} a((40+20+5+10)-(20+10+15+5))+(1+6+11+11-1-16-6-6) \equiv 29-29 \equiv 0$$
So it's done. Now, in the case of the number 99, you can divide the space in sets of 5 consecutive numbers as here:
$$\{1,2,3,4,5\}, \{6,7,8,9,10\},\dots ,\{96, 97, 98, 99 \}$$
and as the numbers of the form $5k$ are $\equiv_{25}0$ when they're at 4th power, the other ones fulfil the identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
$\epsilon-\delta$ proof of $\lim_{ x\to 5} \frac{1}{x-4}=1$ $$\lim_{x\rightarrow 5} \frac{1}{x-4}=1$$
So far I have started the proof
For every $\epsilon > 0$ there exists a $\delta>0$ such that $\left|\frac{1}{x-4}-1\right|< \epsilon$, whenever $0 < |x-5| < \delta.$
I am having trouble figuring out what I need to factor out and getting to that step.
| We see $$\left \lvert \frac{1}{x-4} - 1\right \rvert = \left \lvert \frac{1}{x-4} - \frac{x-4}{x-4}\right \rvert = \left \lvert \frac{5-x}{x-4} \right \rvert.$$ For fixed $\epsilon > 0$, take $\delta = \min\{1/2, \epsilon / 2 \}$. Then for $\lvert x - 5 \rvert < \delta$ we have $\lvert x - 4 \rvert > 1/2$ so $\frac{1}{\lvert x -4 \rvert} < 2$. Thus $$\left \lvert \frac{1}{x-4} - 1\right \rvert =\frac{1}{\lvert x-4 \rvert} \lvert x - 5 \rvert < 2\delta < \epsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the fractional equation: $\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$
Prove that $$\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$$ where $\sqrt 3 = 1.b_1b_2b_3 \dots _{(2)}$.
(Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)
I am not sure how to prove this result because of the fractional part. Should I expand $\sqrt{3}$ and then multiply by $2^{n-1}$?
| From
$$
\sqrt 3 = 1.b_1b_2b_3 \cdots _{\,(2)} \tag1
$$ one gets, using $b_n \in \{0,1\}$,
$$
\begin{align}
\left\{2^{n-1}\sqrt 3\right\} &=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}+\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\}
\\\\&=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}\right)+2^{n-1}\left(\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\}
\\\\&=\left\{\underbrace{2^{n-1}+2^{n-2}b_1+2^{n-3}b_2+\cdots+b_{n-1}}_{\text{integer part}} +\overbrace{\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots}^{\text{fractional part}}\right\}
\\\\&=\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots
\\\\&=0.b_nb_{n+1}\cdots_{\,(2)}
\end{align}
$$ as announced.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
| Here is another way, a few lines shorter,$$\sqrt\frac{1}{(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt{3}})^2}=\frac{1}{\sqrt{4}}=\frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$ $n \ge 1$
$F_n$; Fibonacci numbers
$\frac{1+\sqrt5}{2}=\phi$
Prove
$$1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$$
I can't go any further can anyone help please?
Expand the equation
Let
$A=\phi^{4n+1}F_{2n}F_{2n+1}$
$B=\phi^{2n+1}F_{2n}$
$C=\frac{F_{2n+1}}{\phi{F_{2n}}}$
$$\left(1+\frac{1}{\sqrt{A}+B}\right)^2=C$$
$$1+\frac{2}{\sqrt{A}+B}+\frac{1}{(\sqrt{A}+B )^2}=C$$
This method doesn't seem to be working.
Can A, B and C be simplify more further?
I try and rationalise the denominator
$$\frac{2(\sqrt{A}-B)(A-B)+(\sqrt{A}-B)^2}{(A-B^2)^2}=C$$
Still nothing is been simplified.
| notice that $A=B^2C$, and since $B>0$, $\sqrt A = B\sqrt C$.
Since $\sqrt A + B > 0$, $1 + 1/(\sqrt A + B) = \sqrt C \iff \sqrt A + B + 1 = \sqrt C(\sqrt A + B)$.
But $\sqrt A+B+1 = B\sqrt C + B + 1$ and $\sqrt C(\sqrt A+B) = BC + B\sqrt C$, so $ \sqrt A + B + 1 = \sqrt C(\sqrt A + B) \iff B+1 = BC \iff \phi^{2n+1}F_{2n}+1 = \phi^{2n}F_{2n+1}$
Now if this is true you should be able to prove it by induction or by using a formula for $F_n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\sqrt{-4S^2-2P-\frac q S}$$
$$p=\frac{8ac-3b^2}{8a^2}$$
$$q= \frac{b^3-4abc+8a^2d}{8a^3}$$
$$S=0.5\sqrt{-\frac{2p} 3 + \frac{Q+\Delta_0} Q}{3a}$$
$$Q=\left(\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3} \cdot 0.5\right)^{1/3}$$
$$\Delta_0=c^2-3bd+12ae$$
$$\Delta_1=2c^3-9bcd+27eb^2+27ad^2-72ace$$
Those formula does not work. I ended up with complex roots.
Please help me to use those formulas to solve quartic equation. Any help will be very much appreciated.
| We have
$$x^4-3x^3-11x^2+3x+10$$
Using rational root test:
$$(x-1)(x^3-2x^2-13x-10)$$
And rational root test again:
$$(x-1)(x+1)(x^2-3x-10)$$
Factoring quadratic:
$$(x-1)(x+1)(x-5)(x+2)$$
Thus, roots are $x=\pm1,\; x=-2,\; x=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Least value of $b$ for which inequality is true.
find the least natural number $b$ for which $x+bx^{-2}>2,\forall \space x\in (0,\infty)$
What I got confused with
In my book they have done this: rearranged the function as $$f(x)=x^3-2x^2+b$$ an $f(x)>0$
They have told to find least value of $b$, $f(x)$ should be minimum. This is what i didn't understand. Why should $f(x)$ be minimum $\implies$ $b$ is minimum.
| $$
g(x) = x + b/x^2 > 2 \quad (x > 0) \quad (*) \iff \\
x^3 + b > 2 x^2 \quad (x > 0) \iff \\
f(x) = x^3 - 2 x^2 + b > 0 \quad (x > 0) \quad (**)
$$
A $b$ that will keep $(*)$ true, will also keep $(**)$ true and vice versa.
Thus we are are talking the same set
$$
B = \{ b \mid g(x) > 2, x > 0 \} = \{ b \mid f(x) > 0, x > 0 \}
$$
here.
To find such a $b$ we are looking for a minimum of $f$ on $(0, \infty)$:
$$
0 = f'(x) = 3x^2 - 4 x
$$
which vanishes there for $x = 4/3$, the other candidate $x=0$ was outside our interval of interest. Because $f''(x) = 6x - 4$ and $f''(4/3) = 24/3 - 4 = 4 > 0$ it is a minimum.
The value of $f$ there is
$$
f(4/3)
= (4/3)^3 - 2(4/3)^2 + b
= 64/27 - 32/9 + b
= 64/27 - 96/27 + b
= -32/27 + b
$$
If we adjust it to touch the $x$-axis we get $b = 32/27 = 1.19$ so we choose $b = 2$.
Because the minimum of $f$ is positve, the other values of $f$ on $(0,\infty)$ (which are larger) will be too. So $f$ at the minimum location was the most difficult customer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
| Write $x=y^{3/2}$ and substitute in the first equation to get $y^{3y/2}=y^{y^{3/2}}$; if $y \neq 0,1$, then the exponents must be equal so that $3y/2=y^{3/2}$, which gives $y=\dfrac{9}{4}$ and $x=\dfrac{27}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
$8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ in terms of $a$ and $b$. if $8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ using $a$ and $b$.
My Attempt:
if $8^a=3$ and $3^b=5$ then we can say that $8^{ab}=5$ and then we have $2^{3ab}=10^c$ but i cant solve this equation.
| We have $$2^{3a}=3\implies 2^{3ab}=5\implies 2\times 2^{3ab}=10\implies 2^{3ab+1}=10$$
Thus $$10^c=2^{c(3ab+1)}$$ But we know that $2^{3ab}=5$ so we are asked to solve $$c(3ab+1)=3ab\implies c=\frac {3ab}{3ab+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Comparing this with,the given equation,we get,
$(p=-a),q=(\frac{3a^2-d^2}{3}),r=-a(a^2-d^2)$.
But,after this I have no idea how to eliminate $a$ and $d$ from the equation and find a relation between $p,r$ and $r$.
Thanks for any help!!
| $(p=-a),q=(\frac{3a^2-d^2}{3}),r=-a(a^2-d^2)\\
r=p(a^2-d^2)\\
\frac rp=(a^2-d^2)\\
3q=(3a^2-d^2)\\
3q=2p^2 + \frac rp\\
0=2p^3 - 3pq + r$
For the answer given, it is not correct.
Check it against $(x+1)(x+2)(x+3) = x^3 + 6x^2 + 11x + 6$
$-2p^2 - 3pq + r = -24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Factorization of $x^5+x$ I need to make the decomposition in $\mathbb{R}$ of:
$$x^5+x$$
Here my steps:
$$x(x^4+1)$$
$$x\big[ (x^2)^2+1 \big]$$
$$x\big[(x^2+1)^2-2x^2\big]$$
how should I proceed?
| Another (less ingenious but more systematic?) way is to first work over complex numbers:
$$\begin{align}x^4+1&=(x^2)^2-i^2=(x^2-i)(x^2+i)\\
&=(x+\frac{1+i}{\sqrt2})(x-\frac{1+i}{\sqrt2})\cdot(x+\frac{-1+i}{\sqrt2})(x-\frac{-1+i}{\sqrt2})\\
&=(x-\frac{1+i}{\sqrt2})(x+\frac{-1+i}{\sqrt2})\cdot(x-\frac{-1+i}{\sqrt2})(x+\frac{1+i}{\sqrt2})\\
&=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)\end{align}$$
So this obtains the decomposition by permuting the two hidden factors over complex numbers.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{1}+\frac{1}{2}\right) = \left(\sum_{n=1}^{\infty}\frac{1}{n}\right)-\frac{3}{2}$$
but I can't get a relationship between $a_n$ and $b_n$ of these series.
If I go to the root of the comparsion criterion, I know that:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots+\frac{1}{n}> \frac{2}{4}+\frac{4}{8}+\cdots+g_n$$ for some $g_n$ that I'm lazy to calculate.
Therefore, by the properties of limits, since the righthand sum $p_n$ diverges, the lefthand sum $s_n$ diverges too, because $s_n>p_n$. But I don't know if my teacher would accept that, I think she would only accept that I use the argument for the $a_n$ of the sum, not for the partial sum.
I also thought about rewriting:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = \sum_{n=3}^{\infty} \frac{1}{n}$$
but I don't see how it helps because the indexes are different.
All I need is an argument that will work with the $a_n$ of the $\sum_{n=1}^{\infty}a_n$, in relation with $b_n$ from $\sum_{n=1}^{\infty} b_n$
| For all $n$, $n + 2 \leq n + 2n$, so
$$
\sum_{n=1}^\infty \frac{1}{3n} \leq \sum_{n=1}^\infty\frac{1}{n + 2},
$$
and $\sum_{n=1}^\infty \frac{1}{3n}$ diverges since it is a constant multiple of a divergent series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such that $\gamma\alpha\gamma^{-1}=\beta$
I know that $\alpha=(1,2,5)(4,6)(3)$ and $\beta=(2,3,4)(1,5)(6)$ also if $\sigma_1=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 5 & 4 & 6 \end{smallmatrix} \bigr)$ then $\sigma_1(1,2,5)\sigma_1^{-1}=(2,3,4)$; if $\sigma_2=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix} \bigr)$ then $\sigma_2(4,6)\sigma_2^{-1}=(1,5)$ and if $\sigma_3=(3,6)$ then $\sigma_3(3)\sigma_3^{-1}=(6)$.
But I don't know how to build $\gamma$ such that $\gamma\alpha\gamma^{-1}=\beta$. Can you help me, please?
| $\;\gamma^{-1}\;$ will be the permutation mapping each cycle of $\;\alpha\;$ to a corresponding cycle of $\;\beta\;$ of the same type, thus for example:
$$\gamma^{-1}:=\begin{cases}1\to2\\2\to3\\5\to4\\.......\\4\to1\\6\to5\\.......\\3\to6\end{cases}\implies\gamma=(123654)$$
and we can now check that
$$\gamma^{-1}\alpha\gamma=(123654)(125)(46)(3)(145632)=(15)(234)(5)=\beta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
| $$\sum\limits_{k = 1}^n {\left( {4k - 3} \right)} = 4\sum\limits_{k = 1}^n k - 3 \cdot \sum\limits_{k = 1}^n 1 = 4\frac{{n\left( {n + 1} \right)}}{2} - 3 \cdot n = 2n^2 - n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 5
} |
Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negative.
2 - conclude
Since $f(1)=1$ and the derivative is negative we have showed that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$
Question: is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
| $f(x)=x-\sqrt{x^2-1}= \frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\frac{x^2-x^2+1}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}$
From this, it should be easy to deduct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Showing that $\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$ Showing that
(1)
$$\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$$
see here
(2)
$$(\arcsin(x))^2=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
Look very much similar to (1). How can I make the use of (2) to solve (1)?
Let try and differentiate (1)
$$\frac{d}{dx}\arcsin(x))^2=\frac{d}{dx}\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
$$\arcsin(x)\cdot\frac{1}{\sqrt{1-x^2}}=2\sum_{n=1}^{\infty}\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
Getting close but not near yet
I am stuck, any help please?
| Hint. By using the Euler beta function, one has
$$
\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}=\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx,\quad n \geq 0. \tag1
$$ Then, one may obtain
$$\begin{align}
&\sum_0^\infty\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}
\\\\&=\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx\\\\
&=\int_0^1\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}(2x(1-x))^ndx \:dx\\\\
&=\int_0^1\left(\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}\right)dx\\\\
&=\underbrace{\int_0^1\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}\,dx}_{\large \color{blue}{8-6\pi}}+\underbrace{4\int_0^1-\frac{\log(1-2 x+2 x^2)}{(1-x) x}\,dx}_{\large \color{red}{\pi^2}}
\\\\&=\color{blue}{8-6\pi}+\color{red}{\pi^2}
\\\\&=(\pi-2)(\pi-4)
\end{align}
$$
as announced.
Some details.
Observe that
$$
\sum_{n=0}^\infty(2n-3)y^n=\frac{5 y-3}{(1-y)^2},\quad |y|<1, \tag{A}
$$
$$
\sum_{n=0}^\infty\frac1{n+1}y^n=-\frac{\log(1-y)}y,\quad |y|<1, \tag{B}
$$ One has
$$
\frac{4(n-1)(2n+1)}{n+1}=4(2 n-3)+\frac8{1+n} \tag{C}
$$ then, using $(\text{A})$, $(\text{B})$ and $(\text{C})$ with $y:=2x(1-x)$ gives
$$
\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}(2x(1-x))^n=\frac{-40x^2+40x-12}{(1-2 x+2 x^2)^2}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}
$$ One may classically integrate the fraction by parts. One may then rewrite the $\log$-term as
$$
-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}=\frac{16}{1-u^2}\log \left(1-\frac{1-u^2}2\right) \tag{D}
$$ with $u:=2x-1$, expanding in power series of $(1-u^2)$, integrating termwise to recognize a classic binomial series giving $\pi^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$ I would like to evaluate the following integral:
$$
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}
$$
I tried various methods but without success.
| Hint. One may write
$$
\begin{align}
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} &=\int_{-1}^0 \frac{dx}{(e^x+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)}
\\\\&=\int_0^1 \frac{dx}{(e^{-x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)}
\\\\&=\int_0^1 \frac{e^x\:dx}{(e^{x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)}
\\\\&=\int_0^1 \frac{(e^x+1)\:dx}{(e^{x}+1)(x^2+1)}
\\\\&=\int_0^1 \frac{dx}{(x^2+1)}
\end{align}
$$ then it is easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$ $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}{x^2}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}
{\sqrt{u}}$$
Convert to partial fractions
$$\int_{0}^{\infty}\frac{1}{\sqrt{u}}-\frac{2}{u+1}+\frac{1}{(u+1)^2}du$$
$$=\left.2\sqrt{u}\right|_{0}^{\infty}-\left.2\ln(1+x)\right|_{0}^{\infty} -\left.\frac{1}{1+u}\right|_{0}^{\infty}$$
Where did I went wrong during my calculation?
| By subbing $x=\sqrt{z}$, then $\frac{1}{z^2+1}=u$
$$\int_{0}^{+\infty}\frac{2x}{(x^4+1)^2}\,dx = \int_{0}^{+\infty}\frac{dz}{(z^2+1)^2}=\frac{1}{2}\int_{0}^{1}u^{1/2}(1-u)^{-1/2}\,du$$
that by Euler's beta function and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ equals:
$$ \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\,\Gamma(2)} = \color{red}{\frac{\pi}{4}}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
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how many strings of length 21 from letters {D,L,O} with repetition that have DOLL appearing only once The word DOLL has length $4$. There will be $18$ possible positions where 'DOLL' can appear, each with $3^{17}$ choices. So there will $18 \cdot 3^{17}$ cases where it can appear at least once. But I also know that there will be many duplicates in this approach. How do I go about removing them.
I can count those
*
*With DOLL appearing at least / only $5$ times as $\binom{6}{5}$ each with $3^{21-4 \cdot 5} = 3$ choices.
