Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why is $\int \sec x\,dx$ equal to $\ln | \sec x + \tan x |$? Everyone knows that $\int \sec x\,dx = \ln | \sec x + \tan x |$?
But how to reach it through a conscious deduction, through a clear and objective way?
| It's almost always correct to revert any trig function in an integral to its definition in terms of $\sin x$ and $\cos x$.
So write $$\sec x = \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} = \frac{\cos x}{1-\sin^2 x}$$ Then substitute $u=\sin x$ and $du=\cos x\,dx$. That leads to $$\int \frac{du}{1-u^2}$$ which you then apply partial fractions to get an expression in terms of $\ln$. $$\frac{1}{1-u^2}=\frac{1}{2}\left(\frac 1{1-u} + \frac{1}{1+u}\right)$$
So this gives you $$\frac{1}{2}\left(\ln\left|1+u\right|-\ln\left|1-u\right|\right)=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|$$
Substituting $u=\sin x$ into $\frac{1+u}{1-u}$ and multiply numerator by denominator by $1+\sin x$ we get:
$$\frac{1+u}{1-u} = \frac{(1+\sin x)^2}{1-\sin^2 x} = \left(\frac{1+\sin x}{\cos x}\right)^2=\left(\sec x +\tan x\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/717528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the limit $\lim_{x \to 3}\frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}$ with no l'Hopital rule Solve the limit without using de l'Hopital rule: $$\lim_{x \to 3}\frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}$$
It looks like classical example of squeeze theorem, but I completely can't develop the intuition how to estimate lower and upper bounds. Any help would be appreciated.
Edit (solution):
According to hint of user @Omnomnomnom, we notice that the limit is derivative.
$$x=h+3$$
$$\lim_{h \to 0} \frac{\sqrt[3]{4+h}-\sqrt[3]{4}}{h}$$
It is derivative of $\sqrt[3]{x}$ in the point $x=4$. Thus:
$$\frac{d}{dx}x^{\frac{1}{3}} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$$
Substituting the point $x=4$
$$\frac{1}{3\cdot4^{\frac{2}{3}}} = \frac{1}{3\cdot2^{\frac{4}{3}}} = \frac{1}{6\sqrt[3]{2}}$$
| Let us put $y = x + 1$ so that $y \to 4$ and then we have $$\begin{aligned}L &= \lim_{x \to 3}\frac{\sqrt[3]{x + 1} - \sqrt[3]{4}}{x - 3}\\
&= \lim_{y \to 4}\frac{y^{1/3} - 4^{1/3}}{y - 4}\\
&= \frac{1}{3}\cdot 4^{(1/3) - 1} = \frac{1}{3\cdot 4^{2/3}} = \frac{1}{3\sqrt[3]{16}}\end{aligned}$$ We have used the formula $$\lim_{y \to a}\frac{y^{n} - a^{n}}{y - a} = na^{n - 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/717647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How find this sum $\sum_{k=0}^{\left[\frac{n}{2}\right]}\frac{(-1)^k\binom{n-k}{k}}{n-k}$ How find this sum
$$\sum_{k=0}^{\left[\dfrac{n}{2}\right]}\dfrac{(-1)^k\binom{n-k}{k}}{n-k}$$
My try:since
$$\dfrac{(-1)^k}{n-k}=\int_{-1}^{0}x^{n-k-1}dx$$
then I can't
Thank you very much!
| Following the well-known technique by Wilf we introduce the ordinary
generating function $f(z) = \sum_{n\ge 1} f_n z^n$ where $f_n$ is our
sum so that
$$ f(z) = \sum_{n\ge 1} z^n
\sum_{k\ge 0} (-1)^k {n-k \choose k} \frac{1}{n-k}$$
and re-arrange terms to get
$$ f(z) = \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k
\sum_{n\ge 2k} {n-k\choose k} \frac{z^n}{n-k}$$
which is
$$ f(z) = \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k
\sum_{m\ge 0} {m+k\choose k} \frac{z^{2k+m}}{m+k}
\\= \log\frac{1}{1-z} + \sum_{k\ge 1} (-1)^k z^{2k}
\sum_{m\ge 0} \frac{m+k}{k} {m+k-1\choose k-1} \frac{z^m}{m+k}
\\= \log\frac{1}{1-z} + \sum_{k\ge 1} (-1)^k \frac{z^{2k}}{k}
\sum_{m\ge 0} {m+k-1\choose k-1} z^m
\\ = \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k \frac{z^{2k}}{k} \frac{1}{(1-z)^k}$$
which finally becomes
$$ \log\frac{1}{1-z} + \log \frac{1}{1+z^2/(1-z)}
= \log\frac{1}{1-z} + \log \frac{1-z}{1-z+z^2}
= \log \frac{1}{1-z+z^2}.$$
Now the roots of the denominator of the fractional term are
$$\rho_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}i}{2}$$
and in particular we observe that $$1-z+z^2 = (1-z/\rho_1)(1-z/\rho_2)$$
(which is not true for the general quadratic), so we have
$$f(z)
=
- \log(1-z/\rho_1)
- \log(1-z/\rho_2)
= \log\frac{1}{1-z/\rho_1} + \log\frac{1}{1-z/\rho_2}.$$
Extracting coefficients from this formula we obtain the following
exact expression for $f_n:$
$$f_n =
\frac{\rho_1^{-n}}{n} + \frac{\rho_2^{-n}}{n}
= \frac{\rho_1^n}{n} + \frac{\rho_2^n}{n}$$
because $\rho_1\rho_2 = 1.$
Here we have made repeated use of the following expansion which is popular in combinatorics because it is the generating function of the species of cycles:
$$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}.$$
Two similar calculations can be found at this MSE link.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int \frac{\operatorname d \! x}{\sin^4{x}+\cos^4{x}+\sin^2{x}\cos^2{x}}$ How do you integrate
$$\frac{1}{\sin^4{x}+\cos^4{x}+\sin^2{x}\cos^2{x}}$$
or simply
$$\frac{1}{1-\left(\frac{\sin{2x}}{2}\right)^2}.$$
| $$I=\int\frac{dx}{\sin^4x+\cos^4x+\sin^2x\cos^2x}=\int\frac{(1+\tan^2x)\sec^2x}{1+\tan^2x+\tan^4x}dx$$
Setting $\tan x=u,$
$$I=\int\frac{1+u^2}{1+u^2+u^4}du $$
Now as $\displaystyle 1+u^2+u^4=(1+u^2)^2-u^2=(1+u+u^2)(1-u+u^2),$
$\displaystyle\frac{2(1+u^2)}{1+u^2+u^4}=\frac{(1+u+u^2)+(1-u+u^2)}{(1+u+u^2)(1-u+u^2)}=\frac1{1+u+u^2}+\frac1{1-u+u^2}$
Now, $\displaystyle\frac1{1+u+u^2}=\frac4{(2u+1)^2+(\sqrt3)^2}$
Using Trigonometric substitution, set $\displaystyle2u+1=\sqrt3\tan\phi$
Similarly, for $\displaystyle\frac1{1-u+u^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\binom{n+2}{k+2} = \binom{n}{k+2} + 2\binom{n}{k+1} + \binom{n}{k}$ I have to prove the following statement
$$\binom{n+2}{k+2} = \binom{n}{k+2} + 2\binom{n}{k+1} + \binom{n}{k}$$
I know, I have to use the following fact
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$
but I can't seem to figure out how. Some hints would be appreciated
Thanks
| I kept in mind the recursive characteristics of the Pascal triangle to do this one, together with the relation you had to consider:
$$\binom{n+2}{k+2} = {\color{red}{\binom{n+1}{k+2}}} + {\color{blue}{\binom{n+1}{k+1}}} \\
= {\color{red}{\binom{n}{k+2} + \binom{n}{k+1}}} + {\color{blue}{\binom{n}{k+1} + \binom{n}{k}}}\\
={\color{red}{\binom{n}{k+2}}} + {\color{purple}{2\binom{n}{k+1}}} + {\color{blue}{\binom{n}{k}}}$$
(The one in the middle is purple because red plus blue equals purple)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Shortest distance between ellipse and a line I was trying to find the shortest distance between the ellipse
$$\frac{x^2}{4} + y^2 = 1$$
and the line $x+y=4$. We have to find the point on the ellipse where
its tangent line is parallel to $x+y=4$ and find the distance between those two points.
However, when I used the implicit differentiation, I get
$$\frac{x}{2} + 2y\frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = \frac{-x}{4y}$$
If it's parallel to $x+y=4$, then we need $x=4y$. Do I just plug it into ellipse equation and solve for it and calculate the distance between the point and a line or am I doing it wrong? I just wanted to clarify. Any help would be appreciated. Thanks!
| Simpler solution:
Scale the horizontal axis such that $u=\frac{x}{2}$. Then, the ellipse equation becomes:
$$u^2+y^2=1$$
Likewise, the equation of the line becomes:
$$2u+y=4$$
Now we simply must find the closest distance between that line and the unit circle. This can be done fairly easily using analytic geometry. First find the equation of a line that goes through the origin and is perpendicular to the other line:
$$y=\frac{1}{2}u$$
Now find the intersection point of the two lines. I.e., solve the system:
$$2u+y=4$$
$$y=\frac{1}{2}u$$
The solutions are $u=\frac{8}{5}$ and $y=\frac{4}{5}$. The second line intercepts the unit circle at $u=\frac{2}{\sqrt{5}}$ and $y=\frac{1}{\sqrt{5}}$ (trigonometry not even required – use simple substitution).
So the shortest distance between the circle and the line is the distance between $(\frac{8}{5},\frac{4}{5})$ and $(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})$. If we rescale our coordinate plane by substituting back in: $u=\frac{x}{2}$, then we will find the shortest distance between our ellipse and line as being simply the distance between $(\frac{16}{5},\frac{4}{5})$ and $(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}})$. This is simply given by:
$$\sqrt{\Big(\frac{16}{5}-\frac{4}{\sqrt{5}}\Big)^2+\Big(\frac{4}{5}-\frac{1}{\sqrt{5}}\Big)^2}=\boxed{\frac{\sqrt{17(21-8\sqrt{5})}}{5}}$$
No calculus required!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many different 4 digit combinations will include at least one 7, assuming numbers cannot repeat I cannot get the correct answer - $2016$.
What I have tried so far is thus: the number $7$ can occur $1, 2, 3,$ or $4$ times. Since it is a combination we do not care if the number starts with zero or not since it is a combination and not a number. Therefore:
$(1*9*8*7 + 1*1*8*7 + 1*1*1*7 )*4 = 2268$
where $4$ is the order it can appear.
What should i do instead to get it correct?
| If there has to be exactly one 7, that 7 can go into 4 positions, and the other 3 digits can be arranged in $9 \cdot 8 \cdot 7$ ways. In all, $4 \cdot 9 \cdot 8 \cdot 6$ ways. If starting with 0 is not allowed, the ways of starting 0 in the above is $3 \cdot 8 \cdot 7$ ways, so you get $4 \cdot 9 \cdot 8 \cdot 6 - 3 \cdot 8 \cdot 7 = 1848$
If the 7 can repeat, you have to select 1, 2, 3, or 4 positions for the 7s; the others have to be arranged into the rest of the positions (here $x^{\underline{n}} = x (x - 1) \cdots (x - n + 1)$, for $n$ factors). By the same reasoning above:
$$
\binom{4}{1} \cdot 9^{\underline{3}}
+ \binom{4}{2} \cdot 9^{\underline{2}}
+ \binom{4}{3} \cdot 9^{\underline{1}}
+ \binom{4}{4} \cdot 9^{\underline{0}}
- \binom{3}{1} \cdot 8^{\underline{2}}
- \binom{3}{2} \cdot 8^{\underline{1}}
- \binom{3}{3} \cdot 8^{\underline{0}}
= 1957
$$
It seems the place you've got this from is mistaken.
| {
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"url": "https://math.stackexchange.com/questions/722749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$
my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^2+x^2z^2=q^2-2pr$$and
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)=(p^2-2q)^2-2(q^2-2pr)$$
so
$$\Longleftrightarrow 2[(p^2-2q)^2-2(q^2-2pr)]+2r+7\ge 5(p^2-2q)$$
$$\Longleftrightarrow 2p^4-8p^2q+4q^2+8pr-5p^2+10q+2r+7\ge 0$$
then I can't
This link has a similar problem:
see this
Maybe this problem can use AM-GM inequality,But I can't.Thank you
| Here is my solution.
You want to prove that:
$$
2(x^4+y^4+z^4)+2xyz-5(x^2+y^2+z^2)+7\geq0,
$$
for $x,y,z\geq0$. Let's rewrite this inequality in spherical coordinates:
$$
x=r\sin(\theta)\cos(\varphi),\\
y=r\sin(\theta)\sin(\varphi),\\
z=r\cos(\theta),
$$
then:
$$
a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7\geq0,
$$
for $r\geq0$, $\theta,\varphi\in[0,\pi/2]$, where
$$
a(\theta,\varphi)=2\{\cos^4(\theta)+\tfrac{1}{4}[3 + \cos(4\varphi)] \sin^4(\theta)\},
$$
and
$$
b(\theta,\varphi)=\tfrac{1}{2}\sin(\theta)\sin(2\theta)\sin(2\varphi).
$$
These functions $a(\theta,\varphi)$ and $b(\theta,\varphi)$ are simply obtained by writing $2(x^4+y^4+z^4)$ and $2xyz$ in spherical coordinates and using standard trigonometric relations such as $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ and $\cos(\alpha)^4+\sin(\alpha)^4=\tfrac{1}{4}[3 + \cos(4\alpha)]$.
Note that in the region $\theta,\varphi\in[0,\pi/2]$, we have $b(\theta,\varphi)\geq0$.
Next, the derivative of the polynomial $p(r)=a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7$ equals $0$ in three points,
$$
0,\frac{-3 b(\theta,\varphi) \pm \sqrt{160 a(\theta,\varphi) + 9 b^2(\theta,\varphi)}}{8 a(\theta,\varphi)}.
$$
By computing the second derivative we immediately obtain that $r=0$ is a maximum. On the other hand, since $b(\theta,\varphi)\geq0$ in $\theta,\varphi\in[0,\pi/2]$, the other possible solution (which has to be a minimum) is
$$
r_{\rm min}(\theta,\varphi)=\frac{-3 b(\theta,\varphi) + \sqrt{160 a(\theta,\varphi) + 9 b^2(\theta,\varphi)}}{8 a(\theta,\varphi)}.
$$
So we have,
$$
a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7\geq a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7.
$$
By plotting $a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7$ in the region $\theta,\varphi\in[0,\pi/2]$ with any standard program of numerical plotting we obtain something like this
Hence, certainly the minimum of $a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7$ in the region $\theta,\varphi\in[0,\pi/2]$ is $0$, which in rectangular coordinates is reached at the point $x=y=z=1$. That proves the required inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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If we take 5 shots with a basketball, what's the probability that you make 2 given you make at least 1? Probability of making a shot is $\displaystyle \frac{3}{5}$.
So $2$ shots would be $\displaystyle \dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$ right?
| To find the probability in question, you have to first check what the probability of making at least $1$ shot is, and also the probability of making exactly $2$ shots, and then divide the latter by the former. That is, the probability is $$\frac{\text{(probability of landing 2 shots)}}{\text{(probability of landing at least 1 shot)}}$$
Probability of landing at least $1$ shot:
This is the same as $1$ minus the probability of missing all $5$ shots. The answer, therefore, is $$1 - \left(\frac{2}{5}\right)^5$$
Probability of landing exactly $2$ shots:
This is the answer that you had, i.e. $$\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3$$
Final answer:
$$\frac{\dbinom{5}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^3}{1-\left(\frac{2}{5}\right)^5} = \boxed{\frac{240}{1031}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Prove $\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$ How to prove:
$\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
Please help.
| Common denominator:
$$\frac{\sin^2\theta+1+\cos^2\theta+2\cos\theta}{\sin\theta(1+\cos\theta)}$$
$$=\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{2}{\sin\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that:$ BC^2= 3CM^2 + AC^2 $from this triangle problem? In the triangle $\triangle ABC$, angle $\angle A$ is larger than angle $\angle B$.
We choose points $M$ and $N$ at $AB$ such that $AM=MN=NB$.
How to prove that: $BC^2= 3CM^2 + AC^2$?
Which theorem(s) should I use to prove this problem?
Thanks
| Points $A,M,N$ and $B$ are equally spaced on $(AB)$.
Set a coordinate system with origin at $A$ and $(AB)$ as first
axis.
Then the coordinates of $A,M,N$ and $B$ will be :
$$
A(0,0) \ M(t,0) \ N(2t,0) \ B(3t,0)
$$
for some constant $t$. Now, if we denote by
$(x,y)$ the coordinates of $C$, one has
$$
\begin{array}{lcl}
BC^2 &=& (x-3t)^2+y^2=x^2+y^2-6xt+9t^2 \\
3CM^2 &=& 3((x-t)^2+y^2)=3x^2+3y^2-6xt+3t^2 \\
AC^2 &=& (x-0)^2+y^2=x^2+y^2 \\
\end{array}
$$
and hence
$$
BC^2-(3CM^2+AC^2)=-3x^2-3y^2+6t^2
$$
so your equality holds iff $x^2+y^2=2t^2$. You probably forgot
a condition on $C$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any other method to solve $\sin x -\sqrt 3 \cos x=1$? Consider
$$
\sin x -\sqrt 3 \cos x=1
$$
for $0\leq x \leq 2\pi$.
I solved it by converting $\sin x -\sqrt 3 \cos x$ to $2\sin(x-\pi/3)$ as follows
\begin{align*}
\sin x -\sqrt 3 \cos x &= r\sin(x-\theta)\\
&= r\sin x\cos\theta -r \cos x\sin\theta\\
\end{align*}
we have
\begin{align*}
r\cos \theta &= 1\\
r\sin \theta &=\sqrt 3
\end{align*}
where $r=2$ and $\theta=\frac{\pi}{3}$ are the solution.
So the equation becomes
$$
2\sin (x-\pi/3) =1
$$
and the solution are $x=\pi/2$ and $x=7\pi/6$.
Is there any other method to solve it?
| We know that $\cos{\pi\over3}={1\over2}$ and $\sin{\pi\over3}={\sqrt3\over2}$, so
$$\begin{align}
\sin x-\sqrt3\cos x=1&\implies{1\over2}\sin x-{\sqrt3\over2}\cos x={1\over2}\\
&\implies\sin x\cos{\pi\over3}-\cos x\sin{\pi\over3}={1\over2}\\
&\implies\sin(x-{\pi\over3})={1\over2}\\
&\implies x-{\pi\over3}={\pi\over6}\quad\text{or}\quad x-{\pi\over3}=\pi-{\pi\over6}\\
&\implies x={\pi\over2}\quad\text{or}\quad x={7\pi\over6}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove the limit using definition Prove the limit by using the definition
$$\lim_{(x,y) \rightarrow (0,0)} \frac{x^{2}y^{2}}{ \sqrt{y^2 + 1}-1 } = 0 $$
| First multiply the fraction by $\frac{\sqrt{y^2+1} + 1}{\sqrt{y^2+1} + 1}$ in order to simplify the denominator then switch to polar coordinates $x\mapsto r\cos \theta, \; y \mapsto r\sin \theta$ and take the limit $r \to 0$.
In other words,
\begin{eqnarray*}
\lim_{(x,y)\to 0} \frac{x^2y^2}{\sqrt{y^2+1}-1} =& \lim_{(x,y)\to 0} \frac{x^2y^2}{\sqrt{y^2+1}-1} \frac{\sqrt{y^2+1} + 1}{\sqrt{y^2+1} + 1} \\
=& \lim_{(x,y)\to 0} \frac{x^2y^2(\sqrt{y^2+1} + 1)}{y^2+1-1} \\
=& \lim_{(x,y)\to 0} \frac{x^2y^2(\sqrt{y^2+1} + 1)}{y^2} \\
=& \lim_{r \downarrow 0} \frac{r^4(\cos^2 \theta \sin^2 \theta)(\sqrt{r^2 \sin^2 \theta+1} + 1)}{r^2\sin^2 \theta} \\
=& \lim_{r \downarrow 0} r^2\cos^2 \theta (\sqrt{r^2 \sin^2 \theta+1} + 1) \\
=& 0.
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Showing function is injective? I want to show that this function is injective -
$f(x) = \frac{x}{1 - x^2}$
So when $f(x) = f(y)$ I should have $x = y$
$\frac{y}{1 - y^2} = \frac{x}{1 - x^2}$
$x - xy^2 = y - x^2y$
$x - y = xy^2 - x^2y$
$x - y = xy(y - x)$
$\frac{x-y}{y-x} = xy$
$\frac{x-y}{x-y} = -xy$
$-xy = 1$
$x = -\frac{1}{y}$
Where am I going wrong?
| $x - y = xy(y - x)$
$(x-y)(1+xy)=0$
$x=y$ or $x=-1/y$.
