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Trouble in mapping of möbius transformation Question:-
Show that the transformation $$ w = \frac{2z+3}{z-4}$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= \frac{4w+3}{w-2} $$
Now substituting the value of $z$ in $(1)$
$$\frac{|3w+1|}{|w-2|} =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3v\iota|=2|u-2+v\iota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated
Thankyou
| If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$\Big|{4w+3 -2w+4\over w-2}\Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-{3\over 4}$ or $$4w+3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| Introduce:
$$1+a=x^2,\ \ 1+b=y^2$$
Obviously:
$$2\le x<y\le3$$
Notice that:
$$4\le y + x \le 6\tag{1}$$
Inequality now becomes:
$$\frac{y^2-x^2}{6}\le y-x\le\frac{y^2-x^2}{4}$$
$$\frac{y+x}{6}\le 1\le\frac{y+x}{4}$$
...which is true because of (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of the base of the prism formed by three planes
The three planes $P_1: kx + y+ z=2$,$P_2:x+y-z=3$,$P_3: x+2z=2$ form a triangular prism and area of the normal section (where the normal section of the triangular prism is the plane parallel to the base of the triangular prism) be $k_1$. Then the value of $2\sqrt{14}(k.k_1)$ is?
Since the three planes form a prism, that means the normal vectors have to be co-planar. So, $[\vec{n_1} \vec {n_2} \vec {n_3}] = 0$ (scalar triple product)
Where, $\vec{n_i}$ is the normal vector of the plane $P_i$
Solving that gave me the value $k=2$
The problem arises in evaluating $k_1$. Assuming that the fourth plane is the base of the prism, how do I find the area of the base?
By taking the cross product of $\vec{n_2}$ and $\vec{n_3}$, we get the vector perpendicular to the base of the prism. The magnitude of this vector may or may not give me the area of the base because it may give me a multiple of the area and not the true area.
Any help would be appreciated.
| The three plane equations are $2x+y+z-2=0, x+y-z-3=0, x+2z-2=0$.
Consider the family of planes passing through the line of intersection of the first two planes.
$2x+y+x-2 + \lambda(x+y-z-3)=0$. The plane part of this family, and parallel to the third plane is:
$x+2z+1=0$.
The perpendicular distance between this plane and the third plane is $\frac{3}{\sqrt{5}}$.
The angle between the first and third plane , and between the second and third plane respectively is $\sin^{-1}(\frac{\sqrt{14}}{\sqrt{30}})$ and $\sin^{-1}(\frac{\sqrt{14}}{\sqrt{15}})$.
So we can easily find two side lengths of the triangle enclosed as $\frac{3\sqrt{30}}{\sqrt{75}}$ and $\frac{3\sqrt{15}}{\sqrt{70}}$.
The angle between these two sides is the angle between the first and second plane is $\sin^{-1}(\frac{\sqrt{14}}{3 \sqrt{2}})$.
So the area of the triangle is $k_1=\frac{1}{2} \times \frac{3\sqrt{15}}{\sqrt{70}} \times \frac{3\sqrt{30}}{\sqrt{70}} \times (\frac{\sqrt{14}}{3 \sqrt{2}})=\frac{9}{2\sqrt{14}}$.
So the final answer is $2\sqrt{14} \times 2 \times \frac{9}{2\sqrt{14}} = \boxed{18}$.
PS: I got the solution after placing the bounty lol :)
| {
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"url": "https://math.stackexchange.com/questions/3233592",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Mathematical induction proof $n^4-1$ is divisible by $16$ for all odd integers $n$ I'm stuck towards the end of proving this, here's my attempt:
$P(3) = 80/16 = 5$, True
$P(k) = k^4-1$
$P(k+1)= (k+1)^4-1$
Expanded $= k^4+4k^3+6k^2+4k+1-1$
This is where I am stuck at.
Sorry for the sloppy formatting im still reading how to properly write formulas on this site.
Thanks in advance for your help!
| Without loss of generality we proved the statement for positive odd integers.
Note that a positive odd integer has the form of $2k-1$ where $k\ge 1$
We proceed with induction on k.
Let $P(k)$ be $(2k-1)^4-1$ is a multiple of $16$
For $k=1$, $(2k-1)^4-1=0$ which is a multiple of $16$ so $P(1)$ is true.
Assume that $(2k-1)^4-1$ is a multiple of $16$
We need to show that $(2k+1)^4-1$ is also a multiple of $16$
Note that the binomial theorem implies $$(2k+1)^4 = [(2k-1)+2]^4 = (2k-1)^4+8(2k-1)^3 +24(2k-1)^2+32(2k-1)$$
Thus $$(2k+1)^4 - (2k-1)^4 =8(2k-1)^3 +24(2k-1)^2+32(2k-1)=$$
$$ 8(2k-1)[(2k-1)^2+3(2k-1)+4)]$$
Note that the expression in square bracket is an even integer, so the result is multiple of $16$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=1.
\end{aligned}
\end{equation}
Please give me some ideas,thank you!
| I have a solution that utilises the digamma function:
\begin{align} F(\frac{1}{n}) &= \sum_{k=n}^\infty \int_{\frac{1}{k+1}}^{\frac{1}{k}} (\frac{1}{t} - k)dt = \\
&= \sum_{k=n}^\infty \int_{k}^{k+1} \frac{s - k}{s^2} ds = \\
&= \sum_{k=n}^\infty \int_{0}^{1} \frac{s}{(s+k)^2} ds = \\
&= \sum_{k=0}^\infty\int_{0}^{1} s \frac{1}{(s+k+n)^2} ds = \\
&= \int_{0}^{1} s \,\psi'(s+n) ds = \\
&= [s\psi(s+n)]_{s=0}^1 - \int_{0}^{1} \psi(s+n) ds = \\
&= \psi(n+1) - [\ln \Gamma(s+n)]_{s=0}^1 = \\
&= \psi(n+1) - \ln \frac{\Gamma(n+1)}{\Gamma (n)} = \\
&= \psi(n+1) - \ln n\end{align}
For large $n$ we have Stirling formula $\ln\Gamma(n+1) = (n+\frac12)\ln n - n + \mathcal O(n^{-1})$, so $\psi (n+1) = \ln n + \frac{1}{2n} + \mathcal O(n^{-2}) $, which gives
$$ \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
We still need to show that $\lim_{x\rightarrow 0} \frac{F(x)}{x} = \frac{1}{2}$ also when we approach by a sequence with $x\neq\frac{1}{n}$. Let $n=\lfloor 1/x\rfloor$, so that $\frac{1}{n+1}<x\le\frac{1}{n}$, that is $n\le\frac{1}{x}<\frac{1}{n+1}$. We have
\begin{align} \left|\frac{F(1/n)}{1/n} - \frac{F(x)}{x}\right| &\le \left|\frac{F(1/n)}{1/n} - \frac{F(1/n)}{x}\right| + \left|\frac{F(1/n)}{x} - \frac{F(x)}{x}\right| \le \\
&\le F(1/n) \left|n-\frac{1}{x}\right| + \frac{1}{n} |F(1/n)-F(x)| \le \\
&\le F(1/n) + \frac{1}{n} \int_{x}^{\frac{1}{n}} \Big(\frac{1}{t}- \lfloor\frac{1}{t}\rfloor\Big) dt \le \\
&\le F(1/n) + \frac{1}{n} \int_{x}^{\frac{1}{n}} 1 \,dt = \\
&\le F(1/n) + \frac{1}{n}(\frac{1}{n}-\frac{1}{n+1}) \rightarrow 0\end{align}
which proves that $$ \lim_{x\rightarrow 0} \frac{F(x)}{x} = \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Cubic root of $3 \times 3$ identity matrix Is there a real $3 \times 3$ matrix $A$, such that: $A^3 = I_3$ and has at most one zero entry? If so, how can I find it?
| Since $A=(a_{ij})_{(3\times3)}$ satisfies $A^3-I=0$, so the characteristic equation of $A$ is $x^3-1=0$
Now $x^3-1=0\implies (x-1)(x^2+x+1)=0\implies\begin{vmatrix}
1-x & 0 & 0 \\
0 & -x & 1 \\
0 & -1 & -1-x
\end{vmatrix}=0$
Construct a matrix $A=\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & -1
\end{pmatrix}$
Clearly $(x-1)(x^2+x+1)=0$ is characteristic equation of $A$ and so it also satisfy $A^3-I=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate a Primitive Polynomial LFSR I tried to search on the internet, to read my course multiple times, but the only thing I see are definitions of the primitive polynomials for an LFSR.
I have an exercise:
Find the primitive polynomial of the LFSR of width 4 with longest possible period.
And I just don't know where to begin from.
By all I read, I know that it is irreducible and that the period will be $2^4-1$, so 15.
How do I find it?
| This is a problem that is small enough that we can do it by brute force. The quartic polynomial must have $x^4$ and $1$ terms in it, and so we need to look only at $8$ quartics. Of these,
$x^4 + 1, x^4+x^2+1$ can be eliminated as obvious squares, while $x^4+x^3+x^2+1$, $x^4+ x^3+x+1$, $x^4+x^2+x+1$ have an even number of terms in them and thus have $1$ as a root (i.e., are divisible by $x+1$). That leaves $x^4+x+1$, $x^4+x^3+1$, and $x^4+x^3+x^2+x+1$ as possible candidates of which we can eliminate $x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1}$ as having period $5$ and not $15$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating sum of converging series $\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$ I have a trouble with calculating the sum of this series:
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$$
I tried to split it into three separate series like this:
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n} =2+\sum_{n=1}^{\infty}\frac{2}{3n+1}+\sum_{n=1}^{\infty}\frac{1}{3n-1}-\sum_{n=1}^{\infty}\frac{1}{n} $$ but I'm not able to continue, can you give me some tips?
| For the direct evaluation of the limit, you have received the good solution from J.G.
You could also consider the partial sum using, as you did, partial fraction decomposition
$$\frac{1-n}{9n^3-n}=\frac{1}{3 n-1}+\frac{2}{3 n+1}-\frac{1}{n}$$ which makes
$$S_p=\sum_{n=1}^{p}\frac{1-n}{9n^3-n}=\frac{1}{3} \left(\psi \left(p+\frac{2}{3}\right)-\psi
\left(\frac{2}{3}\right)\right)+\frac{2}{3} \left(\psi
\left(p+\frac{4}{3}\right)-\psi \left(\frac{4}{3}\right)\right)-H_p$$ where appear the digamma function and harmonic number.
Now, using asymptotics and Taylor series you should arrive to
$$S_p=-\left(\gamma +\frac{1}{3}\psi\left(\frac{2}{3}\right)+\frac{2}{3} \psi
\left(\frac{4}{3}\right)\right)+\frac{1}{9 p}-\frac{1}{9
p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.
Now, the particular values
$$\psi\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$
$$\psi\left(\frac{4}{3}\right)=3-\gamma -\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ make
$$S_p=\left(-2+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\right)+\frac{1}{9
p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$
For $p=10$, the exact value is $-\frac{477820712081}{12033629407800}\approx -0.039707$ while the above truncated expansion gives $-\frac{199}{100}+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\approx -0.039782$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
C)$\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$
D)$\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
E) The limit exists, but it is none of the above
The given answer is D). How does one arrive at this result?
| I’m lazy and prefer not to do tedious matrix inversions and multiplications if I can avoid it. Other answers have explained how to quickly eliminate the given possible solutions based on properties of Markov chains and their associated transition matrices, but one can also reason directly from the eigenvalues of the matrix.
It’s often worth examining a matrix for obvious eigenvectors and eigenvalues, especially in artificial exercises, before plunging into computing and solving the characteristic equation. From the first column of $A$, we see that $(1,0,0)^T$ is an eigenvector with eigenvalue $\frac12$. The rows of $A$ all sum to $1$, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The remaining eigenvalue $\frac12$ can be found by examining the trace.
$A$ is therefore similar to a matrix of the form $J=D+N$, where $D=\operatorname{diag}\left(1,\frac12,\frac12\right)$ and $N$ is nilpotent of order no greater than 2. (If $A$ is diagonalizable, then $N=0$.) $D$ and $N$ commute, so expanding via the Binomial Theorem, $(D+N)^n=D^n+nND^{n-1}$. In the limit, $D^n=\operatorname{diag}(1,0,0)$ and the first column of $N$ is zero, so the second term vanishes. Thus, if $A=PJP^{-1}$, then $\lim_{n\to\infty}A^n=P\operatorname{diag}(1,0,0)P^{-1}$, but the right-hand side is just the projector onto the eigenspace of $1$. Informally, repeatedly multiplying a vector by $A$ leaves that vector’s component in the direction of $(1,1,1)^T$ fixed, while the remainder of the vector eventually dwindles away to nothing.
Since $1$ is a simple eigenvalue, there’s a shortcut for computing this projector that doesn’t require computing the change-of-basis matrix $P$: if $\mathbf u^T$ is a left eigenvector of $1$ and $\mathbf v$ a right eigenvector, then the projector onto the right eigenspace of $1$ is $${\mathbf v\mathbf u^T\over\mathbf u^T\mathbf v}.$$ (This formula is related to the fact that left and right eigenvectors with different eigenvalues are orthogonal.) We already have a right eigenvector, and a left eigenvector is easily found by inspection: the last two columns both sum to $1$, so $(0,1,1)$ is a left eigenvector of $1$. This gives us $$\lim_{n\to\infty}A^n = \frac12\begin{bmatrix}1\\1\\1\end{bmatrix}\begin{bmatrix}0&1&1\end{bmatrix} = \begin{bmatrix}0&\frac12&\frac12\\0&\frac12&\frac12\\0&\frac12&\frac12\end{bmatrix}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x \to 0}[\frac{\ln(x+\sqrt{1+x^2})}{x}]^{1/x^2}$. Let $\ln(x+\sqrt{1+x^2})=:y$，then $x=\dfrac{1}{2}(e^y-e^{-y}).$ Therefore
\begin{align*}
\lim_{x \to 0}\left[\frac{\ln(x+\sqrt{1+x^2})}{x}\right]^{\frac{1}{x^2}}&=\lim_{y \to 0}\left(\frac{2ye^y}{e^{2y}-1}\right)^{\frac{4}{e^{2y}+e^{-2y}-2}}\\
&=\lim_{y \to 0}\left(1+\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\right)^{\frac{e^{2y}-1}{2ye^y-e^{2y}+1}\cdot\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\cdot\frac{4}{e^{2y}+e^{-2y}-2}}\\
&=\exp \lim_{y \to 0}\left(\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\cdot\frac{4}{e^{2y}+e^{-2y}-2}\right)\\
&=\exp \lim_{y \to 0}\left(\frac{2ye^y-e^{2y}+1}{2y}\cdot\frac{4e^{2y}}{e^{4y}-2e^{2y}+1}\right)\\
&=\exp \left(2\lim_{y \to 0}\frac{2ye^y-e^{2y}+1}{ye^{4y}-2ye^{2y}+y}\right)\\
&=\exp \left[2\lim_{y \to 0}\frac{2e^y(y-e^y+1)}{(e^{2y}-1)(e^{2y}(4y+1)-1)}\right]\\
&=\exp \left[2\lim_{y \to 0}\frac{y-e^y+1}{e^{2y}(4y^2+y)-y}\right]\\
&=\exp \left[2\lim_{y \to 0}\frac{1-e^y}{e^{2y}(8y^2+10y+1)-1}\right]\\
&=\exp \left[-2\lim_{y \to 0}\frac{y}{e^{2y}(8y^2+10y+1)-1}\right]\\
&=\exp \left[-2\lim_{y \to 0}\frac{1}{e^{2y}(16y^2+36y+12)}\right]\\
&=\exp\left(-\frac{1}{6}\right).
\end{align*}
Please correct me if I'm wrong! Are there simpler solutions?
| Using $\ln(x + \sqrt{x^2 + 1}) = \sinh^{-1}(x) = x - \frac{x^3}{6} + \frac{3 x^5}{40} + \cdots$ then
$$\ln\left(\frac{\sinh^{-1}(x)}{x}\right) = - \frac{x^2}{6} + \frac{11 x^4}{180} - \frac{191 x^6}{5670} + \cdots$$
which leads to
\begin{align}
\lim_{x \to 0} \left(\frac{\sinh^{-1}(x)}{x}\right)^{1/x^2} &= \lim_{x \to 0} e^{\frac{1}{x^2} \, (\ln(\sinh^{-1}(x)) - \ln(x))} \\
&= \lim_{x \to 0} e^{- \frac{1}{6} + \frac{11 x^2}{180} - \frac{191 x^4}{5670} + \cdots} \\
&= e^{-1/6}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
| These are symmetric equations, so denoting $s=x+y,p=xy$ we have
$$s^2-2p=2p^2$$
$$s(1+p)=4p^2$$
Now eliminate $s$:
$$s^2=2p^2+2p=\left(\frac{4p^2}{1+p}\right)^2$$
$$2p(1+p)^3=16p^4$$
If $p=0$ then $s=0$ and obviously $x=y=0$. Otherwise, divide by $p$:
$$(1+p)^3=(2p)^3$$
Assuming we are only solving in the reals:
$$1+p=2p\qquad p=1,s=2$$
By Viète's formulas, $x$ and $y$ are the roots of $t^2-2t+1$, whereby we get $x=y=1$.
| {
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IMO 1984: Prove that $0 ≤ yz + zx +xy −2xyz ≤ \frac {7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$ I tried to solve this inequality because I find exotic. Actually, I didn't look at the right solution. Because before I look at the right solution, I want to know if my solution is right or not.
Prove that $0 ≤ yz + zx +xy -2xyz≤\frac{7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$
Attempts:
If $x=0$. The left side of the inequality is correct. So, I can accept $x,y,z≠0$.
It is enough to prove $\frac 1x+\frac1y+\frac1z≥2$
We have,
$x+y+z≥3\sqrt[3]{xyz}\Longrightarrow xyz≤\frac1{27}$
$\frac 1x+\frac1y+\frac1z≥\frac3{\sqrt[3]{xyz}}≥9≥2$
The left side proved.
It is obvious at least one of the numbers is less than $\frac 12.$
So, we can choose $y$, such that $y≤\frac 12$.
Therefore, we have
$yz + zx +xy −2xyz ≤ \frac7{27}$
$x(y+z)+yz(1-2x)-\frac7{27}≤0$
$x(1-x)+y(1-x-y)(1-2x)-\frac7{27}≤0$
$x-x^2+y-xy-y^2-2xy+2x^2y+2xy^2-\frac7{27}≤0$
$x^2-x-y+xy+y^2+2xy-2x^2y-2xy^2+\frac{7}{27}≥0$
$x^2(1-2y)+x(3y-2y^2-1)+(y^2-y+\frac7{27})≥0$
$2(\frac12-y)\left(x+\frac{y-1}{2}\right)^2+\frac{1}{108} (3y-1)^2(6y+1)≥0$
Of course, I'm not sure the solution is correct. Can you verify the solution?
Thank you.
| I checked your solution. Your solution is right.
I like the following way.
The homogenization helps.
By AM-GM
$$xy+xz+yz-2xyz=(x+y+z)(xy+xz+yz)-2xyz\geq9xyz-2xyz=7xyz\geq0.$$
Also, $$xy+xz+yz-2xyz\leq\frac{7}{27}$$ it's
$$(xy+xz+yz)(x+y+z)-2xyz\leq\frac{7}{27}(x+y+z)^3$$ or
$$\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\geq0,$$ which is true by Schur and AM-GM.
| {
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"question_score": "3",
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expression for the
length of the hypotenuse (in appropriate units).
Below are the choices
$(a - b)^2$
$\sqrt(a^4 + 4a^2b^2 -b^4)$
$a^2 + b^2$
$\sqrt(a^2 + 2ab -b^2)$
$(a + b)^2$
When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:
$(a^2 - b^2)^2 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$
$a^4 + 2a^2b^2 + b^4 = x^2$
$\sqrt(a^4 + 2a^2b^2 + b^4) = x$
Please help! How to get the correct solution?
| $$a^4+2a^2b^2+b^4 = (a^2+b^2)^2$$
| {
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Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$ By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$
How to find the closed form for general series:
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$
| Note that
$$\sum_{k=1}^\infty \frac{(4x)^n}{n^2{{2n}\choose n}}=2\arcsin^2(\sqrt{x}).$$
Hence, for $p=3$ we have the integral form
$$\sum\limits\limits_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\int_{0}^1\frac{2\arcsin^2(\sqrt{x})}{x}\,dx.$$
and you should be able to recover the result $\pi^2\ln(2)-\frac{7}{2}\zeta(3)$.
As regards the case $p=4$,
$$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=\int_0^1\frac{1}{t}\int_{x=0}^t\frac{2\arcsin^2(\sqrt{x})}{x}\,dx\,dt$$
which, according to ykcaZ's comment below, leads to
$$8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$
that is equal to
$$8\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3}\ln^4(2)+\frac{2\pi^2}{3}\ln^2(2)-\frac{19\pi^4}{360}$$
(see tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ ).
More generally, for $p\geq 2$,
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}
=\frac{(-2)^p}{(p-2)!}\int_0^\frac{\pi}{2} x\ln^{p-2}(\sin x)\,dx.$$
Look through the paper Sums of reciprocals of the central binomial coefficients by R. Sprugnoli for more references. See also On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals
| {
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$
I have not come up with any ideas to solve the problem yet. I will probably in the near future but right now, I can't.
| By rearrangement inequality,
$$\sum_{cyc} \frac{a^2}{b+c} \ge \sum_{cyc} \frac{b^2}{b+c}.\tag{1}$$
So
$$2\sum_{cyc}\frac{a^2}{b+c} \ge \sum_{cyc}\frac{a^2+b^2}{b+c}.\tag{2}$$
Now, by C-S
$$\sum_{cyc}(b+c) \sum_{cyc}\frac{a^2+b^2}{b+c} \ge \left(\sum\sqrt{a^2+b^2}\right)^2.\tag{3}$$
It's also clear that
$$\sum_{cyc}(b+c) \le \sqrt2 \sum_{cyc}\sqrt{a^2+b^2}\tag{4}.$$
From (2), (3), and (4), we conclude that $\sum \frac{a^2}{b+c}$ attains its minimum at $a=b=c$.
| {
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nth root of $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}$ Let $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}, n\geq 1$, then $S_n^{\frac{1}{n}}$ converges to
*
*$e/2$
*$e/4$
*$e/8$
*$0$
Clearly $S_n^{\frac{1}{n}}=\frac{n}{(n+1)^{\frac{1}{n}}(n+2)^{\frac{1}{n}}...(n+n)^{\frac{1}{n}}}$, then how can we proceed
| We have $$(n+1)^n \lt (n+1)(n+2)...(n+n) \lt(2n)^n$$
and therefore
$$\frac12=\frac{n}{2n}\le S_n^{\frac{1}{n}}\le \frac{n}{n+1}\le 1$$
and further
$$\frac12 \le \lim\limits_{n\to 1}S_n^{\frac{1}{n}}\le 1$$
if that limit exists.