*With DOLL appearing at least $4$ times as
*
*$21$ positions to place DOLL first time
*$17$ positions to place DOLL second time
*$13$ positions to place DOLL third time
*$9$ positions to place DOLL fourth time
*all with $3^{5}$ choices, so this gives $21 \cdot 17 \cdot 13 \cdot 9 \cdot 3^5$ possibilities
*if I subtract all the choices that have $5$ occurrences of DOLL ($\binom{6}{5} \cdot 3$)from this, I will get only those that have $4$ occurrences of DOLL
But from here on how do I proceed for calculating $3$, $2$ and only $1$ occurrences. Do I apply the inclusion exclusion principle at each step or only when calculating for at most $1$ occurrence of DOLL.
| (This is simply a supplement to the two previous answers.)
As shown in N.F. Taussig's answer, the number of words with DOLL appearing at least once is given by
$\hspace{.5 in}\displaystyle\binom{18}{1}3^{17} - \binom{15}{2}3^{13} + \binom{12}{3}3^9 - \binom{9}{4}3^5 + \binom{6}{5}3=2,161,418,679$
Similarly, the number of words in which DOLL appears at least twice is given by
$\hspace{.5 in}\displaystyle\binom{15}{2}3^{13} -\binom{2}{1} \binom{12}{3}3^9 +\binom{3}{1} \binom{9}{4}3^5 -\binom{4}{1} \binom{6}{5}3=158,835,177$
Subtracting, we get $2,002,583, 502$ words in which DOLL appears exactly once,
as shown in joriki's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's tedious.
Any ideas for a much shorter solution?
| Suppose that none of $a,b,c$ is divisible by $7$.
Write $a^3+b^3 \equiv -c^3$, square both sides, and use Fermat:
$ a^6 + 2a^3b^3 + b^6 \equiv c^6 $
$ 1+2a^3b^3 + 1\equiv 1 $
$ (ab)^3 \equiv 3$
But $3$ is not a cube mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int x^8\sqrt{x^2+1}\ dx$
Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$
$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$
So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$
Now How can I solve it after that, Help Required, Thanks
| HINT:
Integrating by parts,
for $m+1\ne0,$
$$I_m=\int x^m\sqrt{x^2+1}\ dx=\sqrt{x^2+1}\int x^m\ dx-\int\left(\dfrac{d\sqrt{x^2+1}}{dx}\int x^m\ dx\right)dx$$
$$(m+1)I_m=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^{m+2}}{\sqrt{x^2+1}}dx$$
$$=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^m(1+x^2)-x^m}{\sqrt{x^2+1}}dx$$
$$(m+2)I_m=x^{m+1}\sqrt{x^2+1}+I_{m-2}-\int\dfrac{x^{m-2}}{\sqrt{x^2+1}}dx$$
The last part can be reduced further as $x^{m-2}=x^{m-4}(1+x^2)-x^{m-4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 2
} |
Exponential Equations with Fractions I have had some issues with the following two equations:
$$ \frac{3^{n-2}}{9^{1-n}}=9$$
$$\frac{5^{3n-3}}{25^{n-3}}=125$$
If anyone could work them out step by step that would be awesome. I keep arriving at the wrong answer and believe it could be related to the addition of exponents towards the end of the question.
note:$n$ is different in each equation.
| $\frac{3^{n-2}}{9^{1-n}}=9$
$\implies\frac{3^{n-2}}{3^{2(1-n)}}=9$
$\implies\frac{3^{n-2}}{3^{2-2n}}=9$
$\implies3^{(n-2)-(2-2n)}=3^{2} [a^{(m-n)}=\frac{a^m}{b^m}]$
$\implies3^{3n-4}=3^2$
$\implies{3n-4}=2$
$\implies n=2$
$\frac{5^{3n-3}}{25^{n-3}}$
$=\frac{5^{6-2}}{25^{2-3}} [put(n=2)]$
$=\frac{5^4}{5^{-1}}$
$=5^{4-1}$
$=5^3=125$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$ I did the following:
$$f(1-0) + 2f(0) = 3\cdot 0$$
$$f(1) + 2f(0) = 0$$
This reminds me of the equation of the straight line in the plane, then:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{f(1)}\\
{f(0)}
\end{pmatrix} \right> =0$$
$(1,2)$ is a normal vector to $(f(1),f(0))$, then It's possible that: $(f(1)=-2,f(0)=1)$ because:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{-2}\\
{1}
\end{pmatrix}\right> =-2+2=0$$
With generality, It is possible that for all $\alpha$:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{\alpha\cdot (-2)}\\
{\alpha \cdot 1}
\end{pmatrix}\right> =
\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\alpha \begin{pmatrix}
{ -2}\\
{ 1}
\end{pmatrix}\right> =1\cdot -2\alpha+2\alpha \cdot 1=-2\alpha + 2 \alpha =0$$
I tagged with "functional equations" because it seems to be related. I don't know if my solution is correct, what I'm saying with this is that there are infinite solutions but I don't know if assuming arbitrary values for $f(1),f(0)$ can be made because of $3x$.
EDIT: I know that It is possible to solve:
$$f(1) + 2f(0) = 0$$
$$f(0) + 2f(1) = 3$$
And get the solution. But why do I have to use the latter instead of the former?
| Just in case if you want to solve for the actual function next time:Substituting $x$ with $1-x$
$$f(1-(1-x))+2f(1-x)=3(1-x)$$$$f(x)+2f(1-x)=3-3x\tag{1}$$Now multiply the original equation by 2:$$2f(1-x)+4f(x)=6x\tag{2}$$
Subtract: $(2)-(1)$ We get
$$3f(x)=9x-3$$Therefore$$f(x)=3x-1$$$$f(0)=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Five white balls, five black balls There are 5 white balls and 5 black balls in a box. A visually-challenged man takes 5 balls from the box... What's the probability that all the balls which this individual picked are white if we know for sure that there 3 white balls among those 5 balls he took?
Using the formula $P(A | B) = P(A \cap B)/P(B)$, I get a probability of $1/100$.
| Your answer is correct if you are doing "three white balls among five chosen balls", but is incorrect if the problem is looking for "at least three white balls".Stricktly according to your formula, $P(A | B) = \frac{P(A \cap B)}{P(B)}$
$B=$ at least three white balls among five chosen balls, $P(B)=\frac{\begin{pmatrix}5\\3\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}+\begin{pmatrix}5\\4\end{pmatrix}\begin{pmatrix}5\\1\end{pmatrix}+\begin{pmatrix}5\\5\end{pmatrix}\begin{pmatrix}5\\0\end{pmatrix}}{\begin{pmatrix}10\\5\end{pmatrix}}=\frac{126}{252}$
$P(A \cap B)=\frac{\begin{pmatrix}5\\5\end{pmatrix}\begin{pmatrix}5\\0\end{pmatrix}}{\begin{pmatrix}10\\5\end{pmatrix}}=\frac{1}{252}$
So $P(A | B) =\frac{\frac{1}{252}}{\frac{126}{252}}=\frac{1}{126}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $\sum\limits_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}} \geq \frac{(\sum\limits_{i=1}^{n}a_i)^2}{\sum\limits_{i=1}^{n}b_i}$ So I have the following problem, which I'm having trouble solving:
Let $a_1$ , $a_2$ , ... , $a_n$ be real numbers. Let $b_1$ , $b_2$ , ... , $b_n$ be positive real numbers. Prove
$$ \frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}} \geq \frac{(a_{1}+a_{2}+\cdot \cdot \cdot+a_{n})^2}{b_{1}+b_{2}+\cdot \cdot \cdot+b_{n}} $$
I was thinking that I somehow could use the Cauchy–Schwarz inequality, but with no success.
Any help would be very appreciated
| We will use induction.
First, suppose that $n=2$, the given inequality is
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\geq \frac{(a_1+a_2)^2}{b_1+b_2}$$
This can be proven directly by subtracting, then factoring the equation into a complete square.
Suppose the given inequality holds for $n=k$
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_1}+\cdots+\frac{a_k^2}{b_k}\geq \frac{(a_1+a_2+\cdots+a_k)^2}{b_1+b_2+\cdots+b_k}$$
Add $\frac{a_{k+1}^2}{b_{k+1}}$ on both sides,
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_1}+\cdots+\frac{a_k^2}{b_k}+\frac{a_{k+1}^2}{b_{k+1}}\geq \frac{(a_1+a_2+\cdots+a_k)^2}{b_1+b_2+\cdots+b_k}+\frac{a_{k+1}^2}{b_{k+1}}$$
$$\geq \frac{(a_1+a_2+\cdots+a_k+a_{k+1})^2}{b_1+b_2+\cdots+b_k+b_{k+1}}$$
Last inequality holds from the case $n=2$. Thus the given inequality is true for $n=k+1$.
We are done!
Equality holds when
$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int_c\frac{1}{\sin\frac{1}{z}}\text{d}z$ over $C= \{z\big\vert|z|=\frac{1}{5}\}$ Evaluate
$$\int\limits_{|z|=\frac{1}{5}} \frac{1}{\sin\frac{1}{z}}\text{d}z$$
My attempt: I know that this function has non isolated singularity at $0$, and simple poles at $\frac{1}{n \pi}$. Which ever methods I know are applicable only when all the singularities are isolated. So I tried with Residue theorem but then I realised that the theorem is not applicable here. So now I am stuck. Can any one help please?
I can observe that there are only two poles outside the curve c, one at $1/\pi$ and the other at infinity. Can this be used?
Thanks in advance.
| $$f(z) = \frac1{\sin{\left ( \frac1{\zeta+\frac1{n \pi}} \right )}} $$
where $\zeta = z-\frac1{n \pi}$. Then
$$\begin{align}f(z) &= \frac1{\sin{\left ( \frac{n \pi}{1+ n \pi \zeta} \right )}} \\ &= \frac1{\sin{\left [ n \pi \left (1-n \pi \zeta + n^2 \pi^2 \zeta^2-\cdots \right )\right ]}}\\ &= \frac1{\sin{n \pi} \cos{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)} - \cos{n \pi} \sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{\sin{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)}}\\ &= \frac{(-1)^{n+1}}{(n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)-\frac16 (n^2 \pi^2 \zeta - n^3 \pi^3 \zeta^2 +\cdots)^3+\cdots}\end{align}$$
The residue is the coefficient of $\frac1{\zeta}$, which is $(-1)^{n+1}/(n^2 \pi^2)$. Within $|z|=1/5$, we only consider $|n| \ge 2$. Thus away from $z=0$ the contribution to the integral is
$$i 2 \pi (2) \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^2 \pi^2} = -i \frac{4}{\pi} \left (1-\frac{\pi^2}{12} \right)$$
There is also an essential singularity at $z=0$. $f$ may be written as
$$f(z) = \frac1{\frac1z - \frac16 \frac1{z^3} + \frac1{120} \frac1{z^5} - \cdots} $$
and thus has no residue here. Thus, the integral is
$$-i \left (\frac{4}{\pi} - \frac{\pi}{3} \right ) $$
(Thanks to @achillehui for pointing out a crucial mistake.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
| Alternatively, you can simply consider the following cases:
*
*$n\equiv 1\pmod{16} \implies n^4\equiv 1^4\equiv 1\equiv1\pmod{16}$
*$n\equiv 3\pmod{16} \implies n^4\equiv 3^4\equiv 81\equiv1\pmod{16}$
*$n\equiv 5\pmod{16} \implies n^4\equiv 5^4\equiv 625\equiv1\pmod{16}$
*$n\equiv 7\pmod{16} \implies n^4\equiv 7^4\equiv2401\equiv1\pmod{16}$
*$n\equiv 9\pmod{16} \implies n^4\equiv 9^4\equiv(16-7)^4\equiv(-7)^4\equiv7^4\equiv\ldots$
*$n\equiv11\pmod{16} \implies n^4\equiv11^4\equiv(16-5)^4\equiv(-5)^4\equiv5^4\equiv\ldots$
*$n\equiv13\pmod{16} \implies n^4\equiv13^4\equiv(16-3)^4\equiv(-3)^4\equiv3^4\equiv\ldots$
*$n\equiv15\pmod{16} \implies n^4\equiv15^4\equiv(16-1)^4\equiv(-1)^4\equiv1^4\equiv\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 13,
"answer_id": 8
} |
How can I solve this hard system of equations? Solve the system below
\begin{align}
&\sqrt {3x} \left( 1+\frac {1}{x+y} \right) =2\\
&\sqrt {7y} \left( 1-\frac{1}{x+y} \right) =4\sqrt{2}
\end{align}
Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I hope you can help me in approaching this problem.
| Squaring 1st equation you have $3x(x+y+1)^2=4(x+y)^2$, squaring second you have $7y(x+y-1)^2=32(x+y)^2$. Put $s=x+y,y=s-x$ and we get $3s^2x+6sx+3x-4s^2=0,7s^3-7s^2x+14sx-46s^2+7s=0$. The first of these gives $x=\frac{4s^2}{3s^2+6s+3}$.
Substituting in the second we get $$s(21s^4-124s^3-178s^2-124s+21)=0$$ or $$s(3s^2-22s+3)(7s^2+10s+7)=0$$ It is clear from the original equations that $s\ne0$ and $7s^2+10s+7=0$ has no real solutions, so we have $s=\frac{1}{3}(11\pm4\sqrt7)$.
Using the equation for $x$ we get $x=\frac{1}{21}(11+4\sqrt7),y=\frac{2}{7}(11+4\sqrt7)$ or $x=\frac{1}{21}(11-4\sqrt7),y=\frac{22}{7}(11-\frac{8}{\sqrt7})$. But checking with the original equations the second solution fails (because $s-1$ is negative).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| If $7+4\sqrt{3}$ is a perfect square this means $7=a^2+3b^2$ is the sum of two squares in $\Bbb{Z}[\sqrt{3}]$ and $4\sqrt{3}=2ab\sqrt{3}$. So the possibilities are
$$a=\pm 1,b=\pm 2\\a=\pm 2,b=\pm 1$$
Only the latter gives $a^2+3b^2=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 2
} |
Evaluate $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$ $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
My Attempt: I tried to use the regular way.
$A=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
$A^2=1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}$
But then I saw that nothing happened twice and I stopped. Any hints?
It's better to don't use limits but if you used no problem. If there is an answer that doesn't use limits will accept.
update1: The question that look likes this wants to find the limit but I want a much easier way to solve it. But if there isn't any answer easier answer the limition one will accept.
update2: You solve the problem when you know the answer is 3 think that you don't know that the answer is 3 then solve.
| Setting the Scene
Evaluate $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }}$
Let $$f_n(0)=\sqrt{1+n}$$also that $$f_n(k)=\sqrt{1+(n-k)f_n(k-1)}.$$
Then clearly $0<f_n(0)<n+1$ when $n>0$. Now if we assume that $f_n(k)<n+1-k$ and by induction that we see that
$$
f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1)
$$
for all k. The expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$ and thus converges. Hence
$\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }}=3$
EDIT
Showing that $3$ is the limit can be achieved by writing
\begin{align}
3 &= \sqrt{9} \\
&= \sqrt{1 + 8}\\
&= \sqrt{1 + 2 \cdot 4} \\
&= \sqrt{1 + 2\sqrt{16}}\\
&= \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}\\
& = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} \\
&= \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}\\
&= \ldots
\end{align}
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a
Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a<b<c$
Rearranging the equation gives: $$ab+bc+ac=abcd$$
What can we do with this?
| First observe that$$d \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}<2$$Thus $d=1$. And notice that if $a \ge 3$ then$$d \le \frac{1}{3}+\frac{1}{4}+\frac{1}{5}<1$$ So we have $a \le 2$. Its easy to see that $a \neq 1$, so $a=2$. Equation reduces to$$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$$Again if $b \ge 4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{1}{4}+\frac{1}{5}<\frac{1}{2}$$Therefore $b \le 3$. $b=2$ doesn't work. So $b=3$ and finally $c=6$.
Thus the only solution is $(a, b, c, d)=(2, 3, 6, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the Fourier Sine Transform of $\frac{1}{x(a^2+x^2)}$? (Please check my solution) My textbook says that the answer is $\frac{\pi}{2a^2}(1-e^{-as})$ while my answer is $\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$. Here is my solution:
\begin{align}
\hat{f_s} & = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{1}{t(a^2+t^2)} \sin{st}dt \\
& = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{t}{a^2}[\frac{1}{t^2} -\frac{1}{(a^2+t^2)}] \sin{st}dt \\
& = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \int_0^\infty [\frac{\sin{st}}{t} -\frac{t}{(a^2+t^2)} \sin{st}] dt \\
& = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \left( \frac{\pi}{2} - \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \right)
\end{align}
Let this integral be $y$. Therefore,
\begin{align}
y & = \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \\
& = \int_0^\infty \frac{1}{t}[1 -\frac{a^2}{(a^2+t^2)}] \sin{st}dt \\
& = \int_0^\infty [\frac{\sin{st}}{t} -\frac{a^2}{t(a^2+t^2)} \sin{st}] dt \\
& = \left( \frac{\pi}{2} - \int_0^\infty \frac{a^2}{t(a^2+t^2)} \sin{st} dt \right)
\end{align}
Differentiating $y$ w.r.t. $s$ twice, we get
\begin{align}
y'' & = a^2\int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \\
& = a^2 y \\
\end{align}
This is a second order differential equation, whose solution is
$$y= C_1e^{-as}+C_2e^{as}$$
Put $s=0$ in $y$ and $y'$, we get $C_1 = \frac{\pi}{4} = - C_2$. Thus, we get $$y = \frac{\pi}{4}(e^{-as}-e^{as})$$ or
$$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{2}{\pi}}\left(\frac{\pi}{2} - \frac{\pi}{4}(e^{-as}-e^{as})\right)$$ which is nothing but $$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$$
| If you differentiate,
\begin{align}
\left(-\frac{d^2}{ds^2}+a^2\right)\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)dt
&= \int_{0}^{\infty}\frac{\sin(st)}{t}dt \\
&=\int_{0}^{\infty}\frac{\sin(st)}{(st)}d(st) \\
&=\int_{0}^{\infty}\frac{\sin(u)}{u}du=\frac{\pi}{2}.
