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How to prove the identity involving Sinc-series? Here the Sinc function is defined as: ${\rm sinc} (x):= \sin(x)/x$.
I found the following identity by numerical experiments, but how to prove it?
Let:$$
f(x;x_0):={\rm sinc}\ x_0+\sum_{n=1}^\infty {\rm sinc}\ (nx+x_0)+\sum_{n=1}^\infty {\rm sinc}\ (nx-x_0), 0<x<2\pi, x_0\in\mathbf{R}
$$
We have:
$$
f(x;x_0)\equiv f(x;0)
$$
which means the function $f(x;x_0)$ has nothing to do with $x_0$ at all!
Anyone can help me?
| I believe instead that $f$ depends on the value of $x_0$. Here is what I thought: let $x = \pi \in (0,2\pi)$ and let $x_0 \in (0, \pi)$.
For $N \geq 1$, we have:
\begin{align}
\sum_{k = 1}^N \frac{\sin (k \pi + x_0)}{k \pi + x_0} + \frac{\sin (k \pi - x_0)}{k \pi - x_0}
&= \cos x_0 \cdot \left( \sum_{k = 1}^{N} ({-1})^{k} \frac{1}{k \pi + x_0} - \sum_{k = 1}^N ({-1})^k \frac{1}{k \pi - x_0} \right) \\
&= - 2 x_0 \cdot \cos x_0 \cdot \sum_{k = 1}^N ({-1})^k \frac{1}{k^2 \pi^2 - x_0^2} \\
&= - 2 x_0 \cdot \cos x_0 \cdot S_N (x_0)
\end{align}
For all $k \geq 1$, the convergent series for even and odd indices satisfy
\begin{align}
0 < \sum_{k \geq 0} \frac{1}{(2k + 1)^2 \pi^2 -x_0^2} &< \sum_{k \geq 0} \frac{1}{(2k + 1)^2 \pi^2} = \frac{1}{8} \\
0 < \sum_{k \geq 1} \frac{1}{4k^2 \pi^2 - x_0^2} &< \sum_{k \geq 1} \frac{1}{4k^2 \pi^2} = \frac{1}{24}
\end{align}
so that
\begin{equation}
\frac{1}{8} > S_N (x_0) > - \frac{1}{8}
\end{equation}
Taking $y \in (0, \pi)$ close enough to $\pi$ so that $1 / 8 < \mathrm{sinc}\ y < 1 - 1 / 8$, we find that:
\begin{align}
0 &< f (\pi ; y) < 1 \\
f (\pi ; 0) &= \mathrm{sinc}\ 0 = 1
\end{align}
so that $f (\pi ; y) \neq f (\pi ; 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \frac{(x^2-1)^4-3x^2(x^2-1)^2+x^4}{x[(x^2-1)^3-x^2(x^2-1)]}=1$
how can i solve 8 degree equation
| Notice that $f$ is an odd function $f(-x)=-f(x)$.
We also have the relation $f(\frac 1x)=-f(x)=f(-x)$
From these two we can deduce $$f(f(f(-\frac 1x)))=f(f(f(x)))$$
Whenever $x$ is solution of $f^{\circ[3]}(x)=1$ then $-\frac 1x$ is solution too, so we can restrict our search to only positive roots.
The equation $f(x)=a$ is just a quadratic with $\Delta=a^2+4>0$ so the two roots of $f(x)=a$ are real and are given by $h(a)$ and $-\dfrac 1{h(a)}$.
Where $h(a)=\dfrac{a+\sqrt{a^2+4}}2\quad$ gives the positive root of the equation.
We can notice that $h(1)=\phi$ and $-\dfrac{1}{h(1)}=-\frac 1{\phi}=1-\phi$
Similarly we have $-\dfrac 1{h(a)}=a-h(a)$, this will be useful to express the roots of $g$ in a rationalized form (i.e. without square roots on the denominator).
We have the equation $$f(f(f(x)))=1\iff \begin{cases} f(z)=1\\f(y)=z\\f(x)=y\end{cases}$$
The $8$ solutions are $\{a,b,c,d,-\frac 1a,-\frac 1b,-\frac 1c,-\frac 1d\}$
$\begin{cases}
a=h(h(h(1)))\\
b=h(h(-1/h(1)))\\
c=h(-1/h(h(1)))\\
d=h(-1/h(-1/h(1)))
\end{cases}\quad$ or in the "rationalized" form $\quad\begin{cases}
a=h(h(h(1)))\\
b=h(h(1-h(1)))\\
c=h(h(1)-h(h(1)))\\
d=h(1-h(1)-h(1-h(1)))
\end{cases}$
This is more than enough to get numerical values for the roots:
$$a \approx 2.495944000\quad b\approx 1.434666438\quad c\approx 0.789447954\quad d\approx 0.530245300$$
Using the rationalized form you can get these formal solutions while plugging them into a CAS:
$\begin{cases}
a = \frac 18\left(1+\sqrt{5}+\sqrt{22+2\sqrt{5}}+\sqrt{92+4\sqrt{5}+2\sqrt{22+2\sqrt{5}}+2\sqrt{5}\sqrt{22+2\sqrt{5}}}\right)\\
b = \frac 18\left(1-\sqrt{5}+\sqrt{22-2\sqrt{5}}+\sqrt{92-4\sqrt{5}+2\sqrt{22-2\sqrt{5}}-2\sqrt{5}\sqrt{22-2\sqrt{5}}}\right)\\
c = \frac 18\left(1+\sqrt{5}-\sqrt{22+2\sqrt{5}}+\sqrt{92+4\sqrt{5}-2\sqrt{22+2\sqrt{5}}-2\sqrt{5}\sqrt{22+2\sqrt{5}}}\right)\\
d = \frac 18\left(1-\sqrt{5}-\sqrt{22-2\sqrt{5}}+\sqrt{92-4\sqrt{5}-2\sqrt{22-2\sqrt{5}}+2\sqrt{5}\sqrt{22-2\sqrt{5}}}\right)\\
\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$
I have no idea where to go from there.
| Hint:
$$\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2$$
$$1+\cos3x=2\cos^2\dfrac{3x}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}$$
Let $x=\frac{P}{\sqrt{a}}$, $y=\frac{P}{\sqrt{b}}$,$z=\frac{P}{\sqrt{c}}$
Then we have $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
By $AM \ge HM$
$$\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Hence
$$x+y+z \ge 9$$
Any way to proceed further?
| $$\sqrt{b}+\sqrt{c}=\sqrt{b+c+2\sqrt{bc}}>\sqrt{b+c}>\sqrt{a},$$
which says that all denominators are positives.
Now, by C-S $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}\frac{a}{\sqrt{ab}+\sqrt{ac}-a}\geq$$
$$\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sum\limits_{cyc}(\sqrt{ab}+\sqrt{ac}-a)}=\frac{\sum\limits_{cyc}(a+2\sqrt{ab})}{\sum\limits_{cyc}(2\sqrt{ab}-a)}\geq3$$
because the last inequality it's just $$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\geq0.$$
The equality occurs for $a=b=c$, which says that $3$ is a minimal value.
We can use also the Rearrangement.
Indeed, the triples $(\sqrt{a},\sqrt{b},\sqrt{c})$ and $\left(\frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}},\frac{1}{\sqrt{a}+\sqrt{c}-\sqrt{b}},\frac{1}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)$ are the same ordered.
Thus, $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\geq\frac{1}{2}\sum_{cyc}\left(\frac{\sqrt{b}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{\sqrt{b}+\sqrt{c}-\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\right)=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$ and we are done again.
Another way:
$$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-3=\sum_{cyc}\left(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-1\right)=$$
$$=\sum_{cyc}\frac{\sqrt{a}-\sqrt{b}-(\sqrt{c}-\sqrt{a})}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}(\sqrt{a}-\sqrt{b})\left(\frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-\frac{1}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\right)=$$
$$=2\sum_{cyc}\frac{(\sqrt{a}-\sqrt{b})^2}{(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| $$\int \left( 2x+3 \right) \sqrt { 3x+1 } dx=\\ 3x+1={ t }^{ 2 }\\ x=\frac { { t }^{ 2 }-1 }{ 3 } \\ dx=\frac { 2t }{ 3 } dt\\ \int { \left( \frac { 2{ t }^{ 2 }-2 }{ 3 } +3 \right) } { t }\frac { 2t }{ 3 } dt=\frac { 2 }{ 9 } \int { \left( 2{ t }^{ 2 }+7 \right) { t }^{ 2 }dt } =\frac { 4 }{ 9 } \int { { t }^{ 4 }dt } +\frac { 14 }{ 9 } \int { { t }^{ 2 }dt } =\\ =\frac { 4 }{ 9 } \cdot \frac { { t }^{ 5 } }{ 5 } +\frac { 14 }{ 9 } \cdot \frac { { t }^{ 3 } }{ 3 } +C=\frac { 4 }{ 45 } { t }^{ 5 }+\frac { 14 }{ 27 } { t }^{ 3 }+C=\frac { 4 }{ 45 } { \left( 3x+1 \right) }^{ \frac { 5 }{ 2 } }+\frac { 14 }{ 27 } { \left( 3x+1 \right) }^{ \frac { 3 }{ 2 } }+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$ I have to solve this series transforming it into a telescopic sequence
$$\sum_{k=1}^\infty\frac{1}
{k(k+1)(k+2)(k+3)}$$
But I'm lost in the calculation!
| There is a general solution for this type of problem: $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}.$$
Observe that
$$\sum_{k\geq 1} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}=\sum_{k\geq 1} \frac{(k-1)!}{(k+N)!}=\frac{1}{N!}\sum_{k\geq 1}\frac{\Gamma(k)\Gamma(N+1)}{\Gamma(k+N+1)}=\frac{1}{N!}\sum_{k\geq 1} B(k,N+1).$$
Consequently,
\begin{align}
\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)} & = \frac{1}{N!}\sum_{k\geq 1}\int_0^1 x^{k-1}(1-x)^{N}\,dx \\
& =\frac{1}{N!}\int_0^1\left(\sum_{k\geq 1}x^{k-1}\right)(1-x)^N\, dx \\
&=\frac{1}{N!}\int_0^1 (1-x)^{N-1}\, dx \\
& =\frac{1}{N\cdot N!}.