Only one of the proposed solution satisfies these inequalities. From $e=2.7\ldots$ follows $2.7<e<2.8$ and therefore
*
*$\frac{2.7}2 < \frac e 2 < \frac{2.8}{2}\implies 1.35 <\frac e2<1.4$
*$\frac{2.7}4 < \frac e 4 < \frac{2.8}{4}\implies 0.675 <\frac e4<0.7$
*$\frac{2.7}8 < \frac e 8 < \frac{2.8}{8}\implies 0.3375 <\frac e8<0.35$
*$0$
So if this limit exists and if it is one of these four numbers then it must be $\frac e 4$.
| {
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"source": "stackexchange",
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Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
| Changing $x=zt, y=zt+u$, on can express
\begin{align}
I&=\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x\\
&=\int_{-1}^1\cos\frac{\pi t}{2}\,dt\int_{z(1-t)}^\infty \exp\left( -\frac{\pi^2}{16} u^2 \right)\,du
\end{align}
It can be integrated by parts,
\begin{align}
I=\frac{2}{\pi}&\left[\sin \frac{\pi}{2}\int_{0}^\infty \exp\left( -\frac{\pi^2}{16}t^2 \right)\,dt \right. \\
-&\sin \left( \frac{-\pi}{2} \right)\int_{2z}^\infty \exp\left( -\frac{z^2\pi^2}{16}t^2 \right)\,dt\\
-&z\left.\int_{-1}^1\sin \frac{\pi t}{2}\exp\left( -\frac{z^2\pi^2}{16}(1-t)^2 \right)\,dt
\right]
\end{align}
or,
\begin{equation}
I=\frac{2}{\pi}\left[\frac{2}{\sqrt{\pi}}\left( 2-\operatorname{erf}\left( \frac{z\pi}{2} \right) \right)-z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds\right]
\end{equation}
For $z\to\infty$, one can evaluate the integral using the Laplace method,
\begin{equation}
z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds\sim \frac{2}{\sqrt{\pi}}-\frac{2}{\sqrt{\pi}z^2}+\frac{1}{\sqrt{\pi}z^4}+O\left( z^{-6} \right)
\end{equation}
while (DLMF)
\begin{equation}
\frac{2}{\sqrt{\pi}}\left( 2-\operatorname{erf}\left( \frac{z\pi}{2} \right) \right)\sim \frac{2}{\sqrt{\pi}}+O\left( z^{-1}e^{-z^2\pi^2/4} \right)
\end{equation}
and thus
\begin{equation}
I\sim \frac{4}{\pi^{3/2}}\frac{1}{z^2}\left[1-\frac{1}{2z^2}+O\left( z^{-4} \right)\right]
\end{equation}
Alternatively, by expressing the $\cos$ in complex and completing the exponent, the integral can be expressed as
\begin{equation}
z\int_{0}^2\cos \frac{\pi s}{2}\exp\left( -\frac{z^2\pi^2}{16}s^2 \right)\,ds=\frac{\exp(-1/z^2)}{\sqrt{\pi}}\left[
\operatorname{erf}\left( \frac{z^2\pi+2i}{2z} \right)+\operatorname{erf}\left( \frac{z^2\pi-2i}{2z} \right)
\right]
\end{equation}
which gives a closed form expression for the integral.
| {
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Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3. Find the value of the sum S(x)=$\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$
I was able to solve 1 and 2 but I found some difficulty in 3. I will write how I solved the first two questions and if possible I would like to know if I have made a mistake.*
1. Using the ratio test with $U_n=\frac{1}{n(n+2)}$
$$\frac{U_{n+1}}{U_n}=\frac{n(n+2)}{(n+1)(n+3)}=\frac{n^2+2n}{n^2+4n+3}$$
$$L=\lim\limits_{n \to \infty}(\frac{n^2+2n}{n^2+4n+3})=\lim\limits_{n\to \infty}(\frac{\frac{1}{n^2}(1+\frac{2}{n})}{\frac{1}{n^2}(1+\frac{4}{n}+\frac{3}{n^2})})=1$$
Radius of converge R=1/L $\rightarrow$ R = 1
2.
$$\frac{1}{1-x}=-\frac{dy}{dx}\ln(1-x)$$
and since it is known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $x \in]-1,1[$ (too lazy to write proof) we can take the integral of $x^n$ to get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
$$-\int_{0}^{x} x^n = -\frac {x^{n+1}}{n+1} \rightarrow -\sum_{n=0}^\infty\frac {x^{n+1}}{n+1}= -\sum_{n=1}^\infty\frac {x^n}{n}$$
3.
$$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
I took
$$U_n=\frac{1}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$$
I got
$$U_n=\frac{1}{n(n+2)}=\frac{1}{2n}+\frac{-1}{2(n+2)}$$
$\therefore$
$$S(x)=\sum_{n=1}^\infty (\frac{1}{2n}+\frac{-1}{2(n+2)})x^n$$
First, I started to solve $\sum_{n=1}^\infty (\frac{1}{2n})x^n$, and by using the proof of question n.2 I got: $$\sum_{n=1}^\infty (\frac{1}{2n})x^n=\frac{1}{2}\sum_{n=1}^\infty (\frac{1}{n})x^n=\frac{1}{2}(-ln(1-x))=ln(\frac{1}{\sqrt{1-x}})$$
However, I was not able to solve the second part: $$\sum_{n=1}^{\infty}\frac{-1}{2(n+2)}x^n$$
How can I find the sum of the second part to find S(x)? I appreciate the help and I would be glad to know if I have made any mistakes too.
| $$-\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$
multiply by $x$
$$-x\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}$$
$$\int_{0}^{x}-x\log(1-x)dx=\int_{0}^{x}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}dx$$
$$\frac{x(x+2)}{4}-\frac{(x^2-1)}{2}\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+2}}{n(n+2)}$$
divide by $x^2$
$$\frac{x(x+2)}{4x^2}-\frac{(x^2-1)}{2x^2}\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+2)}$$
| {
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"source": "stackexchange",
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Find maximum and minimum of $\sin x + \sin y$ I am working on my scholarship exam practice but I am stuck on finding the minimum. Pre-university maths background is assumed.
When $x + y = \frac{2\pi}{3}, x\geq0, y\geq0$, the maximum of
$\sin x+\sin y$ is ....., and the minimum of that is .....
Let me walk you through what I have got.
$\sin x+\sin y = 2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
By substituting $x + y = \frac{2\pi}{3}$ into the sine function, we have
$\sin x+\sin y = 2\sin (\frac{2\pi}{3\cdot2})\cos (\frac{x-y}{2})$
$\sin x+\sin y = \sqrt{3}\cos (\frac{x-y}{2})$
To find the maximum and minimum, we know that
$-1 \leq\cos (\frac{x-y}{2})\leq1$
$-\sqrt{3} \leq\sqrt{3}\cos (\frac{x-y}{2})\leq\sqrt{3}$
Hence, the maximum is $\sqrt{3}$ which is correct and in accordance with the answer key.
However, it seems that the minimum equals to $-\sqrt{3}$ is incorrect. The answer key provided is $\frac {\sqrt{3}}{2}$. Could you please elucidate how I can get to this answer? My guess is something to do with the condition $x\geq0$ and $y\geq0$ given by the question.
| Since $f(x)=\sin{x}$ is a concave function on $\left[0,\frac{2\pi}{3}\right]$ and $\left(\frac{2\pi}{3},0\right)\succ(x,y),$ where $x\geq y,$
by Karamata we obtain:
$$\sin{x}+\sin{y}\geq\sin(x+y)+\sin0=\frac{\sqrt3}{2}.$$
The equality occurs for $x=\frac{2\pi}{3}$ and $y=0$, which says that we got a minimal value.
The maximal value we can get by Jensen:
$$\sin{x}+\sin{y}\leq2\sin\frac{x+y}{2}=\sqrt3,$$
where the equality accurs for $x=y$.
The first inequality we can prove also by the following way.
$$\sin{x}+\sin{y}-\sin(x+y)=\sin{x}(1-\cos{y})+\sin{y}(1-\cos{x})\geq0.$$
| {
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$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Since $\sin\theta$ and $\cos\theta$ are two solutions of the equation,
Then we have,
$\sqrt{2} \sin^2\theta - \sqrt{3} \sin\theta +k=0$ .....Equation (1)
$\sqrt{2} \cos^2\theta - \sqrt{3} \cos\theta +k=0$ .....Equation (2)
Add (2) to (1),
$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
The answer key provided is $\frac{\sqrt{2}}{4}$. I think I am probably on the right track here but not sure how I should proceed with $\sin\theta$ and $\cos\theta$ next. Please help.
| I'll use $y$ in place of $\theta$ . We know that sum of roots of $ax^2+bx+c$ is $\frac{-b}{a}$
$$siny+cosy=\frac{\sqrt{3}}{\sqrt{2}}$$ also $siny+cosy=\sqrt{2}sin(y+\frac{\pi}{4})$ thus $sin(y+\frac{\pi}{4})=\frac{\sqrt{3}}{2}=sin(\frac{\pi}{3})=sin(\frac{2\pi}{3})$ thus $y=\frac{\pi}{12}$ or $\frac{5\pi}{12}$ thus $sin(\frac{\pi}{12})cos(\frac{\pi}{12})=\frac{1}{2}sin(\frac{\pi}{6})=\frac{k}{\sqrt{2}}$ thus $k=\frac{sqrt{2}}{4}$
| {
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Given remainders from other polynomial divisions. Find the remainder in a polynomial division Let $F(x)$ be a polynomial.
If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$
and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.
What is the remainder if $F(x)$ is divided by $x^2(x-1)$?
My solution :
$F(x) = (x-1)^2 P(x) + x+1$, substitute $x=1$ and get $F(1) = 2$.
Differentiate it and get $F'(x) = (x-1)^2 P'(x)+P(x)(2)(x-1)+1$ then $F'(1)=1$.
And do the same to the other equation $( F(0)=3, F'(0) = 2 )$.
Let the remainder of $F(x)$ divided by $x^2(x-1)$ be $R(x)$. $( R(x) = ax^2+bx+c)$
so $F(x) =x^2(x-1)A(x) + R(x)$
$F(0) = R(0) = 3$
$F(1) = R(1) = 2$
$F'(x) = (x)^2(x-1) A'(x)+x^2A(x)+2x(x-1)A(x) + R'(x)$
$F'(0) = R'(0) = 2$
from $R(0) = 3$ then $c=3$
from $R'(0) = 2$ then $R'(x) = 2ax+b$ then $R'(0) = b = 2$
$R(x) = ax^2+2x+3$ from $R(1) = 2$, will get $a+5 = 2 , a=-3$
so the remainder will be $-3x^2+2x+3$
but one said that the answer can be $-3x^2+2x+1$ too, so what's correct answer.
| $\!\!\overbrace{\bmod x\!-\!1\!:}^{\large x\ \equiv\ 1}\,\ \ f\equiv \overbrace{1\!+\!x}^{\Large 2} \equiv \overbrace{2x\!+\!3\! +\! a x^2}^{\Large 2\ +\ 3\ +\ a\ \ \ }\!\iff\! a=-3\!\iff\! f\equiv\overbrace{ 2x\!+\!3\!-\!3x^2}^{\large\rm your\ answer\ \checkmark}\!\!\pmod{\!x^2(x\!-\!1)}$
Update You mention $\!\bmod\!$ is unknown, so we translate the above into more elementary language. $\ f = 2x+3 + ax^2\ $ and $\,1+x\,$ have equal remainders when dividied by $x-1\,$ iff $\,a=-3,\ $ which is easily verified since a polynomial $\, g(x)\, $ has remainder $g(1)$ when divided by $\,x\!-\!1\,$ by the Polynomial Remainder Theorem. Said more simply and explicitly, with full details, we have:
$ f = \color{#0a0}{\,x\!+\!1 + q\cdot (x\!-\!1)^2}\ $ by hypothesis, and also by hypothesis we have
$ f = 2x\!+\!3 + \color{#c00}g\cdot x^2.\, $ Substitute $\,\color{#c00}{g = a + (x\!-\!1)\,h}\ $ (by dividing $\,g\,$ by $x\!-\!1)\,$ to get
$ f = 2x\!+\!3 + \color{#c00}a\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h}\, =\, \color{#0a0}{x\!+\!1 + q\cdot (x\!-\!1)^2} $
Eval at $\,x=1\:\Rightarrow\: 2+3+\color{#c00}{a+0}\: =\: \color{#0a0}{1+1 +0}\ $ so $\ \color{#c00}{a = -3},\, $ which substituted above yields
$ f = 2x\!+\!3 \color{#c00}{-3}\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h},\, $ which agrees with your result.
| {
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How do I solve quadratic double inequalities? I have two questions involving quadratic double inequalities.
Firstly, what are the steps to get the solution for the following?
$0\le(x+2)^2\le4$
My my thought was to separate the inequality into
$0\le(x+2)^2$ and $(x+2)^2\le4$
Which would then allow me to take the square root of each side of the and.
$0\le x+2$ and $0\ge x+2$ and $x+2 \le 2$ and $x+2 \ge -2$
Which could be simplified to $x+2 \le 2$ and $x+2 \ge -2$
And combined into
$-2 \le x+2 \le 2$
Is this the proper way to think about/solve this problem? Is there a better way to approach it?
Secondly, what strategy could I use to square both sides of
$-2\le x+2\le2$ to get back $0\le(x+2)^2\le4$
Thanks!
| For real $x,$
$$(x+2)^2\ge0$$
So, the problem reduces to $$(x+2)^2\le4\iff x(x+4)\le0$$
$\implies$
either
$x\ge0$ and $x+4\le0\iff 0\le x\le-4$ which is impossible
or $x\le0$ and $x+4\ge0\implies -4\le x\le0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Checking my proof of finding a limit I'm trying to find all the $a\in \mathbb R$ values which we get the following limit:
$$L=\lim_{x\to a} \frac{x^2+ax-2a}{(x^2-1)(x+a)}$$
to be finite. I don't see any problem just inserting $a$ into the limit so we get $\frac{1}{a+1}$ so for $a\neq -1$ we get a finite limit. Is it that easy?
| In this case, as long as we don't have $\frac{0}{0}$ or $\frac{\infty}{\infty}$, yes it is that easy and inserting $a$ is enough. And the answer becomes $a \in \mathbb{R}-\{-1\}$.
EDIT: By the warning from fleablood, we should also consider the case where we can have $\frac{0}{0}$. Here, note that in order for nominator to be $0$, we must have $2a^2-2a = 0 \implies a = 0 \lor a = 1$.
Case 1: $a = 0$. In this case, we have
$$\lim_{x \to 0}\frac{x^2}{(x^2-1)x} = \lim_{x \to 0}\frac{x}{(x^2-1)} = 0 \ne \frac{1}{a+1} = 1$$
Case 2: $a = 1$. In this case, we have
$$\lim_{x \to 1}\frac{x^2+x-2}{(x^2-1)(x+1)} = \lim_{x \to 1}\frac{(x+2)(x-1)}{(x^2-1)(x+1)} = \lim_{x \to 1}\frac{(x+2)}{(x+1)^2} = \frac{3}{4} \ne \frac{1}{a+1} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate the sum of a series, whose next value is dependent on the previous value? Consider the below scenario
A -> Amount ; C->Constant; V1 -> Value of first term ; V2 -> Value of 2nd term and so on up to n terms
Now, I need to get the sum of the below series up to n terms. For e.g.
A*C + (A-V1)*C + (A-V2)*C + (A-V3)*C + .... n terms
Where, V1 = A*C ; V2 = (A-V1)*C ; V3 = (A-V2)*C and so on ...
I tried to do it manually on paper, but am getting huge confusing equations. I am sure there must be an easy way around.
| Noting that:
$$
\begin{aligned}
& \frac{V_0}{A\cdot C} = \frac{0}{A \cdot C} = 0 \\
& \frac{V_1}{A\cdot C} = \frac{\left(A - V_0\right)\cdot C}{A \cdot C} = 1 \\
& \frac{V_2}{A\cdot C} = \frac{\left(A - V_1\right)\cdot C}{A \cdot C} = 1 - C \\
& \frac{V_3}{A\cdot C} = \frac{\left(A - V_2\right)\cdot C}{A \cdot C} = 1 - C + C^2\\
& \dots \\
& \frac{V_m}{A\cdot C} = \frac{\left(A - V_{m-1}\right)\cdot C}{A \cdot C} = 1 - C + C^2 + \dots + C^{m-1} \\
\end{aligned}
$$ multiplying both members by $1 + C$, we get:
$$
\begin{aligned}
& \frac{1 + C}{A\cdot C}\,V_0 = 0 \\
& \frac{1 + C}{A\cdot C}\,V_1 = 1 + C \\
& \frac{1 + C}{A\cdot C}\,V_2 = 1 - C^2 \\
& \frac{1 + C}{A\cdot C}\,V_3 = 1 + C^3\\
& \dots \\
& \frac{1 + C}{A\cdot C}\,V_m = 1 - (-C)^m \\
\end{aligned}
$$
ie:
$$V_m = \frac{A \cdot C}{1 + C} \cdot \left(1 - (-C)^m\right).$$
This done, all that remains is to calculate the following summation:
$$\sum_{m = 0}^n V_m = \frac{A \cdot C}{1 + C} \cdot \left(\sum_{m = 0}^n 1 - \sum_{m = 0}^n (-C)^m\right)$$
ie:
$$S_n = \frac{A \cdot C}{1 + C} \cdot \left(1 + n - \frac{1 - (-C)^{1 + n}}{1 + C}\right),$$
which is what is desired.
In particular, fixed $A = 700\,000$ and $C = 0.015$, tabulating $n$ and $S_n$ we obtain:
If the formula of the partial sums of the geometric progression is not remembered:
$${S^*}_n = 1 + q^1 + q^2 + q^3 + \dots + q^n$$
then multiplying both members by $q$, we get:
$$q \cdot {S^*}_n = q^1 + q^2 + q^3 + q^4 + \dots + q^{1+n}$$
and subtracting the two equations member to member, we have:
$$(1 - q) \cdot {S^*}_n = 1 - q^{1+n}$$
ie:
$${S^*}_n \equiv \sum_{m = 0}^n q^m = \frac{1 - q^{1 + n}}{1 - q}\,,$$
as we wanted to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$
Find $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$$
My work
$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\frac{\sum_{m=1}^n\frac{\sqrt m}{m+1}}{\sqrt n}$$
The series
$$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$
does not converge so can I say $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$ does not exist?
|
The series
$$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$
does not converge so can I say $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$ does not exist?
Because Zacky deleted his answer, I'll repeat the useful observation that divergence of a numerator does not mean the fraction necessarily diverges...
As for finding the limit; you can rewrite towards a Riemann sum:
$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{k=1}^{n}\frac{\sqrt k}{(k+1)\sqrt n}=\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}} \tag{$\star$}$$
Now you have an upper bound:
$$(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}}\color{blue}{\le}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k}{n}}}} = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{blue}{2}$$
but also a lower bound:
$$\begin{align}(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}}
=\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt\frac{k^2+2k+1}{{kn}}}
& \color{red}{\ge}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k+3}{n}}}}\\
& =\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{m=4}^{n+3}\frac{1}{{\sqrt {\frac{m}{n}}}}\\[5pt]
& = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{red}{2}\end{align}$$
So we have:
$$\boxed{\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n} = 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Exploring an inequality between $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} $ and $\frac{3}{1+(abc)^{1/3}}$ if $a,b, c>0$ An AM-HM inequality for three positive numbers leads to
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge \frac{3}{{1+\frac{a+b+c}{3}}}. ~~~~(1)$$
Next, the well known AM-GM inequality
$$\frac{a+b+c}{3} \ge (abc)^{1/3} ~~~~(2)$$ becomes incompatible with (1) to
claim any inequality between $$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} ~ \mbox{and} ~ \frac{3}{1+(abc)^{1/3}}.$$
Hence, the question is: Can one establish an inequality between these two expressions?
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}~ \mbox{and}~ \frac{3}{1+(abc)^{1/3}} ~\mbox{if}~ a,b, c>0$$.
| If $f''(x)>0$ in a domain $D$, the Jensen's inequality claims that the mean of functions is greater or equal to the function of mean:
$$\frac{f(x)+f(y)+f(z))}{3} \ge f\left(\frac{x+y+z}{3} \right) , \forall ~x,y,z \in D.~~~~(1)$$ The sign of the inequality reverses if $f''(x) < 0, \forall x \in D$.
Consider $f(x)=\frac{1}{1+e^{x}}$, then $f''(x)=e^{x}\frac{e^{x}-1}{(e^x+1)^2}>0,~ \mbox{if}~ x>0;$
then for three positive numbers $x,y,z$ from (1), we can write
$$\frac{1}{1+e^x}+ \frac{1}{1+e^y}+\frac{1}{1+e^z} \ge \frac{3}{1+e^{(x+y+z)/3}}.~~~~(2)$$
Letting $e^x, e^y, e^z$ equal to $a,b,c \ge 1$, respectively, we establish an interesting inequality from (1), that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge\frac{3}{1+(abc)^{1/3}},~~~ \mbox{if}~ a,b,c \ge 1. ~~~~(3).$$ Next, when $0 < a,b,c \le 1$, the sign of the inequality reverses. For other combinations of $a,b,c>0$; the inequality (3) may or may not hold. Equality holds when $a=b=c=1.$ We have not included $a,b,c=0$ because these are attained at $x,y,z=-\infty$, however, the equality hold also at $a=b=c=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| Let $\sqrt{3x+7}=a,\sqrt{x+2}=b$
$\implies a,b\ge0$ and $a-b=1$
and $a^2-3b^2=1$
Or $(b+1)^2-3b^2=1$
| {
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"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proofs involving strict inequalities let $a,b \in R$
Prove that if $3 \lt a \lt 5$ and $b= 2 + \sqrt{a-2}$ then, $3 \lt b \lt a$
My approach was simply to start with the first inequality and transform it into b and see what happens.
Subtracting 2 gives: $1 \lt a-2 \lt 3$
Taking the square root gives: $1 \lt \sqrt{a-2} \lt \sqrt{3}$
Adding two gives: $3 \lt 2 + \sqrt{a-2} \lt \sqrt{3}+2$
thus $3 \lt b \lt \sqrt{3}+2$
So the interval on b is in fact smaller than the interval on a but that just doesn't seem like it is enough. That's not a very convincing argument. Isn't it plausible that b could equal 3.11 and a could equal 3.10 or something like that. Those numbers fall in the inside the allowed intervals.
Plus this method didn't work on the other 4 questions in the problem.
Suppose $ a \gt 2$ and $b = 1 + \sqrt{a-1}$ using the same technique lands
$b$ as $b \gt 2$
| You are asked to prove that if $3<a<5$ and $b=2+\sqrt{a-2}$ then $3<b<a$.
You proved $3<b$ (and $b<\sqrt3+2$), but you didn't prove $b<a$.
