Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Can anyone explain to me this square root? step by step? $$\begin{align}
v(p_1, p_2, w)
& = \sqrt{\frac w{p_1^2\left(\frac1{p_1}+\frac1{p_2}\right)}}
+ \sqrt{\frac w{p_2^2\left(\frac1{p_1}+\frac1{p_2}\right)}}
\\
& = \sqrt{\frac w{\left(\frac1{p_1}+\frac1{p_2}\right)}}
\left(\sqrt{\frac1{p_1^2}}+\sqrt{\frac1{p_2^2}}\right)
\\
& = \sqrt{\frac w{p_1}+\frac w{p_2}}
\end{align}$$
So as the title says, can anyone explain to me this square root?
(Original scan)
| Step by step:
Rewrite the original expression as follows:
$$\sqrt{\frac{w}{p_{1}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} + \sqrt{\frac{w}{p_{2}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Factoring, this becomes
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Note that
$$\sqrt{\frac{1}{p_{1}^{2}}} = \frac{1}{p_{1}}$$
and
$$\sqrt{\frac{1}{p_{2}^{2}}} = \frac{1}{p_{2}}$$
So
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)$$
This can be rewritten as
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right) = \sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}$$
And finally, we can simplify this to
$$\sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}} = \sqrt{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)} = \sqrt{\frac{w}{p_{1}} + \frac{w}{p_{2}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the simplest proof of the pythagorean theorem you know? Maybe enough so to explain it to children.
| Consider the right figure and ignore the left one!
The area of the square of side $a+b$ is
\begin{align*}
(a+b)(a+b)
&= a^2 + ab + ab + b^2\\
&= a^2 + 2ab + b^2
\end{align*}
It is equal to the sum of 4 triangles and 1 square. This sum is
\begin{gather*}
4\times \frac{ab}{2} + c^2\\
2ab+ c^2
\end{gather*}
By equating both, we have
\begin{align*}
a^2 + 2ab + b^2
&= 2ab +c^2\\
a^2+b^2=c^2
\end{align*}
Now consider a single triangle so you have proven that $a^2+b^2=c^2$. Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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"answer_id": 5
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How prove $A=B=C$? in $\Delta ABC$,
such
$$\sin{A}+\cos{B}+\tan{C}=\dfrac{3\sqrt{3}+1}{2}$$
prove that
$$A=B=C=\dfrac{\pi}{3}$$
My try:
use
$$\sin{x}+\sin{y}=2\sin{\dfrac{x+y}{2}}\cos{\dfrac{x-y}{2}}$$
then
\begin{align*}&\sin{A}+\cos{B}\\
&=\sin{A}+\sin{(\dfrac{\pi}{2}-B)}=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{A+B-\dfrac{\pi}{2}}{2}}\\
&=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{\pi-2C}{4}}
\end{align*}
my idea is take $\sin{A}+\cos{B}\le f(C)$?and if only if $A=B$,BUt I can't ,Thank you someone can help me
| Fix $A \in (0, \frac{\pi}{2}), A \neq \frac{\pi}{3}$. Let $f: [0, \frac{\pi}{2}) \to \mathbb R$ be
$$f(x)= \sin(A) +\tan(x)+\cos(\pi-A-x) \,.$$
As $f$ is continuous, $f(0)=\sin(A)+\cos(\pi-A) \leq 2 < \dfrac{3\sqrt{3}+1}{2} $ and $\lim_{x \to \frac{\pi}{2}^-} f(x)= +\infty$, by the Intermediate Value Theorem there exists some $x$ so that
$$f(x)=\dfrac{3\sqrt{3}+1}{2} \,.$$
Then $A=A, B= \pi-A -x, C=x$ is a counterexample to your problem. So you cannot.
P.S. We actually prove something stronger: the angle $A$ can take any value in $(0, \frac{\pi}{2})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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In how many ways can $7^{13}$ be represented as product of $3$ natural numbers? How i solved it:
all possible non-distinct groups $(a,b,c)$ are,
$a = 0 \Rightarrow (b,c) = (0,13)(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$
$a = 1 \Rightarrow (b,c) = (1,11)(2,10)(3,9)(4,8)(5,7)(6,6)$
$a = 2 \Rightarrow (b,c) = (2,9)(3,8)(4,7)(5,6)$
$a = 3 \Rightarrow (b,c) = (3,7)(4,6)(5,5)$
$a = 4 \Rightarrow (b,c) = (4,5)$
Thus, $7^{13}$ can be written as product of $3$ natural numbers in $7+6+4+3+1 = 21$ ways.
Though this gives the required solution, it takes time and is a lengthy way. Is there an alternate method to tackle such problems? (May be by using the "bars and stars" method with some adjustments? I tried but failed to get the correct answer that way.) Please help me out and share your method!
Thanks a lot!
| What you are doing, essentially, is partitioning the exponent $13$, and this amounts to solving the equation $a + b + c = 13$, where $(a, b, c)$ is a 3-tuple of non-negative integers.
Hence, the classic Stars-and-Bars Problem comes into play:
Using the fact that there are $$\binom{n + k - 1}{k}$$ distinct n-tuples of non-negative integers whose sum is $k$, we calculate the number of distinct 3-tuples of non-negative integers whose sum is $13$. In the case at hand, the number of distinct 3-tuples given by $(a, b, c)$ such that $7^a\times 7^b \times 7^c = 7^{a + b + c} = 7^{13}$ is given by $$\binom{3 + 13 - 1}{13} = \binom{15}{13} = \binom{15}{2} = \frac{15!}{13!2!}= 105$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
| Assume that $x<0$ and $y>0$, your statement can be written (even if we change $x$ to $-x$),
If $x,y$ are strictly positive such that $(\sqrt{y^2+x}+x)(\sqrt{x^2+y}-y)=y$ then $x=y$.
This equality becomes, $$\overbrace{\dfrac{x+\sqrt{x+y^2}}{y+\sqrt{y+x^2}}}^{A}=\overbrace{\dfrac{y}{x^2+y-y^2}}^{B}.$$ Let us start with $x>y$, clearly $B<1$. Moreover $x\mapsto(x-\sqrt{y+x^2})$ is increasing and $x\mapsto \sqrt{y+x^2}$ strictly increasing, we conclude that $x\mapsto(x+\sqrt{x+y^2})-(y+\sqrt{y+x^2})$ is strictly increasing and thus $A>0$ when $x>y$. We deduces that $A>1$ and therefore $A$ cannot be equal to $B$.
The case $x<y$ can be treated in the same way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can we represent a symmetric curve by a parameter with symmetry?
Question : Can we represent the following curve $C$ by one parameter $t$ as $x=f(t),y=g(t),z=h(t)$ with symmetry?
The curve $C$ in the $xyz$ space is defined as
$$\begin{cases}
x^2+y^2+z^2=1 \\
x^3+y^3+z^3=0 \\
\end{cases}$$
where $x,y,z\in\mathbb R.$
Motivation : I've already got the following example without symmetry :
For $0\le t\lt 2\pi$,
$$f(t)=\frac{\cos t}{i(t)}, g(t)=\frac{\sin t}{i(t)}, h(t)=\frac{\sqrt[3]{\cos^3t+\sin^3t}}{i(t)}\ \text{where $i(t)=\sqrt{1+(\cos^3 t+\sin^3 t)^{2/3}}$}.$$
Since the given equations are symmetrical, I expect that there would exist an example with symmetry. Then, I reached the above question, but I'm facing difficutly. Can anyone help?
| I've just been able to get an example.
First, note that
$$\sin t+\sin\left(t+\frac{2\pi}{3}\right)+\sin\left(t+\frac{4\pi}{3}\right)=0.$$
For any $r\in\mathbb R, 0\le t\lt 2\pi$, letting
$$x=r\sqrt[3]{\sin t}, y=r\sqrt[3]{\sin\left(t+\frac{2\pi}{3}\right)},z=r\sqrt[3]{\sin\left(t+\frac{4\pi}{3}\right)},$$
we know that these satisfy $x^3+y^3+z^3=0.$ Also we know that every point on $x^3+y^3+z^3=0$ can be represented by $r$ and $t$ in this way. Substituting these for $x^2+y^2+z^2=1$, we get
$$x=\frac{\sqrt[3]{\sin t}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}},$$
$$y=\frac{\sqrt[3]{\sin\left(t+\frac{2\pi}{3}\right)}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}},$$
$$z=\frac{\sqrt[3]{\sin\left(t+\frac{4\pi}{3}\right)}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}}$$
as desired.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving the minimum value of (x+a)(x+b)/(x+c) Show that the minimum value of $\frac {(x+a)(x+b)}{(x+c)}$, where a$\gt$c, b$\gt$c, is $(\sqrt{a-c}+\sqrt{b-c})^{2}$ for real values of x$\gt-c$.
I did $$\frac {(x+a)(x+b)}{(x+c)}=y$$ and then took its discriminant greater than zero.
This led me to $$y^2-2(a+b-2c)y+(a-b)^2\gt0$$
I also tried differentiating the expression as follows.
$$y'= \frac {(x+c)[2x+(a+b)]-[x^2+(a+b)x+ab]}{(x+c)^2}=0$$
$$\therefore x^2+2cx+(a+b)c-ab=0$$
I am unable to proceed after this. Please help.
| I'm going to avoid using calculus to solve this question. Instead, I have a different approach which uses simple algebra.
Let's call this given expression $z$.
Put $y = c + x$
$$⇒z = \frac {(y-c+a)(y-c+b)}{y}$$
Simplifying this by opening the brackets we get,
$$⇒z = \frac {(a-c)(b-c) + y(a-c) + y(b-c) + y^2}{y}$$
$$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y$$
Now, here we do a bit of manipulation. We add and subtract $2 \sqrt{(a-c) (b-c)}$. Why? Keep looking.
$$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y + 2 \sqrt{(a-c) (b-c)} - 2 \sqrt{(a-c) (b-c)}$$
Here, we club $\frac {(a-c)(b-c)}{y}, y$ and $2 \sqrt{(a-c) (b-c)}$ to get a squared term. That's why I added and subtacted $2 \sqrt{(a-c) (b-c)}$.
$$⇒z = \frac {(a-c)(b-c)}{y} + y + 2 \sqrt{(a-c) (b-c)} + (a-c) + (b-c) - 2 \sqrt{(a-c) (b-c)}$$
$$⇒ z = (\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2+ (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$
Now, since $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 ≥ 0$$
This is because it is a squared term. It can never be negative, it is only either positive or at the minimum, zero.
So,the expression $z$ is minimum only when $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 = 0$$
Hence, the minimum value of $z$ is, i.e, z reduces to
$$⇒ z = (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$
Which upon simplification is,
$$⇒ z = (\sqrt{a-c} + \sqrt{b-c})^2$$
Which is the required minimum value.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$
My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$
So
$$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$
Though this method might solve, I think this problem has nicer methods. Thanks.
| The interesting thing (which I cannot explain) is that if you have two odd functions
$$f(x)=x+a_3x^3+a_5 x^5+a_7 x^7+?x^9,\quad g(x)=x+b_3x^3+b_5 x^5+b_7 x^7+?x^9$$
with $f'(0)=g'(0)=1$ (or $=-1$) then they "commute up to order 5", i.e.,
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)= ?x^7\qquad(x\to0)\ .$$
To prove this we do the computation for $f\bigl(g(x)\bigr)$:
$$\eqalign{f\bigl(g(x)\bigr)&=x+(a_3+b_3)x^3+(a_5+3a_3b_3+b_5)x^5 + \cr &\qquad\qquad(a_7+5a_5b_3+3a_3b_3^2+3a_3b_5+b_7)x^7\ +\ ?x^9\ .\cr}\tag{1}$$
We see that the coefficients of $x^3$ and $x^5$ both are symmetric in $a$ and $b$, and in addition $a_7+b_7$ will cancel when forming $f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)$. From inspection of $(1)$ we therefore can deduce that
$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)=\bigl(2(a_5 b_3-a_3 b_5)+3(a_3b_3^2-a_3^2 b_3)\bigr)x^7\ +\ ?x^9\ .$$
Inserting here the known coefficients for $\sin$ and $\tan$ we find that the limit in question is $-{1\over30}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the Integral: $\int_0^\infty \frac{( \frac{1}{2} - \cos x )}{x} dx$ Evaluating the Integral:
$\int_0^\infty\left[\frac{1}{2} - \cos\left(x\right)\right]\,{\rm dx \over x}$
I came upon this limit: $\lim_{x\rightarrow\infty} -Ci(x) + Ci(1/x) +\ln(x)$, is it $\gamma$ ?
Here $ Ci(x) = \gamma + \ln x + \int_0^x \frac{\cos t -1}{t} dt $ is the cosine integral and $\gamma$ is the Euler constant. The Limit and the Integral appear to be equal.
| $$\int_0^{+\infty}\left(\frac 1 2 - \cos x\right)\frac {dx} x = \gamma$$
\begin{align*}\int_{1/n}^n \left(\frac 1 2 - \cos x\right)\frac {dx} x &=
\int_{1/n}^1 \left(1 - \cos x\right)\frac {dx} x
-\int_1^n \cos x\frac {dx} x+
\frac 1 2 \left(\int_1^n \frac {dx} x -\int_{1/n}^1 \frac {dx} x\right) \\
&=\int_{1/n}^1 (1-\cos x) \frac {dx} x - \int_1^n \cos x \frac {dx} x
\end{align*}
$$\bbox[0.2ex,border:0.5pt solid black]{\int_0^1 \frac {1-\cos(y)}y\,dy
-\int_1^{+\infty} \frac {\cos(y)} y\,dy=\gamma}$$
Here is the proof of Gronwall, 1918. For $n$ a positive integer
$$\begin{align}
A(n)
&= \int_0^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\
&= \int_0^1 \frac{1-\cos (y)}{y} dy + \int_1^{n\pi} \frac{1-\cos (y)}{y} dy - \ln ( n\pi ) \\
&= \int_0^1 \frac{1-\cos(y)}{t} dy - \int_1^{n\pi} \frac{\cos(y)}{y} dy
\end{align}$$
With a change of variable $x = 2 \pi ny$:
$$\begin{align}
\int_0^{n\pi} \frac{1-\cos(x)}{x} xy
&= \int_0^{\frac{1}{2}} \frac{1 - \cos (2\pi ny)}{y} dy \\
&= \pi \int_0^{\frac{1}{2}} \frac{1 - \cos (2 \pi ny)}{\sin(\pi y)} dy + \int_0^{\frac{1}{2}} g(y) dy - \int_0^{\frac{1}{2}} g(y) \cos ( 2 \pi ny) dy
\end{align}$$
$g(y) = \frac{1}{y} - \frac{\pi}{\sin (\pi y)}$ which is continuous on $[0,\frac{1}{2}]$ with $g(0) = 0$.
With Riemann-Lebesgue theorem
$$\int_0^\frac{1}{2} g(y) \cos ( 2 \pi n y ) dy \rightarrow 0 \text{ for } n \rightarrow + \infty$$
$$\begin{align}
\int_0^\frac12 g(y) dy
&= \lim_{s \to 0^+} \int_s^\frac12 \left( \frac 1y - \frac \pi{\sin(\pi y)} \right) dy \\
&= \lim_{s \to 0^+} \left[ \ln \left( \frac{y}{\tan ( \pi/2\;y )} \right) \right]_s^{\frac{1}{2}} = \ln ( \pi ) - 2 \ln (2)
\end{align}$$
On the other hand
$$\begin{align}
\pi \int_{0}^{\frac{1}{2}} \frac{1 - \cos ( 2 \pi n y)}{\sin ( \pi y )} dy
&= 2\pi \int_{0}^{\frac{1}{2}} \sum_{k=1}^{n} \sin ( (2k-1) \pi y ) dy \\
&= \sum_{k=1}^{n} \frac{2}{2k-1}
={\ln (n) + 2 \ln (2) + \gamma + o(1)}
\end{align}$$
Finally
$$A(n) = \gamma + o(1)$$
Thanks very much to Sladjan Stankovik for providing insight into this problem for me.
| {
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Throwing dice: Probability Total is Divisible by 3
Suppose I throw a fair die $1995$ times. What is the probability that
the total is divisible by 3?
I tried to attack this problem inductively, storing the total in a variable $t \mod 6$, and then showing that for any total $t$ there is a $1/3$ probability of the next throw of the dice creating a total $t+n$ divisible by three. However, I've had difficulty formalizing the proof and I'm wondering if there isn't a better way to do this.
| Let $X_n$ be the $n$th number rolled by a fair die (modulo $3$) and let $S_n = X_1 + \ldots + X_n \pmod 3$ be the sum of the first $n$ numbers modulo $3$. We will prove by induction on $n \in \mathbb N$ that for any $r \in \{0,1,2\}$, we have that $P(S_n = r) = 1/3$. The base case is clear; we will focus on the induction step.
Induction Hypothesis: Assume that the claim holds true for $n = k$.
It remains to prove that the claim holds true for $n=k+1$. Indeed, for any $r \in \{0,1,2\}$, observe that:
\begin{align*}
&P(S_{k+1} = r) \\
&= P(S_k = 0)P(X_{k+1} = r) + P(S_k = 1)P(X_{k+1} = r-1) + P(S_k = 2)P(X_{k+1} = r-2) \\
&= P(S_k = 0)\cdot\frac{1}{3} + P(S_k = 1)\cdot\frac{1}{3} + P(S_k = 2)\cdot\frac{1}{3} \\
&= \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} \qquad\text{by the induction hypothesis}\\
&= 3\left(\frac{1}{3^2}\right)\\
&= \frac{1}{3}\\
\end{align*}
as desired. Note that $r-1$ and $r-2$ are taken modulo $3$.
| {
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} |
How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$ let $a,b,c\ge 0$,and such $abc=1$,show that
$$a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$$
My solution: Without loss of generality,assume that
$a=\max{(a,b,c)}$, since $abc=1$,we have
$a\ge 1$,
we will show that
$$f(a,b,c)\ge f(a,t,t)\ge 0, t=\sqrt{bc},0<t\le 1$$
since $$f(a,b,c)-f(a,t,t)=(\sqrt{b}-\sqrt{c})^2[(\sqrt{b}+\sqrt{c})^2+8a-10]$$
then equivalent to
$$(\sqrt{b}+\sqrt{c})^2+8a\ge 10$$
which is true because
$$(\sqrt{b}+\sqrt{c})^2+8a\ge 4\sqrt{bc}+8a=4(a+\sqrt{bc})+4a\ge 8\sqrt{a\sqrt{bc}}+4a=8\sqrt[4]{a}+4a\ge 12$$
Now,since $a=\dfrac{1}{t^2}$,we have
$$f(a,t,t)=f(\dfrac{1}{t^2},t,t)=\dfrac{(10t^4-7t^2+2t+1)(t-1)^2}{t^4}$$
which is clearly nonnegative because
$$10t^4-7t^2+2t+1=(3t^2-1)^2+t(1-t)+t^4+t>0$$
Have other nice methods? Thank you
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is a linear inequality of $v^2$, which says that
it's remains to prove the last inequality for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
Hence, we need to prove that
$$(a-1)^2(10a^4-7a^2+2a+1)\geq0,$$
which is obvious.
| {
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"source": "stackexchange",
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How can I do this integral? How can I compute this integral?
$$\int\frac{x}{\sqrt{1+x^2}-x^2-1}dx$$
I tried to rationalise the denominator and got:
$$\int\frac{x(-\sqrt{1+x^2}-x^2-1)}{x^4+x^2}dx=\int\frac{-\sqrt{1+x^2}-x^2-1}{x^3+x}dx$$
But even still I find separating the integrals does not help me much in determining how to compute them, apart from:
$$\int\frac{-x^2}{x^3+x}dx$$
For the others I see no clear subsitution or trick, so any help would be great
| $$\int\frac{x}{\sqrt{1+x^2}-x^2-1}dx = \frac x{\sqrt{x^2 + 1} - (x^2 + 1)} \,dx $$
Rationalizing the denominator gives us $$\int \frac{x(\sqrt{x^2 + 1}+ (x^2 + 1)}{(x^2 + 1) - (x^2 + 1)^2}\,dx = \int \frac{x(\sqrt{x^2 + 1}+ x^2 + 1)}{-x^2 - x^4}\,dx$$
$$= \int \frac{\sqrt{x^2 + 1} + x^2 + 1}{-x(x^2 + 1)}\,dx = \int \frac{-1}{x\sqrt{x^2 + 1}} \,dx - \int \frac 1x\,dx$$
| {
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Using power series to evaluate another sum I have that $$\frac{1}{\sqrt{1-z}}=\sum_{k=0}^{\infty}\frac{z^k}{2^{2k}}\binom{2k}{k}$$
And I want to use this to evaluate the sum $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{6k}}\binom{4k}{2k}$$ I'm just not sure how I would get the binomial coefficients to match, I can see that the numbers are doubled, but is there a neat trick/identity to get it to match the coefficient above.
Thanks
| Start by re-writing your sum as follows:
$$\sum_{k\ge 0} \frac{(i/2)^{2k}}{2^{4k}}{4k\choose2k}$$
where $i$ is the imaginary unit.