\end{align}
Therefore, $f(s)=\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)dt$ is a solution of
$$
(-D^2+a^2)f=\frac{\pi}{2} \\
D(-D^2+a^2)f=0.
$$
The general solution of this ODE is
$$
f(s)=Ae^{as}+Be^{-as}+C
$$
Plugging into the ODE gives $a^2C=\frac{\pi}{2}$.
And $f(0)=0$ gives
$$ A+B +C = 0 \\
A+B = -\frac{\pi}{2a^2}
$$
Also, $t=au$ gives
$$
f'(0) = \int_{0}^{\infty}\frac{1}{t^2+a^2}dt=\frac{1}{a}\int_{0}^{\infty}\frac{du}{1+u^2}=\frac{\pi}{2a} \\
\implies A-B= \frac{\pi}{2a^2}.
$$
Adding and subtracting the equations in $A$,$B$ gives
$$
A = 0, \;\; B=-\frac{\pi}{2a^2}
$$
Therefore
$$
f(s) = C+Be^{-as} = \frac{\pi}{2a^2}(1-e^{-as}).
$$
Typically, the sin transform is $\sqrt{\frac{2}{\pi}}f(s)$, which would give
$$
\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\frac{1}{t(a^2+t^2)}\sin(st)dt = \sqrt{\frac{\pi}{2}}\frac{1}{a^2}(1-e^{-as})
$$
Interchanging Differentiation and Integration: You can differentiate with respect to $s$ for $s > 0$ because the improper Riemann integral becomes an alternating series where the error term is bounded by the first neglected term in the series. So you get uniform convergence of the alternating series, its first derivative, and it's second derivative in $s$, provided $s$ remains bounded away from $0$ by any small amount. To be more precise,
\begin{align}
\mathcal{S}\{f\}(s) & =\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)dt \\
& =
\int_{0}^{\infty}\frac{1}{(u/s)((u/s)^2+a^2)}\sin(u)\frac{du}{s} \\
& = s^2\int_{0}^{\infty}\frac{\sin(u)}{u(u^2+s^2a^2)}du \\
& = s^2\sum_{n=1}^{\infty}(-1)^{n-1}\int_{(n-1)\pi}^{n\pi}\frac{|\sin(u)|}{u(u^2+s^2a^2)}du \\
& = s^2\sum_{n=1}^{\infty}(-1)^{n-1}f_n(s)
\end{align}
This is an alternating series where $f_n(s) \ge 0$ for $n=1,2,3,\cdots$, and
$$
\frac{2}{(n\pi)(n^2\pi^2+s^2a^2)}\le f_n(s)
\le\frac{2}{((n-1)\pi)((n-1)^2\pi^2+s^2a^2)} \\
\implies 0 < f_{n+1}(s) \le f_{n}(s),\;\;\; n=1,2,3,\cdots.
$$
This gives you an alternating series for the improper integral, and, for any fixed $\delta > 0$, the series converges uniformly for $s \ge \delta > 0$ because the alternating series satisfies
$$
\left|\sum_{n=1}^{\infty}(-1)^{n-1}f_n(s)-\sum_{n=1}^{N}(-1)^{n-1}f_n(s)\right| \le f_{N+1}(s) \rightarrow 0 \mbox{ uniformly for $s \ge \delta > 0$ as $N\rightarrow\infty$}.
$$
The functions $f_n(s)$ are infinitely differentiable in $s$, and
$$
f_n'(s) = -2sa^2\int_{(n-1)\pi}^{n\pi}\frac{|\sin(u)|}{u(u^2+s^2a^2)^2}du
$$
So the series of derivatives of $f_n'(s)$ is also an alternating series which converges uniformly for $0 < \delta \le s < \infty$. You can do the same thing for second derivatives as well. By classical theorems of differentiation for such series,
$$
\frac{d}{ds} \sum_{n}(-1)^nf_n(s)=\sum_{n}(-1)^nf_n'(s), \\
\frac{d^2}{ds^2}\sum_{n}(-1)^nf_n(s)=\sum_{n}(-1)^nf_n''(s).
$$
The end result is that you may interchange differentiation and the improper Riemann integrals:
$$
\frac{d}{ds}\mathcal{S}\{f\}(s)=\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\frac{d}{ds}\sin(st)ds \\
\frac{d^2}{ds^2}\mathcal{S}\{f\}(s)=\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\frac{d^2}{ds^2}\sin(st)dt.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $\frac{d^ny}{dx^n}$ if $y = \frac{7}{1-x}$ I am wondering what is $\frac{d^n}{dx^n}$ if $y = \frac{7}{1-x}$
Basically, I understand that this asks for a formula to calculate any derivative of f(x) (correct me if I'm wrong). Is that related to Talylor's theory? How do I end up with such a formula?
Thanks!
| Let $f(x)= \sum_{n\geq 0} a_n(x-c)^n$
Then $f^{(n)}(c)=n!a_n $,
\begin{align*}
\frac{7}{1-x} &= \frac{7}{1-c-(x-c)} \\
&= \frac{7}{1-c}\frac{1}{1-\frac{x-c}{1-c}}\\
&= \frac{7}{1-c}\sum_{n\geq 0 } \frac{1}{(1-c)^n} (x-c)^n\\
&=\sum_{n\geq 0 } \frac{7}{(1-c)^{n+1}} (x-c)^n
\end{align*}
Therefore $$f(c)= \frac{7(n!)}{(1-c)^{n+1}}\, \forall c \in (\mathbb{R}-\{1\})$$
| {
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"url": "https://math.stackexchange.com/questions/1832611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute antiderivative of $x \frac{3}{4} \sqrt{ -x}$ I need to compute the area under $x \times\frac{3}{4} \times \sqrt{|x|}$ from x=-1 to x=1. Therefore I need to compute the antiderivative of $x \times \frac{3}{4} \times \sqrt{x} $ for $0 \le x \le 1$, which is straight forward, $ \frac{3}{10} \times x^{2.5}$. And for $ -1 \le x \le 0 $ I need to compute the antiderivative of $x \frac{3}{4} \sqrt{ -x}$. I am stuck on how to rewrite $x $ and $ \sqrt{ -x}$ as a single term.
| $x\frac34\sqrt{|x|}$ is an odd function - so the integral from -1 to 1 is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $(a+b-c)/(a+b+c)$
Given $$2a^2 +17b^2 + 8c^2 -6ab -20bc =0 \\ abc \neq 0.$$
Find the value of $(a+b-c)/(a+b+c)$.
I tried factorizing the equation, but still can't find a solution
| $$0=2\cdot LHS=(2a-3b)^2+(5b-4c)^2$$
and so $2a=3b$ and $5b=4c$. This gives $$\frac{a+b-c}{a+b+c}=\frac{3b/2+b+5b/4}{3b/2+b-5b/4}=1/3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A common term for $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \text{if } n\ \text{ is odd. } \end{cases}$ When I was answering a question here, I found a sequence as a recursive one as given below.
$a_1=1$, and for $n>1$,
$$a_n=\begin{cases}
2a_{n-1} & \text{if } n\ \text{ is even, }\\
2a_{n-1}+1 & \text{if } n\ \text{ is odd. }
\end{cases}$$
I need to find a common term for this sequence. For example, for the sequence $a_1=2$ and $a_n=2a_{n-1}$, for $n>1$, the common term is $a_n=2^n$.
I appreciate any answer or hint in advance.
| It can often be helpful to just write out the first couple of terms:
$$\begin{align}
a_1 &= 1 \\
a_2 &= 2 \\
a_3 &= 5 = 2^2 + 1\\
a_4 &= 10 = 2^3 + 2 \\
a_5 &= 21 = 2^4 + 5 = 2^4 + 2^2 + 1\\
a_6 &= 42 = 2^5 + 10 = 2^5 + 2^3 + 2\\
a_7 &= 85 = 2^6 + 21 = 2^6 + 2^4 + 5 = 2^6 + 2^4 + 2^2 + 1\\
a_8 &= 170 = 2^7 + 42 = 2^7 + 2^5 + 10 = 2^7 + 42 = 2^7 + 2^5 + 2^3 + 2
\end{align}
$$
Do you see a pattern? You basically just need to find an expression changing between $1$ and $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Seeking closed form for infinite sum $\sum \limits_{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ $\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ is approximately $.5229461921333351$ but I've been assured that there is a closed form for this infinite series that involves the TriGamma and Zeta functions but it is not unique. I found one candidate result which is:
$$ \frac{ \pi \sqrt{3}}{72} \Big[3 \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{5}{6} \right) \Big]- \frac{4}{3} \zeta(3) $$
My friend (tormentor) has challenged me to find other such expressions. I know how to transform the look of such expressions if the Gamma function is involved by using functional equations like $\Gamma(x) = \Gamma(1+x)/x$ so I suspect he knows a similar result for the PolyGamma functions. I just can not find the information. By trial and error, I did find that $3 \psi_{1}(1)=\psi_{1}(1/2)$
| The series can be expressed in the form
\begin{align}
\sum_{n=1}^{\infty} \frac{(n!)^2}{n^3 \, (2n)!} &= \frac{1}{2} \, {}_{4}F_{3}\left(1,1,1,1; 2,2,\frac{3}{2}; \frac{1}{4} \right) \\
&= \frac{1}{2} \, \left[ - \frac{8}{3} \, \zeta(3) + \frac{\pi}{4 \sqrt{3}} \, \psi^{(1)}\left(\frac{1}{3}\right) - \frac{\pi}{12 \sqrt{3}} \, \psi^{(1)}\left(\frac{5}{6}\right) \right]
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Derivative of $\arcsin \frac{x-1}{x+1}$ I was looking at a question that asks for the derivative of $\arcsin (\frac {x+1}{x-1}) $.
The solution starts by saying $y = \frac{x+1}{x-1}$, so $1-y^2= \frac{4x}{(x+1)^2}$ and $\frac{1}{\sqrt{1-y^2}}$ and thus $\frac{x+1}{2\sqrt{x}} $
However, I don't see why $1-y^2= \frac{4x}{(x+1)^2} $
| $1-y^2=1-\left(\dfrac{x-1}{x+1}\right)^2=1-\dfrac{x^2-2x+1}{x^2+2x+1}=\dfrac{(x^2+2x+1)-(x^2-2x+1)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}=\dfrac{4x}{(x+1)^2}$
| {
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Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method plz explain here , Thanks
| Using $c=-a-b$,
$\displaystyle \hspace{.2 in}\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right) \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\displaystyle=\left(\frac{a}{a+2b}-\frac{b}{2a+b}-\frac{a+b}{a-b}\right)\left(\frac{a+2b}{a}-\frac{2a+b}{b}-\frac{a-b}{a+b}\right)$
$\displaystyle=\left(\frac{2(a^2-b^2)}{(a+2b)(2a+b)}-\frac{a+b}{a-b}\right)\left(\frac{2(b^2-a^2)}{ab}-\frac{a-b}{a+b}\right)$
$\displaystyle=(a+b)\left(\frac{2(a-b)}{(a+2b)(2a+b)}-\frac{1}{a-b}\right)(a-b)\left(-\frac{2(a+b)}{ab}-\frac{1}{a+b}\right)$
$\displaystyle=(a+b)\left(\frac{-9ab}{(a+2b)(2a+b)(a-b)}\right)(a-b)\left(-\frac{(a+2b)(2a+b)}{ab(a+b)}\right)=\color{red}{9}$
| {
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"timestamp": "2023-03-29T00:00:00",
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'Strange' trigonometric roots of $x^5-4x^4+2x^3+5x^2-2x-1$ - could someone explain? This quintic equation has $5$ real roots:
$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$
The roots are, from left to right:
$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$
$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$
$$x_3=\frac{\cos \frac{7}{22} \pi}{\cos \frac{5}{22} \pi}$$
$$x_4=\frac{\cos \frac{1}{22} \pi}{\cos \frac{7}{22} \pi}$$
$$x_5=\frac{\cos \frac{5}{22} \pi}{\cos \frac{9}{22} \pi}$$
I found these roots numerically, using ISC. The equation was found on Wikipedia in a different form (for $x-4/5$), no solutions were provided.
I can derive this equation for each root individually. But I don't see how do all five roots 'fit' together.
Why ${1,5,7,9,19}$? Why there is no $3/22, 13/22$ or $17/22$ in any of the arguments?
In any case, I would be grateful for the explanation for how these roots all fit together.
| Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get
$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\
x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\
x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\
x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\
x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi}
$$
Note that the identity
$$\cos(3x)= \cos(x) [2 \cos(2x)-1]$$
gives you a nicer form for the roots. With this form, after the proper substitution you should be able to reduce your polynomial to the minimal polynomial of $\cos(\frac{\pi}{11})$, which will explain where the roots are coming.
| {
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How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative? No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative.
This is what I have tried:
$(a * b) * c = a * (b * c) [Associativity];
$(a^2 - ab + b^2) * c = a * (b^2 - bc + c^2)$;
I have tried using $(a - b)^2$: $(a - b)^2 * c = a * (b - c)^2$;
Like other associativity problems, I could prove that but reaching an equality after expanding both sides, but that doesn't seems to be the case here. What am I doing wrong? Is this even an associative operation?
| Just do it:
$(a*b)*c = (a^2 - ab + b^2)*c = (a^2 - ab + b^2)^2-a^2c +abc-b^2c+c^2=a^4 -2a^3b + 3a^2b^2 -2ab^3+b^4-a^2c+abc-b^2c+c^2$
$a*(b*c) = a^2-a (b^2-bc+c^2) +(b^2 - bc + c^2)^2$
It's pretty clear these need not be equal. One is "dominated" by $a^4$ and $b^4$ while the other is dominated by $c^4$ and $b^4$.
Let $a=100$, $b=c=0$. Then $(a*b)*c = (a*0)^2 -(a*0)0+0^2=(a*0)^2=(100^2-0*100+0)^2=100^4$. Whereas $a*(b*c)=100^2 -100 (0*0)+(0*)^2=100^2-100 (0^2-0*0+0^2)+(0^2-0*0+0^2)^2=100^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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hard integral problems to solve I'm practicing harder integration using techniques of solving with special functions
I have difficulties with these two hard integrals; don't even know how to start,
$$\int_0 ^\infty x^p e^{-\frac{\theta}{x}+Bx}dx$$
where $\theta,B>0$
$$pv\int_0 ^\infty \frac{x^p e^{\cos{\theta x}} \cos(\frac{\pi p}{2} - \sin\theta x)}{1-x^2}dx$$
please help me to start of giving your solutions! thank you so much and have a good day/night
| Let $\theta=b$ and $B = -a \lt 0$. Then the first integral is a generalization of this integral. Using the same substitutions:
$u=a x+b/x$. Then $a x^2-u x+b = 0$, and therefore
$$x = \frac{u}{2 a} \pm \frac{\sqrt{u^2-4 a b}}{2 a}$$
$$dx = \frac1{2 a}\left (1 \pm \frac{u}{\sqrt{u^2-4 a b}}\right ) du $$
Now, it should be understood that as $x$ traverses from $0$ to $\infty$, $u$ traverses from $\infty$ down to a min of $2 \sqrt{a b}$ (corresponding to $x \in [0,\sqrt{b/a}]$), then from $2 \sqrt{a b}$ back to $\infty$ (corresponding to $x \in [\sqrt{b/a},\infty)$). Therefore the integral is
$$\frac1{2 a} \int_{\infty}^{2 \sqrt{a b}} du \, \left (\frac{u}{2 a} - \frac{\sqrt{u^2-4 a b}}{2 a} \right )^p \left (1 - \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} \\+ \frac1{2 a} \int_{2 \sqrt{a b}}^{\infty} du \, \left (\frac{u}{2 a} + \frac{\sqrt{u^2-4 a b}}{2 a} \right )^p \left (1 + \frac{u}{\sqrt{u^2-4 a b}}\right ) e^{-u} $$
which simplifies, subbing $u=2 \sqrt{a b} \cosh{v}$, to
$$\begin{align}\int_0^{\infty} dx \, x^p e^{-\left (a x+\frac{b}{x} \right )} &=2 \left (\sqrt{\frac{b}{a}} \right )^{p+1} \int_0^{\infty} dv \, \cosh{[(p+1) v]} \, e^{-2 \sqrt{a b} \cosh{v}} \\ &=2 \left (\sqrt{\frac{b}{a}} \right )^{p+1} K_{p+1} \left ( 2 \sqrt{a b}\right )\end{align}$$
| {
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Why can't we use the law of cosines to prove Fermat's Last Theorem? In investigating approaches to Fermat's Last Theorem I came across the following and I can't figure out where I am going wrong. Any input would be greatly appreciated.
We want to show that $a^n + b^n = c^n$ cannot hold for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$. Assuming by way of contradiction that we have $a^n + b^n = c^n$ we must have $a$, $b$, and $c$ forming the sides of a triangle since $(a+b)^n > c^n$ so $a+b>c$. Therefore the law of cosines can apply and we can write:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
where $C$ is the angle opposite to side $c$. If we add and subtract $2ab$ on the right-hand side we get
$$c^2 = {(a+b)}^2 -2ab(\cos{C}+1)$$
Now, $a+b$ and $c$ share a common factor since $(a+b) | (a^n+b^n)$ for odd $n$ and $c^n = a^n+b^n$. (Here $x | y$ means as usual, "$x$ divides $y$").Therefore, they share the same factor with $2ab(\cos{C}+1)$.
Now, $\cos{C} + 1$ must be a rational number since $a$, $b$, and $c$ are all integers. So let $\cos{C} +1 = \frac{r}{s}$ where $r$ and $s$ are integers and $(r,s)=1$. (i.e. $\frac{r}{s}$ is a reduced fraction). (Here, $(r,s)$ means as usual the greatest common divisor of $r$ and $s$.)