$b<a$ means $\sqrt{a-2}+2<a$ or $\sqrt{a-2}<a-2$.
This suggests letting $x=a-2$ and proving $\sqrt x<x$.
Well, if $x>1$, which is true since $a>3$, then $x^2>x>0$ so $x>\sqrt x$.
That means $a-2>\sqrt{a-2}$ or $b=\sqrt{a-2}+2<a.\quad$ QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the limit $\lim\limits_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$ without using L'Hopital (with my attempts) Help me solve this limit, using simple limit work: $$\lim_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$$
I tried extracting $\frac {e^x}{x}$ but that was dead end, then $\frac {e^x-1}{x}+\frac{1}{x}-\frac{1}{x} \frac{x}{\arctan{x}}$ but I got stuck there.
| Everything is locally a power series, so,
as $x \to 0$,
$\begin{array}\\
\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)
&=\left(\frac {1+x+O(x^2)}{x}-\frac {1}{x-x^3/3+O(x^5)}\right)\\
&=\left(\frac {(1+x+O(x^2))(1-x^2/3+O(x^4))-1}{x(1-x^2/3+O(x^4))}\right)\\
&=\left(\frac {(1+x+O(x^2))-1}{x(1+O(x^2))}\right)\\
&=\left(\frac {x+O(x^2)}{x+O(x^3))}\right)\\
&=1+O(x)\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-substitution:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2+\alpha^2
$$
where $u=x+\frac{b}{2}$ and $\alpha=\frac{1}{2}\sqrt{4c-b^2}$. The book says: "... this is possible because $4c-b^2>0$."
Eagerly I tried an example, using the quadratic $x^2-8x+1$.
Then let $a=1, b=-8, c=1$ and:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2 + \alpha^2
$$
where $u = x+b/2 = x-4$ but we run into a problem:
$$\alpha=\frac{1}{2}\sqrt{4c-b^2} = \frac{1}{2}\sqrt{(4)(1) - (-8)^2} = \frac{1}{2}\sqrt{4-64}$$
So $\alpha$ doesn't satisfy $4c-b^2>0$. Maybe I'm missing something obvious? Or is the book missing a caveat that this method doesn't always work. Because in the book they make it sound like "... this is possible because $4c-b^2>0$." is always true.
| Check your book properly. For them to make the statement
...this is possible because $4c-b^2>0$
must mean that they have said somewhere above that they are considering this method subject to that restriction.
When such discussions are made, they're usually split into cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor a fourth degree polynomial I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:
$$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$
I've tried to see if this equation can be factored by grouping or by long division to no avail. Any help would be greatly appreciated.
| Let
$$f (x) = x^4 + 10 x^3 + 39 x^2 + 70 x + 50$$
Converting to a depressed quartic see here, we see that the $x$ term drops out as well
$$f\left(x-\frac{5}{2}\right)=x^4+\frac{3 x^2}{2}+\frac{25}{16}=\left(x^2+\frac{5}{4}\right)^2-x^2=\left(x^2+\frac{5}{4}+x\right)\left(x^2+\frac{5}{4}-x\right)$$
The factorization of $f$ is obtained by replacing $x$ with $x+\frac{5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions. My attempt: I rewrote it as $2x^2=5y^2+7. 2x^2$ is always even, so in order for the RHS to be even, this means that $5y^2$ must be odd since an odd number plus $7$ is even.
If I evaluate when y is odd, so if $y=2k+1$ for some integer $k$, I get: $2x^2=20k^2+20k+12$. This is the same as $x^2=10k^2+10k+6$, which implies that $x^2$ is congruent $6$ (mod $10$).
Here, I arrive at an issue because if $x=4$, then I get that $x^2$ is congruent to $6$ (mod $10$), but I am supposed to show that the equation does not have a $6$ (mod $10$) congruency.
| We can write the equation in this form too:
$$2(x^2+y^2) = 7(y^2+1)$$
This means that $7 | x^2+y^2$, also its easy to show that $\gcd(x, y) = 1$, becouse if $\gcd(x, y) = d > 1$, then $d | 7$. Now suppose that $d = 7$, then for $x=7a$ and $y=7b$:
$$2x^2 - 5y^2 = 2\times 49 \times a^2 - 5 \times 49 \times b^2 = 49(2a^2 - 5b^2) = 7 $$
Its contradiction. So $d=1$.
But $x^2+y^2 = 7k$ has no solution where $x$ and $y$ are coprime, . So the given diophantine equation has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ratio of area of intersection of two circles and enclosing rectangle Consider two circles with radius $r$ with distance $d < 2r$ from their centres. The area of their intersection is given by:
$A = r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) + r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) -
\frac{1}{2}\sqrt{(-d+2r)(d^2)(d+2r)}$
Now consider the smallest rectangle enclosing this intersection and call its area $B$. I have two questions:
*
*What is $B/A$?
*$B/A = 4/\pi$ when $d=0$. Is this its maximum?
| The area of a circular sector is given by:
$$A_{\text{circular sector}} = \frac{r^2}{2} (\theta - \sin(\theta))$$
The intersection of two circles is just twice this:
$$A_{\text{intersection}} = r^2 (\theta - \sin(\theta))$$
The height of the rectangle can be found using simple trigonometry:
$$h = 2r \sin\left(\frac{\theta}{2}\right)$$
And the width of the rectangle is twice the height of the circle segment.
$$w = 2 r \left(1 - \cos\left(\frac{\theta}{2}\right)\right)$$
$\theta$ itself can be found also using trigonometry.
$$\theta = 2 \arccos\left(\frac{d}{2r}\right)$$
So the ratio between the area of the intersection and its bounding rectangle is:
$$\begin{split}
R &= \frac{4r^2\left(1 - \cos\left(\frac{\theta}{2}\right)\right)\sin\left(\frac{\theta}{2}\right)}{r^2 \left(\theta - \sin\left(\theta\right)\right)} \\
&= \frac{4 \left(1 - \cos\left(\arccos\left(\frac{d}{2r}\right)\right)\right) \sin\left(\arccos\left(\frac{d}{2r}\right)\right)}{2\arccos\left(\frac{d}{2r}\right) - \sin\left(2\arccos\left(\frac{d}{2r}\right)\right)} \\
\end{split}$$
It is true that $\sin( \arccos(x)) = \sqrt{1 - x^2}$ and $\sin(2 \arccos(x)) = 2x \sqrt{1-x^2}$ so:
$$\begin{split}
R &= \frac{4 \left(1 - \frac{d}{2r}\right)\sqrt{1 - {\left(\frac{d}{2r}\right)}^2}}{2 \left(\arccos\left(\frac{d}{2r}\right) - \frac{d}{2r} \sqrt{1-{\left(\frac{d}{2r}\right)}^2} \right)} \\
&= \frac{2\left(1 - \frac{d}{2r}\right)\sqrt{1 - {\left(\frac{d}{2r}\right)}^2}}{\arccos\left(\frac{d}{2r}\right) - \frac{d}{2r} \sqrt{1-{\left(\frac{d}{2r}\right)}^2}} \\
\end{split}$$
We can take $\frac{d}{2r}$ to be $u$, a dimensionless parameter that describes the relationship between the size of the circles and the distance between them.
$$R = \frac{2 \left(1 - u\right)\sqrt{1 - u^2}}{\arccos(u) - u \sqrt{1-u^2}}$$
This function is monotonically increasing on the interval $u \in [0, 1)$ so $R$ has a maximum in the limit when $u = 1$ (and in particular, the complete overlap case of $u = 0$ where $R = \frac{4}{\pi}$ is actually the minimum).
This value can be found via L'Hopital's rule:
$$\begin{split}
\lim_{u \rightarrow 1} \, &\frac{2 \left(1 - u\right)\sqrt{1 - u^2}}{\arccos(u) - u \sqrt{1-u^2}} \\
&=\lim_{u \rightarrow 1} \, \frac{2\frac{2u^2 - u - 1}{\sqrt{1 - u^2}}}{-2 \sqrt{1 - u^2}} \\
&=\lim_{u \rightarrow 1} \, -\frac{2u^2 - u - 1}{1 - u^2} \\
&=\lim_{u \rightarrow 1} \, \frac{(2u + 1)(1-u)}{(1 + u)(1-u)} \\
&=\lim_{u \rightarrow 1} \, \frac{(2u + 1)}{(1 + u)} \\
&= \frac{3}{2}
\end{split}$$
| {
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A finite product : $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$ Find the maximum and minimum of the following products :
$A)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$
$B)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n+1-k})$
My idea is :
$n-1+k>k$ then : $\frac{1}{n-1+k}<\frac{1}{k}$
We obtain :
$\prod_{k=0}^{n-1}(1-\frac{1}{k})$
But I don't have ideas to complete my work , and is my attempt correct ?
| Let us note that for $a,b\in\mathbb N$, $2\le a \le b$ we have $$ \prod_{m=a}^b m = \frac{\prod_{m=1}^b m}{\prod_{m=1}^{a-1} m} = \frac{b!}{(a-1)!}$$
For the product A) we have
$$ \prod_{k=0}^{n-1} \left(1-\frac{1}{n-1+k}\right) = \prod_{k=0}^{n-1}\frac{n-2+k}{n-1+k} = \frac{\prod_{k=0}^{n-1}(n-2+k)}{\prod_{k=0}^{n-1}(n-1+k)} = \frac{\prod_{m=n-2}^{2n-3} m}{\prod_{m=n-1}^{2n-2} m} = \\ = \frac{\frac{(2n-3)!}{(n-3)!}}{\frac{(2n-2)!}{(n-2)!}} = \frac{(n-2)! \cdot (2n-3)!}{(n-1)! \cdot (2n-2)!} = \frac{n-2}{2n-2}$$
For the product B) we have
$$ \prod_{k=0}^{n-1} \left(1-\frac{1}{n+1-k}\right) = \prod_{k=0}^{n-1}\frac{n-k}{n+1-k} = \frac{\prod_{k=0}^{n-1}(n-k)}{\prod_{k=0}^{n-1}(n+1-k)} = \frac{\prod_{m=1}^{n} m}{\prod_{m=2}^{n+1} m} = \\ = \frac{m!}{\frac{(m+1)!}{1!}} = \frac{1! \cdot n!}{(n+1)!} = \frac{1}{n+1}$$
You can use such methods to solve other similar products.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Shortest distance around a pyramid
Transcript:
The diagram shows a square based pyramid with base PQRS and vertex O. All the edges are length 20 meters. Find the shortest distance, in meters, along the outer surface of the pyramid from P to the midpoint of OR.
The only way I have been able to solve this question is using a computer and the paper is non calculator, so there must be a faster, better solution.
My solution is creating a point X on OQ and labelling the midpoint of OR as M, and setting $\theta = OPQ$. I then calculated $PX + PM$ in terms of $\theta$ and found the minimum point of this function, in order to find the shortest possible distance. Finding the minimum point would be nowhere near possible under time constraint without a computer.
I have included the correct answer, so it is the working I am looking for.
Answer:
D $10\sqrt7$
| Draw the black and blue lines as shown below:
$\hspace{3cm}$
The length of blue line (using Cosine theorem):
$$c+d=\sqrt{20^2+y^2-2\cdot 20y\cos 60^\circ}+\sqrt{10^2+y^2-2\cdot 10y\cos 60^\circ}=\\
\sqrt{y^2-20y+400}+\sqrt{y^2-10y+100}$$
Set its derivative to zero:
$$(c+d)'=\frac{y-10}{\sqrt{y^2-20y+400}}+\frac{y-5}{\sqrt{y^2-10y+100}}=0 \Rightarrow \\
(y-10)^2(y^2-10y+100)=(5-y)^2(y^2-20y+400) \Rightarrow \\
3y^2-20y=0 \Rightarrow \\
y=\frac{20}{3}$$
Hence:
$$c+d=\sqrt{(\frac{20}{3})^2-20\cdot \frac{20}{3}+400}+\sqrt{(\frac{20}{3})^2-10\cdot \frac{20}{3}+100}=10\sqrt{7},$$
which is intuitively less since it passes through the two triangular faces while the black line passes through the quadratic base with larger area and the triangular face.
Anyway, to verify it, consider the length of the black line (using the Cosine theorem):
$$a+b=\sqrt{(20\sqrt{2})^2+x^2-2\cdot 20\sqrt{2}y\cos 45^\circ}+\sqrt{10^2+x^2-2\cdot 10x\cos 60^\circ}=\\
\sqrt{x^2-40x+800}+\sqrt{x^2-10x+100}$$
Set its derivative to zero:
$$(a+b)'=\frac{x-20}{\sqrt{x^2-40x+800}}+\frac{x-5}{\sqrt{x^2-10x+100}}=0 \Rightarrow \\
(x-20)^2(x^2-10x+100)=(5-x)^2(x^2-40x+800) \Rightarrow \\
13x^2-40x-800=0 \Rightarrow \\
x=\frac{20}{13}(1+3\sqrt{3})$$
Hence:
$$a+b=\sqrt{x^2-40x+800}+\sqrt{x^2-10x+100}=10\sqrt{7+2\sqrt{3}},$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
} |
Unsure about solution to $\int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n}\:dx$ Spurred on by a question posed on MSE, I was hoping to resolve the following definite integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n}\:dx
\end{equation}
Where $n \in \mathbb{N},\: n\geq 2$. The approach I've taken is valid (I believe), but the final solution seems invalid. Here I used employ Feynman's Trick by introducing the following function:
\begin{equation}
J(p) = \int_0^\frac{\pi}{4} \frac{\ln\left|1 + \tan(x)\right|}{p + \tan(x)}\:dx
\end{equation}
Where $p \in \mathbb{R}$, $ 0\leq p \leq 1$. We observe that:
\begin{equation}
I_n = \frac{(-1)^{n - 1} J^{n - 1}(1)}{(n - 1)!}
\end{equation}
Where $J^m(p)$ is the $m$-th derivative of $J(p)$. To resolve $J(p)$ we first let $u = \tan(x)$ to yield:
\begin{align}
J(p) &= \int_0^1 \frac{\ln\left|1 + u\right|}{\left(u + p\right)\left(u^2 + 1\right)}\:du = \int_0^1 \frac{\ln\left|1 + u\right|}{p^2 + 1}\left[ \frac{1}{u + p} + \frac{p}{u^2 + 1} - \frac{u}{u^2 + 1} \right]\:du \nonumber \\
&=\frac{1}{p^2 + 1}\left[A(p) + pB - C \right]
\end{align}
Now $B$ and $C$ can be resolved without non-elementary functions using the self-similar substitution:
\begin{equation}
B = \int_0^1 \frac{\ln\left| u + 1\right|}{u^2 + 1}\:du = \frac{\pi}{8}\ln(2), \qquad C = \int_0^1 \frac{u}{u^2 + 1}\ln\left| u + 1\right|\:du = \frac{\pi^2}{96} + \frac{\ln^2(2)}{8}
\end{equation}
$A(p)$ can be resolved into the Dilogarithm with some simple substitions and repositioning. We first let $w = u + p$ to yield:
\begin{align}
A(p) &= \int_0^1 \frac{\ln\left|1 + u\right|}{u + p}\:du = \int_p^{p + 1} \frac{\ln\left| 1 + w - p \right|}{w}\:dw = \int_p^{p + 1} \frac{\ln\left|\left(1 - p\right)\left( \frac{w}{1 - p} + 1\right) \right|}{w}\:dw \nonumber \\
&= \ln\left|1 - p\right|\int_p^{p + 1} \frac{1}{w}\:dw + \int_p^{p + 1} \frac{\ln\left|\frac{w}{1 - p} + 1\right|}{w}\:dw \nonumber \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \int_p^{p + 1} \frac{\ln\left|\frac{w}{1 - p} + 1\right|}{w}\:dw
\end{align}
For the final integral, let $u = \frac{w}{1 - p}$:
\begin{align}
A(p) &= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \int_{\frac{p}{1 - p}}^{\frac{1 + p}{1 - p}} \frac{\ln\left|u + 1\right|}{u}\:du \nonumber \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \bigg[ -\operatorname{Li}_{2}(-u)\bigg]_{\frac{p}{1 - p}}^{\frac{ 1 + p}{1 - p}} \nonumber \\
&= \ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \left[ \operatorname{Li}_{2}\left(\frac{p}{p - 1}\right) - \operatorname{Li}_{2}\left(\frac{p + 1}{p - 1}\right) \right]
\end{align}
Thus $J(p)$ becomes:
\begin{align}
J(p) &= \frac{1}{p^2 + 1}\bigg[\ln\left|1 - p\right|\ln\left| \frac{1 + p}{p} \right| + \left[ \operatorname{Li}_{2}\left(\frac{p}{p - 1}\right) - \operatorname{Li}_{2}\left(\frac{p + 1}{p - 1}\right) \right] \nonumber \\
&\quad + \frac{\pi}{8}\ln(2)p- \left(\frac{\pi^2}{96} + \frac{\ln^2(2)}{8}\right) \bigg]
\end{align}
My concern is evaluating this at $p = 1$. Have I fallen prey to an invalid use of the Linearity property of continuous functions? Is my method valid?
| This is not an answer but a different method to find a closed form of ${{I}_{n}}$
For $x={{\tan }^{-1}}\left( u \right)$ we have:
$$
{{I}_{n}}=\int_{0}^{1}{\frac{\ln \left( 1+u \right)}{{{\left( 1+u \right)}^{n}}\left( 1+{{u}^{2}} \right)}du}
$$
Now using this result (thanks to Sangchul Lee) :
$$
\frac{1}{{{(1+x)}^{n}}(1+{{x}^{2}})}=\left( \sum\limits_{k=1}^{n}{\frac{\sin (k\pi /4)}{{{2}^{k/2}}}}\frac{1}{{{(x+1)}^{n+1-k}}} \right)+\frac{\cos (n\pi /4)-x\sin (n\pi /4)}{{{2}^{n/2}}(1+{{x}^{2}})}
$$
Multiply both sides by $\ln \left( 1+x \right)$ and integrate from $0$ to $1$ you get (separate the last term in the sum):
$$
\begin{align}
& {{I}_{n}}=\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{(x+1)}}+\sum\limits_{k=1}^{n-1}{\left[ \frac{\sin (k\pi /4)}{{{2}^{k/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{{{(x+1)}^{n+1-k}}}} \right]} \\
& \quad +\frac{\cos (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{(1+{{x}^{2}})}}-\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\int_{0}^{1}{\frac{x\ln \left( 1+x \right)dx}{(1+{{x}^{2}})}} \\
& \\
\end{align}
$$
at this point we have every thing except:
$$
\int_{0}^{1}{\frac{\ln \left( 1+x \right)dx}{{{(x+1)}^{n+1-k}}}}=\frac{1+{{2}^{k-n}}\left( k-n \right)\ln 2-{{2}^{k-n}}}{{{\left( k-n \right)}^{2}}},\quad k<n$$
Finally
$$
\begin{align}
& {{I}_{n}}={{\ln }^{2}}2\frac{\sin (n\pi /4)}{{{2}^{n/2+1}}}+\sum\nolimits_{k=0}^{n-1}{\left[ \frac{\sin (k\pi /4)}{{{2}^{k/2}}}\frac{1+{{2}^{k-n}}\left( k-n \right)\ln 2-{{2}^{k-n}}}{{{\left( k-n \right)}^{2}}} \right]} \\
& \quad +\pi \ln 2\frac{\cos (n\pi /4)}{{{2}^{n/2+3}}}-\frac{\sin (n\pi /4)}{{{2}^{n/2}}}\left( \frac{{{\pi }^{2}}}{96}+\frac{{{\ln }^{2}}2}{8} \right) \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| The feasible set is a (compact) ellipse $E$ in the $(x,y)$-plane. We have to consider the Lagrangian
$$\Phi:=(x+5)(y+5)-\lambda(x^2+xy+y^2-3)\ .$$
The equations
$$\Phi_x=y+5-\lambda(2x+y)=0,\qquad\Phi_y=x+5-\lambda(x+2y)=0\ ,$$
or
$$\left\{\eqalign{2\lambda x+(\lambda-1)y&=5 \cr (\lambda-1)x+2\lambda y&=5\cr}\right.$$
are symmetric in $x$ and $y$, hence the solution is given by
$$x=y={5\over3\lambda-1}\ .\tag{1}$$
Plugging this into the equation $x^2+xy+y^2=3$ of $E$ gives $(3\lambda-1)^2=25$, or $3\lambda-1=\pm5$. From $(1)$ we then get two conditionally stationary points of $f$ on $E$, namely $(1,1)$ and $(-1,-1)$. From $f(1,1)=36$ and $f(-1,-1)=16$ it then follows that the minimum of $f$ on $E$ is $16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
| Since you got that $x^3+y^3+z^3=1+3xyz$, easy to understand that you meant $x^2+y^2+z^2=1,$ which gives $xy+xz+yz=0.$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=0$ and we obtain: $$0\leq\prod_{cyc}(x-y)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}w^3-w^6\right),$$ which gives
$$-\frac{4}{27}\leq w^3\leq0$$ and we obtain:
$$x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+3w^3\geq1-\frac{4}{9}=\frac{5}{9}.$$
The equality occurs for example for $(x,y,z)=\left(-\frac{1}{3},\frac{2}{3},\frac{2}{3}\right),$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving an integral equation
Question:
I have had a hard time solving and understanding the solution to this problem.
Given Solution:
Questions like why (and how) this step:
Or why $5I_{3}$ = $8I_{1}$
Have been puzzling me. Can anyone please give a clear answer to why and how each step is done. Thanks!
| The point is that, since
$$
\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right] = 4n x^{n-1}\sqrt{4-x^2} - (n+3)x^{n+1}\sqrt{4-x^2},
$$
and the fudamental theorem of calculus tells us that
$$
\int_0^2\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right]dx = 2^n(4-2^2)^{3/2}-0^n(4-2^2)^{3/2} = 0,
$$
we must have that
$$
\int_0^2\left[4n x^{n-1}\sqrt{4-x^2} - (n+3)x^{n+1}\sqrt{4-x^2}\right]dx = 0.
$$
Since integration is linear, the integral can be separated into
$$
4n\int_0^2 x^{n-1}\sqrt{4-x^2}dx - (n+3)\int_0^2 x^{n+1}\sqrt{4-x^2}dx = 0,
$$
that is,
$$
(n+3)\int_0^2 x^{n+1}\sqrt{4-x^2}dx = 4n\int_0^2 x^{n-1}\sqrt{4-x^2}dx,
$$
and so we have $(n+3)I_{n+1} = 4n I_{n-1}$.
$5I_3 = 8I_1$ just comes from setting $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find integers $x$ and $y$ such that $|5^x - 2^y| = 1$.
Find integers $x$ and $y$ such that $$\large |5^x - 2^y| = 1$$
Below is a graph of the equation $|5^x - 2^y| = 1$. As can clearly be seen, $(0, 1)$ and $(1, 2)$ are the solutions. I don't know the other answers, maybe there could be none.
| $$5^x - 2^y = 1$$
If $y=1$, then there is no integer $x$ such that $5^x=3$.