So the function is not injective.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Help with induction proof for formula connecting Pascal's Triangle with Fibonacci Numbers I am in the middle of writing my own math's paper on the topic of Pascal's Triangle. During the investigation I have came up with a formula for counting elements of Fibonacci Sequence using the entries from Pascal's Triangle (binomial coefficients). I know that there is a general formula for that (including floor of n), which I have explained, but what I also wanted to do in my work, was to create two formulas for counting even entries of Fibonacci Sequence and the odd ones. What I am struggling with however, is how to prove it using the induction. I attach the screenshot of the page that deals directly with ODD numbers. If you guys could help me with the proof either for even numbers, for odd or for the general formula, I would greatly appreciate it.
| General term: $s_n = \displaystyle \sum_{0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n-1-r}{r} \text{ for } n \in \mathbb{N} \tag{1}$
I think you struggled because the binomial sum $s_{n+2}$ has a decomposition that involves both its even and odd numbered predecessors, and not just any one of them, as you were trying in the inductive step.
Decomposition: $s_{n+2} = s_{n+1} + s_{n} \tag{2}$
This is easy to prove by using the combinatorial identity $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$. See here and here.
We can then use this property along with strong induction to prove:
$P(n): s_n = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \text{ true } \forall n \in \mathbb{N}\tag{3}$
Proof-sketch:
Base case: Show $P(1)$ and $P(2)$ to be true (by evaluating both sides).
Inductive step: Assume $P(n)$ and $P(n+1)$ true.
Then $\begin{aligned}s_{n+2} & = s_{n+1} + s_{n} \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+1} + \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n\left(\frac{3 + \sqrt{5}}{2}\right) - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n\left(\frac{3 - \sqrt{5}}{2}\right) \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+2} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+2} \\ & \implies P(n+2) \text{ true }\end{aligned}$
This completes our proof. $ \blacksquare$
Note: The formulas for odd $n$ and even $n$ follow automatically from $(3)$ by the removal of the floor function in $s_n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating $\int_0^{2 \pi} \sin^4 \theta\: \mathrm{d} \theta$ Evaluate the following integral:
$$\int_0^{2 \pi} \sin^4 \theta \:\mathrm{d} \theta$$
My approach: Parametrize and obtain $$\frac{1}{(2i)^4} \int_{|z|=1} \left (z-\frac{1}{z} \right)^4 \frac{1}{iz}\:\mathrm{d}z=\frac{1}{(2i)^4} \int_{|z|=1} \left (\frac{(z+1)(z-1)}{z} \right)^4 \frac{1}{iz}\:\mathrm{d}z$$
Can I directly use the residue theorem from here with a residue at $z=0$?
| $$
I=\int \sin^4xdx=\int\sin^3x\sin xdx=\int \sin^3x(-\cos x)'dx=-\sin^3x\cos x+\int3\sin^2x\cos^2dx=-\sin^3x\cos x+3\int(\frac{1}{2}\sin 2x)^2dx
$$
$$
\cos 4x=1-2\sin^22x\Rightarrow\sin^22x=\frac{1-\cos4x}{2}\\
$$
$$
I=-\sin^3x\cos x+\frac{3}{4}\int\frac{1-\cos4x}{2}dx=-\sin^3x\cos x+\frac{3}{8}\int (1-\cos4x)dx=\\=-\sin^3x\cos x+\frac{3}{8}x-\frac{3}{32}\sin4x+c\\
\Rightarrow\int_0^{2\pi} \sin^4xdx=\frac{3\pi}{4}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$\lfloor (25x-2)/4 \rfloor =(13x+4)/3$ This problem is designated for those who have the basic knowledge of floor brackets
I know the answer will be 6. But tell me how.
| It is clear that $\dfrac{13x + 4}{3}$ must be an integer so $x$ must have the form of $2 + 3k$.
Substitute this into the equation and now we must solve:
$$\left\lfloor\frac{48+75k}{4}\right\rfloor = \left\lfloor 12 + \frac{75k}{4}\right\rfloor = 12 + \left\lfloor \frac{75k}{4}\right\rfloor = 13k + 10$$
$$ \left\lfloor \frac{75k}{4}\right\rfloor = 13k - 2$$
Find an approximate solution by solving $\frac{75k}{4} = 13k - 2$ and reaching $k = -\frac{8}{23}$.
We will now look for solutions on intervals around this point.
$$ \left.\left\lfloor \frac{75k}{4}\right\rfloor\right|_{k = -8/23} = -7$$
Let $13k -2 = -8,$ $k = -\dfrac{6}{13}.$
Let $13k -2 = -7,$ $k = -\dfrac{5}{13}.$
Let $13k -2 = -6,$ $k = -\dfrac{4}{13}.$
Let $13k -2 = -5,$ $k = -\dfrac{3}{13}.$
From these possible values of $k$ we find that $x = \dfrac{8}{13}, \dfrac{11}{13}, \dfrac{14}{13}, \dfrac{17}{13}$.
Plugging in to the original equation to verify we find two extraneous solutions and that $\boxed{x = \dfrac{14}{13}, \dfrac{17}{13}}$.
This method will not work well in all cases. Because $\frac{75}{4} = 18.75 \not\approx 13$ I assumed that there are a maximum of four solutions. In a case where we must solve $ \lfloor \frac{75k}{4}\rfloor = 19k - 2$ there are five solutions. In $ \lfloor \frac{75k}{4}\rfloor = 18.75k - 2$ there will be infinitely many.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Compute the condition number of matrix A=$\left(\begin{array}{cc} 1 & -1\\0&-1\end{array}\right)$ in norm $||.||_2$ Compute the condition number of matrix A=$\left(\begin{array}{cc} 1 & -1\\0&-1\end{array}\right)$ in norm $||.||_2$.
Since $A^{-1}$ is same as $A$. How to dela with that?
| The condition number of $A$ is the ratio of the maximum and minimum singular values of $A$, and the singular values of $A$ are the square roots of the eigenvalues of $A^T A = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$.
The characteristic polynomial of $A^T A$ is $\lambda^2 - 3 \lambda + 1$, so the eigenvalues of $A^T A$ are $\frac{3 \pm \sqrt{5}}{2}$. Thus the condition number of $A$ is
\begin{equation}
\kappa = \sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}} = \frac{3 + \sqrt{5}}{2}.
\end{equation}
Alternatively, we could note that the norm of $A$ is the maximum singular value of $A$:
\begin{equation}
\|A\| = \sqrt{\frac{3 + \sqrt{5}}{2}}.
\end{equation}
Using your observation that $A^{-1} = A$,
we have $\kappa = \|A^{-1}\| \|A\| = \|A\|^2 = \frac{3 + \sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/736034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find an orthogonal matrix P and a diagonal matrix D such that $P^TAP$=D The symmetric matrix $A$ below has distinct eigenvalues $−6, −12$ and $−18$. Find an orthogonal matrix $P$ and a diagonal matrix $D$ such that $P^TAP=D$
$$
A
= \left[
\begin{array}{ccc}
-13 & 1 & 4 \\
1 & -13 & -4 \\
4 & -4 & -10 \\
\end{array}
\right]
$$
I know my $D$ matrix will be the eigenvalues along the diagonal. But how do I find $P$? I am having trouble with this computation. I spent 15 minutes doing the computation but I get it wrong all the time.
| Find the orthogonal eigenvectors $v_1,v_2,v_3$ corresponding to your $3$ eigenvalues, put them as columns into the matrix $P$ and that will do it.
Here are the details:
$$
A - (-6)I
= \left[
\begin{array}{ccc}
-7 & 1 & 4 \\
1 & -7 & -4 \\
4 & -4 & -4 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & -7 & -4 \\
-7 & 1 & 4 \\
1 & -1 & -1 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & -7 & -4 \\
0 & -6 & -3 \\
0 & 6 & 3 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & 0 & -1/2 \\
0 & 1 & 1/2 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
so a normalized evector for $\lambda_1 = -6$ is $v_1 = \frac{1}{\sqrt{6}}(1,-1,2)$,
$$
A - (-12)I
= \left[
\begin{array}{ccc}
-1 & 1 & 4 \\
1 & -1 & -4 \\
4 & -4 & 2 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & -1 & -4 \\
0 & 0 & 0 \\
0 & 0 & 18 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & -1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
so a normalized evector for $\lambda_2 = -12$ is $v_2 = \frac{1}{\sqrt{2}}(1,1,0)$, and
$$
A - (-18)I
= \left[
\begin{array}{ccc}
5 & 1 & 4 \\
1 & 5 & -4 \\
4 & -4 & 8 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & 5 & -4 \\
5 & 1 & 4 \\
1 & -1 & 2 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & 5 & -4 \\
0 & 6 & -6 \\
0 & -6 & 6 \\
\end{array}
\right]
\sim \left[
\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
so a normalized evector for $\lambda_3 = -18$ is $v_3 = \frac{1}{\sqrt{3}}(-1,1,1)$. Finally,
$$
P = \frac{1}{\sqrt{6}}\left[
\begin{array}{ccc}
1 & \sqrt{3} & -\sqrt{2} \\
-1 & \sqrt{3} & \sqrt{2} \\
2 & 0 & \sqrt{2} \\
\end{array}
\right]
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find this range of the fucntion $f(a,b)=\left(\frac{1}{a}+\frac{1}{b}+1\right)(a-3b+15)$ let $a,b>0$,and such $a+2b=3$,find range of the follow function
$$f(a,b)=\left(\dfrac{1}{a}+\dfrac{1}{b}+1\right)(a-3b+15)$$
My idea: since
\begin{align*}f(a,b)&=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{a+2b}{3}\right)(a-3b+5(a+2b))
=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{a+2b}{3}\right)(6a+7b)\\
&=6+\dfrac{7b}{a}+\dfrac{6a}{b}+7+\dfrac{1}{3}(6a^2+19ab+14b^2)
\end{align*}
Follow I want use AM-GM inequality,But I can't .Thank you
| As mentioned above, as $b \to 0_+$, we have $f(a, b) \to \infty$, so there is no maximum. For finding the minimum, it seems expressing it as a function of one variable (using the constraint) and calculus seems straightforward though messy. If you really must use inequalities, here is an approach.
$f(1, 1) = 39$, so the minimum cannot be higher than this. By rewriting the inequality and Cauchy-Schwarz, we have
$$\left(\frac{1}{a}+\frac{1}{b}+1\right)(a-3b+15) = \left(\frac{1}{a}+\frac{1}{b}+1\right)(5a+5b+3)\ge \left(\sqrt5+\sqrt5+\sqrt3\right)^2 \approx 38.49$$
As the equality cannot be achieved, this value is never achieved, but it shows that the minimum must be higher than this. So this gives us for the range that $f \in [m, \infty)$, where $38.49 < m \le 39$.
If the bound isn't sufficient and you need the exact value, note that we can write
$$a-3b+15 = a-3b+15-3k + k(a+2b) = (k+1)a+(2k-3)b+15-3k$$
So for any real $k \in [\frac32, 5]$, we can have by CS inequality:
$$f(a, b) = \left(\frac{1}{a}+\frac{1}{b}+1\right)\left((k+1)a+(2k-3)b+(15-3k \right)\ge \left(\sqrt{k+1}+\sqrt{2k-3}+\sqrt{15-3k}\right)^2 $$
We can find the minimum only if we can have equality condition for some $k$, i.e. we must have a solution for allowable $a, b, k$ s.t.
$$(k+1)a^2= (2k-3)b^2= 15-3k$$
While it looks terrible to try solving analytically, numerically, $a \approx 0.967588, b \approx 1.01621, k \approx 3.57291$ works, giving a minimum of $f\ge m \approx 38.9836$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $(x-2)^4+(x-3)^4=1$? I'm trying my hand on these types of expressions. how to solve $x$ and $y$ in $\displaystyle(x-2)^4+(x-3)^4=1$ and $\displaystyle (y-1)(y-2)(y-3)(y-4)=1$?
please write step by step solution ! thank u
| HINT: For the first case set $$z=\frac{x-2+x-3}2\iff 2z=2x-5$$
$$2^4=(2x-4)^2+(2x-6)^2=(z+1)^4+(z-1)^4=2\left(z^4+1+\binom42z^2\right)$$
Rearrange to form a Quadratic Equation in $z^2$
For the second set $\displaystyle a=\frac{y-1+y-2+y-3+y-4}4$
| {
"language": "en",
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"source": "stackexchange",
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Solving of an integral $$\int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx$$
Could anyone help me calculate this integral? Thanks in advance.
| Here's another way:
$$\int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx=\int \left(x^2 + \arctan(x)\right) d(\arctan x)=\\
\left(x^2 + \arctan(x)\right)\arctan x-\int \arctan xd(x^2 + \arctan(x))=\\
\left(x^2 + \arctan(x)\right)\arctan x-\int \arctan xd(x^2)-\int \arctan xd(\arctan(x))=\\
\left(x^2 + \arctan(x)\right)\arctan x-\dfrac{(\arctan x)^2}{2}-\int \arctan xd(x^2)$$
$$\text{Now, }\int \arctan xd(x^2)=x^2\arctan x-\int \dfrac{x^2}{x^2+1}dx$$
$$\text{This gives }\dfrac{(\arctan x)^2}{2}+\int \dfrac{x^2+1-1}{x^2+1}dx=\dfrac{(\arctan x)^2}{2}+x-\arctan x+C\\
\implies \int \left(\frac{x^2 + \arctan(x)}{1 + x^2}\right) dx=\dfrac{(\arctan x)^2}{2}+x-\arctan x+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use generating functions to find the number of partitions of $n>1$ that have an odd number of even parts $k=1,...,10$ Here are some examples where we find $f(n)$ - the number of partitions that satisfy our condition:
$\boldsymbol{2} = (1+1) \rightarrow \boldsymbol{f(2)=0} $
$\boldsymbol{3} = (1+1+1) = \boldsymbol{(1+2)} \rightarrow \boldsymbol{f(3)=1} $ (we have one '2' in (1+2), which satisfies the "odd number of even parts" condition)
$\boldsymbol{4} = (1+1+1+1) = \boldsymbol{(1+1+2)} = (2+2) = (1+3) \rightarrow \boldsymbol{f(4)=1} $
$\boldsymbol{5} = (1+1+1+1+1) = \boldsymbol{(1+1+1+2)} = (1+2+2) = (1+1+3) = \boldsymbol{(1+4)} \rightarrow \boldsymbol{f(5)=2} $
$\boldsymbol{6} = (1+1+1+1+1+1) = \boldsymbol{(1+1+1+1+2)} = (1+1+2+2) = \boldsymbol{(2+2+2)} = (1+1+1+3) = (3+3) = \boldsymbol{(1+2+3)} = \boldsymbol{(1+1+4)} = (2+4) = (1+5)\rightarrow \boldsymbol{f(6)=4} $
Now how can we do this for any number $n$, keeping in mind that all parts should be in range $1,...,10$?
| The generating function in the oeis entry provided by Wouter M. can be modified.
For all the parts to be in $1,2,\ldots ,10$, the g.f. is:
$\displaystyle
P(x)=\frac{1}{2}\left(1-\prod_{i=1}^{5}\dfrac{1-x^{2\, i}}{1+x^{2\, i}}\right)\left(\prod_{j=1}^{10}\dfrac{1}{1-x^i}\right)
$
If the partion is to have more than one part, subtract that by $\frac{x^2}{1-x^2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Several (advanced) L'Hospital problems Problems :
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{E}.\ \lim_{x\rightarrow0}\left(a^x+b^x\right)^\frac{1}{x}
\end{align}
$$
Here are my solutions concerning A~E.
A. $$\ln y=\tan\left(\frac{\pi x}{2}\right)\ln(x-2)=\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}$$
$$\lim_{x\rightarrow 1}\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}=_H\lim_{x\rightarrow1}\frac{\frac{\pi}{2}\sec^2(\frac{\pi x}{2})}{x-2}=\infty$$so, $$A\rightarrow \infty$$
Is this conclusion right?
C.
$$=_H\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{\frac{1}{1+x^2}-1}=_H\lim_{x\rightarrow0}\frac{\frac{-1}{2}\frac{1}{\sqrt{(1-x^2)^3}}(-2x)}{-\frac{2x}{(1+x^2)^2}}=\frac{1}{-2}=-\frac{1}{2}$$
| Do you have to do that with hopital?
Because all those limits are trivial if you use taylor approximations;
$\sin x \sim x - \frac{x^3}{6}$, $\cot(x) \sim \frac{1}{x} - \frac{x}{3}$ $\arcsin x \sim x + \frac{x^3}{6}$, $(1 + x)^\alpha \sim 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2} x^2$ etc..
Just substitute and you're good.
| {
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Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$ Compute the integral.
$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$
The answer at the back of the book is
$$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
| Supplementing Fly by Night's answer with a trick sometimes handy when calculating the residues at a root of unity.
Write $P(z)=z^4$, $Q(z)=z^8+1$, $f(z)=P(z)/Q(z)$.
*
*The poles in the upper half plane are $z=z_k=e^{(2k+1)\pi i/8},k=0,1,2,3$. Their complex conjugates are the other four poles, so they are all simple. A trick is to use the fact that $z_k^8=-1$. Therefore
$$ Res(f;z_k)=\frac{P(z_k)}{Q'(z_k)}=\frac{z_k^4}{8z_k^7}=\frac{z_k^5}{8z_k^8}=-\frac18z_k^5.$$
*For further simplification it is helpful to observe that for all the poles $z_k$ we have $z_k^4=\pm i$. The choices of the signs work out nicely in that the contributions of $z_0$ and $z_3$ (resp. $z_1$ and $z_2$) sum up to simple real values of a trig function. The known values
$$
\cos\frac{\pi}8=\frac12\sqrt{2+\sqrt2},\quad\sin\frac{\pi}8
=\frac12\sqrt{2-\sqrt2}
$$
come in handy, and
$$
\begin{aligned}
\int&=2\pi i\sum_{k=0}^3Res(f,z_k)\\
&=-\frac{\pi i}4(z_0^5+z_1^5+z_2^5+z_3^5)\\
&=-\frac{\pi i}4(iz_0-iz_1+iz_2-iz_3)\\
&=\frac{\pi}4[(z_0-z_3)-(z_1-z_2)]\\
&=\frac{\pi}4[2\sin\frac{3\pi}8-2\sin\frac{\pi}8]\\
&=\frac{\pi}4[2\cos\frac{\pi}8-2\sin\frac{\pi}8]\\
&=\frac{\pi}4\left(\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}\right).
\end{aligned}
$$
| {
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Prove that positive $x,y$ satisfy $\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{1}{1+xy}$.
Prove that positive $x,y$ satisfy $$\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{1}{1+xy}$$
My teacher claims this lemma is often useful. How to prove it?
I've tried using $a^2+b^2\ge 2ab$ and $a^2+b^2\ge \frac{(a+b)^2}{2}$ and $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\ldots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$.
| One possible approach: we can divide it into cases.
Case 1: $x,y \geq 1$ or $x,y \leq 1$
Using $a^2+b^2\ge 2ab$, which yields
$$
\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{2}{1+x+y+xy}
$$
We must then show that
$$
\frac{2}{1+x+y+xy}\geq \frac{1}{1 + xy} \iff\\
2 + 2xy \geq 1 + x + y + xy \iff\\
1 - x - y + xy \geq 0 \iff\\
(x-1)(y-1) \geq 0
$$
Case 2: Otherwise
Not sure what to do here yet. Hopefully, this partial answer can be useful.
| {
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evaluation of a limit with trigonometric and expontential components I'm trying to evaluate the limit $$\lim_{x\to 0} \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x}$$
I see that $\large\frac{1}{e^x-x-1}\to +\infty$, since $e^x-x-1>0,\, \forall x\in \mathbb{R^*}$. Also $\large\frac{x^3-\sin^3x}{\sin^3x} \to 0$. But this of course leads nowhere.
On the other hand, $$\left| \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x} \right|=\frac{1}{e^x-x-1}\cdot\left| \frac{x^3-\sin^3x}{\sin^3x} \right|= \frac{1}{e^x-x-1}\cdot\left| \left(\frac{x}{\sin x} \right)^3-1\right| $$
Could this lead anywhere by bounding? Any hints would be appreciated.
| $$\lim_{x\to 0} \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x}$$
$$=\lim_{x\to 0} \frac{1-\frac{\sin^3x}{x^3}}{(e^x-x-1)\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(1-\frac{\sin x}{x})\left(1+\frac{\sin x}{x}+\frac{\sin^2 x}{x^2}\right)}{(e^x-x-1)\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(1-\frac{\sin x}{x})}{(e^x-x-1)}\lim_{x\to 0}\frac{\left(1+\frac{\sin x}{x}+\frac{\sin^2 x}{x^2}\right)}{\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(x-\sin x)}{(xe^x-x^2-x)}3$$
$$=\lim_{x\to 0} \frac{(1-\cos x)}{(e^x(x+1)-2x-1)}3$$
$$=\lim_{x\to 0} \frac{(\sin x)}{(e^x(x+2)-2)}3$$
$$=\lim_{x\to 0} \frac{(\cos x)}{(e^x(x+3))}3=1$$
| {
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Tangent planes perpendicular at each point of intersection Find the set of all points $(a,b,c)$ in 3-space for which the two spheres $(x-a)^2+(y-b)^2+(z-c)^2=1$ and $x^2+y^2+z^2=1$ intersect orthogonally.( Their tangent planes should be perpendicular at each point of intersection.)