\end{align}
Here we have $N=3$. Hence, the answer is $\frac{1}{3\cdot 3!}=\frac{1}{18}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Taylor series of $\ln(1+x)$ bounded for 2nd order
Prove that for all $x > 0 \Rightarrow $ $x - \frac{x^2}{2} < \ln(1+x) < x- \frac{x^2}{2} + \frac{x^3}{3}$
My Attempt -
I want to show the boundaries of the Maclaurin Series of $\ln(1+x)$ up to $x^2$. As such -
$P_n(x) = f(x) + \dfrac{f'(x)}{1!}(x)+\dfrac{f''(x)}{2!}(x)^2 + R_n(x)$
$P_n(x) = x-\dfrac{x^2}{2} + R_n(x)$
I know that $ 0<R_n(x) \le \dfrac{f^{(3)}(c)(x^3)}{3!} \,\, = \, \dfrac{x^3}{3}$ (That is i am not sure how to justify instead of $f$ being differentiable and continuous, and is it greater or greater than or equal ?).
Therefore we can conclude that due to $x>0 $ and $R_n(x) $ boundaries$ \,\,\Rightarrow $ $$x - \frac{x^2}{2} <\ln(1+x) < x- \frac{x^2}{2} + \frac{x^3}{3}.$$
Thank you very much in advance, would love to hear you thoughts about this.
| Let $f (x)=\ln (1+x) $.
$$f'(x)=\frac {1}{1+x} $$
$$f''(x)=\frac {-1}{(1+x)^2} $$
$$f^{(3)}(c)=\frac {2}{(1+c)^3} $$
we know that $0 <c <x $, so
$$0<f^{(3)}(c)<2$$
and $$0<R_n (x)<\frac {2x^3}{6} $$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integer solutions of $a^6 + 4 b^3 = 1$ This question is simplified case of
Integer solutions of $a^6 + 4 b^3 = c^6$.
Is there integer solutions for
$$a^6 + 4 b^3 = 1$$
The way I am trying to prove is based on Fermat's Theorem, but I can not get the final result.
| From $a^6+4b^3=1$, we see that $a$ must be odd. Let's rewrite the equation as
$$(a^3-1)(a^3+1)=-4b^3$$
Since $a^3$ is odd, we have $\gcd(a^3-1,a^3+1)=2$, so we have
$$\begin{align}
a^3-1&=2c^3\\
a^3+1&=2d^3
\end{align}$$
where $cd=-b$ with $\gcd(c,d)=1$. But this implies
$$a^3=c^3+d^3$$
which, as Fermat observed, has no nontrivial integer solutions. So we're left with $(a,b)=(\pm1,0)$ as the only integer solution.
Remark (added later): As the OP brilliantly observes, there is really no need to rely on Fermat. If you subtract the two equations instead of adding them, you get $d^3-c^3=1$, which immediately eliminates anything nontrivial, since cubes have a hard time differing by $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove: For $\triangle ABC$, if $\sin^2A + \sin^2B = 5\sin^2C$, then $\sin C \leq \frac{3}{5}$. We have a triangle $ABC$. It is given that $\sin^2A + \sin^2B = 5\sin^2C$. Prove that $\sin C \leq \frac{3}{5}$.
Let's say that $BC = a$, $AC=b$, $AB=c$.
According to the sine law,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$ then
$\sin A = \frac{a}{2R}\implies \sin^2A = \frac{a^2}{4R^2}$
$\sin B = \frac{b}{2R}\implies \sin^2B = \frac{b^2}{4R^2}$
$\sin C = \frac{c}{2R}\implies \sin^2C = \frac{c^2}{4R^2}$
Then we get:
$$\frac{a^2}{4R^2} + \frac{b^2}{4R^2} = 5\frac{c^2}{4R^2}.$$
Since $4R^2 > 0$, we get that $a^2 + b^2 = 5c^2$
Guys, is that correct? Even if it is, do you have any ideas what shall I do next?
| By the law of sines and AM-GM we obtain $$5c^2=a^2+b^2\geq2ab,$$ which gives
$$\frac{c^2}{ab}\geq\frac{2}{5}.$$
In another hand, by the law of cosines we obtain:
$$\cos{C}=\frac{a^2+b^2-c^2}{2ab}=\frac{2c^2}{ab}\geq\frac{4}{5}.$$
Id est, $$\sin{C}=\sqrt{1-\cos^2C}\leq\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of Indefinite Integralsl involving Trigonometric function
Finding $\displaystyle \int\frac{\ln(\cot x)}{\bigg(\sin^{2009}x+\cos^{2009} x\bigg)^2}\cdot (\sin^{2008}(2x))dx$
Try: $$\int \frac{\ln(\cot x)}{\bigg(1+\tan^{2009}(x)\bigg)^2}\cdot \tan^{2008}(x)\cdot \sec^{2010}(x)dx$$
Now substuting $\tan^{2009} x=t$ and $\displaystyle \tan^{2008}(x)\cdot \sec^2(x)dx=\frac{1}{2009}dt$
So $$ -\frac{1}{(2009)^2}\int\frac{\ln(t)}{(1+t)^2}\cdot (1+t^{\frac{2}{2009}})^{1005}dt$$
Could some help me to solve it, thanks
| Using $\sin(2x)=2\sin x\cos x$, we have
$$\begin{align}&\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot (\sin^{n}(2x))\ \mathrm dx\\\\&=\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot 2^n\sin^nx\cos^nx\ \mathrm dx\\\\&=-2^n\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot \sin^{2n+2}x\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\\\\&=-2^n\int\frac{\ln(\cot x)}{(1+\cot^{n+1} x)^2}\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\end{align}$$
Let $s=1+\cot^{n+1}x$. Then, since $\frac{\mathrm ds}{n+1}=\frac{-\cot^nx}{\sin^2x}\ \mathrm dx$, we get
$$\begin{align}&-2^n\int\frac{\ln(\cot x)}{(1+\cot^{n+1} x)^2}\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\\\\&=-2^n\int\frac{\frac{1}{n+1}\ln(s-1)}{s^2}\cdot\frac{\mathrm ds}{n+1}\\\\&=\frac{2^n}{(n+1)^2}\int (s^{-1})'\ln(s-1)\ \mathrm ds\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)-\int\frac{1}{s(s-1)}\ \mathrm ds\right)\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)+\int\left(\frac{1}{s}-\frac{1}{s-1}\right)\ \mathrm ds\right)\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)+\ln s-\ln(s-1)\right)+\mathrm C\\\\&=\frac{2^n}{(n+1)^2}\left(\frac{\ln(\cot^{n+1}x)}{1+\cot^{n+1}x}+\ln(1+\cot^{n+1}x)-\ln(\cot^{n+1}x)\right)+\mathrm C\end{align}$$
Therefore, for $n=2008$, we have
$$\int\frac{\ln(\cot x)}{(\sin^{2009}x+\cos^{2009} x)^2}\cdot (\sin^{2008}(2x))\ \mathrm dx$$$$=\frac{2^{2008}}{2009^2}\left(\frac{\ln(\cot^{2009}x)}{1+\cot^{2009}x}+\ln(1+\cot^{2009}x)-\ln(\cot^{2009}x)\right)+\mathrm C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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convergence of the series $\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$
I'm studying the convergence of the series
$$\sum_{n=1}^\infty \frac {\sqrt{n^2+1}-n}{\sqrt{n}}$$
*
*$\frac {\sqrt{n^2+1}-n}{\sqrt{n}}>0, \forall n \ge 1$
*$\lim_{n \to +\infty}\frac {\sqrt{n^2+1}-n}{\sqrt{n}}=\lim_{n \to +\infty}\frac {n^2+1-n^2}{\sqrt{n}*(\sqrt{n^2+1}+n)}=0$
*The suggested solution in my book says that the series converges but:
$\frac {\frac {\sqrt{n^2+1}-n}{\sqrt{n}}}{(\frac {1}{n})^{- \frac {1}{2}}}= \frac {n* \sqrt {1+ \frac {1}{n^2}}}{ \sqrt {n}}* \frac {1}{\sqrt {n}} \rightarrow 1 $ for $n \rightarrow \infty $
| Now, $$\frac{1}{\sqrt{n}(\sqrt{n^2+1}+n)}<\frac{1}{2\sqrt{n^3}}$$ and since $\frac{3}{2}>1$ it converges.