Now observe that by cancellation of odd powers, we have that
$$\frac{1}{2}\left(\frac{1}{\sqrt{1-z}}+\frac{1}{\sqrt{1+z}}\right) =
\sum_{k\ge 0} \frac{z^{2k}}{2^{4k}}{4k\choose2k}.$$
It follows that your sum is
$$\frac{1}{2}\left(\frac{1}{\sqrt{1-i/2}}+\frac{1}{\sqrt{1+i/2}}\right).$$
Now let $\theta = \arctan(1/2,1)$ and start simplifying, getting
$$\frac{1}{2}
\left(\frac{\sqrt{1+i/2}}{\sqrt{1+1/4}}+\frac{\sqrt{1-i/2}}{\sqrt{1+1/4}}\right)
= \frac{1}{\sqrt{5}}
\left(\sqrt{1+i/2}+\sqrt{1-i/2}\right)\\
=\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} 2\cos(1/2\times\theta).$$
Now we have
$$\cos(1/2\times\theta) = \sqrt{\frac{1+\cos\theta}{2}}.$$
To find $\cos\theta$ consider the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1/2)$. Scale by $\sqrt{5}/2$ to obtain a point on the unit circle, getting $(0,0)$, $(2/\sqrt{5},0)$ and $(2/\sqrt{5},1/\sqrt{5}).$ It is now immediate that $$\cos\theta = 2/\sqrt{5}.$$
This finally yields
$$\cos(1/2\times\theta) = \frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}}$$
which gives for our sum the value
$$\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} \times 2\times
\frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}} =
\sqrt{\frac{2}{5}} \sqrt{1+\frac{\sqrt{5}}{2}}.$$
| {
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Finding the interval of convergence I need to find the interval of convergence of the power series $$\sum_{n=1}^\infty (-1)^n\frac {{n^2}{x^n}}{2^n}$$
I have deduced that the radius of convergence is $1$, but am unsure of how to then evaluate the series at $x = -1$ and $x = 1$ to find the interval of convergence. Any help is appreciated.
| That's not quite correct: The radius of convergence is $2$, not $1$. To see this, apply the ratio test:
\begin{align*}
\left|\frac{a_{n + 1}}{a_n}\right| &= \frac{(n + 1)^2 x^{n + 1} / 2^{n + 1}}{n^2 x^n / 2^n} \\
&= \frac{(n + 1)^2 x}{2n^2} \\
&= \left(\frac{n + 1}{n}\right) \frac{x}{2}
\end{align*}
Now take a limit as $n \to \infty$.
Now if $x = \pm 2$, our series is
$$\sum_{n = 1}^{\infty} (-1)^n \frac{n^2(\pm 2)^n}{2^n} = \sum_{n = 1}^{\infty} (\mp1)^n n^2$$
The sequence terms do not converge to $0$, and the sum is divergent in either case.
| {
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How to find the intersection of three spheres (full solutions)? The three equations of spheres are given
$(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$
$(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$
$(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$
How do I find $(x,y,z)$ analytically?
| To get you started:
*
*$(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$
*$(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$
*$(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$
Use 1.+2.
*
*$x^2-2xx_1+x_1^2+y^2-2yy_1+y_1^1+z_1^2-2zz_1+z_1^2=a^2$
*$x^2-2xx_2+x_2^2+y^2-2yy_2+y_2^2+z_2^2-2zz_2+z_2^2=b^2$
Now 1. - 2.
$$ -2xx_1+2xx_2+x_1^2-x_2^2-2yy_1+2yy_2+2zz_1-2zz_2+y_1^2-y_2^2+z_1^2-z_2^2=a^2-b^2$$
Simplify
$$2x(x_2-x_1)+2y(y_2-y_1)+2z(z_2-z_1)+(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)=a^2-b^2$$
$$\Leftrightarrow x=\dfrac{a^2-b^2-2y(y_2-y_1)-2z(z_2-z_1)-(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)}{2(x_2-x_1)}$$
Now put $x$ into 1. or 2. and solve the quadratic for $y$.
Use at last Ari's hint.
| {
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Why divide numerator and denominator by highest power in a rational function? $$\lim_{x \to \infty}\frac{x^2+3}{x^3+1} \;\;\;\;\;\;\;\; \textrm{or}\;\;\;\;\;\;\;\; \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x}$$
Why do we proceed by first dividing each term in the numerator and denominator by the highest power of $x$ in the denominator?
In the first function, we would divide each term by $x^3$, and in the second function, we would divide each term by $x^3$.
Why divide by $x$ with the highest power in the denominator when determining the limit?
Thank you.
| Because terms of the form $\frac{1}{x^n}$ converge to zero as $x \to \infty$.
In your example above, $ \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x} = \lim_{x \to \infty}\frac{10}{2+3\frac{1}{x}+6\frac{1}{x^2}} = \frac{10}{2} = 5$.
| {
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Rational sequence with $a_{n+1}=2a_n^2-1$ Suppose we start with a rational number $a_0$, and define $a_{n+1}=2a_n^2-1$ for $n\geq 0$. For what $a_0$ will it be the case that $a_i=a_j$ for some $i\neq j$?
We can start with something like $a_0=1$, then $a_1=1$ so $a_0=a_1$.
If $a_0=0$, we get $0, -1, 1, 1, \ldots$
Likewise if $a_0=-1$, we get $-1,1,1,\ldots$.
But how can we find all $a_0$?
| Let $a_{0} = \cos \theta$. Then it is easy to check that $a_{n} = \cos (2^{n}\theta)$. So if $a_{i} = a_{j}$ for some $i \neq j$, then we must have
\begin{align*}
\cos(2^{i}\theta) = \cos(2^{j}\theta)
&\quad \Longleftrightarrow \quad 2^{i}\theta = 2n\pi \pm 2^{j}\theta, \quad n \in \Bbb{Z} \\
&\quad \Longleftrightarrow \quad \theta = \frac{2n\pi}{2^{i} \pm 2^{j}}, \quad n \in \Bbb{Z}
\end{align*}
Thus the problem reduces to find the condition of $(i, j, n, \pm)$ such that
$$ \cos \left( \frac{2n\pi}{2^{i} \pm 2^{j}} \right) \in \Bbb{Q}. $$
Referring to this posting, this is possible if and only if
\begin{align*}
\theta \equiv 0, \pm \frac{\pi}{3}, \pm \frac{\pi}{2}, \pm \frac{2\pi}{3} \pmod{2\pi}
\end{align*}
This corresponds to $a_{0} \in \{0, \pm \frac{1}{2}, \pm 1 \}$.
| {
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fourth powers as sums of squares Is it possible to have a fourth power that is the sum of two squares in four different ways, e.g., $w^4 = a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = g^2 + h^2$ with the added restriction that $e = a+c$ and $g = a-c$ ? What is the lowest example, or why is it impossible?
| Since $e=a+c$ and $g=a-c$, then the OP wishes to solve the system,
$$a^2 + b^2= c^2 + d^2\tag1$$
$$a^2 + b^2= (a+c)^2 + f^2\tag2$$
$$a^2 + b^2= (a-c)^2 + h^2\tag3$$
$$a^2 + b^2= w^4\tag4$$
It turns out that, using an elliptic curve, there is an infinite number of primitive integer solutions to the first $3$ conditions. However, it is highly doubtful the $4$th can be satisfied as well.
The first was completely solved by Brahmagupta as,
$$a,\;b,\;c,\;d = p r + q s,\; p s - q r,\; p r - q s,\; p s + q r$$
The second and third become,
$$-(3 p^2 - q^2) r^2 + (p^2 + q^2) s^2 = f^2$$
$$(p^2 + q^2) r^2 + (p^2 - 3q^2) s^2 = h^2$$
For example, let $p=1,\;q=2$, then,
$$r^2+5s^2=f^2\tag5$$
$$5r^2-11s^2=h^2\tag6$$
with initial rational point $r=178,\; s=19$. Two quadratic polynomials to be made a square and that has a rational point can be transformed into an elliptic curve, so $(5),(6)$ has infinitely many primitive solutions.
Thus, one small solution to the first three conditions is then,
$$a,\;b,\;c,\;d = 216,\; 337,\; 140,\; 375$$
$$f,\;h =183, \;393 $$
The problem is the fourth condition which becomes,
$$a^2+b^2=(p^2+q^2)(r^2+s^2) = w^4\tag7$$
A third quadratic that is to be made not just a square but a fourth power, makes it highly unlikely that $(5),(6),(7)$ can be solved simultaneously.
| {
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determinant calculation This question is in my assignment. We are not allowed to use any symbol to represent any elementary row and column operations used in the solution. We must solve it step-by-step. Please help me to check my solution word by word including my spelling and grammar.
Question:
Given that
$$\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=2$$
find
$$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}.$$
Solution:
We interchange the second and third rows of the matrix $\begin{pmatrix}a& b& c\\ d& e& f\\ g& h& i\end{pmatrix}$ to get the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ and we have
$$\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=-2.$$
We interchange the first and third columns of the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ to get the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=-\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-(-2)=2.$$
We multiply the second column of the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=3\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=(3)(2)=6.$$
We add $(-2)$ times the third column of the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ to its first column to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=6.$$
We add $(-3)$ times the first row of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ to its second row to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=6.$$
Finally, we multiply the first column of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{pmatrix}$ and we have
$$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}=3\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=(3)(6)=18.$$
Thank you.
| Very well done, down to the meticulous details!
Your solution is very easy to read and to follow; the rationale and your calculations are correct, and I couldn't find any grammatical or spelling errors.
Remark: I wish all posters were as industrious as you in taking the time to carefully format your work, not just to hand in, but also here in your post at MSE!
| {
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Proving an inequality by induction, how to figure out intermediate inductive steps? I'm working on proving the following statement using induction:
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
Fair enough. I'll start with the basis step:
Basis Step: (n=1)
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
$$ \frac{1}{1^2} \le \frac{2}{1+1} $$
$$ 1 \le 1 \checkmark $$
Inductive Step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} $$
My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.
I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} $$
However, I'm not quite sure what needs to done to obtain this after attempting a few times.
| For the inductive step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} +\frac{1}{(n+1)^2}\le \frac{2n(n+1)+1}{(n+1)^2}=\frac{2n^2+2n+1}{n^2+2n+1}=\frac{2n+2+\frac{1}{n}}{n+2+\frac{1}{n}}\leq \frac{2(n+1)}{(n+1)+1}$$
| {
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Integrating trigonometric function problem $\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx$ \begin{eqnarray*}
\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &=& \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\
&=& \int \frac{3\tan x +2}{2\tan x +3} dx\\
&& u = \tan x \text{ and } du = \sec^2 x \ dx
\end{eqnarray*}
Am I going in the right direction with this one? It seems like not.
| Since the integrand has the form
$$\frac{a\sin(x) + b\cos(x)}{c\sin(x) + d\cos(x)},$$
if we write it as
$$\frac{P(x)}{Q(x)} = \frac{a\sin(x) + b\cos(x)}{c\sin(x) + d\cos(x)},$$
then we should be able to find constants $A$ and $B$ such that $P(x) \equiv A Q(x) + B Q'(x)$, so that the integral is simply
$$\int A + B\frac{Q'(x)}{Q(x)} \, dx,$$
where the second term $Q'(x)/Q(x)$ contributes $\log\left (Q(x)\right )$ to the integral.
In our case, we have
$$3\sin(x) + 2\cos(x) \equiv A\left ( 2\sin(x) + 3\cos(x) \right ) + B\left ( 2\cos(x) - 3\sin(x) \right ).$$
Equating coefficients of $\sin(x)$ and $\cos(x)$ yields $2A - 3B = 3$ and $3A + 2B = 2$, and thus $A = \frac{12}{13}$ and $B=-\frac{5}{13}$ (you should check this).
Therefore the integral is just
$$\int \frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)} \, dx = \frac{12}{13}x - \frac{5}{13}\log \left ( 2\sin(x)+3\cos(x) \right ) + C.$$
| {
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recurrence relation for squares of fibonacci numbers I have a problem finding a proof that the squares of the Fibonacci numbers satisfy the recurrence relation $a_{n+3} - 2*a_{n+2} - 2*a_{n+1} + a_n = 0$ and solving this recurrence relation.
Some help would be great!
| Using this on $\displaystyle a_{n+3}-2a_{n+2}-2a_{n+1}+a_n=0$
we have
$\displaystyle a_n=A(-1)^n+B\left(\frac{3+\sqrt5}2\right)^n+C\left(\frac{3-\sqrt5}2\right)^n\ \ \ \ (1)$ where $A,B,C$ are arbitrary constants
Now using Binet's formula, the $n$th Fibonacci term $\displaystyle F_n=\frac{a^n-b^n}{a-b}$ where $a,b$ are the roots of $t^2-t-1=0$
$\displaystyle\implies a+b=1, ab=-1, (a-b)^2=(a+b)^2-4ab=5$
$\displaystyle\implies a^2=\frac{3+\sqrt5}2,b^2=\frac{3-\sqrt5}2;$
$$\displaystyle\implies F_n^2=\frac{a^{2n}+b^{2n}-2(ab)^n}{(a-b)^2}=\frac{(a^2)^n+(b^2)^n-2(-1)^n}5=\frac{\left(\frac{3+\sqrt5}2\right)^n+\left(\frac{3-\sqrt5}2\right)^n-2(-1)^n}5 $$ which is clearly a special form of $(1)$
| {
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Closed form for integral $\int_{0}^{\pi} \left[1 - r \cos\left(\phi\right)\right]^{-n} \phi \,{\rm d}\phi$ Is there a closed form for $$I_n =\int_{0}^{\pi} \frac{\phi}{(1 - r \cos\phi)^n} \,{\rm d}\phi $$ for $\left\vert\,r\,\right\vert < 1$ real and $n > 0$
integer ?
The solution to this integral would give a closed form solution for this integral, which describes the interaction energy of vector resonant relaxation in astrophysical dynamics.
Using Mathematica and analytic methods I have derived the following result for $n=\{1,2,3,4,5,6\}$:
$$\tag{1} I_n = -\frac{a_n r}{s^{2n-2}} + \frac{b_n}{s^{2n-1}}\left[\chi_2(q) +\left(-\frac{c_n}{b_n}s + {\rm arctanh}(q) \right){\rm arctanh}(r) \right]$$
where $s=\sqrt{1-r^2}$, $q=\sqrt{(1-r)/(1+r)}$, $\chi_2(x)$ is the Legendre chi function, $a_n$, $b_n$, and $c_n$ are constants given by
\begin{align}
a_1 &= 0, \quad b_1 = 4, \qquad c_1 = 0,\\
a_2 &= 0, \quad b_2 = 4, \qquad c_2 = 2,\\
a_3 &= 1, \quad b_3 = 4 + 2r^2, \qquad c_3 = 3,\\
a_4 &= \frac{7}{3}, \quad b_4 = 4 + 6r^2, \qquad c_4= \frac{11}{3} + \frac{4}{3}r^2,\\
a_5 &= \frac{23}{6}+\frac{11}{12}r^2, \quad b_5 = 4 + 12 r^2 + \frac{3}{2}r^4,
\qquad c_5=\frac{25}{6} + \frac{55}{12}r^2,\\
a_6 &=\frac{163}{30} + \frac{47}{12}r^2, \quad b_6 = 4 + 20 r^2 + \frac{15}{2}r^4,
\qquad c_6=\frac{137}{30} + \frac{607}{60}r^2 + \frac{16}{15}r^4.
\end{align}
Is the integral $I_n$ in the general closed form given by (1) for all $n$? If so, what are the constants $a_n$, $b_n$, and $c_n$?
| First lets make the integral from $-\pi/2$ to $\pi/2$. Set $x = \phi-\pi/2$. We then get
$$I = \int_{-\pi/2}^{\pi/2}(1+r\sin(x))^{-n}(x+\pi/2)dx$$
Now recall that
$$(1+r\sin(x))^{-n} = \sum_{k=0}^{\infty} (-1)^k \dbinom{n+k}k r^k \sin^k(x)$$
Hence,
$$I = \sum_{k=0}^{\infty}(-r)^k \dbinom{n+k}k \int_{-\pi/2}^{\pi/2} (x+\pi/2)\sin^k(x)dx$$
Now from here, we can obtain
$$\int_{-\pi/2}^{\pi/2} \sin^k(x) dx \text{ and }\int_{-\pi/2}^{\pi/2} x\sin^k(x) dx$$ Lets call them $I_k$ and $J_k$ respectively. Hence,
$$I = \dfrac{\pi}2\sum_{k=0}^{\infty}r^{2k} \dbinom{n+2k}{2k} I_{2k} - \sum_{k=0}^{\infty}r^{2k+1} \dbinom{n+2k+1}{2k+1} J_{2k+1} \tag{$\star$}$$
Expression for $J_{2k+1}$ and convergence of the above summation:
We have
$$J_{2k+1} = -\int_{-\pi/2}^{\pi/2}x \sin^{2k}(x) d(\cos(x)) = \int_{-\pi/2}^{\pi/2} \cos(x) \sin^{2k}(x) dx + 2k \int_{-\pi/2}^{\pi/2} x \cos^2(x) \sin^{2k-1}(x) dx$$
$$\int_{-\pi/2}^{\pi/2} \cos(x) \sin^{2k}(x) dx = \int_{-1}^1 t^{2k} dt = \dfrac2{2k+1}$$
$$\int_{-\pi/2}^{\pi/2} x \cos^2(x) \sin^{2k-1}(x) dx = J_{2k-1} - J_{2k+1}$$
Hence,
$$(2k+1)J_{2k+1} = \dfrac2{2k+1} + 2k J_{2k-1}$$
Using this recurrence you can obtain $J_{2k+1}$. Also, it is easy to show that $\left \vert J_{2k+1} \right \vert \leq 2$ using induction.
Similarly, $I_{2k} = \dfrac1{4^k} \dbinom{2k}k \pi \leq \pi$.
Now, for $\vert r \vert < 1$, we have
$\displaystyle \sum_{k=0}^{\infty} \dbinom{n+2k}{2k} r^{2k}$ and $\displaystyle \sum_{k=0}^{\infty} \dbinom{n+2k+1}{2k+1} r^{2k}$ converges.
This ensures $\star$ makes perfect sense.
| {
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Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| A slight generalization:
$$
\begin{align}
\sum_{k=0}^n\binom{k}{a}\binom{n-k}{b}
&=\sum_{k=0}^n\binom{k}{k-a}\binom{n-k}{n-k-b}\tag{1}\\
&=\sum_{k=0}^n(-1)^{n-a-b}\binom{-a-1}{k-a}\binom{-b-1}{n-k-b}\tag{2}\\
&=(-1)^{n-a-b}\binom{-a-b-2}{n-a-b}\tag{3}\\
&=\binom{n+1}{n-a-b}\tag{4}\\
&=\binom{n+1}{a+b+1}\tag{5}
\end{align}
$$
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $\binom{n}{k}=(-1)^k\binom{k-1-n}{k}$
$(3)$: Vandermonde's Identity
$(4)$: $\binom{n}{k}=(-1)^k\binom{k-1-n}{k}$
$(5)$: $\binom{n}{k}=\binom{n}{n-k}$
Use the identity above to show that
$$
\sum_{n=0}^m\binom{n}{2}\binom{m-n}{0}=\binom{m+1}{2+0+1}=\binom{m+1}{3}
$$
| {
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Solve: $2\log_{3}(x)-\log_{3}(x+6)=1$ just getting going with logarithms, having trouble with this question.
$$2\log_{3}(x)-\log_{3}(x+6)=1$$
$$\log_{3}x^2-\log_{3}(x+6)=1$$
Stuck at this point: What do I do next?
$$\log_{3}\bigg({\dfrac{x^2}{x+6}}\bigg)=1$$
Edit: Got it!
$$\dfrac{x^2}{x+6}=3^1$$
$$\dfrac{x^2}{x+6}=3$$
$${x^2}=3(x+6)$$
$$x^2=3x+18$$
$$x^2-3x-18=0$$
$$(x-6)(x+3)=0$$
$x=6 $ and cannot equal $\ne3$ as $2\log_{3}(-3) < 0$.
| Just use exponents. If $\log_xy=z$, then $y=z^x$.
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Why does $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ have new divisors $59$ and $509$ all of a sudden? I am a noob when it comes to math so please bear with me. Why $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ has $2$ new divisors $59$ and $509$. I mean, all of its divisors are prime factors and by no way $59$ nor $509$ can be reached when multiplying any of its divisors.
How does the remainder $1$ affect the divisibility of a number?
| You can understand this much better if you know the concept of 'co-primes'.
Two numbers are said to be co-primes, if they do not have common factor other than 1.
Also note that for any positive integer $n$ where $(n\neq1)$, $n$ and $n+1$ are always co-primes as they will not have any common factor.Hence they are co-primes.
Let $a= 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
Let $b = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
As $a$ and $b$ are co-primes, none of the factors which divide $a$ will divide $b$.
Also note that any number of the form $p_1 \cdot p_2 \cdot \cdot \cdot \cdot p_n + 1$,where $p_1,p_2,\cdot\cdot\cdot p_n$ are all the primes till $p_n$, is either a prime or divisible by a prime not included in the list.