If $y=2$, then $5^x=5$ implies $x=1$.
If $y\ge 3$, then$$5^x\equiv 1\pmod 8$$
So, we see that $x$ has to be even.
Then, we have
$$1-(-1)^y\equiv 1\pmod 3\implies (-1)^y\equiv 0\pmod 3$$
which is impossible.
$$5^x - 2^y = -1$$
If $y=1$, then there is no integer $x$ such that $5^x=1$.
If $y=2$, then there is no integer $x$ such that $5^x=3$.
If $y\ge 3$, then we get $$5^x\equiv 7\pmod 8$$which is impossible.
Therefore, the only solution for $$|5^x-2^y|=1$$
is $(x,y)=(1,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
| First notice that
$$\lim_{x\to 0} \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{2\sin^2\frac{x}2}{x} = \lim_{x\to 0} \left(\frac{\sin^2\frac{x}2}{\frac{x}2}\right)\cdot\sin\frac{x}2 = 1 \cdot 0 = 0$$
so
$$\lim_{x\to 0}\frac{ \cos{x^2}-\cos{x}}{x} = \lim_{x\to 0} \left(\frac{\cos x^2-1}{x^2}\right)x + \lim_{x\to 0} \frac{1-\cos x}{x} = 0$$
Finally we get
$$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} =\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = -\cos 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe that when $n = 2k + 1$ ($k\in \mathbb{Z}, k \geq 0$) that,
\begin{equation}
I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0
\end{equation}
This can be easily shown by noticing that the integrand is odd over the region of integration about $x = \frac{\pi}{4}$. Thus, we need only resolve the cases when $n = 2k$, i.e.
\begin{equation}
I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx
\end{equation}
Here I have isolated two methods.
Method 1:
Let $u = \tan(x)$:
\begin{equation}
I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du
\end{equation}
We note that:
\begin{equation}
\ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}
\end{equation}
By Leibniz's Integral Rule:
\begin{align}
I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\
&= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\
&=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0}
\end{align}
Method 2:
We first observe that:
\begin{align}
\ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)
\end{align}
By the linearity property of proper integrals we observe:
\begin{align}
I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\
& = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0)
\end{align}
Where
\begin{equation}
F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx
\end{equation}
Utilising the same identity given before, this becomes:
\begin{align}
F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\
&= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\
&= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]
\end{align}
Thus,
\begin{equation}
I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)}
\end{equation}
So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?
| $$\begin{split}
I_{2k} &= \int_0^\infty\frac{\ln^{2k}u}{u^2 + 1}du \\
&= \int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_1^{+\infty}\frac{\ln^{2k}\left(u\right)}{u^2 + 1}du \\
&=\int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_0^{1}\frac{\ln^{2k}t}{t^2 + 1}dt \,\,\,\left(\text{by } u\rightarrow \frac
1 t\right)\\
&=2\int_0^1\frac{\ln^{2k}u}{u^2 + 1}du\\
&=2\sum_{n\in\mathbb N}(-1)^n\int_0^1u^{2n}\ln^{2k} (u) du
\end{split}$$
Now, let $$J_{p,q}=\int_0^1u^p\ln^q (u)du$$
If $q\geq 1$, by integration by parts,
$$J_{p, q}=\left. \frac{u^{p+1}}{p+1}\ln^q u\right]_0^1-\int_0^1\frac{u^{p+1}}{p+1}q\ln^{q-1}(u)\frac{du}u=-\frac q{p+1}J_{p,q-1}$$
Consequently, if $q\geq 1$, $$J_{p,q} = (-1)^{q}\frac{q!}{(p+1)^{q+1}}$$
We conclude that
$$I_{2k}=2\cdot (2k)!\sum_{n\in\mathbb N}\frac{(-1)^n}{(2n+1)^{2k+1}}$$
Following Zachy's suggestion, the last sum is known as the Dirichlet Beta function
$$I_{2k}=2\cdot (2k)!\beta(2k+1)$$
Finally, values of $\beta$ at odd numbers are known in terms of Euler's numbers and we get
$$\boxed{I_{2k}=2\frac{(-1)^kE_{2k}\pi^{2k+1}}{4^{k+1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Given three positive numbers $a,b,c\in R_{+}^{*}$ prove the following inequality
If $\ abc=1$ then prove that
$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤1$$
where $a,b,c>0$
I think this inequality can be proved by holder ?
My attempt using $\ am-gm$
$$3-a+a^{6}≥3-a \quad(etc)$$
$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤\displaystyle\sum_{cyc}\frac{1}{3-a}$$
Now I will get tow case if $a>3$ and if $a<3$.
If $a>3,$ the inequality is true.
Now if $a<3$
Using : $3-a>0$ (etc)
So: $$\sum_{cyc}\frac{1}{3-a}≤1$$
Is my work correct?
| $$\sum_{cyc}\frac{1}{a^6-a+3}\leq1$$ it's
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}-\frac{5}{9}\ln{a}\right)\geq0.$$
Let $f(x)=\frac{1}{3}-\frac{1}{x^6-x+3}-\frac{5}{9}\ln{x}.$
Thus, $$f'(x)=\tfrac{(1-x)(5x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45)}{9x(x^6-x+3)^2}.$$
We see that the polynomial $x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45$
has only one changing of coefficients sign, which by the Descartes's rule https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
says that this polynomial has unique positive root $x_1$ and easy to see that $x_1>1$.
For $x=1$ our $f$ has a local minimum and for $x=x_1$ our $f$ has a local maximum,
$f(x_1)>0$ and $f$ decreases on $[x_1,+\infty).$
Since $\lim\limits_{x\rightarrow+\infty}f(x)=-\infty,$ we obtain that $f$ has an unique root $x_0$ on $[x_1,+\infty)$.
By calculator easy to see that $x_0=1.696...$ and since $f(1)=0$,
we got that our inequality is proven for $\max\{a,b,c\}\leq1.696.$
Now, let $a>1.696.$
Thus, by AM-GM $$\sum_{cyc}\frac{1}{a^6-a+3}<\frac{1}{1.696^6-1.696+3}+\frac{2}{3-\frac{5}{6\sqrt[5]6}}=0.867...<1$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$.
Given positive $a, b, c$, prove that $$\large (a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$
As a starting point,
$$(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$
$$\iff \frac{a^2 + b^2 + c^2}{ab + bc + ca} \ge \frac{(a^3 + b^3 + c^3)(a + b + c)}{(a^2 + b^2 + c^2)^2}$$
$$\iff \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2(ab + bc + ca)} \ge \frac{ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2}{(a^2 + b^2 + c^2)^2} \tag 1$$
$$\iff \sum_{cyc}(a - b)^2\left[\frac{1}{2(ab + bc + ca)} - \frac{ab}{(a^2 + b^2 + c^2)^2}\right] \ge 0$$
$$\iff \frac{1}{2} \cdot \sum_{cyc}\frac{[a^4 + b^4 + c^4 + 2(ca)^2 + 2(bc)^2 - 2a^2bc - 2b^2ca](a - b)^2}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$
$$\iff \frac{\displaystyle (a^4 + b^4 + c^4) \cdot \sum_{cyc}(a - b)^2 + 2\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$
Please note for $(1)$ that $$(a^2 + b^2 + c^2) - (ab + bc + ca) = \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2}$$ and $$(a^3 + b^3 + c^3)(a + b + c) - (a^2 + b^2 + c^2)^2 = ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2$$
Moreover, evaluating $\displaystyle \sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]$, we have that $$\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2] = \sum_{cyc}ca(ca - b^2)(c^2 + a^2 + 2b^2 - 2bc - 2ab)$$
$$ = \sum_{cyc}[(ca)^2 - b^2ca](c^2 + a^2) + 2bca \cdot \sum_{cyc}(ca - b^2)(b - c - a)$$
$$ = \sum_{cyc}b^2[(ab)^2 - c^2ab + (bc)^2 - a^2bc] - 2bca \cdot \sum_{cyc}b[(ab - c^2) + (bc - a^2) - (ca - b^2)]$$
I'm going to stop here. There was an attempt, one that failed tremendously.
| Your SOS works!
Indeed, by your work we obtain:
$$(a^2+b^2+c^2)^3-(a^3+b^3+c^3)(ab+ac+bc)(a+b+c)=$$
$$=(a^2+b^2+c^2)^2(ab+ac+bc)\left(\frac{a^2+b^2+c^2}{ab+ac+bc}-\frac{(a^3+b^3+c^3)(a+b+c)}{(a^2+b^2+c^2)^2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)^2-2ab(ab+ac+bc))\geq$$
$$\geq\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)(ab+ac+bc)-2ab(ab+ac+bc))=$$
$$=\frac{1}{2}(ab+ac+bc)\sum_{cyc}(a-b)^2((a-b)^2+c^2)\geq0.$$
Also, we can use $uvw$, which gives a solution immediately.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $a^2+b^2+c^2$, $ab+ac+bc$ and $a+b+c$ dot't depend on $w^3$ and since $$a^3+b^3+c^3=27u^3-27uv^2+3w^3,$$ it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Since, our inequality is homogeneous, it's enough to assume $b=c=1$, which gives
$$(a-1)^2(a^4-2a+4)\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of an Infinite Sequence
If $$ S = \frac{1}{1\cdot3\cdot5} + \frac{1}{3\cdot 5 \cdot 7} +
\frac{1}{5\cdot 7 \cdot 9} \cdots $$
$$S =\, ? $$
My Attempt
Let the general term be $ a_n $.
Then, $$ a_n = \frac{1}{(2n-1)(2n+1)(2n+3)} $$
$$ a_n = \frac{1}{4} \left[\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)} \right]$$
$$\sum_{n=1}^{∞}a_n = \frac{1}{4} \left[\frac{1}{1\cdot3} - \frac{1}{(2n+1)(2n+3)}\right]$$
As n $\to$ ∞,
$$ \sum_{n=1}^{∞}a_n = \frac{1}{12}$$
Is this the correct?
| You are right but more methodically:
For the series $$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Let $$V_n=\frac{1}{(2n-1)(2n+1)},$$ then $$T_n=\frac{1}{4}[V_{n}-V_{n+1}].$$ By telescopic summing we get $$S_N=\sum_{n=1}^{N}=\frac{1}{4} [V_1-V_{N+1}]=\frac{1}{4} \left[\frac{1}{1.3}-\frac{1}{(2N+1)(2N+3)}\right].$$ Hence $$S_{\infty}=\frac{1}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ dividing by $x^2-1$ have remainder $2x$ I need help in this problem.
Problem: Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ divided by $x^2-1$ has remainder $2x$ and suppose $f(x)=x^3+ax^2+bx+c$.
| Let $f(x) = ax^3 + bx^2 + cx + d$
and also $f(x) = (px+q)(x^2-1) + 2x$
Comparing the coefficients of $f(x)$, we have
$a = p$, $b = q$, $c = 2 - p$, $d = -q$
Also $f(2) = 0 \implies 8a + 4b + 2c + d = 0$
Substituting the values of $a, b, c$ in this equation and simplifying
$6p + 3q + 4 = 0$
You can find infinite number of solutions to the above equation.
For example, if you choose $p = 1$ then $q = -\frac{10}{3}$
and $f(x) = x^3 -\frac{10}{3}x^2 + x + \frac{10}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find modulus and argument of $\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}}$ A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.
Given that:
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}} $$
To prove:
$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = \theta $$
My work so far;
Proof:
$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$
I'm not sure about the above, if there's an easy explanation or method to go through would be great as questions like these are giving me a bit of trouble.
| Note that\begin{align}\bigl(1-\cos(4\theta)\bigr)^2+\sin^2(4\theta)&=2-2\cos(4\theta)\\&=2-2\bigl(1-2\sin^2(2\theta)\bigr)\\&=4\sin^2(2\theta)\\&=16\sin^2(\theta)\cos^2(\theta)\end{align}and that\begin{align}\sin^2(2\theta)+4\cos^4(\theta)&=4\sin^2(\theta)\cos^2(\theta)+4\bigl(1-\sin^2(\theta)\bigr)^2\\&=4\sin^2(\theta)-4\sin^4(\theta)+4-8\sin^2(\theta)+4\sin^4(\theta)\\&=4\bigl(1-\sin^2(\theta)\bigr)\\&=4\cos^2(\theta)\end{align}and that therefore$$\lvert\zeta\rvert^2=\frac{\bigl(1-\cos(4\theta)\bigr)^2+\sin^2(4\theta)}{\sin^2(2\theta)+4\cos^4(\theta)}=4\sin^2(\theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding a closed form for recurrence relations $a_n=na_{n-1}+1$ and $a_n=na_{n-1}+n$ Consider the sequence defined by
$$
\begin{cases}
a_0=1\\
a_n=n\cdot a_{n-1}+1 & \text{if }n\ge 1
\end{cases}
$$
Find a closed form for $a_n$.
Second case is as follows:
$$
\begin{cases}
a_0=1\\
a_n=n\cdot a_{n-1}+n & \text{if }n\ge 1
\end{cases}
$$
Find a closed form for $a_n$.
| Try exponential generating functions (and another link here). For example, let's look at the first recurrence
$$f(x)=\sum\limits_{n=0}\color{red}{a_n}\frac{x^n}{n!}=
1+\sum\limits_{n=1}a_n\frac{x^n}{n!}=\\
1+\sum\limits_{n=1}(n\cdot a_{n-1} + 1)\frac{x^n}{n!}=
1+\sum\limits_{n=1}a_{n-1} \frac{x^n}{(n-1)!} + \sum\limits_{n=1}\frac{x^n}{n!}=\\
x\left(\sum\limits_{n=1}a_{n-1} \frac{x^{n-1}}{(n-1)!}\right) + \sum\limits_{n=0}\frac{x^n}{n!}=\\
x\left(\sum\limits_{n=0}a_{n} \frac{x^{n}}{n!}\right) + \sum\limits_{n=0}\frac{x^n}{n!}=\\
x\cdot f(x) + e^x$$
or (applying series multiplication, below the 2nd defition here)
$$f(x)=\frac{e^x}{1-x}=\left(\sum\limits_{n=0}\frac{x^n}{n!}\right)\left(\sum\limits_{n=0}x^n\right)=\\
\sum\limits_{n=0}\left(\sum\limits_{i=0}^n\frac{1}{i!}\cdot 1\right)x^n=
\sum\limits_{n=0}\color{red}{\left(\sum\limits_{i=0}^n\frac{n!}{i!}\right)}\frac{x^n}{n!}$$
and finally
$$a_n=\sum\limits_{i=0}^n\frac{n!}{i!}=
n!\left(\sum\limits_{i=0}^n\frac{1}{i!}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve inequality Given positive numbers $a,b,c$ satisfying $a^2+b^2+c^2=1$, prove the following inequality
$$\frac{a}{\sqrt{1-bc}} + \frac{b}{\sqrt{1-ac}}+\frac{c}{\sqrt{1-ab}}\le\frac{3}{\sqrt{2}}$$
Thanks
I have tried using CS, try to make use of $a+b+c\leq\sqrt3$, $abc\leq\frac{1}{3\sqrt3}$, but got nowhere –
| By C-S
$$\sum_{cyc}\frac{a}{\sqrt{1-bc}}\leq\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{1-bc}}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{a}{1-bc}\leq\frac{9}{2(a+b+c)},$$ which is true by SOS:
$$\frac{9}{2(a+b+c)}-\sum_{cyc}\frac{a}{1-bc}=\sum_{cyc}\left(\frac{3}{2(a+b+c)}-\frac{a}{1-bc}\right)=$$
$$=\frac{1}{2(a+b+c)}\sum_{cyc}\frac{3(a^2+b^2+c^2-bc)-2a(a+b+c)}{1-bc}=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{2a^2+6b^2+6c^2-6bc-4ab-4ac}{1-bc}=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(c-a)(6c-a-3b)-(a-b)(6b-a-3c)}{1-bc}=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}(a-b)\left(\frac{6a-b-3c}{1-ac}-\frac{6b-a-3c}{1-bc}\right)=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2(7-c(a+b)-3c^2)}{(1-ac)(1-bc)}=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2(4c^2-c(a+b)+7(a^2+b^2))}{(1-ac)(1-bc)}=$$
$$=\frac{1}{4(a+b+c)}\sum_{cyc}\frac{(a-b)^2\left(\left(2c-\frac{a+b}{4}\right)^2+7(a^2+b^2)-\frac{(a+b)^2}{16}\right)}{(1-ac)(1-bc)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of trees $T = (V,E) \;$ such that $V = \left \{ 1,2,3,4,5,6,7 \right \} \;$, $\deg(1) = 2$ , $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ I tried summing all of the cases (I found 4), but somehow i'm getting the wrong answer.
For ex. the number of trees on 7 vertices such that the following sequence is the sequence of their degrees:
My solution (wrong) using Cayley's formula:
The only possible degree sequences are:
$(2,1,1,2,2,2,2) = \binom{5}{1}*\binom{6}{4}$
$(2,3,2,2,1,1,1) = \binom{5}{2}*6\binom{5}{2}$
$(2,3,3,1,1,1,1) = \binom{5}{2,2}*\binom{6}{2}$
$(2,4,2,1,1,1,1) = \binom{5}{3}*6*4$
What am I missing here?
| The unlabelled trees with $7$ vertices are
We have to rule out the last three which have a vertex of degree $\geq 5$ or no vertex of degree $2$.
So for each of the $8$ cases we compute the number of labelled trees such that
$\deg(1) = 2$ and $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ and we add them all together:
$$5\cdot \frac{6!}{2}+3\cdot\frac{6!}{2}+3\cdot 6!+3\cdot\frac{6!}{3!}+1\cdot \frac{6!}{2^3}+1\cdot\frac{6!}{2}+2\cdot\frac{ 6!}{3!}+2\cdot\frac{6!}{2^2}=6450.$$
Each term is obtained as the number of vertices of degree $2$ (positions for the vertex $1$) multiplied by $6!$ (for arranging the remaining $6$ vertices) and finally divided by the number of symmetries.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$\omega = \frac{1 + \sqrt3 i}{2}$ , $ \omega^5 = ? $
$\omega = \frac{1 + \sqrt3 i}{2} $, $ \omega^5 = ? $
$\omega^3 = 1$ by definition?
So, $\omega^5 = \omega^2$
But why do i get wrong answer?
| Not by definition, just check that
$$\omega=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i=(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}))=e^{i\frac{\pi}{3}}$$
If you make $\omega^3$, then you have $e^{i\frac{\pi}{3}\cdot 3}=e^{i\pi}=-1$.
So $\omega^5=\omega^3\omega^2=-\omega^2$ is your mistake
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} = \frac{4}{3}$$
Then the minimum value is $$24 \times \frac{5}{4} \times \frac{5}{3} = 50$$
Is there any mistake with the solution ?
| Your mistake in the following.
You proved that $$\frac{9}{\cos^2x}+\frac{16}{\sin^2x}\geq\frac{24}{|\sin{x}\cos{x}|}=\frac{48}{|\sin2x|},$$
but you did not find a minimal value.
After your first step we see that $$\frac{48}{|\sin2x|}\geq48,$$ but it does not give a minimal value because the value $48$ does not occur.
One of right solutions is the following.
By C-S
$$\frac{9}{\cos^2x}+\frac{16}{\sin^2x}=(\cos^2x+\sin^2x)\left(\frac{9}{\cos^2x}+\frac{16}{\sin^2x}\right)\geq(3+4)^2=49.$$
The equality occurs for $(\cos{x},\sin{x})||\left(\frac{3}{\cos{x}},\frac{4}{\sin{x}}\right),$ which says that we got a minimal value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a function with given conditions I want to find the equation of a function $f(x)$ such that the following conditions are met:
$(i)\ f(3) = 0, f(0) =1$.
$(ii)\ f(x)$ is an even function.
$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.
$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.
Now since the function is even, I get $x=-3$ as another root and then I can graph the function. But how do I find its equation? Any hints? Thank you.
In general is there some method or software to find the equation of a function given its graph?
| I am assuming you want to find a rather than all functions that satisfy that property. You can approach this with a sort of an ad hoc method.
First lets find some function $f(x)$ that satisfies $(i)$. This should be easy enough, the polynomial $x(x-3)$ does the job. (This i incorrect as I misread the condition to be $f(0) = 0$, this is fixed in the edit below)
Now for $(ii)$, we want $f(x) = f(-x)$. Notice squaring a number makes its sign redundant so we try incorporating that into our function to get $x^2(x^2-9)$. (always check back to see you havent lost $(i)$)
$(iii)$ requires us to divide by $0$ when $x = \pm 4$. So let $$f(x) = \frac{x^2(x^2-9)}{(x-4)(x+4)} = \frac{x^2(x^2-9)}{(x^2-16)}$$
Notice how this still has properties $(i)(ii)$
Finally, we need the horizontal asymptotic condition. Notice right now that $\lim_{x \rightarrow \infty} f(x) = \infty$ since the numerator is of degree $4$ while the denominator is of degree $2$. We can first let $f(x)$ converge when $x$ goes to infinity simply by squaring the denominator. This results in
$$f(x) = \frac{x^2(x^2-9)}{(x^2-16)^2}$$
This gives us a horizontal asymptote of $y=1$. Multiplying our function by $2$ then fixes this
$$f(x) = 2\frac{x^2(x^2-9)}{(x^2-16)^2}$$
we are done.
EDIT (Sorry for the error)
we remove the $x^2$ and the square on the denominator to get $\frac{2(x^2-9)}{x^2 - 16}$ Notice at $x = 0$ this evaluates to $9/8$. We want to add a term that disappears at $x = \pm 3$ and as $x \rightarrow \infty $ and evaluates to $-1/8$ when $x = 0$. Following similar line of reasoning as above we get $\frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2}$ so our function becomes
$$f(x) = \frac{2(x^2-9)}{x^2 - 16} + \frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2} = (\frac{x^2-9}{x^2 -16})(2+ \frac{32}{9(x^2 -16)})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximizing area of rectangle inscribed in circle sector of radius 2 Question:
A rectangle is inscribed in a circle sector. The top two corners of the rectangle lies on the radius of the circle sector and the bottom two corners lie on the arc of the circle sector.
The radius is 2 and angle is $\frac{2\pi}{3}$.
Find the maximum area of the rectangle that has these properties and are located exactly like this using calculus-based optimization.
Attempted solution:
Let the shorter side of the rectangle be labelled x and the longer be labelled y. The distance from the center of the circle to the top of the rectangle is b.
Now, create a dashed-line triangle from the center of the circle to the bottom right of the rectangle. Call the top angle $\alpha$.