Because
$$\sum_{n=1}^{+\infty}\frac{\sqrt{n^2+1}-n}{\sqrt{n}}<\sum_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$$ and since $\sum\limits_{n=1}^{+\infty}\frac{1}{2\sqrt{n^3}}$ converges, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find a power series for $\frac{z^2}{(4-z)^2}$. Find a power series for $$\frac{z^2}{(4-z)^2}$$
What I did is:
Since
$$\sum^{\infty}_{k=0}z^k=\frac{1}{1-z}$$
Take the derivative, so
$$\sum^{\infty}_{k=1}kz^{k-1}=\frac{1}{(1-z)^2}$$
So
$$\frac{z^2}{(4-z)^2}=\frac{z^2}{4^2(1-\frac{z}{4})^2}=\frac{z^2}{4^2}\sum^{\infty}_{k=1}k\left(\frac{z}{4}\right)^{k-1}=\sum^{\infty}_{k=1}k\left(\frac{z}{4}\right)^{k+1}$$
However, the solution is
$$\sum^{\infty}_{k=0}\frac{k+1}{2\cdot 4^{k}}z^{k+2}$$
Thanks for any comments~
| For $|z|<4$ we have:$$\frac{1}{4-z}=\sum_{n=0}^{\infty}\frac{z^n}{4^{n+1}}\to \frac{1}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^n}{4^{n+2}}\to\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^{n+2}}{4^{n+2}}$$
and for $|z|>4$ similarly: $$\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)4^n}{z^{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Changing modulus in modular arithmetic Is it true that
$$a\equiv b\pmod{m}\implies\frac{a}{n}\equiv\frac{b}{n} \pmod{\frac{m}{n}},$$ where $a, b, m, n, \frac{a}{n}, \frac{b}{n}, \frac{m}{n}\in\mathbb{N}$? If so, how do I prove it?
| From the definition of modulo:
$$a \equiv b \pmod m \implies a=km+b \tag{1}$$
$$\frac{a}{n} \equiv {\frac{b}{n}} \pmod {\frac{m}{n}} \implies \frac{a}{n}=k \frac{m}{n} + \frac{b}{n} \tag{2}$$
Observe $(2)$.
\begin{align}
\frac{a}{n}&=k \frac{m}{n} + \frac{b}{n} \tag {2} \\
\frac{a}{n} \cdot n&=k \frac{m}{n} \cdot n + \frac{b}{n} \cdot n \\
a & = km+b \tag{1}
\end{align}
By multiplying $(2)$ by $n$, we get $(1)$, so the expressions are the same. Hence $a \equiv b \pmod m \implies \frac{a}{n} \equiv \frac{b}{n} \pmod{ \frac{m}{n}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I simplified the equation as $$ \int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\tan^2(x))dx$$
Then I simplified $\tan^2(x)\equiv\frac{\sin^2(x)}{\cos^2(x)}, \sin^2(x)\equiv1-\cos^2(x)$ so it becomes, $\tan^2(x)\equiv\frac{1-\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}-1$ making the overall integral
$$\int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\frac{1}{\cos^2(x)}-1)dx$$
$$=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx-\int_\frac{-π}{3}^{\frac{π}{3}}1dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx$$
I know that $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$ but that's off by heart and not because I can work it out. Since $x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}=0$, the final equation becomes
$$\int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx=\tan(x)|_{\frac{-π}{3}}^{\frac{π}{3}}=2 \sqrt3$$
Is what I did correct because I feel like I've made a mistake somewhere but can't find it. Also why does $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
EDIT - Made an error in $\int\frac{1}{\cos^2(x)}dx=\tan^2(x)+c$, it's actually $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
| To simplify, try this approach of $\tan^2 x = \sec^2 x - 1$ to get
$$\int_{-\pi/3}^{\pi/3}(2-\sec^2 x)dx = 2 \cdot \int^{\pi/3}_{0}(2-\sec^2 x)dx$$
To get $2 \cdot[2x-\tan x]_{0}^{\pi/3}$ = $\frac{4\pi}{3} - 2\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Young Tableaux generating function The number of young tableaux of $n$ cells is known to satisfy the recurrence $a_{n+1} = a_{n} + na_{n-1}$. I am trying to find the generating function but I keep getting something dependent on $n$. Here's what I did so far:
Denote by $f(x) = \sum_{n\geq 1}a_nx^n$. We have $\sum_{n \geq 1} a_{n+1}x^n = f(x) + nxf(x)$ (if we assume $a_1 = 1, a_2 = 2$). We can infer that $\sum_{n \geq 1} a_{n+1}x^n = \frac{f(x)}{x} - 1$.
| Using $a_{n+1} = a_{n} + n \, a_{n-1}$ with $a_{0} = 1$ then the following exponential generating function can be obtained.
Let
$$B(t) = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \, t^{n}$$
then
\begin{align}
\sum_{n=0}^{\infty} \frac{a_{n+1}}{n!} \, t^{n} &= \sum_{n=0}^{\infty} \frac{(n+1) \, a_{n+1}}{(n+1)!} \, t^{n} = \frac{d}{dt} \, \sum_{n=1}^{\infty} \frac{a_{n}}{n!} \, t^{n} \\
&= \frac{d}{dt} \left( B(t) - a_{0} \right) = B'(t)
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty} \frac{a_{n+1}}{n!} \, t^{n} &= \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \, t^{n} + \sum_{n=0}^{\infty} \frac{n \, a_{n-1}}{n!} \, t^{n} \\
B' &= B + t \, B
\end{align}
which can be seen as
$$ \frac{d}{dt} \, \ln(B) = 1 + t $$
and leads to
$$\sum_{n=0}^{\infty} \frac{a_{n}}{n!} \, t^{n} = e^{t + \frac{t^{2}}{2}}.$$
The linear form can be obtained in a similar manor. Let
$$A(t) = \sum_{n=0}^{\infty} a_{n} \, t^{n}$$
for:
\begin{align}
\sum_{n=0}^{\infty} a_{n+2} \, t^{n} &= \sum_{n=0}^{\infty} a_{n+1} \, t^{n} + \sum_{n=0}^{\infty} (n+1) \, a_{n} \, t^{n} \\
\frac{1}{t^2} \, (A - a_{0} - a_{1} \, t) &= \frac{1}{t} \, (A - a_{0}) + \frac{d}{dt} \left( t \, A \right) \\
t^3 \, A' + (t^2 + t -1) \, A &= (a_{0} - a_{1}) \, t - a_{0}.
\end{align}
Solving this equation leads to
$$A(t) = \frac{1}{t} \, e^{\frac{1}{t} - \frac{1}{2 \, t^2}} \, \left( \int_{1}^{t} e^{-\frac{1}{x} + \frac{1}{2 \, x^2}} \, \left(\frac{- a_{0} + (a_{0} - a_{1}) x}{x^2}\right) \, dx + c_{1} \right).$$
From here it is a matter of determining the constants.
| {
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"url": "https://math.stackexchange.com/questions/2613328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Arc length of curve of intersection between cylinder and sphere
Given the sphere $x^2+y^2+z^2 = \frac{1}{8}$ and the cylinder $8x^2+10z^2=1$, find the arc length of the curve of intersection between the two.
I tried parametrizing the cylinder (the task specifies this as a hint). My attempt:
$$x(t) = \frac{1}{\sqrt{8}} \sin(t)$$
$$z(t) = \frac{1}{\sqrt{10}} \cos(t)$$
Plugging this into $x^2+y^2+z^2 = \frac{1}{8}$, I solve for $y$ to get
$$y = \sqrt{\frac{\cos(2t)+1}{4\sqrt{5}}}$$
I then tried integrating $|x(t), y(t), z(t)|$ from $0$ to $2\pi$
with no luck. I suspect my parametrization is wrong as my expression for $y$ looks rather ugly. Any ideas?
| Apparently I was right.
If we define $r(t) = (x, y, z)$ where
$$x = \frac{1}{\sqrt{8}}sin(t)$$
$$z = \frac{1}{\sqrt{10}}cos(t)$$
$$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$
we find
$$|r(t)| = \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} = \frac{1}{2\sqrt{2}}$$
$$\int_{0}^{2\pi} |r(t)| dt = \int_{0}^{2\pi} \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} dt $$
$$ = \int_{0}^{2\pi} \frac{1}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$
The correct answer is $\pi\sqrt{2}$, I suspect the reason lies in the fact that $$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$, meaning we must also take into account the negative value here (seems we must multiply our integral by 2). I used wolfram to my advantage to figure out the integral, my approach by hand was a failure as it got quite messy.
This is from my calculus 2 class, I'm not familiar with elliptic integrals yet, but I thank you for the assistance Satish.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I have a inequality, I don't know where to start Show that for $x,y,z > 0$ the inequality is true:
$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$
I have tried Holder, but i had no luck. Please give me a hint of how to start.
| The hint.