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Evaluating Definite Integral $\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx$ How can I prove that
$$\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3})$$
Can someone help me, please?
| I found an answer from the solution of my another question that is here.
$$\begin{align}\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)\,dx&=\int_1^2\arcsin\left(\frac{\sqrt{x^2-1}+2+2-4\sqrt{x^2-1}}{\sqrt{5}\sqrt{5}x}\right)\,dx\\
&=\int_0^{\pi/3}\arcsin(\sin(2\text{arcsec}(\sqrt{5})+\alpha))\tan(\alpha)\sec(\alpha)\,d\alpha\\
&=\int_0^{\pi/3}(\pi-2\text{arcsec}(\sqrt{5})-\alpha)\tan(\alpha)\sec(\alpha)\,d\alpha\\
&=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3}).\end{align}$$
| {
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Sylow subgroups of $2\times 2$ invertible matrices with elements in $\mathbb Z_5$. Let $G$ be the group of all $2 \times 2$ invertible matrices with elements in $\mathbb Z_5$. This group has order of $480=2^5\cdot 3 \cdot 5$.
I want to know if there is a quick and efficient strategy to find an example of each type of Sylow subgroup. For example, I know that there could be $1, 3, 5,$ or $15$ $2$-Sylow subgroups with $32$ elements.
| You want matrices such that their order is $1\le2^n\le32$. So try finding "roots" of $I$:
$$\begin{align} \begin{pmatrix}a&b\\c&d\end{pmatrix}^2&\equiv I\pmod{5}\\ \begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}&\equiv I\pmod{5}\\
a^2+bc\equiv d^2+bc&\equiv 1\pmod 5\\
b(a+d)\equiv c(a+d)&\equiv 0\pmod 5
\end{align}$$
Since $a^2-d^2=(a+d)(a-d)\equiv0\pmod 5$, $5|(a+d)$ or $5|(a-d)$.
Case $1$: $5|(a+d)$
If $a\equiv d\equiv0\pmod 5$, then $bc\equiv 1$, so we have solutions $\begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&2\\3&0\end{pmatrix},\begin{pmatrix}0&3\\2&0\end{pmatrix},\begin{pmatrix}0&4\\4&0\end{pmatrix}$
Otherwise, if $(a,d)=(1,4)\lor(4,1)\Longrightarrow bc\equiv0$ and if $(a,b)=(2,3)\lor(3,2)\Longrightarrow bc\equiv 2$. In the former case, we get solutions $(b,c)=(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(2,0),(3,0),(4,0)$ while the latter gets $(b,c)=(1,2),(2,1),(3,4),(4,3)$.
Case 2: $5|(a-d)$
We already covered $a\equiv d\equiv 0$, so we can move on
Since $a\equiv d\not\equiv0\pmod 5$, $a+d\equiv 2a\not\equiv 0$, so $b\equiv c\equiv 0\pmod 5$. We then have $a^2\equiv d^2\equiv 1$, so the only solutions in this case are $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}4&0\\0&4\end{pmatrix}$
So that's all $32$ roots of $I$, and these will generate more than one $2$-Sylow (you can check that the roots aren't closed). In general $n^{th}$ roots of $e$ for groups is a good start, but it depends on the group's structure.
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Proving by induction inequalities that lack the variable on the right side: $\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$ Doing proof by induction exercises with inequalities, I got stuck on one that is a bit different from the others. There is no $n$ term on the rightmost part of the inequality:
Prove that the following holds for all $n \ge 1$:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
All the other proofs I did before had some $n$ involved there, but now there is none. I wonder how will this difference affect my standard attempt:
Test for $n = 1$:
$$\frac{1}{3} \le \frac{5}{6} \implies 6 \le 15$$
We have to prove that it holds for $n + 1$, that is:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
Asssume
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
We start off with
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
And now I see the problem. Normally, I apply the hypothesis first on this part:
$$\color{blue}{\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
But all it yields is this:
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} + \frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
If I move $\frac{1}{(n + 1)+(n+2)}$ to the other side, all I get is
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} \le \frac{5}{6} - \frac{1}{(n + 1)+(n+2)}$$
Which cannot be guaranteed.
How can I prove this, then?
| we assume
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
looking for $n+1$ case, we see :
$$\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n+1)+(n+2)}=\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}-\frac{1}{n+2}\leq \frac{5}{6}-\frac{1}{n+2}\leq \frac{5}{6}$$
| {
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Does a matrix $A$ of $\pm1$'s of order $11$ exist with $\det A >4000$? How to prove or disprove that statement: there exists a square matrix of size 11 having all entries $ \pm 1$ and its determinant greater than 4000?
| Lets walk backwards. Pick any such matrix with the extra
$$A=\begin{pmatrix}
1&1&1&1&1&1&1&1&1&1&1 \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
-1&*&*&*&*&*&*&*&*&*&* \\
\end{pmatrix}$$
Adding the first row to the others, and getting a two from the other rows we get
$$\det(A)=2^{10} \det\begin{pmatrix}
1&1&1&1&1&1&1&1&1&1&1 \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
0&*&*&*&*&*&*&*&*&*&* \\
\end{pmatrix}=\\2^{10} \det\begin{pmatrix}
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
*&*&*&*&*&*&*&*&*&* \\
\end{pmatrix}$$
where this time every $*$ is $0$ or $1$. Note that starting from this last $0,1$ matrix you can backtrack your row reduction and get a $-1,1$ matrix.
Thus the problem reduced to the following:
Find a $10\times 10$ matrix with all entries $0,1$ so that the determinant is at least $4$, which is easy to solve. An easy way to produce such a matrix is the following:
$$B=\begin{pmatrix}
0&1&1&1&1 \\
1&0&1&1&1 \\
1&1&0&1&1 \\
1&1&1&0&1 \\
1&1&1&1&0 \\
\end{pmatrix}$$
Then $B$ is invertible, and the determinant is a multiple of $4$ [just add all the rows to the first one.]
Construct now the $0,1$ matrix
$$\begin{pmatrix}
B&0 \\
0&I_5 \\
\end{pmatrix}$$
Backtracking back to, any $1$ in $\begin{pmatrix}
B&0 \\
0&I_5 \\
\end{pmatrix}$ stays a $1$, any $0$ becomes $-1$ and you add the first row and column of $A$.
| {
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Questions regarding minors of a positive definite matrix In my lectures on Matrix computations, there's a section titled Gaussian elimination and Cholesky decomposition. It is a follows:
suppose $A=A^T$ is positive definite, $a_{11>0}$ and $A_{22}$ is positive definite as well, then we have:
$
A=\begin{bmatrix}
a_{11} & A_{21}^T\\
A_{21} & A_{22}\\
\end{bmatrix}
$
$
E_{1}=\begin{bmatrix}
1 & 0\\
-A_{21}.a_{11}^{-1} & I_{n-1}\\
\end{bmatrix}
$
$\implies$
$
E_1A=\begin{bmatrix}
a_{11}&A_{21}^T\\
0&A_{22}-A_{21}.a_{11}^{-1}.A_{21}^T
\end{bmatrix}
=A^{(2)}
$
I don't seem to understand why $A_{22}$ has to be positive definite. And from there, why is $A_{22}-A_{21}.a_{11}^{-1}.A_{21}^T$ positive definite, I don't get it.
| For the other expression see that for $u = (x, Y^T)^T \neq 0$ where $Y\in \mathbb R^{n-1}$ and $x\in\mathbb R$ we have
$$\begin{align*}u^T A u & = (x\quad Y^T) \left(\begin{matrix} xa_{11}+ A_{21}^T Y \\ xA_{21} + A_{22} Y\end{matrix}\right) \\
& = x^2a_{11} + x A_{21}^T Y + Y^T A_{21} x + Y^T A_{22} Y \\
& =x^2a_{11} + 2xA_{21}^T Y + Y^T A_{22} Y > 0 \end{align*}$$
If you chose $x = 0$ and $Y \neq 0$ you have $A_{22} \succ 0$.
For $x = -A_{21}^TY \cdot a_{11}^{-1}$ and $Y \neq 0$ we get
$$\begin{align*}
\underbrace{Y^T A_{21} a_{11}^{-2} A_{21}^T Y}_{=x^2} \cdot a_{11} \underbrace{-2a_{11}^{-1} (A_{21}^TY)^2}_{=2 x A_{21}^T Y} + Y^T A Y & > 0 \\
-a_{11}^{-1} (A_{21}^TY)^2 + Y^T A_{22} Y & > 0 \\
Y^T (A_{22} - a_{11}^{-1} A_{21} A_{21}^T)Y & > 0\end{align*}$$
And you're done.
| {
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Using Laplace transform to solve an IVP The original problem is:
$y''+9y = e^{-t} \sin (t)$
and the initial conditions are: $y(0)=-1$ and $y'(0)=1$.
I've taken the laplace of both sides and got:
$Y(s^2+9) = \frac{1}{(s+1)^2 +1} -s +1$
Once I divide through by $s^2 + 9$ the problems begin. The last two terms are straightforward but I'm having trouble finding an inverse laplace for the first term which is $\frac{1}{(s^2+9)(s^2+2s+2)}$.
I've tried to decompose it into the addition of two fractions by partial fraction decomposition by I keep failing to find two terms for A and B. What should i try or have i made a mistake so far in the steps i've taken?
| Everything up to this point looks good:
$$Y(s) = \frac{1}{(s^2+9)((s+1)^2+1)} - s + 1$$
To do partial fractions on $\dfrac{1}{(s^2+9)((s+1)^2+1)}$, note that
$$\frac{1}{(s^2+9)(s^2+2s+2)} = \frac{As+B}{s^2+9} + \frac{Cs+D}{s^2+2s+2}$$ since each of $s^2+9$ and $s^2+2s+2$ are irreducible quadratics.
I leave it to you to show that
$$\dfrac{1}{(s^2+9)(s^2+2s+2)} = -\dfrac{2s+7}{85(s^2+9)} + \dfrac{2s+11}{85(s^2+2s+2)}$$
EDIT: Alternatively, you could note that
$$\mathcal{L}^{-1}\left\{\frac{1}{(s^2+9)((s+1)^2+1)}\right\} = \mathcal{L}^{-1}\left\{\frac{1}{s^2+9}\right\}\ast\mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2+1}\right\} = \frac{1}{3}\sin(3t)\ast e^{-t}\sin t$$
where $f(t)\ast g(t) = \displaystyle\int_0^t f(t-\tau)g(\tau)\,d\tau$ denotes convolution. The computation of the convolution can be found in the spoiler below.
$$\begin{aligned}&\phantom{=} \frac{1}{3}\sin(3t)\ast e^{-t}\sin t\\ &= \frac{1}{3}\int_0^t \sin(3t-3\tau)e^{-\tau}\sin\tau \,d\tau\\ &= \frac{1}{6}\int_0^te^{-\tau}(\cos(3t-3\tau-\tau)-\cos(3t-3\tau+\tau))\,d\tau\\ &= \frac{1}{6}\left[\int_0^te^{-\tau}\cos(3t-4\tau)\,d\tau -\int_0^t e^{-\tau}\cos(3t-2\tau)\,d\tau\right]\\ &= \frac{1}{6}\left[\left.\left[\frac{e^{-\tau}}{17}(-4\sin(3t-4\tau)-\cos(3t-4\tau)\right]\right|_0^t - \left.\left[\frac{e^{-\tau}}{5}(-2\sin(3t-2\tau)-\cos(3t-2\tau)\right]\right|_0^t\right]\\ &= \frac{1}{6}\left[\frac{1}{17}\left[e^{-t}(4\sin t-\cos t)+4\sin(3t)+\cos(3t)\right]\right. \\ &\phantom{= \frac{1}{6}\left[\frac{1}{17}\left[e^{-t}(4\sin t-\cos t)\right]\right.}\left.+\frac{1}{5}\left[e^{-t}(2\sin t+\cos t)-2\sin(3t)-\cos(3t)\right]\right] \\ &=\frac{1}{6}\left[\frac{1}{85}\left[e^{-t}(54\sin t+12\cos t) - 14\sin(3t)-12\cos(3t)\right]\right]\\ &= \frac{1}{255}\left[e^{-t}(27\sin t+6\cos t) - 7\sin(3t)-6\cos(3t)\right]\end{aligned}$$
Where
$$\int e^{ax}\cos(bx+c)\,dx = \frac{e^{ax}}{a^2+b^2}(a\cos(bx+c)+b\sin(bx+c))+C$$
can easily be shown using integration by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve$(log_{2}(x+1))^2=4$ $$(log_{2}(x+1))^2=4$$
$$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$
$$x^{2}+2x-15=0$$
$$(x+1)*(x+1)=16$$
$$x^{2}+2x+1=16$$
$$x^{2}+2x-15=0$$
$$(x+5)(x-3)=0$$
$$x_1=-5; x_2=3$$
The solution is only $x_1=3$. Is this correct?
| $$ (\log_2(x+1))^2 = 4 \Rightarrow \log_2(x+1) = 2 , - 2 $$
$$ \Rightarrow x + 1 = 2^2 , 2^{-2} \Rightarrow x = 3 \space \textrm{or} \space -\frac{3}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/597119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If sum of $N$ natural numbers is less than $N+2$ then each of these numbers is less than $3$. Prove by contradiction.
If sum of $N$ natural numbers is less than $N+2$ then each of these numbers is less than $3$.
Attempt:
Assume if the sum of $N$ natural numbers is less than $N+2$ then each of these numbers is greater than $3$.
Let $x$ be the sum of $N$ natural numbers
$x < N + 2 $
| Try this:
Assume the sum of $N$ natural numbers is less than $N+2 \implies$ at least one of these numbers is not less than $3$.
Call that number $Y$. But then $Y+2 \ge 3+2=5.$ But $5 > N+2 \implies$ Contradiction!
$\therefore$ the sum of $N$ natural numbers is less than $N+2 \implies$ each of these numbers is less than $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/597545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $\lim_{x \rightarrow 0}\frac{9}{x}\bigl(\frac{3}{(x+3)^3}-\frac{1}{9}\bigr)$ I have to determine the following:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)$$
I've got so far:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)= \lim_{x \rightarrow 0}\left(\frac{27}{x(x+3)^3}-\frac{1}{x}\right)=\lim_{x \rightarrow 0} \left(\frac{27-(x+3)^3}{x(x+3)^3}\right)=\cdots$$
How to go on? I've got $\frac{\infty}{0}...$
| Let $L$ = the limit in question, then use definition of derivative for $f(x) = \frac{3}{(x + 3)^3}$, then $L = 9f'(0) = 9\left(\frac{-9}{(0 + 3)^4}\right) = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expected Value of a Sum Let $f(x) = |x-1|, 0 \leq x \leq 2$ be the probability density function of a random variable $X$. Find $E(X^2+X)$.
The book gives a different answer from what I got, and I have no idea why.
$E(X) = \int_{0}^2x|x-1|dx = \int_0^1 x(1-x)dx + \int_1^2 x(x-1)dx = 1$, and similarly $E(X^2)= \int_0^2x^2|x-1|dx = \int_0^1x^2(1-x)dx + \int_1^2x^2(x-1)dx = 3/2$.
So, $E(X^2+X) = E(X^2) + E(X) = 3/2 + 1 = 5/2$ is what I got, but the answer in the book is given as $13/6$. What am I doing wrong here?
| Firstly, $E(X) = (1/2-1/3)+(8/3-1/3)+(1/2-2) = 1/6 + 14/6 - 9/6 = 1$. So this is right.
Now, $E(X^2) = (1/3-1/4)+(4-1/4)+(1/3-8/3) \\ = 1/12 + 45/12 -28/12 = 18/12 = 3/2.$
So this is also right...
We should check that the probability density function is normalised.
So, $\int_{0}^{2} |x-1| dx = 2\cdot \int_{0}^{1} (1-x)\; dx = 2\cdot(1(1-0) - 1/2(1-0))= 1$.
So you should get $\frac{5}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/600526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this integral $I=\int\frac{1}{\sin^5{x}+\cos^5{x}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx$$
my solution: since
\begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\
&=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\
&=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x})
\end{align*}
then let $$\sin{x}+\cos{x}=t\Longrightarrow \sin{x}\cos{x}=\dfrac{1}{2}(t^2-2)$$
then
$$\sin^5{x}+\cos^5{x}=t[1-\dfrac{1}{2}(t^2-2)-\dfrac{1}{4}(t^2-2)^2]=-\dfrac{1}{4}t(t^4-2t^2+4)$$
$$x+\dfrac{\pi}{4}=\arcsin{\dfrac{t}{\sqrt{2}}}\Longrightarrow dx=\dfrac{1}{\sqrt{2-t^2}}dt$$
so
$$I=-4\int\dfrac{dt}{t\sqrt{2-t^2}(t^4-2t^2+4)}=-2\int\dfrac{1}{u\sqrt{2-u^2}(u^2-2u+4)}du$$
where $u=t^2$
Then I can't,
because this wolf can't http://www.wolframalpha.com/input/?i=1%2F%28xsqrt%282-x%5E2%29%28x%5E2-2x%2B4%29%29dx
Thank you for you help
| since
\begin{align*}\sin^5{x}+\cos^5{x}&=(\sin^2{x}+\cos^2{x})(\sin^3{x}+\cos^3{x})-\sin^2{x}\cos^2{x}(\sin{x}+\cos{x})\\
&=(\sin{x}+\cos{x})(1-\sin{x}\cos{x})-(\sin{x}+\cos{x})(\sin^2{x}\cos^2{x})\\
&=(\sin{x}+\cos{x})[1-\sin{x}\cos{x}-\sin^2{x}\cos^2{x}]
\end{align*}
so
\begin{align*}\int\dfrac{1}{\sin^5{x}+\cos^5{x}}dx&=\dfrac{1}{(\sin{x}+\cos{x})(1-\cos{x}\sin{x}-\cos^2{x}\sin^2{x})}dx\\
&=\int\dfrac{1}{\sqrt{1-2y}(1+2y)(1-y-y^2)}dy (y=\cos{x}\sin{x})\\
&=\dfrac{4}{5}\int\dfrac{1}{\sqrt{1-2y}(1+2y)}dy+\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy
\end{align*}
note
$$\int\dfrac{1}{\sqrt{1-2y}(1+2y)}=\int\dfrac{1}{z^2-2}dz=\dfrac{1}{2\sqrt{2}}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+C_{1} (z=\sqrt{1-2y})$$
and we
$$\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=-4\int\dfrac{z^2-2}{z^4-4z^2+1}dz$$
and
\begin{align*}
\int\dfrac{z^2-2}{z^4-4z^2+1}dz&=\int\dfrac{z^2-1}{z^4-4z^2+1}dz-\int\dfrac{1}{z^4-4z^2+1}dz\\
&=\int\dfrac{d(z+\dfrac{1}{z})}{(z+\dfrac{1}{z})^2-6}-\int\dfrac{1}{z^4-4z^2+1}dz
\end{align*}
we can easy have
$$\int\dfrac{1}{z^4-4z^2+1}dz=\dfrac{1}{2}\left(\dfrac{1+z^2}{z^4-4z^2+1}+\int\dfrac{1-z^2}{z^4-4z^2+4}\right)$$
so
$$\dfrac{1}{5}\int\dfrac{1+2y}{\sqrt{1-2y}(1-y-y^2)}dy=\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C_{2}$$
so
$$I=\dfrac{\sqrt{2}}{5}\ln{\left|\dfrac{z-\sqrt{2}}{z+\sqrt{2}}\right|}+\dfrac{1}{5}\left(\dfrac{1}{\sqrt{2}}\ln{\left|\dfrac{z^2-\sqrt{2}z+1}{z^2+\sqrt{2}z+1}\right|}-\dfrac{1}{\sqrt{6}}\ln{\left|\dfrac{z^2-\sqrt{6}z+1}{z^2+\sqrt{6}z+1}\right|}\right)+C$$
where $z=\cos{x}-\sin{x},C=\dfrac{4}{5}C_{1}+C_{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
negative exponent problem $$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$
Does this equal =
$$
\begin{align*}
& \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\
=&\sqrt{1 + 3 + 9 + 27 + 81} \\
=&\sqrt{121} \\
=&11.
\end{align*}
$$
The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.