Now assuming $a$, $b$, and $c$ are relatively prime we must have $(ab) |s$ for otherwise $c$ and $2ab$ would share a common factor. Even moreso we must have $ab=s$ since otherwise $\frac{2abr}{s}$ would not be an integer. (Since $c - a - b$ is even, we don't need $2 | s$). So we can write:
$$\cos{C}+1 = \frac{r}{ab}$$ or equivalently $$\cos{C} = \frac{r - ab}{ab}$$
Now we had from the law of cosines:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
so making the substitution $\cos{C} = \frac{r - ab}{ab}$ we get
$$c^2 = a^2 + b^2 - 2r + 2ab$$
If we subtract $a^2$ to both sides and factor out the $b$ on the right-hand side, we get:
$$c^2 - a^2 = b(b + 2a) - 2r$$
Now, $(c - a) | (c^2 - a^2)$ and also $(c-a) | (c^n - a^n)$. Then we must have $((c-a),b) >1$ since $b^n = c^n - a^n$. From the equation above, we must therefore also have $(b,2r) > 1$. Similarly we can show that we must have $(a,2r) > 1$.
However, both of these conclusions are problematic since $r$ was initially assumed to be relatively prime to $s = ab$. The only other option is that $a$ and $b$ are both even, but this is also problematic since $a$ and $b$ are assumed to be relatively prime.
Thus we cannot have $a^n + b^n = c^n$ for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$.
I'm sure someone has thought of this approach before so where am I going wrong?
| How do you get to the conclusion that $ab\vert s$? I honestly can't see it. The way I see it you have:
$$\dfrac{r}{s}=\dfrac{(a+b)^2-c^2}{2ab}. $$
Now $(a+b)^2-c^2$ is even. You can check this case by case, when $a,b$ are odd then $c$ has to be even and so forth. So at least one of them is even but by your assumption maximal one is even and therefore $(a+b)^2-c^2$ is even. Therefore
$$ \dfrac{r}{s}=\dfrac{\dfrac{(a+b)^2-c^2}{2}}{ab}. $$
But there is no apparent (at least not to me) reason why this shouldn't reduce further. If it does your argument breaks down at this point.
Here is an actual counter example: of course I can't give an example of $a,b,c$ with $a^n+b^n=c^n$ but your argument that $ab\vert s$ only uses that $a,b,c$ are coprime. So let $a=13, b=15$ and $c=22$ than you have that $a,b,c$ are relatively prime and furthermore:
$$\dfrac{r}{s}=\dfrac{(a+b)^2-c^2}{2ab}=\dfrac{10}{13}, $$
therefore $s\neq ab=195$.
| {
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Evaluate $\int_0^1\frac{1}{\lfloor1-\log_{2}(1-x)\rfloor}dx$ Problem:
$$\int_0^1\frac{1}{1-\log_{2}(1-x)}dx$$
Which I simplified down to this:
$$2\log(2)\int_0^1\frac{1}{2^xx} dx$$
Now I am stuck as the answer is supposed to be computed without the use of a calculator, meaning non standard use of Ei(x) would not be applicable, methods such as Feynmann don't seem to work in this case either, please could someone assist me. Thanks
Edit : The denominator is floored , sorry
$$\int_0^1\frac{1}{\lfloor1-\log_{2}(1-x)\rfloor}dx$$
I am not sure what this now means, any help?
| The answer given by @clark is correct for the originally given integral. Things change quite a bit when we bring in the floor function.
In a comment on the question you asked what the floor function means in this instance. It means the same thing it always does: $\lfloor y \rfloor$ is the greatest integer less than or equal to $y$. Therefore $\lfloor 1 - \log_2(1-x) \rfloor$ is the greatest integer less than or equal to $1 - \log_2(1-x)$.
Note that the integral is from $0$ to $1$. Therefore we only care about the integrand on the interval $[0,1]$. Also, $1-\log_2(1-x)$ is a strictly increasing function of $x$ on $(0,1)$, because its derivative is $\frac{1}{1-x}$, which is positive on $(0,1)$. And since we have $1 - \log_2(1-0) = 1$, then the integral (if it exists, which it does) must be positive.
So now we need to better understand $\lfloor 1 - \log_2(1-x) \rfloor$ before we can do the integral. Of course we know that $\lfloor 1 - \log_2(1-0) \rfloor = \lfloor 1 \rfloor = 1$. But for which value of $x$ does this expression equal 2? 3? 4? etc.?
$\lfloor 1-\log_2(1-x) \rfloor = 2$ when $1-\log_2(1-x) \in [2,3)$. This basically gives us two inequalities:
$$
1-\log_2(1-x) \ge 2 \qquad \text{and} \qquad 1-\log_2(1-x) < 3
$$
Solving these inequalities gives us $\dfrac{1}{2} \le x < \dfrac{3}{4}$.
More generally:
$\lfloor 1-\log_2(1-x) \rfloor = n$ when $1-\log_2(1-x) \in [n,n+1)$. This gives us two inequalities:
$$
1-\log_2(1-x) \ge n \qquad \text{and} \qquad 1-\log_2(1-x) < n+1
$$
Solving gives us $1-\dfrac{1}{2^{n-1}} \le x < 1 - \dfrac{1}{2^n}$.
So what we see then is
$$
\lfloor 1 - \log_2(1-x) \rfloor = n \quad \text{if} \quad 1-\dfrac{1}{2^{n-1}} \le x < 1 - \dfrac{1}{2^n}.
$$
Note also that
$$
[0,1] = \bigcup_{n=1}^{+\infty} \left[1-\frac{1}{2^{n-1}}, 1-\frac{1}{2^n}\right].
$$
Therefore we have:
\begin{align*}
\int_0^1 \frac{dx}{\lfloor 1 - \log_2(1-x)\rfloor}
&= \sum_{n=1}^{+\infty}\int_{1-1/2^{n-1}}^{1-1/2^n} \frac{1}{n} \, dx\\[0.3cm]
&= \sum_{n=1}^{+\infty} \left(\frac{1}{n} \cdot x \bigg|_{x=1-1/2^{n-1}}^{x=1-1/2^n}\right)\\[0.3cm]
&= \sum_{n=1}^{+\infty} \frac{1}{n2^n}\\[0.3cm]
&= \ln 2
\end{align*}
| {
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Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| using shorthand
$$ s = \sin \theta; c = \cos \theta; s2 = \sin 2 \theta; c2 = \cos 2 \theta; $$
Square and add
$$ s^2 +1/s^2 + c^2 + 1/c^2 = m^2+n^2$$
or
$$ sc= \frac{1}{\sqrt{m^2n^2-1}}, \, s2= \frac{2}{\sqrt{m^2+n^2-1}} $$
$$c2 = \sqrt{1-s2^2}=\sqrt{ \frac{ m^2+n^2-5}{m^2+n^2-1}}$$
$$s = \sqrt { \frac{ 1-c2}{2}}$$
Plug into first given relation
$$\sqrt { \frac{ 1-c2}{2}} + \sqrt { \frac{2}{ 1-c2}} = m $$
Simplify. There are two signs for every $\pm$, so there are eight eliminants to be considered.
| {
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What is the actual meaning of power which is less than 1? $3^4 = 3 \cdot 3 \cdot 3 \cdot 3$
$3^2 = 3 \cdot 3$
$3^1 = 3$
But my problem is for below 1:
Example:
$3^{0.7} = \ ?$
Can you please explain it to clear my doubt?
| Exponents less than 1 are better understood in terms of fractions rather than decimals:
$3^4=3^{2+2}=3^2•3^2$
Therefore:
$3^1=3^{\frac{1}{2}+\frac{1}{2}}=3^\frac{1}{2}•3^\frac{1}{2}=(3^\frac{1}{2})^2$
What value, when multiplied by itself, is equal to 3? The only possibility is ±$\sqrt{3}$. This is why we understand an exponent of $\frac{1}{2}$ to be equivalent to the square root. Similarly, an exponent of $\frac{1}{3}$ is the cube root. Any value in the numerator is the equivalent of an integer exponent.
To determine the value of $3^{0.7}$, convert the exponent to a fraction: $0.7=\frac{7}{10}$. So $3^{0.7}=3^\frac{7}{10}=(3^7)^{1/10}=\sqrt[10]{3^7}≈2.16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solving $\cos^{-1}x+\cos^{-1}2x=-\pi$ gives the invalid answer $x=0$ To solve the inverse trigonometric equation
$$\cos^{-1}x+\cos^{-1}2x=-\pi,$$
I use the normal cosine addition
\begin{align}
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=\cos(-\pi)\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\\
(2x^2+1)^2&=(1-x^2)(1-4x^2)\\
4x^4+1+4x^2&=1-4x^2-x^2+4x^4\\
9x^2&=0\\
x&=0.
\end{align}
Putting $x=0$ in the equation gives LHS $\ne$ RHS:
$$\cos^{-1}0+\cos^{-1}0=-\pi \\\pi=-\pi.$$
Since the equation has no solution, why does solving it give zero as a solution? Is my method wrong or is there something else which gives up one solution (i.e. $x=0$ on solving algebraically)?
|
\begin{align}
\cos^{-1}x+\cos^{-1}2x=-\pi\tag{*}\\
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\tag1\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\tag2\\
(2x^2+1)^2&=(1-x^2)(1-4x^2)\tag3\\
4x^4+1+4x^2&=1-4x^2-x^2+4x^4\tag4\\
x&=0\tag5.
\end{align}
Note that $$\cos^{-1}x+\cos^{-1}2x=\begin{cases}2\pi-&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[-0.5,0)\\&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[0,0.5]\end{cases},$$ so step $(1)$ is actually invalid. Fortunately—replacing step $(1)$—the following argument/working is valid:
$$\begin{align}
&\cos^{-1}x+\cos^{-1}2x =-\pi\tag{*}\\
&\implies \cos(\cos^{-1}x+\cos^{-1}2x) =\cos(-\pi)\\
&\implies 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}=-1\tag2\\
&\implies (2x^2+1)^2=(1-x^2)(1-4x^2)\tag3\\
&\implies 4x^4+1+4x^2=1-4x^2-x^2+4x^4\tag4\\
&\implies x=0\tag5.
\end{align}$$
Therefore, $$\text{equation } (*)\implies x=0.$$
Now, as pointed out by Crostul, since $\arccos$ is a nonnegative function, the given equation $(*)$ is trivially inconsistent (has no solution). Thus, $x{=}0$ is an extraneous solution (it was created in step $(2)\,).$
Since the equation has no solution, why does solving it give zero as a solution?
When a given equation $(A)$ implies equation $(B),$ $(A)$'s solution set (containing the actual solutions) is a subset of $(B)$'s solution set (containing the candidate solutions).
In particular, none of the candidate solutions being an actual solution is necessarily due to deductive explosion, which is necessarily due to equation $(A)$ being inconsistent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Exact value of $\cos^2(\frac{\pi}{8})+\sin^2(\frac{15\pi}{8})?$ I tried separating it into $\cos^2(\frac{\pi}{8})+\sin^2(\frac{\pi}{8}+\frac{14\pi}{8})?$ and using the angle sum identity but it didn't help.
| Notice that $\frac{15\pi}{8} = 2\pi - \frac{\pi}{8}$. So $\sin \frac{15\pi}{8} = -\sin \frac{\pi}{8}$. Your expression then reduces/evaluates down to:
$$\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
point of maximal curvature on $y=\ln x$ While attempting the problem below, I got the cross product of the function's derivative and second derivative to equal 0. If this happens, what does this mean for the curvature formula?
My work:
$$y = f(x)$$
$$\rm{curvature}(x) = {f''(x)\over(1+(f'(x)^2)^{3/2}}$$
$$f'(x) = {1\over x}\text{ and }f''(x) = -{1\over2x^2}$$
$$\rm{curvature}(x) = (1+1/x^2)^{3/2}/(-2x^2)$$
setting the curvature equal to 0 gives a maximum of 0...is this correct?
Determine the point on the plane curve f(x) = ln x where the curvature is maximum. You may need to review max-min methods from Calculus I to do this problem. Be sure to check that the curvature is max at the critical point.
| For $x>0$,
\begin{align*}
y &= \ln x \\
y' &= \frac{1}{x} \\
y'' &= -\frac{1}{x^2} \\
\kappa &= \frac{|y''|}{(1+y'^2)^{3/2}} \\
&= \frac{\frac{1}{x^2}}{\left( 1+\frac{1}{x^2} \right)^{3/2}} \\
&= \frac{x}{(1+x^2)^{3/2}} \\
\frac{d\kappa}{dx} &=
\frac{1}{(1+x^2)^{3/2}}-
\frac{x\times 2x \times \frac{3}{2}}{(1+x^2)^{5/2}} \\
&= \frac{1+x^2-3x^2}{(1+x^2)^{5/2}} \\
&= \frac{1-2x^2}{(1+x^2)^{5/2}} \\
\end{align*}
Now $\kappa'=0$ when $\displaystyle x=\frac{1}{\sqrt{2}}$ and $\kappa' \lessgtr 0$ when $\displaystyle x \gtrless \frac{1}{\sqrt{2}}$,
hence $\displaystyle x=\frac{1}{\sqrt{2}}$ is the position with maximal curvature.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$
Is the proof valid?
| Use Titus lemma:
$$\frac{x^2}{a}+\frac{y^2}b\ge \frac{(x+y)^2}{a+b}$$
Then $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\frac{\sqrt{x}^2}{\sqrt{y}}+\frac{\sqrt{y}^2}{\sqrt{x}}\ge \frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})}=\sqrt{x}+\sqrt{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are:
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$
Is it right?
| For another perspective:
Let $r_1, r_2, r_3, r_4$ be the zeroes of
$$f(x) = x^4 + ax^3 + bx^2 - 8x + 4.$$
Since
$$x^4 + ax^3 + bx^2 - 8x + 4 = (x^2 + cx + d)^2,$$
$r_1, r_2, r_3, r_4$ are also the zeroes of
$$g(x) = (x^2 + cx + d)^2.$$
Using the quadratic formula, we get (without loss of generality) that
$$r_1 = r_2 = \dfrac{-c - \sqrt{c^2 - 4d}}{2}$$
and
$$r_3 = r_4 = \dfrac{-c + \sqrt{c^2 - 4d}}{2}.$$
Now, we have the equations
$$4 = {r_1}\cdot{r_2}\cdot{r_3}\cdot{r_4}$$
$$-(-8) = {r_1}{r_2}{r_3} + {r_1}{r_2}{r_4} + {r_1}{r_3}{r_4} + {r_2}{r_3}{r_4}$$
$$b = {r_1}{r_2} + {r_1}{r_3} + {r_1}{r_4} + {r_2}{r_3} + {r_2}{r_4} + {r_3}{r_4}$$
$$-a = r_1 + r_2 + r_3 + r_4,$$
which can then be solved for $a, b, c, d$.
In particular, from
$$4 = {r_1}\cdot{r_2}\cdot{r_3}\cdot{r_4}$$
we get
$$4 = \left(\dfrac{c^2 - (c^2 - 4d)}{4}\right)^2 = \left(\dfrac{4d}{4}\right)^2 \iff 4 = d^2 \iff d = \pm 2.$$
I leave the other computations as an exercise to the interested reader. =)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\[6pt]
& \frac{14x^2-64x-9x^2+21x+30}{(x+1)(x-4)} < 0 \\[6pt]
& \frac{5x^2-43x+30}{(x+1)(x-4)}<0
\end{align}
using quadratic formula, roots of $5x^2-43x+30$ comes $7.83$(approx.) and $0.763$(approx.)
so rewriting the equation as
$$\frac{(x-7.83)(x-0.763)}{(x+1)(x-4)}<0$$
I then did the plotting of zeroes and poles on number line for finding the values for $x$ but I donot get 2 integral values (which is the answer). Can anyone tell where I did wrong?
| Correcting your initial mistake will lead to the inequality
\begin{equation}
\dfrac{(x+1)(x-6)}{(x+1)(x-4)}<0
\end{equation}
which is positive on the interval $(6,\infty)$ since the product of $x$ coefficients is positive, and the expression changes sign at each zero and pole as you move left along the $x$-axis since each of the four linear factors is raised to an odd power. Thus it is negative only on the intervals $(-1,1)$ and $(4,6)$.
So the only integral solutions are $x=0$ and $x=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$
What is the coefficient of coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$?
Using summation of G.P., this is equivalent to finding the coefficient of $x^{41}$ in
$$\left(x^5 \left(\frac{1-x^5}{1-x}\right)\right)^5$$
and thus finding coefficient of $x^{16}$ in $(\frac{1-x^5}{1-x})^5$. How to proceed after this?
| It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^{16}]\left(\frac{1-x^5}{1-x}\right)^5
&=[x^{16}](1-x^5)^5\sum_{k=0}^\infty \binom{-5}{k}(-x)^k\tag{1}\\
&=[x^{16}](1-5x^5+10x^{10}-10x^{15})\sum_{k=0}^\infty \binom{k+4}{4}(-x)^k\tag{2}\\
&=\binom{20}{4}-5\binom{15}{4}+10\binom{10}{4}-10\binom{5}{4}\tag{3}\\
&=4845-5\cdot 1365+10\cdot 210-10\cdot 5\\
&=70
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion of $\frac{1}{(1-x)^5}$
*In (2) it is sufficient to expand to binomial up to $x^{15}$ since we only need coefficients of powers of $x^{16}$. We also use the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k
\end{align*}
*In (3) we select the coefficients of the series accordingly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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where is my mistake with this equation ot hyperbolic problem? Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure
Thanks help
| I know it's rude to give complex solutions to real problems, but I think we can easily see where the second solution $x=2-\sqrt{5}$ came from.
We want to find solutions for: $$\begin{cases} y^2=4x \\ x^2-y^2=1 \end{cases}$$ Let's try find them in $\mathbb{C}$.