If we consider the equations of the two spheres as level surfaces and take the gradients the we get - $$\nabla f_1 = 2(x-a)i+2(y-b)j+2(z-c)k;$$ $$\nabla f_2 = 2xi+2yj+2zk.$$
Since the two spheres intersect orthogonally at $(x,y,z)$, we must have $\nabla f_1 \cdot \nabla f_2=0$. This gives $$x(x-a)+y(y-a)+z(z-a)=0.$$
Can anyone suggest how to proceed from here to obtain the set of points (a,b,c)?
| If $(x,y,z)$ belongs to the sphere $x^2+y^2+z^2=1$, then we have
$$\nabla f_1\cdot \nabla f_2=0 \iff x(x-a)+y(y-b)+z(z-c)=0 \\ \iff x^2+y^2+z^2-(ax+by+cz)=0 \\ \iff ax+by+cz =1$$
On the other hand, if $(x,y,z)$ also belongs to the sphere $(x-a)^2+(y-b)^2+(z-c)^2=1$, we have $$x^2+y^2+z^2-2(ax+by+cz)+a^2+b^2+c^2=1$$
And we know $x^2+y^2+z^2=1=ax+by+cz$ so
$$1-2(1)+a^2+b^2+c^2=1 \\ \text{i.e.}\;\; a^2+b^2+c^2=2$$
Therefore, spheres $x^2+y^2+z^2=1$ and $(x-a)^2+(y-b)^2+(z-c)^2=1$ intersect orthogonally iff $(a,b,c)$ lies in the sphere $x^2+y^2+z^2=2$.
| {
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Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Furthermore, another approach.
We can show that $4 \sin \theta + 3 \cos \theta$ has its maximum at $y = 5$. Or, more generally, $a \sin \theta + b \cos \theta$ has a maximum at $y = \sqrt{a^2 + b^2}$.
We can manipulate the expression $a \sin \theta + b \cos \theta$ to become $\sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin \theta + \frac{b}{\sqrt{a^2 + b^2}} \cos \theta\right)$
Notice that $\frac{a}{\sqrt{a^2 + b^2}}$ and $\frac{b}{\sqrt{a^2 + b^2}}$ are equivalent to $\cos \alpha$ and $\sin \alpha$ for $\alpha = \tan^{-1} \frac ba$.So now we have $\sqrt{a^2 + b^2} \left(\cos \alpha \sin \theta + \sin \alpha \cos \theta \right) = \sqrt{a^2+b^2} \sin(\alpha + \theta)$, which is maximized when $\alpha + \theta = \pi/2 \rightarrow \theta = \pi / 2 - \alpha$
Thus, we know that $4 \sin \theta + 3 \cos \theta$ is maximized since it equals $5$. Continuing to work generally, we should substitute $\theta = \pi/2 - \alpha$ into $a \cos \theta - b \sin \theta: a \cos(\pi/2 - \alpha) - b \sin (\pi/2 - \alpha) \rightarrow a \sin \alpha - b \cos \alpha$. Recall that $\alpha = \tan^{-1} \frac{b}{a}$, so it follows that $\alpha = \sin^{-1} \frac{b}{\sqrt{a^2 + b^2}} = \cos^{-1} \frac{a}{\sqrt{a^2 + b^2}}$. So finally we have $a \sin \alpha - b \cos \alpha = \frac{ab}{\sqrt{a^2 + b^2}} - \frac{ba}{\sqrt{a^2+b^2}} = \boxed{0}$
| {
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How to find $\int\sqrt{(26x-x^2)}dx $ How do I find $\int \sqrt{(26x-x^2)} dx $
Is this an integration by parts question?
Thanks,
--Nick
| Integrate as follows
\begin{align}
\int \sqrt{26x-x^2} dx &=
\int \frac{\sqrt{26x-x^2}}{2(x-13)}d[(x-13)^2]\\
&=\frac12(x-13)\sqrt{26x-x^2}+ \frac{169}2\int\frac1{ \sqrt{26x-x^2}}dx\\
&= \frac12(x-13)\sqrt{26x-x^2}+\frac{169}2\sin^{-1}{\frac {x-13}{13}}+C\\
\end{align}
| {
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Multiplying in GF(128) I know that in GF(128) $a + b = a \oplus b$.
I have an multiplication table for GF(128). In this table $7\cdot 5 = 27$. How can I create table like this?
| In general the multiplication depends on what defining polynomial you use. For the example that you give that is not immediately needed as the factors have such a low degree. My educated guess is that whoever gave that table is internally using a monomial presentation, and converts the sequence of coefficients to integers. Thus $7=111_2$ truly means the coset of the polynomial
$$1\cdot x^2+1\cdot x+1=x^2+x+1$$
(modulo some ideal $I$ of the ring $\Bbb{Z}_2[x]$). Similarly
$$5=101_2=1\cdot x^2+0\cdot x+1=x^2+1.$$
Therefore their product is the coset of
$$
\begin{aligned}
"7"\cdot"5"&=(x^2+1)(x^2+x+1)=(x^4+x^3+x^2)+(x^2+x+1)\\
&=x^4+x^3+2x^2+x+1\\
&=x^4+x^3+x+1,
\end{aligned}
$$
because the arithmetic of the coefficiente of polynomials is done modulo two. Note that
$x^4+x^3+x+1$ corresponds to $11011_2=27$.
You get the field $GF(128)$ if you do all the arithmetic as polynomials of degree at most six modulo two and reduce the high degree ($\ge7$) terms using the defining polynomial of degree seven (that seven comes from $128=2^7$). I usually use $x^7+x^3+1$ when I need $GF(128)$. This means that (for example)
$$
x^4\cdot x^3=x^7=x^7+(x^7+x^3+1)=2x^7+x^3+1=x^3+1,
$$
so $"16"\cdot"8"=1001_2="9"$, and
$$
x^4\cdot x^5=x^9=x^9+x^2(x^7+x^3+1)=2x^9+x^5+x^2=x^5+x^2,
$$
so $"16"\cdot"32"=x^5+x^2=100100_2="36"$.
In other words you declare that $x^7=x^3+1$, use that and all its consequences such as
$x^8=x^4+x$, $x^9=x^5+x^2$ et cetera.
| {
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If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
This is what I have done:
Let $y = 3^x$
$$9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$$ $$\implies9y^2 + (t^2 - 4t - 2)y + 1 > 0$$
For the LHS to be greater than zero, $b^2 - 4ac$ has to be $\lt 0$, since coefficient of $y^2$ is greater than $0$(which will give us an upward opening parabola).
$$(t^2 - 4t - 2)^2 - 36< 0$$
The answer that I get is different from the correct answer. The correct answer to this is $t \in \mathbb{R} - \{2\}$. What did I do wrong?
| Alternately, by positivity of $3^x$ and AM-GM:
$$(t^2-4t-2)3^x+3^{2x+2}+1\ge (t^2-4t-2)3^x+2\sqrt{3^{2x+2}} = (t^2-4t+4) \cdot 3^x$$
with equality possible when $x=-1$. So we need $t^2-4t+4 = (t-2)^2 > 0$ which means $t \neq 2$.
| {
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Need help Proving Identities Prove the Identity: $$\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta + 1}{\sin \theta \cos \theta}. $$
| I will assume you meant to write the following:
$$
\frac{1+ \cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta + 1}{\sin \theta \cos \theta}
$$
Therefore, you can combine the fractions on the left side to get the following:
$$
\frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos \theta + 1}{\sin \theta \cos \theta}
$$
Then, using the identity $\cos^2 \theta + \sin^2 \theta = 1$, we get:
$$
\frac{\cos \theta + 1}{\sin \theta \cos \theta} = \frac{\cos \theta + 1}{\sin \theta \cos \theta}
$$
Therefore, this is true.
| {
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How do you find L of LU for this LU factorization? $$
\begin{bmatrix}
3 & -7 & -2 \\
-3 & 5 & 1 \\
6 & -4 & 0 \\
\end{bmatrix}
$$
The method my book gave me of doing this is:
divide col1 by first pivot position, then I am completely lost, as my $U$ doesn't help for whatever reason I have:
$$
\begin{bmatrix}
3 & -7 & -2 \\
0 & -2 & 1 \\
0 & 3 & 2 \\
\end{bmatrix}
$$
How do I turn $-2$ & $3$ into
$$
\begin{bmatrix}
1 & 0 & 0 \\
-1 & 1 & 0 \\
2 & -5 & 1 \\
\end{bmatrix}
$$
| We can use Doolittle's Method:
$$\begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix} = \begin{bmatrix}
3 & -7 & -2 \\
-3 & 5 & 1 \\
6 & -4 & 0 \\
\end{bmatrix}$$
Solving for each of the variables, in the correct order yields:
*
*$u_{11} = 3, u_{12} = 7, u_{13} = -2$
*$l_{21} = -1, u_{22} = -2, ...$
*$l_{31} = 2, l_{32} =-5, ...$
*$\ldots $
So, we arrive at:
$$A = \begin{bmatrix}3 & -7 & -2 \\-3 & 5 & 1 \\6 & -4 & 0 \\\end{bmatrix} = LU = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2& -5 & 1 \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 & -2 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix}$$
We could have also used Crout's or Choleski's Method for the $LU$ approach. See: what are pivot numbers in LU decomposition? please explain me in an example
Please note that sometimes an LU decomposition is not possible, and sometimes, when it is, we have to resort to using permutation matrices and other approaches.
| {
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Greatest prime factor of $4^{17}-2^{28}$ I have seen the solution to this problem.
What is the greatest prime factor of $ \ 4^{17} - 2^{28} \ $?
Answer: 7
$$ 4^{17}-2^{28} \ = \ 2^{34}-2^{28} \ = \ 2^{28} \ (2^6-1) \ = \ 2^{28} \ \cdot \ 63 \ = \ 2^{28}\ \cdot \ 3^2 \ \cdot \ 7 $$
I understand the problem up to this part: $2^{28} \ (2^6-1) \ $ .
Why did this occur? I can't explain it.
Thanks!
| $$4^{17} = (2^2)^{17} = 2^{2 \cdot 17} = 2^{34} = 2^{28 + 6} = 2^{28} \cdot 2^6$$
Now recognize the common factor of $2^{28}$ and factor it out.
| {
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Labeling the vertices of a polygon with 0's and 1's Suppose $P_n$ is the regular polygon with n vertices ($n\geq 5$). Let $V=\{v_1,\ldots,v_n\}$ be the vertex set. I would like to define a labeling function $\ell:V\to \{0,1\}$ so that $\sum_{i=1}^{n}\ell(v_i)$ is even and $\geq 4$. This means that a labeling assigns a 0 or a 1 to each vertex and that this must be done so the number of 1's is even and is at least 4.
Two such labelings are equivalent if there is a automorphism of $P_n$ that sends one labeling to the other. How many equivalence classes of such labelings are there are on $P_n$?
| What you would want to do is find the Polya inventory polynomial in two colors for the Dihedral group $D_{n}$. Begin with the cycle index of $D_{n}$, $Z(D_{n})$:
$$Z(D_{n}) = \frac{1}{2n} \sum_{d|n} \phi(d) a_d^{n/d} + \left\{
\begin{array}{lr}
\frac{1}{2}a_1 a_2^{(n-1)/2} &n \textrm{ odd}\\
\frac{1}{4}\left(a_2^{n/2} + a_1^2 a_2^{(n-2)/2}\right) &n \textrm{ even}
\end{array}\right.$$
Where $\phi$ is Euler's phi function. Next replace $a_i$ by $(b^i + w^i)$ to get the Polya inventory polynomial:
$$P(b,w) = \frac{1}{2n} \sum_{d|n} \phi(d) (b^d + w^d)^{n/d} + \left\{
\begin{array}{lr}
\frac{1}{2}(b+w)(b^2 + w^2)^{(n-1)/2} &n \textrm{ odd}\\
\frac{1}{4}\left((b^2 + w^2)^{n/2} + (b+w)^2 (b^2+w^2)^{(n-2)/2}\right) &n \textrm{ even}
\end{array}\right.$$
The coefficient of $b^i w^j$ in this polynomial counts the number of distinct bracelets with $n$ beads, $i$ of which are black and $j$ of which are white. To relate this to your polygons, let the black beads be the ones labeled with 1, and the white beads labeled 0. Then summing the coefficients of each $b^i w^j$ in $P(b,w)$ with $i$ even and $i\geq 4$ will give you the number of such polygons that you are looking for.
As you can see this value depends on the factorization of $n$. You may be able to get some nice results for certain $n$. For example, if $n=p$, where $p$ is an odd prime, the polynomial will be:
$$\begin{array} \\
P(b,w) &= \frac{1}{2p}\left((b+w)^p + (p-1)(b^p + w^p)\right) + \frac{1}{2}(b+w)(b^2 + w^2)^{(p-1)/2} \\
&= \frac{1}{2p}\left(\sum_{i=0}^p \binom{p}{i}b^i w^{p-1} + (p-1)(b^p + w^p)\right) + \frac{1}{2}(b+w)\sum_{i=0}^{(p-1)/2} \binom{(p-1)/2}{i} b^{2i} w^{p-1+i} \\
\end{array}$$
Taking only coefficients of even powers of $b$ at least 4, we get that the number of labelings is:
$$\begin{array} \\
&= \frac{1}{2p}\sum_{i=2}^{\lfloor p/2 \rfloor} \binom{p}{2i} + \frac{1}{2}\sum_{i=2}^{(p-1)/2} \binom{(p-1)/2}{i} \\
&= \frac{1}{2p}\left(2^{p-1} - 1 - \frac{p(p-1)}{2}\right) + \frac{1}{2}\left(2^{(p-1)/2} - 1 - \frac{p-1}{2}\right) \\
&= \frac{1}{2p}\left(2^{p-1} - 1\right) - \frac{p-1}{4} + 2^{(p-3)/2} - \frac{1}{2} - \frac{p-1}{4} \\
&= \frac{1}{2p}\left(2^{p-1} - 1\right) + 2^{(p-3)/2} - \frac{p}{2}
\end{array}$$
This is a nice quick intro to the topic of Polya enumeration if you're interested in more.
| {
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Partial fraction decomposition of a rational function The form of the partial fraction decomposition of a rational function is given below.
$$\frac{x−3x^2−26}{(x+1)(x^2+9)} = \frac{A}{x+1}+ \frac{Bx+C}{x^2+9}$$
What are the values of $A,B$ and $C$?
So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$
I also found the indefinite integral, which equals $$\frac13\tan^{-1}\left(\frac x3\right)-3 \log(1+x) + C$$
| An alternative technique called Heaviside cover-up makes the process simpler.
It gives $\begin{align}A = \frac{x-3x^2 - 26}{x^2 + 9}\Bigg|_{x=-1} = -3\end{align}$
And then requires that $\dfrac{A}{x+1}$ be subtracted from both sides of the partial fraction decomposition identity. Hence, we get
$\begin{align} \frac{Bx+C}{x^2+9} &=\frac{x−3x^2−26}{(x+1)(x^2+9)} + \frac{3}{x+1}\\ &=\frac{x−3x^2−26 + 3x^2 + 27}{(x+1)(x^2+9)}\\&=\frac{1}{x^2+9}\end{align}$
$\therefore B = 0, C = 1$
The key feature of this technique is that it eliminates the need to solve a system of linear equations to obtain the coefficients, which can often be tedious.
See here and here for some more examples.
| {
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Finding the Maclaurin Series for $\sqrt{1+x^2}$ I can't find the Maclaurin series for $\sqrt{1+x^2}$. Every time it try to find it I get the Maclaurin series for $\sqrt {1+x}$. Can someone explain it to me?
Thanks!
| The easiest way is to use the binomial theorem. For $n \ge 1$ you have that:
\begin{align}
\binom{1/2}{n}
&= \frac{(1/2)^{\underline{n}}}{n!} \\
&= \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)
\ldots \left(\frac{1}{2} - n + 1\right)}{n!} \\
&= \frac{1 \cdot (-3) \cdot (-5) \cdot \ldots \cdot (- 2 n - 3)}{2^n n!} \\
&= \frac{(-1)^{n - 1} (2 n - 2)!}{n! (n - 1)!} \\
&= (-1)^{n - 1} \frac{1}{2^{2 n - 1} n}\binom{2 n - 2}{n - 1}
\end{align}
Thus:
$$
\sqrt{1 + x^2}
= 1 + \sum_{n \ge 1}
(-1)^{n - 1} \frac{1}{2^{2 n - 1} n}\binom{2 n - 2}{n - 1} x^{2 n}
$$
To prove the (generalized) binomial theorem in turn is rather easy:
$$
\frac{\mathrm{d}^n}{\mathrm{d} u^n} (1 + u)^a
= a^{\underline{n}} (1 + u)^{a - n} \\
$$
Evaluate at $u = 0$, all that is left is $a^{\underline{n}} = a (a - 1) \ldots (a -n + 1)$
| {
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Prove: $\text{if }a,b,c\in\mathbb{Z} \text{ and } a^2+b^2=c^2\text{ then }3\mid ab$ This is for a homework assignment. I am supposed to prove,
$$\text{if }a,b,c\in\mathbb{Z} \text{ and } a^2+b^2=c^2\text{ then }3\mid ab$$
I've tried using direct proofs and proofs by contrapositive, but I can't seem to get anywhere. Is there a good approach to prove this, preferably without modulus?
| If $a$ is not a multiple of $3$, then one of $a+1$, $ a-1$ is. Hence $3\mid (a+1)(a-1)=a^2-1$. Similarly, if $b$ is also not a multiple of $3$, we have $3\mid b^2-1$.
Hence if neither $a$ nor $b$ is a multiple of $3$, then $3\mid a^2-1+b^2-1=c^2-2$. Then $c^2$ (and $c$) is not a multiple of $3$, so $3\mid c^2-1$. But then also $3\mid (c^2-1)-(c^2-2)=1$. Contradiction. We conclude that at least one of $a,b$ must be a multiple of $3$.
| {
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Integrating around pie-slice domain We are asked to show $$\int_0^{\infty}\frac{\log(x)}{x^3+1}dx=-\frac{2\pi^2}{27}$$, and $$\int_0^{\infty}\frac{1}{x^3+1}dx=\frac{2\pi}{3\sqrt{3}}$$
By integrating around a pie slice with angle $\frac{2\pi}{3}$. If we denote $\Gamma_R$ as the outer arc of this slice with radius $R$, and $\gamma_\epsilon$ as the inner arc of this sector with radius epsilon, we see these vanish as $R\rightarrow\infty, \epsilon \rightarrow 0$.