From this, we can get a value for $\frac{y}{2}$, since this is the side opposite the angle:
$$\sin \alpha = \frac{\frac{y}{2}}{2} \Rightarrow y = 4 \sin \alpha$$
Using the same triangle but with the adjacent side x + b:
$$\cos \alpha = \frac{x+b}{2} \Rightarrow x = 2 \cos \alpha - b$$
Then we use half of the top triangle above the rectangle to get a value for b in terms of y:
$$\tan \frac{2\pi}{6} = \frac{\frac{y}{2}}{b} \Rightarrow b = \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}}$$
Combining these to get a value for x:
$$x = 2 \cos \alpha - b = 2 \cos \alpha - \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}} = 2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}}$$
The area becomes:
$$A(\alpha) = yx = 4 \sin \alpha (2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}})$$
Taking the derivative:
$$A'(\alpha) = 8(-sin^2 \alpha + cos^2 \alpha - 2 \sin \alpha \cos \alpha)$$
Setting to zero:
$$A'(\alpha) = 0 \Rightarrow a = \frac{\pi}{2} - \frac{3\pi}{8}$$
Putting this into the formula for A:
$$A( \frac{\pi}{2} - \frac{3\pi}{8}) = 4(\sqrt{2} -1)$$
However, this does not match the expected answer of $\frac{4}{\sqrt{3}}$.
What went wrong? What are some productive ways to finish this off?
| I got $$A(\alpha)=\frac{\sin(2\alpha)}{2}-\frac{\sin^2(\alpha)}{\tan(60^{\circ})}$$ so $$A'(\alpha)=8\left(\cos(2\alpha)-\frac{\sin(2\alpha)}{\tan(60^{\circ})}\right)$$
and
$$A'(\alpha)=0$$ if $$\tan(2\alpha)=\tan(60^{\circ})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given the sum and the product of two variables; what's the sum of their reciprocals? I'm sorry if it's a simple problem, but I need an explanation..
If
$$ x + y = 2 $$
$$ xy = 3$$
Then:
$${1\over x} + {1\over y} = z $$
$${z\in \{ \Bbb R}\}$$
My attempt was as simple like that:
$$x=2-y$$
$$(2-y)y=3$$
$$y^2 -2y + 3 = 0$$
$$(y^2 -2y +1) +2 = 0$$
$$(y-1)^2 = -2$$
$$(y-1) = \pm \sqrt{-2} = \pm i\sqrt{2}$$
$$y= 1\pm i\sqrt{2}$$
$$x=2-(1\pm i\sqrt{2})$$
$$x= 1\pm i\sqrt{2}$$
Since $$x = y$$ Therefore:
$${1\over x} + {1\over y} = {2\over 1\pm i\sqrt{2}}$$
My question is about how to proceed after? or maybe I did a mistake before? How can I reach the final real solution?
Thank you.
| Your approach is correct up to where you solved for the values of $y$ and $x$. The sum $\frac{1}{x} + \frac{1}{y}$ will always be equal to $\frac{1}{1 - i \sqrt 2} + \frac{1}{1 + i \sqrt 2}$. Cross multiplying gives you $\frac{1 + i \sqrt 2 + 1 - i \sqrt 2}{(1 - i \sqrt 2)(1 + i \sqrt 2)} = \frac{2}{1 + 2} = \frac{2}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number theory congruence
Prove that $n^{12} - a^{12}$ is divisible by $91$ if $n$ and $a$ are co prime to $91$
What I tried:
*
*$91$ divides $n^{12}-a^{12}$
*Therefore $n^{12}-a^{12}=91k$
*But $91=13\times7$
*$n^{12}-a^{12}=(13\times7)k$
*$n^{12}=(13\times7)k+a^{12}$
*$n^{12}$ divides $91$
*Therefore $n$ is co prime to $91$
| Two things should leap to mind. And a third thing comes with practice.
1) $91 = 13*7$ so to prove $91|M$ it will work to prove a) $7|M$ and b) $13|M$. If you can prove a) and b) you are done.
2) $n^{12} - a^{12} = (n^6 - a^6)(n^6 + a^6) = (n^3 - a^3)(n^3 + a^3)(n^6 + a^6)=....$. The expression $n^{12}-a^{12}$ is "very" factorable. If we can prove that of the many factors that $7$ and/or $13$ must divide at least one of the factors that will help. There are so many factors something working out is likely.
And with practice:
3) $12 = 13 -1$ and $13$ is prime. So Euler's Theorem/Fermats Little Theorem will probably come into play.
.....
Factor $n^{12} -a^{12}$ completely.
$n^{12}-a^{12} = (n-a)(n^11 + n^10a + .... +na^{11} + a^{12})=$
$(n^2 - a^2)(n^{10} + n^8a^2 + .... + n^2a^8 + a^{10})=$
$(n^3 - a^3)(n^9 + n^6a^3 + n^3a^6 + a^9)=$
$(n^4-a^4)(n^8 + n^4a^4 + a^8)=$
$(n^6 - a^6)(n^6 + a^6) =$
$(n^6+a^6)(n^3 + a^3)(n^2+a^2)(n+a)(n-a)$.
Those probably have different remainders when divided by $7$ or $13$. One of them can probably be proven to be $0$. but... that may be hard to figure out. Let's put a pin in it and look at 3)
3) Fermat's little theorem states if $\gcd(k,13)=1$ then $k^{12}\equiv 1 \pmod {13}$. And $n,a$ are coprime to $91$ so they are coprime to $13$ so
$a^{12} \equiv 1 \pmod {13}$ and $n^{12} \equiv 1 \pmod {13}$ so $n^{12}-a^{12}\equiv 1-1 \equiv 0 \pmod 13$ so $13|(n^{12} - a^{12})$.
So now we just need to prove $7|n^{12} - a^{12}$....
Back to the factoring... oh, wait!....
Fermat's Little Theorem states $n^6\equiv 1\pmod 7$ and $a^6\equiv \pmod 7$ so $n^{12}-a^{12} = (n^6- a^6)(n^6 + a^6)\equiv (1-1)(n^6 + a^6)\equiv 0 \pmod 7$ so $7|n^{12}-a^{12}$.
And we are done!
We didn't need to do that factoring stuff at all.
Still it'd be interesting to see if we can do it by considering just the factors. But we don't have to.
| {
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Value of a and P(x) when P(x) is a rational number when satisfies a certain equation This is the question as I still don't have permission to post picture, but it is basically when P(x) = x^3+x^2+ax+1, when a is a rational number, P(X) is also rational number for every x that satisfy x^2+2x-2=0
Consider the integral expression in $x$ $$ P= x^3 + x^2 + ax +1 $$ where $a$ is a rational number. At $a=a_0$ the value of $P$ is a rational number for any $x$ which satisfies the equation $x^2 + 2x -2 = 0$, and n this case the value of $P$ is $P_0$.
Find the values of $a_0$ and $P_0$
| Just another way without explicit root finding: $x^2=2-2x\implies x^3=2x-2x^2=2x-2(2-2x)=6x-4$, so in this case, $P(x)=(6x-4)+(2-2x)+ax+1=(a+4)x-1$.
As $x^2+2x-2$ has no rational roots ($\pm1,\pm2$ are not roots); to have $P$ rational, we must have $a_0=-4 \implies P_0=-1$
| {
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Prove that $\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ I have to prove that
$$
I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
$$
I know that
$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{8}\log(2)
$$
and I can write
$$
I= \int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx+\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$
So I have only to evaluate $$I_{-}=\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$
Using the expansion of the logarithm valid in the range of integration I can also write
$$I_{-}=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x^2}dx
$$
but then I can go any further because I can't find a simple solution that depend only on $n$ for the rational (and solvable) integral under the sum sign .
Maybe is simple to considerate instead of $ I_{-} $ the integral $\int_{0}^{1}\frac{\log(1-x^2)}{1+x^2}dx=I$ because it seems to me that $\int_{0}^{1}\frac{x^{2n}}{1+x^2}dx= \sum_{k=0}^{n-1} (\frac{(-1)^{n+k}}{2k+1} + \frac{\pi}{4}(-1)^n)$ but again I can go further.
Any help would be appreciated.
Thanks
| \begin{align}
I&=\int_{-1}^1\frac{\ln(1+x)}{1+x^2}\ dx\overset{x=\frac{1-y}{1+y}}{=}\int_0^\infty\frac{\ln2-\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\int_0^\infty\frac{\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\left(\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\underbrace{\int_1^\infty\frac{\ln(1+y)}{1+y^2}\ dy}_{\large y\mapsto 1/y}\right)\\
&=\frac{\pi}{2}\ln2-\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy-\int_0^1\frac{\ln(1+y)-\ln y}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\int_0^1\frac{\ln y}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-2\left(\frac{\pi}{8}\ln2\right)-G\\
&=\frac{\pi}{4}\ln2-G
\end{align}
.
\begin{align}
\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&\overset{y=\frac{1-u}{1+u}}{=}\int_0^1\frac{\ln2-\ln(1+u)}{1+u^2}\ du\\
&=\frac{\pi}{4}\ln2-\int_0^1\frac{\ln(1+u)}{1+u^2}\ du\\
2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{4}\ln2\\
\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{8}\ln2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Help me out with the sum $\sum_{n= 0}^{N-1} \frac{ \left(a-b \cos{\left(\frac{2 \pi n}{N} \right)} \right)^2}{a^2 + b^2 -2ab\cos{\frac{2\pi n}{N}}}$ I am trying to find an analytical expression for the summation below
$$
\sum_{n= 0}^{N-1} \frac{ \left(a-b \cos{\left(\frac{2 \pi n}{N} \right)} \right)^2}{a^2 + b^2 -2ab\cos{\frac{2\pi n}{N}}}
$$
with $a>b$.
By trying it in MATLAB, I found that for large $N$ it gives a number which is
proportional to $N$. Therefore I am certain there must be analytical solution for this summation.
Can anyone help out?
Thank you
| My method gets the wrong answer, which may be because I approximate it at the start by an integral.
$$\int_0^N \frac{(a-b\cos(2\pi n/N))^2}{a^2+b^2-2ab\cos(2\pi n/N)}dn$$
Substitute $z=\exp(2\pi i n/N)$ to get
$$\frac N{2\pi i}\oint\frac{(a-(z+z^{-1})b/2)^2} {a^2+b^2-ab(z+z^{-1})}\frac{dz}z\\
=\frac N{8\pi i}\oint\frac{(2az-b(z^2+1))^2}{(az-b)(a-bz)}\frac{dz}{z^2}$$
This has poles at $0$, $b/a$ and $a/b$, and the integral is around the unit circle. The residue at $b/a$ is $1-b^2/a^2$, and at $0$ is $3-b^2/a^2$, so the final answer is
$$\frac N4(4-2b^2/a^2)$$
The other solutions had $a^N-b^N$ in the denominator. I lost that when I approximated the sum by an integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$ I tried to prove this by induction:
Base case for n = 1 satisfies $\because$ $1+x$ = $1+x$
I.H: $(1+x)^k \geq 1 + kx$
Inductive Step for $k+1$:
$(1+x)^{k+1} \geq 1 + (k+1)x$
$(1+x)^k (1+x) \geq 1 + kx + x$
$(1+kx) (1+x) \geq 1 + kx + x$ by I.H.
$kx^2 + kx + x + 1 \geq 1 + kx + x$
$kx^2 \geq 0$ which is true for $x>-1$ therefore proved?
Is this the correct way to prove by induction?
| You can save your proof by added "is implied by" ($\Leftarrow$) marks between lines to indicate the desired direction:
Inductive Step for k+1
:
$(1+x)^{k+1}≥1+(k+1)x\Leftarrow$
$(1+x)^k(1+x)≥1+kx+x\Leftarrow$ by I.H.
$(1+kx)(1+x)≥1+kx+x\Leftarrow$
$kx^2+kx+x+1≥1+kx+x\Leftarrow$
$kx^2≥0$
which is true as $k \ge 1 > 0$.
This would be a correct induction step.
But my advice on a stylish and better induction step would be:
......
We are assuming that $(1+x)^k \ge 1 + kx$ and wish to show this implies $(1+x)^{k+1} \ge 1 + (k+1)x$.
$(1+x)^k \ge 1+kx$ (we are presuming this) and as $x > -1$ we know $1+x > 0$ and thus:
$(1+x)^k(1+x) \ge (1+kx)(1+x)$ so
$(1+ x)^{1+x} \ge 1 + kx + x + kx^2= 1+(k+1)x + kx^2$
And as $k$ and $x^2$ are non-negative, $kx^2 \ge 0$, so
$(1+x)^{1+x} \ge 1+(k+1)x + kx^2 \ge 1+(k+1)x$.
......
but ultimately the proof will need to be written by you.
......
I just realized no-one has yet addressed that claiming $(1+x)^k(1+x) \ge (1+kx)(1+x)$ requires knowing that $1+x > 0$. This is why the condition $x > -1$ is required.
| {
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Equation of a line after being reflected in another line using the transformation matrix For a specific case, consider the line $ y = 3x + 1 $. How can I find the equation of the new line when this is reflected in the line $ y = 2x $ ? I would like to solve this using solely matrices and not considering angles of lines.
My attempt:
The general matrix transformation for a reflection in $ y = mx $ is
$ \dfrac{1}{1+m^2}\begin{bmatrix}
1-m^2 & 2m \\
2m & m^2-1
\end{bmatrix} $ so when $ m = 2 $, $ \mathbf{T} = \dfrac{1}{5} \begin{bmatrix}
-3 & 4 \\
4 & 3
\end{bmatrix} $
Using the vector equation of a line, $ y = 3x+1 $ can be written as
$ \mathbf{r} = \begin{bmatrix}
0\\
1
\end{bmatrix} +\lambda \begin{bmatrix}
1\\
3
\end{bmatrix} = \begin{bmatrix}
\lambda \\
3\lambda +1
\end{bmatrix} $
Applying the transformation,
$ \mathbf{r}' = \mathbf{T} \ \mathbf{r} = \dfrac{1}{5}\begin{bmatrix}
-3 & 4\\
4 & 3
\end{bmatrix}\begin{bmatrix}
\lambda \\
3\lambda +1
\end{bmatrix} = \dfrac{1}{5}\begin{bmatrix}
9\lambda +4\\
13\lambda +3
\end{bmatrix} $
Converting back to Cartesian,
$ x = 9\lambda + 4 \rightarrow \lambda = \dfrac{x-4}{9} $
$ y = 13 \left ( \dfrac{x-4}{9} \right ) + 3 $
$ 9y = 13x-25 $.
However some graphing on Desmos shows this is not correct, and the right answer looks more like $ 9y = 13x-5 $.
Any help in finding where is my mistake is much appreciated.
| There is a much faster way to make the computations: remember that a reflection is its own inverse, so if $\mathbf r'$ has coordinates $(x',y')$, we have
$$\mathbf r=T\mathbf r'\iff \begin{pmatrix}x\\y\end{pmatrix}=T\begin{pmatrix}x'\\y'\end{pmatrix}=\tfrac 15 \begin{pmatrix}-3x'+4y'\\4x'+3y'\end{pmatrix}$$
and plugging this relation in the equation of the line $y=3x+1$, we obtain
\begin{align}
\frac 15(4x'+3y')&=\frac 35(-3x'+4y')+1\iff 4x'+3y'=-9x'+12 y'+5\\&\iff13x'-5=9y'.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the sum of all values of $x$ which satisfy the following systems of equations: Systems of equations are as follows:
$$\left\lbrace\begin{aligned}
y&=x^2-5x+5\\
z&=x^2-12x+35\\
y^z&=1\\
x,&y,z \in\mathbb{R}\\
\end{aligned}\right.$$
I got the obvious $x=5,7,1$ and $4$ but apparently there are other answers as the sum should equal to $20$ and the answers I got sum to $17$; and if $z=0$, $y$ can be anything and if $y=1$, $z$ can be anything.
| There are infinitely many values of $x$ satisfying this system. For if you write the second equation as $z=x^2-5x+5-7x+30,$ and substitute for $x^2-5x+5$ using the first equation, we get $$z=y-7x+30.$$ Finally, the last equation implies $z=0$ or $y=1.$
If $z=0,$ then $y=7x-30$ and if $y=1,$ then $z=31-7x.$ Thus we have the families of solutions $(\lambda,7\lambda-30,0)$ and $(\mu, 1, 31-7\mu).$ The last case is when $z=0$ and $y=1,$ which gives the singleton $(31/7, 1, 0).$ Thus, you literally can't add all the values of $x$ satisfying the relation as $\lambda$ and $\mu$ can be any real number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I know how to get that answer. Where is the mistake in this method though?
| It is $$\lim_{x\to \infty}\frac{1}{-\sqrt{1+\frac{4}{x^6}}}=…$$
| {
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If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| Maple:
simplify(a^4+b^4+c^4, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 209/2
simplify(a^5+b^5+c^5, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 334
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove $9 \mathrel| (4^n+6n-1)$ by induction I know that this question was already answered, but I would like to know if the second step of induction its okay the way I did it.
This question is different from Induction proof for $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$ and Let $n ∈ N, n ≥ 1$. Prove that $4^n + 6n - 1$ is divisible by $9$. since it is just about to check a part from a proof that I did (if the way I thought is correct or not).
Suppose that $9 \mathrel| (4^k+6k-1)$ is true for $k\in \mathbb{N}, k\geq1$. I want to prove $P(k+1)$: $ 9 \mathrel| 4^{(k+1)} +6(k+1) - 1$
Since $9 \mathrel| (4^k+6k-1)$, I can rewrite as:
$4^k+6k-1 = 9q$, where $q \in \mathbb{Z} $. By the definion of Divisibility.
*
*Multiplying the equation on both sides by 4, I'll get:
$4^{(k+1)} +24k-4 = 36q$
*
*Adding 3 on both sides
$4^{(k+1)} +24k - 1 = 36q + 3$
*
*Subtracting $18k$ on both sides
$4^{(k+1)} +6k - 1 = 36q + 3 -18k$
*
*Adding 6 in order to have $6(k+1)$ as a factor
$4^{(k+1)} +6k + 6 - 1 = 36q + 3 -18k + 6$
$4^{(k+1)} +6(k+1) - 1 = 36q + 9 -18k $
Note that $36q + 9 -18k$ can be written like this: $9(4q + 1 - 2k)$. Thus $9Q$, $Q=(4q + 1 - 2k), Q \in \mathbb{Z}$.
Therefore:
*
*$4^{(k+1)} +6(k+1) - 1 = 9Q$. Which means that $4^{(k+1)} +6(k+1) - 1$ is also divided by 9.
Does my proof look fine? If there is a problem, please tell me. All help is appreciated.
| If $n \equiv 0 \pmod 3$, then for an integer $k$:
$$4^n + 6n-1 \equiv 4^{3k} + 18k - 1 \equiv 1^k + 0 - 1 \equiv 0 \pmod 9.$$
If $n \equiv 1 \pmod 3$, then:
$$4^n + 6n-1 \equiv 4 \cdot4^{3k} + 18k + 6- 1 \equiv 4+0+6-1 \equiv 0 \pmod 9.$$
Finally, if $n \equiv 2 \pmod 3$, we have:
$$4^n + 6n-1 \equiv 16 \cdot4^{3k} + 18k + 12- 1 \equiv 16+0+12-1 \equiv 0 \pmod 9.$$
Since an integer can only congruent to either $0 \pmod 3, 1 \pmod 3$ or $2 \pmod 3$, this completes the proof for all integer $n$.
| {
"language": "en",
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What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers? I have got these triplet solution $(x,y,z)=(1,1,1),(4,1,2)$ for this equation:
$$2^x 3^y+1=7^z$$
with $x, y , z$ are integers, But i can't get general solution of it, I have attempted to use Gausse theorem for the solution of $ ax+by= c $, with $a, b, c$ are integers but my problem i can't transfer the titled equation to that of Gausse as a linear form, any way ?
| As Aqua notes, we must have $z\gt0$, so let's write $z=c+1$. Then
$$2^x3^y=7^{c+1}-1=(7-1)(1+7+7^2+\cdots+7^c)$$
It follows that $x,y\gt0$ as well, so let's write $x=a+1$ and $y=b+1$. We now have
$$2^a3^b=1+7+7^2+\cdots+7^c$$
If $b\gt0$, then, since $7\equiv1$ mod $3$, the number of terms on the right hand side must be a multiple of $3$, in which case we have
$$1+7+7^2+\cdots+7^c=(1+7+7^2)(1+7^3+7^6+\cdots+7^{3k})$$
for some $k\ge0$. But $1+7+7^2=57=3\cdot19$, and $19$ is not divisible by $2$ or $3$. So we must have $b=0$.
If $a=0$, we have the solution $(x,y,z)=(1,1,1)$. If $a\gt0$, we must have an even number of terms in $1+7+7^2+\cdots+7^c$, so that
$$1+7+7^2+\cdots+7^c=(1+7)(1+7^2+7^4+\cdots+7^{2k})$$
for some $k\ge0$. Thus $a\gt0$ implies $a=3+a'$ for some $a'\ge0$, with $2^{a'}=1+7^2+7^4+\cdot+7^{2k}$. Now if $a'\gt0$, we must have an even number of terms in $1+7^2+7^4+\cdots+7^{2k}$, in which case we have
$$1+7^2+7^4+\cdots+7^{2k}=(1+7^2)(1+7^4+7^8+\cdots+7^{4k'})$$
for some $k'\ge0$. But $1+7^2=50=2\cdot5^2$ is not a power of $2$. So we must have $a'=0$, and this leaves $(x,y,z)=(4,1,2)$ as the only other solution.
Remark: The prime $19$ plays the same role here as it does in Aqua's answer, but it is obtained in a somewhat different manner (with a mod $3$ argument rather than a mod $9$ observation). The approach here can be modified to solve the equation $2^x3^y5^z+1=31^w$.
| {
"language": "en",
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"answer_count": 3,
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} |
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi-a}{4}\right)$ and $v:= \tan\left(\dfrac{\pi-b}{4}\right)$ the inequality reduces to proving
$$u^2+v^2+\dfrac{(1-uv)^2}{(u+v)^2} \ge 1\quad\quad (*)$$ ($$u,v\in (0,1)$$)
( using $a+b+c=\pi$ and the formula for $\tan(x+y)$ and the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ )
I'm having trouble in proving that last inequality.
Any suggestions are welcome.
Thanks.
Edit : is the following reasoning to prove the inequality (*) sound ?
(*) is obvious when $u^2 + v^2 \ge 1$ so we only have to deal with the case $u^2 + v^2 \le 1$ which we assume true in what follows.
(*) $\iff (u^2 + v^2)(u + v)^2 \ge (u+v)^2+(1-uv)^2$
Setting $x:= u^2 + v^2$ and $a:=uv$ we get
(*) $\iff x^2+(2a-1)x \ge a^2+4a-1$
Now some calculus $(x^2+(2a-1)x)' = 2x + 2a-1$
the function $\phi:x\mapsto x^2+(2a-1)x$ then has a minimum at $\dfrac 12 - a$ which is $\phi\left(\dfrac 12 - a\right) = -a^2+a-\dfrac 14$
It then suffices to have $-a^2+a-\dfrac 14 \ge a^2+4a-1$
This last ineq is equivalent to
$8a^2+12a-3 \le 0$
which, in turn, is equivalent to
$a \in \left[\dfrac{-6-\sqrt{60}}{8}, \dfrac{-6+\sqrt{60}}{8}\right]$
Recall that we're working under the assumption $u^2 +v^2 \le 1$, that yields in particular that $uv \le \dfrac 12$.
since $ \dfrac{-6-\sqrt{60}}{8} \le 0 \le a := uv \le \dfrac 12 \le \dfrac{-6+\sqrt{60}}{8}$, we're done.