We got, as you said before, the equation $x^2-4x-1=0$ that gives us the solutions you found. We will call them $x_1=2+\sqrt{5}$ and $x_2=2-\sqrt{5}$.
But now we have to find $y$. Let's use $x_1$ in the first equation to find $y^2=8+4\sqrt{5}$ which will give us $y_1=\sqrt{8+4\sqrt{5}}$ and $y_2=-\sqrt{8+4\sqrt{5}}$.
You can see that the points you have in the picture are $A=(2+\sqrt{5},\sqrt{8+4\sqrt{5}})$ and $B=(2+\sqrt{5},-\sqrt{8+4\sqrt{5}})$, and that solves part of the contraddiction.
But where does $x_2$ come from? Let us see that $x_2=2-\sqrt{5}<0$ (in fact you noticed $x_1x_2=-1<0$) so the equation $y^2=8-4\sqrt{5}$ has no real solution! Instead in $\mathbb{C}$ we could get $y_3=i\sqrt{8+4\sqrt{5}}$ and $y_4=-i\sqrt{8+4\sqrt{5}}$ which define other two points $C=(2-\sqrt{5},+i\sqrt{8+4\sqrt{5}})$ and $D=(2-\sqrt{5},-i\sqrt{8+4\sqrt{5}})$. These two points are not seen in the picture because they are complex points (they've got imaginary $y$ coordinate) so thay cannot be represented in a picture on the $\mathbb{R}^2$ plane.
In your problem you want to discuss only real intersections of the curves, so you don't notice there are two complex intersections (and you don't want to because you're working with real numbers, so the second equation we solved in $\mathbb{C}$ doesn't give any solution in $\mathbb{R}$). Then the second solution for the equation in $x$ is to be rejected because cannot provide a real solution for the system.
Hope I helped
| {
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Finding the inverse matrix I have these matrices:
Find the inverse matrices:
\begin{bmatrix}
1 & 1 & 0& 0&\dots & 0& 0\\0 & 1 & 1& 0&\dots & 0& 0 \\0 & 0 & 1& 1&\dots & 0& 0 \\\dots & \dots & \dots& \dots&\dots & \dots& \dots\\0 & 0 & 0& 0&\dots & 1& 1 \\0 & 0 & 0& 0&\dots & 0& 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & a & a^2& a^3&\dots & a^{n-1}& a^n\\0 & 1 & a& a^2&\dots & a^{n-2}& a^{n-1} \\0 & 0 & 1& a&\dots & a^{n-3}& a^{n-2} \\\dots & \dots & \dots& \dots&\dots & \dots& \dots\\0 & 0 & 0& 0&\dots & 1& a \\0 & 0 & 0& 0&\dots & 0& 1 \\
\end{bmatrix}
These are examples of exam problems I will be facing soon. I want to understand how to solve these type of problems
PS I am new in math and I couldn't find any better resourse if you can link me a good source for linear algebra, that would be great.
| Using the usual algorithm, you can calculate that $$\begin{pmatrix}
1&1&0&0&\mid&1&0&0&0\\
0&1&1&0&\mid&0&1&0&0\\
0&0&1&1&\mid&0&0&1&0\\
0&0&0&1&\mid&0&0&0&1
\end{pmatrix}\xrightarrow{R_3\rightarrow R_3-R_4}\begin{pmatrix}
1&1&0&0&\mid&1&0&0&0\\
0&1&1&0&\mid&0&1&0&0\\
0&0&1&0&\mid&0&0&1&-1\\
0&0&0&1&\mid&0&0&0&1
\end{pmatrix}\\ \xrightarrow{R_2\rightarrow R_2-R_3} \begin{pmatrix}
1&1&0&0&\mid&1&0&0&0\\
0&1&0&0&\mid&0&1&-1&1\\
0&0&1&0&\mid&0&0&1&-1\\
0&0&0&1&\mid&0&0&0&1
\end{pmatrix}\\ \xrightarrow{R_1\rightarrow R_1-R_2}
\begin{pmatrix}
1&0&0&0&\mid&1&-1&1&-1\\
0&1&0&0&\mid&0&1&-1&1\\
0&0&1&0&\mid&0&0&1&-1\\
0&0&0&1&\mid&0&0&0&1
\end{pmatrix}.$$
Do this calculation yourself and try to understand the pattern, then show this pattern for a general matrix of the desired form. The other matrix can be dealt with in much the same fashion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.
| Recall the identity $$\sin \alpha \cos \beta = \frac{1}{2} \left( \sin(\alpha+\beta) + \sin(\alpha - \beta) \right).$$ Then
$$\begin{align*} \sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right)
&= \frac{1}{2} \left( \sin \frac{4\pi}{7} + \sin \frac{6\pi}{7} + \sin \frac{-2\pi}{7} + \sin \frac{10\pi}{7} + \sin \frac{-6\pi}{7} \right), \\
\end{align*}$$
and observing that $$\sin \frac{10\pi}{7} = \sin \left(2\pi - \frac{4\pi}{7}\right) = \sin \frac{-4\pi}{7} = -\sin \frac{4\pi}{7}, \\
\sin \frac{-2\pi}{7} = - \sin \frac{2\pi}{7}, \\
\sin \frac{-6\pi}{7} = - \sin \frac{6\pi}{7},$$
we have
$$\sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right) = -\frac{1}{2} \sin \frac{2\pi}{7},$$ where upon dividing both sides by $\sin 2\pi/7$, we obtain the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\cos 36^\circ - \cos 72^\circ$ without the aid of a calculator I have a quick question about a difficult trigonometric functions problem that I have been assigned. The problem is as follows: Evaluate $$\cos36° - \cos72°$$ without the aid of a calculator. In terms of my attempts at the problem, I have converted $\cos 36°$ into 2$\cos^{2}18°$ - 1, and I have converted $\cos72°$ into 1 - 2$\sin^{2}36°$. By the property that $\sin x$ = $\cos(90° - x)$, the latter expression becomes 1 - 2$\cos^{2}54°$. Unfortunately, I've hit a roadblock and don't really know what to do from here. Would, perhaps, adding the two determined expressions yield anything of use? Thanks for all advice.
EDIT: Or, perhaps, would it be wiser to convert $\cos72°$ into 2$\cos^{2}36°$ - 1 and to convert $\cos36°$ into 1 - 2$\sin^{2}18°$?
| First write it as $$\cos 36°-\cos 72°=2\sin { \frac { 72°-36° }{ 2 } } \sin { \frac { 36°+72° }{ 2 } } =2\sin { 18°\sin { 54° } } =2\sin { 18° } \cos { 36° } $$
then let calculate $\sin { 18° } $
$$\cos { 36° } =\sin { 54° } \\ \cos { 2\cdot 18°= } \sin { 3\cdot 18° } \\ \theta =18°\\ \cos { 2\theta } =\sin { 3\theta } \\ 1-2\sin ^{ 2 }{ \theta } =3\sin { \theta -4\sin ^{ 3 }{ \theta } } \\ 4\sin ^{ 3 }{ \theta -2\sin ^{ 2 }{ \theta } -3\sin { \theta } +1 } =0\\ \left( \sin { \theta } -1 \right) \left( 4\sin ^{ 2 }{ \theta +2\sin { \theta } -1 } \right) =0$$
from here we get $\sin { 18°=\frac { \sqrt { 5 } -1 }{ 4 } } $
so
$$\cos 36°-\cos 72°=2\sin { \frac { 72°-36° }{ 2 } } \sin { \frac { 36°+72° }{ 2 } } =2\sin { 18°\sin { 54° } } =2\sin { 18° } \cos { 36° } =\\ =2\sin { 18°\left( 1-2\sin ^{ 2 }{ 18° } \right) } =2\sin { 18°-4\sin ^{ 3 }{ 18° } } =2\left( \frac { \sqrt { 5 } -1 }{ 4 } \right) -4{ \left( \frac { \sqrt { 5 } -1 }{ 4 } \right) }^{ 3 }\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
| If $a=1$, we obtain $2\cos x=3$, which is invalid (unless you want to consider complex values). Set $X=\cos x$ and $Y=\sin x$ to get
$$
\begin{cases} (a+1)X+(a-1)Y=2a+1 \\[6px] X^2+Y^2=1 \end{cases}
$$
Therefore $(a-1)Y=2a+1-(a+1)X$ and therefore
$$
(a-1)^2X^2+(2a+1-(a+1)X)^2=(a-1)^2
$$
With standard steps we find
$$
(2a^2-4a+2)X^2 - 2(2a^2-a-1)X + 3a^2 + 6a=0
$$
and it's possible to factor as
$$
2(a-1)^2X^2-2(a-1)(2a+1)X+3a^2+6a=0
$$
Now we have
$$
(a-1)^2(2a+1)^2-2(a-1)^2(3a^2+6a)=(a-1)^2(1-8a-2a^2)
$$
and therefore
$$
X=\frac{(a-1)(2a+1)\pm(a-1)\sqrt{1-8a-2a^2}}{2(a-1)^2}=
\frac{2a+1\pm\sqrt{1-8a-2a^2}}{2(a-1)}
$$
Now compute the corresponding values for $Y$.
By the way, the condition for the equation to have real solutions is that
$$
\frac{|2a+1|}{\sqrt{(a+1)^2+(a-1)^2}}\le1
$$
hence
$$
\frac{-2-\sqrt{6}}{2}\le a\le\frac{-2+\sqrt{6}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
So,
$$A=0$$
$$2A+B=0$$
$$A+B=1$$
but that does not make sense, because if $A=0$, then $2A+B$ can't be zero, could you please tell me what's the problem?
| You can integred it directly without partial fraction as $$\int { \frac { dx }{ { \left( x+1 \right) }^{ 2 } } =\int { \frac { d\left( x+1 \right) }{ { \left( x+1 \right) }^{ 2 } } } =-\frac { 1 }{ x+1 } +C } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts :
$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
My attempt :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^3}{-3}\cdot\frac{(1+x^2)^2e^x-e^x\cdot2(1+x^2)\cdot2x}{(1+x^2)^4}dx$$
Now it is little bit confusing. How can I simplify this ? Is there another method to do it differently ?
| $$\int \frac { (1-x)^{ 2 }\cdot e^{ x } }{ (1+x^{ 2 })^{ 2 } } dx=\int { \frac { { e }^{ x } }{ (1+x^{ 2 }) } dx-2\int { \frac { x{ e }^{ x } }{ (1+x^{ 2 })^{ 2 } } dx= } } \\ =\int { \frac { d{ e }^{ x } }{ 1+x^{ 2 } } } -2\int { \frac { x{ e }^{ x } }{ (1+x^{ 2 })^{ 2 } } dx= } \left( \frac { { e }^{ x } }{ 1+x^{ 2 } } +2\int { \frac { { xe }^{ x } }{ { \left( 1+x^{ 2 } \right) }^{ 2 } } dx } \right) -\\-2\int { \frac { x{ e }^{ x } }{ (1+x^{ 2 })^{ 2 } } dx=\frac { { e }^{ x } }{ 1+x^{ 2 } } +C } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Geometry problem related to right angled triangles.
In the given figure $AC = 12 cm, AE = 6 cm$ and $CD = 8 cm$. CD is perpendicular to $AD$ and $BE$ is perpendicular to $AC$. How can we find the value of $BE$?
| This is a method without using similar triangles:
First of all, we can find $\text{AD}$ by using the Pythagorean Theorem:
$$\text{AD}=\sqrt{\text{AC}^2-\text{CD}^2}\to\text{AD}=\sqrt{12^2-8^2}=4\sqrt{5}\space\text{cm}$$
So, for $\text{DE}$:
$$\text{DE}=\text{AD}-\text{AE}\to\text{DE}=\left(4\sqrt{5}-6\right)\space\text{cm}$$
Now, we can find $\text{CE}$ using again the Pythagorean Theorem:
$$\text{CE}=\sqrt{\text{CD}^2+\text{DE}^2}\to\text{CE}=\sqrt{8^2+\left(4\sqrt{5}-6\right)^2}=2\sqrt{45-12\sqrt{5}}\space\text{cm}$$
So, we also can say that:
$$
\begin{cases}
\text{BE}^2+\left(12-\text{AB}\right)^2=180-48\sqrt{5}\\
\text{BE}^2+\text{AB}^2=36
\end{cases}\Longleftrightarrow
\begin{cases}
\text{BE}^2=180-48\sqrt{5}-\left(12-\text{AB}\right)^2\\
\text{BE}^2+\text{AB}^2=36
\end{cases}
$$
Now, we can conclude:
$$\left(180-48\sqrt{5}-\left(12-\text{AB}\right)^2\right)+\text{AB}^2=36\Longleftrightarrow\text{AB}=2\sqrt{5}\space\text{cm}$$
So, for $\text{BE}$ we find:
$$\text{BE}^2=180-48\sqrt{5}-\left(12-2\sqrt{5}\right)^2=16\to\text{BE}=4\space\text{cm}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factors of -2 that sum to I'm trying to simplify this equation: $x^2+x-2$. I put it on WolframAlpha and it gives me this:
The factors of -2 that sum to are $2$ and $-1$. So, $x^2+x-2 = (x+2)(x-1)$
Can someone please explain to me what does this mean? How is this technique called?
Thanks!
| Alternative way to factor
$$x^{ 2 }+x-2=x^{ 2 }+x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } -2={ \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 9 }{ 4 } =\left( x+\frac { 1 }{ 2 } -\frac { 3 }{ 2 } \right) \left( x+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right) \\ =\left( x-1 \right) \left( x+2 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
| We start from $\left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right)$:
$ \left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right) = \frac{((n+1)^2 - n)(n+2)}{(n+1)^3}$
We then set up the inequality and simplify in order to see if we can get a trivial inequality.
$ \frac{((n+1)^2 - n)(n+2)}{(n+1)^3} \geq 1 \Leftrightarrow (n+2)((n+1)^2 - n) \geq (n+1)^3$
$(n+2)((n+1)^2 - n) = (n+1 + 1)((n+1)^2 - n) = (n+1)^3 - (n)(n+1) + (n+1)^2 - n$
$(n+1)^3 - (n)(n+1) + (n+1)^2 - n \geq (n+1)^3 \Leftrightarrow -n(n+1) + (n+1)^2 - n \geq 0$
$-n(n+1) + (n+1)^2 - n \geq 0 \Leftrightarrow (n+1)^2 = n^2 + 2n + 1\geq n + n(n+1) = n^2 + 2n$
$n^2 + 2n + 1\geq n^2 + 2n \Leftrightarrow 1 \geq 0$
Since $1 \geq 0$ the inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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A pair of sequences defined by mutual addition/multiplication Define sequences $\{a_n\},\,\{b_n\}$ by mutual recurrence relations:
$$a_0=b_0=1,\quad a_{n+1}=a_n+b_n,\quad b_{n+1}=a_n\cdot b_n.\tag1$$
The sequence $\{a_n\}$ begins:
$$1,\,2,\,3,\,5,\,11,\,41,\,371,\,13901,\,5033531,\,69782910161,\,...\tag2$$
It appears in OEIS as $A003686$ under the name "Number of genealogical $1$-$2$ rooted trees of height $n$." (note that indexing starts with $1$ rather than $0$ in OEIS).
It seems that for $n\to\infty$,
$$\ln\ln a_n=n\cdot\ln\phi+c+o(1),\tag3$$
where $\phi=\frac{1+\sqrt5}2$ is the golden ratio, the last term $o(1)$ exponentially decays in absolute value (and seems to have alternating signs for $n\ge8$), and the constant
$$c\approx-1.11328370529375397170010672464407271138948509227239...\tag4$$
(see more digits here)
The equivalent asymptotics is given in OEIS without proof.
How can we prove $(3)$?
Is there a closed form or some alternative representation (integral, series, product, continued fraction, etc) for the constant $c$? Is it irrational?
Is there an efficient way to numerically calculate $c$ to a higher precision?
| Given a sequence $\{A_n\}$ with the Fibonacci recurrence relation
$$A_{n+2} = A_{n+1} + A_n$$
we get the well-known asymptotic
$$A_n \sim C\phi^n$$
where $\phi$ is the golden ratio and $C$ depends on the initial terms $A_1$ and $A_2$. For example if $A_1 = A_2 = 1$ we get the Fibonacci sequence and $C = 1/\sqrt{5}$. But if we start with different initial values we get a difference constant $C$.
Now consider a more general recurrence:
$$A_{n+2} = A_{n+1} + A_n + g_n$$
where $g_n \rightarrow 0$ as $n \rightarrow \infty$. Then it seems clear that we'll also get the asymptotic
$$A_n \sim C\phi^n$$
where now the constant $C$ depends also on the function $g_n$.
In particular, for our sequence we have
$$\log a_{n+2} = \log a_{n+1} + \log a_{n} + \log \left(1 + \frac{1}{a_n} - \frac{a_n}{a_{n+1}}\right)$$
where the last term goes to zero pretty quickly, so we have
$$\log a_n \sim C\phi^n$$
for some $C$, but I am doubtful that we can get a closed formula for $C$.
DETAILED PROOF ADDED LATER:
LEMMA: Let
$$A_{n+2} = A_{n+1} + A_n + c$$
be a recurrent sequence with constant $c$ and initial conditions $A_1$ and $A_2$. Then
$$A_n = \frac{\phi^{n-1}-\psi^{n-1}}{\sqrt{5}} A_2 + \frac{\phi^{n-2}-\psi^{n-2}}{\sqrt{5}} A_1 + \left[\left(\frac{5+3\sqrt{5}}{10}\right)\phi^{n-2} + \left(\frac{5-3\sqrt{5}}{10}\right)\psi^{n-2} -1 \right]c$$
where $\phi = (1+\sqrt{5})/2$ and $\psi = (1-\sqrt{5})/2$.