There is only one residue in this domain, namely $e^{i\pi/3}$. The Residue here equals:
$$Res[\frac{\log(x)}{x^3+1},e^{i\pi/3}]=\frac{\log(e^{i\pi/3})}{e^{i2\pi/3}+e^{i2\pi/3}+e^{i2\pi/3}}=\frac{1+{i\pi/3}}{3e^{i2\pi/3}}$$
So our total integral over this domain equals:
$$2\pi i\frac{1+{i\pi/3}}{3e^{i2\pi/3}}=\int_\epsilon^{R}\frac{\log(x)}{x^3+1}dx+\int_R^{\epsilon}\frac{\log(x)}{x^3+1}dx+\int_R^{\epsilon}\frac{i\frac{2\pi}{3}}{x^3+1}dx$$ by making the change of variables $z=re^{i\theta}, 0\le\theta\le\frac{2\pi}{3}$ for the upper line segment of the pie slice... but clearly there is a problem somewhere since the two log integrals cancel. What is the issue?
| Integrating $ \displaystyle f(z) = \frac{\log z}{z^{3}+1}$ around the contour you described and using the principal branch of the logarithm,
$$ \begin{align} \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx + \int_{\infty}^{0} \frac{\log x + \frac{2 \pi i}{3}}{(xe^{2 \pi i /3})^{3} + 1} \ e^{2 \pi i /3} \ dx &= 2 \pi i \ \text{Res}[f(z), e^{i \pi/3}] \\ &= 2 \pi i \lim_{z \to e^{i \pi /3}} \frac{\log z}{3z^{2}} \\ &= -\frac{2 \pi^{2}}{9} e^{-2 \pi i /3}. \end{align}$$
Rearranging, we have
$$ (1-e^{2 \pi i/3}) \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx - \frac{2 \pi i}{3} e^{2 \pi i/3} \int_{0}^{\infty} \frac{1}{x^{3}+1} \ dx = -\frac{2 \pi^{2}}{9} e^{-2 \pi i /3}. $$
And multiplying both sides of the equation by $e^{-i \pi /3}$,
$$ -2i \sin \left( \frac{\pi}{3} \right) \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx - \frac{2 \pi i}{3} e^{i \pi /3} \int_{0}^{\infty} \frac{1}{x^{3}+1} \ dx = \frac{2 \pi^{2}}{9}$$
or
$$ -i \sqrt{3} \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx +\left(\frac{\pi \sqrt{3}}{3} - \frac{i\pi}{3} \right) \int_{0}^{\infty} \frac{1}{1+x^{3}} \ dx = \frac{2 \pi^{2}}{9}.$$
Now equate the real and imaginary parts on both sides of the equation and solve the system of equations to obtain the answers.
| {
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Can someone help me to evaluate this integral: Can someone help me with evaluating this integral:
$$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx$$
I tried using integration by parts, integration by substitution....but nothing...
| Hint: $$\int\frac{2x^2-1} {\sqrt{x^2-1}}\, dx = x\sqrt{x^2-1}+C$$ thus $$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx=2\sqrt{4-1}-1\sqrt{1-1}\\=2\sqrt3$$
You can substitute $x^2-1=z$ to get $$\int\frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \frac12\int\frac{2z+1} {\sqrt{z}}\, dz \\= \frac12\left(\int \sqrt{z}\, dz+\int \frac{1}{\sqrt{z}}\, dz\right)\\=\frac13\sqrt{z^3}+\sqrt{z}+C=\sqrt{z}\left(\frac{z}{3}+1\right)+C$$
So, making the final passage
$$\int_{1}^{2} \frac{2x^2-1} {\sqrt{x^2-1}}\, dx \\= \int_{0}^{3} \frac12\int\frac{2z+1} {\sqrt{z}}\, dz\\=\sqrt{3}\left(\frac{3}{3}+1\right)-\sqrt{0}\left(\frac{0}{3}+1\right)=2\sqrt3$$
| {
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Fairly simple trig question Three points at coordinates $(0,c)$, $(p,q)$, $(0,d)$ respectively. The angle at $(p,q)$ between $(0,c)$ and $(0,d)$ is $θ$. Find $d$.
P.s. This isn't homework.
| Hint :
Let $A(0,c)$, $B(0,d)$, and $C(p,q)$. Length $AB=c-d$,
\begin{align}
AC^2=p^2+(c-q)^2
\end{align}
and
\begin{align}
BC^2=p^2+(q-d)^2
\end{align}
Using cosine formula, we get
\begin{align}
AB^2&=AC^2+BC^2-2\cdot AC\cdot AB\cos\theta\\
(c-d)^2&=p^2+(c-q)^2+p^2+(q-d)^2-2\sqrt{(p^2+(q-d)^2)(p^2+(q-d)^2)}\cos\theta\\
\end{align}
| {
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Fibonacci polynomials and factorization redux I recently asked a question about factorizing a certain expression involving Fibonacci and Lucas polynomials. That question had a very simple and nice answer, but now I've come across another similar question which has me stumped.
Conjecture: The polynomial $4- F_n(x)^4(x^2+3)$ factors as $H_n(x)H_n(-x)$ for some integer polynomial $H_n(x)$, where $F_n(x)$ is the nth Fibonacci polynomial.
Any ideas?
(In case anybody is wondering why I care about these question, they are related to calculating the Conway polynomial of certain families of knots.)
Edit: As requested, here are the first few examples. Let $G_n(x)=4- F_n(x)^4(x^2+3)$. Then
*
*$G_1(x)=(1+x)(1-x)$
*$G_2(x)=(1+x)(1-x)(2+x^2)^2$
*$G_3(x)=(-1 - 3 x + 2 x^2 - 4 x^3 + x^4 - x^5) (-1 + 3 x + 2 x^2 + 4 x^3 +
x^4 + x^5)$
*$G_4(x)= (2 - 4 x + 4 x^2 - 10 x^3 + 4 x^4 - 6 x^5 + x^6 - x^7) (2 + 4 x +
4 x^2 + 10 x^3 + 4 x^4 + 6 x^5 + x^6 + x^7)$
*$G_5(x)=(-1 - 5 x + 6 x^2 - 20 x^3 + 11 x^4 - 21 x^5 + 6 x^6 - 8 x^7 + x^8 -
x^9) (-1 + 5 x + 6 x^2 + 20 x^3 + 11 x^4 + 21 x^5 + 6 x^6 + 8 x^7 +
x^8 + x^9)$
Edit 2: Following Will Jagy's ideas, define a family of polynomials $H_n(x)$ by $H_0(x)=2$, $H_1(x)=-1+x$ and then $$H_{n+2}(x)=(x^2+2)H_{n+1}(x)-H_n(x)+(-1)^n(6+2x^2)$$ for $n\geq 0$. Computations show that $$H_n(x)H_n(-x)=G_n(x)$$ for $n\leq 20$. What remains is to prove that this is a valid formula in general.
| This looks a little better, $H_1(x) = 1 + x, \; H_2(x) = 2 + 2x + x^2 + x^3, \; H_3(x) = -1 + 3 x + 2 x^2 + 4 x^3 + x^4 + x^5, $ then
$$ H_{n+2}(x) = (2+ x^2) H_{n+1}(x) - H_n(x) + OTHERERROR_{n+2}(x), $$
where this way the error stays quadratic, by hand calculations the first three error terms come out
$$ -4 + x^2, \; 6 + 2 x^2, \; -6 - 2 x^2... $$
| {
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How to find the full Taylor expansion of the following: I need to find the full Taylor expansion of $$f(x)=\frac{1+x}{1-2x-x^2}$$
Any help would be appreciated. I'd prefer hints/advice before a full answer is given. I have tried to do partial fractions\reductions. I separated the two in hopes of finding a known geometric sum but I could not.
Edit: I guess you could say that I did not have the.... insight to take the path with the partial decomposition mentioned. I have done some work (I had to go to the gym that is why it took a while)
$$\frac{1+x}{1-2x-x^2}=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)}$$ I am going to work with this to go further.
I got to this:
$$\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right) $$ But I think this is wrong for some reason.
Edit: Figured it out.
$$\begin{align*}
\implies\frac{1+x}{1-2x-x^2}&=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)} \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a-x}-\frac{1}{b+x}\right) \mbox{where $a=\sqrt{2}-1$ and $b=\sqrt{2}+1$}. \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a} \frac{1}{1-\frac{x}{a}}-\frac{1}{b}
\frac{1}{1-\frac{x}{-b}}\right) \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a}\sum_{n=0}^\infty \frac{1}{a^n}x^n-\frac{1}{b}\sum_{n=0}^\infty\frac{1}{(-b)^n}x^n\right) \\[2mm]
&=\frac{1}{2}\left(\frac{1}{\sqrt{2}-1}\sum_{n=0}^\infty \frac{1}{(\sqrt{2}-1)^n}x^n-\frac{1}{\sqrt{2}+1}\sum_{n=0}^\infty\frac{1}{(-\sqrt{2}-1)^n}x^n\right) \\[2mm]
&=\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right) \\
&=1+3x+7x^2+17x^3+\ldots
\end{align*}$$
| Hint For the function
$$g(x) = \frac{1}{1-x} \text{ over } x \in (-1,1),$$
the Taylor series is the ordinary geometric series
$$\frac{1}{1-x} = 1 + x + x^2 + \ldots = \sum_{k=0}^\infty x^k.$$
If the partial fraction decomposition is given by
$$
\frac{A}{x-B} + \frac{C}{x-D}
$$
how can you get the decomposition you are seeking?
| {
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Circle geometry problem to find a relation between angles
In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is $θ$ radians. The shaded region is bounded by the circumference of the circle and the arc with centre A joining B and C. The area of the shaded region is equal to half the area of the circle.
Show that $$\cos2θ = \frac{2\sin2θ - π}{4θ}$$
I had trouble showing this. I said that the area of the shaded region is equal to area of the BAC sector + $2*$ area of BOA sector - $2*$ area of BOA triangle and equated this to $\frac{1}{2}πr^2$, but I keep getting the same result of $2\sin2θ=\frac{1}{2}π-θ$, and I do not know how to continue.
| The radius of the arc $BC$ is $|AB|=2r\cos\theta$, and $\measuredangle AOB=\pi-2\theta$.
\begin{align}
\text{Dashed area}&= \text{Area of sector }BAC+2(\text{Area of sector } BOA-\text{Area of triangle} BOA)\\
&=\frac{1}{2}|AB|^2(2\theta)+2\left[\frac{1}{2}r^2(\pi-2\theta)-\frac{1}{2}r^2\sin(\pi-2\theta)\right]
\end{align}
On the other hand he area of the dashed surface is $\frac{1}{2}\pi r^2$, then
\begin{align}
\frac{1}{2}|AB|^2(2\theta)+2\left[\frac{1}{2}r^2(\pi-2\theta)-\frac{1}{2}r^2\sin(\pi-2\theta)\right]&=\frac{1}{2}\pi r^2 \\
(2r\cos \theta)^2(\theta)+r^2\left[\pi-2\theta-\sin(\pi-2\theta)\right]&=\frac{1}{2}\pi r^2\\
4\theta\cos^2\theta+\pi-2\theta-\sin2\theta&=\frac{1}{2}\pi \\
2\theta(2\cos^2\theta-1)&=\sin 2\theta-\frac{1}{2}\pi \\
2\theta\cos 2\theta&=\frac{2\sin 2\theta - \pi}{2}\\
\cos 2\theta&=\frac{2\sin 2\theta - \pi}{4\theta}
\end{align}
| {
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Prove this identity: $\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$ I am stuck on this identity
$$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$
I began working on the left side trying to get things to cancel out or equal one by the Pythagorean identities. I am stuck and can't get it to reduce anymore.
| Let's take this one step at a time.
First, move the denominator to the right side.
1) $$ 2sin^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Use the identity $ sin^2(x) = 1-cos^2(x) $ to rewrite $ sin^4(x) = (1-cos^2(x))^2$.
2) $$ 2(1-cos^2(x))^2 + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Now foil out $2(1-cos^2(x))^2 = 2(1 - 2cos^2(x) + cos^4(x))$ and distribute the 2.
3) $$ 2-4cos^2(x) + 2cos^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1$$
Now there are three things we can do here
First we have $ 2cos^4(x) $ and $ -2cos^4(x) $ so those will cancel out. Second we have $ -4cos^2(x) + cos^2(x) = -3cos^2(x)$. And last we can move the 2 from the left side of the equation to the right.
4) $$ -3cos^2(x) = 3sin^2(x) - 3 $$
Now factor out the 3 and it cancels out
5) $$-cos^2(x) = sin^2(x) - 1 $$
And just rearrange it and we get
6) $$ 1 = sin^2(x) + cos^2(x) $$
Hope that helps!
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Prove that integral is independant of its parameter We are given intergral $\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}}$ and task is to prove that it's independent of $\alpha$.
Task was too complicated for me, so I had to stick with solution, recommended in the textbook. It goes on like that:
$\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_0^1 {\frac {dx} {(1+x^2)(1+x^\alpha)}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = J_1 + J_2$
In $J_1$ we do replacement $y = \frac 1 x$
$J_1 = \int_1^\infty \frac {dy} {-(1+y^2)(1+y^{-\alpha})}$
Than textbook says that next step is
$J_1+J_2 = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}) \frac {dx}{1+x^2}}$
From where it's obvious that inital statement is true. Problem is, I completely fail to understand how that summation of $J_1+J_2$ was done.
Edit
To be more explicit, I can't get this: $\int_1^\infty{\frac {dy}{-(1+y^2)(1+y^{-\alpha})}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}) \frac {dx}{1+x^2}}$
So I would appreciate some explanations and pointers.
| Add the two fractions together and you get
$$J_1+J_2 = \int_1^\infty {\left(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}\right) \frac {dx}{1+x^2}}=\int_1^{\infty}\frac{1+x^{\alpha}}{1+x^{\alpha}}\frac {dx}{1+x^2}=\int_1^{\infty}\frac {dx}{1+x^2}$$
| {
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How prove this $(xy+yz+xz)\left(\frac{xy}{z^2+1}+\frac{yz}{x^2+1}+\frac{zx}{y^2+1}\right)\le\frac{1}{10}$
let $x,y,z>0$ and such $x+y+z=1$, show that
$$(xy+yz+xz)\left(\dfrac{xy}{z^2+1}+\dfrac{yz}{x^2+1}+\dfrac{zx}{y^2+1}\right)\le\dfrac{1}{10}$$
my idea:
$$\dfrac{xy}{z^2+1}=\dfrac{xy}{z^2+(x+y+z)^2}=\dfrac{xy}{2z^2+2xy+2yz+2xz+x^2+y}$$
Maybe this is old inequality,and It is said can use Cauchy-Schwarz inequality to solve it
Thank you
| By C-S and AM-GM $$\sum_{cyc}\frac{xy}{z^2+1}=\sum_{cyc}\frac{xy}{2(x+z)(y+z)+x^2+y^2}\leq\sum_{cyc}\frac{xy}{(8+2)^2}\left(\frac{8^2}{2(x+z)(y+z)}+\frac{2^2}{x^2+y^2}\right)=$$
$$=\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}+\frac{1}{50}\sum_{cyc}\frac{2xy}{x^2+y^2}\leq\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}+\frac{3}{50}.$$
Thus, it remains to prove that
$$\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}\leq\frac{(x+y+z)^2}{10(xy+xz+yz)}-\frac{3}{50}$$ or
$$\frac{8}{25}\sum_{cyc}\frac{xy}{(x+z)(y+z)}\leq\frac{\sum\limits_{cyc}(5x^2+7xy)}{50(xy+xz+yz)}$$ or
$$\sum\limits_{cyc}(5x^2+7xy)\prod_{cyc}(x+y)\geq16(xy+xz+yz)\sum_{cyc}(x^2y+x^2z)$$ or
$$\sum_{cyc}(5x^4y+5x^4z-4x^3y^2-4x^3z^2-8x^3yz+6x^2y^2z)\geq0$$ or
$$5\sum_{cyc}(x^4y+x^4z-x^3y^2-x^3z^2)+\sum_{cyc}(x^3y^2+x^3z^2-2x^3yz)-6xyz\sum_{cyc}(x^2-xy)\geq0$$ or
$$\sum_{cyc}(x-y)^2(5xy(x+y)+z^3-3xyz)\geq0,$$
which is true by AM-GM:
$$5xy(x+y)+z^3-3xyz\geq xy(x+y)+z^3-3xyz\geq$$
$$\geq2\sqrt{x^3y^3}+z^3\geq3\sqrt[3]{x^3y^3z^3}-3xyz=0.$$
Done!
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Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question. Part (a) was the following:
Write the sum: $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)$ using summation notation.
It was simple enough : $\sum k(k+2)$.
For this question, the base case $(n=1)$ holds, as $1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6$.
Induction step: Assume the above holds for all $n = k$, prove that it holds for all $n = k+1$
I'm a bit lost from here, help?
| Begin with $$\frac{n(n+1)(2n+7)}{6} = \sum^n_{i=1}n(n+2)$$ and then add $(n+1)(n+3)$ to both sides and try to rewrite the left side in form $$\frac{(n+1)(n+2)(2(n+1)+7)}{6}.$$
| {
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$\sqrt{\frac{15}4+\sum\cos(A-B)}\ge\sum\sin A$ in a triangle? How can I prove that ( $\small{\sum}$ denotes cyclic sum here), for any triangle $ABC$:
$$\sqrt{\frac{15}4+\sum\cos(A-B)}\ge\sum\sin A$$
I don't see where to begin even. Any hints would be appreciated :)
| Squaring both sides leads to,
$\displaystyle \dfrac{15}{4} + \sum\limits_{cyc}\cos(A-B) \ge (\sum\limits_{cyc} \sin A)^2 = \sum\limits_{cyc} \sin^2 A + 2\sum\limits_{cyc}\sin A\sin B$
$\implies \displaystyle \dfrac{15}{4} + \sum\limits_{cyc}(\cos A \cos B + \sin A \sin B) \ge 3 - \sum\limits_{cyc} \cos^2A + 2 \sum\limits_{cyc}\sin A\sin B \implies \sum\limits_{cyc} \cos^2A + \sum\limits_{cyc}(\cos A \cos B - \sin A \sin B) + \dfrac34 \ge 0 \implies \sum\limits_{cyc} \cos^2A + \sum\limits_{cyc}\cos (A+B) + \dfrac34 = \sum\limits_{cyc}\left( \cos^2A - \cos A +\frac14\right)=\sum\limits_{cyc}\left(\cos A - \frac12\right)^2 \ge 0$
which is true.
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Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ I simplified to get $6x^4-25x^3+12x^2+25x+6=0$
Or $6 (x^2+1)^2+25x(1-x^2)=0$
Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$
I can't find a way to proceed further.
| Divide the whole equation by $x^2$ and simplifing:
$$6(x^2+\frac{1}{x^2})-25(x-\frac{1}{x})+12=6((x-\frac{1}{x})^2+2)-25(x-\frac{1}{x})+12=0$$
Put $x-\frac{1}{x}=y$ to solve the resulting equation:
$6y^2-25y+24=0$
| {
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Areas in a rectangle Suppose $P,Q, R$, and $S$ are the midpoints of the sides $AB, BC, CD$, and $DA$, respectively of rectangle $ABCD$. If the area of the rectangle is $\delta$, then prove that the area of the figure bounded by the straight lines $AQ, BR, CS$, and $DP$ is $\frac{\delta}{5}$.
I began by imposing a coordinate system but couldn't find a way to relate the given area with the area of the rectangle.
How should I begin?
| Without loss of generality, make $ABCD$ a square with $A(0,0)\ B(0,2) \ C(2,2)\ D(2,0)$.
Using the lines $y = -2x + 2,y= -2x+4, y=x/2,y= x/2 + 1$. Two of these intersections are $(\frac{2}{5}, \frac{6}{5}), (\frac{4}{5},\frac{2}{5})$. The distance between them is a side of the small square and is $\frac{2}{\sqrt{5}}$. The area of the small square is $\frac{4}{5}$, divided by the total area of $4$ gives you a $\frac{1}{5}$ area. Transformations to a rectangle preserve the areas because everything is proportional.
| {
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$\lim_{x\to\infty}\frac{\sqrt[104]{x}}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}$ $$\lim_{x\to\infty}\frac{\sqrt[104]{x}}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}$$
I need to take this limit. I suceed in proving that:
$$\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}>\sqrt[105]{x}$$
I tought this would help me, but in the end, I have:
$$\frac{1}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}<\frac{1}{\sqrt[105]{x}} \implies$$
$$\frac{\sqrt[104]{x}}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}<\frac{\sqrt[104]{x}}{\sqrt[105]{x}} \implies$$
$$\lim_{x\to\infty}\frac{\sqrt[104]{x}}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}<\lim_{x\to\infty}\frac{\sqrt[104]{x}}{\sqrt[105]{x}}<\lim_{x\to\infty}\sqrt[10920]{x} = \infty$$
Wich does not help me. Is there a way to solve this by comparsion? It would be better to me. If not, is there a way to fator these roots out?
| For $x$ big enough we have $x+17<2^7x$, so $$\sqrt[7]{x}<\sqrt[7]{x+17}<2\sqrt[7]{x}.$$
Hence for $x$ big enough $6+\sqrt[7]{x+17}<6+2\sqrt[7]x<2^5\cdot \sqrt[7]{x} $ and then $$\sqrt[35]{x}<\sqrt[5]{6+\sqrt[7]{x+17}}<2\sqrt[35] x.$$
Hence for $x$ big enough $7+\sqrt[5]{6+\sqrt[7]{x+17}}<7+2\sqrt[35] x<2^3\sqrt[35]x$ and then
$$ \sqrt[105]x<\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}<2\sqrt[105]x.$$
We conclude
$$ \frac{\sqrt[104]x}{\sqrt[3]{7+\sqrt[5]{6+\sqrt[7]{x+17}}}}>\frac{\sqrt[104]x}{2\sqrt[105]x}=\frac12\sqrt[10920]x$$
for $x$ sufficiently big.
| {
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Integral $\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16\sqrt 2}$ This integral below
$$
I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2}
$$
is what I am trying to prove. Thanks.
We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$. Next I wrote
$$
I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\
\int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx,
$$
however only the middle integral is convergent. I am not sure how to go about solving this problem. Thank you
| A related problem. Recalling the Mellin transform of a function $f$
$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x)dx \implies F''(s)=\int_{0}^{\infty} \ln(x)^2x^{s-1}f(x)dx .$$
Now the whole problem boils down to finding the Mellin transform of $\frac{1+x^2}{1+x^4}$, differentiating twice, and then taking the limit as $s \to 1$.