The rule I'm using, incorrectly, is this:
$$\frac{1}{3^{-2}} = 3^2 = 9.$$
| When you have a doubt with such expressions (I use here you specific problem), inside the square root, multiply both numerator and denominator by the reverse of the most negative power (here : 3^4) and perform the multiplications by (3^4) for each term of the denominator. Then, perform the additions and simplifications.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$ The following problem(p.668, 7) is from Integrals and Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] states that
$$\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$$
How one can show that?
| \begin{align}
\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}&=-\sum_{k=0}^\infty \frac{1}{k+1/2}+\frac{2}{3}\sum_{k=0}^{\infty} \frac{1}{k+1/4}+\frac{1}{3} \sum_{k=0}^\infty \frac{1}{k+1} \\
&=\frac{1}{3}\sum_{k=0}^\infty \frac{1}{k+1}-\frac{1}{k+1/2}+\frac{2}{3} \sum_{k=0}^{\infty} \frac{1}{k+1/4}-\frac{1}{k+1/2} \\
&=-\frac{1}{6}\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+1/2)}+\frac{1}{6}\sum_{k=0}^\infty \frac{1}{(k+1/4)(k+1/2)} \\
\text{Using Gauss's Digamma Theorem, }\\
&=-\frac{1}{6}\cdot \frac{\psi(1)-\psi(1/2)}{1/3-1/2}+ \frac{1}{6}\cdot \frac{\psi(1/4)-\psi(1/2)}{1/4-1/2} \\
&=-\frac{1}{6}\cdot \log 16+\frac{1}{6}\cdot 2(\pi+\log4)=\frac{\pi}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/601231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Prove an trigonometric identity. Can someone help me by solving it? $$1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x}=\sin x \cos x.$$
I cannot come to the result. I make that like $1-\dfrac{\sin^2 x}{1+\frac{\cos x}{\sin x}}$
| $$
\begin{align}
1-\frac{\sin^2(x)}{1+\cot(x)}-\frac{\cos^2(x)}{1+\tan(x)}
&=\frac{(1+\tan(x))-\sin^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{1}\\
&=\frac{1+\cos^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{2}\\
&=\frac{\cos^2(x)\tan(x)+\sin^2(x)}{1+\tan(x)}\tag{3}\\
&=\frac{\cos^2(x)\tan(x)+\cos^2(x)\tan^2(x)}{1+\tan(x)}\tag{4}\\[4pt]
&=\cos^2(x)\tan(x)\tag{5}\\[12pt]
&=\sin(x)\cos(x)\tag{6}
\end{align}
$$
Explanation:
$(1)$: put everything over a common denominator
$(2)$: $\cos^2(x)=1-\sin^2(x)$
$(3)$: $\sin^2(x)=1-\cos^2(x)$
$(4)$: $\sin^2(x)=\cos^2(x)\tan^2(x)$
$(5)$: factor out $1+\tan(x)$ from the numerator and cancel
$(6)$: $\sin(x)=\cos(x)\tan(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/604115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$? $$\lim _{x \rightarrow 0} \left(\frac{
\sin x}{x}\right)^{1/x}$$
I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
| Notice by Young's Inequality since $(\frac{1}{x} + (1 - \frac{1}{x})) = 1$, then
$$ (\frac{\sin x}{x} )^{1/x}=(\frac{\sin x}{x} )^{1/x} 1^{1 - \frac{1}{x}} \leq \frac{\sin x}{x^2} + 1 - \frac{1}{x} = \frac{\sin x - x}{x^2} + 1$$
Now, for positive $x$ and for $a \leq 1$, we have that $a^x \leq x +1 $. Hence $a^{1/x} \geq \frac{1}{x+1}$. Now, since $\frac{\sin x}{x} \leq 1$, we apply this inequality with $a = \frac{\sin x}{x} $ to obtain
$$ (\frac{\sin x}{x} )^{1/x} \geq \frac{1}{x+1}$$. Hence we have
$$ \frac{1}{x+1} \leq (\frac{\sin x}{x} )^{1/x} \leq \frac{\sin x - x}{x^2} + 1$$.
Now, since $\lim{ \frac{1}{x+1} } = 1 $ and
$$ \lim (\frac{\sin x - x}{x^2} + 1 ) = \lim ( \frac{\sin x - x}{x^2} ) = 1 + \lim ( \frac{\cos x - 1}{2x} ) = 1 + \lim ( \frac{- \sin x}{2} ) = 1 $$
Now, result follows by the squeeze trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/605202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.
If $abc=1$ and $a,b,c$ are positive real numbers, prove that $${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1\,.$$
The whole problem is in the title. If you wanna hear what I've tried, well, I've tried multiplying both sides by 3 and then using the homogenic mean. $${3 \over a+b+1} \le \sqrt[3]{{1\over ab}} = \sqrt[3]{c}$$ By adding the inequalities I get $$ {3 \over a+b+1} + {3 \over b+c+1} + {3 \over c+a+1} \le \sqrt[3]a + \sqrt[3]b + \sqrt[3]c$$ And then if I proof that that is less or equal to 3, then I've solved the problem. But the thing is, it's not less or equal to 3 (obviously, because you can think of a situation like $a=354$, $b={1\over 354}$ and $c=1$. Then the sum is a lot bigger than 3).
So everything that I try doesn't work. I'd like to get some ideas. Thanks.
| I have other nice Cauchy-Schwarz inequality solve it.
since
$$\dfrac{1}{1+a+b}=1-\dfrac{a+b}{1+a+b}$$
so the original inequality can be written
$$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge2$$
use Cauchy-Schwarz inequaliy and the AM-GM inequality,we have
$$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge\dfrac{(\sum\sqrt{a+b})^2}{\sum(a+b+1)}=\dfrac{2p+2\sum\sqrt{(a+b)(a+c)}}{2p+3}\ge\dfrac{2p+2\sum(a+\sqrt{bc})}{2p+3}=\dfrac{4p+2\sum\sqrt{bc}}{2p+3}\ge 2$$
because use AM-GM inequality
$$\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\ge 3\sqrt[3]{abc}=3$$
where $p=a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Solving an indefinite integral in the middle of a question In the process of solving an integral, I came across the following step and couldn't proceed: $$\int \sqrt{\frac{2t^2-1}{1-2t^2+t^4}}dt$$I know that I should use partial fractions but I don't know how to apply that here. Any suggestions?
| Let's call
$$
f(t)=\sqrt{\frac{2t^2-1}{1-2t^2+t^4}}=\frac{\sqrt{2t^2-1}}{|1-t^2|}
$$
which is even because $f(-t)=f(t)$. Let's call $\varphi(t)$ the function defined for $|t|>\frac{1}{\sqrt 2}$ and $|t|\ne 1$ as
$$
\varphi(t)=\frac{\sqrt{2t^2-1}}{1-t^2}
$$
so that the function $f(t)$, for $|t|>\frac{1}{\sqrt 2}$, may be written as
$$
f(t)=\begin{cases} \varphi(t) & |t|<1\\ -\varphi(t) & |t|>1\end{cases}
$$
Then, the integral which we have to evaluate is
$$
I=\int \sqrt{\frac{2t^2-1}{1-2t^2+t^4}}\operatorname{d}t=\int f(t)\operatorname{d}t=\begin{cases} \int\varphi(t) \operatorname{d}t & \frac{1}{\sqrt 2}<|t|<1\\ -\int\varphi(t)\operatorname{d}t & |t|>1\end{cases}
$$
So we can evaluate the integral
$$
J=\int\varphi(t) \operatorname{d}t=\int \frac{\sqrt{2t^2-1}}{1-t^2} \operatorname{d}t.
$$
Let's put
$$
x=\frac{\sqrt 2 t}{\sqrt{2t^2-1}}
$$
so that $$\operatorname{d}t=-\frac{1}{\sqrt 2}\frac{1}{(x^2-1)^{3/2}}\operatorname{d}x$$
and
$$
\begin{align}
\sqrt{2t^2-1}&=\frac{1}{(x^2-1)^{1/2}} &
1-t^2&=\frac{x^2-2}{2(x^2-1)}
\end{align}
$$
Putting all together we find
$$
J=\int \frac{1}{(x^2-1)^{1/2}}\frac{2(x^2-1)}{x^2-2}\left(-\frac{1}{\sqrt 2}\right)\frac{1}{(x^2-1)^{3/2}}\operatorname{d}x
$$
and simplifying we obtain
$$
J=-\sqrt 2 \int \frac{1}{(x^2-1)(x^2-2)}\operatorname{d}x.
$$
Expanding the integrand function in partial fraction one has
$$
\begin{align}
\frac{1}{(x^2-1)(x^2-2)}&=\frac{1}{x^2-2}-\frac{1}{x^2-1}\\
&=\frac{1}{2\sqrt 2}\left(\frac{1}{x-\sqrt 2}-\frac{1}{x+\sqrt 2}\right)-\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)
\end{align}
$$
Integrating term by term and recalling that $\int \frac{1}{u}\operatorname{d}u=\log u + c$ we find
$$
J=-\frac{1}{2}\log\left(\frac{x-\sqrt 2}{x+\sqrt 2}\right)+\frac{1}{\sqrt 2}\log\left(\frac{x-1}{x+1}\right)+C
$$
Finally, substituting back $x=\frac{\sqrt 2 t}{\sqrt{2t^2-1}}$ and simplifying, we obtain
$$
J=-\frac{1}{2}\log\left(\frac{t-\sqrt{2t^2-1}}{t+\sqrt{2t^2-1}}\right)+\frac{1}{\sqrt 2}\log\left(\frac{\sqrt 2 t-\sqrt{2t^2-1}}{\sqrt 2t+\sqrt{2t^2-1}}\right)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $A + B + C = \pi$, then show that $\sin(A) + \sin(B) + \sin(C) = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ So i have
$A + B + C = \pi$
$$\frac{A}{2} + \frac {B}{2} + \frac{C}{2} = \frac{\pi}{2}$$
$$4\cos\left(\frac{-B-C + \pi}{2}\right)\cos\left(\frac{-A -C + \pi}{2}\right)\cdots$$
And I doubt this leads to anywhere.
So then I tried,
$\sin\left(\frac{-B-C + \pi}{2}\right)\cdots$ and this didn't go anywhere either.
I don't know what to try, and I've seen other people's solutions and they do something like:
$\sin(C) = \sin(A + B)$, $\cos(C/2) = \sin(\frac{A + B}{2})$ but i don't see where they got this part from. Other people use Euler's formula or whatever but I haven't learned that yet so I can't use it.
| $$\sin(A) + \sin(B) + \sin(C)= 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A - B}{2}\right) + 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right) $$
$$\sin(A) + \sin(B) + \sin(C)= 2\cos\left(\frac{C}{2}\right)\cos\left(\frac{A - B}{2}\right) + 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$\sin(A) + \sin(B) + \sin(C)= 2\cos\left(\frac{C}{2}\right)\cos\left(\frac{A - B}{2}\right) + \cos\left(\frac{A + B}{2}\right)$$
$$\sin(A) + \sin(B) + \sin(C)= 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $\sum_{k=1}^{n}{3^kk}$ I need hint to find this sum $$\sum_{k=1}^{n}{3^kk}$$ $\sum_{k=1}^{n}{3^k}$ can be easily calculated as it is sum of geometric progression.
| An alternative solution:$$\begin{align}\sum 3^kk=3+2\cdot3^2+3\cdot3^3+4\cdot3^4&\ldots\\=3+3^2+3^3+3^4&\ldots\\+3^2+3^3+3^4&\ldots\\ +3^3+3^4&\ldots\\\vdots\\=\sum3^k+3\sum3^k+3^2\sum3^k&\ldots\end{align}$$
So if $S_n=\sum_{k=1}^{n}3^kk$, then $$\begin{align}S_n&=\sum_{k=1}^n3^k\sum_{i=0}^{n-k}3^i\\&=\sum_{k=1}^n3^{k}\frac{3^{n-k+1}-1}{2}\\&=\frac12n3^{n+1}-\frac12\sum_{k=1}^n3^{k}\\&=\frac12n3^{n+1}-\frac32\frac{3^{n}-1}{2}\\
&=\frac{2n3^{n+1}-3^{n+1}+3}{4}\\
&=\frac34\left((2n-1)3^{n}+1\right)\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/612219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction that $3^{4n + 1} + 5^{2n + 1}$ is divisble by $8$ This is a homework problem:
Prove that: $$ 3^{4n+1} + 5^{2n+1}$$ is divisible by $8$ for every natural number $n$.
Base case: $$n = 0$$
$$ 3^{0 + 1} + 5^{0 + 1} = 8$$
$$8\bmod8 = 0 $$ Base case verified.
Inductive Hypothesis: Assume for some $n$, $$ 3^{4n+1} + 5^{2n+1} =8k$$ where $k$ is any non-negative integer.
Prove that: $$ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$$
Indeed: $$ 3^{4n+4+1} + 5^{2n+2+1} = 3^4\cdot3^{4n+1} + 5^2\cdot5^{2n+1}$$
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (80+1)\cdot3^{4n+1} + (24+1)\cdot5^{2n+1} $$
$$ = 80\cdot3^{4n+1} + 24\cdot5^{2n+1} + 3^{4n+1} + 5^{2n+1} $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1}) + 8k $$
$$ = 8(10\cdot3^{4n+1} + 3\cdot5^{2n+1} + k) $$
Is this wrong in any particular way? My answer is different from the textbook, but I feel like I sufficiently proved that $ 3^{4(n+1)+1} + 5^{2(n+1)+1} =8l$.
Edit: What the book did:
$$ = 81\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = (56+25)\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25\cdot3^{4n+1} + 25\cdot5^{2n+1} $$
$$ = 56\cdot3^{4n+1}+25(3^{4n+1} + 5^{2n+1}) $$
$$ = 56\cdot3^{4n+1}+25(8k) $$
$$ = 8(7\cdot3^{4n+1}+25k) $$
| The two methods differ only in how they rearrange the summands to show dvisibility by $8$
$$\begin{eqnarray} 81a+25b &=&\,80a+24b+\color{#c00}{a+b} &=&\, (10a+3b+\color{#c00}k)\,\color{#c00}8\ \ {\rm by}\ \ \color{#c00}{a+b = 8k} \\ 81a+25b &=&\, 56a + 25(\color{#c00}{a+b}) &=&\, (7a+25\color{#c00}k)\,\color{#c00}8\\
\end{eqnarray}$$
More simply: $\ {\rm mod}\ 8\!:\ 81\equiv 1\equiv 25\ \Rightarrow\ 81a + 25b\equiv \color{#c00}{a+b \equiv 0}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the remainder when $1.1!+2.2!+3.3!+ ... +10.10! +2$ is divided by $11!$ Find the remainder when $1.1!+2.2!+3.3!+ ... +10.10! +2$ is divided by $11!$
An attempt: Rearranging:
$$\frac{1}{11!}+\frac{2.2!}{11}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}+\frac{2}{11!}$$
$$\frac{1}{11!}+\frac{2}{11!}+\frac{2.2!}{11}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{(2+1)2!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{3.2!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$\frac{1}{11!}+\frac{(3+1)3!}{11!}+\frac{3.3!}{11} \cdots +\frac{10.10!}{11}$$
$$.$$
$$.$$
$$.$$
$$\frac{1}{11!}+\frac{(10+1)10!}{11!}$$
$$\frac{11!+1}{11!}$$
What is the remainder?
| $$\sum_{k=0}^n k.k!=(n+1)!$$.this summation should be used in your question
| {
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Value of $ \frac{a^2 + b^2 + c^2}{ac^2 - ab^2} $ If $a + b + c = 0$ , then value of $ \frac{a^2 + b^2 + c^2}{ac^2 - ab^2} $ is?
According to me if $ a + b + c = 0$, then $a + b = -c$
This implies $ (a + b)^2 = c^2$
This implies $a^2 + b^2 = c^2 - 2ab$
Putting this value in $ \frac{a^2 + b^2 + c^2}{ac^2 - ab^2} $ we get
$ \frac {c^2 - 2ab + c^2}{ac ^ 2 - ab} = \frac {2c^2 - 2ab}{a(c^2 - b)} $
But I'm stuck after this. Am I right in the above steps, if yes, then how should I continue?
| Note that in the denominator
$$ ac^2-ab^2=a(c+b)(c-b)=a^2(b-c)$$
for further simplification. Actually, since $a$ plays a special role in the original expresion, I suggest to simply eliminate $a$ by replacing it with $-(b+c)$ throughout, e.g.
$$\frac{a^2+b^2+c^2}{ac^2-ab^2}=\frac{2(b^2+bc+c^2)}{(b+c)(b^2-c^2)} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve using polar form $x^4-x^3+x^2-x+1=0$. This equation has to be solved in complex numbers using polar form. I tried grouping the power of x terms but it is not working. Any other idea of solving the equation?
| Note that
\begin{align}
x^4-x^3+x^2-x+1=0 &\,\,\,\Longleftrightarrow\,\,\, (x^4-x^3+x^2-x+1)(x+1)=0 \quad\text{and}\quad x\ne -1 \\ &\,\,\,\Longleftrightarrow\,\,\, x^5+1=0 \quad\text{and}\quad x\ne -1.
\end{align}
Now, solve $x^5+1=0$ using polar form and of course exclude the solution $x=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
how do i write $y = 2\sin2(x + \frac{\pi}{4}) - \cos2(x + \frac{\pi}{4} )$ in format $y = A\sin(kx) + B\cos(kx)$ The problem is the double angles. I tried to simplify them and change them around but no luck,
$$\begin{align} y&=A\sin(kx)+B\cos(ky)\\
y&=2\sin2(x+\pi/4)-\cos2(x+\pi/4)\\
&=2\left(2\sin(x+\pi/4)\cos(x+\pi/4)\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\cos\frac\pi 4+\cos x\sin\frac\pi4\right)\left(\cos x\cos\frac\pi 4-\sin x\sin\frac\pi4\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=\frac 4{\sqrt 2}\left(\sin x+\cos x\right)\left(\cos x-\sin x\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=2\sqrt 2(\cos^2x-\sin^2x)+\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)^2-\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)^2\\
&=2\sqrt 2(\cos^2x-\sin^2x)+4\left(\frac1{\sqrt2}\sin x\frac1{\sqrt2}\cos x\right)\\
&=2\sqrt 2(\cos^2x-\sin^2x+\sin x\cos x) \end{align}$$
| Don't forget that double-angle formulas work in both directions.
| {
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Posssible pentagons in 3D A non-planar pentagon in $\mathbb{R}^3$ has equal sides and four right angles. What are the possible values for the fifth angle?
My attempt
It was quite easy to find an example for $60^\circ$: $A=(1, 0, 1), B=(1, 1, 1), C=(0, 1, 1), D=(0, 0, 1)$ and then $\triangle DEA$ is equilateral. I couldn't find any other.
We can assume WLOG that the first three vertices are the same as A,B,C in the above example, by rotating the pentagon (which is an isometry and therefore preserves angles), but there are two vertices left to determine, which are $6^\circ$ of freedom - so analytic solution won't work.
| Let the pentagon be formed by placing five unit vectors $v_1, \ldots, v_5 \in \mathbb{R}^3$ head to tail, with right angles between two successive vectors. Then $$v_1+v_2+v_3+v_4+v_5 = 0.$$
Let $v_{km} = \langle v_k, v_m \rangle$ denote the inner product between two vectors. In particular $$v_{11}=v_{22}=v_{33}=v_{44}=v_{55}=1 \\
v_{12} = v_{23} = v_{34} = v_{45} = 0.
$$
From $0 = \langle v_k, 0 \rangle = \langle v_k, v_1+\ldots+v_5 \rangle$ we get the linear system
$$
\begin{pmatrix}
1&1&1&1&0&0&0\\
1&0&0&0&1&1&0\\
1&1&0&0&0&0&1\\
1&0&1&0&1&0&0\\
1&0&0&1&0&1&1
\end{pmatrix} \begin{pmatrix}
1\\
v_{13}\\
v_{14}\\
v_{15}\\
v_{24}\\
v_{25}\\
v_{35}
\end{pmatrix} = 0.
$$
The kernel of this matrix is spanned by the columns of $$\begin{pmatrix}-2&0\\1&0\\0&1\\1&-1\\2&-1\\0&1\\1&0\end{pmatrix}.$$
Therefore the Gram matrix $(v_{km})$ must be of the form
$$\begin{pmatrix}
1&0&-\tfrac{1}{2}&x&-\tfrac{1}{2}-x\\
0&1&0&-1-x&x\\
-\tfrac{1}{2}&0&1&0&-\tfrac{1}{2}\\
x&-1-x&0&1&0\\
-\tfrac{1}{2}-x&x&-\tfrac{1}{2}&0&1
\end{pmatrix}
$$
for some $x\in\mathbb{R}$. This matrix must be positive semi-definite and of rank three. This is the case if and only if $x=0$ or $x=-\tfrac{6}{7}$. The fifth angle $\alpha$ satisfies $\cos(\alpha) = -v_{15}$. So $\cos(\alpha)=\tfrac{1}{2}$ or $\cos({\alpha}) = -\tfrac{5}{14}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\omega^2+\omega+1$divides a polynomial The question is
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?
| Let $$g(n)=n^2+n+1$$ and $$h(n)=n^3-n+1$$.
Clearly $$f(n)=g(n) \cdot h(n)$$
Now $h(1)=1$ and $h'(n) = 3 n^2-1 > 0$ for $n\ge 1$. So, for $n>1$ both $g(n)$ and $h(n)$ are greater than $1$. So
$f(n)$ **is composite for all $n>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Help with Evaluating triple integral Evaluate $\int \int \int_B xyz^2 dV$, where B is a cuboid bounded by the regions $ 0 \le x \le 1 $, $ -1 \le y \le 2 $, $ 0 \le z \le 3 $.
I keep getting $ \frac{27}{4}$ as my answer but apparently it's incorrect...Any help would be appreciated.
| Just as another check on this, we can use the "odd symmetry" about $ \ y = 0 \ $ of the integrand function $ \ xyz^2 \ $ to cancel the contributions from the portions of the volume over $ \ -1 \ \le \ y \ \le \ 0 \ $ and $ \ 0 \ \le \ y \ \le \ 1 \ , $ leaving integration only over $ \ 1 \ \le \ y \ \le \ 2 \ . $ I won't even bother "separating" the variables; we still find
$$ \int_0^3 \int_1^2 \int_0^1 \ xyz^2 \ \ dx \ dy \ dz \ \ = \ \ \int_0^3 \int_1^2 \ \left(\frac{1}{2}x^2yz^2 \right) \vert_{x=0}^{x=1} \ \ dy \ dz $$
$$ = \ \ \frac{1}{2} \ \int_0^3 \int_1^2 \ yz^2 \ \ dy \ dz \ \ = \ \ \frac{1}{2} \ \int_0^3 \ \left(\frac{1}{2}y^2z^2 \right) \vert_{y=1}^{y=2} \ \ dz $$
$$ = \ \ \frac{1}{2} \cdot \frac{3}{2} \ \int_0^3 \ z^2 \ \ dz \ \ = \ \ \frac{1}{2} \cdot \frac{3}{2} \ \left(\frac{1}{3}z^3 \right) \vert_{z=0}^{z=3} \ \ = \ \ \frac{1}{2} \cdot \frac{3}{2} \cdot 9 \ \ . $$
[So, 'nother county heard from...]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$
Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.