It's $$\sum_{cyc}\left(\frac{x^2}{y}-2x+y\right) \geq \sum_{cyc}\left(\frac{(y+z)^2}{z+x}-2(y+z)+(z+x)\right)$$ or
$$\sum_{cyc}\frac{(x-y)^2}{y}\geq\sum_{cyc}\frac{(x-y)^2}{z+x}$$ or
$$\sum_{cyc}(x-y)^2S_z\geq0,$$
where $S_z=\frac{1}{y}-\frac{1}{z+x}.$
Now, by C-S $$\sum_{cyc}S_x=\sum_{cyc}\left(\frac{1}{x}-\frac{1}{x+y}\right)=\sum_{cyc}\left(\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)-\frac{1}{x+y}\right)\geq$$
$$\geq\sum_{cyc}\left(\frac{1}{2}\cdot\frac{4}{x+y}-\frac{1}{x+y}\right)=\sum_{cyc}\frac{1}{x+y}>0$$ and
$$\sum_{cyc}S_xS_y=\sum_{cyc}\left(\frac{1}{z}-\frac{1}{x+y}\right)\left(\frac{1}{x}-\frac{1}{y+z}\right)=$$
$$=\sum_{cyc}\left(\frac{1}{xy}-\frac{1}{z(y+z)}-\frac{1}{x(x+y)}+\frac{1}{(x+y)(x+z)}\right)=$$
$$=\sum_{cyc}\left(\frac{1}{xy}-\frac{1}{y(x+y)}-\frac{1}{x(x+y)}+\frac{1}{(x+y)(x+z)}\right)=$$
$$=\sum_{cyc}\frac{1}{(x+y)(x+z)}>0.$$
Now, since $$\sum_{cyc}S_x>0,$$ we can assume that $S_y+S_z>0$.
Also, we have
$$\sum_{cyc}(x-y)^2S_z=(x-y)^2S_z+(x-y+y-z)^2S_y+(y-z)^2S_x=$$
$$(S_y+S_z)(x-y)^2+2(x-y)(y-z)S_y+(S_x+S_y)(y-z)^2$$ and it's enough to prove that
$$S_y^2-(S_z+S_y)(S_x+S_y)\leq0,$$
which is $\sum\limits_{cyc}S_xS_y\geq0$.
Done!
| {
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"url": "https://math.stackexchange.com/questions/2614390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the integral $\int \sqrt{1+\sin x}\, dx$. I want to calculate the integral $\int \sqrt{1+\sin x}\, dx$.
I have done the following:
\begin{equation*}\int \sqrt{1+\sin x}\, dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{1-\sin x}}\, dx=\int \sqrt{\frac{1-\sin^2 x}{1-\sin x}}\, dx=\int \sqrt{\frac{\cos^2x}{1-\sin x}}\, dx=\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx\end{equation*}
We substitute $$u=\sqrt{1-\sin x} \Rightarrow du=\frac{1}{2\sqrt{1-\sin x}}\cdot (1-\sin x)'\, dx \Rightarrow du=-\frac{\cos x}{2\sqrt{1-\sin x}}\, dx \\ \Rightarrow -2\, du=\frac{\cos x}{\sqrt{1-\sin x}}\, dx $$
We get the following:
\begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=\int(-2)\, du=-2\cdot \int 1\, du=-2u+c\end{equation*}
Therefore \begin{equation*}\int \frac{\cos x}{\sqrt{1-\sin x}}\, dx=-2\sqrt{1-\sin x}+c\end{equation*}
In Wolfram the answer is a different one. What have I done wrong?
| Rewrite the given integral using trigonometric/hyperbolic substitutions :
$$\int \sqrt{1+\sin x} dx ={\displaystyle\int}\sqrt{2}\cos\left(\dfrac{2x-{\pi}}{4}\right)\,\mathrm{d}x$$
Apply the substitution :
$$u=\dfrac{2x-{\pi}}{4} \to dx = 2du$$
which means the integral becomes equal to :
$$=\class{steps-node}{\cssId{steps-node-1}{2^\frac{3}{2}}}{\displaystyle\int}\cos\left(u\right)\,\mathrm{d}u =2^\frac{3}{2}\sin\left(u\right)$$
Undo now the substitution for $u$ and get :
$$=2^\frac{3}{2}\sin\left(\dfrac{2x-{\pi}}{4}\right)$$
which means that :
$$\int \sqrt{1+\sin x} dx =2^\frac{3}{2}\sin\left(\dfrac{2x-{\pi}}{4}\right) + C =2\sin\left(\dfrac{x}{2}\right)-2\cos\left(\dfrac{x}{2}\right)+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough?
| It is true, but formally I think it would be more elegant to say
$\frac{\sqrt{n-2\sqrt{n}}}{\sqrt{n}}=\sqrt{\frac{n-2\sqrt{n}}{n}}=\sqrt{1-\frac{2}{\sqrt{n}}}\longrightarrow1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The general problem of finding $\int \frac{dx}{(ax^2 + bx + c)^2}$ I know that $\displaystyle\int \dfrac1{ax^2+bx+c}\,\mathrm dx$ is easily solvable using completing the square, but my question is how would would find
$$\displaystyle\int\frac{1}{\left(ax^2+bx+c\right)^2}\,\mathrm dx$$
I have tried using the same approach as before but its not getting me anywhere... any help would be great!
| If the quadratic factor ($a \neq 0$) that appears in the denominator can be factored into linear factors over the reals, then a partial fraction decomposition can be used. I will assume this is doable for you so will only concentrate on the case where the quadratic factor in the denominator is irreducible over the reals. In this case $a \neq 0$ and $b^2 < 4ac$.
As you noted, one begins by completing the square. Doing so yields
$$I = \int \frac{dx}{(a x^2 + bx + c)^2} = \frac{1}{a^2} \int \frac{dx}{\left [\left (x + \frac{b}{2a} \right )^2 + \left (\frac{c}{a} - \frac{b^2}{4a^2} \right )\right ]^2}.$$
Now let
$$k^2 = \frac{c}{a} - \frac{b^2}{4a^2} > 0,$$
and is the case since $b^2 < 4ac$. Using a substitution of $u = x + b/(2a)$ we have
\begin{align*}
I &= \frac{1}{a^2} \int \frac{du}{(u^2 + k^2)^2}\\
&= \frac{1}{a^2 k^2} \int \frac{(u^2 + k^2) - u^2}{(u^2 + k^2)^2} \, du\\
&= \frac{1}{a^2 k^2} \int \frac{du}{u^2 + k^2} - \frac{1}{a^2 k^2} \int \frac{u^2}{(u^2 + k^2)^2}\\
&= \frac{1}{a^2 k^2} \left (I_1 - I_2 \right ).
\end{align*}
The first of these integrals can be readily found. The result is
$$I_1 = \frac{1}{k} \tan^{-1} \left (\frac{u}{k} \right ) + C_1 = \frac{1}{k} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + C_1.$$
For the second integral, if we note that
$$\int \frac{2u}{(u^2 + k^2)^2} \, du = -\frac{1}{u^2 + k^2} + C,$$
integrating by parts we have
\begin{align*}
I_2 &= \int \frac{u^2}{(u^2 + k^2)^2} \, du\\
&= \frac{1}{2} \int \frac{2u}{(u^2 + k^2)^2} \cdot u \, du\\
&= -\frac{u}{2(u^2 + k^2)} + \frac{1}{2} \int \frac{du}{u^2 + k^2}\\
&= -\frac{u}{2(u^2 + k^2)} + \frac{1}{2k} \tan^{-1} \left (\frac{u}{k} \right ) + C_2\\
&= -\frac{2ax + b}{4(ax^2 + bx + c)} + \frac{1}{2k} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + C_2.
\end{align*}
Thus
$$\boxed{\int \frac{dx}{(ax^2 + bx + c)^2} = \frac{1}{2a^2 k^3} \tan^{-1} \left (\frac{2ax + b}{2ak} \right ) + \frac{2ax + b}{4a^2 k^2(ax^2 + bx + c)} + C}$$
for the irreducible case where $a \neq 0$, $b^2 < 4ac$ such that $k = \sqrt{\dfrac{c}{a} - \dfrac{b^2}{4a^2}}$.
Comment
The above procedure can be readily generalised to the case
$$\int \frac{dx}{(ax^2 + bx + c)^n},$$
where $n \in \mathbb{N}$ and the factor appearing in the denominator is irreducible over the reals.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove via induction $\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$ I have to prove by induction, that
$$\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$$
$\begin{align}
\sum_{k=2}^{n+1}{\frac{k-1}{k!}} &= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{(n+1)-1}{(n+1)!}} \\
&= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{n}{(n+1)!}} \\
&= \frac{n!-1}{n!} + \frac{n}{(n+1)!} \\
&= \frac{[n!-1](n+1)!}{n!(n+1)!} + \frac{n! \cdot n}{n!(n+1)!} \\
&= \frac{[n!-1](n+1)! + n! \cdot n}{n!(n+1)!} \\
&= \frac{n!(n+1)!-(n+1)!+n!n}{n!(n+1)!} \\
\end{align}$
Question: How should I go on ? Do I have a mistake until now?
| From your third line:
\begin{align}
\frac{n!-1}{n!} + \frac{n}{(n+1)!} &= \frac{(n!-1)(n+1)}{n!(n+1)}+\frac{n}{(n+1)!} \\
&= \frac{(n+1)!-(n+1)+n}{(n+1)!} \\
&=\frac{(n+1)!-1}{(n+1)!}
\end{align}
Don't forget to check base case.
To go on from where you stop, divide numerator and denominator by $n!$.
| {
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If $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Say $a,b,c$ are integers, $a>0$. Suppose $ax^2+bx+c=0$ has $2$ different solutions in $(0,1)$ then prove $a\geq 5$. Find an example for $a=5$.
I am struggling with this for some time with no success.