Can you finish it?
| {
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how did we find the Legendre symbol? The Legendre symbol from a prime $p>2$ and $(a,p)=1$ is:
$$\left ( \frac{a}{p} \right )=\left\{\begin{matrix}
1, & \text{if a is a quadratic residue} \\
-1& \text{if a is a non-quadratic residue}
\end{matrix}\right.$$
According to my notes for $p=11$:
$$\left ( \frac{1}{11} \right )=\left ( \frac{3}{11} \right )=\left ( \frac{4}{11} \right )=\left ( \frac{5}{11} \right )=\left ( \frac{9}{11} \right )=1$$
$$\left ( \frac{2}{11} \right )=\left ( \frac{6}{11} \right )=\left ( \frac{7}{11} \right )=\left ( \frac{8}{11} \right )=\left ( \frac{10}{11} \right )=-1$$
How did we find the Legendre symbol?
| What Dietrich said! :)
Alternatively, there are a number of ways to evaluate a Legendre symbol.
For starters you can apply its definition.
Since $x^2 \equiv 1 \pmod {11}$ has the solution $x\equiv 1 \pmod {11}$, it follows that:
$$\left({1\over 11}\right) = 1$$
Or alternatively you can apply the property:
$$\left({a\over p}\right) \equiv a^{\frac{p-1}2} \pmod p$$
We can apply it and find for instance:
$$\left({3\over 11}\right) \equiv 3^{\frac{11-1}2} \equiv 3^5 \equiv 1 \pmod{11}$$
| {
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Log Log Integrals Evaluate the integrals
\begin{align}
I_{1} &= \int_{0}^{1} \ln^{2n}(x) \ \ln\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
and
\begin{align}
I_{2} = \int_{0}^{1} \ln^{2n}(x) \ \ln^{2}\left(\ln\left(\frac{1}{x}\right)\right) dx.
\end{align}
From the resulting values of the integrals above is it possible to evaluate the integral
\begin{align}
I_{3} = \int_{0}^{1} \left( x^{a} + \frac{1}{x^{a}} \right) \ \ln^{p}\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
where $p=1,2$ in a compact form? Please show all work in in the solutions.
| $I_j$ with $j = 1,2$ can be calculated as follows:
Substitute $u = -\ln x$, $\,dx = e^{-u} \,du$ to obtain
\begin{align}
I_1 &=
\int_0^{\infty} x^{2n}e^{-x} \ln x \,dx = \partial_{\mu}|_0\int_0^{\infty} x^{2n+\mu}e^{-x}\,dx \\
&=\partial_{\mu}|_0 \,\Gamma(2n+\mu+1) \\
&= \Gamma(2n+1) \ \psi(2n+1)
\end{align}
and similarly, by differentiating twice before setting $\mu=0$, we obtain
\begin{align}
I_{2} = \Gamma(2n+1) \ \left[\psi(2n+1)^2+\psi_1(2n+1)\right].
\end{align}
Now we will use these results (with $n = 0$) to prove some lemmas.
First notice that for $b>0$, we have
\begin{align}
\int_0^{\infty} e^{-bx} \ln x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x} (\ln x - \ln b)\,dx \\
&= - \frac{\ln b}{b}+ \frac{1}{b}\int_0^{\infty} x^0e^{-x} \ln x \,dx \\
&= - \frac{\ln b}{b}+ \frac{1}{b}\psi(1)\Gamma(1) \\
&= - \frac{\ln b+\gamma}{b}
\end{align}
where I used the known value $\psi(1) = -\gamma$.
Also note that
\begin{align}
\int_0^{\infty} e^{-bx} \ln^2 x \,dx &= \frac{1}{b}\int_0^{\infty} e^{-x} \left(\ln^2 x - 2 \ln b \ln x + \ln^2 b\right)\,dx \\
&= \frac{1}{b}\int_0^{\infty} e^{-x} (\ln^2 x - 2 \ln b \ln x + \ln^2 b)\,dx \\
&= \frac{1}{b} \left(\pi^2/6+\gamma^2 + 2 \gamma \ln b + \ln^2 b \right) \\
&= \frac{\pi^2/6+(\gamma+\ln b)^2}{b}
\end{align}
where I used that $\psi_1(1) = \zeta(2) = \pi^2/6$.
Now substitute $u = -\ln x$ to obtain for $p = 1$
\begin{align}
\int_0^{1} \left(x^a + x^{-a} \right) \ln\left( \ln \frac{1}{x} \right) \,dx
&= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x} \right) \ln x \,dx \\
&= \frac{2 \gamma}{1-a^{2}} - \frac{\ln (1-a)}{1-a}- \frac{\ln (1+a)}{1+a},
\end{align}
where $a \neq 1$, and for $p = 2$
\begin{align}
\int_0^{1} \left(x^a + x^{-a} \right) \ln^2\left( \ln \frac{1}{x} \right) \,dx &= \int_0^{\infty} \left(e^{-(1+a)x} + e^{-(1-a)x} \right) \ln^2 x \,dx \\
&= \frac{\pi^{2}}{3(1-a^{2})} + \frac{(\gamma+\ln (1-a))^2}{1-a} + \frac{(\gamma+\ln (1+a))^2}{1+a},
\end{align}
where $a \neq 1$.
| {
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Calculate simple integral undefined I'm not being able to calculate $ \large{\int{\sqrt{\frac{x}{a-x}}} dx} $ , someone could help me? I tryed to use integration by parts, but i achieved $0 = 0$.
Thanks in advance.
| Let $\displaystyle u^2=\frac{x}{a-x}$. Then $x=\displaystyle\frac{au^2}{1+u^2}$ and so $\displaystyle du=\frac{2au\, du}{(1+u^2)^2}$. Then,
$$\int \sqrt{\frac{x}{a-x}}dx = \int \sqrt{u^2}\frac{2au}{(1+u^2)^2}du=\int \frac{2au^2\, du}{(1+u^2)^2}. $$
Now, let $u=\tan t$ and $\displaystyle\cos t=\frac{1}{\sqrt{1+u^2}}$. Then $du=\sec^2 t\,dt$ and we have
$$ \int \sqrt{\frac{x}{a-x}}dx =\int \frac{2au^2\, du}{(1+u^2)^2}=\int \frac{2a\tan^2 t\sec^2 t\, dt}{(\sec^2t)^2}=2a\int \frac{\tan^2t}{\sec^2t}dt=$$
$$=2a\int \frac{\sin^2t}{\cos^2t}\cdot \cos^2t\, dt=2a\int \sin^2t dt=2a\int \frac{1-\cos 2t}{2}dt= $$
$$=a\int dt-a\int \cos 2t \, dt = at-a\frac{1}{2}\sin 2t +c= $$
$$=at-a\sin t\cdot \cos t +c = a\arctan u-a\frac{u}{\sqrt{1+u^2}}\cdot \frac{1}{\sqrt{1+u^2}}+c= $$
$$= a\arctan u -\frac{au}{1+u^2}+c =$$
$$=a\arctan\sqrt{\frac{x}{a-x}}-\frac{\sqrt{\frac{x}{a-x}}}{1+\frac{x}{a-x}} +c.$$
| {
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Evaluate $\lim_{x\to 0}\frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}$
Evaluate
$$ \displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg).$$
I tried to use L'Hopital's rule but it got very messy. Moreover I also tried to analyze from graphs, but I was getting the limit $= 0$ by observing it. However, the answer given in my book is $\frac{1}{4}$. Is there any method to do without Taylor series and L' Hopital's rule (like using special limits). We are given that the limit exists. Any help will be appreciated.
Thanks!
| All right, let's go for the Taylor series. We'll be needing terms up to $O(x^6)$.
We'll start by writing $\cos x^{\sin x}$ as $\left(1 + \varepsilon\right)^{\sin x}$, with $\varepsilon \to 0$ as $x \to 0$. The lowest order term in $\sin x$ is x, while the lowest in $\varepsilon$ is $x^2$. To get sufficient detail, we'll therefore need the first two terms in the general binomial formula:
$$(1 + \varepsilon)^{\sin x} = 1 + \varepsilon\sin x + \frac{1}{2!}\varepsilon^2\sin x (\sin x -1) + O(x^7).$$
To the necessary order in $x$, we can write $\varepsilon = -\frac{x^2}{2} + \frac{x^4}{4!} + O(x^6)$ and $\sin x = x - \frac{x^3}{3!} + O(x^5)$. Inserting then gives
$$\begin{align} (1 + \varepsilon)^{\sin x} &= 1 + \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)\left(x - \frac{x^3}{3!}\right)\\ &\quad + \frac{1}{2} \left(-\frac{x^2}{2} + \frac{x^4}{4!}\right)^2\left(x - \frac{x^3}{3!}\right)\left(x - \frac{x^3}{3!} -1\right) + O(x^7)\\
&= 1 - \frac{x^3}{2} + \frac{x^5}{12} + \frac{x^5}{24} - \frac{x^5}{8} + \frac{x^6}{8} + O(x^7)\\
&= 1 -\frac{x^3}{2} + \frac{x^6}{8} + O(x^7)
\end{align} $$
And for the square root part of the numerator we have
$$
\sqrt{1 - x^3} = 1 - \frac{x^3}{2} -\frac{x^6}{8} + O(x^9). $$
Combining gives a numerator of $\frac{1}{4}x^6 + O(x^7)$, which divides by the denominator to give $\frac{1}{4} + O(x)$, and so the limit as $x$ tends to zero is indeed $\frac{1}{4}$.
| {
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The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
| Extract from my note Derivatives of generalized power functions that appeared in the Reader Reflections column of Mathematics Teacher [Volume 103, Number 9; May 2010; pp. 630-631]:
Regarding the editor's note on higher derivatives of $x^{x},$ let $n$ be a positive integer, ${n \choose k}$ be the usual binomial coefficient, $f(x) = x^{x},$ and $g(x) = 1 + \ln{x}.$ Kulkarni (1984) gave the recursion formula
$$ f^{(n+1)}(x) \;\; = \;\; f^{(n)}(x)g(x) \;\; + \;\; \sum\limits_{k=1}^{n} \left[ {n \choose k} (-1)^{k-1}(k-1)! \right] f^{(n-k)}(x)x^{-k} $$
by observing $f^{(n+1)}(x)$ is the $n$th derivative of $f'(x) = x^x(1 + \ln{x})$ and then writing down the Leibniz formula for the $n$th derivative of a product, using the fact that the $k$th derivative of $1 + \ln x$ is $(-1)^{k-1}(k-1)!x^{-k}.$
S. B. Kulkarni, Solution to Problem 3977, School Science and Mathematics 84 #7 (November 1984), 629-630.
| {
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$\epsilon-N$ sequence convergence proof verification
Prove that the sequence $\displaystyle{\frac{n^3+4}{3n^3-2n^2+1}}$ converges to $\displaystyle{\frac{1}{3}}$
Here's my (somewhat terse) proof. Can someone please verify it?
Let $\epsilon > 0$.
Let $N = \displaystyle{\frac{13}{9\epsilon}+\frac{2}{3}}$, and let $n > N$.
Now, $\displaystyle{n > \frac{13}{9\epsilon} + \frac{2}{3}}$.
$\displaystyle{3n - 2 > \frac{13}{3\epsilon}}$
$\displaystyle{\frac{13}{3n-2}} < 3\epsilon$
$\displaystyle{\frac{13n^2}{3n^3-2n^2}}<3\epsilon$
$\displaystyle{\frac{13n^2}{3n^3-2n^2+1}} < 3\epsilon$
$\displaystyle{\frac{2n^2+11}{3n^3-2n^2+1} < 3\epsilon}$
$\displaystyle{\left|\frac{n^3+4}{3n^3-2n^2+1} - \frac{1}{3}\right| < \epsilon}$
| I don't see any flaw in your development.
You can take shortcuts, though:
$\lim_{n\rightarrow\infty}\frac{n^3+4}{3n^3-2n^2+1}=\lim_{n\rightarrow\infty}\frac{1+\frac4{n^3}}{3-\frac2n+\frac1{n^3}}=\frac{\lim_{n\rightarrow\infty}1+\frac4{n^3}}{\lim_{n\rightarrow\infty}3-\frac2n+\frac1{n^3}}=\frac13$
| {
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How to prove $\frac{1+\sin{6^\circ}+\cos{12^\circ}}{\cos{6^\circ}+\sin{12^\circ}}=\sqrt{3}$? I found this interesting result. Show that
$$\dfrac{1+\sin{6^\circ}+\cos{12^\circ}}{\cos{6^\circ}+\sin{12^\circ}}=\sqrt{3}.$$
See this Wolfram Alpha output.
My attempt is very ugly. We know that
$$\sin{6^\circ}=\dfrac{1}{8}(-1-\sqrt{5})+\dfrac{1}{4}\sqrt{\dfrac{3}{2}(5-\sqrt{5})}$$
$$\cos{6^\circ}=\dfrac{1}{4}\sqrt{\dfrac{1}{2}(5-\sqrt{5})}+\dfrac{1}{8}\sqrt{3}(1+\sqrt{5})$$
$$\sin{12^\circ}=2\sin{6^\circ}\cos{6^\circ}$$$$\cos{12^\circ}=2\cos^2{6^\circ}-1$$ and it is very ugly, maybe there exist simple methods.
Thank you.
| \begin{align}
x&=\frac{1+\sin6+\cos12}{\cos6+\sin12} \\
&=\frac{2\sin30+\sin78+\sin6}{\cos78+\cos6} \\
&=\frac{2(\sin54-\sin18)+2\sin42\cos36}{2\cos42\cos36} \\
&=\frac{4\cos36\sin18+2\sin42\cos36}{2\cos42\cos36} \\
&=\frac{2\cos36(2\sin18+\sin42)}{2\cos42\cos36} \\
&=\frac{2\sin18+\sin42}{\cos42} \\
&=\frac{\sin42+\sin18+\sin18}{\cos42} \\
&=\frac{2\sin30\cos12+\cos72}{\cos42} \\
&=\frac{\cos72+\cos12}{\cos42} \\
&=\frac{2\cos42\cos30}{\cos42} \\
&=2\cos30 \\
&=\sqrt 3
\end{align}
| {
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Probability that $5 \mid x^4 - y^4$ for random $x, y$ Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3,\cdots,100\}$. Find the probability that $x^4 - y^4$ is divisible by $5$.
I don't know how to proceed with this problem. So any help would be appreciated.
| HINT:
If $\displaystyle 5|(x^4-y^4)$
Case $1:$
$\displaystyle 5|x^4$
as $5$ is prime, it must divide $x$ and subsequently $5|y^4\iff 5|y$
Case $2:$
As $5$ is prime, if $5\nmid x,(5,x)=1$
Now, $\displaystyle x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2-4+5)=(x-1)(x+1)(x-2)(x+2)+5(x^2-1)$
If $5\nmid x,5$ must divide exactly one of $(x-1),(x+1),(x-2),(x+2)$
$\displaystyle\implies 5\mid(x^4-1)$
Similarly, $5\nmid y,(5,y)=1$
Now there are $\displaystyle\frac{100}5=20$ multiples of $5$ in the given set
We need $$P\{xy, (xy,5)=1 \text{ or } (5|x\text{ and } 5|y)\}$$
Can you take it from here?
| {
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Series expansion of $\frac{x^2}{1+ \sin x}$ For the series expansion at $x=0$ for
$\dfrac{x^2}{1+ \sin x}$ WolframAlpha gives $$x^2 -x^3 +x^4-\frac{5x^5}{6}+\frac{2x^6}{3}-\frac{61x^7}{120}+O(x^8)$$
But I'm missing something in the simplification, and I can't quite see it. How do I move from
$\dfrac{x^2}{1+(x-\frac{x^3}{6}+O(x^5))}$ to that final form?
| Note that $\frac{1}{1+x} = 1 - x + x^2 - ...$ as $x \rightarrow 0$. So, since $\sin(x) \rightarrow 0$ as $x \rightarrow 0$, we can write it as follows:
$\frac{x^2}{1+\sin(x)} = x^2\cdot (1 - \sin(x) + \sin^2(x) - \sin^3(x) + ...) \\
= x^2\cdot (1 - x + x^3/6 + x^2 - x^3 + \mathcal{O}(x^4)) \\
= x^2 - x^3 + x^4 - 5x^5/6 + \mathcal{O}(x^6).$
If more terms are desired, expand as required.
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Matrix that moves the graph of a polynomial. Consider the vector space of all polynomials of degree $\leq2$. Find the matrix which describes: $$p(t)\mapsto p(t-1)$$
So $t^2-2t+1$ for example would change to $(t-1)^2-2(t-1)+1$.
(The graph moves one unit to the right.)
First I set up this equation for the example above:
$$
\begin{pmatrix}
\cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot
\end{pmatrix}
\begin{pmatrix}
t^2\\
-2t\\
1\\
\end{pmatrix}
=\begin{pmatrix}
(t-1)^2\\
-2(t-1)\\
1\\
\end{pmatrix}=
\begin{pmatrix}
t^2-2x+1\\
-2t+2\\
1\\
\end{pmatrix}
$$
I figured out that the matrix has to be
\begin{pmatrix}
1 & 1 & 1\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{pmatrix}
Which looks very nice, so I thought I was done. Just to be sure, I took another example:
$$t^2+t+1 ~~~ \longmapsto ~~~(t-1)^2+(t-1)+1$$
But in this case, the matrix seems to be
\begin{pmatrix}
1 & -2 & 1\\
0 & 1 & -1\\
0 & 0 & 1\\
\end{pmatrix}
Finally, I multiplied out the first example to $t^2-4t+4$, took the representing vector and the matrix for that transformation had only entries on the diagonal (obviously, but still) - so which one is it/what am I doing wrong?
| \begin{align}
p(t) & = at^2+bt+c \\
p(t-1) & = a(t-1)^2+b(t-1)+c \\
& = at^2 + (b-2a)t+ (c+a-b).
\end{align}
$$
\begin{bmatrix} a \\ b \\ c \end{bmatrix} \mapsto \begin{bmatrix} a \\ b-2a \\ c+a-b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\
-2 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/806146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving a recursion relation: $a_{n+1}=-3a_n+4a_{n-1}+2$ I'm having trouble solving this recursion relation. I believe it's a non-homogeneous one.
Here it is:
$$a_{n+1}=-3a_n+4a_{n-1}+2$$
Really, I am just having trouble with the particular solution.
The initial values are not specified.
Thanks!
| My favorite way to solve these is with linear algebra:
$$\begin{align}
a_{n + 1}= -3a_n + 4a_{n-1} + 2 \\
a_{n} = 1a_n + 0a_{n-1} + 0 \\
\end{align}$$
$$\begin{bmatrix} a_{n+1} \\ a_n \end{bmatrix} =
\begin{bmatrix} -3 & 4 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix}
+ \begin{bmatrix} 2 \\ 0 \end{bmatrix}
$$
$$\begin{bmatrix} a_{n+1} \\ a_n \\ 1 \end{bmatrix} =
\begin{bmatrix} -3 & 4 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} a_n \\ a_{n-1} \\ 1 \end{bmatrix}
$$
$$\begin{bmatrix} a_{n+1} \\ a_n \\ 1 \end{bmatrix} =
\begin{bmatrix} -3 & 4 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}^n
\begin{bmatrix} a_1 \\ a_0 \\ 1 \end{bmatrix}
$$
Although I prefer leaving it like this, you can use jordan decomposition:
$$\begin{cases}
P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & -\frac 14 \\ 0 & \frac 52 & 0 \end{bmatrix} \\
D = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix} \\
\begin{bmatrix} a_{n+1} \\ a_n \\ 1 \end{bmatrix} =
(P\,D\,P^{-1})^n
\begin{bmatrix} a_1 \\ a_0 \\ 1 \end{bmatrix}
\end{cases}
$$
$$\begin{bmatrix} a_{n+1} \\ a_n \\ 1 \end{bmatrix} =
P\,\begin{bmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (-4)^n \end{bmatrix}\,P^{-1}
\begin{bmatrix} a_1 \\ a_0 \\ 1 \end{bmatrix}
$$
$$
a_n = \frac 1{25}\bigg( a_1\,5\left(1 - (-4)^{n+1}\right) + a_0\,5\left(4 + (-4)^{n+1}\right) + 2(-4)^{n+1} + 10n + 8 \bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\lfloor (3 + \sqrt{5})^{34} \rfloor \pmod {100}$ The problem is to evaluate $\lfloor (3 + \sqrt{5})^{34} \rfloor \pmod {100}$
No calculators are allowed.