Thanks for taking time to check the correctness of the above proof.
| Using the inequality for concave upward graph of $y\,=({tan\,(x)})^2$.
That is $\frac{(\sum_{}^{}{tan}^2{\frac{180-A}{4})}}{3} \;$ $ \ge\,$ ${tan}^2(({\frac{(180-A)+(180-B)+(180-C)}{4}} )/3)$ =1/3
Hence $(\sum_{}^{}{tan}^2{\frac{180-A}{4}})$ $\ge\,1$
| {
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} |
Inverse of rational function $y= \frac{3-x}{1+x^2}$ I have the function $$y= \frac{3-x}{1+x^2}$$ and I want to find the inverse of this function.
I know that
$$x= \frac{1 \pm \sqrt{1-4y(3-y)}}{2y}$$
My question is how do I find the domain where the function is
$$x= \frac{1- \sqrt{1-4y(3-y)}}{2y}$$
and
$$x= \frac{1+ \sqrt{1-4y(3-y)}}{2y}$$
respectively?
| I guess you got the idea, you just need to be more careful with the signs.
To find the inverse of:
$$inverse\:\frac{3-x}{1-x^2}$$
We'll need to find the inverse function in the form $y=g(x)$, given the above $f(x)$.
First, we need to get the quadratic equation solution general formula:
given:
$$Ax^2+Bx+C=0$$
$y_1(x)$ and $y_2(x)$ are roots such that:
$$y_1(x)=\frac{-B+\sqrt{B^2-4AC}}{2A}$$
and
$$y_2(x)=\frac{-B-\sqrt{B^2-4AC}}{2A}$$
To get the inverse of $f(x)$, we do two steps:
1-Interchange the variable name $x$ with $y$.
2-Solve the resulting equation for $y$.
1-Interchange the variable name $x$ with $y$ given:
$$f(x)=y=\frac{3-x}{1-x^2}$$
You get:
$$x=\frac{3-y}{1-y^2}$$
2-Solve for $y$:
$$ x(1-y^2) = 3-y$$
$$x-xy^2 = 3-y$$
$$x-xy^2-3+y=0$$
$$(-x)y^2+(1)y+(x-3)
Here $A=-x$, $B=1$, $C=x-3$
$$y_1(x)=\frac{-1+\sqrt{1^2-4(-x)(x-3)}}{2x}$$
Simplifying to get:
$$y_1\left(x\right)=\frac{-1+\sqrt{1+4\left(x\right)\left(x-3\right)}}{2x}$$
You can do the same for $y_2(x)$
$$y_2\left(x\right)=\frac{-1-\sqrt{1+4\left(x\right)\left(x-3\right)}}{2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The Riccatti equation for The Cox-Ingerson-Ross Model. (My Question)
I went through the calculations halfway, but I cannot find out how to calculate the following Riccatti equation. Please tell me how to calculate this The Riccatti equation with its computation processes. If you have other solutions, please let me know.
*
*If B(s) satisfies the following O.D.E (The Riccatti equation),
\begin{eqnarray}
B'(s) + \beta B(s) + \frac{1}{2} \sigma^2 B(s)^2 =1
\end{eqnarray}
*
*the answer must be below. (Please show the computation processes.)
\begin{eqnarray}
B(s)= \frac{ 2 \left( \exp(\gamma s) -1 \right) }{2\gamma +(\beta +\gamma)\left( \exp(\gamma s) -1 \right) } \qquad \mbox{with} \ \mbox{ $\gamma=\sqrt{ \beta^2+2\sigma^2}$}
\end{eqnarray}
(Thank you for your help in advance.)
(Cross-link)
I have posted the same question on https://quant.stackexchange.com/questions/47311/the-riccatti-equation-for-the-cox-ingerson-ross-model
(Original Questions)
(1) Write down the bond pricing P.D.E for the function
\begin{eqnarray}
P(t, T) = E^{ \mathbb{Q} } \left[ \exp \left( - \int^T_t r_s ds \right) \middle| r_t=x \right]
\end{eqnarray}
(2) and show that in case $\alpha =0 $ the corresponding bind price $P(t, T)$ equals
\begin{eqnarray}
P(t, T) = \exp \left( - B(T-t) r_s \right)
\end{eqnarray}
where $t \in [0, T] $ and
\begin{eqnarray}
B(x)= \frac{ 2 \left( \exp(\gamma x) -1 \right) }{2\gamma +(\beta +\gamma)\left( \exp(\gamma x) -1 \right) }
\end{eqnarray}
with $\gamma=\sqrt{ \beta^2+2\sigma^2}$.
(1) My answer
*
*Since the Cox-Ingerson-Ross Model has the below S.D.E, its corresponding P.D.E (namely the bond pricing P.D.E) comes to the following equation by Feynman-Kac Theorem (or by Exercise 4.1.(1) ). Besides, the terminal condition is $F(T, x)=1$.
\begin{eqnarray}
dr_t= (\alpha - \beta r_t ) dt + \sigma \sqrt{r_t} dB_t
\end{eqnarray}
*which models the variations of the short rate process $r_t$, where $\alpha, \beta, \sigma $ and $r_0$ are positive parameters. When the model is the Ho-Lee Model, $dr_t = \theta dt + \sigma dB_t$, its P.D.E is below.
\begin{eqnarray}
\partial_t F(t, x) + \theta \partial_x F(t, x) + \frac{1}{2} \sigma^2 \partial_{xx} F(t, x) -xF(t, x) =0
\end{eqnarray}
*Then the Cox-Ingerson-Ross Model has the following P.D.E.
\begin{eqnarray}
\partial_t F(t, x) + (\alpha - \beta x ) \partial_x F(t, x) + \frac{1}{2} \sigma^2 x\partial_{xx} F(t, x) -xF(t, x) =0
\end{eqnarray}
*When $\alpha=0$, $dr_t= - \beta r_t dt + \sigma \sqrt{r_t} dB_t$, it comes to below.
\begin{eqnarray}
\partial_t F(t, x) - \beta x \partial_x F(t, x) + \frac{1}{2} \sigma^2 x\partial_{xx} F(t, x) -xF(t, x) =0
\end{eqnarray}
*Here, if the S.D.E is the Generalized Affine Model, it comes to below
\begin{eqnarray}
dr_t= \left( \eta_t + \lambda_t r_t \right) dt + \sqrt{ \delta_t + \gamma_t r_t} dB_t
\end{eqnarray}
*The S.D.E of the Generalized Affine Model yields a bond pricing formula of the form:
\begin{eqnarray}
P(t, T) = \exp \left( A(T-t) +C(T-t)r_t\right)
\end{eqnarray}
*Comparing the conditional bond pricing formula, $P(t, T) = \exp \left( - B(T-t) r_s \right) $, to the above formula, one reaches below.
\begin{eqnarray}
&& A(T-t)=0 \\
&&C(T-t)r_t = - B(T-t) r_s
\end{eqnarray}
*Let $F(t, x)=\exp \left( - B(T-t) x \right) $.
\begin{eqnarray}
\partial_t F(t, x) &=& B'(T-t) x F(t, x) \\
\partial_x F(t, x) &=& -B(T-t) F(t, x) \\
\partial_{xx} F(t, x) &=&B(T-t)^2 F(t, x)
\end{eqnarray}
*The P.D.E comes to below.
\begin{eqnarray}
&&\partial_t F(t, x) - \beta x \partial_x F(t, x) + \frac{1}{2} \sigma^2 x\partial_{xx} F(t, x) -xF(t, x) \\
&&\qquad \qquad = B'(T-t) x F(t, x) - \beta x (-B(T-t) F(t, x)) \nonumber \\
&& \qquad \qquad\qquad + \frac{1}{2} \sigma^2 x B(T-t)^2 F(t, x) -xF(t, x)\\
&& \qquad \qquad = B'(T-t) x F(t, x) + \beta x B(T-t) F(t, x) \nonumber \\
&& \qquad \qquad\qquad + \frac{1}{2} \sigma^2 x B(T-t)^2 F(t, x) -xF(t, x)\\
&& \qquad \qquad\qquad =0
\end{eqnarray}
$\square$
(2) My Answer
*
*Since $F(t, x) \neq 0$ and $x \neq 0$, the above equation comes to below.
\begin{eqnarray}
&& B'(T-t) x F(t, x) + \beta x B(T-t) F(t, x) \nonumber \\
&& \qquad\qquad \qquad\qquad + \frac{1}{2} \sigma^2 x B(T-t)^2 F(t, x) -xF(t, x)\\
&& \qquad \qquad = B'(T-t) x + \beta x B(T-t) + \frac{1}{2} \sigma^2 x B(T-t)^2 -x \\
&& \qquad \qquad = B'(T-t) + \beta B(T-t) + \frac{1}{2} \sigma^2 B(T-t)^2 -1 \\
&&\qquad \qquad =0
\end{eqnarray}
*Let $T-t=s$, one reaches the following equation.
\begin{eqnarray}
B'(s) + \beta B(s) + \frac{1}{2} \sigma^2 B(s)^2 =1
\end{eqnarray}
*One finds out it is the Riccatti equation because of $A(s)=0$.
(Thank you for your help in advance.)
$\square$
| I solved by myself. The following is this solution.
*
*Let $T-t=s$, one reaches the following equation.
\begin{eqnarray}
B'(s) + \beta B(s) + \frac{1}{2} \sigma^2 B(s)^2 =1
\end{eqnarray}
*One finds out it is the Riccatti equation because of $A(s)=0$. Therefore, one reaches the following equation.
\begin{eqnarray}
B' = - \frac{1}{2} \sigma^2 B^2 - \beta B +1
\end{eqnarray}
*Since this is the Riccatti equation, one finds out the special solution. Let $B'=0$. Then, one reaches the following equations.
\begin{eqnarray}
&& - \frac{1}{2} \sigma^2 B^2 - \beta B +1 =0 \\
&& \sigma^2 B^2 + 2 \beta B - 2 =0 \\
&& B = \frac{-\beta \pm \sqrt{ \beta^2 + 2 \sigma^2} }{\sigma^2} \\
&& B = \frac{-\beta \pm \gamma}{\sigma^2}
\end{eqnarray}
*Use $B=(-\beta -\gamma)/\sigma^2$. Let $K=(-\beta -\gamma)/\sigma^2$. Moreover, let $B=u+K$.
\begin{eqnarray}
B^2 &=& u^2 + 2 K u + K^2 \\
B'&=&u'\\
&=& - \frac{1}{2} \sigma^2 B^2 - \beta B +1 \\
&=& - \frac{1}{2} \sigma^2 ( u^2 + 2 K u + K^2 ) - \beta (u+K) +1\\
&=& - \frac{1}{2} \sigma^2 u^2 - \sigma^2 K u - \beta u + \left( - \frac{1}{2} \sigma^2 K^2 - \beta K +1\right) \\
&=& - \frac{1}{2} \sigma^2 u^2 - \sigma^2 K u - \beta u + 0 \\
&=& - \frac{1}{2} \sigma^2 u^2 - \sigma^2 K u - \beta u \\
u' &=& - \frac{1}{2} \sigma^2 u^2 - \sigma^2 K u - \beta u
\end{eqnarray}
*Let $u=1/z$. Besides, $u'=-z'/z^2$.
\begin{eqnarray}
u' &=& - \frac{1}{2} \sigma^2 u^2 - \sigma^2 K u - \beta u \\
-\frac{z'}{z^2} &=& - \frac{1}{2} \sigma^2 \frac{1}{z^2} - ( \sigma^2 K + \beta ) \frac{1}{z} \\
z' &=& \frac{\sigma^2}{2} + ( \sigma^2 K + \beta ) z
\end{eqnarray}
*Let $M=\sigma^2/2$ and $N= ( \sigma^2 K + \beta ) $. Therefore, with Integral constant $C$,
\begin{eqnarray}
z&=& C e^{Nt} - \frac{M}{N} \\
z&=& \frac{1}{u}
= \frac{1}{B-K}
= C e^{Nt} - \frac{M}{N}
=\frac{C N e^{Nt} - M}{N} \\
B&=&\frac{N}{C N e^{Nt} - M} +K %= \frac{N}{C N e^{Nt} - M} + \frac{C N e^{Nt} - KM}{C N e^{Nt} - M}
= \frac{C N K e^{Nt} - KM +N}{C N e^{Nt} - M}
\end{eqnarray}
*Let $t=0$, since $B=0$.
\begin{eqnarray}
&& \frac{C N K e^{0} - KM +N}{C N e^{0} - M} = 0 \\
&& \frac{C N K - KM +N}{C N - M} = 0
\end{eqnarray}
*Here, one reaches the following condition.
\begin{eqnarray}
C &\neq& \frac{M}{N} = \frac{\sigma^2/2}{\sigma^2 K + \beta}= \frac{\sigma^2/2}{ -\beta - \sqrt{ \beta^2 + 2 \sigma^2}+ \beta} = -\frac{\sigma^2}{2 \gamma}
\end{eqnarray}
*One computes the numerator while paying attention to the above conditions.
\begin{eqnarray}
C &=& \frac{KM-N}{KN}
\end{eqnarray}
*One reaches the following equations.
\begin{eqnarray}
K&=& \frac{- \beta - \gamma}{\sigma^2} \\
M&=& \frac{\sigma^2}{2} \\
KM&=& \frac{- \beta - \gamma}{2} \\
N&=& \sigma^2 K +\beta = \beta - \gamma + \beta= - \gamma
\end{eqnarray}
*Substitute the above results into $C$.
\begin{eqnarray}
C &=& \frac{KM-N}{KN} = \frac{\frac{-\beta - \gamma}{2}+ \frac{2}{2} \gamma}{ \frac{- \beta - \gamma}{\sigma^2} ( - \gamma) }
= \frac{ \frac{-\beta + \gamma}{2}}{ \gamma \frac{ \beta + \gamma}{\sigma^2} }
= - \frac{ (\beta - \gamma) \sigma^2 }{ \gamma ( \beta + \gamma ) 2 } \\
CN&=&\frac{ \beta - \gamma }{ \beta + \gamma } \frac{ \sigma^2 }{2} \\
CNK&=& \frac{ \beta - \gamma }{ \beta + \gamma } \frac{ \sigma^2 }{2} \left(\frac{- \beta - \gamma}{\sigma^2} \right) = - \frac{\beta - \gamma}{2}
\end{eqnarray}
*Substitute the above results into $B$.
\begin{eqnarray}
B&=& \frac{C N K e^{Nt} - KM +N}{C N e^{Nt} - M} = \frac{ - \frac{\beta - \gamma}{2} e^{- \gamma t} + \frac{ \beta + \gamma}{2} - \gamma }{ \frac{ \beta - \gamma }{ \beta + \gamma } \frac{ \sigma^2 }{2} e^{- \gamma t} - \frac{\sigma^2}{2} } \\
&=& \frac{ - \frac{\beta - \gamma}{2} e^{- \gamma t} + \frac{ \beta - \gamma}{2} }{ \frac{ \beta - \gamma }{ \beta + \gamma } \frac{ \sigma^2 }{2} e^{- \gamma t} - \frac{\sigma^2}{2} }
= \frac{ - \left( \frac{\beta - \gamma}{2} \right) \left( e^{- \gamma t} -1\right) }{ \frac{\sigma^2}{2} \left( \frac{ \beta - \gamma }{ \beta + \gamma } e^{- \gamma t} -1 \right)} \\
&=& - \frac{ ( \beta - \gamma )( \beta + \gamma ) \left( e^{- \gamma t} -1\right) }{ \sigma^2 \left( ( \beta - \gamma ) e^{- \gamma t} - ( \beta + \gamma ) \right)}
=- \frac{ ( \beta^2 - \gamma^2 ) \left( e^{- \gamma t} -1\right) }{ \sigma^2 \left( ( \beta - \gamma ) e^{- \gamma t} - ( \beta + \gamma ) \right)} \\
&=& \frac{ 2\sigma^2 \left( e^{- \gamma t} -1\right) }{ \sigma^2 \left( ( \beta + \gamma ) e^{- \gamma t} - ( \beta + \gamma ) - 2 \gamma e^{- \gamma t} \right)} \\
&=& \frac{ 2 \left( e^{- \gamma t} -1\right) }{ ( \beta + \gamma ) \left( e^{- \gamma t} - 1 \right) - 2 \gamma e^{- \gamma t} }
= \frac{ 2 \left( 1 - e^{ \gamma t} \right) }{ ( \beta + \gamma ) \left( 1 - e^{ \gamma t} \right) - 2 \gamma } \\
B(t) &=& \frac{ 2 \left( \exp(\gamma t) -1 \right) }{2\gamma +(\beta +\gamma)\left( \exp(\gamma t) -1 \right) }, \qquad \mbox{ with $\gamma=\sqrt{ \beta^2+2\sigma^2}$.}
\end{eqnarray}
$\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so far:
$$\cos 3 \beta + \sin 3 \beta = ( \cos \beta - \sin \beta)(1+4\sin\beta\cos\beta)$$
Which trigonometric identity should I use to proceed?
| Hint:
$$\cos3\beta+\sin3\beta=4(\cos^3\beta-\sin^3\beta)-3(\cos\beta-\sin\beta)$$
$$(\cos\beta-\sin\beta)^3=\cos^3\beta-\sin^3\beta-3\cos\beta\sin\beta(\cos\beta-\sin\beta)$$
$$y^2=?$$
Replace the values of $\cos\beta\sin\beta,\cos\beta-\sin\beta$
| {
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How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$ This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and compute the nominator with the value of square root 2 and 3 memorized. And as to $e\cdot 2^{1/3}$, I just don't see how to compare it to other two items without take cubic and compute. This whole process is very time consuming.
I have seen some tricks to compare say $2^\pi,\pi^2$. But the technique does not seem to apply here.
| If you raise them to the third power you get
$\pi^3 \approx (3+\frac 17)^3 \approx 3^3 + 3*3^2*\frac 17 \approx 27*(1\frac 17)$
$e^3*2 \approx 2(3-0.29)^3 \approx 2(3^3 - 3*3^2*0.29)\approx 27*2*(0.71)\approx 27*1.42$ so $e*2^{\frac 13} > \pi$. (And as a weird unexpected bonus I get that $e^3*2 \approx 27*\sqrt 2$.)
$(\frac {1 +\sqrt 2}{\sqrt 3-1})^2 =\frac {3+2\sqrt 2}{4- 2\sqrt 3}$ and
$(\frac {1 +\sqrt 2}{\sqrt 3-1})^3=\frac {3+2\sqrt 2}{4- 2\sqrt 3}\frac {1 +\sqrt 2}{\sqrt 3-1}=\frac {3+4+5\sqrt 2}{-6-4+6\sqrt 3}$
$=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}\frac {6\sqrt 3 + 10}{6\sqrt 3 + 10}$
$=\frac{30\sqrt 6 +42*\sqrt 3 + 50\sqrt 2 +70}{36*3-100}$
$\approx \frac 18(30*1.7*1.4 + 42*1.7 + 50*1.4 + 70)\approx$
$\frac 18(51*1.4 + 7*10.2 + 70 + 70)\approx$
$\frac 18(70*4+1.4+1.4) \approx \frac {70.7}2 \approx 35.35$.
$\approx 27 + 8 \approx 27(1\frac 14)$
And $1\frac 17 < 1\frac 14 < 1.42$ so
So $\pi < \frac {1 +\sqrt 2}{\sqrt 3-1} < e*2^{\frac 13}$
.... which.... was really way too much work for too little result but.... well, it's one way to do it.
| {
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$24$ is the largest integer divisible by all integers less than its square root
Show that $24$ is the largest integer divisible by all integers less
than its square root.
This is what I have done :
Let $m$ be the greatest integer such that $m^2\leq n$, so $i\mid n$ for all i $\in \{1,2,\cdots m-1,m\}$ so lcm$(1,2,\cdots,m-1,m)\,\mid n $.
But how do i show that this is not possible when $n\geq 25$ or $m\geq 5$.
| Here's a hint / observation:
The number $25$ is not divisible by $2, 3,$ or $4$, so $25$ doesn't work.
If $n > 25$, the number must be divisible by $2^2 \cdot 3 \cdot 5 = 60$. But $\sqrt{60} > 7$, so the number would need a factor of $7$ also. Now we're looking at $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$, which means we need $2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 = 232792560$, which means ... :)
| {
"language": "en",
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Show that $\int_0^1 \frac{x \text{csch}(a x)}{\sqrt{\cosh (2 a)-\cosh (2 a x)}} \mathrm dx=\frac{\pi \sin ^{-1}(\tanh (a))}{2 \sqrt{2} a^2 \sinh (a)}$ Gradshteyn & Ryzhik $3.535$ states that
$$\int_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh (2 a)-\cosh (2 a x)}} \mathrm dx=\frac{\pi \sin ^{-1}(\tanh (a))}{2 \sqrt{2} a^2 \sinh (a)}$$
Holds for $a>0$. I found this result neat enough but can't prove it so far. Any help is appreciated!
| Let $f \colon (0,\infty) \to (0,\infty)$,
\begin{align}
f(a) &= \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh(2a) - \cosh(2ax)}} \, \mathrm{d} x = \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{2 \left[\frac{1}{1-\tanh^2(a)} - \frac{1}{1-\tanh^2(ax)}\right]}} \, \mathrm{d} x \\
&=\frac{1}{\sqrt{2} \sinh(a)} \int \limits_0^1 \frac{x}{\tanh(ax) \sqrt{1 - \frac{\tanh^2(ax)}{\tanh^2(a)}} \cosh^2(ax)} \, \mathrm{d} x \\
&\!\!\!\!\!\!\!\stackrel{u = \frac{\tanh(ax)}{\tanh(a)}}{=} \frac{1}{\sqrt{2} a^2 \sinh(a)} \int \limits_0^1 \frac{\operatorname{artanh}(\tanh(a) u)}{u \sqrt{1-u^2}} \, \mathrm{d} u \equiv \frac{g(\tanh(a))}{\sqrt{2} a^2 \sinh(a)} \, .