I try Vieta's formula $0<x_1+x_2 = -{b\over a}<2$ and $x_1x_2 ={c\over a}<1$ so $c<a$ and $0<-b<2a$.
| Let $z=-b$. We know that the minimum is going to be at $(\frac{z}{2a},c-\frac{z^2}{4a})$. (For proof plug $\frac{z}{2a}$, the extremum, into $ax^2-zx+c$). We must find values of $a,b,c$ such that $a>c>0,z>0,a+c>z,c-\frac{z^2}{4a}<0$. Rewriting the last one, we get $z^2>4ac$. However, because everything is in integers we can change $A>B$ to $A\geq B+1$. We get $a>c\geq 1,z\geq 1,a+c\geq z+1,z^2\geq 4ac+1$. Let $w=z+1$, and we see that $c\geq 1,a+c\geq w,\frac{z^2-1}{4}=\frac{(z+1)(z-1)}{4}=\frac{w^2-2w}{4}\geq ac$.
We see that these form the boundaries of the region of solutions in the $a$-$c$ plane. In order to have a non-empty solution set, the intersection of $a+c\geq w$ and $\frac{w^2-2w}{4}\geq ac$ must be above the line $c=1$. We do the following:
$$
a=-c+w=\frac{w^2-2w}{4c}\\
-c^2+wc-\frac{w^2-2w}{4}=0\\
c=\frac{-w \mp \sqrt{w^2-4(-1)(-\frac{w^2-2w}{4})}}{-2}=\frac{w \pm \sqrt{w^2-(w^2-2w)}}{2}=\frac{w \pm \sqrt{2w}}{2}.
$$
Because the original equations were symmetric in $a$ and $c$, we know that they will share these roots. As $a>c$, $a$ will take the $+$ and $c$ will take the $-$, so $c=\frac{w-\sqrt{2w}}{2}$.
This means that $c=\frac{w-\sqrt{2w}}{2}\geq 1$, which we now simplify.
$$
\frac{w-\sqrt{2w}}{2}=1\\
w-\sqrt{2w}=2\\
w-2=\sqrt{2w}\\
z-1=\sqrt{2z+2}\\
(z-1)^2=z^2-2z+1=2z+2\\
z^2-4z-1=0\\
z\geq \frac{4+\sqrt{4^2-4*(-1)*1}}{2}=2+\sqrt{4-(-1)}=2+\sqrt{5}\approx 4.236\\
z\geq 5
$$
When $z=5$ (or $b=-5$) and $c=1$, you get that $a=5$. If you increase in z, c grows so slowly that a is forced to grow faster in order to meet up with z. This means $a$ can never get lower than $5$, and is only $5$ when $b=-5$ and $c=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
| If you do have to prove it with the ε-$n_0$ method, it's much simpler that that:
First rewrite $a_n$ as
$$a_n=\frac{3n+5}{2n+6}=\frac32-\frac2{n+3}.$$
Then, if $m,n>n_0$,
$$\bigl|a_m-a_n\bigr|=2\,\biggl|\frac1{m+3}-\frac1{n+3}\biggr|=\frac{
2\,| m-n|}{(m+3)(n+3)}\le 2\biggl|\frac1m-\frac1n\biggr|<2\biggl(\frac1m+\frac1n\biggr)<\frac4{n_0},$$
and this one is less than $\varepsilon$ as soon as
$\;n_0>\dfrac 4\varepsilon$.
| {
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$\arg(\frac{z_1}{z_2})$ of complex equation If $z_1,z_2$ are the roots of the equation $az^2 + bz + c = 0$, with $a, b, c > 0$; $2b^2 > 4ac > b^2$; $z_1\in$ third quadrant; $z_2 \in$ second quadrant in the argand's plane then, show that $$\arg\left(\frac{z_1}{z_2}\right) = 2\cos^{-1}\left(\frac{b^2}{4ac}\right)^{1/2}$$
| Since $a,b,c$ are real then the roots are $\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}$.
Given the quadrants locations, $z_1$ is the root with $-$ sign.
Thus $\dfrac{z_1}{z_2}=\dfrac{-b-i\sqrt{4ac-b^2}}{-b+i\sqrt{4ac-b^2}}=\dfrac{(b+i\sqrt{4ac-b^2})^2}{4ac}=\dfrac{(b^2-2ac)+ib\sqrt{4ac-b^2}}{2ac}$
It is quite easy to see that this complex is of module $1$.
[ $\require{cancel}(b^2-2ac)^2+b^2(4ac-b^2)=\cancel{b^4}-\cancel{4b^2ac}+4a^2c^2+\cancel{4b^2ac}-\cancel{b^4}=4a^2c^2$ ]
So $\cos(\theta)=\dfrac{b^2-2ac}{2ac}=\dfrac{b^2}{2ac}-1\iff 1+cos(\theta)=2\cos(\frac{\theta}2)^2=\dfrac{b^2}{2ac}$
Given the conditions of positivity for $a,b,c$ and $b^2-2ac$, we get that $\theta$ is in the first quadrant and the formula given in the problem.
| {
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Proving the area of the circle using sticks. I was just trying to prove the area of the circle but couldn't reach any conclusion.so here i went----- I know
For 2 identical sticks making a regular polygon we have the regular polygon as square i.e with 4 sides. For 3 identical sticks which are able to make aregular polygon we have hexagon similarly for n identical sticks we have 2n sided polygon. Now if n tends to infinity we get a circle. Now i am stuck in connecting the length of the stick and the equal angle between them with the side of the regular polygon. I cant generalise. Please help. Please note- all the sticks intersect each other at a single point i.e the centre of the geometrical figure. We get the polygons by joining the terminal end points of the sticks.
| I suppose the diameter is the length of the stick, let it be $d$, then we need to prove that the area of the circle with infinite sticks is $\pi\, \frac{d^2}{4}$.
The angle of the internal triangle formed with $n$ sticks is $\frac{n-1}{n}\times \frac{\pi}{2}$
Hence,
\begin{align*}
\text{base} &= d \, \cos{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)}\\
\text{height} &= \frac{d}{2}\, \sin{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)}\\
\end{align*}
Hence, the total area of the polygon is
\begin{align*}
A &= \frac{1}{2}\cdot d \, \cos{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)} \cdot \frac{d}{2}\, \sin{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)} \times 2\, n\\
\end{align*}
Then, take the limit
\begin{align*}
A &= \lim_{n\to\infty} \frac{d^2}{4} \sin{\left(\frac{n-1}{n}\cdot \pi\right)} \times n\\
&= \frac{d^2}{4}\cdot \pi
\end{align*}
| {
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Difference between minus one and plus one induction? I recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, however my professor has insisted I solve it by using $P(n-1)$ as opposed to $P(n+1)$, which I've done below.
Basis
$$\frac{1(1+1)(2*1+1)}{6} = 1$$
Inductive Step $(n-1)$
$$1^2+2^2+\ldots + (n-1)^2 = \frac{(n-1)(n)(2(n-1)+1)}{6}$$
Which Simplifies to
$$\frac{(n-1)(n)(2n-1)}{6} \rightarrow \frac{2n^3-3n^2+n}{6}$$
Add $6\frac{n^2}{6}$ to both sides and we've proven by induction.
My question is do there exists any mathematical proofs for which solving by Induction with $n+1$ and $n-1$ are not interchangeable and should I petition my professor to be able to use them interchangeably. I am aware that solving using $n-1$ and $n+1$ is identical, at least for every scenario I've come across (we're working with positive integers so I'm not expecting any variance from that), however given the overwhelming amount of resources, I can't for the life of me figure out why I am being instructed to use a method opposite what seems to be the norm for any other reason besides my teacher's personal preference.
| I think the only reason s/he did it this way is s/he wanted to have the final simplification end in the form $\frac {2n^3 + 3n^2 + n}6$
Had he done $P(n) \implies P(n+1)$ it would have involve a lot of factoring to get in the form $\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$.
Try it:
$1 + 2 + .... + n^2 = \frac {2n^3 + 3n^2 + n}6$
So $1 + 2 + ...... + n^2 + (n+1)^2 = \frac {2n^3 + 3n^2 + n}6 + n^2 + 2n + 1$
$= \frac {2n^3 + 9n^2 + 13n + 6}6$
... and we have to somehow get that to .... $=\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$
... which isn't impossible ...
$\frac {2n^3 + 9n^2 + 13n + 6}6= \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 7n + 4}6$
$=\frac {2(n+1)^3 + 3n^2 + 6n + 3 + n + 1}6$
$= \frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$
... but.... why schlep? That factoring and working backwords isn't the point.
It's easier to follow it and to go to the conclusion if you you work toward simplifying.
=====
I suppose an easier compromise would be to do:
$\frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6} = \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 6n + 3 + n+1}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$
first.
Then:
$1+2 + ..... + n^2 + (n+1)^2 =$
$\frac {2n^3 + 3n^2 + n}6 + (n+1)^2 = \frac {2n^3 + 3n^2 + n}6 +n^2 + 2n + 1=$
$\frac {2n^3 + 3n^2 + n + 6n^2 + 12n + 6}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$
$= \frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6}$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim\limits_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$ Can anyone help me solve this?
I know the answer is 4, but I don't really know how do I find the biggest power of $x$ when there's a square root.
$$\lim_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$$
| As an alternative, note that
$$\frac{\sqrt{x^4}+3x^2}{x^2-5x}=\frac{4x^2}{x^2-5x}\le\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\le\frac{\sqrt{(x^2+2)^2}+3x^2}{x^2-5x}=\frac{4x^2+2}{x^2-5x}$$
thus for squeeze theorem
$$\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\to4$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c$ be in Arithmetic Progression, If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.