I think I have to get rid of $\sqrt{5}$ somehow since it is irrational and would make it hard to find the floor without doing it numerically. But I'm not sure how.
| This answer inspired by the Fibonacci sequence, in that $$F_n = \frac{\varphi^n - \overline{\varphi}\,^n}{\sqrt 5} = \frac{\lfloor{\varphi^n}\rfloor + 1}{\sqrt 5} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n\,_{2,1}$$
which gives us an easy way to compute $\lfloor \varphi^n \rfloor$ using the above matrix. So attempting to find a matrix that allows the computation of $\lfloor\phi^n\rfloor$ for $\phi = 3 + \sqrt 5$ and $\overline \phi = 3 - \sqrt 5$, consider the matrix:
$$M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \tag 1$$
It has characteristic equation
$$P(x) = x^2 - (D+A)x + AD - BC \tag 2$$
which to have the desired eigenvalues, the characteristic equation must also be:
$$P(x) = (x - \phi)(x - \overline \phi) = x^2 - 6x + 4 \tag 3$$
Eliminating variables $C$ and $D$ between (1) (2) and (3), it leaves:
$$M = \begin{bmatrix} A & B \\ \frac{(6-A)A - 4}{B} & 6 - A \end{bmatrix} \tag 4$$
Now any choice of $A$ and $B$ will work, but to choose something easy, consider $A = 3$ to make the diagonal nice and $B=1$ to make the rest nice:
$$M = \begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}$$
with the eigenvalue decomposition being:
$$M =
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}
\begin{bmatrix} 3 + \sqrt{5} & 1 \\ 0 & 3 - \sqrt{5}\end{bmatrix}
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}^{-1}
$$
and it follows:
$$M^n =
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}
\begin{bmatrix} \phi^n & 1 \\ 0 & \overline \phi\,^n\end{bmatrix}
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}^{-1}
=
\begin{bmatrix}
\frac{\phi^n + \overline \phi\,^n}{2} & \frac{\phi^n + \overline \phi\,^n}{2\sqrt{5}} \\
\frac{\phi^n + \overline \phi\,^n}{2 / \sqrt{5}} & \frac{\phi^n + \overline \phi\,^n}{2} \\
\end{bmatrix}
$$
So we have $2 M^n\,_{1,1} = \phi^n + \overline \phi\,^n = \lfloor \phi^n \rfloor + 1$, altogether:
$$\lfloor(3 + \sqrt{5})^n\rfloor = 2 {\begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}^n}_{1,1} - 1$$
$$\downarrow$$
$$\boxed{\lfloor(3 + \sqrt{5})^n\rfloor \equiv 2 {\begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}^n}_{1,1} - 1 \pmod{100}} \tag{!!}$$
Matrix modular exponentiation with integers is fairly easy to do with repeated squaring:
$$
\begin{array} {|c|c|} \hline
M^1 & \begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix} \\ \hline
M^2 = (M^1)^2 & \begin{bmatrix} 14 & 6 \\ 30 & 14 \end{bmatrix} \\ \hline
M^4 = (M^2)^2 & \begin{bmatrix} 76 & 68 \\ 40 & 76 \end{bmatrix} \\ \hline
M^8 = (M^4)^2 & \begin{bmatrix} 96 & 36 \\ 80 & 96 \end{bmatrix} \\ \hline
M^{16} = (M^8)^2 & \begin{bmatrix} 96 & 12 \\ 60 & 96 \end{bmatrix} \\ \hline
M^{32} = (M^{16})^2 & \begin{bmatrix} 36 & 4 \\ 30 & 36 \end{bmatrix} \\ \hline
M^{34} = M^{32}M^2 & \begin{bmatrix} 24 & 70 \\ 60 & 24 \end{bmatrix} \\ \hline
\end{array}$$
So $\lfloor(3 + \sqrt{5})^{34}\rfloor \equiv 2 \times 24 - 1 \equiv 47 \pmod{100}$
It seems that using the above style argument, one can say in general that if $0 < X - \sqrt{Y} < 1$, then the follow identity holds:
$$\lfloor (X + \sqrt{Y})^n \rfloor = 2{\begin{bmatrix} X & Y \\ 1 & X \end{bmatrix}^n}_{1,1} - 1$$
I wonder if this is a well known identity?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sin^4(x)\cos^2(x) + \sin^2(x)\cos^4(x) - \sin^2(x)\cos^2(x)=0.$ Using the following expression:
$$\sin^4(x)\cos^2(x) + \sin^2(x)\cos^4(x) - \sin^2(x)\cos^2(x).$$
The above expression is supposed to evaluate to zero, but how?
| \begin{align}
\sin^4 x \cos^2 x + \sin^2 x\cos^4 x - \sin^2 x \cos^2 x&=\sin^2 x \cos^2 x\ (\sin^2 x + \cos^2 x - 1)\\
&=\sin^2 x \cos^2 x\ (1-1)\\
&=\large\color{blue}0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
An identity of an Elliptical Integral
Suppose $0<k<1$ and $\displaystyle
K(k)=\int_0^1\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-k^2x^2)}}$. Let
$\tilde{k}$ be $\tilde{k}^2=1-k^2$. Show that $$\displaystyle
K(k)=\frac{2}{1+\tilde{k}}K\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)$$
There's a hint in Stein's Complex Analysis which is this change of variable : $x=\dfrac{2t}{1+\tilde{k}+(1-\tilde{k})t^2}$.
| For me, the only way I can remember this sort of complicated identities is through the relationship between the
complete elliptic integral of the first kind
$K(k)$ and the corresponding arithmetic-geometric mean.
$$K(k) = \int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \frac{\pi}{2\text{AGM}( 1, \sqrt{1-k^2})}\tag{*1}$$
Consider following integral
$$I(a,b) = \int_0^{\frac{\pi}{2}}
\frac{d\theta}{\sqrt{a^2\cos^2\theta + b^2\sin^2\theta}}$$
Introduce $x = b\tan\theta$, we can rewrite it as
$$I(a,b) = \int_0^{\frac{\pi}{2}} \frac{d\tan\theta}{
\sqrt{(1+\tan^2\theta)(a^2+b^2\tan^2\theta})}
= \int_0^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2+b^2)}}\tag{*2}$$
Substitute $x$ by $\sqrt{ab} t$, we have
$$I(a,b)
= \frac{1}{\sqrt{ab}}\int_0^\infty \frac{dt}{\sqrt{t^4 + \left(\frac{a}{b}+\frac{b}{a}\right)t^2 + 1}}
= \frac{1}{\sqrt{ab}}\int_0^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t}
$$
Notice the last integrand is invariant under transform $\displaystyle\;t \leftrightarrow \frac{1}{t}$. If we introduce two more variables $s$ and $y$ such that
$$s = \frac12 (t - t^{-1}) = \frac{y}{\sqrt{ab}}$$ and using the fact $$\frac{dt}{t} = \frac{d(t - t^{-1})}{t + t^{-1}} = \frac{ds}{\sqrt{s^2+1}}$$
We can rewrite $I(a,b)$ as
$$
\frac{2}{\sqrt{ab}}\int_1^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t}
= \frac{2}{\sqrt{ab}}\int_0^\infty
\frac{ds}{\sqrt{\left(4s^2 + \frac{(a+b)^2}{ab}\right)(s^2+1)}}\\
= \int_0^\infty
\frac{dy}{\sqrt{\left(y^2 + \left(\frac{a+b}{2}\right)^2\right)(y^2 + ab)}}
$$
Compare this with $(*2)$, we obtain an important identity:
$$I(a,b) = I\left(\frac{a+b}{2}, \sqrt{ab}\right)$$
This means $I(a,b)$ is invariant if we replace $(a,b)$ by their AM and GM.
Start with any pair of numbers $a,b$, it is well known if you repeat taking AM/GM of them,
the pairs will ultimately converge to a single number. This is called the arithmetic geometric mean of $a$ and $b$ and usually denoted as $\text{AGM}(a,b)$. If one replace $a$, $b$ by this AGM in the definition of $I(a,b)$, we obtain
$$I(a,b) = \frac{\pi}{2\text{AGM}(a,b)}$$
Together with the obvious identity $K(k) = I(1,\sqrt{1-k^2})$, we immediately obtain $(*1)$.
Using these tools and notice $\text{AGM}(a,b)$ is homogenous. i.e.
$$\text{AGM}(\lambda a, \lambda b) = \lambda \text{AGM}(a,b) \quad\implies\quad I(\lambda a, \lambda b) = \frac{1}{\lambda} I(a,b),$$
the desired identity follows immediately.
$$
K(k)
= I(1,\tilde{k}) = I\left(\frac{1+\tilde{k}}{2},\sqrt{\tilde{k}}\right) = \frac{2}{1+\tilde{k}} I\left(1,2\frac{\sqrt{\tilde{k}}}{1+\tilde{k}}\right)\\
= \frac{2}{1+\tilde{k}} K\left( \sqrt{1 - \frac{4\tilde{k}}{(1+\tilde{k})^2}} \right)
= \frac{2}{1+\tilde{k}} K\left( \frac{1-\tilde{k}}{1+\tilde{k}} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
If $AB = BA^2$ and $B^5 = I,$ Then how can we prove $A^{31} = I.$
If $A$ and $B$ are two non singular matrices, $AB = BA^2$ and $B^5 = I,$ then how can we prove $A^{31} = I$?
$\bf{My\; Trial::}$ Using $B^5 = I\Rightarrow B^5A^5 = IA^5 = A^5\Rightarrow B^4BA^2A^3 = A^5$
Now Using $BA^2 = AB$, we get $B^4ABA^3 = A^5\Rightarrow B^4ABA^2A=A^5\Rightarrow B^4A^2BA=A^5$
I did not understand How can I prove it.
plz Help me
Thanks
| Your first equation can be written as
$$B^{-1}AB=A^2$$
If we conjugate by $B$ again we get
$$B^{-2}AB^2=B^{-1}A^2B=B^{-1}AB B^{-1}AB=A^2 A^2=A^4$$
Iterating this we get
$$B^{-5}AB^5=A^{2^5}$$
or
$$A=A^{32}$$ and this gives
$$A^{31}=I.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$
Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$
What I did:
We can prove:
$${\cos x -1 + {x^2\over 2} \over {x^4\over 24}} < 1$$
Lets define:
$f(x) = \cos x -1 + {x^2\over 2}$ and $g(x)= {x^4\over 24}$
By LMVT:
$${{f(x) - f(0)} \over {g(x) - f(0)}} = {f'(y)\over g'(y)} = {{-\sin y + y} \over {4x^3\over 24}} = {{-\sin y + y} \over {x^3\over 6}}\text{ where }y\in(0,x)$$
I tried to show the last expression is smaller than $1$, but without success.
What's the trick? Maybe the $f(x), g(x)$ are wrong?
| \begin{align}
f(x) & =1-\dfrac{x^2}{2} + \dfrac{x^4}{24}, & f(0) =1& =\cos0 \\[6pt]
f'(x) & = -x + \frac{x^3}{6}, & f'(0) =0 & =\cos'0 \\[6pt]
f''(x) & = -1 + \frac{x^2}{2}, & f''(0)=-1 & =\cos''0 \\[6pt]
f'''(x) & = x, & f'''(0) = 0 & = \cos'''0 \\[6pt]
f''''(x) & = 1, & f''''(0) = 1 & =\cos''''0
\end{align}
We have $f''''(x)\ge\cos''''x$ for all values of $x$, because $f''''$ remains equal to $1$ while $\cos''''=\cos$ oscillates between $1$ and $-1$.
Thus the derivative of $f'''$ is everywhere greater than the derivative of $\cos'''$ and their values at $0$ are equal. Hence the mean value theorem tells us that $f'''\ge\cos'''$ on $[0,\infty)$ and $f'''\le \cos'''$ on $(-\infty,0]$
Since the derivative of $(f-\cos)''$ is $\ge0$ on $[0,\infty)$ and $\le0$ on $(-\infty,0]$ and $=0$ at $0$, we must have $(f-\cos)''\ge0$ everywhere. Hence the mean value theorem tells us that the graph of $f-\cos$ is concave upward everywhere. If it is concave upward everywhere at equal to $0$ at $0$, then it is non-negative everywere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proof by induction that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ I am trying to understand how to do proof by induction for inequalities. The step that I don't fully understand is making an assumption that n=k+1. For equations it is simple. For example:
Prove that 1+2+3+...+n = $ \frac {n(n+1)}{2} $ is valid for $ i \ge 1 $
1) Base case for n=1. (...)
2) Assume that equation is true for n=k. (...)
3) Proof for n=k+1.
1+2+3+...+k+(k+1) = $ \frac {(k+1)[(k+1)+1]}{2} $
So we add +1 to k on both sides.
For inequalities:
Prove that $ 4n \lt 2^n $ is valid for $ n \ge 5 $
1) Base case. (...)
2) n=k assumption. (...)
3) n=k+1 proof.
How I would do this(add 1 to k on both sides):
$ 4(k+1) < 2^{k+1} $
How it's done in tutorials:
$ 4(k+1) < 2^k + 4 $
$ 4k+4 < 2^k + 4 $
Why they add extra 4 to RHS instead of increasing k by 1?
| For the first part, you haven't done anything, you just rewrote everything. Most importantly, you didn't prove the statement. Induction works by first showing that $P(1)$ holds then assuming $P(n)$ and proving $P(n+1)$. From there, it follows that the $n$ such that $P(n)$ is true must be all of $\mathbf{N}$.
In our case $P(n)$ is the statement that $1+2+\dots+n=\frac{n(n+1)}{2}$. If we look at $n=1$, we see that $\frac{n(n+1)}{2}=\frac{1\cdot 2}{2}=1$. So that $P(1)$ holds.
Now, we let $n$ be such that $P(n)$ holds and prove that $P(n+1)$ must also be true. Let's consider $1+2+\dots+n+(n+1).$ Recall that we know $P(n)$ is true, so, we can substitute $\frac{n(n+1)}{2}$ in for $1+\dots+n$. Therefore, $$1+2+\dots+n+(n+1)=\frac{n(n+1)}{2}+n+1=\frac{n^2+n+2n+2}{2}=\frac{(n+1)((n+1)+1)}{2}.$$ Notice that the far right hand side is exactly what we wanted to show, that is, that $P(n+1)$ holds.
So, we've shown that the set of $n$ such that $P(n)$ holds is all of $\mathbf{N}$, that is, all $n\geq 1$.
Now, for the second question. We let $Q(n)$ be the statement $4n<2^n$ for $n\geq 5$ (I'm assuming yours is a typo, clearly $8=4\cdot 2< 2^2=4$ is not true.)
So, let's do the base case: $2^5=32$ and $4\cdot 5=20$ so clearly we have $4\cdot 5< 2^5$. Thus, $Q(5)$ is true.
Now, let $n$ be such that $Q(n)$ holds. We show that $Q(n+1)$ holds too. Observe that $4(n+1)=4n+4<2^n+4$ by the assumption. However, let's recall that $4=2^2<2^n$ for $n> 2$. Therefore, $$4(n+1)<2^n+2^n=2(2^n)=2^{n+1}.$$
Notice that we've only shown that $Q(n)$ implies $Q(n+1)$ when $n>2$. But, since $5>2$, it follows that $Q(n)$ is true for all $n\geq 5$. Which is exactly what we wanted to show.
Hopefully this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/815963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\sqrt[4]{161-72 \sqrt{5}}$ $$\sqrt[4]{161-72 \sqrt{5}}$$
I tried to solve this as follows:
the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:
$$a^4+30 a^2 b^2+25 b^4=161$$
$$4 a^3 b+20 a b^3=-72$$
In an attempt to solve this, I first tried to factor stuff and rewrite it as:
$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$
$$4 a b \left(a^2+5 b^2\right)=-72$$
Then letting $p = a^2 + 5b^2$ and $q = ab$ you get
$$4 p q=-72$$
$$p^2+10 q^2=161$$
However, solving this yields messy roots. Am I going on the right path?
| Another approach. We can apply twice the following general algebraic identity involving nested
radicals
\begin{equation*}
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{
a^{2}-b}}{2}}\tag{1}
\end{equation*}
to get
\begin{equation*}
\sqrt[4]{161-72\sqrt{5}}=\sqrt[4]{161-
\sqrt{25\,920}}=\sqrt{5}-2.
\end{equation*}
The numerical computation can be carried out as follows:
\begin{eqnarray*}
\sqrt[4]{161-72\sqrt{5}} &=&\left( \sqrt{\frac{161+\sqrt{161^{2}-25\,920}}{2}
}-\sqrt{\frac{161-\sqrt{161^{2}-25\,920}}{2}}\right) ^{1/2} \\
&=&\left( \sqrt{\frac{161+1}{2}}-\sqrt{\frac{161-1}{2}}\right) ^{1/2} \\
&=&\sqrt{9-\sqrt{80}} \\
&=&\sqrt{\frac{9+\sqrt{9^{2}-80}}{2}}-\sqrt{\frac{9-\sqrt{9^{2}-80}}{2}} \\
&=&\sqrt{\frac{9+1}{2}}-\sqrt{\frac{9-1}{2}}\\
&=&\sqrt{5}-2.
\end{eqnarray*}
ADDED. Note: If the radical were of the form $\sqrt{a+\sqrt{b}}$, then the applicable identity would be
\begin{equation*}
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{
a^{2}-b}}{2}}.\tag{2}
\end{equation*}
Proof (from Sebastião e Silva, Silva Paulo, Compêndio de Álgebra
II, 1963). To find two rational numbers $x,y$ such that
\begin{equation*}
\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y},\text{ with }a,b\in \mathbb{Q},
\end{equation*}
we square both sides and rearrange the terms
\begin{equation*}
2\sqrt{xy}=a-x-y+\sqrt{b}.
\end{equation*}
Squaring again yields
\begin{equation*}
4xy=\left( a-x-y\right) ^{2}+2\left( a-x-y\right) \sqrt{b}+b.
\end{equation*}
Since $x,y\in \mathbb{Q}$, $a-x-y=0$, which means that $x,y$ satisfy the system of equations
\begin{equation*}
x+y=a,\qquad xy=\frac{b}{4}.
\end{equation*}
Consequently they are the roots of
\begin{equation*}
X^{2}-aX+\frac{b}{4}=0,
\end{equation*}
i.e.
\begin{eqnarray*}
x &=&X_{1}=\frac{a+\sqrt{a^{2}-b}}{2} \\
y &=&X_{2}=\frac{a-\sqrt{a^{2}-b}}{2}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
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} |
Real $2\times 2$ matrix $X$ such that $X^2 + 2X= -5I$ Find a real $2\times 2$ matrix $X = \left(\begin{matrix} a& b\\ c & d\end{matrix}\right)$ such that $X^2 + 2X = -5I.$
With this question I'm kinda lost with the $2X$ part but a full explanation or a little bit of a lead would be greatly appreciated.
| Expanded you get that
\begin{matrix}
5 + a (2 + a) + b c & b (2 + a + d) \\
c (2 + a + d) & 5 + b c + d (2 + d) \\
\end{matrix}
equals 0.
Solving using Mathematica 9
X = {{a, b}, {c, d}};
Id = IdentityMatrix[2];
Solve[ X.X + 2 X == -5 Id, {a, b, c, d}]
Atleast these are possible answers:
{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 - 2i},
{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 + 2i},
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 - 2i},
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 + 2i}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
A sine integral The integral
\begin{align}
\int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta
\end{align}
is claimed to not have a closed form expression. In this view find the series solution of the integral as a series involving of $n$.
Editorial note:
As described in the problem several series may be obtained, of which, all seem to hold validity. As a particular case, from notes that were made a long while ago, the formula
\begin{align}
\int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta = \sum_{r=1}^{\infty} (-1)^{r-1} \ \ln\left(\frac{2r+1}{2r-1}\right) \ \sin(r n \pi)
\end{align}
is stated, but left unproved. Can this formula be proven along with finding other series dependent upon $n$?
| In the case that $n=2k$ is an even positive integer we have the identity $$\frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$
Therefore,
$$ \begin{align} \int_{0}^{\pi /2} \frac{\sin 2kx}{\sin x} \ dx &= 2 \int_{0}^{\pi/2} \Big(\cos x + \cos 3x + \ldots + \cos(2k-1) \ x \Big) \ dx \\ &= 2 \left(\sin x + \frac{1}{3} \sin 3x + \ldots + \frac{1}{2k-1} \sin(2k-1) x \right)\Bigg|^{\pi/2}_{0} \\ &= 2 \left(1- \frac{1}{3} + \ldots + \frac{(-1)^{k+1}}{2k-1} \right). \end{align}$$
EDIT:
To prove that $$ \frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x,$$ notice that
$$\begin{align} \frac{\sin kx}{\sin x} &= \frac{e^{ikx}-e^{-ikx}}{e^{ix}-e^{-ix}} \frac{e^{-ix}}{e^{-ix}} \\ &= \frac{e^{i(k-1)x} - e^{i(-k-1)x}}{1-e^{-2ix}} \\ &= e^{i(k-1)x} \frac{1-e^{-2ikx}}{1-e^{-2ix}} \\ &= e^{i(k-1)x} \left(1 + e^{-2ix} + e^{-4ix} + \ldots + e^{-2i(k-1)x} \right) \\ &= e^{i(k-1)x} + e^{i(k-3)x} + \ldots + e^{i(-k+1)x}. \end{align}$$
Therefore,
$$ \frac{\sin 2kx}{\sin x} = e^{i(2k-1)x} + e^{i(2k-3)x} + \ldots + e^{ix} + e^{-ix} + \ldots + e^{i(-2k+1)x}$$
$$ = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to integrate $\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$?