In particular, as $n \rightarrow \infty$,
$$A_n = \left(\frac{\phi A_2 + A_1}{\sqrt{5}} + \frac{5+3\sqrt{5}}{10} c\right)\phi^{n-2} - c + o(1).$$
PROPOSITION: Let
$$A_{n+2} = A_{n+1} + A_n + c_n$$
be a recurrent sequence with initial conditions $A_1$ and $A_2$ and $c_n$ a bounded sequence. Then
$$A_n \sim C \phi^n$$
for some constant $C$ and $\phi = (1+\sqrt{5})/2$, i.e. the limit
$$\lim_{n \rightarrow \infty} \frac{A_n}{\phi^n}$$
exists.
Proof of LEMMA: Write recurrence in matrix form and take matrix powers.
Proof of PROPOSITION: We can use the Lemma to bound our sequence by sequences with constants $c = M$ and $c = -M$ to see that $A_n/\phi^n$ is bounded, say $|A_n/\phi^n| \leq B.$
We will further prove that $A_n/\phi^n$ is a Cauchy sequence, and therefore converges to a limit. We have $|c_n| < M$ for some $M$. Let $\epsilon > 0$. Choose $N$ large enough so that $M / \phi^N < \epsilon$. For $n \geq N$, we again use the Lemma to bound our sequence by sequences with constants $c = M$ and $c = -M$, but now starting from at initial values $A_{N-1}$ and $A_N$. So we have
$$\left| A_n - \left(\frac{\phi^{n-N+1}-\psi^{n-N+1}}{\sqrt{5}} A_N + \frac{\phi^{n-N}-\psi^{n-N}}{\sqrt{5}} A_{N-1} \right)\right|$$
$$\leq \left(\left(\frac{5+3\sqrt{5}}{10}\right)\phi^{n-N} + \left(\frac{5-3\sqrt{5}}{10}\right)\psi^{n-N} -1 \right)M.$$
We divide through by $\phi^n$ to get
$$\left| \frac{A_n}{\phi^n} - \left(\phi \frac{1 - (\psi/\phi)^{n-N+1}}{\sqrt{5}} \frac{A_N}{\phi^N} + \frac{1}{\phi}\frac{1-(\psi/\phi)^{n-N}}{\sqrt{5}} \frac{A_{N-1}}{\phi^{N-1}} \right)\right|$$
$$\leq \left(\left(\frac{5+3\sqrt{5}}{10}\right) + \left(\frac{5-3\sqrt{5}}{10}\right)\left(\frac{\psi}{\phi}\right)^{n-N} - \frac{1}{\phi^{n-N}} \right) \frac{M}{\phi^N} < 2\epsilon.$$
Now choose $N_2 > N$ large enough so that $|\psi/\phi|^{N_2-N} B < \epsilon$. Then for $n \geq N_2$ we have
$$\left| \frac{A_n}{\phi^n} - \left(\phi \frac{1}{\sqrt{5}} \frac{A_N}{\phi^N} + \frac{1}{\phi}\frac{1}{\sqrt{5}} \frac{A_{N-1}}{\phi^{N-1}} \right)\right| < 3 \epsilon.$$
By the triangle inequality, when $n \geq N_2$, we have
$$\left| \frac{A_n}{\phi^n} - \frac{A_{N_2}}{\phi^{N_2}}\right| < 6 \epsilon,$$
proving that $A_n/\phi^n$ is a Cauchy sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Neglected constant curvature difference surfaces What are some surfaces where $ \kappa_1-\kappa_2$ is constant? On a sphere where all are umbilical points.. is a special case.
For the $ \kappa_1+\kappa_2$ = constant case we have DeLaunay and Minimal surfaces.
$ \kappa_1,\kappa_2$ are the principal curvatures.
EDIT1:
In the limited number of numerical integrations done so far as a surface of revolution I obtain a symmetrical U shaped meridian between vertical asymptote planes in one case and two cuspidal ring edges in another.
| Another trivial case is a cylinder with $\kappa_{1}=\dfrac{1}{r}$ and $\kappa_{2}=0$.
How about another variation with $2\kappa_{1}-\kappa_{2}=0$?
Mylar balloon
\begin{align*}
\left(
\begin{array}{c}
x \\
y \\
z \\
\end{array}
\right)
&=
r\left(
\begin{array}{c}
\text{cn} \left( u,\frac{1}{\sqrt{2}} \right) \cos v \\[5pt]
\text{cn} \left( u,\frac{1}{\sqrt{2}} \right) \sin v \\[5pt]
\sqrt{2} \left[
E \left(
\text{am} \left( u,\frac{1}{\sqrt{2}} \right) ,\frac{1}{\sqrt{2}}
\right)
-\frac{1}{2} u
\right]
\end{array}
\right) \\
\kappa_{1} &= \frac{1}{2r} \, \text{cn} \left( u,\frac{1}{\sqrt{2}} \right) \\
\kappa_{2} &= \frac{1}{r} \, \text{cn} \left( u,\frac{1}{\sqrt{2}} \right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve the following using AM-GM inequality The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is
Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$
Now my question start from here .
Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$
| No, I think. Because does not exist $a>0$, for which $3\sqrt[3]a>3\sqrt[3]a$.
For the original problem we see that since for $4ax^2=\frac{1}{2x}$ we get an equality and we obtain $a\geq\frac{1}{27}$, so the answer is $\frac{1}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\lim_{x\rightarrow c}x^2=c^2$ What is the common proof for $\lim_{x\rightarrow c}x^2=c^2$.
I tried like this
Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$.
Therefore $$-\varepsilon<x^2-c^2<\varepsilon$$
$$c^2-\varepsilon<x^2<\varepsilon+c^2$$
$$-c-\sqrt{c^2+\varepsilon}<x-c<-c+\sqrt{c^2+\varepsilon}$$
Choose $\delta=-c+\sqrt{c^2+\varepsilon}$which proves the problem
Is this proof is correct ? If there is any mistake in it give a correct proof please .thanks
| $$lim_{x \to c} x^2 = c^2$$
Let $\epsilon>0$ , we know that
$$x^2-c^2=(x-c)(x+c)$$
$$|x^2-c^2|=|x-c||x+c|$$
If
$$|x-c|<2 \implies -2<|x|-|c|<2 (by\ \ triangular \ \ property)$$
$$|c|-2<|x|<2+|c|$$
By triangular property, we have
$$|x+c|\le |x|+|c|<2+|c|+|c|$$
$$|x+c|<2+2|c|$$
So,
$$|x^2-c^2|<2.(2+2|c|)$$
$$|x^2-c^2|<(4+4|c|)$$
$$|x^2-c^2|<(2+2|c|)(|x-c|)$$
If $$|x-c|<\frac{\epsilon}{2+2|c|}$$
Then
$$|x^2-c^2|<\epsilon$$
Define
$$\delta (\epsilon) = inf \{2, \frac{\epsilon}{2+2|c|}\}$$
Then if $|x-c|< \delta \implies |x^2-c^2|<\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove $\tan(A+B)$ using $\cos(A-B)$ and $\sin(A-B)$ Use $\cos(A-B)$ and $\sin(A-B)$ to prove $$\tan(A+B)=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$$
It seems like we cannot simply change $A+B$ to $A+(-B)$ to prove it? Any ideas?
| $$\begin{align*}\tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)} &= \frac{\sin A\cos B + \sin B\cos A}{\cos A\cos B - \sin A\sin B} \div \frac{\cos A\cos B}{\cos A\cos B} \\ & = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}} \\ & = \frac{\tan A + \tan B}{1 - \tan A \tan B}\end{align*}$$
Note that if you know $\sin (A-B)$ then use $B \to -B$ to get $\sin (A+B)$ with the usual odd sine and even cosine, same goes for $\cos(A-B)$.
| {
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"url": "https://math.stackexchange.com/questions/1869555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using power series representation find an approximation to $\int_{0}^{0.1} \arctan(2x)dx$ Using power series representation find an approximation to
$$\int_{0}^{0.1} \arctan(2x)dx$$
This is my solution:
$$\frac{d}{dx}\arctan(2x)=\ \frac{2}{1+4x^2} $$
$$\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$= \int_{0}^{0.1}2\frac{1}{1-(-4x^2)} \, dx =\int_{0}^{0.1} \sum_{n=0}^\infty 2(-1)^n (4x^2)^n \, dx= \sum_{n=0}^{\infty}\frac{2(-1)^{n}4^{n}0.1^{2n+1}}{2n+1}$$
When I input the formula that I got in the calculator I get a value of 0.1973. But when I input the integral of arctangent I get 0.0099. I guess I missed something and can't find what. Where is my mistake?
| We have:
$$ I=\int_{0}^{\frac{1}{10}}\arctan(2x)\,dx = \frac{1}{2}\int_{0}^{\frac{1}{5}}\arctan(x)\,dx \stackrel{IBP}{=} \frac{\arctan\frac{1}{5}}{10}-\frac{\log\left(1+\frac{1}{25}\right)}{4}\tag{1}$$
where, for any $x\in(0,1)$,
$$ \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)},\qquad \log(1+x^2)=\sum_{n\geq 1}\frac{(-1)^{n+1} x^{2n}}{n}.\tag{2} $$
Due to Leibniz' test,
$$ \left|\arctan\frac{1}{5}-\sum_{n=0}^{1}\frac{(-1)^n}{5^{2n+1}(2n+1)}\right|<\frac{1}{5^6},\qquad \left|\log\left(1+\frac{1}{25}\right)-\sum_{n=1}^{2}\frac{(-1)^{n+1}}{5^{2n}n}\right|<\frac{1}{3\cdot 5^6} $$
hence it follows that:
$$ I \approx \frac{1}{10}\left(\frac{1}{5}-\frac{1}{3\cdot 5^3}\right)-\frac{1}{4}\left(\frac{1}{5^2}-\frac{1}{2\cdot 5^4}\right) = \color{red}{\frac{149}{15000}}=0.00993333\ldots \tag{3}$$
and the approximation error is less than $2\cdot 10^{-5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the derivative of $f(x) = \frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}$ Question:
Find the derivative of:
$$f(x) = \frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}$$
Attempted Solution
The most productive approach seems to be logarithmic differentiation:
$$\ln |f(x)| = \ln \left|\frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}\right|$$
Distributing the natural logarithm function gives addition instead of multiplication and subtraction instead of division:
$$\ln \left|\frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}\right| = \ln |e^{x^2}| + \ln |(\arcsin x)^2| + \ln |x|$$
$$+ \ln |\sqrt{\cos x}| - \ln |(\ln x)^6| - \ln |\sin^2 x|$$
Taking the derivative of both sides gives us:
$$f'(x) = f(x) \left( \frac{2 e^{x^2}}{e^{x^2}} + \frac{2 \arcsin x}{(\arcsin x)^2} \frac{1}{\sqrt{1-x^2}} + \frac{1}{x} + \frac{-\sin x}{2 \cos x} + \frac{6 (\ln x)^5}{(\ln x)^6} \frac{1}{x} - \frac{2 \sin x \cos x}{\sin x}\right) $$
Simplifying gives:
$$f'(x) = f(x) \left( 2 + \frac{2}{\arcsin x \sqrt{1-x^2}} + \frac{1}{x} -\frac{1}{2} \tan x + \frac{6}{x \ln x} + 2\cos x \right)$$
However, this is not the correct answer. In particular, the first and last terms are wrong. Where and how did it go wrong?
| Take the natural log to get $$\ln f = \ln e^{x^2} + 2\ln \arcsin x + \ln x + \frac{1}{2} \ln \cos x - 6\ln \ln x - 2\ln \sin x$$
And then simplifying that a bit, i.e: $\ln e^{x^2} = x^2$ and differentiating gives $$f' = f\left(2x + \cdots - 2 \frac{\cos x}{\sin x}\right)$$
So that gives you the correct first and last terms. Remember that logs do more than just convert multiplication to addition, they turn powers to coefficients, making differentiation a lot easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Minimum of f(x,y,z) let $x,y,z> -1$,then how can we find the minimum of this function
$$f(x,y,z)=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}$$
I think that we can use a famous inquality but I can not find it.
So thanks
| For every $t\in\mathbb{R}$, we have
$$t\le\frac{1+t^2}{2}$$
thus
$$\frac{1+x^2}{1+y+z^2}\ge\frac{1+x^2}{1+z^2+\frac{1+y^2}{2}}$$
$$\frac{1+y^2}{1+z+x^2}\ge\frac{1+y^2}{1+x^2+\frac{1+z^2}{2}}$$
$$\frac{1+z^2}{1+x+y^2}\ge\frac{1+z^2}{1+y^2+\frac{1+x^2}{2}}$$
Set $a=1+x^2$ , $b=1+y^2$ and $c=1+z^2$, so
$$f(x,y,z)\ge \frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}$$
By application of Cauchy inequality, we have
$$\left(\frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}\right)\left[\,a(2c+b)+b(2a+c)+c(2b+a)\,\right]\ge 2(a+b+c)^2$$
thus
$$\left(\frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}\right)\ge \frac{2(a+b+c)^2}{3ab+3bc+3ac}\ge 2$$
finally
$$\color{red}{f(x,y,z)\ge 2}$$
Note
If $u=\left(\sqrt{\frac{a}{2c+b}}\ ,\, \sqrt{\frac{b}{2a+c}}\, ,\,\sqrt{\frac{c}{2b+a}}\right)$ and $v=\left(\sqrt{a(2c+b)}\,,\,\sqrt{b(2a+c)}\,,\,\sqrt{c(2b+a)}\,\right)$
then
$$\|u\|^2\,\|v\|^2\ge(u.v)^2$$
and
$$\frac{2(a+b+c)^2}{3ab+3bc+3ac}\ge 2\Leftrightarrow (a-b)^2+(a-c)^2+(b-c)^2\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathcal o \left(1/x+\sqrt{1/x^3}\right)}-1)=...$$
Main question: Now wouldn't it have to be $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ instead? So I could write $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ or $\mathcal O(1/x+\sqrt{1/x^3})$?
Are there other ways to calculate the limit?
| Use formula $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ in order to get:
$$L=\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}.$$
Now divide numerator and denumerator with $\sqrt{x+\sqrt{x}}$ and calculate the limit:
$$L=\lim_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1+\frac{\sqrt{x+\sqrt{x}}}{x}}{1+\frac{\sqrt{x}}{x}}}+\sqrt{\frac{1}{1+\frac{1}{\sqrt{x}}}}}=\frac{1}{\sqrt{\frac{1+0}{1+0}}+\sqrt{\frac{1}{1+0}}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Closed form for the integral $\int_0^\infty \frac{dx}{\sqrt{(x+a)(x+b)(x+c)(x+d)}}$ Let's consider the function defined by the integral:
$$R(a,b,c,d)=\int_0^\infty \frac{dx}{\sqrt{(x+a)(x+b)(x+c)(x+d)}}$$
I'm interested in the case $a,b,c,d \in \mathbb{R}^+$. Obviously, the function is symmetric in all four parameters.
This function has some really nice properties.
$$R(ka,kb,kc,kd)=\frac{1}{k} R(a,b,c,d)$$
Thus:
$$R(a,a,a,a)=\frac{1}{a}$$
Moreover:
$$R(a,a,b,b)=\frac{\ln a-\ln b}{a-b}$$
This is the reciprocal of the logarithmic mean of the numbers $a$ and $b$.
For the case of $R(a,a,b,c)$ we can use the Euler substitution to obtain:
$$R(a,a,b,c)=2 \int_{\sqrt{bc}-a}^{\frac{b+c}{2}-a} \frac{dt}{t^2-(a-b)(a-c)}=$$
$$=\frac{1}{\sqrt{(a-b)(a-c)}} \left( \ln \frac{\sqrt{(a-b)(a-c)}+\sqrt{bc}-a}{\sqrt{(a-b)(a-c)}-\sqrt{bc}+a}-\ln \frac{\sqrt{(a-b)(a-c)}+\frac{b+c}{2}-a}{\sqrt{(a-b)(a-c)}-\frac{b+c}{2}+a} \right)$$
As far as I see, this function is deeply related to various means (you can see arithmetic, geometric and logarithmic means represented in the above expressions).
Is there a closed form for the general case of $R(a,b,c,d)$ for $a,b,c,d \in \mathbb{R}^+$? Perhaps, in terms of hypergeometric functions or elliptic integrals?
|
Hint:
Assume that $0<a<b<c<d$. Using the magical linear-fractional transformation,
$$\frac{\left(d-b\right)\left(x+a\right)}{\left(d-a\right)\left(x+b\right)}=t,$$
we obtain:
$$\begin{align}
R{\left(a,b,c,d\right)}
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{\left(x+a\right)\left(x+b\right)\left(x+c\right)\left(x+d\right)}}\\
&=\small{\int_{\frac{a\left(d-b\right)}{b\left(d-a\right)}}^{\frac{d-b}{d-a}}\frac{\mathrm{d}t}{\sqrt{\frac{\left(d-a\right)\left(b-a\right)t}{\left(d-b\right)-\left(d-a\right)t}\cdot\frac{\left(d-b\right)\left(b-a\right)}{\left(d-b\right)-\left(d-a\right)t}\cdot\frac{\left(d-a\right)\left(d-b\right)\left(1-t\right)\left[\left(d-b\right)\left(c-a\right)-\left(d-a\right)\left(c-b\right)t\right]}{\left[\left(d-b\right)-\left(d-a\right)t\right]^{2}}}}\cdot\frac{\left(d-a\right)\left(d-b\right)\left(b-a\right)}{\left[\left(d-b\right)-\left(d-a\right)t\right]^{2}}}\\
&=\int_{\frac{a\left(d-b\right)}{b\left(d-a\right)}}^{\frac{d-b}{d-a}}\frac{\mathrm{d}t}{\sqrt{t\left(1-t\right)\left[\left(d-b\right)\left(c-a\right)-\left(d-a\right)\left(c-b\right)t\right]}}\\
&=\frac{1}{\sqrt{\left(d-b\right)\left(c-a\right)}}\int_{\frac{a\left(d-b\right)}{b\left(d-a\right)}}^{\frac{d-b}{d-a}}\frac{\mathrm{d}t}{\sqrt{t\left(1-t\right)\left[1-\frac{\left(d-a\right)\left(c-b\right)}{\left(d-b\right)\left(c-a\right)}t\right]}}.\\
\end{align}$$
The rest is pretty straightforward. Can you take it from there?
| {
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"url": "https://math.stackexchange.com/questions/1882179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Sequence and Series bounded Problem: Let $ x_n $ be a sequence defined by $x_n = \displaystyle {\sum_{k=n+1}^{ 2n}\frac{1}{k}}$. Show that $ x_n$ converge.