My Attempt:
$a,b,c$ are in AP so
$$b=\dfrac {a+c}{2}$$
$b,c,a$ are in HP so
$$c=\dfrac {2ab}{a+b}$$
Multiplying these relations:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$
| Since $a,b,c$ are in arithmetic progression, we get
\begin{align*}
&c-b=b-a\\[4pt]
\implies\;&a = 2b - c\\[4pt]
\end{align*}
Since $b,c,a$ are in harmonic progression, we get
\begin{align*}
&\frac{1}{a}-\frac{1}{c} = \frac{1}{c}-\frac{1}{b}\\[4pt]
\implies\;&a = \frac{bc}{2b-c}\\[4pt]
\implies\;&a = \frac{bc}{a}\\[4pt]
\implies\;&\frac{a}{c}=\frac{b}{a}\\[4pt]
\end{align*}
hence, $c,a,b$ are in geometric progression, as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the tangent is $2x$ so the gradient of the normal is $-\frac{1}{2}x$ and the equation of the normal will be $-\frac{1}{2}x + v$
Where $x=\sqrt{c}$
then $y = -\frac{1}{2}\sqrt{c} + v$
We know $y = 2c$ therefore $2c = -\frac{1}{2}\sqrt{c} + v$ and so
$v = 2c + \frac{1}{2}\sqrt{c}$
So the equation of the normal is $y = -\frac{1}{2}x +2c +\frac{1}{2}\sqrt{c}$
We can tidy this up to get $2y = -x + 4c + \sqrt{c}$
Unfortunately the book tells me the answer is $2y\sqrt{c} = -x+\sqrt{c}(4c+1)$
It looks like I lost a $\sqrt{c}$ somewhere? Where did I go wrong?
Thank you
Gary
| HINT:
The gradient is $2x$ but you have that $x=\sqrt{c}$ so the gradient is equal to $2\sqrt c$
| {
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Probability that $a^2+b^2+c^2$ divisible by $7$
Three numbers $a,b,c\in\mathbb{N}$ are choosen randomly from the set of natural numbers. The probability that $a^2+b^2+c^2$ is divisible by $7$ is
Try:any natural number when divided by $7$ gives femainder $0,1,2,3,4,5,6$
So it is in the form of $7k,7k+1,7k+2\cdots ,7k+6,$ where $k\in \mathbb{W}$
Could some help me to how to solve it, thanks
| What remainders do the squares of numbers give when divided by 7 these are $$(7k)^2 \rightarrow 0 \\ (7k+1)^2 \rightarrow 1 \\ (7k+2)^2 \rightarrow 4 \\ (7k+3)^2 \rightarrow 2 \\ (7k+4)^2 \rightarrow 2 \\ (7k+5)^2 \rightarrow 4 \\(7k+6)^2 \rightarrow 1 \\$$
Now the sum of remainders must add upto a multiple of 7, this can happen if remainders are $(0,0,0), (1,2,4)$
The first case of remainders can be chosen in $\frac{1}{7^3}$
Edit
Doing corrections by mike and almagest
The second case can be written in $3!$ ways, and probability of each being $\frac8{343}$, the probability of second case is $3!\cdot\frac8{343}=\frac{48}{343}$
And total probability comes to be $\frac17$
| {
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Largest constant for an inequality with powers of three variables Let real $(a,b,c)$, with $a+b+c=0$, and let real positive $p$ with $p\ne1$ and $p\ne2$.
With some constant $C = C(p)$, the following inequality must hold:
$$
(a^2)^p + (b^2)^p + (c^2)^p \ge C \cdot (a^{2} + b^{2} + c^{2} )^p
$$
It is understood that $ (a^2)^p$ is nonnegative, likewise for $b$ and $c$.
Find the largest constant $C(p)$ for which the inequality holds.
Note the power mean inequality, without regard of the condition $a+b+c=0$, gives, for $p \ge 1$, a (too small) constant $\tilde C = (\frac{1}{3})^{p-1}$ so this appears to be the right behavior w.r.t. (large) $p$.
| By homogeneity, let $a^2 + b^2 + c^2 = 1$. Then we need to establish
$$
(a^2)^p + (b^2)^p + (c^2)^p \ge C(p)
$$
The two conditions $a^2 + b^2 + c^2 = 1$ and $a+b+c=0$ form a unit circle in $(a,b,c)$-space. Let's parametrize this circle as follows:
$$
a = \frac{-2}{\sqrt{6}}\cos(\phi)\\
b = \frac{2}{\sqrt{6}}\cos(\phi + \pi/3)\\
c = \frac{2}{\sqrt{6}}\cos(\phi - \pi/3)
$$
One can check that the two conditions are obeyed.
Then we have to show
$$
f(\phi) = \Big(\frac{2}{3}\Big)^p\Big((\cos^2(\phi))^p + (\cos^2(\phi + \pi/3))^p + (\cos^2(\phi - \pi/3))^p\Big) \ge C(p)
$$
$
f(\phi)$ is periodic in $\phi$ with period length $\pi/3$. Further, note $
f(\phi)= f(-\phi)$ hence there is an extremum at $\phi =0$.
Notice $
f(\phi = 0 ) =f(\phi = \pi/3 ) = \Big(\frac{2}{3}\Big)^p\Big(1 + 2(\cos^2(\pi/3))^p \Big) = \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p \Big)$
Another extremum in the first period (and hence, the same value in all other periods) is obtained at $\phi = \pi/6$. This can be seen by putting $\phi = \pi/6 + \alpha$, leading to
$$
f(\alpha) = \Big(\frac{2}{3}\Big)^p\Big((\cos^2(\pi/6 + \alpha))^p + (\sin^2(\alpha))^p + (\cos^2(\pi/6 - \alpha))^p\Big)
$$
Hence $f(\alpha) = f(-\alpha)$ which shows that $\alpha =0$ is an extremum, and
$$
f(\phi = \pi/6 ) = \Big(\frac{2}{3}\Big)^p\Big(2 (\cos^2(\pi/6))^p\Big) = 2 \Big(\frac{2}{3}\Big)^p\Big(\frac{3}{4}\Big)^p = \Big(\frac{1}{2}\Big)^{p-1}
$$
Now it is easy to see which of the two extrema is the minimum. We have $f(\phi = 0 ) -f(\phi = \pi/6 ) = \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p -2\Big(\frac{3}{4}\Big)^p \Big) {\ge \atop \le} 0$ for ${{0 \le p \le 1 \; {\rm{and}} \; {p \ge 2 }}\atop {1 \le p \le 2 }} $, with equality for $p=1$ and $p=2$.
So we have, for $0 \le p \le 1 \; {\rm{and}} \; {p \ge 2 }$:
$$
f(\phi ) \ge f(\phi = \pi/6 ) = \Big(\frac{1}{2}\Big)^{p-1} = C(p)
$$
which corresponds to a minimum at $(a,b,c) = (0,-1,1)$ and multiples and permutations.
For ${1 \le p \le 2 }$, we have
$$
f(\phi ) \ge f(\phi = 0 ) = \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p \Big) = C(p)
$$
which corresponds to a minimum at $(a,b,c) = (-2,1,1)$ and multiples and permutations. $\qquad \Box$
| {
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Solve $x^2 + x = y^4 + y^3 + y^2 + y$ over integers. I am trying to solve solve $x^2 + x = y^4 + y^3 + y^2 + y$ over the integers. So far I have decomposed it into $x(x + 1) = y(y + 1)(y^2 + 1)$ and noticed that both sides of the equation are nonnegative. Furthermore $GCD(x, x + 1) = GCD(y, y + 1) = GCD(y, y^2 + 1) = GCD(y + 1, y^2 + 1) = 1$.
| Hint: If you multiply both sides by $4$ and add $1$, you get
$$(2x+1)^2=4y^4+4y^3+4y^2+4y+1=(2y^2+y+1)^2-(y^2-2y).$$
This results in two squares that are very close together. Can you prove that they are too close together when $y$ is sufficiently large?
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with $\sin 90^\circ =1$ We know
$$\sin\theta = \frac{\text{perpendicular}}{\text{hypotenuse}}$$
then $\sin 90^\circ = 1$ refers in this case perpendicular = hypotenuse. But, then, the base becomes $0$ according to the Pythagorean triplet law. How is this possible?
| The Pythagorean Theorem asserts that $a^2 + b^2 = c^2$ such that $c$ represents the hypotenuse, namely the side of a triangle that is directly opposite the $90^\circ$ angle. There has to be a $90^\circ$ angle because this theorem only applies to right triangles. If $\sin 90^\circ = 1$, since $$\sin(\cdot) = \frac{\text{opposite}}{\text{hypotenuse}}$$ then we have $\text{opp} = \text{hyp}$. Let $a = \text{opp}$ then we have $$a^2 + b^2 = a^2$$ and therefore $b^2 = 0\Leftrightarrow b = 0$. This means that we don’t have a right triangle because the side $b$ has no length (it has length $0$) and so it does not exist. We simply have a line $a = c$. And since $a/c = 1$ then $\sin 90^\circ = 1$.
Consider for example, $$x^2 + y^2 = z^2.$$ Let $z = y + 1$, then we have $y = \dfrac{x^2 - 1}{2}$. If $x = z$, then $$y = \frac{(y+1)^2 - 1}{2}\Leftrightarrow 1 = \frac{y + 2}{2}$$ and therefore $y = 0$. We can’t escape the fact that $y = 0$ or $b = 0$ and so this is simply just a special case of $\sin\theta$ such that $\theta=90^\circ$.
| {
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Polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$ given some values what is $f(1)$? Let $f$ be a polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$, the minimum of f is attained at $3$, and $f(0)=3$, Then $f(1)$ equals.