How to integrate $$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$$ ?
I have:
$$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx = \int \frac{\cos x}{\sqrt{2\sin x\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\cos x}{\sqrt{\sin x}\sqrt{\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\cot x} \,dx \\
t = \sqrt{\cot x} \implies x = \cot^{-1} t^2 \implies \,dx = -\frac{2t\,dt}{1 + t^4}$$
so I have:
$$-\sqrt2 \int \frac{t^2 \,dt}{1 + t^4}$$
I tried partial integration on that but it just gets more complicated. I also tried the substitution $t = \tan \frac{x}{2}$ on this one: $\frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx$
$$= \frac{1}{\sqrt2}\int \sqrt{\frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1+t^2}}} \frac{2\,dt}{1+t^2} = \frac{1}{\sqrt2}\int \sqrt{\frac{1 - t^2}{2t}} \frac{2\,dt}{1+t^2} = \int \sqrt{\frac{1 - t^2}{t}} \frac{\,dt}{1+t^2}$$
... which doesn't look very promising.
Any hints are appreciated!
| $$I=\int \frac{\cos x}{\sqrt{2\sin x\cos x}}dx$$
$$I=\int \frac{\cos x}{\sqrt{(sin x+\cos x)^2-1}}dx=\int \frac{\cos x}{\sqrt{1-(\sin x-\cos x)^2}}dx$$
$$2I=\frac12\left[ \int \frac{\cos x-\sin x}{\sqrt{(sin x+\cos x)^2-1}}dx +\int \frac{\cos x+\sin x}{\sqrt{1-(\sin x-\cos x)^2}}dx\right]$$
$$I=\frac14\left[ \int \frac{d(\sin x+\cos x)}{\sqrt{(sin x+\cos x)^2-1}} +\int \frac{d(\sin x-\cos x)}{\sqrt{1-(\sin x-\cos x)^2}}\right]$$
$$I=\frac14\left[ \cosh^{-1}(\sin x+\cos x)+\sin^{-1}(\sin x-\cos x) \right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Non-existence of natural numbers such that $\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}$ Show that for any $n\in\mathbb{N}$ there does not exist natural numbers $x,y$ such that $$\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}.$$
| Hint: Assume on the contrary that there is such $x, y$. Show $x+y \geq 2n+1$. Write $x+y=2n+1+k, k \geq 0$ and show $(2n+1-k)^2-1<4xy<(2n+1-k)^2$.
Hint 2: To show the last part, square the original inequality, shift some terms, and square again. Proving the lower bound will require slightly more work than the upper bound.
Edit: Since you appear to have so much difficulty with the lower bound, I guess it would be best to explain in more detail.
As mentioned in the second hint, square both sides and subtract $x+y$ to get
$$2n+1+2\sqrt{n}\sqrt{n+1}-(x+y)<2\sqrt{xy}<4n+2-(x+y)$$
Since $x+y=2n+1+k$,
$$2\sqrt{n}\sqrt{n+1}-k<2\sqrt{xy}<2n+1-k$$
The upper bound trivially implies $k<2n+1$, so $k \leq 2n$, so $k<2\sqrt{n}\sqrt{n+1}$.
Thus the lower bound is positive, and we may square again.
$$(2\sqrt{n}\sqrt{n+1}-k)^2<4xy<(2n+1-k)^2$$
For the lower bound,
\begin{align}
& (2\sqrt{n}\sqrt{n+1}-k)^2-((2n+1-k)^2-1)\\
& =(4n(n+1)+k^2-4k\sqrt{n}\sqrt{n+1})-(4n^2+4n+1+k^2-2k(2n+1)-1)\\
&=2k(2n+1-2\sqrt{n}\sqrt{n+1}) \\
&\geq 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Rotation and surface in space Which surfaces in the space represent this equation
$$
x^2+2y^2+z^2+4xy+2xz-4yz = 1 ?
$$
I calculate the characteristic polynomial.
$$
\begin{bmatrix}
1-\lambda & 2 &1 \\
2 & 2-\lambda &-2 \\
1 &-2 & 1-\lambda \\
\end{bmatrix}
$$
$$
A=
\begin{vmatrix}
1& 2& 1 \\
2& 2& -2 \\
1& -2& 1 \\
\end{vmatrix}
$$
$$
\begin{bmatrix}
1-\lambda & 2 &1 \\
2 & 2-\lambda &-2 \\
1 &-2 & 1-\lambda \\
\end{bmatrix}
= (1-\lambda)^2(2-\lambda)-4-4-((2-\lambda)+8(1-\lambda))
$$
$$
=(1-\lambda)^2(2-\lambda)-18-\lambda
$$
$$
=(2-\lambda)(4-\lambda)(2+\lambda)
$$
$\lambda_1 = 2$
$$
(A-\lambda I)=
\begin{vmatrix}
-1& 2& 1 \\
2& 0& -2 \\
1& -2& -1 \\
\end{vmatrix}
\Rightarrow
v_1 =
\begin{bmatrix}
1\\
0\\
1\\
\end{bmatrix}
$$
$\lambda_2 = 4$
$$
(A-\lambda I)=
\begin{vmatrix}
-3& 2& 1 \\
2& -2& -2 \\
1& -2& -3 \\
\end{vmatrix}
\approx
\begin{vmatrix}
-3& 2& 1 \\
-1& 0& -1 \\
0& -2& -4 \\
\end{vmatrix}
\Rightarrow
v_2 =
\begin{bmatrix}
1\\
2\\
-1\\
\end{bmatrix}
$$
$\lambda_3 = -2$
$$
(A-\lambda I)=
\begin{vmatrix}
-1& 2& 1 \\
2& 0& -2 \\
1& -2& -1 \\
\end{vmatrix}
\Rightarrow
v_3 =
\begin{bmatrix}
-1\\
1\\
1\\
\end{bmatrix}
$$
$$
D=
\begin{bmatrix}
2 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & -2\\
\end{bmatrix}
P= \frac{1}{\sqrt{6}}
\begin{bmatrix}
\sqrt{3} & 1& -\sqrt{2}\\
0 & 2 & \sqrt{2}\\
\sqrt{3} & -1 & \sqrt{2}\\
\end{bmatrix}
$$
Where shoud i see what is this function and its rotation?
| The short version is that you correctly found $P^T A P = D,$ with $P^T P = P P^T = I.$ If your new triple of coordinates is called $u,v,w,$ you now have $2 u^2 + 4 v^2 - 2 w^2 = 1,$ which is a hyperboloid of one sheet.
| {
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"url": "https://math.stackexchange.com/questions/822286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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compute taylor series about $x=0$ of $\arctan(e^x -1 )$ hello I am having some issue and need a little guidance with this taylor expansion
$$f(x)=arctan(e^x -1)$$
the terms i should get are $x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{11 x^4}{24}-\frac{5 x^5}{24}$ but I am having some trouble with the expansion
should I use the definition of $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ and truncate the first two terms $ 1 + x$ or should do three terms instead $1 + x +\frac{x^2}{2}$ after which taking the derivative of $$arctan(x)= \frac{1}{1 + x^2}*\frac{dy}{dx} $$ and then applying the the integration of the geometric series where by $$\int\frac{1}{1-x}= \int\sum_{k=0}^\infty {x^k}$$
or should i approach the question differently all together?
| All roads lead to rome! Let's proceed by your first idea. We have
$$\frac1{1+u^2}=1-u^2+u^4+O(u^6)$$
hence using that $\arctan(0)=0$ we get
$$\arctan u=u-\frac{u^3}{3}+\frac{u^5}{5}+O(u^6)$$
moreover we have
$$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+O(x^6)$$
now we replace $u$ by $x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}$ and we only retain the terms with degree less or equal $5$ we get the desired result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$ Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$?
The correct answer should be 4.
| $\tan9^{\circ} + \tan81^{\circ} = \dfrac{\sin9^{\circ}}{\cos9^{\circ}} + \dfrac{\sin81^{\circ}}{\cos81^{\circ}} = \dfrac{\sin(9^{\circ} + 81^{\circ})}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\sin9^{\circ}} = \dfrac{2}{\sin18^{\circ}}$. Similarly:
$\tan27^{\circ} + \tan63^{\circ} = \dfrac{1}{\cos27^{\circ}\cos63^{\circ}} = \dfrac{1}{\cos27^{\circ}\sin27^{\circ}} = \dfrac{2}{\sin54^{\circ}}$.
Now let $x = \sin18^{\circ}$, then from $\cos36^{\circ} = \sin54^{\circ}$ we have:
$1 - 2x^2 = 3x - 4x^3 \to 4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, since $x \neq 1$, $x = \dfrac{\sqrt{5} - 1}{4} = \sin18^{\circ}$, and $\sin54^{\circ} = 1 - 2x^2 = 1 - 2\left(\dfrac{\sqrt{5} - 1}{4}\right)^2 = \dfrac{\sqrt{5} + 1}{4}$. Thus:
$S = \dfrac{2}{\dfrac{\sqrt{5} - 1}{4}} -\dfrac{2}{\dfrac{\sqrt{5} + 1}{4}} = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving an expression is $4$
Is $\displaystyle\sqrt[\huge3]{\frac{1}{2} \left(56-\sqrt{\frac{84640}{27}}\right)}+\sqrt[\huge 3]{\frac{1}{2} \left(\sqrt{\frac{84640}{27}}+56\right)}=4$ true ?
This was asked during an oral examination where calc and CAS are forbidden.
Mathematica seems to say it's true.
Can someone find a nice proof ?
| What are these guys sadists ? How is one supposed to answer this in an oral exam ?
Anyway here is another solution, we recognise the expression as the form of Cardano's solution to the cubic $x^3+px+q=0$
$$\frac{1}{3}\left( \sqrt[3]{-\frac{27}{2}q +
\frac{3\sqrt{-3D}}{2}}+
\sqrt[3]{-\frac{27}{2}q -
\frac{3\sqrt{-3D}}{2}} \right)$$ where $D=-27q^2-4p^3$
if we multiply the $\frac{1}{3}$ through we have
$$ \sqrt[3]{\frac{1}{2}(-q +
\sqrt{ \frac{-3D}{27} })}+
\sqrt[3]{\frac{1}{2}(q - \sqrt{\frac{-D}{27}}} )$$
So we recognise immediately that $q=-56$ and $-D=84640$ so
$$27q^2+4p^3=84640$$ and this gives $p=-2$
So the cubic in question is $x^3-2x-56$ which has the root $4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Solving $x^2 - x - 1 > 0$ I am having problems understanding how to solve:
$ x^2 - x - 1 > 0 $.
Any help would be much appreciated.
| Hint
Completing the square, we obtain
$$
x^2-x-1
= \left(x-\frac{1}{2}\right)^2 - \frac{1}{4} - 1
= \left(x-\frac{1}{2}\right)^2 - \frac{5}{4}
=0.
$$
Then, rearrange terms and apply the square root to both side to get
$$
x - \frac{1}{2}
=\pm \frac{\sqrt{5}}{2}
\implies x=\frac {1\pm \sqrt{5}}{2}.
$$
Since the coefficient of $x^2$ is positive, then the expression is positive outside the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of the following Legendre Symbols and Use Gausslemma to compute each of the Legendre symbols below Find the value of the following Legendre Symbols:
a) $(\frac{18}{43})$
b) $(\frac{19}{23})$
Use Gauss' Lemma to compute each of the Legendre symbols below:
a) $(\frac{8}{11})$
b) $(\frac{6}{31})$
| $a)$ $$\left ( \frac{18}{43} \right )=\left ( \frac{2 \cdot 3^2}{43} \right )=\left ( \frac{2}{43} \right )\left ( \frac{3}{43} \right )^2=(-1)^({43^2-1}{8})=(-1)^{(\frac{42}{8} )\cdot (\frac{44}{8})}=[(-1)^6]^{\frac{11}{2}}=(-1)^{3 \cdot 11}=-1$$
$b)$ $$\displaystyle{\left ( \frac{19}{23} \right )=(-1)^{{9}\cdot{11}} \left ( \frac{23}{19} \right )=(-1)\left ( \frac{4}{19} \right )=(-1)\left ( \frac{2}{19} \right )^2=(-1) \cdot (-1)^{\frac{(19-1) \cdot (19+1)}{8}}= (-1) \cdot (-1)=(-1) \cdot (-1)^{45}=1}$$
The Gauss' Lemma is:
$$\text{Let prime } p>2 \text{ and } p \nmid a. \\If \ m \text{ from the numbers } \\
a \cdot 1, a \cdot 2, \dots , a \cdot \frac{p-1}{2} \text{ have a remainder,when they are divided by } p, \text{ that is between } \frac{p+1}{2} \text{ and } p-1, \text{ then } \left ( \frac{a}{p} \right )=(-1)^m $$
Can you apply it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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Oops, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$ Oops, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$
I almost slipped and just assumed $\sqrt{C}=C$ but when you take the square root of both sides, you are really ending up with $y=\sqrt{2x^2+C}$ And you can't just take the C out of the radical. Right?
| First, note that it's not true that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}, \; \forall a,b \in \mathbb{R}$, as, for example: $$\underbrace{\sqrt{1+1}}_{=\sqrt{2}} \neq \underbrace{\sqrt{1} + \sqrt{1}}_{=2}$$
hence, in general, $\sqrt{2x^2+C} \neq \sqrt{2x^2} + \sqrt{C}$, except for some specific values of $x$ and $C$.
There is also this common misconception that $\sqrt{x^2}=x,\; \forall x \in \mathbb{R}$. This equality is only true if $x \geq 0$ !
What we know is that
$$\sqrt{x^2}=| x |,\; \forall x \in \mathbb{R}.$$
So, when you take the square root of both sides of $y^2=2x^2+C$, you end up with:
$$|y|=\sqrt{2x^2+C}.$$
Thus, the answer to your question is: no, $y^2=2x^2+C$ is not the same as $y=\sqrt{2x^2}+C$ and it's not even the same as $y=\sqrt{2x^2+C}$.
When in doubt, you can always have a look at some plots: (I used Mathematica, but you could get similar pictures using WolframAlpha)
*
*Plot of $y^2=2x^2+C$, for $C=2$
*
*Plot of $|y|=\sqrt{2x^2+C}$, for $C=2$
*
*Plot of $y=\sqrt{2x^2+C}$, for $C=2$
*
*Plot of $y=\sqrt{2x^2}+C$, for $C=2$
| {
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Given an ellipse's center, focus and point, find its equation. Given an ellipse's center is $(2,1)$, focus is (2,4) and point is (3,-3), we have
Plug in center:
$\frac{(x-2)^2}{a^2}+\frac{(y-1)^2}{b^2} = 1$
Use focus:
$4^2=a^2-b^2$
$16=a^2-b^2$
Use point:
$\frac{(3-2)^2}{a^2}+\frac{(-3-1)^2}{b^2} = 1$
$\frac{1}{a^2}+\frac{16}{b^2} = 1$
$b^2+16a^2 = a^2b^2$
Is this right?
Combine equations:
$b^2+16a^2 = a^2b^2$
$16=a^2-b^2$
So, we have:
$0 = b^2 - b^2 - 16^2$
$b^2 = \frac{1}{2} (1+5 \sqrt{41})$
and
$a^2 = 16 + \frac{1}{2} (1+5 \sqrt{41})$
| This is wrong. The answer is $\frac{(x-2)^2}{9}+\frac{(y-1)^2}{18}=1$. Thanks for the WolframAlpha suggestion, @alexqwx . I never knew WA could do that.
| {
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Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
show that
$$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$
My attempt:
Since
$$\sqrt{4+\dfrac{i}{n^2}}-2=\dfrac{\dfrac{i}{n^2}}{\sqrt{4+\dfrac{i}{n^2}}+2}=\dfrac{1}{\sqrt{4n^2+i}+2}\dfrac{i}{n}$$
then I can't work it. Thank you.
| $\sqrt{4+\frac k{n^2}}-2=2(1+\frac k{8n^2}+o(\frac k{8n^2})-1)=\frac k{4n^2}(1+o(1))$
(use the formula $\sqrt{1+x}=1+\frac x2+\dots$)
| {
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Inequality problem of mathematics If $x,y,z$ are unequal positive quatities then prove that ,
$(1+x^3)(1+y^3)(1+z^3)>(1+xyz)^3$
| lemma: if $a,b,c,x,y,z,p,q,r\ge 0$ then we have
$$(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)\ge (axp+byq+czr)^3$$
poof: note and Use AM-GM inequality
\begin{align*}3&=\sum_{cyc}\dfrac{a^3}{a^3+b^3+c^3}+\sum_{cyc}\dfrac{x^3}{x^3+y^3+z^3}+\sum_{cyc}\dfrac{p^3}{p^3+q^3+r^3}\\
&\ge\sum_{cyc}\dfrac{3axp}{\sqrt[3]{(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)}}\\
\end{align*}
so
$$axp+byq+czr\le \sqrt[3]{(a^3+b^3+c^3)(x^3+y^3+z^3)(p^3+q^3+r^3)}$$
| {
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Derivative of a definite integral (exercise)
Let, $$G(x)= \int_{x}^{\frac{1}{x}}\left (s+\frac{1}{s}\right)^9 ds$$
Find the derivative of $G(x)$.
Here is my work:
$$\frac{d}{dx}G(x)= \frac{d}{dx}\int_{x}^{\frac{1}{x}}\left (s+\frac{1}{s}\right)^9 ds=$$
$$=-\frac{d}{dx}\int_{0}^{x}\left (s+\frac{1}{s}\right)^9 ds+\frac{d}{dx}\int_{0}^{\frac{1}{x}}\left (s+\frac{1}{s}\right)^9 ds=$$
Now let $u=\frac{1}{x}$, and so $du=-\frac{1}{x^2}dx$.
$$=-\frac{d}{dx}\int_{0}^{x}\left (s+\frac{1}{s}\right)^9 ds+\frac{d}{du}\int_{0}^{u}\left (s+\frac{1}{s}\right)^9 ds=$$
Now applying the chain rule for differenciation:
$$=-\left (x+\frac{1}{x}\right)^9+\left (u+\frac{1}{u}\right)^9 \cdot \frac{du}{dx}=-\left (x+\frac{1}{x}\right)^9+\left (\frac{1}{x}+x\right)^9 \cdot \frac{d}{dx}\frac{1}{x}=$$
$$=-\left (x+\frac{1}{x}\right)^9-\left (\frac{1}{x}+x\right)^9 \cdot \frac{1}{x^2}= G'(x)$$
Is this right? Thanks
| Let $f(s)=(s+\frac{1}{s})^{9}$ and let $F(s)$ be a primitive function of $f(s)$.
$G(x)=F(1/x)-F(x)$
$G'(x)=f(1/x)\left(-\frac{1}{x^{2}}\right)-f(x)=$
$=-(x+\frac{1}{x})^{9}\cdot (\frac{1}{x^{2}}+1)$
wich corresponds to your answer.
| {
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Find all positive integer solutions $(x,y,z)$ that satisfy $5^x \cdot 7^y +4= 3^z$? This is another contest math-problem.
The only problem that I cannot find the way to tackle this problem.
Can anybody try to provide the solution to solve this problem?
Thanks
| Since $x,y,z \in \mathbb{N}$ we get that $5|5^x7^y$, so we have: $3^z \equiv 4 \mod 5$. Since $ord_5(3) = 4$ and $3^2 \equiv 4 \pmod 5$ we get $\fbox{$z = 4n + 2$}$. Obviously $z$ is even so we have $\fbox{$z=2k$}$. Then we have:
$$5^x7^y = 3^{2k} - 4 = (3^k - 4)(3^k + 4)$$
Now let $d = (3^k - 4,3^k + 4)$, then $d$ divides their diference, so $d\mid 8$. But both numbers are obviously odd, hence $(3^k - 4,3^k + 4) = 1$. Since we have 2 different prime factors on the left side we have $4$ cases:
Case 1:
\begin{cases} 3^k - 4 = 5^x \\ 3^k + 4 = 7^y \end{cases}
Working wrt modulo 5 and 7 respectively we get $k=4n + 2$ and $k=6n + 1$, an obvious contradiction.