\end{align}
Here we have introduced $g \colon [0,1] \to [0,\infty)$,
$$ g(b) = \int \limits_0^1 \frac{\operatorname{artanh}(b u)}{u \sqrt{1-u^2}} \, \mathrm{d} u \, . $$
We can differentiate under the integral sign to obtain
\begin{align}
g'(b) &= \int \limits_0^1 \frac{\mathrm{d}u}{\sqrt{1-u^2}(1-b^2 u^2)} \stackrel{u = \frac{1}{\sqrt{1+v^2}}}{=} \int \limits_0^\infty \frac{\mathrm{d} v}{1 - b^2 + v^2} = \frac{\pi}{2 \sqrt{1-b^2}}
\end{align}
for $b \in (0,1)$, which implies
$$ g(b) = g(0) + \int \limits_0^b g'(c) \mathrm{d} c = 0 + \frac{\pi}{2} \int \limits_0^b \frac{\mathrm{d} c}{\sqrt{1-c^2}} = \frac{\pi}{2} \arcsin(b) $$
for $b \in [0,1]$. Therefore,
$$ f(a) = \frac{g(\tanh(a))}{\sqrt{2} a^2 \sinh(a)} = \frac{\pi \arcsin(\tanh(a))}{2 \sqrt{2} a^2 \sinh(a)}$$
holds for $a > 0$.
| {
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"source": "stackexchange",
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} |
Question about the particular part in a non homogeneous recurrence Question about the particular part in the following non homogeneous recurrence :
$$a_n - 6a_{n-1} + 9a_{n-2} = n * 3^n $$
I have the following particual part : $$ a_n = n * 3^n$$
Now the solution of the homogenous part is $$x_1 = 3, x_2 = 3$$ and is of the form $$a_n = (An * B)* 3^n$$
What im struggling with is understanding how this helps me solve for the particual part. The solution to the particual part is: $$ a_n = (Cn^3 + Dn^2) * 3^n$$ how i got to it was $$n^2 * (Cn + D) * 3^n$$ now the part i dont understand is why is there $$(Cn + D)$$ and not just $ Cn * 3^n$
Edited
| Another, general, take is using generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
- 6 \sum_{n \ge 0} a_{n + 1} z^n
+ 9 \sum_{n \ge 0} a_n z^n
&= \sum_{n \ge 0} n \cdot 3^n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
- 6 \frac{A(z) - a_0}{z}
+ 9 A(z)
&= z \frac{d}{d z} \frac{1}{1 - 3 z} \\
&= \frac{3 z}{(1 - 3 z)^2}
\end{align*}$
Solve for $A(z)$:
$\begin{align*}
A(z)
&= \frac{(9 a_1 - 54 a_0 + 3) z^3 - (6 a1 - 45 a_0) z^2 + (a_1 - 12 a_0) z + a_0}
{(1 - 3 z)^4}
\end{align*}$
You want the coefficient of $z^n$. Use the generalized binomial theorem:
$\begin{align*}
(1 + u)^{-m}
&= \sum_{n \ge 0} \binom{-m}{n} u^n \\
&= \sum_{n \ge 0} (-1)^n \binom{n + m - 1}{m - 1} u^n
\end{align*}$
This gives:
$\begin{align*}
[z^n] A(z)
&= \left(
(9 a_1 - 54 a_0 + 3) [z^{n - 3}]
- (6 a1 - 45 a_0) [z^{n - 2}]
+ (a_1 - 12 a_0) [z^{n - 1}]
+ a_0 [z^n]
\right) (1 - 3 z)^{-4} \\
&= \left(
(9 a_1 - 54 a_0 + 3) \binom{n - 3 + 4 - 1}{4 - 1}
- (6 a1 - 45 a_0) \binom{n - 2 + 4 - 1}{4 - 1}
+ (a_1 - 12 a_0) \binom{n - 1 + 4 - 1}{4 - 1}
+ a_0 \binom{n + 4 - 1}{4 - 1}
\right) \cdot 3^n
\end{align*}$
Now the binomial coefficient $\binom{n}{m}$ is a polynomial of degree $m$ in $n$, expanding those you get the full solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can i do this to an infinite series? $$let , Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$$
$$Then , Y= \sqrt{2Y}$$
$$Y^2 = 2Y$$
$$Y^2 - 2Y = 0$$
$$Y = 0 , Y =2$$
Now y can't be zero so ,
$$Y = 2$$
Is This Correct?
| If you assume the number exists and has a value then that is true.
But it could be that the number doesn't exist.
Consider $1 + 2 + 4 + 8 + 16+..... = M$ then $2M = 2+4 + 8 + 32 + ....$ and $2M + 1 = 1 + 2 + 4+ 8+16 + 32 + .... = M$ so $2M + 1 = M$ and $M = -1$.
That's obviously not true.
ANd this fails because $\lim\limits_{n\to \infty}\sum\limits_{k=0}^n 2^k=\infty$.
However If $a_0 =\sqrt{2}$ and $a_1= \sqrt {2\sqrt 2}$ and $a_{k+1} =\sqrt {2a_k}$ then IF $\{a_n\}$ converges so $Y = \lim\limits_{n\to \infty} a_n\ne \infty$, we would be okay. And we could say $Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$ and $Y= 2$ by your argument.
We can prove $\lim\limits_{n\to \infty} a_n\ne \infty$ by noting that $a_0=\sqrt 2 < 2$ and if $a_k <2$ we would have $a_{k+1} =\sqrt {2a_k}< \sqrt {2*2} = 2$. so by induction all $a_n < 2$ so $\lim\limits_{n\to \infty} a_n\ne \infty$.
Furthermore $a_{k+1} = \sqrt{2a_k} > \sqrt{a_k*a_k} = a_k$.
So we have $\sqrt{2} =a_1 < a_2 < a_3 < ....... < 2$. So $\lim\limits_{n\to \infty} a_n = Y$ exists.
| {
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"source": "stackexchange",
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How would I go about computing this finite sum? How would I go about computing the sum
$$
\sum_{k=1}^{n} \dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}.
$$
I have tried partial fractions but have gotten stuck trying to find the coefficients. I decomposed it like this:
$$
\dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a_1}{k^2} + \frac{a_2}{k+1} + \frac{a_3}{(k+1)^2}
$$.
| Hint:
Without the need for solving the coefficients:
$\dfrac{-k^2+2k+1}{k^2(k+1)^2} =\dfrac{k^2+2k+1-2k^2}{k^2(k+1)^2}=\dfrac{(k+1)^2}{k^2(k+1)^2}-\dfrac{2}{(k+1)^2}=\dfrac{1}{k^2}-\dfrac{2}{(k+1)^2}.$
Thus, the sum is reduced to
$$\sum_{k=1}^{n}\dfrac{2^k}{k^2}-\sum_{k=1}^{n}\dfrac{2^{k+1}}{(k+1)^2}.$$
And then note that
$$\sum_{k=1}^{n}\dfrac{2^{k+1}}{(k+1)^2}=\sum_{k=2}^{n+1}\dfrac{2^{k}}{k^2}=-2+\dfrac{2^{n+1}}{(n+1)^2}+\sum_{k=1}^{n}\dfrac{2^{k}}{k^2}.$$
The rest is for you to finish.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does an elementary indefinite integral for 2^sin(x) exist? Wolframalpha says no, and I've had little luck elsewhere, maybe because search engines aren't the best at making sense of the "^" symbol.
And if not, are any numerical methods known for solving this integral?
| $\int 2^{\sin x}~dx$
$=\int e^{\ln2\sin x}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
For $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
For $n$ is any non-negative integer,
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
$=x+\sum\limits_{n=1}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343239",
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"question_score": "1",
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An inequality involving $|a|^p$: how can I prove this? How can I prove the following lemma?
Let $1<p<\infty$ and let $\epsilon >0$. Then there exists a constant $C \geq 0$ (may depend on $p$ and $\epsilon$) such that for all $a, b \in \mathbb{C}$, $||a+b|^p - |b|^p| \leq \epsilon |b|^p +C|a|^p $.
I am suffering since I cannot use any homogeneity argument because of the $|a+b|^p$ term. I tried to use the fact that $|t|^p$ is a convex function for each fixed $p>1$, but it does not seem to work, as far as I tried.
Thanks!
| Let
$$x =|a+b|,\quad y = |b|,$$
then the issue inequality takes the form of
$$|x^p-y^p| \le \varepsilon y^p + C|x+e^{i\varphi} y|^p,\tag1$$
where the phase $\varphi$ depends of the phases of $(a+b)$ and $b.$
Since for the arbitrary $u,v\in\mathbb C$
\begin{cases}
||u|-|v|| \le |u+v|\\
||u|-|v|| \le |u-v|,
\end{cases}
then the worst case of $(1)$ is the case
$$|x^p-y^p| \le \varepsilon y^p + C|x - y|^p.\tag2$$
Inequality $(2)$ is homogenuis by $x$ and $y,$ so the least value of $C$ should provide the inequality
$$|z^p-1| \le \varepsilon + C|z - 1|^p,\tag3$$
where $z\in[0,\infty).$
If $\color{brown}{z=0},$ then $C_{L0} =1-\varepsilon.$
If $\color{brown}{z=1},$ then $C_{L1} =0.$
If $\color{brown}{z\to \infty},$ then $C_{L\infty} = 1.$
Denote
$$f(C,z) = |z^p-1| - \varepsilon - C|z - 1|^p.\tag4$$
The least value of $C$ can be defined from the condition
$$\max\limits_{z\in[0,\infty)} f(C,z) = 0.$$
The inner maxima can be achived only if $f'_z(C,z)=0.$
If $\color{brown}{z\in(0,1)}$ then
$$f(C,z) = 1-z^p-\varepsilon - C(1-z)^p,$$
\begin{cases}
-pz^{p-1}+Cp(1-z)^{p-1} = 0\\
1-z^p-\varepsilon - C(1-z)^p = 0,
\end{cases}
\begin{cases}
\dfrac z{1-z}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\\
1 - \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\right)^p
+ C\left(\dfrac{1}{\sqrt[p-1]C+1}\right)^p,
\end{cases}
$$1 - \varepsilon = \dfrac{C\left(\sqrt[p-1]C+1\right)}{\left(\sqrt[p-1]C+1\right)^p}
= \dfrac{C}{\left(\sqrt[p-1]C+1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1-\varepsilon},$$
$$C_{L1-}= \left(\dfrac z{1-z}\right)^{p-1} = \dfrac{1-\varepsilon}{\left(1-\sqrt[p-1]{1-\varepsilon}\right)^{p-1}}.$$
If $\color{brown}{z\in(1,\infty)}$ then
$$f(C,z) = z^p-1-\varepsilon - C(z-1)^p,$$
\begin{cases}
pz^{p-1}-Cp(z-1)^{p-1} = 0\\
z^p-1-\varepsilon - C(z-1)^p = 0,
\end{cases}
\begin{cases}
\dfrac z{z-1}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\\
1 + \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\right)^p
- C\left(\dfrac{1}{\sqrt[p-1]C-1}\right)^p,
\end{cases}
$$1 + \varepsilon = \dfrac{C\left(\sqrt[p-1]C-1\right)}{\left(\sqrt[p-1]C-1\right)^p}
= \dfrac{C}{\left(\sqrt[p-1]C-1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1+\varepsilon},$$
$$C_{L1+}= \left(\dfrac z{z-1}\right)^{p-1} = \dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}.$$
Since
$$\dfrac{C_{L1-}}{C_{L1+}}
= \dfrac{1-\varepsilon}{1+\varepsilon}\left(\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1}
= \left(\dfrac{\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}}\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1}$$
$$ = \left(\dfrac{\sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1}
= \left(1+\dfrac{2\ \sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}-\sqrt[p-1]{1+\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1} \le1,$$
then the least value of the constant $C$ is
$$C_L = \max(C_{L0},C_{L1},C_{L\infty},C_{L1-},C_{L1+}) = C_{L1+}
= \color{brown}{\dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}}.$$
| {
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Is $SU(3)$ generated by the exponentials of the Gell-Mann matrices? It is well known that the lie algebra of $SU(3)$ is spanned by the (half) Gell-Mann matrices $T_i=\lambda_i/2$, and that these generate $SU(3)$ via the map $\theta_iT_i \rightarrow e^{\theta_iT_i}$.
However, the above map involves taking a linear combination of the generators and then taking the exponential. Do we generate $SU(3)$ if we take the exponential of this basis first? In other words, is it true that if we take the set $\{{e^{T_i}}\}, i=1,...8$, that these matrices generate $SU(3)$?
| For SU$(3)$, the following set $F_m$ and its algebraic relations differ from that of the eight Gell Mann generators $g_m$.
The $F_m$ have no simple relation to the $g_m$. For example, in the usual representation, $7$ of the $8$ generators $g_m$ have a diagonal element that is zero;
Is the set of the eight finite SU(3) elements $F_m=e^{\pi \cdot i \cdot g_m / 2}$ $-$ defined by the eight SU(3) generators $g_m$.
Here are the Gell-Man matrices $g_n$:
$g_1=\left ( \begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_2=\left ( \begin{array}{ccc} 0 & -i & 0\\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_3=\left ( \begin{array}{ccc} -1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right )$, $g_4=\left ( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right )$, $g_5=\left ( \begin{array}{ccc} 0 & 0 & -i\\ 0 & 0 & 0 \\ i & 0 & 0 \end{array} \right )$, $g_6=\left ( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right )$, $g_7=\left ( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & -i \\ 0 & i & 0 \end{array} \right )$, $g_8=\left ( \begin{array}{ccc} 1/\sqrt{3} & 0 & 0\\ 0 & 1/\sqrt{3} & 0 \\ 0 & 0 & -2/\sqrt{3} \end{array} \right )$,
And these are the first matrices $F_n$:
$F_1=\left ( \begin{array}{ccc} 0 & i & 0\\ i & 0 & 0 \\ 0 & 0 & 1\end{array} \right )$, $F_2=\left ( \begin{array}{ccc} 0 & 1 & 0\\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right )$, $F_3=\left ( \begin{array}{ccc} i & 0 & 0\\ 0 & -i & 0 \\ 0 & 0 & 1 \end{array} \right )$, $F_4=\left ( \begin{array}{ccc} 0 & 0 & i\\ 0 & 1 & 0 \\ i & 0 & 0 \end{array} \right )$, $F_5=\left ( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{array} \right )$, $F_6=\left ( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & i \\ 0 & i & 0 \end{array} \right )$, $F_7=\left ( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right )$, Now $ F_8$ is less simple.
| {
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"answer_count": 1,
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} |
Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\begin{align}
\cos x \cos y &=4/15\\
\sin x \sin y &=-1/15
\end{align}
Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have
$$
225 \sin^4 y -210 \sin^2 y +1=0
$$
with its solution
$\sin^2 y = \frac{7\pm4\sqrt3}{15}$.
Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.
\begin{align}
\sin^2(2y) &= 4\cos^2 y\sin^2 y\\
&= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\
\sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)}
\end{align}
$\sin 2y$ must be negative.
Edit
Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.
By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is.
| You might note that $7+4\sqrt3=(2+\sqrt3)^2$ and $2(2+\sqrt3)=(1+\sqrt3)^2$, and that your $\cos^2y$ has opposite sign to $\sin^2y$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\sum_{k=1}^n k2^{n-k}$ I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:
$$\begin{align}
&\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} & &\text{(1)} \\
&k2^{-k-1} = -\frac{d}{dk}2^{-k} & &\text{(2)}
\end{align}$$
Plugging $(2)$ to $(1)$,
$$\begin{aligned}
2^{n+1}\sum_{k=1}^n k2^{-k-1} &= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\
&= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\
&= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\
&= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\
&= -2^{n+1}\cdot 0 \\
&= 0 \end{aligned}$$
Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks!
| Actually, the answer is quite clean. It is $2^{n+1}-n-2$.
You can prove this recursively.
Start with, let's say $n=2$.
This sum is equal to $1(2)+2(1)=2^3-2-2$.
Now, for $n=3$, the sum is equal to $1(4)+2(2)+3(1)$.
How do we get there? First we multiply $1(2)+2(1)$ by $2$, to get $1(4)+2(2)$, and then we add $3$ to get $1(4)+2(2)+3(1)$.
Now, doing the same steps, for $f(n)=2^{n+1}-n-2$, we have that multiplying by $2$ and adding $n+1$ gives $$f(n+1)=2(2^{n+1}-n-2)+n+1$$ $$=2^{n+2}-n+1-4$$ $$=2^{n+2}-(n+1)-2$$ which works by the rule we just declared.
| {
"language": "en",
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"question_score": "2",
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Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1\right)$. Can this be evaluated analytically? Any help will be appreciated.
| An alternative approach. By termwise integration of a Maclaurin series
$$ I=\int_{0}^{1}\frac{x^2}{\sqrt{1+x^4}}\,dx = \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n}{4n+3} $$
and due to the fact that $P_{2n}(0)=\left[\frac{1}{4^n}\binom{2n}{n}\right](-1)^n$, the RHS of the previous line is the value at $x=\frac{1}{2}$ of the following function:
$$ f(x) = \sum_{n\geq 0}\frac{1}{2n+3} P_{n}(2x-1). $$
Now we may exploit three well known Fourier-Legendre expansions:
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{2}{(1-2n)(2n+3)}P_n(2x-1), $$
$$ K(x) = \sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1), $$
$$ E(x) = \sum_{n\geq 0}\frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
(the argument $x$ stands for the elliptic modulus, like in Mathematica's notation) and deduce from partial fraction decomposition that
$$ I = \left.\frac{1}{2}K(x)-E(x)+\sqrt{1-x}\right|_{x=1/2}. $$
$K\left(\frac{1}{2}\right)$ and $E\left(\frac{1}{2}\right)$ are well known to be related to $\Gamma\left(\frac{1}{4}\right)^2$, and related to each other via Legendre's identity.
| {
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The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we reached at this point
$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
Please help someone in solving this question. I am very grateful to you.
| This second sum can be written as $$\sum_{n=2}^{\infty}\frac{1}{n}\left(\frac{1}{5}\right)^n = \frac{1}{25}\sum_{n=0}^{\infty}\frac{1}{n+2}\left(\frac{1}{5}\right)^n$$
This is the Lerch Transendent, so $$=\frac{1}{25}\phi\left(\frac{1}{5}, 1,2\right)$$
By the identities $\phi(z,s,a) = \frac{1}{z}\left(\phi(z,s,a-1) - \frac{1}{((a-1)^2)^{s/2}}\right)$, $\frac{1}{z}\mathrm{Li}_n(z)=\phi(z,n,1)$ and $\mathrm{Li}_1(z)=-\ln(1-z)$:
$$\begin{align} &=\frac{1}{25}\cdot5\left(\phi\left(\frac{1}{5},1,1\right)-1\right) \\ &= \frac{1}{5}\left(5\,\mathrm{Li}\left(\frac{1}{5}\right)-1\right) \\ &= -\ln\left(\frac{4}{5}\right)-\frac{1}{5}\end{align}$$
Therefore, the final answer to your entire sum is $$\begin{align} &= \frac{1}{20} - \left(-\ln\left(\frac{4}{5}\right)-\frac{1}{5}\right)\\ &= \frac{1}{4}+\ln\left(\frac{4}{5}\right)\\ &\approx 0.026856\end{align}$$
More Reading:
Lerch Transendent
Lerch Identity
Polylogarithm
| {
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$
thus $r=-1$ and $s = 2$
so $x = 2\times5\times8 -1\times9\times1=80 - 9=71\mod 180$
But the problem is that $71\equiv8\mod9$ and $71\equiv1\mod5$, but $71 \ne 2 \mod 4$. What is wrong?
| From $x\equiv2\mod4$, let $x=4a+2$ that $a$ is an integer. Then sub it into the second equation,$$4a+2\equiv8\mod9\\4a\equiv6\mod9\\2a\equiv3\mod9\\2a\equiv12\mod9\\a\equiv6\mod9$$ After that, we do the same thing: let $a=9b+6$ that $b$ is an integer, $x=4\left(9b+6\right)+2=36b+26$. Then sub it into the third equation, $$36b+26\equiv1\mod5\\36b\equiv0\mod5\\b\equiv0\mod5$$ Then, let $b=5k$ that $k$ is an integer. $x=36\left(5k\right)+26=180k+26$, and that is the answer.
Conclusion: The answer is $180k+26$ for $k\in\mathbb{Z}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks
| The product of three consecutive numbers is divisible by $3$.
If $n$ is even, then $4\mid n^2$. Otherwise $n^2-1$ and $n^2 + 1$ are both even.
Now the hard part:
\begin{array}{c|c}
n \pmod 5 & n^2 \pmod 5\\ \hline
0 & 0 \\
1 & 1 \\
2 & 4 \\
3 & 4 \\
4 & 1
\end{array}
or, without modular notation,
\begin{array}{c|c}
n & n^2 & \text{The remainder when dividing by five}\\ \hline
5k & 25k^2 & 5(5k)\\
5k+1 & 25k^2 + 10k + 1 & 5(5k^2 + 2k) + 1 \\
5k+2 & 25k^2 + 20k + 4 & 5(5k^2 + 4k) + 4 \\
5k+3 & 25k^2 + 30k + 9 & 5(5k^2 + 6k + 1) + 4 \\
5k+4 & 25k^2 + 40k + 16 & 5(5k^2 + 8k + 3) + 1 \\
\end{array}
No matter what $n$ is, $n^2$ is either $-1, 0$, or $1 \pmod 5$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Tree diagram for Conditional probability problem of two sons being disease carriers The question : A woman has a 50% chance of carrying hemophilia. She also has two sons If she is a carrier, each son independently has 0.5 probability of having the disease. If she is not a carrier, her sons will independently be normal (i.e., will not have the disease).
(c) If 1st son is normal, what is the probability that the second son is also normal?
(d) If both sons are normal, what is the probability that she is a carrier?
This is the link to the tree diagram i drew enter image description here
Would the above tree diagram be a correct conceptualization of the given problem?
My understanding of (c) is that since both sons being normal are independent events then
P (2nd son normal | 1st son normal) = P(2nd son normal) = 0.75 using Partition Rule.
Also would it be correct to say part (d) can be solved via Bayes Rule?
| Let $C$ denote the event that the woman is a carrier.
Let $N_{1}$ denote the event that the first son is normal.
Let $N_{2}$ denote the event that the second son is normal.