$(A) \ 1$
$(B) \ 2$
$(C) -1$
$(D) -2$
I am not sure how to deal with $f''(x) \rightarrow2$ as $x\rightarrow\infty$ . Give me a hint to try.
EDIT : Work after hint
Let $f(x)=ax^2+bx +c $
$f''(x)=2\implies 2=2a \implies a=1 $
$f(0)=3=c$
$f(x)=x^2+bx +3 $
Using the fact that minima attained at 3 we have
$f(3)=12+3b=3 \implies b=-3$
$f(x) = x^2-3x+3$
$f(1) = 1-3+3 = 1$
| Dave showed that the degree of the polynomial is $2$. So we have $f(x)=a_2x^2+a_1x+a_0$.
We know that $f''(x)=2a_2=2\implies a_2=1\implies f(x)=x^2+a_1x+a_0$
We also know that there is a minima at $x=3\implies f'(3)=0\implies f'(3)=2\overbrace x^3+a_1=6+a_1=0\implies a_1=-6$ hence we have $f(x)=x^2-6x+a_0$.
We also know that $f(0)=3$ thus $f(0)={\overbrace x^0}^2-6\overbrace x^0+a_0=a_0=3$
And the final answer is $f(x)=x^2-6x+3$, with this we get $f(1)=1^2-6\times 1+3=\boxed{-2}$
| {
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Is it possible to know whether the following series converges without its formula? Given this series:
$$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots$$
Is it possible to know if the series converges without its formula? I tried using the comparison test with both $1/n$ and $1/n^2$ but these don't yield results. The formula for this series is $1/[2(2n+1)]$, which I can use in other convergence tests, but if I didn't know that or couldn't find it, is it still possible to determine convergence without the formula?
Thanks
| $$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots > \\
\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{16} + \dfrac{1}{20} + \ldots = \\
\dfrac{1}{4}\left(\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \ldots \right)$$
| {
"language": "en",
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"source": "stackexchange",
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Finding the maximum value without using derivatives
Find the maximum value of $$f(x)=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1}$$ without using derivatives.
The domain of $f(x)$ is $x \in [1,\infty)$. Then, using derivatives, I can prove that the function decreases for all $x$ from $D(f)$ and the maximum value is $f(1)= 2 - \sqrt{2}$. However, this uses derivatives.
| Some other manipulation perhaps?
$$\begin{align}
f(x)&=2\sqrt{x}-\sqrt{x+1}-\sqrt{x-1} \\
&=\sqrt{x}\left[2-\frac{\sqrt{x+1}}{\sqrt{x}}-\frac{\sqrt{x-1}}{\sqrt{x}}\right] \\
&=\sqrt{x}\left[2-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)\right]
\end{align}$$
With $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right) \to 2$ shows that $\left[2-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)\right]$ is very close to $0$ for larger $x \ \ $ , letting you look at smaller values that take away the diminishing effect of $\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right) $
| {
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"source": "stackexchange",
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How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by 11 How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by $11$
I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is $5! = 120$.
| Hint. By the divisibility rule by $11$ we have to count the arrangements $d_1,d_2,d_3,d_4,d_5$ of the digits $2,3,4,5,6$ such that $d_1+d_3+d_5-(d_2+d_4)$ is divisible by $11$. Notice that
$$-2=2+3+4-(5+6)\leq d_1+d_3+d_5-(d_2+d_4)\leq 4+5+6-(2+3)=10$$
therefore we should have $d_1+d_3+d_5=d_2+d_4=\frac{2+3+4+5+6}{2}=10$.
In how many ways we can do that?
| {
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All possible Jordan normal forms with a given minimal polynomial
Suppose $A \in M_{n\times n}(\Bbb R)$ has minimal polynomial $$(\lambda-1)^2(\lambda+1)^2$$ and $n\leq 6$. Find all possible Jordan normal forms.
If we denote the $n \times n$ Jordan block
$$J_\alpha (n) = \begin{bmatrix}\alpha&1&\cdots&\cdots\\0&\alpha&1&\cdots\\\vdots&\vdots\\0&0&\cdots&\alpha\end{bmatrix}$$
Then for the $n=6$ case: the characteristic polynomial is
$$(\lambda-1)^3(\lambda+1)^3$$
or
$$(\lambda-1)^4(\lambda+1)^2$$
or
$$(\lambda-1)^2(\lambda+1)^4$$
So are the possible Jordan normal forms the following?
$$\left\{ \begin{bmatrix}J_1(3)&O\\O&J_{-1}(3)\end{bmatrix}, \begin{bmatrix}J_{-1}(3)&O\\O&J_{1}(3)\end{bmatrix}, \begin{bmatrix}J_1(4)&O\\O&J_{-1}(2)\end{bmatrix}, \\ \begin{bmatrix}J_{-1}(2)&O\\O&J_{1}(4)\end{bmatrix}, \begin{bmatrix}J_1(2)&O\\O&J_{-1}(4)\end{bmatrix}, \begin{bmatrix}J_{-1}(4)&O\\O&J_{1}(2)\end{bmatrix} \right\}$$
| Any of the Jordan blocks cannot be of size $3$ (or more) because it would then correspond to a factor $(\lambda\pm 1)^{3\text{ or more}}$ in the minimal polynomial. However, to get to the minimal polynomial with factors $(\lambda\pm 1)^2$, there must be at least one Jordan block of size 2. Therefore, the solutions for $n=6$ are:
$$\left[\begin{array}{cc|cc|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&1&0&0\\0&0&0&1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$
$$\left[\begin{array}{cc|c|c|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&0&0&0\\\hline0&0&0&1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$
$$\left[\begin{array}{cc|cc|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&-1&1&0&0\\0&0&0&-1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$
$$\left[\begin{array}{cc|cc|c|c}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&-1&1&0&0\\0&0&0&-1&0&0\\\hline0&0&0&0&-1&0\\\hline0&0&0&0&0&-1\end{array}\right]$$
$$\left[\begin{array}{cc|c|cc|c}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&0&0&0\\\hline0&0&0&-1&1&0\\0&0&0&0&-1&0\\\hline0&0&0&0&0&-1\end{array}\right]$$
| {
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Understanding quadratic reciprocity I'm reading this proof and I don't think I understand the notion of quadratic reciprocity properly.
$q$ is a prime. $p = 2q + 1$.
I'm referring to section (6) which claims:
$p\equiv 2\pmod{5}$ so by quadratic reciprocity $5$ is a quadratic non-residue modulo $p$.
I know that $$\left( \frac{p}{5} \right) \cdot \left( \frac{5}{p} \right) = (-1)^{\frac{(p-1)(5-1)}{4}} = 1$$
How can deduce that $5$ is quadratic non residue from this?
Thanks
By the way,
As a side question: Is there a quick way finding out that if $q\equiv \pm 2 \pmod {5}$ then $p\equiv 0,2 \pmod{5}$?
I did this:
*
*If $q\equiv 2 \pmod{5}$ then $p = 2(5k+2) +1 = 10k + 5 \equiv 0 \pmod{5}$
*If $q \equiv -2 \pmod{5}$ then $p = 2(5k-2) + 1 = 10k - 3 \equiv 2 \pmod{5}$
| As $p\equiv 2\mod 5$, $\biggl(\dfrac p5\biggr)=\biggl(\dfrac 25\biggr)=-1$, so by the quadratic reciprocity law $\biggl(\dfrac 5p\biggr)=-1$.
Second point: if $q\equiv \pm2\mod 5$ and $p=2q+1$, then $p\equiv 2\cdot2+1\equiv 0$ or $p\equiv 2(-2)+1\equiv1+1=2$.
| {
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How do I evaluate $\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$?
$$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$$
This is the answer WolframAlpha gives:
substituting it with polar coordinates makes it worse..
| You just need to evaluate the integrals from inside. When you are integrating with respect to $x$, the other variables ($y$ and $z$) are constants:
$$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z=$$$$\int_{1}^{2} \int_{0}^{z}\left.-\frac{1}{2(x+y+z)^2}\right|_{x=0}^{x=y+z}\text{d}y\text{d}z=$$$$-\frac{1}{2}\int_{1}^{2} \int_{0}^{z}\frac{1}{(2y+2z)^2}-\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$-\frac{1}{2}\int_{1}^{2} \int_{0}^{z}\frac{1}{2^2}\frac{1}{(y+z)^2}-\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$\frac{3}{8}\int_{1}^{2} \int_{0}^{z}\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$\frac{3}{8}\int_{1}^{2}\left.-\frac{1}{y+z}\right|_{y=0}^{y=z}\text{d}z=$$$$-\frac{3}{8}\int_{1}^{2}\frac{1}{2z}-\frac{1}{z}\text{d}z=$$$$\frac{3}{16}\int_{1}^{2}\frac{1}{z}\text{d}z=$$$$\frac{3}{16}\log(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Real Fundamental- System/Matrix of a Differential equation We consider:
$$y''' - 2y'' + 2y' - y = 0$$
The real solution to this equation is:
$$y(x) = c_3e^{x} + c_2e^{x/2}sin\left(\frac{\sqrt{3}x}{2}\right) + c_1e^{x/2}cos\left(\frac{\sqrt{3}x}{2}\right)$$
How do we now represent it as a fundamental- system/matrix ?