Case 2:
\begin{cases} 3^k + 4 = 5^x \\ 3^k - 4 = 7^y \end{cases}
Working modulo $3$ we get: $5^x \equiv 2^x \equiv 1 \pmod 3 \implies \fbox{$x=2m$}$
So we get: $3^k = (5^m - 2)(5^m + 2)$. As previously we get that the two factors on the right side are comprime. Since $5^m + 2 > 1$ we get that the only possibilty is:
\begin{cases} 5^m - 2 = 1 \\ 5^m + 2 = 3^k \end{cases}
For the first equation we have: $5^m = 3$ a contradiction.
Case 3:
\begin{cases} 3^k + 4 = 1 \\ 3^k - 4 = 5^x7^y \end{cases}
An obvious contradiction for the first equation.
Case 4:
\begin{cases} 3^k + 4 = 5^x7^y \\ 3^k - 4 = 1 \end{cases}
From the second equation we get $3^k = 5$ a contradiction, because $k \in \mathbb{N}$
Hence this equation doesn't have a solution in $\mathbb{N}$
| {
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What is the value of $c$ in this problem of sequence and series? $a,b,c$ is in A.P. and $a^2,b^2,c^2$ is in G.P.If $a<b<c$ & $a + b + c = \dfrac {3}{2}$ ,what is the value of $c$?[When I wanted to do it,I found $a + d = \dfrac {1}{2}$,$d$ being the common difference,say.Thus $b$ = $\dfrac{1}{2}$ . Now I get $a + c = 1$.As $a^2,b^2,c^2$ is in G.P. we get $b$ =$\sqrt{ac}$
& finally get $a = c$ But it will be a contradiction.]
| Note that the geometric progression is in the squares of $a,b,c$ so that you have $a^2c^2=b^4$ whence $ac=\pm b^2$
You have $a+c=1$ from the AP condition and $ac=\pm b^2=\pm\frac 14$ from the GP condition. With the positive sign we know that $a,c$ are roots of the quadratic $$x^2-x+\frac 14=\left(x-\frac 12\right)^2=0$$Which is the case you have considered. But with the negative sign:$$x^2-x-\frac 14=0=\left(x-\frac 12\right)^2-\frac 12$$ and $x=\cfrac {1\pm \sqrt 2}{2}$ giving the arithmetic progression $$\frac {1- \sqrt 2}2,\frac 12, \frac {1+ \sqrt 2}2$$ with common difference $\cfrac {\sqrt 2}2$ and the geometric progression $$\frac {3-2\sqrt 2}4, \frac 14, \frac {3+2\sqrt 2}4$$ with common ratio $3+2\sqrt 2$
| {
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$f(x) = ab^x$ where $a$ and $b$ are positive constants. $f(5) = 96, f(7) = 384$. What is the value of $a$? What I've figured out so far:
$$
\begin{align}
ab^5 &= 96\\
ab^7 &= 384\\
\frac{384}{96} &= 4\\
4(ab^5) &= ab^7\\
4a\cdot 4b^5 &= ab^7\\
3a\cdot 4b^5 &= b^7\\
3a\cdot 4 &= b^2\\
12a &= b^2
\end{align}
$$
?????
| Your error is in the line
$$3a \cdot 4b^5 = b^7$$
It appears you've subtracted $a$ from both sides, rather than dividing.
We have
$$\frac{384}{96} = \frac{ab^7}{ab^5} = b^2 \implies b^2 = 4$$
Do you see how to finish from here?
| {
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$2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$ How can i prove that $2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$.
I tried $log(2^x + 3^x - 4^x + 6^x - 9^x) = log (1296^x) = x log(1296)$ i don't know if im correct here i stuck some help please
| Note that
$$9^x-6^x+4^x-3^x-2^x+1=\frac{1}{2}\left(( 3^x-1)^2+(2^x-1)^2+(3^x-2^x)^2\right)>0$$
| {
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Proof that doesn't exists a rational $s$ such that $s^2 = 6$ Well, I solved it, and I would like to know if there is anything that can be corrected or improved here. I think that the proof ended up too long, and with too many letters. Surely there is a better way to write it. Alternate solutions are welcome too. Thank you.
Proof: Suppose by contradiction that exists $s \in \Bbb Q$ such that $s^2 = 6$. Then, we have $s = \dfrac{p}{q}$, with $p,q \in \Bbb Z, q \neq 0, \gcd(p,q) = 1$. The standard strategy is to contradict the part about the $\gcd$. We have: $$\begin{align} \left(\frac{p}{q}\right)^2 &= 6 \\ p^2 &= 6q^2 \\ p^2 &= 2(3q^2)\end{align}$$
so $p^2$ is even, and it follows that $p$ is even, and so exists $m \in \Bbb Z$, with $p = 2m$. Proceeding, we have: $$\begin{align} (2m)^2 &= 2(3q^2) \\ 4m^2 &= 2(3q^2) \\ 2m^2 &= 3q^2\end{align}$$
From here, we have that $3q^2$ is even. If $q$ is also even, $\gcd(p,q) \neq 1$ and we're finished. Let's see the case that $q$ is odd. Then exists $\ell \in \Bbb Z$, with $q = 2 \ell + 1$. Proceeding in this case: $$\begin{align} 2m^2 &= 3(2 \ell + 1)^2 \\ 2m^2 &= 3(4 \ell^2 + 4\ell + 1) \\ 2m^2 &= 12 \ell^2 + 12 \ell + 3 \\ 2m^2 &= 2(6 \ell^2 + 6 \ell + 1) + 1 \end{align}$$
which is a contradiction, because the left-hand side is even, and the right-hand side is odd. Therefore $q$ must be even, and $\gcd(p,q) \neq 1$. Hence, there is no rational number whose square is $6$.
| This is getting really old.
Copy and paste from
another answer of mine.
If $n$ is a positive integer that is
not a square of an integer,
then $\sqrt{n}$ is irrational.
Let $k$ be such that
$k^2 < n < (k+1)^2$.
Suppose $\sqrt{n}$ is rational.
Then there is a smallest positive integer $q$ such that
$\sqrt{n} = p/q$.
Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k}
= \frac{n-k\sqrt{n}}{\sqrt{n}-k}
= \frac{n-kp/q}{p/q-k}
= \frac{nq-kp}{p-kq}
$.
Since $k < \sqrt{n} < k+1$,
$k < p/q < k+1$,
or $kq < p < (k+1)q$,
so $0 < p-kq < q$.
We have thus found a representation of
$\sqrt{n}$ with a smaller denominator,
which contradicts the specification of $q$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$, then find the numerical value of $\frac{x}{y}$ If $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$, then find the numerical value of $\frac{x}{y}$
My try:
$2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$
$log_e{(x-2y)^2} = log_e{xy}$
$(x - 2y)^2 = xy$
But expanding the LHS doesn't give the result. What should I do further?
| If you are asked to calculate $\left(\frac {x}{y} \right) $ start with that and assume it to $t$.
Then, $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$ becomes
$2 \cdot log_e{(ty -2y)} = log_e{y} + log_e{ty}$
$\implies$${y^2}{{(t-2)}^2}=ty^2$
$\implies$${t^2}-4t+4=t$
$\implies$$t^2-5t+4=0$
$\implies$$(t-4)(t-1)=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Predicting eigenvalues of bigger matrices Consider the following $(3 \times 3)$ matrix:
$K_3 = \left( \begin{array}{ccc}
a & -1 & 0 \\
-1 & a+1 & -1 \\
0 & -1 & a \end{array} \right)$
The question has a quantum physics context, so we'll assume that $a$ is such that $K$ is hermitian. This matrix has eigenvalues $-1 + a$, $a$, and $a+1$. Now consider growing the matrix to a $(4\times4)$ in the following way:
$K_4 = \left( \begin{array}{ccc}
a & -1 & 0 & 0 \\
-1 & a+1 & -1 & 0 \\
0 & -1 & a+1 & -1 \\
0 & 0 & -1 & a \end{array} \right)$
Its eigenvalues are $-1 + a$, $1+a$, $1-\sqrt{2} + a$, and $1+\sqrt{2} + a$. There seems to be at least some structure in this. I was wondering whether there's a way to predict what the eigenvalues of increasingly bigger matrices of this form will be. Specifically, is it possible to find the eigenvalues of:
$K_n = \left( \begin{array}{ccccccc}
a & -1 & 0 & \dots & 0 & 0 & 0 \\
-1 & a+1 & -1 & \dots & 0 & 0 & 0 \\
0 & -1 & a+1 & \dots & 0 & 0 & 0 \\
0 & 0 & -1 & \dots & -1 & 0 & 0 \\
0 & 0 & 0 & \dots & a+1 & -1 & 0 \\
0 & 0 & 0 & \dots & -1 & a+1 & -1 \\
0 & 0 & 0 & \dots & 0 & -1 & a \end{array} \right)$
I can imagine the expressions become rather monstrous if it's at all possible, so taking something like $a = 1$ is fine for my purposes, if it simplifies matters.
With the help of mookid I managed to find a general form for the characteristic polynomials of $K - Ia$. Defining $P_n$ as the characteristic polynomial of the $(n \times n)$ matrix $K - Ia$, we have:
$P_n = -\lambda Q_{n-1} - Q_{n-2}$
Where $\lambda$ are the requested eigenvalues and $Q_n$ are the characteristic polynomials of the same matrix, where the zero on the $nn$-th entry has been replaced by a one:
$Q_n = (1-\lambda) Q_{n-1} - Q_{n-2}$
With boundary conditions $Q_1 = -\lambda$ and $Q_2 = \lambda^2 - \lambda - 1$.
Quoting wikipedia, with regards to invertibility of tridiagonal matrices: "Closed form solutions can be computed for special cases such as symmetric matrices with all off-diagonal elements equal". I'm hoping that means there also exists an analytic solution for the roots of these polynomials (and hence the eigenvalues for arbitrary $n$). Does anyone know if this is possible, and/or how to go about finding the solution?
| You just have to go for the $a=0$ case; in general, the eigenvalues are
$a+\lambda, \lambda$ eigenvalue of $\left( \begin{array}{ccccccc}
0 & -1 & 0 & \dots & 0 & 0 & 0 \\
-1 & 1 & -1 & \dots & 0 & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 & 0 \\
0 & 0 & -1 & \dots & -1 & 0 & 0 \\
0 & 0 & 0 & \dots & 1 & -1 & 0 \\
0 & 0 & 0 & \dots & -1 & 1 & -1 \\
0 & 0 & 0 & \dots & 0 & -1 & 0 \end{array} \right)$
Now refering to wikipedia, the sequences of characteristic polynomials is
$$P_2 = X^2 - 1, P_3 = -X^3+X
\\
P_{n+1} = X\times P_n - (-1)\times(-1)\times P_{n-1}
=-XP_n - P_{n-1}$$
| {
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Derivative of $(5x-2)^3$ How is the derivative of $(5x-2)^3$ equal to $15(5x-2)^2$ and not $3(5x-2)^2$. According to $\frac{df}{dx} = nx^{n-1}$, it has to be $3(5x-2)^2$ right. Please explain.
| You simply forgot to remember using the chain rule:
\begin{align}
\frac{d}{dx}(5x-2)^3 = 3(5x-2)^2 \frac{d}{dx}(5x-2) =3(5x-2)^2 \cdot 5 = \boxed{15(5x-2)^2}
\end{align}
| {
"language": "en",
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Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$ I am trying to solve this:
$$\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh x+\frac{ix}{2}\right)\,dx$$
I don't have much ideas about the problem. I thought of writing $\cos x=\dfrac{e^{ix}+e^{-ix}}{2}$ but couldn't proceed after that.
$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{ -(ab\cosh x-iac \sinh x)-\frac{x}{2}}+e^{ -(ab\cosh x+iac \sinh x)+\frac{x}{2}}\right)\,dx$$
$\displaystyle
\begin{aligned}
ab\cosh x-iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)-\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\
&=a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)\\
\end{aligned}$
Similarly,
$\displaystyle
\begin{aligned}
ab\cosh x+iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)+\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\
&=a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)\\
\end{aligned}$
where $\alpha=\arcsin\left(\dfrac{b}{\sqrt{b^2+c^2}}\right)$
So the integral can be simplified to:
$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{-a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)-\frac{x}{2}}+e^{-a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)+\frac{x}{2}}\right)\,dx$$
A solution without using contour integration is appreciated. Thanks!
| *
*Using parity properties, the integral can be rewritten as
$$I=\frac12\int_{-\infty}^{\infty}e^{-ab\cosh x+iac\sinh x-\frac{x}{2}}dx$$
*Note that $b\cosh x-ic\sinh x=\sqrt{b^2+c^2}\cosh\left(x-i\phi\right)$ with
$$\cos\phi=\frac{b}{\sqrt{b^2+c^2}},\qquad\sin\phi=\frac{c}{\sqrt{b^2+c^2}}.$$ Hence, shifting $x$ by $i\phi$ [for that we need to assume that $\phi\in(-\pi/2,\pi/2)$], we obtain
$$I=\frac{e^{-i\phi/2}}{2}\int_{-\infty}^{\infty}e^{-a\sqrt{b^2+c^2}\cosh x-\frac{x}{2}}dx$$
*The last integral can be expressed in terms of the Macdonald function $K_{\nu}(r)$, which has integral representation
$$K_{\nu}(r)=\frac12\int_{-\infty}^{\infty}e^{-r\cosh x\pm\nu x}dx.$$
In addition, for $\nu=\frac12$ this function reduces to an elementary one:
$\displaystyle K_{1/2}(x)=\sqrt{\frac{\pi}{2x}}e^{-x}$.
Therefore the final result is
$$\boxed{\displaystyle I=
\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}\;\exp\left\{-a\sqrt{b^2+c^2}-\frac{i}2\arctan\frac{c}{b}\right\}}$$
| {
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Solving a recurrence relation ${}$ I feel I'm wasting my time trying to solve this
$a_0$ is given
$\displaystyle a_{n+1}=\frac{n-1}{n+2}(a_n-n-2)$
Mathematica found a closed form but there's a problem when evaluating for $n=0$
$\displaystyle-\frac{30 C-n^5+5 n^3-4 n}{5 n-5 n^3}$
Context: https://puzzling.stackexchange.com/questions/1820/paying-the-unfair-troll-toll
| As correctly stated in the comments, we can start from the fact that, by definition, for $n=0$ the answer is $a_0$.
The recurrence relation can be rewritten to express $a_n$ as a function of $a_{n-1}$. Substituting $n$ with $n-1$ we get$$\displaystyle a_{n}=\frac{n-2}{n+1}(a_{n-1}-n-1)$$
For $n=1$, the formula gives $a_1=-\frac{1}{2}(a_0-1-1)=1-a_0/2$, whereas for $n=2$ the formula gives $a_2=\frac{0}{3}[(1-a_0/2)-2-1]=0$. This means that, for the successive terms where $n\geq3$, the term $a_0$ is canceled out and will not longer appear in the results.
Now let us study the results of the recurrence formula for $n\geq3$. In doing this, for each $n$, we can write the result separately writing the fraction and the quantity in brackets. Also, we can express the quantity in brackets as fractions with equal denominator.
For $n=3$, the formula gives $a_3=\frac{1}{4}(-3-1)=\frac{1}{4}(-4)=\frac{1}{4}(-20/5)$. The result is $-1$.
For $n=4$, the formula gives $a_4=\frac{2}{5}(-1-4-1)=\frac{2}{5}(-6)=\frac{2}{5}(-30/5)$. The result is $-12/5$.
For $n=5$, the formula gives $a_5=\frac{3}{6}(-12/5-5-1)=\frac{3}{6}(-42/5)$. The result is $-21/5$.
For $n=6$, the formula gives $a_6=\frac{4}{7}(-21/5-6-1)=\frac{4}{7}(-56/5)$. The result is $-32/5$.
Continuing in this way, it is not difficult to show that the numerators of the quantities in brackets ($20, 30, 42, 56...$) represent the progression given by $(n+1)(n+2)$. Hence, the general formula can be rewritten as$$\displaystyle a_{n}=-\frac{1}{5}\frac{(n-2)}{(n+1)}(n+1)(n+2)=-\frac{1}{5}(n-2)(n+2)=-\frac{1}{5}(n^2-4)$$
which gives a closed formula for the calculation of $a_n$. As a counterproof:
$a_3=-\frac{1}{5}(3^2-4)=-1$
$a_4=-\frac{1}{5}(4^2-4)=-12/5$
$a_5=-\frac{1}{5}(5^2-4)=-21/5$
$a_6=-\frac{1}{5}(6^2-4)=-32/5$
and so on.
Lastly, note that the formula given by Mathemathica and reported in the question may be correct if we neglect the term $30C$. In fact: $$\displaystyle-\frac{-n^5+5 n^3-4 n}{5 n-5 n^3}=-\frac{-n^4+5 n^2-4}{5-5 n^2}=-\frac{1}{5}\frac{n^4-5 n^2+4}{n^2-1}=-\frac{1}{5}\frac{(n^2-4)(n^2-1)}{n^2-1}=-\frac{1}{5}(n^2-4)$$
| {
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How to evaluate $\int \frac{x^6}{x^4-1} \, \mathrm{d}x.$ Evaluate the integral:
$$\int \frac{x^6}{x^4-1} \, \mathrm{d}x$$
After a lot of help I have reached this point:
$x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D$
But now I don't really know how to solve for $A, B, C$, and $D$. Please help!
| Rewrite the integrand as
\begin{align}
\frac{x^6}{x^4-1}&\stackrel{\color{red}{[1]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{x^2}{x^4-1}}\\
&=\color{darkgreen}{x^2}+\color{blue}{\frac{x^2}{(x^2-1)(x^2+1)}}\\
&\stackrel{\color{red}{[2]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac{1}{x^2-1}}\right)}\\
&=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac{1}{(x-1)(x+1)}}\right)}\\
&\stackrel{\color{red}{[3]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac12\left[\frac1{x-1}-\frac1{x+1}\right]}\right)}\\
&=x^2+\color{red}{\underbrace{\color{black}{\frac{1}{2(x^2+1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ x=\tan\theta}}}+\color{red}{\underbrace{\color{black}{\frac1{4(x-1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ u=x-1}}}-\color{red}{\underbrace{\color{black}{\frac1{4(x+1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ v=x+1}}}
\end{align}
Notes :
$\color{red}{[1]}\;\;\;$Polynomial long division
$\color{red}{[2]}\text{ and }\color{red}{[3]}\;\;\;$Partial fractions decomposition
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute radius of convergence Someone can help me with this radius of convergence please?
$$
\sum_{n=0}^{\infty}{\frac{3^{2}(5n^{7}+2)}{2^{n}(5n^{3}-1)}}x^{n}
$$
I tried
$$
r = \lim_{n \to \infty } \frac{3^2(5n^7+2)(2^{n+1})(5(n+1)^{3}-1)(x^{n})}{2^{n}(5n^3-1)3^2(5(n+1)^{7}+2)(x^{n+1})}
$$
After to operate, I can't get to answer
| If you want to use the ratio test I recommend writing the polynomial parts this way:
$${\frac{5n^7+2}{5n^3-1}}=n^4\frac{5+\frac{2}{n^7}}{5-\frac{1}{n^3}}$$
because then you have
$$\left({\frac{5n^7+2}{5n^3-1}}\right)\bigg/
\left({\frac{5(n+1)^7+2}{5(n+1)^3-1}}\right)
=\left(\frac{n}{n+1}\right)^4
\left(\frac{5+\frac{2}{n^7}}{5-\frac{1}{n^3}}\right)\bigg/
\left(\frac{5+\frac{2}{(n+1)^7}}{5-\frac{1}{(n+1)^3}}\right)$$
and all the limits you need are very easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Algebraic Symbol Manipulation While Finding the Sum of a Series In a precalc text, in the chapter on geometric progressions and series, we're told of the formula
$S=\frac{a(1-r^{n+1})}{1-r}$
and that:
$S=\frac{a}{1-r}$ is valid for $|r|<1$
We're then asked to find the sum of $3+\sqrt{3}+1+...$
Literally that's all the information given, including the "...".
The text gives the answer as: $\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$ but I'm struggling to derive the calculations that get us there.
I set $a=3$ and $r = \frac{\sqrt{3}}{3}$ which, by the 2nd formula above, gives me $\frac{3}{1-\frac{\sqrt{3}}{3}}$
I clear the radical sign in the denominator by multiplying by the conjugate:
$\frac{3}{1-\frac{\sqrt{3}}{3}}\cdot\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}$ which gives me $\frac{3+\frac{3\sqrt{3}}{3}}{\frac{2}{3}}\Rightarrow3+\frac{3\sqrt{3}}{3}\cdot\frac{3}{2}\Rightarrow...?$
It's at this point the head scratching begins.
Hints?
| First thing is you should have cancelled $3$ in the numerator and denominator, so $\frac{3\sqrt3}3$ becomes $\sqrt3$. So
$$\dfrac{3+\frac{3\sqrt3}3}{\frac23}=\frac32(3+\sqrt3)=\frac32(\sqrt3)(\sqrt3+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.