I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$
So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.
The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.
| You have
$
\sin x=\dfrac{\sqrt{2}}{2}
$
so
$
\cos x=\sqrt{1-(\sin x)^{2}}=\dfrac{\sqrt{2}}{2}
$
and
$
\tan x =\dfrac{\sin x}{\cos x}\ = 1
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$ If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$
Then prove that:
$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.
I have no idea how to go about this.
Please help.
| Let's introduce the shorthand notations
$$
\begin{align}
a &= \frac{-\alpha+\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\alpha}{2},\\
b &= \frac{\alpha-\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\beta}{2},\\
c &= \frac{\alpha+\beta-\gamma}{4} = \frac{\pi}{4} - \frac{\gamma}{2},
\end{align}
$$
then
$$
a+b = \frac{\gamma}{2},\quad a+c = \frac{\beta}{2},\quad b+c = \frac{\alpha}{2},
$$
$$
a+b+c = \frac{\pi}{4},
$$
$$
\sin a\sin b \sin c = \cos a\cos b\cos c.\tag{1}
$$
Now
$$
\cos(a+b+c) = \cos(a+b)\cos(c) - \sin(a+b)\sin c= \frac{\sqrt{2}}{2},
$$
so that
$$
\cos(a+b)\cos(c) - \sin a\sin c\cos b - \sin b\sin c\cos a = \frac{\sqrt{2}}{2}
$$
add $(2\cos a\cos b\cos c)$ to both sides, and we get
$$
\cos(a+b)\cos(c) + \cos(a+c)\cos b+ \cos(b+c)\cos(a)= \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c,
$$
which reduces to
$$
\begin{multline}
\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) +
\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\
\cos\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c. \tag{2}
\end{multline}
$$
On the other hand,
$$
\sin(a+b+c) = \sin(a+b)\cos c + \cos(a+b)\sin c = \frac{\sqrt{2}}{2},
$$
so that,
$$
\sin(a+b)\cos c + \cos a\cos b\sin c = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{3}
$$
Similarly, if we take $\sin(a+b+c) = \sin((a+c)+b)$, we get
$$
\sin(a+c)\cos b + \cos a\cos c\sin b = \frac{\sqrt{2}}{2} + \sin a\sin b\sin c.\tag{4}
$$
If we add eqs. $(3)$ and $(4)$, we get
$$
\sin(a+b)\cos c + \sin(a+c)\cos b + \sin(b+c)\cos a = \sqrt{2} + 2\sin a\sin b\sin c,
$$
which reduces to (using also $(1)$)
$$
\begin{multline}
\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) +
\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\
\sin\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \sqrt{2} + 2\cos a\cos b\cos c. \tag{5}
\end{multline}
$$
Finally, we subtract eq $(5)$ from eq $(2)$:
$$
\begin{multline}
\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)
\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \ldots = - \frac{\sqrt{2}}{2},
\end{multline}
$$
which is
$$
\begin{multline}
\frac{\sqrt{2}}{2}\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)\left(\cos\left(\frac{\alpha}{2}\right)+\sin\left(\frac{\alpha}{2}\right)\right) + \ldots + \frac{\sqrt{2}}{2} =0,
\end{multline}
$$
so that finally
$$
\cos\alpha +\cos\beta+\cos\gamma + 1 = 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this $H_{2n}-H_{n}+\frac{1}{4n}>\ln{2}$
Show that, for every positive integer $n$,
$$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}+\dfrac{1}{4n}>\ln{2}$$
I know this
$$\lim_{n\to\infty}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}=\ln2$$
and use this
$$\ln{(1+\dfrac{1}{n})}<\dfrac{1}{n}$$ is not useful
But for this inequality I can't. Thank you
| Write $$u_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n}$$
$$ = [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{4n}] - [\frac{1}{1} + \frac{1}{2} + \dots+ \frac{1}{n}] - [\frac{1}{2n+1} + \frac{1}{2n+2} + \dots +\frac{1}{4n -1}]$$
$$= [\gamma_{4n} + \log(4n)] - [\gamma_n + \log(n)] - [\gamma_{4n -1} + \log(4n-1) - \gamma_{2n} - \log(2n)]$$
Now take $\lim_{n \rightarrow \infty} u_n$.
All $\gamma_{n}$, $\gamma_{4n - 1}$, $\gamma_{2n}$, $\gamma_{n}$ etc will be $\gamma$ (Euler's constant) and will be cancelled out.
What remains?
$\lim_{n \rightarrow \infty}[\log{\frac{4n}{4n - 1}} + \log{\frac{2n}{n}}] = \log 2$
Now see $u_{n+1} - u_n = \frac{1}{4(n+1)} - \frac{1}{4n} + \frac{1}{2(n+1)} - \frac{1}{n+1} < 0$
Thus the sequence $\{u_n\}$ is monotone decreasing and converges to $\log 2$. So for all $n$, $u_n > \log 2$
Thus ultimately we are getting for any $n$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{4n} > \log 2$$
| {
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Integral Question $$
\int _{\large{ -\frac { \pi }{ 2 } } }^{\large{ \frac { \pi }{ 2 } } }{ \frac { \mathrm{d}x }{ \sin x-2\cos x+3 } }
$$
Wolframalpha suggests to evaluate indefinite integral by substituting $u = \tan(\frac { x }{ 2 })$. Is there a more basic way to evaluate this definite integral?
| There is another way to deal with this, but it's a special-case trick and may never serve you again. But here goes: when you see $$A\sin(x) + B \cos(x),$$ you can rewrite that as
$$
K \sin(x + c)
$$
for some constants $K$ and $c$. How? Pick
$$
K = \sqrt{A^2 + B^2};
$$
in your case, that gives you $K = \sqrt{5}$.
Let
$$
c = \cos^{-1}(A/K).
$$
In your case, you get $c = 1.107...$ radians.
Now check: if $\sin(c)$ has the same sign as $B$, leave $c$ as is; if not, negate $c$.
In your case, $\sin(c) = 0.89...$, so the sign's wrong, and we have to make $c = -1.107...$ radians. But I'm just going to leave it as $-\cos^{-1}(1/\sqrt{5})$, and write it as $c$.
Alternatively, you can let $c = atan2(A, B)$, assuming you know about the atan2 function.
So now your integrand becomes
$$
\frac{dx}{K\sin(x + c) + 3} \\
= \frac{dx}{\sqrt{5}\sin(x + c) + 3}
$$
You can let $u = x + c$ and $du = dx$ to get
$$
\frac{du}{\sqrt{5}\sin(u) + 3}.
$$
Multiply top and bottom by the "conjugate" to get
$$
\frac{(\sqrt{5}\sin(u) - 3 ) du}{5\sin^2(u) - 9}.
$$
Now write $5 \sin^2 u - 9 = 5(\sin^2 u - 1) - 4 = -5 \cos^2(u) - 4$, so the integrand becomes
$$
-\frac{(\sqrt{5}\sin(u) - 3 ) du}{5\cos^2(u) + 4} \\
= -\frac{\sqrt{5}\sin(u)du}{5\cos^2(u) + 4} + \frac{3 du}{5\cos^2(u) + 4}
$$
The first part can be addressed with a substitution $v = \cos u$, which produces a logarithm term.
The second part...well, that's difficult in itself, but only solution is to let $u = \tan^{-1}(x)$. Then $\cos(u) = \dfrac{x}{\sqrt{1 + x^2}}$, so $\cos^2(x) = \dfrac{x^2}{1 + x^2}$, while $du = \frac{1}{1 + x^2} dx$. So the second term becomes
$$
\frac{3 du}{5\cos^2(u) + 4} \\
= \frac{3}{5\dfrac{x^2}{1 + x^2} + 4} \frac{1}{1 + x^2} dx \\
= \frac{3}{5x^2 + 4(1 + x^2)} dx \\
= \frac{3}{9x^2 + 4} dx \\
= \frac{3/4}{\frac{9}{4}x^2 + 1} dx \\
= \frac{3/4}{(\frac{3}{2}x)^2 + 1} dx
$$
Now letting $z = \frac{3}{2} x$, you've got an arctangent integral, and you should be on your way.
After all that, the $\tan(t/2)$ substitution seems pretty nice, doesn't it? :)
Key idea: a linear combination of $\sin(ax)$ and $\cos(ax)$ can be rewritten as $K\sin(ax + c)$ for some $K$ and $c$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $z\in\Bbb C$ such that $|z+1|+ |z-1|=4$ I'd like to find all points of the complex plane which satisfy
$$|z+1| + |z-1| = 4. $$
I know this is an ellipsis with foci $1$ and $-1$, and I know that the answer is
$$3 x^2+4 y^2 \leq 12,$$
but I can't find a correct way of getting there.
First, I write $z$ as $x + i y$ and square both sides of the equation, then divide by 2 and get
$$x^2+y^2+1+\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =8.$$
Pass $x^2+y^2+1$ to the RHS (right hand side), then
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =7-x^2-y^2.\tag{1} $$
Now, I would have to square both sides of the equation like this,
$$((x-1)^2+y^2)((x+1)^2+y^2) = (7-x^2-y^2)^2, \tag{2}$$
but the problem is that I cannot assure that the RHS is not negative, so there could be a value for $z$ such that (2) is satisfied but not (1), i.e it could exist $z=x + i y$ which satisfies
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =-(7-x^2-y^2)\tag{3} $$
in which case also satisfy (2) but not (1)!
So I would get an incorrect solution.
| I get the sense that you want an algebraic approach to transform the equation from the complex plane to Cartesian coordinates. To this end, the key is to first square both sides: $$\begin{align*} 16 &= (|z-1| + |z+1|)^2 \\ &= |z-1|^2 + 2|z-1||z+1| + |z+1|^2 \\ &= (x-1)^2 + y^2 + 2|z^2-1| + (x+1)^2 + y^2 \\ &= 2x^2 + 2 + 2y^2 + 2|(x+iy)^2 - 1| \\ &= 2\left(x^2 + y^2 + 1 + \sqrt{(x^2-y^2-1)^2 + (2xy)^2} \right). \end{align*}$$ Now dividing both sides by 2 and rearranging, we square again: $$(8 - (x^2 + y^2 + 1))^2 = (x^2 - y^2 - 1)^2 + 4x^2 y^2.$$ Put all terms on one side and simplify: $$\begin{align*} 0 &= (x^2-y^2-1)^2 - (x^2+y^2-7)^2 + 4x^2 y^2 \\ &= (-2y^2+6)(2x^2-8) + 4x^2 y^2 \\ &= -4(x^2-4)(y^2-3) + 4x^2 y^2 \\ &= 4(3x^2 + 4y^2 - 12). \end{align*} $$ This immediately gives the desired expression. Note that the locus of points satisfying the equality is itself necessarily an equation, not an inequality: the locus is the boundary of the ellipse, not its interior.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
characteristic polynomial of graph I have 2 questions about how to find the characteristic polynomial of some graphs.
*
*If G is a simple cycle with n vertices and n edges, $C_n$, I need to find the characteristic polynomial of $C_n$ (the characteristic polynomial of the adjacency matrix).
I tryed to find some reccursive equation to the characteristic polynomial of the adjacency matrix:
$$
\begin{pmatrix}
0 & 1 & 0 & & \cdots & 0 & 1 \\
1 & 0 & 1 & 0 & \cdots & 0 & 0\\
0 & 1 & 0 & 1 & 0 & \cdots & 0 \\
\vdots && \ddots & \ddots& \ddots& \ddots \\
0 && &&&&1\\
1 &0&&&0&1&0\\
\end{pmatrix}
$$
but I didn't succeed.
\2. I have a graph G that is k-regular, and I need to prove a connection between characteristic polynomials of G and G complement, $\overline G$:
$$ p_\overline G(x) = (-1)^n{x-n+k+1\over x+k+1}p_G(-x-1) $$
| Jesus revealed this answer to me the following for the second part
I supply the answer for part two
the Let $x_i$ be the eigenvalue of $X$, then $$\Phi (X, x)=\displaystyle \prod _{i=1} ^{n} (x-x_i)= (x-x_1)\prod _{i=2} ^{n} (x-x_i) = (x-k)\prod _{i=2} ^{n} (x-x_i)$$
It follows from this that
$$\Phi (X, -x-1)=\displaystyle (-x-k-1)\prod _{i=2} ^{n} (-x-1- \theta _i)$$
Similarly we calculate
\begin{align*}
\Phi (\overline{X}, x) &=\displaystyle \prod _{i=1} ^{n} (x-x_i)\\ &= (x-x_1)\prod _{i=2} ^{n} (x-x_i)\\ &= (x-(n-k-1))\prod _{i=2} ^{n} (x-(-1-\theta _i))\\ &= (x-n+k+1) \prod _{i=2}^{n}(x+1+\theta_i)\\ &= (x-n+k+1) \frac{n+k+1}{n+k+1} \prod _{i=2}^{n}(x+1+\theta_i)\\ &= (x-n+k+1) \frac{n+k+1}{n+k+1} \left( -1 \right) ^{n-1} \prod _{i=2}^{n}(-x-1-\theta_i)\\ &= \left( -1 \right) ^{n} \frac{n+k+1}{x-n+k+1} (-x-k-1) \prod _{i=2}^{n}(-x-1-\theta_i)\\ &= \left( -1 \right) ^{n} \frac{n+k+1}{x-n+k+1} \Phi (X, -x-1)
\end{align*}\\
Hence we can write $\displaystyle \Phi (\overline{X}, x)= \left( -1 \right) ^{n} \frac{n+k+1}{x-n+k+1} \Phi (X, -x-1) $second part
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this minimun of this $\frac{xy}{x^5+xy+y^5}+\frac{yz}{y^5+yz+z^5}+\frac{xz}{x^5+xz+z^5}$ let $x,y,z>0$ and such
$$x+y+z=1$$
Find minimum of the
$$\dfrac{xy}{x^5+xy+y^5}+\dfrac{yz}{y^5+yz+z^5}+\dfrac{xz}{x^5+xz+z^5}$$
I have find this maximum
note
$$\dfrac{xy}{x^5+xy+y^5}\le -243\dfrac{x+y}{841}+\dfrac{23031}{1682}$$
see
http://www.wolframalpha.com/input/?i=xy%2F%28x%5E5%2Bxy%2By%5E5%29%2B243*%28x%2By%29%2F841-23031%2F1682
But for minimum value,I can't,(maybe use AM-GM) Thank you,
| The max is $\dfrac{81}{29}$ and min is $\dfrac{.25}{2*(0.5)^5+.25}$, a simple method is let two varies equal$(x=y)$ and have a function of $x$, see graphic below:
edit :to prove max, consider $xy \le \dfrac{x^2+y^2}{2}, x^5+y^5 \ge 2(\dfrac{x^2+y^2}{2})^{\frac{5}{2}},\dfrac{xy}{x^5+xy+y^5}\le \dfrac{x^2+y^2}{x^2+y^2+4(\dfrac{x^2+y^2}{2})^{\frac{5}{2}}}=\dfrac{p}{p+4(\dfrac{p}{2})^{\frac{5}{2}}}=f(p)$
$f(p)$ is concave and mono decreasing function and $x^2+y^2+z^2 \ge \dfrac{1}{3}$ ,so we can quickly know the max will be got when $x=y=z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit of logarithm functions How can I calculate limit expressed by:
\begin{equation}
\lim_{x\to \infty} \left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln(x)}
\end{equation}
| Taking ln of both sides, you obtain, for $x\to+\infty$:
\begin{equation}
x\ln(x)\ln\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)\sim x\ln(x)\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}-1\right)=\frac{x\ln(x)}{\ln(x^2+2x+3)}\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)\sim \frac{x\ln(x)}{\ln(x^2+2x+3)}\left(\frac{x^2+3x+4}{x^2+2x+3}-1\right)=
\frac{x\ln(x)}{\ln(x^2+2x+3)}\left(\frac{x+1}{x^2+2x+3}\right)\sim \frac{\ln(x)}{\ln(x^2+2x+3)}\sim \frac{\ln(x)}{\ln(x^2)}\to\frac 12.
\end{equation}
Thus your limit is $\sqrt e$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the second derivative $(x^3+x-1)(x^3+1)$ $$(x^3+x-1)(x^3+1)$$
$$=x^6+x^4-x^3+x^3+x-1$$
$$f'(x)= 6x^5+4x^3-3x^2+3x^2+1$$
Am I suppose to cancel out $-3x^2+3x^2$?
$$f''(x)= 30x^4+12x^2-6x+6x$$
Can someone check my work please? Thank you so much!
| You can cancel the 6x, but yes, that is correct according to wolfram alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $P=\sqrt{2x^2+2y^2-2x+2y+1}+\sqrt{2x^2+2y^2+2x-2y+1}+\sqrt{2x^2+2y^2+4x+4y+4}$ Let $x,y∈R$ . Find the minimum value of this expression:
$P=\sqrt{2x^2+2y^2-2x+2y+1}+\sqrt{2x^2+2y^2+2x-2y+1}+\sqrt{2x^2+2y^2+4x+4y+4}$
We have: $P=\sqrt{(\sqrt{2}x-\frac{1}{\sqrt{2}})^2+(\sqrt{2}y+\frac{1}{\sqrt{2}})^2}+...+\sqrt{(\sqrt{2}x+\sqrt{2})^2+(\sqrt{2}y+\sqrt{2})^2}$
I think use vector or Geometric method to solve this problem. But I don't know how to choose vectors or points logically?
| Hint:
You need $P(x, y)$ to be the Fermat point of the triangle $A(\frac12, -\frac12), B(-\frac12, \frac12), C(-1, -1)$.
Note that $\triangle ABC$ is isosceles, and from properties of the Fermat Point, it should lie on the angle bisector at $C$. This gives us $x=y$. Now there are two good routes,
(i) to use calculus in one variable, or
(ii) to note an additional property of the Fermat point $\angle APB = \angle BPC = \angle CPA = 120^{\circ}$ and use coordinate geometry / vectors.
Added: To check your final answer,
I get the minimum as $P \ge 2+\sqrt3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $(x,y)$ such that $\sin x+\sin y=\sin(x+y)$ and $|x|+|y|=1$
Find all $(x,y)$ such that $$\sin x+\sin y=\sin(x+y)$$ and $|x|+|y|=1$
My work:
We have, $\sin x+\sin y=\sin x\cos y+\cos x \sin y$
By comparing both sides, we have,$\cos y=1,\cos x=1$, but certainly that is not true, as $|x|+|y|=1$.
So,$\sin x(1-\cos y)=\sin y(\cos x-1)$
Now, I am stuck. I cannot break this loop to proceed further. Please help!
| Using Prosthaphaeresis Formulas,
$$\sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2$$
$$\sin(x+y)=2\sin\frac{x+y}2\cos\frac{x+y}2$$
So, $$\sin x+\sin y-\sin(x+y)=2\sin\frac{x+y}2\left(\cos\frac{x-y}2-\cos\frac{x+y}2\right)$$
$$=2\sin\frac{x+y}2\cdot2\sin\frac x2\sin\frac y2 (\text{ again Prosthaphaeresis })$$
Now we know $\displaystyle \sin A=0\implies A=n\pi$ where $n$ is an integer
As $|x|+|y|=1, |x|=1-|y|\le 1$
Case $1a)$ If $\displaystyle \sin\frac x2=0\iff x=2n\pi\implies|2n\pi|\le1\implies|n|<1\implies n=0\implies x=0$
$\displaystyle|y|=1\iff y=\pm1$
Case $1b)$ Similarly, if $\displaystyle \sin\frac y2=0$
Case $2)$ If $\displaystyle \sin\frac{x+y}2=0, x+y=2n\pi$
Using this, $\displaystyle |x+y|\le |x|+|y|=1\implies |2n\pi|\le1\implies |n|<1\implies n=0$
$\displaystyle\implies x+y=0$ and we have $\displaystyle|x|+|y|=1$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the e^{matrix} Can anyone help me out with the following question:
Compute $e^{A}$ for the following matrix:
$A=\begin{bmatrix}
2 & 1 & -1 \\
0 & 4 & -2 \\
0 & 2 & 0 \\
\end{bmatrix}$
I've written $A=\begin{bmatrix}
2 & 0 & 0 \\
0 & 2 & 0\\
0 & 0 & 2 \\
\end{bmatrix}+\begin{bmatrix}
0 & 1 & -1 \\
0 & 2 & -2 \\
0 & 2 & -2\\
\end{bmatrix}=D+N$ with D diagonal and N nilpotent.
Now I thought: $e^{A}=e^{D+N}=e^2 I e^N=e^2[I+N]=\begin{bmatrix}
e^2& e^2 & -e^2 \\
0 & 3e^2 & -2e^2 \\
0 & 2e^2 & -e^2 \\
\end{bmatrix}$.