I can bounded $ x_n <\frac{ n}{ n+1}$, but I can't show that $ x_n$ is increasing, thus would use that es increasing and bounded then is convergent.
| Simply notice that, since $\frac{1}{2n+1} > \frac{1}{2n+2}$, $$x_{n+1} - x_n = \sum_{k=n+2}^{2n+2} \frac{1}{k} - \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{2}{2n+2} - \frac{1}{n+1} = 0.$$
Hence the sequence is increasing. Then apply your bound $x_n \le \frac{n}{n+1} < 1$ to conclude convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| $$\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots$$
When we use the LOG function get (when $n$ is positive):
$$\ln\left(\left(2^n\right)^{\frac{1}{2^n}}\right)=\frac{n\ln(2)}{2^n}$$
And using, when $a$ and $b$ are positive:
$$\ln(ab)=\ln(a)+\ln(b)$$
So, we get that:
$$\ln(\text{P})=\frac{1\ln(2)}{2^1}+\frac{2\ln(2)}{2^2}+\frac{3\ln(2)}{2^3}+\dots=\ln(2)\left[\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots\right]$$
And, now we know that:
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots=\frac{2}{(2-1)^2}=2$$
Using, when $|x|>1$:
$$\sum_{n=1}^{\infty}\frac{n}{x^n}=\frac{x}{(x-1)^2}$$
So:
$$\ln(\text{P})=\ln(2)\left[2\right]=2\ln(2)\Longleftrightarrow\text{P}=4$$
We, find the answer:
$$\color{red}{\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots=4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
Prove the inequalitiy
If $a, b, c$ are positive reals such that:
$\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le
1$ then:
$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge
1$
I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.
| .WLOG, $a <b <c$, then $\frac{1}{b+c} < \frac{1}{a+c} < \frac{1}{a+b}$
Note that $$\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} = (a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) - 3$$
Hence,
$$
(a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \leq 4
$$
Further, by the harmonic mean-arithmetic mean inequality,
$$
\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \geq \dfrac{9}{2(a+b+c)+3}
$$
Now, let $x=\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) $ and $y = a+b+c$. Then, we have: $2yx+3x \geq 9$ and $4 \geq yx + x$. Multiplying by $2$ and changing the sign, $-2yx-2x \geq 8$. Adding ,we get $x \geq 1$. Hence the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?
| When you arrive at
$$
-\tan^2x \frac{2\tan x}{1-\tan^2x} = \frac{2\tan x}{1-\tan^2x}
$$
you can't multiply both sides by $\frac{1-\tan^2x}{2\tan x}$, because this operation is allowed only when the multiplier is nonzero.
You should rather move everything to the right-hand side and collect terms, getting
$$
0=\frac{2\tan x}{1-\tan^2x}(\tan^2x+1)
$$
Since $\tan^2x+1\ne0$, you get
$$
\tan x=0
$$
and so $x=k\pi$.
You can also observe that $\tan3x=\tan x$ means
$$
3x=x+k\pi
$$
so
$$
x=k\frac{\pi}{2}
$$
provided $\tan x$ and $\tan3x$ exist, which is not the case when $k$ is odd. Thus the solutions are $x=(2h)\frac{\pi}{2}=h\pi$ ($h$ integer).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $ $$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$
I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?
| $$I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\tan\left(x - \frac{1}{x}\right)dx = \int_{2}^{\frac{1}{2}}u\cdot \frac{-1}{u^2}\tan\left(\frac{1}{u}-u\right)du = -\int_{\frac{1}{2}}^{2}\frac{1}{u}\tan\left(u - \frac{1}{u}\right)du = -I$$
Where the second step was obtained by letting $\displaystyle x = \frac{1}{u}, dx = \frac{-1}{u^2}du$.
Since $I = -I$, we have $I = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Elementary matrix operations I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is
$$
A=
\begin{pmatrix}
1 & 1 & 0 & 2\\
2 & 1 & 1 &-2\\
2 & 2 & 2 & 1
\end{pmatrix},\quad \tilde{b}=
\begin{pmatrix}
12\\0\\14
\end{pmatrix}
$$
Proceeding with an augmented matrix,
I have
\begin{align*}
\left(
\begin{array}{rrrr|c}
1 & 1 & 0 & 2 & 12\\
2 & 1 & 1 &-2 & 0\\
2 & 2 & 2 & 1 & 14
\end{array}
\right)&\sim
\left(
\begin{array}{rrrr|r}
1 & 1 & 0 & 2 & 12\\
0 &-1 & 1 &-6 & -24\\
0 & 0 & 2 & -3 &-10
\end{array}
\right)\\
&\sim
\left(
\begin{array}{rrrr|r}
1 & 1 & 0 & 2 & 12\\
0 & 1 &-1 & 6 & 24\\
0 & 0 & 1 & -3/2 &-5
\end{array}
\right)
\end{align*}
I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.
Solving this now, I have
\begin{align*}
x_4&=\alpha\in\mathbb{R}\\
x_3&=-5+\frac{3}{2}\alpha\\
x_2&=19-\frac{9}{2}\alpha\\
x_1&=-7+\frac{5}{2}\alpha
\end{align*}
So from this, my solution will be
$$
\tilde{x}=
\begin{pmatrix}
-7+\frac{5}{2}\alpha\\
19-\frac{9}{2}\alpha\\
-5+\frac{3}{2}\alpha\\
\alpha
\end{pmatrix}
=
\begin{pmatrix}
-7\\19\\-5\\0
\end{pmatrix}+\beta
\begin{pmatrix}
5\\-9\\3\\2
\end{pmatrix},\quad\beta\in\mathbb{R}.
$$
This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.
Edit: My professor's solution is
$$
\tilde{x}=
\begin{pmatrix}
3\\1\\1\\4
\end{pmatrix}+\alpha
\begin{pmatrix}
5\\-9\\3\\2
\end{pmatrix},\quad\beta\in\mathbb{R}.$$
Can somebody please let me know what I've done incorrectly?
Thank you.
| Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.
| {
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Help simplify using quadratic formula $\dfrac{4\pm\sqrt{28}}{2}=2\pm\sqrt7$
My Question is how did $\dfrac{4\pm\sqrt{28}}{2}$ become simplified as $2\pm\sqrt7$
Can you help me by explaining the steps clearly
:) Many Thanks
| Notice that $\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\times\sqrt{7}=2\sqrt{7}$.
Hence $\dfrac{4\pm\sqrt{28}}{2}=\dfrac{4\pm2\sqrt{7}}{2}=\dfrac{2(2\pm\sqrt{7})}{2}=2\pm\sqrt{7}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}$$
But I know that is wrong because it looks nothing like the graph of the real part of 1/z^2 on wolfram alpha found here: http://www.wolframalpha.com/input/?i=1%2F(x%2Bi*y)%5E2
Then I thought I could must multiply $1/z^2$ by $z/z$ to get $\frac{x}{z^3}$ and $\frac{iy}{z^3}$ however graphing these again shows that they are not the real and complex parts of $\frac{1}{z^2}$.
| \begin{eqnarray}
f(z)=1/(z^2)=1/((x+iy)^2)&=&1/(x^2-y^2+2ixy) \newline
&=&(1/(x^2-y^2+2ixy))\cdot((x^2-y^2-2ixy)/(x^2- y^2-2ixy)) \newline
&=&((x^2-y^2-2ixy)/((x^2-y^2)^2+4x^2y^2)) \newline
&=&(x^2-y^2-2ixy)/(x^2+y^2)^2
\end{eqnarray}
therefore $\Re(1/z^2)=(x^2-y^2)/(x^2+y^2)^2$ and $\Im(1/z^2)=-2xy/(x^2+y^2)^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$? How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
| There's a trigonometric identity for $\arctan$: $\arctan(\alpha) \pm \arctan(\beta) = \arctan(\frac{\alpha \pm \beta}{1 \mp \alpha\beta})$.
In our case:
$$
\pi-\arctan(\frac{8}{x})-\arctan(\frac{13}{20-x})=\arctan(\frac{x}{8})+\arctan(\frac{20-x}{13})\\
\pi-\arctan(\frac{13}{20-x})-\arctan(\frac{20-x}{13})=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\
\pi-(\arctan(\frac{13}{20-x})+\arctan(\frac{20-x}{13}))=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\
\pi-\arctan(\frac{\frac{13}{20-x}+\frac{20-x}{13}}{1-(\frac{13}{20-x}\cdot \frac{20-x}{13})})=\arctan(\frac{\frac{8}{x}+\frac{x}{8}}{1-(\frac{8}{x}\cdot \frac{x}{8})})\\
\pi-\arctan(\frac{13^2+(20-x)^2}{0})=\arctan(\frac{8^2+x^2}{0})\\
$$
Note that the angle for $lim_{x\rightarrow \infty} \arctan(x) =\frac{\pi}{2}$ (in case $x<0$ we get $-\frac{\pi}{2})$.
hence, we get: $\pi - \frac{\pi}{2}=\frac{\pi}{2}$.
| {
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Solution of inequality Question:
$x$ and $y$ are different integers and both $x$ and $y$ are greater than $0$. If $x^2-y^2<8$ and $x+y>3$, what is the maximum value of $x$?
My solution:
I used plugged in number method, and found for $x=4$ and $y=3$, $4^2-3^2=7$ and $4+3=7$, satisfies both equations. but, if $x=5$ and $y=4$, values doesn't satisfy first equation. hence, maximum possible value of $x= 4$.
Is there any other easy way than this to solve this problem?
| As $x^2 - y^2 = (x-y)(x+y) < 8 $, and $x+y > 3$ we know $x-y<2$.
So, if there is a additional condititon $x \neq y$ (I will asume that, also that $x>y$), then $x=1+y$.
So $(y+1)^2 - y^2 < 8$, so $2y+1<8 $, and maximum of $y$ is $3$.
Without the condition $x \neq y$ the value of $y$ is unbounded.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Shortest distance between z axis and the line $x+y+2z=3, 2x+3y+4z+4=0$ z axis is : $x=0=y$
$x + B(y) = 0$ --> (1)
$x+y+2z-3+ A (2x+3y+4z+4 ) = 0$ --> (2)
These are two planes constructed through the two lines. We find the parallel planes so that it is easier to calculate the distance after that.
Therefore, for planes to be parallel :
$\frac{1+2A}{1} $ = $\frac{1+3A}{B}$ = $\frac{2+4A}{0}$
But now no values of A,B will satisfy this.
How to proceed further ?
| Solving the given equations for $x$ and $y$ in function of $z$ gives $x=13-2z$, $y=-10$. It follows that $$\sqrt{x^2+y^2}=\sqrt{(13-2z)^2+100}$$ is minimal when $z={13\over2}$, and the minimal value is $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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When is ${\large\int}\frac{dx}{\left(1+x^a\right)^a}$ an elementary function? Consider the following indefinite integral:
$$F_a(x)=\int\frac{dx}{\left(1+x^a\right)^a}.$$
Here $a\in\mathbb R$ is a parameter, and $x>0$ is a variable. For what values of the parameter $a$ the function $F_a(x)$ is an elementary function of the variable $x$?
Here are some examples (in each case there is implicitly an additive constant of integration).
$$F_1(x)=\int\frac{dx}{\left(1+x\right)}=\ln(1+x)$$
$$F_{-1}(x)=\int\frac{dx}{\left(1+x^{-1}\right)^{-1}}=x+\ln x$$
$$F_2(x)=\int\frac{dx}{\left(1+x^2\right)^2}=\frac12\left(\frac{x}{x^2+1}+\arctan(x)\right)$$
$$F_{1/2}(x)=\int\frac{dx}{\left(1+x^{1/2}\right)^{1/2}}=\frac43\left(x^{1/2}-2\right) \left(1+x^{1/2}\right)^{1/2}$$
$$F_{1/3}(x)=\int\frac{dx}{\left(1+x^{1/3}\right)^{1/3}}=\frac9{40}\left(9-6x^{1/3}+5x^{2/3}\right)\left(1+x^{1/3}\right)^{2/3}$$
$$F_{1+\sqrt2}(x)={\large\int}\frac{dx}{\left(1+x^{1+\sqrt2}\right)^{1+\sqrt2}}=\left(x+\frac{x^{2+\sqrt2}}{\sqrt2}\right)\left(1+x^{1+\sqrt2}\right)^{-\sqrt2}$$
$$F_{1-\sqrt2}(x)={\large\int}\frac{dx}{\left(1+x^{1-\sqrt2}\right)^{1-\sqrt2}}=\left(x-\frac{x^{2-\sqrt{2}}}{\sqrt{2}}\right) \left(1+x^{1-\sqrt{2}}\right)^{\sqrt{2}}$$
$$F_{(1+\sqrt5)/2}(x)={\large\int}\frac{dx}{\left(1+x^{(1+\sqrt5)/2}\right)^{(1+\sqrt5)/2}}=x \left(1+x^{(1+\sqrt5)/2}\right)^{(1-\sqrt5)/2}$$
$$F_{(1-\sqrt5)/2}(x)={\large\int}\frac{dx}{\left(1+x^{(1-\sqrt5)/2}\right)^{(1-\sqrt5)/2}}=x\left(1+x^{(1-\sqrt{5})/2}\right)^{(1+\sqrt{5})/2}$$
All these results can be easily verified by differentiation. For the general case, Mathematica gives the result in terms of the hypergeometric function:
$$F_a(x)=\int\frac{dx}{\left(1+x^a\right)^a}=x\cdot{_2F_1}\left(\frac1a,a;1+\frac1a;-x^a\right).$$
For what values of the parameter $a$ does this hypergeometric function reduce to an elementary function?
| Using integration by parts leads to \begin{align}\int\frac{dx}{(1+x^a)^a}&=\frac x{(1+x^a)^a}+\int\frac{a^2x^adx}{(1+x^a)^{a+1}}\\
&=\frac x{(1+x^a)^a}+\int a^2\left\{\frac1{(1+x^a)^a}-\frac1{(1+x^a)^{a+1}}\right\}dx\\
&=\frac x{(1-a^2)(1+x^a)^a}-\frac{a^2}{1-a^2}\int\frac{dx}{(1+x^a)^{a+1}}
\end{align}
so you can also consider integrals of the form $\int dx/(1+x^a)^{a+n}$.
The substitution $u=1/x$ gives$$F_a(x)=\int\frac{-du}{u^2(1+u^{-a})^a}=-\int\frac{u^{a^2-2}du}{(1+u^a)^a}$$ $$\int\frac{dx}{(1+x^a)^{a+n}}=-\int\frac{u^{a^2+an-2}du}{(1+u^a)^{a+n}}$$which implies $a^2+an-1$ must be a multiple of $a$ (i.e. $a^2-1=k\cdot a$).
Subbing $u=x^a$ turns $F_a(x)$ into $$\frac1a\int\frac{u^{1/a-1}du}{(1+u)^a}$$which if elementary will only have roots, rational functions, and logarithms (specifically, logarithms will only appear when $a\in\mathbb{Z}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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normalization of the nodal cubic Consider the nodal cubic $R = k[x,y]/(y^2 - x^3 + x^2)$, so $x,y$ satisfy $y^2 = x^3 - x^2$.
The idea behind normalization is to adjoin "$\frac{y}{x}$", since then in the fraction field $K$ of $R$, one has $\left(\frac{y}{x}\right)^2 = x-1$.
It's straightforward to show that the subring $S$ of $K$ generated by $R$ and $\frac{y}{x}$ is normal and integral over $R$, hence $S := R[\frac{y}{x}]$ is the normalization/integral closure of $R$.
It's also not hard to show that $S\cong R[t]/(xt - y, t^2-x+1)$.
My question is - do you need both $xt -y$ AND $t^2-x+1$?
Is $S\cong R[t]/(xt - y)$?
Is $S\cong R[t]/(t^2-x+1)$?
| If $xt=y$ and $t^2-x+1=0$, then multiplying the second equation by $x^2$ and using the first one shows that $y^2-x^3+x^2=0$. This means that you forget about this last equation, and $S$ is $k[x,t,y]/(xt-y,t^2-x+1)$.
Now in this quotient, we have $x=t^2+1$ so we can remove the generator $x$, and it is isomorphic to $k[y,t]/(t^2+1)t-y)$. Of course, now we can also remove $y$, and we end up with $S=k[t]$.
Using this description of $S$, can you answer your questions?
Notice that $R[t]/(xt-y)=k[x,y,t]/(y^2-x^3+x^2,xt-y)$. As $y=xt$ here, we can remove the generator $y$, and $R[t]/(xt-y)$ is isomorphic to $k[x,t]/(x^2t^2-x^3+x^2)$, which is clearly not a domain.