$\begin{aligned}P\left(N_{1}\right) & =P\left(N_{1}\mid C\right)P\left(C\right)+P\left(N_{1}\mid C^{\complement}\right)P\left(C^{\complement}\right)\\
& =\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{2}\\
& =\frac{3}{4}
\end{aligned}
\tag1$
and:
$\begin{aligned}P\left(N_{1}\cap N_{2}\mid C\right) & =P\left(N_{1}\mid C\right)P\left(N_{2}\mid C\right)\\
& =\frac{1}{2}\cdot\frac{1}{2}\\
& =\frac{1}{4}
\end{aligned}
\tag2$
Applying $(2)$ we find:
$\begin{aligned}P\left(N_{1}\cap N_{2}\right) & =P\left(N_{1}\cap N_{2}\mid C\right)P\left(C\right)+P\left(N_{1}\cap N_{2}\mid C^{\complement}\right)P\left(C^{\complement}\right)\\
& =\frac{1}{4}\cdot\frac{1}{2}+1\cdot\frac{1}{2}\\
& =\frac{5}{8}
\end{aligned}
\tag3$
Applying $(1)$ and $(2)$ we find:
$\begin{aligned}P\left(N_{2}\mid N_{1}\right)\frac{3}{4} & =P\left(N_{2}\mid N_{1}\right)P\left(N_{1}\right)\\
& =P\left(N_{1}\cap N_{2}\right)\\
& =\frac{5}{8}
\end{aligned}
\tag4$
Applying $(4)$ we find the solution of c): $$P\left(N_{2}\mid N_{1}\right)=\frac{4}{3}\cdot\frac{5}{8}=\frac{5}{6}$$
Applying $(3)$ and $(2)$ we find:
$\begin{aligned}P\left(C\mid N_{1}\cap N_{2}\right)\frac{5}{8} & =P\left(C\mid N_{1}\cap N_{2}\right)P\left(N_{1}\cap N_{2}\right)\\
& =P\left(C\cap N_{1}\cap N_{2}\right)\\
& =P\left(N_{1}\cap N_{2}\mid C\right)P\left(C\right)\\
& =\frac{1}{4}\cdot\frac{1}{2}\\
& =\frac{1}{8}
\end{aligned}
\tag5$
Applying $(5)$ we find the solution of d): $$P\left(C\mid N_{1}\cap N_{2}\right)=\frac{8}{5}\cdot\frac{1}{8}=\frac{1}{5}$$
Now check yourself (I can't read your link properly).
| {
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} |
jacobian of a dot product of 2 functions take the function $f: \mathbb R^3 \rightarrow \mathbb R^2$ an a function $g: \mathbb R^2 \rightarrow \mathbb R^2 $ defined via $$ f
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
x+z^3\\
xyz
\end{pmatrix}
$$
and $$ g
\begin{pmatrix}
s\\
t
\end{pmatrix} =
\begin{pmatrix}
s^2 + t\\
s+t
\end{pmatrix}
$$
compute the vector $$ J(g \circ f) \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$$ , that is the jacobian of the dot product
| I can see that you have no trouble bar silly mistakes in computing the Jacobian. More precisely, we have :
$$
Df(x,y,z) = \begin{pmatrix}
1 & 0 & 3z^2 \\
zy&xz&xy
\end{pmatrix} ;
Dg(s,t) = \begin{pmatrix}
2s & 1 \\
1&1
\end{pmatrix}
$$
Now, the chain rule is the following : (with variables $a,b,c$ to avoid confusion)
$$
D(g \circ f)(a,b,c) = Dg(f(a,b,c)) \times Df(a,b,c)
$$
So, in steps , given a point $(a,b,c)$ :
*
*Find $f(a,b,c)$.
*Find $Df$ evaluated at $(a,b,c)$, and $Dg$ evaluated at $\color{green}{f(a,b,c)}$ , and not $\color{red}{(a,b,c)}$.
*Multiply them as matrices, in the order specified in the formula.
For example, you have $(a,b,c) = (1,2,3)$. We find that $f(1,2,3) = (28,6)$.
Now, we have $Df(1,2,3) = \begin{pmatrix}
1 & 0 & 27 \\
6 & 3 & 2
\end{pmatrix}$.
We also have $Dg(28,6) = \begin{pmatrix}
56&1\\1&1
\end{pmatrix}$
And therefore, multiplying them in that order gives
$$
\begin{pmatrix}
56&1\\1&1
\end{pmatrix} \times \begin{pmatrix}
1 & 0 & 27 \\
6 & 3 & 2
\end{pmatrix} = \begin{pmatrix}
62 & 3 & 1514\\
7 & 3 & 29
\end{pmatrix}
$$
Which is the answer.
| {
"language": "en",
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If $P^{-1}(D + A)P$ is a diagonal matrix, with $D$ a diagonal matrix and $D+A$ hermitian, is $P^{-1}AP$ also diagonal? Suppose $D+A$ is hermitian matrix, and $D$ is a diagonal matrix. Then $A$ is a hermitian matrix as well.
Since $D+A$ is a hermitian matrix, there exists $P$ such that $P^{-1}(D+A)P$ is a diagonal matrix.
The question is, how do I prove that same $P$ also diagonalizes $A$ such that $P^{-1}AP$ is a diagonal matrix? Or is there a counterexample?
| Here is an counter-example. Put \begin{align*} A = \begin{bmatrix} 3 & i \\ -i & 1 \end{bmatrix}, D= \begin{bmatrix} -2 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}
We can find an unitary matrix $U = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix} $ such that $$U^{-1}(D+A)U = \frac{1}{2}
\begin{bmatrix} 1 & -i \\ 1 & i \end{bmatrix} \begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix}= \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$
However, $$U^{-1}AU = \frac{1}{2}
\begin{bmatrix} 1 & -i \\ 1 & i \end{bmatrix} \begin{bmatrix} 3 & i \\ -i & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360663",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the equation for image of a map Consider the morphism $\mathbb A^1 \to \mathbb A^2$ given by $f(t) = (t^n,(1-t)^n)$. What is the defining equation of the image? Is there a systematic way to do this?
| Too long for a comment.
In M2
R=QQ[t,x,y,MonomialOrder=>Eliminate 1]
I1=ideal(x-t,y-(1-t))
toString (gens gb I1)_0_0
-- x+y-1
I2=ideal(x-t^2,y-(1-t)^2)
toString (gens gb I2)_0_0
-- x^2-2*x*y+y^2-2*x-2*y+1
I3=ideal(x-t^3,y-(1-t)^3)
toString (gens gb I3)_0_0
-- x^3+3*x^2*y+3*x*y^2+y^3-3*x^2+21*x*y-3*y^2+3*x+3*y-1
I4=ideal(x-t^4,y-(1-t)^4)
toString (gens gb I4)_0_0
-- x^4-4*x^3*y+6*x^2*y^2-4*x*y^3+y^4-4*x^3-124*x^2*y-124*x*y^2-4*y^3+6*x^2-124*x*y+6*y^2-4*x-4*y+1
The leading terms seem to be $(x+y)^n$ for $n$ odd and $(x-y)^n$ for $n$ even.
In geogebra
| {
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-2=-7 \Leftrightarrow n = -5 $
*$n-2=-1 \Leftrightarrow n = 1 $
*$n-2=7 \Leftrightarrow n = 3 $
*$n-2=-1 \Leftrightarrow n = 9 $
So $S = \{ 1, 3, 9 \}$
I decompose $n+1$ the exact same way:
$$n+1 = 3n+11 - 2(n+5)$$
But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:
*
*$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
*$3n+11 = -1 \Leftrightarrow n = -4$
*$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
*$3n+11 = 1 \Leftrightarrow n = -3$
Any clues?
| Set $n+1=m$
$3n+11=3(m-1)+11=3m+8\equiv8\pmod m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Apostol Theorem 1.35 (Every nonnegative real number has a unique nonnegative square root)
In the above theorem I am not able to understand inequalities in the two marked boxes.
To be specific in the first inequality $0<c<b$, I don't understand why we have $c<b$ when $c=\frac{1}{2}(b+\frac{a}{b})$. In the second inequality how do we get $(b+c)^2<b^2+3bc$?
| At the beginning of that paragraph $c=b - \frac{(b^2-a)}{2b}$ which is clearly $<b$ as we substract from $b$ a positive number (as $b^2-a>0$ in this subcase)
and that $b - \frac{(b^2-a)}{2b}$ equals $\frac{1}{2}(b + \frac{a}{b})$ is a simple fraction manipulation:
$$c= b - \frac{(b^2-a)}{2b}= \frac{2b^2}{2b} - \frac{(b^2-a)}{2b} = \frac{b^2+a}{2b} = \frac{1}{2}(\frac{b^2+a}{b})= \frac12(b + \frac{a}{b}) >0 $$
For the second we choose $0<c<b$ so
$$b^2 + c(2b+ c) < b^2 + c(2b+b) = b^2 + 3bc$$
whiel $c < \frac{a-b^2}{3b}$ implies $3bc < a-b^2$ and so the rest of the line is explained too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.
| Let $$s_n(x)=1+x+x^2+\dots +x^{n-1}.$$
Then $s_{n+1}(x)=x^n+s_n(x)$ and $xs_n(x)+1=s_{n+1}(x)$. Thus
$$x^n+s_n(x)=xs_n(x)+1,$$
that is, $$x^n-1=(x-1)s_n(x).$$
We let $p_n(x)=s_n(-x)=1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1}.$ We then have
$$(x+1)p_n(x)=1+(-1)^{n-1}x^n.$$
Set $n=5$ so that $(-1)^{5-1}=(-1)^4=1$ and
$$\begin{align}
x^5+1&=(x+1)p_5(x)\\
&=(x+1)(1-x+x^2-x^3+x^4)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$
I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$:
$$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$
We need to prove
$$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$
Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much
| Let $c\neq {\rm mid}\{a,b,c\}$ and $c'={\rm mid}\{a,b,c\}.$
Now, let another variables be $a'$ and $b'$.
Thus, since our inequality does not depend on any permutations of $a'$, $b'$ and $c'$, we need to prove that:
$$\frac{a'+b'}{c'}+\frac{b'+c'}{a'}+\frac{c'+a'}{b'}\geq2\sqrt{(a'+b'+c')\left(\frac{a'}{b'c'}+\frac{b'}{c'a'}+\frac{c'}{a'b'}\right)},$$ where $a'$, $b'$ and $c'$ are positives such that $c'={\rm mid}\{a',b',c'\},$ which you made already.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of a sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ Given sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ where a is some natural number.
Show $\lim_{n \to \infty}p_n=0$
| This is a direct consequence of the central limit theorem!
EDITED: With probabilistic intuition, $p_n\rightarrow 0$ is obvious. It is just about tail probability.
Detailed proofs
(Proof using the Chebyshev inequality)
$p_n$ is the probability
\begin{align}
\mathbb{P}\left(X \leq \frac{a\sqrt{n}-1}{2}\right)
\end{align}
where $X$ follows the binomial random variable $B(n,1/2)$. Then, the Chebyshev inequality gives
\begin{align}
\mathbb{P}\left(X \leq \frac{a\sqrt{n}-1}{2}\right) &= \mathbb{P}\left(X-\frac{n}{2} \leq \frac{a\sqrt{n}-1}{2}-\frac{n}{2}\right)\\
&\leq \mathbb{P}\left(\left| X-\frac{n}{2}\right| \geq \frac{n}{2}-\frac{a\sqrt{n}-1}{2}\right)\\
&\leq \frac{n/4}{\left(\frac{n}{2}-\frac{a\sqrt{n}-1}{2}\right)^2} \rightarrow 0.
\end{align}
(Proof using the central limit theorem)
Consider a sequence of i.i.d. Bernoulli random variables $X_1, X_2, \ldots, X_n$. Note that
\begin{align}
p_n &= \mathbb{P}\left(X_1 + X_2 + \ldots +X_n \leq \frac{a\sqrt{n}-1}{2}\right)\\
&= \mathbb{P}\left(\frac{X_1 + X_2 + \ldots +X_n-\frac{n}{2}}{\sqrt{n}} \leq \frac{\frac{a\sqrt{n}-1}{2}-\frac{n}{2}}{\sqrt{n}}\right)
\end{align}
On the other hand, by the central limit theorem, the distribution of
\begin{align}
\frac{X_1 + X_2 + \ldots +X_n-\frac{n}{2}}{\sqrt{n}}
\end{align}
converges to the normal distribution $\mathcal{N}(0,1/2)$. Hence $p_n \rightarrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ for the given conditions. I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.
The proof provides me with the information that $x^2\gt 2$ and $x>0$.
I've taken the following steps to simplify the algebra...to the point where I arrive at:
$\frac{(x^2+2)^2}{(2*x)^2} = \frac {x^4+4*x^2+4}{4*x^2}$
I simplified this further to produce: $1 + \frac {1}{4}*x^2 + \frac{1}{x^2}$
In looking at this simplification, I see that I need to demonstrate (remembering the conditions), $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
Now, $\frac{1}{4}*x^2 \gt \frac{1}{2}$ and $\frac{1}{x^2}\lt \frac{1}{2}$ but I cannot figure out how to conclude that the addition of the two outputs something greater than 1. I imagine that I somehow need to show that the first term $\frac{1}{4}*x^2$ is somehow bigger than $0.5$ than $\frac{1}{x^2}$ is less than $0.5$.
In attempt to address this, I performed the following operation:
$\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} = \frac{1}{4}*x^4$
I know that $x^4$ must be greater than $4$...therefore $\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} \gt 1$.
Have I therefore successfully demonstrated that the addition of the two fractions:
$\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
and can conclude that:
$(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ ?
Edit: While the answers below also provide the solution to my question (albeit using a different strategy from the very beginning), I have figured out how to solve my question by finishing the procedure that I pursued above
Specifically, revisiting $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$, let's subtract $1$ from either side and then multiply through by $4x^2$ to produce:
$x^4-4x^2+4 \gt 0$ (multiplying by $4x^2$ will not change the direction of the inequality because $x\gt0$ )
We can factor this to $(x^2-2)^2$ and because $x \gt 0$, this number will always be greater than $0$.
| Note that
$$\frac{\frac{1}{4}\cdot x^2}{\frac{1}{x^2}} \gt 1 \implies \frac{1}{4}\cdot x^2-\frac{1}{x^2}>0$$
and then
$$\frac{1}{4}\cdot x^2+\frac{1}{x^2}>\frac{2}{x^2}$$
For the solution we can proceed as follows
$$\left(x-\left(\frac{x^2-2}{2 x}\right)\right)^2 \gt 2$$
$$\left(x^2-2x\left(\frac{x^2-2}{2x}\right)+\frac{(x^2-2)^2}{4x^2}\right) \gt 2$$
$$\frac{(x^2-2)^2}{4x^2} \gt 0 $$
which is true for $x^2\neq 2$ and $x\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all five digit number $\overline{abcde}$ such that $\overline{abcde} = \overline{(ace})^2$ Find all five digit number $\overline{abcde}$ such that $$\overline{abcde} = \overline{(ace})^2$$
This question popped in my mind while solving other elementary numbers and I have been trying to solve it ever since but without any luck.
My Take : Since the digit place of $e^2$ should be equal to digit's place of $e$ , So the only possible values of $e$ are $0,1,5$ and $6$
Also since the First digit of the numbers are equal , We can conclude that the the only possible values of $a=1$.
Hence our number can take the following form :
$$(1bcd0),(1bcd1),(1bcd5),(1bcd6)$$
But how do we further solve this?
Also Another interesting part of this question would be to solve for $\overline{abcd}$ such that $$\overline{abcd} = \overline{(bd)}^2$$
| We can minimize trial and error with some clever use of modular arithmetic.
Let $N=100a+10c+e$ be the square root. Thus $N^2\equiv e^2$ and we require also $N^2\equiv e\bmod 10$. Therefore $e^2\equiv e$ forcing $e\in\{0,1,5,6\}$.
We also know that $(100a)^2<10000(a+1)$ or $a^2<a+1$ forcing $a=1$. Then $N^2<20000$ but $145^2>140×150=21000$, therefore $N<145$. This result together with the earlier constraint on $e$ leaves only eighteen candidates, which can be exhaustively searched with little trouble; but we can do even better than that.
Consider the case $e=0$. Then $N=100+10c$ (with $a=1$) and $N^2=10000+2000c+100c^2$. For the hundreds digit in $N^2$ to match $c$ we must then have $c^2\equiv c\bmod 10$. This constraint admits$c\in\{0,1,5,6\}$, but only $0$ and $1$ satisfy the bounty $N<145$ which implies $c\le 4$. Thereby we identify
$100^2=10000$
$110^2=12100$
For $e=1$ we have
$N^2=10000+2000c+100(c^2+2)+20c+1$
With $c\le 4$, $20c+1<100$ and thus the hundreds digit is $\equiv c^2+2\bmod 10$. Therefore we must satisfy
$c^2-c+2\equiv 0\bmod 10$
which has a discriminant that is not a quadratic residue $\bmod 5$. So nobody's home here.
The cases $e=5$ and $e=6$ are left to the reader; they are handled similarly to $e=1$ as described above. For these cases $N<145$ implies $c\le 3$ which will then fix the hundreds digit of $N^2$ as $\equiv c^2+c$ ($e=5$) or $\equiv c^2+c+2$ ($e=6$). We will then get only one additional solution which the reader can find. I list the complete solution set as (with $x$ digits to be filled in):
$100^2=10000$
$1xx^2=1xxxx$
$110^2=12100$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
| Let $a_n=3^{3n}+3^{2n}+3^{n} + 1 = (3^3)^n+(3^2)^n+(3^1)^n + (3^0)^n$.
Since
$$(x-3^3)(x-3^2)(x-3^1)(x-3^0)= x^4 - 40 x^3 + 390 x^2 - 1080 x + 729$$
we have
$$
a_{n+4} =40 a_{n+3} - 390 a_{n+2} + 1080 a_{n+1} - 729a_{n}
$$
The claim that $a_n$ is a multiple of $4$ for all $n$ follows by induction since it is true for $n=0,1,2,3$.
(The actual coefficients in that recurrence do not matter. It only matters that they are integers.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the value of the periodic continued fraction with given terms
Find the value of the periodic continued fraction with the terms
$1, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, . . .$
We see that it starts to be periodic after $1$, i.e, $3,4,3,2$ then $3,4,3,2$ etc...
I know that $x= \frac{A_{k+1}}{B_{k+1}}$ = $\frac{A_{k-1}+x.A_K}{B_{k-1}+x.B_K}$ where $q_{k+1}=x$
I have $$x={3+\cfrac{1}{4+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{x}}}}}$$ for the periodic part. If I use my formula to compute the right-hand side, I will end up with a quadratic in $x$. Solving for $x$ gives me the following quadratic formula: $5x^2-14x-7=0$, then I did $1+\frac1x$ to find the whole continued fraction, and I got the same quadratic equation. Is this correct?
| Yes, you're right. We have $5x^2-14x-7=0$. This is the answer.
\begin{eqnarray}
x&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{2+\dfrac{1}{x}}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{\dfrac{2x+1}{x}}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{x}{2x+1}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{\dfrac{6x+3}{2x+1}+\dfrac{x}{2x+1}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{\dfrac{7x+3}{2x+1}}}\\
&=&3+\dfrac{1}{4+\dfrac{2x+1}{7x+3}}\\
&=&3+\dfrac{1}{\dfrac{28x+12}{7x+3}+\dfrac{2x+1}{7x+3}}\\
&=&3+\dfrac{1}{\dfrac{30x+13}{7x+3}}\\
&=&3+\dfrac{7x+3}{30x+13}\\
&=&\dfrac{90x+39}{30x+13}+\dfrac{7x+3}{30x+13}\\
&=&\dfrac{97x+42}{30x+13}\\
\end{eqnarray}
Now we have
\begin{eqnarray}
x&=&\dfrac{97x+42}{30x+13}\iff\\
30x^2+13x&=&97x+42\iff\\
30x^2-84x-42&=&0\iff\\
5x^2-14x-7&=&0
\end{eqnarray}
Solving this equation we have
$$x_{1,2}=\dfrac{14\pm\sqrt{56}}{10}=\dfrac{14\pm 2\sqrt{14}}{10}=\dfrac{7}{5}\pm \dfrac{1}{5}\sqrt{14}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$. While trying to come up with an answer to this question without using the theorem used in the existing answer, whose proof is non-trivial, I tried to find
$$ \tag{1}
\sup_{a a^* + b b^* = 1} | a^2x + aby |
$$
for some fixed complex numbers $x, y \in \mathbb{C}$
In the question the conditions $|x| = 1$ and $y \ne 0$ are added but I am interested if there's also a more general solution without those constrains.
The task of the question is to prove that $$(1)> |x|.$$
My progress
Aiming to use Lagrange multipliers, I expanded the term $a^2x + aby$ in terms of $n_1 := \Re(n)$ and $n_2 := \Im(n)$ for $n \in \{a,b,x,y\}$ to then use that
$$\tag{2}
|z|
= \sqrt{z_1^2 + z_2^2}
$$ for all $z \in \mathbb{C}$. I ended up with
$$
\Re(a^2x + aby)
= a_1 b_1 y_1 + a_1 b_2 y_2 + a_2 b_1 y_2 - a_2 b_1 y_2 + a_2^1 x_1 + a_2^2 x_1
$$
and
$$
\Im(a^2 x + aby)
= a_1 b_1 y_2 - a_1 b_2 y_1 + a_2 b_1 y_1 + a_2 b_2 y_2 + a_1^2 x_2 + a_2^2 x_2.
$$
Plugging this into (2) yields the square root of more than 40 terms. Is there a nice way to factor this or to approach this?
If this were only a real problem, we'd have
$$
\sup_{a^2 + b^2 = 1} ax^2 + a b y
= \sup_{|a| \le 1} a + a \sqrt{1 - a^2} y
$$
Setting the derivative w.r.t to $a$ equal to zero we obtain
$$
0 = \sqrt{1 - a^2} y + 1 - \frac{a^2 y}{\sqrt{1 - a^2}},
$$
giving
$$
a^2 = \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}.
$$
Plugging the positive root in gives
$$
a + a \sqrt{1 - a^2} y
= \left(\sqrt{\frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}}\right)\left(1 + \sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}} \cdot y\right),
$$
which looks approximately linear in $y$ when plotted with WolframAlpha.
Unfortunately, $\Re\left(\sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}}\right) = 0$ for all $y \in \mathbb{R}$.
| Here is the solution to Christian Blatter's maximization problem, writing $p$ for $\psi$:
We have
$$f(p)
=\cos^2(p)+\cos(p)\sin(p)y
=\frac12(\cos(2p)+1)+\frac12\sin(2p) y
$$
so maximizing this is the same as
$g(q)
:=\cos(q)+\sin(q)y
$
with $p = q/2$.
Let
$\tan(r) = y$,
so
$\sin(r)
=\frac{y}{\sqrt{1+y^2}}
$
and
$\cos(r)
=\frac{1}{\sqrt{1+y^2}}
$.
Then
$\begin{array}\\
g(q)
&=\cos(q)+\sin(q)y\\
&=\sqrt{1+y^2}(\frac1{\sqrt{1+y^2}}\cos(q)+\frac{y}{\sqrt{1+y^2}}\sin(q))\\
&=\sqrt{1+y^2}(\cos(r)\cos(q)+\sin(r)\sin(q))\\
&=\sqrt{1+y^2}\cos(r-q)\\
\end{array}
$
This is maximized when
$r-q=0$
so that
$q = r
=\arctan(y)
$
or
$2p = r
$
or
$p = \frac12\arctan(y)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$
I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?
| Rearrange the equation as
$$\frac{2x}{x^2-25}-\frac{1}{x-5}-\frac{1}{x+5}
=\frac{2x+1}{x^2-25}-\frac{2x}{x^2-25}= \frac{1}{x^2-25}=0$$
which leads to the solutions
$$x=\pm \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3392171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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