But this isn't the same as the answer if I compute it with for example maple. Does anyone see my mistake?
| The usual way to compute the exponential of a matrix is using the Jordan decomposition.
The Jordan decomposition of your matrix is
$$
\begin{bmatrix}
2&1&-1\\0&4&-2\\0&2&0
\end{bmatrix}
=
\begin{bmatrix}
0&-1&0\\1&-2&0\\1&-2&1
\end{bmatrix}
\begin{bmatrix}
2&0&0\\0&2&1\\0&0&2
\end{bmatrix}
\begin{bmatrix}
0&-1&0\\1&-2&0\\1&-2&1
\end{bmatrix}^{-1}
$$
Now, since the matrices commute, we can compute
$$
\begin{align}
\exp\left(\begin{bmatrix}
2&0&0\\0&2&1\\0&0&2
\end{bmatrix}
\right)
&=
\exp\left(\begin{bmatrix}
2&0&0\\0&2&0\\0&0&2
\end{bmatrix}
\right)
\exp\left(\begin{bmatrix}
0&0&0\\0&0&1\\0&0&0
\end{bmatrix}
\right)\\[6pt]
&=
\begin{bmatrix}
e^2&0&0\\0&e^2&0\\0&0&e^2
\end{bmatrix}
\begin{bmatrix}
1&0&0\\0&1&1\\0&0&1
\end{bmatrix}\\
\end{align}
$$
Finally,
$$
\begin{align}
\exp\left(\begin{bmatrix}
2&1&-1\\0&4&-2\\0&2&0
\end{bmatrix}
\right)
&=
\begin{bmatrix}
0&-1&0\\1&-2&0\\1&-2&1
\end{bmatrix}
\exp\left(\begin{bmatrix}
2&0&0\\0&2&1\\0&0&2
\end{bmatrix}\right)
\begin{bmatrix}
0&-1&0\\1&-2&0\\1&-2&1
\end{bmatrix}^{-1}\\
\end{align}
$$
I don't see anything wrong with your computation with the nilpotent matrix. In fact, I get the same result when completing the computations above.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\sum\limits_{cyc}\frac{a^2}{b(a^2-ab+b^2)}\ge\frac{9}{a+b+c}$ let $a,b,c>0$, show that
$$\dfrac{a^2}{b(a^2-ab+b^2)}+\dfrac{b^2}{c(b^2-bc+c^2)}+\dfrac{c^2}{a(c^2-ca+a^2)}\ge\dfrac{9}{a+b+c}$$
My try: since this inequality is homogeneous ,without loss of generality, we assume that
$$a+b+c=3$$
then
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2}{b(a^2-ab+b^2)}\ge 3$$
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2(a+b)}{b(a^3+b^3)}\ge 3$$
then I can't,Thank you
| Proof without words ( partial answer / informal proof ) .
A formal proof has been given in another answer by another author.
Picture on the left: geometry of the conditions $a,b,c > 0$ and $a+b+c=3$ .
Picture in the middle: outside ($< 0$ : olive green) and inside ($\ge 0$ : white) of the function:
$$f(a,b,c) = \sum_{cyc}\frac{a^2}{b(a^2-ab+b^2)} - \frac{9}{a+b+c}$$
as seen in the plane of the red triangle in the picture on the left.
Since there are no green spots inside the triangle, the function is expected to be $\ge 0$ there.
Picture on the right: contour lines of $f(a,b,c)$ inside the triangle at levels $N = 1/2^k \;; \; k=0,\cdots,7$ . Contours are darker at lower level values.The minimum is expected to be $f(1,1,1)=0$ , which is at the center of the triangle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find what the sum of this infinite series is. I have the following series:
$\sum_{n\geq3} \frac{4n-3}{n^3-4n}$
which I've simplified into the following:
$\frac{3}{4}\sum_{n\geq3}{ \frac{1}{n}} + \frac{5}{8}\sum_{n\geq3} \frac{1}{n-2} - \frac{11}{8}\sum_{n\geq3} \frac{1}{n+2}$
And this is where I'm stuck... How do I calculate the total sum?
Thanks in advance!
| $$\sum_{3}^{\infty} \frac{ 4n-3 }{ n^3-4n }=\\
\sum_{3}^{\infty} \frac{ 4n-3 }{ n(n^2-4) }=\\
\sum_{3}^{\infty} \frac{ 4n }{ n(n-2)(n+2) }+\sum_{3}^{\infty} \frac{ -3 }{ n(n-2)(n+2) }=\\
4\sum_{3}^{\infty} \frac{ 1 }{ (n-2)(n+2) }-3\sum_{3}^{\infty} \frac{ +1 }{ n(n-2)(n+2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$.
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$
where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.
Not sure if all these denotions are conventions that always mean what I've mentioned.
Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
| Note that when $i$ is odd then $3|(2^{i}-2)$ and hence $\lfloor \frac {2^{i}}{3} \rfloor = \lfloor \frac {2^{i}-2+2}{3} \rfloor = \lfloor \frac {2^{i}-2}{3} + \frac {2}{3} \rfloor = \frac {2^{i}-2}{3}$.
Also when $j$ is even, then $3|(2^{j}-1)$ and hence $\lfloor \frac {2^{j}}{3} \rfloor = \lfloor \frac {2^{j}-1+1}{3} \rfloor = \lfloor \frac {2^{j}-1}{3} + \frac {1}{3} \rfloor = \frac {2^{j}-1}{3}$.
Therefore,
$$ \sum_{r=0}^{2010} \lfloor \frac {2^{r}}{3} \rfloor = \sum_i \lfloor \frac {2^{i}}{3} \rfloor + \sum_j \lfloor \frac {2^{j}}{3} \rfloor = \sum_i \frac {2^{i}-2}{3} + \sum_j \frac {2^{j}-1}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$ Problem :
The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
If this could have been like this $\cos\frac{\pi}{8} + \cos\frac{3\pi}{8}+\cos\frac{5\pi}{8}+\cos\frac{7\pi}{8}$ then we can take the terms like
$(\cos\frac{3\pi}{8}+\cos\frac{5\pi}{8} ) + (\cos\frac{\pi}{8} +\cos\frac{7\pi}{8})$
and solve further , but in this case due to power of 4 I am unable to proceed please suggest... thanks..
| For any $\alpha \in \mathbb{R}$,
let $\lambda_k = \cos\left[\frac{\pi}{8}(2k-\alpha)\right]$ for $k = 1,2,\ldots,8$. All of them satisfy
$$\cos(8\cos^{-1}\lambda_k) = \cos\left[\pi(2k-\alpha)\right] = \cos\pi\alpha$$
This implies they are the 8 roots of a polynomial of degree 8:
$$T_8(x) -\cos\pi\alpha = 0 \quad\iff\quad 128x^8-256x^6+160x^4-32x^2 + (1-\cos\pi\alpha) = 0 $$
where
$$T_8(x) = \cos(8\cos^{-1}x) = 128x^8-256x^6+160x^4-32x^2 + 1$$
is the Chebyshev polynomial of $1^{st}$ kind with degree $8$. Notice $\lambda_{4+k} = -\lambda_k$ for $k = 1,\ldots, 4$, this gives us
$$\begin{align}
\prod_{k=1}^4(x - \lambda_k^2) = \prod_{k=1}^8(\sqrt{x}-\lambda_k)
&= \frac{1}{128}(T_8(\sqrt{x})-\cos\pi\alpha)\\
&= x^4 - 2 x^3+\frac54 x^2-\frac14 x + \frac{1-\cos\pi\alpha}{128}
\end{align}$$
By comparing the coefficients of $x^3$ and $x^2$, we get
$$\sum_{k=1}^4 \lambda_k^2 = 2\quad\text{ and }\quad
\sum_{1\le k < \ell \le 4} \lambda_k^2\lambda_\ell^2 = \frac54$$
When $\alpha = 1$, $\;\displaystyle\sum_{k=1}^4 \lambda_k^4\;$ reduces to
$$\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$$
and hence the desired sum can be evaluated as
$$\sum_{k=1}^4 \lambda_k^4 = \left(\sum_{k=1}^4 \lambda_k^2 \right)^2 - 2 \left(\sum_{1\le k < \ell \le 4} \lambda_k^2\lambda_\ell^2\right) = 2^2 - 2\cdot\frac54 = \frac32$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration by substitution fail for $\int \frac{1}{(1+\sin x)}\, \mathrm dx$. We tried to do an integral with the universal trigonometric substitution $$\int \frac{1}{(1+\sin x)}\, \mathrm dx$$
Meaning, we substituted: $ t = \tan \frac{x}{2} \Rightarrow$
$$\int \frac{1}{(1+\sin x)}\, \mathrm dx = \int \frac{\frac{2}{1+t^2}}{1+\frac{2t}{1+t^2}}\, \mathrm dt = \int \frac{2}{(1+t)^2}\, \mathrm dt = \frac{-2}{1+t} = \frac{-2}{1+\tan \frac{x}{2}} + C$$
But the answer is:
$$ \tan x - \frac{1}{\cos x} + C $$
What did we do wrong?
| An alternative method of integration first involves multiplying the numerator and denominator by $1-\sin x$: this gives $$\begin{align*} \int \frac{1}{1+\sin x} \, dx &= \int \frac{1 - \sin x}{1 - \sin^2 x} \, dx \\ &= \int \frac{1 - \sin x}{\cos^2 x} \, dx \\ &= \int \sec^2 x - \frac{\sin x}{\cos^2 x} \, dx \\ &= \tan x - \int \frac{-du}{u^2}, \quad u = \cos x, du = -\sin x \, dx \\ &= \tan x - \frac{1}{u} + C \\ &= \tan x - \sec x + C. \end{align*}$$ To see the equivalence of this form with the expression $$-\frac{2}{1+\tan \frac{x}{2}} + C,$$ consider their difference, with $\theta = x/2$: $$\begin{align*} \tan x - \sec x + \frac{2}{1 + \tan \frac{x}{2}} &= \frac{\sin 2\theta - 1}{\cos 2\theta} + \frac{2}{1 + \tan \theta} \\ &= \frac{\sin 2\theta - 1}{\cos 2\theta} + \frac{2 \cos \theta}{\sin \theta + \cos \theta} \\ &= \frac{\sin 2\theta - 1 + 2 \cos\theta(\cos \theta - \sin \theta)}{\cos 2\theta} \\ &= \frac{2 \cos^2 \theta - 1}{\cos 2\theta} \\ &= 1. \end{align*}$$ Thus their difference is constant, and both are antiderivatives.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An upper bound on certain finite trigonometric series given a lower bound
Let $f$ be the function $f(x)=1+a\sin{x}+b\cos x+c\sin{(2x)}+d\cos{(2x)}$, where $a,b,c,d$ are arbitrary real numbers. Prove that if $f(x)>0$ for all $x\in \mathbb R$, then $f(x)<3$ for all $x\in \mathbb R$.
My try:
\begin{align*}f(x)&=1+a\dfrac{e^{ix}-e^{-ix}}{2i}+b\dfrac{e^{ix}+e^{-ix}}{2}+c\dfrac{e^{2ix}-e^{-2ix}}{2i}+d\dfrac{e^{2ix}+e^{-2ix}}{2}\\
&=1+e^{ix}\left(\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{-ix}\left(-\dfrac{a}{2i}+\dfrac{b}{2}\right)+e^{2ix}\left(\dfrac{c}{2i}+\dfrac{d}{2}\right)+e^{-2ix}\left(\dfrac{d}{2}-\dfrac{c}{2i}\right).
\end{align*}
I don't see a way to proceed from there. Thank you very much.
| As obareey mentioned in the comments to the question, a (finite) sum of sine waves of the same frequency is again a sine wave of that frequency. In particular, we have
$$\begin{align}
a \sin x + b \cos x &= \sqrt{a^2 + b^2} \cos(x + \phi_1)
\\
c \sin 2x + d \cos 2x &= \sqrt{c^2 + d^2} \cos(2x + \phi_2)
\end{align}$$
for some $\phi_1, \phi_2 \in \mathbb R$. Denote $\alpha := \sqrt{a^2 + b^2}$, $\beta := \sqrt{c^2 + d^2}$, and
$$
g(x) := \alpha \cos(x + \phi_1) + \beta \cos(2x + \phi_2) = f(x) - 1.
$$
It suffices to show that if there exists an $x^* \in \mathbb R$ such that $g(x^*) \geq 2$, then there exists an $x' \in \mathbb R$ such that $g(x') \leq -1$.
To prove, first note that since $\alpha, \beta \geq 0$ and $\cos \leq 1$, we have $\alpha + \beta \geq g(x^*) \geq 2$. Next, consider the sets
$$\textstyle
S := \{ x \in \mathbb{R} : \cos(x + \phi_1) \leq -\frac{1}{2}\}
= \bigcup_{k \in \mathbb{Z}} \left[2\pi k - \phi_1 + \frac{2\pi}{3}, 2\pi k - \phi_1 + \frac{4\pi}{3}\right]
\\\textstyle
T := \{ x \in \mathbb{R} : \cos(2x + \phi_2) \leq -\frac{1}{2}\}
= \bigcup_{k \in \mathbb{Z}} \left[\pi k - \frac{\phi_2}{2} + \frac{\pi}{3}, \pi k - \frac{\phi_2}{2} + \frac{2\pi}{3}\right].
$$
Observe that $S$ is composed of closed intervals of length $2\pi/3$, while
$$\textstyle
\mathbb R \setminus T = \bigcup_{k \in \mathbb{Z}} \left(\pi k - \frac{\phi_2}{2} - \frac{\pi}{3}, \pi k - \frac{\phi_2}{2} + \frac{\pi}{3}\right)
$$
is composed of open intervals of length $2\pi/3$. Thus $S \not\subseteq \mathbb{R} \setminus T$, meaning that $S \cap T \neq \varnothing$. If we now choose an $x' \in S \cap T$, then $g(x') \leq -\frac{1}{2}(\alpha + \beta) \leq -1$, as desired.
| {
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"url": "https://math.stackexchange.com/questions/646272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}$
I need to find sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}$
I'm getting it as $2\ln2-1$
| The sequence converges absolutely by comparison with $\sum_{n=1}^\infty \frac{1}{n^2}$. Note
$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
Therefore $$\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{2n}-\frac{1}{2n-1}\right)+\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=1+2\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=1-2\ln(2)$$
The second to last equality follows by noticing that all fractions of the forms $\frac{1}{2n}$ and $\frac{-1}{2n-1}$ appear twice, except $-1$, which is counted only once.
| {
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"url": "https://math.stackexchange.com/questions/651031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{x \to 0} \left( \frac{\ln (\cos x)}{x\sqrt {1 + x} - x} \right)$ efficiently I need to evaluate:
$$\lim_{x \to 0} \left( \frac{\ln (\cos x)}{x\sqrt {1 + x} - x} \right)$$
Now, it looked to me like a classic L'Hôpital's rule case. Indeed, I used it (twice), but then things became messy and complicated.
Am I missing the point of this exercise? I mean, there must be a "nicer" way.
Or should I stick with this road?
EDIT:
Regarding Yiorgos's answer: Why is the following true?
$$\ln\left(1- {x^2 \over 2}\right) \approx -{x^2 \over 2}$$
| Elaborating on Martin's answer, the calculation of limit is as follows:
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\frac{\log(\cos x)}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\log(\cos x)}{\cos x - 1}\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}1\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\cos x - 1}{x\left(\sqrt{1 + x} - 1\right)}\\
&= \lim_{x \to 0}\frac{\cos x - 1}{x\left(\sqrt{1 + x} - 1\right)}\cdot\frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}\\
&= \lim_{x \to 0}\frac{\cos x - 1}{x^{2}}\cdot\left(\sqrt{1 + x} + 1\right)\\
&= -\frac{1}{2}\cdot 2 = -1\end{aligned}$
Note that if $y = \cos x - 1$ then $y \to 0$ as $x \to 0$ and hence $\dfrac{\log(1 + y)}{y} = \dfrac{\log \cos x}{\cos x - 1}$ tends to $1$. The other limit $\dfrac{\cos x - 1}{x^{2}}$ is easily done by using $\cos x - 1 = -2\sin^{2}(x/2)$ and then using the fundamental limit $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.
| {
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"source": "stackexchange",
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Triangle Calculation Question In a triangle $ABC$, it is known that $ AC = m$, $AB = 3m$, $BC = Rm$.
Find for which values of $R $ the triangle is:
(i) An acute angle triangle.
(ii) A right angle triangle.
(iii) An obtuse angle triangle.
I have tried several approaches including the law of cosine and the Pythagorean theorem, but it didn't help.
| You can use the law of cosines 9 times, since we are unsure which angle will be the right or obtuse in those cases. For the sake of simplicity, lets let $AB=c, AC=b, BC=a$. Choose angle $A$ to be our first case; by the law of Cosines,
$$a^2=b^2+c^2-2bc\cos{(A)}$$
The three cases are: if $A\lt{90}$ degrees, if $A=90$ degrees, and if $A\gt{90}$ degrees.
If $A\lt{90}$, then $\cos{(A)}$ is positive which means
$$(Rm)^2=m^2+(3m)^2-2m(3m)\cos{(A)}$$
$$R^2m^2=10m^2-6m^2\cos{(A)}$$
Thus
$R=\sqrt{10-6\cos{A}}.$
If $A=90$ degrees, then $\cos{(A)}=0$ which means
$$R^2m^2=m^2+9m^2$$
Thus
$R=\sqrt{10}$.
Finally if $A\gt{90}$ degrees, then $\cos{(A)}$ is negative which
$$R^2m^2=m^2+9m^2+2m(3m)\cos{(A)}$$
$$R^2m^2=10m^2+6m^2\cos{(A)}$$
Thus $R=\sqrt{10+6\cos{(A)}}$
Repeat this argument with $b^2=a^2+c^2-2ac\cos{(B)}$ and $c^2=a^2+b^2-2ab\cos{(C)}$.
EDIT: with the other two cases, you will end up with quadratics in $R$, since the law of cosines with produce an $R^2$ term and an $R$ term. For example,
$$m^2=9m^2+R^2m^2-18R\cos{(B)}$$
Now you must solve
$$R^2-18\cos{(B)}R+8=0$$
When $B=90$ degrees it just becomes solving $R^2+8=0$ which produces complex solutions, not what we are after...
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $a^5 b - b^5 a$ is divisible by $30$ for any integers $a$ and $b$ I am trying to prove that $a^5\times b - b^5\times a$ is divisible by $3$. The actual task is to prove divisibility by $30$ but I have managed to prove that the expression is divisible by $5$ and $2$. However, I am having difficulties proving it's divisible by $3$. I don't want a direct answer but rather some kind of a hint.
| We can write $\displaystyle F=a^5b-ab^5=b(a^5-a)-a(b^5-b)$
Using Fermat's little theorem, $\displaystyle c^5-c\equiv0\pmod5$ for all integer $c$
$\displaystyle\implies F\equiv0\pmod5$ as $a,b$ are integers
Again from this, $\displaystyle a^3-a=(a-1)a(a+1)$ is divisible by $3!=6$
$\displaystyle\implies a^3\equiv a\pmod6\implies a^5b=a^3\cdot a^2b\equiv a\cdot a^2b\pmod6\equiv a^3b\equiv ab$
Similarly, $b^5a\equiv ba\pmod6$
$\displaystyle\implies a^5b-b^5a\equiv?\pmod6$
Now as $F$ is divisible by $5,6$ so it will be divisible by lcm$(5,6)$
Generalization:
Consider $\displaystyle a^nb-b^na=ab\left[\underbrace{a^{n-1}-1}-(\underbrace{b^{n-1}-1)}\right]$
It will be divisible by prime $p$ if $p$ divides at least one of $a,b$
Else $(a,p)=(b,p)=(ab,p)=1$
Now using Fermat's little theorem, each of $a^{n-1}-1,b^{n-1}-1$ will be divisible by $p$ if $\phi(p)=p-1$ divides $n-1$
So if $p-1$ divides $n-1,p$ will divide $\displaystyle a^nb-b^na$ for all integer $a,b,n$
If $p_i$s are distinct primes, $\prod_{1\le i\le r} p_i$ will divide $\displaystyle a^nb-b^na$ if $n-1$ is divisible by $p_i-1$ for each $1\le i\le r$
i.e., if $n-1$ is divisible by lcm$(p_i-1)$ for $1\le i\le r$
Here, we can safely set $p_1=2,p_2=3,p_3=5$ hence, $n-1$ needs to be divisible by lcm$(1,2,4)=4$
| {
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Convergence Proof: $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$ I have to check whether the following expression converges; if yes I have to give the limit.
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
Now I did the following:
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
$$\lim_{x\rightarrow\infty} x\sqrt{\frac{4}{x}+1}- x\sqrt{1+\frac{1}{x}}$$
$$\lim_{x\rightarrow\infty} x \lim_{x\rightarrow\infty}(\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}})$$
That confuses me. The left limit approaches $\infty$ while the right approaches $0$. What is wrong here or what can I conclude from that?
Thank you very muvh for your help in advance!