On the other hand, $R[t]/(t^2-x+1)$ is $k[x,y,t]/(y^2-x^3+x^2,t^2-x+1)$, which, using $x=t^2+1$ to remove $x$, is isomorphic to $k[y,t]/(y^2-(t^2+1)^3+(t^2+1)^2)$. This has some singular points.
| {
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Show that: $nf(\frac{x_1+\cdots+x_n}{n})+n\left(\frac{f(M)+f(m)}{2}-f(\frac{M+m}{2})\right) \ge f(x_1)+\cdots+f(x_n)$ I am looking for a proof of the problem as following:
Let $f(x)$ is a real continuous function that is strictly convex ($f''>0$) on $[m, M]$, let $m \le x_i \le M$, for $i=1,2,\ldots,n$ then show that:
$$nf\left(\frac{x_1+\cdots+x_n}{n}\right)+n\left(\frac{f(M)+f(m)}{2}-f\left(\frac{M+m}{2}\right)\right) \ge f(x_1)+\cdots+f(x_n)$$
Equality holds if only if $m=x_1=x_2=\cdots=x_n=M$
| The inequality does not hold in general. For a simple counterexample, consider the convex function:
$$
f(x) =
\begin{cases}
0 & \text{if $x \lt \frac{2}{3}$} \\
x - \frac{2}{3} & \text{if $x \ge \frac{2}{3}$}
\end{cases}
$$
and $m=0, M=1, n=3, x_1=0, x_2=x_3=1$.
Since $f(0)=f(\frac{1}{2})=f(\frac{2}{3})=0, f(1)=\frac{1}{3}$ the inequality becomes:
$$
3 f(\frac{2}{3}) + 3(\frac{f(0)+f(1)}{2} - f(\frac{1}{2})) \ge f(0) + 2 f(1)
$$
$$
\frac{1}{2} \ge \frac{2}{3}
$$
where the latter is obviously false.
For a counterexample using a strictly convex function, one can choose $f(x) = e^x$, $m = 0 \le M$, $n=3$, $x_1=0, x_2=x_3=M$. The inequality becomes:
$$
3 e^{\frac{2}{3}M} + 3(\frac{1+e^M}{2} - e^\frac{M}{2}) \ge 1 + 2 e^M
$$
$$
- \frac{1}{2} e^M + 3 e^{\frac{2}{3}M} - 3 e^\frac{M}{2} + \frac{1}{2} \ge 0
$$
The latter will fail for large enough $M$ since the dominant term $e^M$ has a negative coefficient.
| {
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Loxodromics (curves with fixed angle with meridians) of a revolution surface I am trying to find the curves with a fixed angle $\phi$ with meridians for the revolution surface given by the revolution of the graph of $z=\frac{1}{x}$ around the $z$ axis. The surface is easily parametrized as
$$(x,\theta)\mapsto(x \cos \theta, x\sin \theta,\frac{1}{x}) \qquad x\in \mathbb R_{>0},\theta \in (0,2\pi)$$
Now, some tedious computations (I hope they are correct) show that the first and the second fundamental form for this parametrization are
$$G_{I}=\begin{pmatrix}1+\frac{1}{x^4} &0 \\ 0 & x^2 \end{pmatrix} \qquad G_{II}=\frac{1}{\sqrt{x^4+1}}\begin{pmatrix}\frac{1}{x} & 0 \\ 0 & -x \end{pmatrix}$$
Recall that the angle $\phi$ between two $C^1$ curves $\gamma_1,\gamma_2$ is given by
$$\cos \phi=\frac{\gamma_1'^TG_I \gamma_2'}{\sqrt{\gamma_1'^T G_I \gamma_1'}\sqrt{\gamma_2'^TG_I\gamma_2'}}$$
Since we are looking for curves $(x(t),\theta(t))^T$ with constant angle with meridians, we want that
$$\frac{\begin{pmatrix}1 & 0 \end{pmatrix}\begin{pmatrix} 1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t)\end{pmatrix}\begin{pmatrix} x'(t) \\ \theta'(t) \end{pmatrix}}{\sqrt{\begin{pmatrix} x(t) & \theta(t)\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} x'(t)\\ \theta'(t) \end{pmatrix}}\sqrt{\begin{pmatrix} 1 & 0\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} 1 \\ 0\end{pmatrix}}}=K$$
which, dropping the dependency on $t$, yields the differential equation
$$x'\left (1+\frac{1}{x^4}\right )=K\sqrt{1+\frac{1}{x^4}}\sqrt{x'^2\left ( 1+\frac{1}{x^4}\right)+x^2\theta'^2}$$
Not looking very good. Through a few manipulations, I brought this to the form
$$x'(t)=K\sqrt{x'^2+\frac{x^6\theta'^2}{x^4+1}}$$
Since this is an exercise from an exam, I would expect the solution to have a nice closed form, but maybe I am mistaken.
Is there any elementary way of solving such differential equation or of finding an explicit equation for the loxodromics?
| After Futurologist's answer, you just need $$I=\int \frac{\sqrt{x^4+1}}{x^3}dx=\int \frac{\sqrt{x^4+1}}{x^4}x\,dx$$ So, define $$x^2=y\implies x=\sqrt y\implies dx=\frac{dy}{2 \sqrt{y}}$$ which make $$I=\frac 12\int \frac{\sqrt{y^2+1}}{ y^2}\,dy=\frac{1}{2} \left(\sinh ^{-1}(y)-\frac{\sqrt{y^2+1}}{y}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Question about Spivak's chapter 5 problem 3 part 3 Question:
Find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$
Relating equations to the problem;
$$f(x)=x^4+\frac{1}{x} \\a=1, L=2 $$
rewriting the equation as the definition of the limits
$$|x^4+\frac{1}{x}-2|<\epsilon$$ if and only if $$0<|x-1|<\delta$$
As x approaching closer and closer to a $x^4$ approach to 1 and $\frac{1}{x}$ approach to 1 ,realizing it's the addition of limits rewriting the definition as;
$$|x^4+\frac{1}{x}-(1+1)|<\epsilon$$ now I got
$|x^4-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_1$ ,and I got $|\frac{1}{x}-1|<\frac{\epsilon}{2}$ if $0<|x-a|<\delta_2$
starting with $x^4$ factoring I got
$$|(x-1)(x^3+x^2(1)+1^2(x)+1^3)|<\frac{\epsilon}{2}$$
choose $\delta_1=min(1)$ , and I got $0<|x-1|<1$, using triangle inequality $|x|-|1|\le|x-1|<1$, setting $|x|-|1|<1$, and I got $|x|<2$
rewriting the expression $(x^3+x^2(1)+(1^2)(x)+1^3)\le|x^3|+|x^2||1|+|1^2||x|+|1^3|$, sub |x| in the equation I got $|2^3|+|2^2||1|+|1^2||2|+|1^3|=15$, and I just need to choose that $\delta_1= min(1, \frac{\epsilon}{30})$
for the next limit of $\frac{1}{x}$, I know I can choose $\delta_2$ such that
if $$0<|x-1|<\delta_2$$ then $$|\frac{1}{x}-1|<min(\frac{|1|}{2},\frac{\epsilon|1^2|}{4})$$
now choose $\delta=min(\delta_1,\delta_2)$ so I got $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4},\frac{\epsilon}{30})$
but when I looked at the solution it is $\delta=min(1,\frac{1}{2},\frac{\epsilon}{4}, \frac{\epsilon}{8\times2\times2})$
So it is $\frac{\epsilon}{32}$ rather than $\frac{\epsilon}{30}$, could anyone check my math and see where I did wrong, and furthermore, in his final answer Spivak ignore $\frac{\epsilon}{4}$, and he got $\delta=min(\frac{1}{2},\frac{\epsilon}{32})$, can anyone give an explanation on why that is so? Is it because $\frac{\epsilon}{4}$ is bigger than $\frac{\epsilon}{32}$, and it did not matter since $\frac{\epsilon}{32}$ is closer to 1?
| There are many ways to solve this. We can do a bit of simplification at the beginning using the triangle inequality, and end up with simpler factors, as follows:
$$\begin{aligned}
\left|x^4 + \frac{1}{x} - 2\right| &= \left|(x^4 - 1) + \left(\frac{1}{x} - 1\right)\right| \\
&\leq |x^4 - 1| + \left|\frac{1}{x} - 1\right| \\
&= |x-1||x+1||x^2+1| + |x-1|\left|\frac{1}{x}\right|\\
\end{aligned}$$
Now if we constrain $|x-1| < 1/2$, we will have $|x| > 1/2$, so $|1/x| < 2$. We will also have $|x| < 3/2$, so $|x+1| \leq |x|+1 < 5/2$ and $|x^2 + 1| \leq |x^2| + 1 < 9/4 + 1 = 13/4$. Therefore we end up with
$$\left|x^4 + \frac{1}{x} - 2\right| < \frac{5}{2}\cdot\frac{13}{4}|x-1| + 2|x-1| = \frac{81}{8}|x-1|$$
So if we choose $\delta = \min(1/2, 8\epsilon/81)$, then for all $|x-1| < \delta$, we will have
$$\left|x^4 + \frac{1}{x} - 2\right| < \epsilon$$
| {
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If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then how to show that $A^2+B^2$ equals $A+B$?
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals ?
(a) $2AB$
(b) $2BA$
(c) $A+B$
(d) $AB$
I tried
$(A+B)^2=A^2+B^2+AB+BA$
or,$A^2+B^2=(A+B)^2-AB-BA$
$=(A+B)^2-A-B$
$
=(A+B)^2-(A+B)=(A+B)(A+B-1)$
How to reach the answer from here?
| More precisely, there is a positive integer $p$ s.t. $A,B$ are simultaneously similar to $A'=\begin{pmatrix}I_p&0\\0&0_{n-p}\end{pmatrix},B'=\begin{pmatrix}I_p&Q\\0&0_{n-p}\end{pmatrix}$ where $Q$ is an arbitrary $(p\times n-p)$ matrix.
Proof. Since $A^2=A$, there is $p$ s.t. $A$ is similar to $A'=\begin{pmatrix}I_p&0\\0&0_{n-p}\end{pmatrix}$; let $B'=\begin{pmatrix}P_p&Q\\R&S_{n-p}\end{pmatrix}$. Then just proceed with identification in the relations $A'B'=B',B'A'=A'$.
| {
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How to simplify this expression of square roots? If one simplifies the infinite nested radical
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
Does one get $$\sqrt x\frac{1+\sqrt5}2$$
Any help would be appreciated!
| You have
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
$$ = \sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
$$ = \sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
The continued square root $\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$ is a well known form of $\phi = \frac{\sqrt{5}+1}{2}$ (see here) and so you end up with:
$$ \frac{\sqrt{x}(\sqrt{5}+1)}{2}$$
as required.
| {
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Spivak's Calculus 4th ed: Absolute value definition p. 11 After defining the absolute value $|a|$ of $a$ as
$$
|a| =
\begin{cases}
a, & a\geq 0\\
-a, & a\leq 0,
\end{cases}
$$
the text provides some examples. While $|-3| = 3$ and $|7| = 7$ are simple enough (by direct application of the above definition), I do not understand the latter two examples, which are
$$
|1 + \sqrt{2} - \sqrt{3}| = 1 + \sqrt{2} - \sqrt{3}
$$
and
$$
|1 + \sqrt{2} - \sqrt{10}| = \sqrt{10} - \sqrt{2} - 1.
$$
Without calculating the actual values of $\sqrt{2}$, $\sqrt{3}$, $\sqrt{10}$ and so on, how can one arrive at these equations?
EDIT: As @Bernard points out, the idea is to just compare the squares. Both of these absolute values use the sum $1 + \sqrt{2}$, we can think of its square, $3 + 2\sqrt{2}$ (which is less than $3 + 2\cdot2 = 3 + 4)$. In the first, we are essentially doing $3 + 2\sqrt{2} - 3$, which is a positive value, so we can just leave it as is. For the second, $3 + 2\sqrt{2} < 3 + 4 < 10$, so we have to re-order the $\sqrt{10}$ to come first, and then subtract the $1$ and $\sqrt{2}$ afterwards. It's more calculating than I'd like (and I'm not sure exactly what Spivak was trying to accomplish by throwing in these examples), but I am satisfied for now.
| For the first absolute value, it is enough to compare the squares: $(1+\sqrt2)^2=3+2\sqrt 2>(\sqrt 3)^2$.
For the second absolute value, $3+2\sqrt 2<3+4<10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a Differential equation using eigenvectors I want to solve the following differential equation using eigenvectors:
$x'=y$
$y' = 2x-5y$
Or using matrices:
$\begin{pmatrix}
x' \\
y'
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
2 & -5
\end{pmatrix} \begin{pmatrix}
x \\
y
\end{pmatrix}$
With the initial conditions $x(0) = 0$ and $y(0) = 2$
Now I know that all I'd have to do is essentially find the exponential of the matrix
$$ A =\begin{pmatrix}0 & 1 \\2 & -5 \end{pmatrix}$$ but my teacher wants us to solve it using eigenvectors.
Let's find A's eigenvectors:
$det(A-\lambda I) = -\lambda(-5-\lambda)-2 = \lambda^2+5\lambda-2$
A's eigenvalues are:
*
*$\lambda_1 = \dfrac{-5+\sqrt{33}}{2}$
*$\lambda_2 = \dfrac{-5-\sqrt{33}}{2}$
Now what do I do to solve the differential equation using A's eigenvector?
| You have done most of the work already. $A's$ eigenvalues are:
$$\lambda_{1, 2} = \dfrac{-5\pm\sqrt{33}}{2}$$
We now have to find the eigenvectors and solve $[A - \lambda_i I]v_i = 0$. So for $\lambda_{1,2}$, we have:
$$[A - \lambda_{1,2} I]v_{1,2} = [A - \dfrac{1}{2} (-5 \pm \sqrt{33}) I]v_{1,2} = \begin{pmatrix}
- \dfrac{1}{2} (-5 \pm \sqrt{33}) & 1 \\
2 & -5 - \dfrac{1}{2}(-5 \pm \sqrt{33})
\end{pmatrix}v_{1, 2} = \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
When we do the RREF of this matrix, we have:
$$\begin{pmatrix}
1 & \dfrac{1}{4} (-5 \mp\sqrt{33}) \\
0 & 0 \end{pmatrix}v_{1, 2}= \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
This gives us:
$$v_{1,2} = \begin{pmatrix}
\dfrac{1}{4}(5 \pm \sqrt{33})\\1
\end{pmatrix}$$
We can now write:
$$X(t) = \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$$
You have two initial conditions to solve for $c_1$ and $c_2$ and should get:
$$X(t) = \begin{pmatrix}
x(t)\\y(t)
\end{pmatrix} =\begin{pmatrix}
\dfrac{2}{\sqrt{33}}~e^{-1/2 (5+\sqrt{33}) t} (e^{\sqrt{33} t}-1) \\ \dfrac{1}{33} e^{-1/2 (5+\sqrt{33}) t} ((33-5 \sqrt{33}) e^{\sqrt{33} t}+33+5 \sqrt{33})
\end{pmatrix}$$
Update For $\lambda_1$, our RREF is:
$$\begin{pmatrix}
1 & \dfrac{1}{4} (-5 -\sqrt{33}) \\
0 & 0 \end{pmatrix}v_1= \begin{pmatrix}
0 \\
0
\end{pmatrix}$$
Let the vector $v_1 = \begin{pmatrix} a \\ b \end{pmatrix}$, so this gives us:
$$a + \dfrac{1}{4} (-5 -\sqrt{33})b = 0 \implies a = \dfrac{1}{4} (5 +\sqrt{33})b$$
We are free to choose whatever $b$ we'd like, so choose $b = 1$ and conclude.
This is how we arrived at:
$$v_1 = \begin{pmatrix}
\dfrac{1}{4}(5 + \sqrt{33})\\1
\end{pmatrix}$$
I should also mention, note how this choice also satisfies the second equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The square of any odd integer is odd. Suppose that $n$ is an odd integer. Then $n = 2k + 1$ for some integer $k$.
Hence $n^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1$.
Since $k$ is an integer, $2k^2 + 2k$ is an integer. Thus $n^2 = 2k' + 1$ for some integer $k'$.
Therefore $n^2$ is odd.
Is this correct?
| you can write more generally that the product of 2 odd integers is odd
let $a = 2k+1$ and $b = 2p+1$
then $ab = (2k+1)(2p+1) = 2(2kp+k+p)+1 $ which is the general format of odd numbers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
| WLOG, assume that $x \ge y \ge z$.
We have
$$\sqrt{1 - xy} \ge \sqrt{1 - (x + y)^2/4}.$$
Also, we have
$$\sqrt{1 - yz} + \sqrt{1 - zx} \ge 1 + \sqrt{1 - yz - zx}.$$
(Note: Squaring both sides, it is just
$2\sqrt{1 - yz - zx + z^2xy} \ge 2\sqrt{1 - yz - zx}$.)
Thus, it suffices to prove that
$$\sqrt{1 - (x + y)^2/4} + 1 + \sqrt{1 - yz - zx} \ge 2$$
or
$$\sqrt{1 - (2 - z)^2/4} + \sqrt{1 - (2 - z)z} \ge 1$$
or
$$\frac12\sqrt{z(4 - z)} + (1 - z) \ge 1$$
which is true (using $z \le 2/3$).
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding the maximum value of the sum of the product of sets of 3 variables given their sum Given $x_{1}+x_{2}+...+x_{2014}=2014$, what is the maximum value of: $x_{1}.x_{2}.x_{3}+x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014}$.
It looks like an application of AM-GM inequality on terms obtained after proper rearranging of the given objective function or restriction, but do not find clear what such rearrangement could be.
| $$\begin{align}
\text{Objective} &= x_{1}.x_{2}.x_{3} + x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014} \\
&\leqslant (x_1 + x_4 + \dots+x_{2014})(x_2 + x_5 + \dots+x_{2012})(x_3 + x_6 + \dots+x_{2013}) \\
&\leqslant \left(\frac{x_1 + x_2 + x_3 + \dots + x_{2014}}{3} \right)^3 =\left( \frac{2014}3 \right)^3
\end{align}$$
However this bound is achieved for e.g. when $x_4 = x_5 = x_6 = \frac{2014}3$ and all other $x_i$ are zero, hence this is the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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