FunkyPeanut
| You get nowhere by reasoning that way. The canonical way to work on this is "rationalizing": you multiply and divide by $\sqrt{4x+x^2}+\sqrt{x^2+x}$. That way you get
$$
\sqrt{4x+x^2}-\sqrt{x^2+x}=\frac{4x+x^2-x^2-x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac{3x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac3{\sqrt{4/x+1}+\sqrt{1+1/x}}\to\frac32
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(\frac{ceiling(x)+n}{m}) = ceiling(\frac{x+n}{m})$ I can't get eq (1) and eq (2)
Question: Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(\frac{ceiling(x)+n}{m}) = ceiling(\frac{x+n}{m})$
Answer:
Let us define the real number $x$ as the sum of an integer ‘$a$’ and a positive real number ‘$b$’ which is lesser than $1$. Therefore, $x = a + b$, where $a$ is an integer and $b$ is a real number lesser than $1$.
Therefore,
$=ceiling(\frac{ceiling(x)+n}{m})$
$=ceiling(\frac{ceiling(a+b)+n}{m})$
$=ceiling(\frac{ceiling(b)+a+n}{m})$
$=ceiling(\frac{1+a+n}{m})$
$=ceiling(\frac{1}{m}+\frac{a+n}{m}) \dots (1)$
Evaluating the other expression,
$= ceiling(\frac{x+n}{m})$
$= ceiling(\frac{a+b+n}{m})$
$= ceiling(\frac{x}{m}+\frac{n}{m})$
$= ceiling(\frac{b}{m}+\frac{a+n}{m}) \dots (2)$
| There can be no integer between $\frac{b+a+n}{m}$ and $\frac{1+a+n}m$ because there is no integer (and even less so a multiple of $m$) between $b+a+n$ and $1+a+n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex Numbers Geometry I'm not sure where to begin on this problem - do I plug in for a and solve for z?
I was also given a hint:
Let z be a point on the line we're trying to describe. We have good tools in complex numbers for collinearity and perpendicularity. Which would be useful here.
Here is the problem:
Let a and b be two complex numbers on the unit circle, i.e. $|a| = |b| = 1.$
(a) Show that the equation of the tangent to the unit circle at a is given by $ z + a^2 \overline{z} = 2a.$
(b) Show that the intersection of the tangents to the unit circle at a and b is $\frac{2ab}{a + b}.$
| (a) Since we know the tangent line and the radius from the origin to $a$ are perpendicular, we can say $$z-a = a(e^{\pi i/2}k) = aki$$ where $k$ is some integer.
Manipulating the equation we get:
\begin{align*}
z-a &= aki \\
z &= a(1+ki) \\
\overline{z} &= \overline{a}(1-ki) \\
a^2\overline{z} &= a^2\overline{a}(1-ki)
\end{align*}
Let's simplify the right side: $$a^2\overline{a}(1-ki) = a\cdot a\overline{a}(1-ki) = a\cdot |a|^2(1-ki) = a(1-ki)$$
Now we can continue:
\begin{align*}
a^2\overline{z} &= a^2\overline{a}(1-ki) \\
a^2\overline{z} &= a(1-ki) \\
a^2\overline{z} &= a-aki \\
a^2\overline{z} &= 2a - a - aki \\
a^2\overline{z} &= 2a - z \\
z+a^2\overline{z} &= 2a
\end{align*}
(b) Let's set $w$ as the intersecting point. Using out solution from part a, we can say: $$w+a^2\overline{w}-2a=0=w+b^2\overline{w}-2b$$
Let's manipulate this equation:
\begin{align*}
w+a^2\overline{w}-2a &= w+b^2\overline{w}-2b \\
a^2\overline{w}-b^2\overline{w} &= 2a-2b \\
\overline{w}(a^2-b^2) & = 2(a-b) \\
\overline{w}(a+b) &= 2 \\
\overline{w} &= \frac{2}{a+b} \\
w &= \frac{2}{\overline{a}+\overline{b}}
\end{align*}
Let's take a step back and figure $\overline{a}$ and $\overline{b}$
\begin{align*}
a\overline{a} = |a|^2 \\
a\overline{a} = 1 \\
\overline{a} = \frac{1}{a}
\end{align*}
Using same logic, we can also derive $\overline{b} = \frac{1}{b}$
Now let's substitute these values in:
\begin{align*}
w &= \frac{2}{\overline{a}+\overline{b}} \\
&= \frac{2}{\frac{1}{a}+\frac{1}{b}} \\
&= \frac{2}{\frac{a+b}{ab}} \\
&= \frac{2ab}{a+b}
\end{align*}
| {
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Prove trig identity: $(\sin \theta + 1)(\sin \theta − 1) = −\cos^2 θ$ This is my attempt:
$(\sin \theta+1)(\sin \theta-1) = \sin\theta^2 - \sin\theta + \sin\theta - 1$
$= \sin^2\theta - 1$
$= -\cos^2\theta$
Is it correct, and can it be improved? Thanks!
| Yes this is correct.You must be knowing that $(x+y)(x-y)=x^2-y^2$. Therefore $(sin\theta+1)(sin\theta-1)=sin^2\theta-1=-cos^2\theta$
| {
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For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
We need to use induction.
The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:
*
*$P(n_o)$
*$( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$
Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.
Suppose,
$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.
$\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$
The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.
$(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$
$\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$
We need to demonstrate that
$(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
holds true for $n \geq 2$
The question is do I just substitute a number that is greater than two after I did the induction?
| Without Induction:
$$
\begin{align}
\prod_{k=2}^n\left(1-\frac1{k^2}\right)
&=\prod_{k=2}^n\left[\left(1-\frac1k\right)\left(1+\frac1k\right)\right]\\
&=\prod_{k=2}^n\left(\frac{k-1}{k}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\
&=\prod_{k=1}^{n-1}\left(\frac{k}{k+1}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\
&=\frac12\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k}{k+1}\right)}\cdot\frac{n+1}{n}\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k+1}{k}\right)}\\
&=\frac{n+1}{2n}
\end{align}
$$
Comment on the Induction:
The formula for the product of $n-1$ terms is is $\frac{n+1}{2n}$.
For $n=1$, the product of $0$ terms is $1$ and that is what the formula gives.
Term $n$ is $1-\frac1{(n+1)^2}$ so the product of $n$ terms is
$$
\begin{align}
\frac{n+1}{2n}\left(1-\frac1{(n+1)^2}\right)
&=\frac{n+1}{2n}\cdot\frac{n(n+2)}{(n+1)^2}\\
&=\frac{n+2}{2(n+1)}
\end{align}
$$
which is the formula for $n$ terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/663101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Time Complexity in $\theta$ Notation $$T(n) = 2T\left(\frac{n}{2}\right) + T\left(\frac{n}{4}\right) + 5$$
What is the time complexity of the given algorithm in $\theta$ notation.
Thanks in advance.
| If $T(n)=2T\left(\frac n2\right)+T\left(\frac n4\right)+5$ then one expansion results in
$$T(n)=4T\left(\frac n4\right)+2T\left(\frac n8\right)+10+2T\left(\frac n8\right)+T\left(\frac n{16}\right)+5+5\\=4T\left(\frac n4\right)+4T\left(\frac n8\right)+T\left(\frac n{16}\right)+20$$
A second expansion brings
$$T(n)=8T\left(\frac n8\right)+4T\left(\frac n{16}\right)+20+8T\left(\frac n{16}\right)+4T\left(\frac n{32}\right)+20\\+2T\left(\frac n{32}\right)+T\left(\frac n{64}\right)+5+20\\
=8T\left(\frac n8\right)+12T\left(\frac n{16}\right)+6T\left(\frac n{32}\right)+T\left(\frac n{64}\right)+65$$
So we have an increasing number of terms per expansion, and apparent-quadratic behavior in the constant term $(5\cdot 1, 5\cdot 4, 5\cdot 13,\dots)$. One more expansion should help clarify these behaviors:
$$T(n)=16T\left(\frac n{16}\right)+8T\left(\frac n{32}\right)+40+24T\left(\frac n{32}\right)+12T\left(\frac n{64}\right)+60\\+12T\left(\frac n{64}\right)+6T\left(\frac n{128}\right)+30+2T\left(\frac n{128}\right)+T\left(\frac n{256}\right)+5+65\\=
16T\left(\frac n{16}\right)+32T\left(\frac n{32}\right)+24T\left(\frac n{64}\right)+8T\left(\frac n{128}\right)+T\left(\frac n{256}\right)+200$$
If we follow the constant term further, we now have $5\cdot 1, 5\cdot 4, 5\cdot 13, 5\cdot 40.$ The pattern of these constants is emerging, and appears to be $O(n^4)$. Continuing with the analysis in this direction should provide a satisfactory conclusion.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
expression for$\sum_{k=0}^{n}(k+1)(k+2)$ what is a closed expression for$\sum_{k=0}^{n}(k+1)(k+2)$? the answer says to look at Generating function $a_k=(k+1)(k+2)$ and derive twice $\dfrac{1}{x}=1+x+x^2+...$ so the solution starts: $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$but isn't $\sum_{k=0}^{\infty}t^k=\dfrac{1}{(1-t)}$? so how the lhs and rhs in $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$ are equal?
| Using the generalization proven in this answer, and the substitutions
$$
\begin{align}
M&\mapsto n\\
m&\mapsto k\\
k&\mapsto 2\\
n&\mapsto 0
\end{align}
$$
we get
$$
\begin{align}
\sum_{k=0}^n(k+1)(k+2)
&=2\sum_{k=0}^n\binom{k+2}{2}\binom{n-k}{0}\\
&=2\binom{n+3}{3}
\end{align}
$$
An Alternate Approach
The recursive definition of the Pascal Triangle says that
$$
\binom{k+2}{2}=\binom{k+3}{3}-\binom{k+2}{3}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^n(k+1)(k+2)
&=\sum_{k=0}^n2\binom{k+2}{2}\\
&=2\left[\sum_{k=0}^n\binom{k+3}{3}-\sum_{k=0}^n\binom{k+2}{3}\right]\\
&=2\left[\sum_{k=0}^n\binom{k+3}{3}-\sum_{k=-1}^{n-1}\binom{k+3}{3}\right]\\
&=2\left[\binom{n+3}{3}-\binom{2}{3}\right]\\
&=2\binom{n+3}{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
} |
Proving that for each two parabolas, there exists a transformation taking one to the other Lets $p_1, p_2$ be two prabolas on the plane. Prove there is a transformation $T:\mathbb R^2 \rightarrow \mathbb R^2$, $T(\vec x)=A \vec x+\vec b$, $A$ being a multiplication of scalar and orthogonal matrices and $\vec b, \vec x \in \mathbb R^2$, taking $p_1$ to $p_2$.
I have no idea how to prove this really... Thanks in advance for any help!
| It is claimed that a parametrization of the general 2-D parabola is given by:
$$
x(t) = \frac{1}{2} a_x.t^2 + v_x.t + s_x \\
y(t) = \frac{1}{2} a_y.t^2 + v_y.t + s_y
$$
Let this be the parabola $p_1$ . And let the proposed linear transformation be given by:
$$
\left[\begin{array}{c} x' \\ y' \end{array}\right] =
\left[\begin{array}{cc} a_{11} & a_{12} \\
a_{21} & a_{22} \end{array}\right]
\left[\begin{array}{c} x \\ y \end{array}\right] +
\left[\begin{array}{c} b_x \\ b_y \end{array}\right]
$$
Giving:
$$
x' = a_{11}\left[ \frac{1}{2} a_x t^2 + v_x t + s_x \right] +
a_{12}\left[ \frac{1}{2} a_y t^2 + v_y t + s_y \right] + b_x\\
y' = a_{21}\left[ \frac{1}{2} a_x t^2 + v_x t + s_x \right] +
a_{22}\left[ \frac{1}{2} a_y t^2 + v_y t + s_y \right] + b_y
$$
Or:
$$
x' = \frac{1}{2} \left[ a_{11} a_x + a_{12} a_y \right] t^2 +
\left[ a_{11} v_x + a_{12} v_y \right] t +
\left[ a_{11} s_x + a_{12} s_y + b_x \right] \\
y' = \frac{1}{2} \left[ a_{21} a_x + a_{22} a_y \right] t^2 +
\left[ a_{21} v_x + a_{22} v_y \right] t +
\left[ a_{21} s_x + a_{22} s_y + b_y \right]
$$
Or:
$$
x' = \frac{1}{2} a'_x t^2 + v'_x t + s'_x \\
y' = \frac{1}{2} a'_y t^2 + v'_y t + s'_y
$$
Let this be the parabola $p_2$ . Then we have:
$$
a_x a_{11} + a_y a_{12} = a'_x \\
v_x a_{11} + v_y a_{12} = v'_x \\
a_x a_{21} + a_y a_{22} = a'_y \\
v_x a_{21} + v_y a_{22} = v'_y \\
$$
From these the matrix coefficients can be solved:
$$
a_{11} = \frac{v_y a'_x - a_y v'_x}{a_x v_y - v_x a_y} \\
a_{12} = \frac{a_x v'_x - v_x a'_x}{a_x v_y - v_x a_y} \\
a_{21} = \frac{v_y a'_y - a_y v'_y}{a_x v_y - v_x a_y} \\
a_{22} = \frac{a_x v'_y - v_x a'_y}{a_x v_y - v_x a_y}
$$
Provided that the denominators $(a_x v_y - v_x a_y)$ are nonzero,
meaning that the parabolas are not degenerated i.e. they are no straight lines. Then, at last, the displacement vector is found by:
$$
b_x = s'_x - \left[a_{11} s_x + a_{12} s_y\right] \\
b_y = s'_y - \left[a_{21} s_x + a_{22} s_y\right]
$$
In the article Parabolic Curves
it is shown how to eliminate the parameter $t$ to get more familiar representation of the parabolas.
Extra. Let's start again with the above parametrization, written as:
$$
x - s_x = v_x \cdot t + a_x \cdot \frac{1}{2} t^2 \\
y - s_y = v_y \cdot t + a_y \cdot \frac{1}{2} t^2
$$
Now put:
$$
\xi = t \quad ; \quad \eta = \frac{1}{2} t^2 \quad \Longrightarrow \quad
\eta = \frac{1}{2} \xi^2
$$
And:
$$
x - s_x = v_x \, \xi + a_x \, \eta \\
y - s_y = v_y \, \xi + a_y \, \eta
$$
Two equations with two unknowns:
$$
\xi = \frac{+ a_y (x-s_x) - a_x (y-s_y)}{v_x a_y - v_y a_x} \qquad ; \qquad
\eta = \frac{- v_y (x-s_x) + v_x (y-s_y)}{v_x a_y - v_y a_x}
$$
Substitute into the "normed parabola" $\;\eta = \frac{1}{2} \xi^2\;$ or $\;\frac{1}{2} \xi^2 - \eta = 0$ :
$$
\frac{1}{2} \left[ \frac{a_y (x-s_x) - a_x (y-s_y)}{a_y v_x - a_x v_y}
\right]^2 + \left[ \frac{v_y (x-s_x) - v_x (y-s_y)}{a_y v_x - a_x v_y}
\right] = 0
$$
Which is the equation of an arbitrary parabola in $x$ and $y$ alone.
Update.
But, if I read the question well, it is required that the transformation
taking the parabola $p_1$ into $p_2$ is represented by a scalar times an
orthogonal matrix. This is most easily achieved by a parameter shift in:
$$
x(t) = \frac{1}{2} a_x.t^2 + v_x.t + s_x \\
y(t) = \frac{1}{2} a_y.t^2 + v_y.t + s_y
$$
As follows:
$$
\overline{x}(t) = \frac{1}{2} a_x(t-\tau)^2 + v_x(t-\tau) + s_x \\
\overline{y}(t) = \frac{1}{2} a_y(t-\tau)^2 + v_y(t-\tau) + s_y
$$
$$
\overline{x}(t) = \frac{1}{2} a_x.t^2+\left(v_x-a_x.\tau\right)\,t
+\left(\frac{1}{2}a_x.\tau^2-v_x.\tau+s_x\right) \\
\overline{y}(t) = \frac{1}{2} a_y.t^2+\left(v_y-a_y.\tau\right)\,t
+\left(\frac{1}{2}a_y.\tau^2-v_y.\tau+s_y\right)
$$
$$
\overline{x}(t) = \frac{1}{2} a_x.t^2 + \overline{v}_x.t + \overline{s}_x \\
\overline{y}(t) = \frac{1}{2} a_y.t^2 + \overline{v}_y.t + \overline{s}_y
$$
Now it's always possible to choose $\tau$ in such a way that $(\overline{v}_x,\overline{v}_y)$ is perpendicular to $(a_x,a_y)$ :
$$
a_x.\left(v_x-a_x.\tau\right) + a_y \left(v_y-a_y.\tau\right) = 0 \quad \Longrightarrow \quad
\tau = \frac{v_x a_x + v_y a_y}{a^2_x + a^2_y}
$$
This means that the local coordinate systems $(\vec{v},\vec{a})$ of our parabolas have become orthogonal.
And a transformation that takes an orthogonal coordinate system into another orthogonal coordinate system is itself orthogonal (apart from some scalar eventually).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Four real roots of $x^4+2x^3+mx^2+2x+1=0$ iff $m$ is... The Equation $x^4+2x^3+mx^2+2x+1=0$ has $4$ different real roots iff:
a) $m<3$;
b) $m<2$;
c) $m<-6$;
d) $1<m<3$;
e) $-6<m<2$
| Notice that $x=0$ is not going to be a solution. Then notice the coefficients are symmetric. This means that if we divide by $x^{4/2}$ we get $$(x^2+1/x^2)+2(x+1/x)+m=0$$
which can be written as
$$(x+1/x)^2+(x+1/x)+m-2=0.$$
Putting $y:=x+1/x$ we get $$y^2+y+m-2=0$$
which has two different real solutions for $1-4(m-2)>0$, i.e. $9>4m$. For these we have $x+1/x=y=\frac{-1\pm\sqrt{1-4(m-2)}}{2}$.
These gives us two polynomials of degree two:
$$x^2-\frac{-1\pm\sqrt{1-4(m-2)}}{2}x+1.$$
To get two different real solutions out of these we need again positive discriminant $\left(\frac{-1\pm\sqrt{1-4(m-2)}}{2}\right)^2-4>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can you prove $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$? Show that $$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$$
This can be shown through expansion but there is a more elegant solution
I cannot discover anything I would consider elegant. Can anyone help?
| $$ a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - bc - ca - ab) $$
Note that this is the determinant of
$$ M \; = \;
\left( \begin{array}{rrr}
a & b & c \\
c & a & b \\
b & c & a
\end{array}
\right),
$$
the matrix being evidently singular when $a=b=c$ or when $a+b+c=0.$
Meanwhile, with the evident possibility of permuting the letters,
$$ (a^2 + b^2 + c^2 - bc - ca - ab) = \frac{1}{4} \left( \, (a+b-2c)^2 + 3 (a-b)^2 \, \right) $$
and so is positive semidefinite only, it comes out to $0$ when $a=b=c.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
$\large \lim\limits_{x\to 2}\frac{x-2}{2 + \log_2(3) - x -\log_2(2x-1)} $ determination The limit of
$\lim\limits_{x\to2}\dfrac{x-2}{\;2 + \log_2(3) - x - \log_2(2x-1)} $ is $\dfrac{-4}{\ln (2e)}$
i need to show that. I had try this:
*
*rearrange the denominator
$\lim\limits_{x\to2}\dfrac{x-2}{-(x-2) + \log_2(3) - \log_2(2x-1)} $
*dividing all stuff by (x-2)
$\lim\limits_{x\to2}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2x-1)}{x-2}} $
*change variable $x-2=u$ so $u\to0$
$\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{\log_2(3) - \log_2(2u+3)}{u}} $
*simplify
$\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{- \log_2(2/3u+1)}{u}} $
5 changing the log base to e
$\lim\limits_{u\to0}\dfrac{1}{-1 + \dfrac{- \ln(2/3u+1)}{2/3u}\times \dfrac{2}{3 \ln 2}}
$
6.the final solution is $\dfrac{1}{-1 - 1\times \dfrac{2}{3 \ln 2}} = (-3/2) \ln2
$
this is not the correct answer. Why? where is my mismatch.
| The limit is much easier to calculate if we recall the definition of derivative: $$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}.$$ Then with $a = 2$, and $f(x) = \log_2 (2x-1),$ we obtain $$\lim_{x \to 2} \frac{\log_2(2x-1) - \log_2 3}{x-2} = \frac{d}{dx}\left[\log_2 (2x-1) \right]_{x=2} = \left[\frac{2}{(2x-1)\log 2}\right]_{x=2} = \frac{2}{3 \log 2}.$$ It follows that $$\begin{align*} \lim_{x \to 2} \frac{x-2}{(2-x) + \log_2 3 - \log_2 (2x-1)} &= -(1 + f'(2))^{-1} \\ &= -\left( 1 + \frac{2}{3 \log 2} \right)^{\!-1} \\ &= - \frac{3 \log 2}{2 + 3 \log 2}. \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667960",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$ First we need to prove the basis. If we let $n=3$, then $(1+ \frac{1}{3})^3 < 3$
$(\frac{3}{3}+ \frac{1}{3})^3 < 3$
$(\frac{4}{3})^3 < 3$
$(\frac{64}{27}) < 3$
The inequality statement is true
For $P(n), (1+ \frac{1}{n})^n < n$
We assume that $(1+ \frac{1}{n})^n < n$ is true for $P(n+1)$
$(1+ \frac{1}{n+1})^{n+1} < n+1$
$(1+ \frac{1}{n+1})^{n})(1+ \frac{1}{n+1})^{1}) < n+1$
And then I'm stuck afterwards. I know that there are a variety of problems that use induction and they have different methods, but I only know the ones that are similar to $1+2+3+...+n = n+2$ or $7^n-8^n$ is divisible by $8$. Is there any technique to tackle this type of problem?
| The following inequality will be needed:
$$\frac{1}{n+1}<\frac{1}{n} \Leftrightarrow 1+\frac{1}{n+1}<1+\frac{1}{n}\\
\Leftrightarrow \left(1+\frac{1}{n+1}\right)^n<\left(1+\frac{1}{n}\right)^n.$$
From the induction hypothesis $\left(1+\frac{1}{n}\right)^n<n$ and the algebraic identity $\left(1+ \frac{1}{n+1}\right)^{n+1} = \left(1+ \frac{1}{n+1}\right)^n\left(1+ \frac{1}{n+1}\right)$,
$$\Rightarrow\left(1+\frac{1}{n+1}\right)^n<n\\
\Leftrightarrow \left(1+ \frac{1}{n+1}\right)^n\left(1+ \frac{1}{n+1}\right)<n\left(1+ \frac{1}{n+1}\right)\\
\Leftrightarrow\left(1+ \frac{1}{n+1}\right)^{n+1}<n+\frac{n}{n+1}$$
Can you take it from there?
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Do there exist 4 rationals satisfying $a^2+b^2+c^2+d^2=1$ and $2a+b+c+d=0$?
Does there exist 4 rationals $(a,b,c,d)$ which satisfy the following two relations?,
$a^2+b^2+c^2+d^2=1$ and $2a+b+c+d=0$?
I spent a lot of time with it and tried the criteria of quadratic equations having rational root but it doesn't seem to be working! Any help or comments would be appreciated , Thanks!
| From the second equation, $a=-(1/2)(b+c+d)$ which may be put into the first, and after multiplying by $4$ we have $(b+c+d)^2+4b^2+4c^2+4d^2=4.$ Now since the $4$ on the right is $2^2$ we may divide the equation by $4$, at the same time resetting $b,c,d$ to respectively $b/2,c/2,d/2.$ [so we do not divide the coefficients $4$ by $4$, rather we divide the variables].Thus we have arrived at
$$(b+c+d)^2+4b^2+4c^2+4d^2=1.\tag{1}$$
Now consider the equation $w^2+4x^2+4y^2+4z^2=1,$ which has the rational solution $(w,x,y,z)=(1,0,0,0).$ We carry out the usual steps to get a rational parametrization of the solutions: Put $w=1-k,\ x=rk,\ y=sk,\ z=tk.$ This leads to $k=2/(1+4r^2+4s^2+4t^2),$ and then if we let $D=1+4r^2+4s^2+4t^2$ we have
$w=(D-2)/D,x=2r/D,y=2s/D,z=2t/D.$ Now comparing to $(1)$ we see we want a solution for which $w=x+y+z,$ which leads to
$$4r^2+4s^2+4t^2-2r-2s-2t=1.\tag{2}$$
Now let $a=4r-1,b=4s-1,c=4t-1,$ and note that $(2)$ is equivalent to $a^2+b^2+c^2=7.$ But this is not possible for rational $a,b,c$ as it leads, on clearing denominators, to an integer equation of the form $p^2+q^2+r^2=7s^2,$ and as is known an integer of the form $7s^2$ is not the sum of three squares.
Added: The usual theorem for a number $t$ to not be the sum of three squares is that $t$ be of the form $4^n(8m+7).$ Now if the above $7s^2$ were the sum of three squares, one could write $s=2^u\cdot w$ with $w$ odd. Then $s^2=4^u\cdot w^2,$ and since $w$ is odd, $w^2$ is $1$ mod 8, which makes $7w^2$ of the form $8m+7$, and thus $7s^2=4^u(8m+7)$ so is not the sum of three squares.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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evaluate $\int\frac{3x}{\sqrt{1-2x}}dx$ I'm trying to evaluate $\int\frac{3x}{\sqrt{1-2x}}dx$
This is what I got so far:
Let $u$ = $1-2x$
$x$ = $\frac{u-1}{-2}$
$du$ = $-2$ $dx$
$\frac{-du}{2}$ = $dx$
Therefore,
$\int\frac{3\frac{u-1}{-2}}{\sqrt{u}}\frac{-du}{2}$
$\int\frac{3(u-1)}{-2\sqrt{u}}\frac{-du}{2}$
$\frac{3}{4}\int\frac{(u-1)}{\sqrt{u}}{du}{}$
$\frac{3}{4}\int\frac{u}{{u^\frac{1}{2}}}-\frac{1}{{u^\frac{1}{2}}}{du}$
$\frac{3}{4}\int u^\frac{1}{2} -u^\frac{-1}{2}{du}$
$\frac{3}{4}[\frac{2}{3}u^\frac{3}{2} -2u^\frac{1}{2}]+C$
$\frac{6}{12}u^\frac{3}{2} -\frac{6}{4}u^\frac{1}{2}+C$
$\frac{1}{2}u^\frac{3}{2} -\frac{3}{2}u^\frac{1}{2}+C$
$\frac{1}{2}u^\frac{3}{2} -\frac{3}{2}u^\frac{1}{2}+C$
$\frac{u^\frac{3}{2}-3u^\frac{1}{2}}{2}+C$
$\frac{\sqrt{u^3}-3\sqrt{u}}{2}+C$
$\frac{\sqrt{(1-2x)^3}-3\sqrt{1-2x}}{2}+C$
The answer I'm supposed to get is $-\sqrt{1-2x} {(x+1)}+C$
| The answer is
$$\sqrt{(1-2x)^3}=\sqrt{(1-2x)(1-2x)^2}=\sqrt{(1-2x)}\sqrt{(1-2x)^2}=\color{red}{\sqrt{(1-2x)}}(1-2x)
\\=\frac{\color{red}{\sqrt{(1-2x)}}(1-2x)-3\color{red}{\sqrt{1-2x}}}{2}+C
\\=\sqrt{1-2x}[\frac{(1-2x)-3}{2}]+C
\\=-\sqrt{1-2x}[x+1]+C$$
this is the correct answer, and your book has a typo.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
For what values of $x$ does $A^{-1}$ exist? Let $A = \begin{bmatrix}4&-1&2\\5&x&7\\x&-1&3\end{bmatrix}$
I'm trying to find the values of $x$ such that $A^{-1}$ exists.
What I have tried:
$4(3x + 7) - (-1)(15)(7x) + (2)(-5)(x^2) = 0$
$-10x^2 + 117x + 28 = 0$
but I don't think that's right!
| Note that
\begin{align*}
\det A&=4\cdot\left(3x+7\right)-(-1)\cdot\left(15-7x\right)+2\cdot\left(-5-x^2\right)\\
&=-2x^2+5x+33 \\
&=-2(x+3)\left(x-\frac{11}{2}\right)
\end{align*}
Hence $A$ is invertible if and only if $x\neq -3$ and $\displaystyle x\neq\frac{11}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra - Determine if the set of matrices in $M_{22}$ is linearly independent or linearly dependent Is the set of matrices $\{\begin{bmatrix}0&1\\0&1\end{bmatrix},\begin{bmatrix}1&1\\2&2\end{bmatrix},\begin{bmatrix}1&4\\2&2\end{bmatrix},\begin{bmatrix}5&4\\2&6\end{bmatrix}\}$ in $M_{22}$ linearly independent or linearly dependent?
For some reason I'm stuck.
What I have done:
$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = 0$
What I was thinking of doing is:
$A_1(det(A_1)) +A_2(det(A_2) + A_3(det(A_3)) + A_4(det(A_4)) = 0$
I don't know if that's right but I do know I'm on the right track! Any help would be much appreciated!
| Looking into the determinant in this way is heading in the wrong direction. The determinant is not preserved under addition. To illustrate:
$$
\overbrace{\begin{bmatrix} 2 & 2 \\ -1 & 2 \\ \end{bmatrix}}^{\text{det}=6}
+
\overbrace{\begin{bmatrix} 0 & -1 \\ 1 & -2 \\ \end{bmatrix}}^{\text{det}=1}
=
\overbrace{\begin{bmatrix} 2 & 1 \\ 0 & 0 \\ \end{bmatrix}}^{\text{det}=0}.$$
To continue, notice that the system of equations $$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$$
is equivalent to
\begin{align*}
A_2+A_3+5A_4 &= 0 \\
A_1+A_2+4A_3+4A_4 &= 0 \\
2A_2+2A_3+2A_4 &= 0 \\
A_1+2A_2+2A_3+6A_4 &= 0.
\end{align*}
We can solve this in the usual way (i.e., Gaussian elimination).
Alternatively, we write the matrices as the rows of a $4 \times 4$ matrix
$$
\begin{bmatrix}
0 & 1 & 0 & 1 \\
1 & 1 & 2 & 2 \\
1 & 4 & 2 & 2 \\
5 & 4 & 2 & 6 \\
\end{bmatrix}.
$$
If the rows of this matrix are linearly independent (which occurs if and only if the matrix has non-zero determinant), then the original set of matrices is linearly independent.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help mathematical induction Prove by mathematical induction. I was hoping if someone could give me a hint on how to solve this problem.
$$\frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{n^2} < 2 - \frac {1}{n} $$ for all integers $n\geq 2$.
| *
*check this for $k=2$
*suppose it is true for $n=k$ i.e.
$$\frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{k^2} < 2 - \frac {1}{k}$$
*Prove it for $n=k+1$ using 2
$$ \implies \frac{1}{1^2}+ \frac {1}{2^2} + ....+ \frac{1}{k^2}+\frac{1}{(k+1)^2} < 2 - \frac {1}{k}+\frac{1}{(k+1)^2}
\\ =2 -( \frac {1}{k}-\frac{1}{(k+1)^2})
\\=2 -\frac{k^2+k+1}{k(k+1)^2}
\\ < 2 -\frac{k^2+k}{k(k+1)^2}
\\= 2 -\frac{1}{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/673220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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computing $\lim_{X \to \infty }((x^3+2x)^\frac{1}{3}-\sqrt{x^2-2x})$ $\lim_{X \to \infty }((x^3+2x)^\frac{1}{3}-\sqrt{x^2-2x})$
Anyone have a nice elegent way wo solve it? i tried L'Hôpital's rule but didn't work.
also i tried to multiple by Conjugation and it failed too.
thanks
| We have:
$$\sqrt[3]{x^3+2x}-x= x\Bigl(\sqrt[3]{1+\frac 2{x^2}}-1\Bigr)\sim \frac 2{3x}\to 0$$
$$\sqrt{x^2-2x}-x= x\Bigl(\sqrt{1-\frac 2x}-1\Bigr)\to -1$$
Consequently:
$$\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=(\sqrt[3]{x^3+2x}-x)-(\sqrt{x^2-2x}-x)\to 1$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spivak's Calculus - Fibonacci Sequence In one of the questions in Spivak's Calculus he defines a Fibonacci sequence $a_1$, $a_2$, $a_3$ as:
$$a_1 = 1$$
$$a_2 = 1$$
$$a_n = a_{n-1} + a_{n-2}$$
for $n\ge3$
and proves $$a_n = \frac{(\frac{1+\sqrt5}{2})^n - (\frac{1-\sqrt5}{2})^n}{\sqrt{5}}$$
According to his proof:
But from his proof I don't quite understand how he got from the second line to the third and from the third to the fourth. What throws me off here is where he got the $(1+\frac{1+\sqrt5}{2})$ from and the sudden n, $(1+\frac{1-\sqrt5}{n})$ under the last fraction. For someone who understands what Spivak did, could you please explain this to me, thanks.
| Going from the second line to the third line, he uses that
\begin{align*}
\left(\frac{1 + \sqrt{5}}{2}\right)^2 &= \frac{1 + 2 \sqrt 5 + 5}{4} \\
&= \frac{4 + (2 + 2 \sqrt 5)}{4} \\
&= 1 + \frac{1 + \sqrt 5}{2}
\end{align*}
Going from line three to four is just using the fact that
$$a^{n - 2} a^2 = a^{n - 2 + 2} = a^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Writing $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\frac1x+\frac1y+\frac1z$ Can we write $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\dfrac1x+\dfrac1y+\dfrac1z$?
What about $x^3+y^3+z^3$?
| Yes, by the Fundamental Theorem of Symmetric Polynomials since $x^2+y^2+z^2$ is symmetric and $xy+yz+xz=xyz*(1/x+1/y+1/z).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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evaluation of $\int\frac{1}{(1+x)(2+x)^2(3+x)^3}dx$ Evaluation of $\displaystyle \int\frac{1}{(1+x)(2+x)^2(3+x)^3}dx$
$\bf{My\; Trial \; Solution::}$ Given $\displaystyle \int\frac{1}{(1+x)(2+x)^2(3+x)^3}dx\;\;\;,$ Let $(3+x)=y\;,$ Then $dx = dy$
So Integral Convert into $\displaystyle \int\frac{1}{(y-2)(y-1)^2y^3}dy\;\;,$ Now Let $\displaystyle y = \frac{1}{z}\;\;,$ Then $\displaystyle dy = -\frac{1}{z^2}dz$
So Integral convert into $\displaystyle \int\frac{z^6}{(2z-1)(z-1)^2}\cdot \frac{1}{z^2}dz = \int\frac{z^4}{(2z-1)(z-1)^2}dz$
Now How can I solve after that
Help Required
Thanks
| Use partial fraction decomposition to decompose the expression into (I used the Wolf):
$$\frac{1}{(x+1) (x+2)^2 (x+3)^3} = \frac{2}{x+2}-\frac{17}{8}\frac{1}{x+3}-\frac{1}{(x+2)^2}-\frac{5}{4}\frac{1}{(x+3)^2}-\frac{1}{2}\frac{1}{(x+3)^3}+ \frac{1}{8}\frac{1}{x+1}$$
and then integrate directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
integral computation $\int_{-\infty}^{\infty} \frac{1}{(1+x+x^2)^2} dx $ Compute the following integral:
$\int_{-\infty}^{\infty} \frac{1}{(1+x+x^2)^2} dx$. Can some one give me some hints on how to do this?
I tried writing $(1+x+x^2)=f(x)$ and then multiplying and dividing by $\frac{d f(x)}{dx}=1+2x$ so we get $\int_{-\infty}^{\infty} \frac{1}{1+2x} \frac{1+2x}{(f(x))^2} dx $ = $\int_{-\infty}^{\infty} \frac{1}{1+2x} \frac{df(x)}{(f(x))^2} $. Here I tried to use integration by parts but I can't get anywhere.
| Let
$$\mathcal{I}(\lambda) = \int_{-\infty}^{\infty} \frac{1}{x^2+x+\lambda}\;{dx} $$
where
$\lambda > 0$. We have
$$x^2+x+\lambda = (x+\frac{1}{2})^2+(\lambda-\frac{1}{4})\ ,$$
thus by setting $t = {x+\frac{1}{2}}$ we have
$$\mathcal{I}(\lambda) = \frac{2}{\sqrt{4\lambda-1}}\int_{-\infty}^{\infty}\frac{1}{1+t^2}\;{dt} = \frac{2\pi}{\sqrt{4\lambda-1}}\ .$$
Differentiating we get
$$\mathcal{I}'(\lambda) = -\frac{4\pi}{\sqrt{(4\lambda-1)^3}}$$
and
$$-\mathcal{I}'(1) = \frac{4\pi}{3\sqrt{3}}\ .$$
See here if you haven't seen this method before.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to prove that $\sum_{n=1}^{+\infty}\frac{1}{n^2+1}=\frac{-1+\pi \coth (\pi)}{2}$? I typed into my Mathematica:$\sum _{n=1}^{\infty } \frac{1}{n^2+1}$ , and the result was: $$\frac{-1+\pi \coth (\pi)}{2}$$ I know how to estimate the aforementioned sum , but I have no idea how to get its closed form.
| Knowing the partial fraction decomposition of the cotangent [a standard (because it's easy to derive and very useful) example in complex analysis texts illustrating the Mittag-Leffler theorem],
$$\pi \cot \pi z = \frac{1}{z} + \sum_{n\neq 0} \frac{1}{z-n} + \frac{1}{n} = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n},$$
we can compute the sum of all series of the form
$$\sum_{n=1}^\infty \frac{1}{n^2 - a^2}$$
where $a$ is not an integer:
$$\begin{align}
\sum_{n=1}^\infty \frac{1}{n^2-a^2}
&= \lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n^2-a^2}\\
&= -\frac{1}{2a}\lim_{N\to\infty} \sum_{n=1}^N \frac{1}{a-n} + \frac{1}{a+n}\\
&= -\frac{1}{2a}\lim_{N\to\infty} \left(-\frac{1}{a} +\sum_{n=-N}^N \frac{1}{a-n}\right)\\
&= -\frac{1}{2a}\left(\pi \cot \pi a - \frac{1}{a}\right)\\
&= \frac{1-\pi a\cot \pi a}{2a^2}.
\end{align}$$
To obtain the terms $\frac{1}{n^2+1}$ we choose $a = i$ and get
$$\sum_{n=1}^\infty \frac{1}{n^2+1} = \frac{1 - \pi i \cot \pi i}{2i^2}.$$
Using $i^2 = -1$ in the denominator and
$$\cot \pi i = \frac{\cos \pi i}{\sin \pi i} = \frac{\cosh \pi}{i\sinh\pi} = \frac{1}{i}\coth \pi,$$
the result becomes
$$\sum_{n=1}^\infty \frac{1}{n^2+1} = \frac{\pi \coth\pi - 1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For the Fibonacci sequence prove that $\sum_{i=1}^n F_i= F_{n+2} - 1$
For the Fibonacci sequence $F_1=F_2=1$, $F_{n+2}=F_n+F_{n+1}$, prove that $$\sum_{i=1}^n F_i= F_{n+2} - 1$$ for $n\ge 1$.
I don't quite know how to approach this problem.
Can someone help and explain please?
| Just for fun, let us try to prove $\sum\limits_{k=1}^n F_k= F_{n+2} - 1$ using Binet's formula.
Recall that we have
$$F_n=\frac{a^n-b^n}{\sqrt5}$$
where $a=\frac{1+\sqrt5}2$ and $b=\frac{1-\sqrt5}2$.
A few identities, that are easy to check and which may be useful:
$$
\begin{align*}
a+b&=1\\
a-b&=\sqrt5\\
ab&=-1\\
1+a&=a^2\\
1+b&=b^2
\end{align*}
$$
(Most of these identities simply follow from noticing that $a$, $b$ are roots of $x^2-x-1$ and using Vieta's formulas).
Now we want to compute
$$\sum\limits_{k=1}^n F_k=
\sum\limits_{k=1}^n \frac{a^k-b^k}{\sqrt5}=
\frac 1{\sqrt5} \left(\sum\limits_{k=1}^n a^k- \sum\limits_{k=1}^n b^k\right)
$$
Using the formula $1+x+\dots+x^n = \frac{x^{n+1}-1}{x-1}$ we get
$$
\begin{gather*}
\frac1{\sqrt5} \left(\frac{a^{n+1}-1}{a-1}-\frac{b^{n+1}-1}{b-1}\right) = \\
\frac1{\sqrt5} \cdot \frac{(a^{n+1}-1)(b-1)-(b^{n+1}-1)(a-1)}{(a-1)(b-1)} = \\
\frac1{\sqrt5} \cdot \frac{a^{n+1}b-a^{n+1}-b-b^{n+1}a+b^{n+1}+a}{ab-a-b+1} = \\
\frac1{\sqrt5} \cdot \frac{-(a^{n+1}-b^{n+1})+ab(a^n-b^n)+(a-b)}{ab-(a+b)+1} = \\
\frac1{\sqrt5} \cdot \frac{-(a^{n+1}-b^{n+1})-(a^n-b^n)+\sqrt5}{-1-1+1} = \\
\frac{-(a^{n+1}-b^{n+1})-(a^n-b^n)+\sqrt5}{-\sqrt5}= \\
\frac{(a^{n+1}-b^{n+1})+(a^n-b^n)-\sqrt5}{\sqrt5}= \\
\frac{a^{n+1}-b^{n+1}}{\sqrt5}+\frac{a^{n}-b^{n}}{\sqrt5}-\frac{\sqrt5}{\sqrt5} = \\
F_{n+1}+F_n-1 = F_{n+2}-1
\end{gather*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$2\sin x=3\cot x$ from $0^\circ$ to $360^\circ$ I used two slightly different approaches to solve this. First approach gives 2 correct solutions, second approach gives 4 solutions of which 2 are correct and 2 wrong, I just cannot figure out why I'm getting 2 extra wrong answers with second approach.
| When you write $$2\sin^2 x=3\sqrt{1-\sin^2 x}$$ $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ you're really saying $$2\sin^2 x=3\sqrt{1-\sin^2 x}\Rightarrow\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x$$ It's false that $$\left(\frac{2\sin^2 x}{3}\right)^2=1-\sin^2 x\Rightarrow 2\sin^2 x=3\sqrt{1-\sin^2 x}$$ because $f(x)=x^2$ isn't an injective function. From that, you get that all the solutions to the first equation are solutions to the last equation, but not all the solutions to the last equation are solutions to the first equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/682003",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.