| Proceeding from
$$X' = Ax = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$
we find the eigenvalues $\lambda_1 = 1, \lambda_2 = \frac{1}{2}\left(1 + i\sqrt{3}\right), \lambda_3 = \frac{1}{2}\left(1 - i\sqrt{3}\right)$, which correspond to the following eigenvectors: $$v_1 = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, v_2 = \begin{pmatrix}\frac{1}{2}\left(-1 - i\sqrt{3}\right)\\
\frac{1}{2}\left(1 - i\sqrt{3}\right)\\ 1\end{pmatrix}, v_3 = \begin{pmatrix}\frac{1}{2}\left(-1 + i\sqrt{3}\right)\\
\frac{1}{2}\left(1 + i\sqrt{3}\right)\\ 1\end{pmatrix}$$
The Real Fundamental System is then:
$$y_1(x) = e^x\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}
\\y_2(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix}
\\y_3(x) = e^{\frac{1}{2}x}\begin{pmatrix}\frac{1}{2}(-\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ \frac{1}{2}(\cos(\frac{\sqrt{3}x}{2})+\sqrt{3}\sin(\frac{\sqrt{3}x}{2}))\\ 1\end{pmatrix}$$
Tell me if there's something wrong or another and better representation for the Real Fundamental System.
| {
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Find the Maclaurin Series for $e^{\arctan x}$ up to $\mathcal{O}(n^5)$ The question is to find the value of the Maclaurin series expansion of $e^{\arctan x}$ up to (but not including) $\mathcal{O}(n^5)$.
I tried using the Maclaurin series for $e^{u}$ then subbing $\arctan x$ into it, and I got $$\sum_{n=0}^{\infty } \frac{(\arctan x)^{n}}{n!}$$
Is this the correct approach to solving this problem or am I missing something? Cause if so I'm not sure when to stop to reach $\mathcal{O}(n^5)$
| https://en.wikipedia.org/wiki/Exponential_formula
\begin{align}
\arctan x & = x - \frac{x^3} 3 + \frac{x^5} 5 - \frac{x^7} 7 + \cdots \\[10pt]
& = a_1 x + a_2 \frac{x^2} 2 + a_3 \frac{x^3} 6 + a_4 \frac{x^4}{24} + \cdots \\[10pt]
a_1 & = 1 \\
a_2 & = 0 \\
a_3 & = -1/3 \\
a_4 & = 0 \\
a_5 & = +1/5 \\
& \,\,\,\vdots \\[10pt]
e^{\arctan x} & = 1 + b_1 x + b_2 \frac{x^2} 2 + b_3 \frac{x^3} 6 + b_4 \frac{x^4} {24} + \cdots \\[10pt]
b_1 & = a_1 \\
b_2 & = a_1^2 + a_2 \\
b_3 & = a_1^3 + 3a_1 a_2 + a_3 \\
b_4 & = a_1^4 + 4a_1 a_3 + 3a_2^2 + 6a_1^2 a_2 + a_4 \\
& \text{etc.}
\end{align}
The pattern is this:
The coefficient of $a_2^2$ is $3$ because there are three ways to partition a set of four objects into two sets of two:
$$
ab/cd, \qquad ac/bd, \qquad ad/bc
$$
The coefficient of $a_1^2 a_2$ is $6$ because there are six ways to partition a set of four objects into two sets of one and a set of two:
$$
a/b/cd, \qquad a/c/bd, \qquad a/d/bc, \qquad b/c/ad, \qquad b/d/ac, \qquad c/d/ab
$$
And so on.
Thus
$$
e^{\arctan x} = 1 + x + x^2 + \frac{x^3} 9 - \frac{x^4}{72} + \cdots
$$
You should probably check my arithmetic.
Next we have
$$
b_5 = a_1^5 + 10a_1^3 a_2 + 15a_1 a^2_2 + 10a_1^2 a_3 + 10a_2 a_3 + 5a_1 a_4 + a_5.
$$
| {
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Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem, i applied AM-GM inequality for $(x \;, \;6-x)$
$$\Biggl (\frac{(x+(6-x)}{2}\Biggr) \ge \sqrt {x(x-6)}$$
Expanding out we get $$(x-3)^2\ge0$$
Holding the equality when $x=3$
We can do the same with $(y \;, \; 6-y)$ and $(z \;, \; 6-z)$ getting $(y-3)^2\ge0$ and $(z-3)^2\ge0$ holding when $y,z =3$ and getting that one solution for the system is $x=y=z=3$ but i don't know if this is enough for proving that those are the only solutions.
| You got till $x(6-x)y(6-y)z(6-x)=9^3$.
Now it just remains to note by AM-GM, $x(6-x)y(6-y)z(6-x) \leqslant \left(\frac16(x+6-x+y+6-y+z+6-z) \right)^6=3^6=9^3$ with equality possible iff $x=6-x=y=6-y=z=6-z$. So $x=y=z=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving general cubic without complicated substitutions I am trying to solve a general cubic without complicated substitutions:
$$
ax^3+bx^2+cx +d = 0
$$
Assuming nonzero $a$, after dividing both sides by $a$, moving the $\frac{d}{a}$ term to the RHS and performing further simplifications I get:
$$
(x + \frac{b}{3a}) [ (x+\frac{b}{3a})^2 - \frac{b^2-3ac}{3a^2} ] = \frac{9abc-27a^2d-2b^3}{27a^3}
$$
On the LHS I seem to have a product of a linear function and a quadratic function in a vertex form, however I am stuck as to how to proceed next because of the RHS term.
I notice that the $b^2-3ac$ term is $\Delta_0$ and the $9abc-27a^2d-2b^3$ term is, in my case, $-\Delta_1$ from Wikipedia's general formula of the cubic function.
How can I take it further from here?
| Consider an equation of degree $3$ over $\mathbb{C}$: $ R(x)=x^3+ux^2+vx+w=0$ where $u,v,w\in\mathbb{C}$.
Putting $ x=y-u/3$, we obtain an equation in the form $ y^3+px+q=0$.
If $ p=0$, then it's easy, else:
Putting $ y=z+b/z$, we obtain an equation in the form
(1) $ z^6+a_4z^4+a_3z^3+a_2z^2+a_0=0$.
The system $ a_4=a_2=0$ gives $ b=-p/3$.
With this value of $ b$, (1) is easily solved: $ 2$ values for $ z^3$, therefore $ 6$ values for $ z$. Finally, the formula $ y=z+b/z$ gives $ 2$ times each root of $ y^3+px+q$ and we are done.
Example: $R(x)=x^3-3x^2+x-1$; $x=y+1$ gives $y^3-2y-2=0$. $y=z+2/(3z)$ gives $z^6-2z^3+8/27=0$, that is $z^3=1\pm (1/9)\sqrt{57}$; finally the $6$ values of $z$ are
$z=\rho (1\pm (1/9)\sqrt{57})^{1/3}$ where $\rho\in\{1,\dfrac{-1\pm i\sqrt{3}}{2}\}$ is a cubic root of $1$.
| {
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Probability of $3+3$ cards, out of $6$ cards drawn from a solitaire
A solitaire consists of $52$ cards. We take out $6$ out of them (wihout repetition). Find the probability there are $3+3$ cards of the same type (for example, $3$ "1" and $3$ "5").
Attempt.
First approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and by the multiplication law of probability, the desired probability is $$\binom{13}{2}\frac{4}{52}\,\frac{3}{51}\,\frac{2}{50}\,
\frac{4}{49}\,\frac{3}{48}\,\frac{2}{47}.$$
Second approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and the desired probability is $$\binom{13}{2}\frac{\binom{4}{3}\binom{4}{3}\binom{4}{0}\ldots\binom{4}{0}}{\binom{52}{6}}.$$
These numbers don't coincide, so I guess (at least) one of them is not correct.
Thanks in advance for the help.
| There are $13$ possible numbers/people. We need to pick $3$, so the probability is $\binom{4}{3} \cdot \binom{13}{2} \cdot \binom{4}{3}$. The total probability of choosing 6 cards is $\binom{13}{2}$. The total probability is $\frac{4 \cdot 78 \cdot 4}{20358520}$
, which equals $\frac{1248}{20358520} \Rightarrow \approx 6.13011162*10^{-5}$
| {
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How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$
| Hint:
As $\dfrac{d(2x-x^2)}{dx}=2-2x$
$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^2-2x+2x-2+2}{\sqrt{2x-x^2}}$$
$$=-\sqrt{1-(x-1)^2}-\dfrac{2-2x}{\sqrt{2x-x^2}}+\dfrac2{\sqrt{1-(x-1)^2}}$$
Now use $\#1,\#8$ of this
| {
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Simplifying tensor equation I was given a following tensor equation to simplify and solve:
$$A^s \cdot (A^s \otimes A^a \otimes I) \cdot A^s$$ where:
$A^s$- symmetric part of the representation; $A^s=\frac{1}{2}(A-A^T)$
$A^a$- anti-symmetric part of the representation; $A^a=\frac{1}{2}(A+A^T)$
$I$ - Kronecker delta, here
$I =\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$
$\otimes$ - Kronecker product
I have already calculated the $A^s$ and $A^a$ from the base tensor $A$ (in the exercise the basis is orthonormal, so $A^T = A_{ji}$):
$A =
\begin{pmatrix}
0 & 0 & 2\sqrt{3} \\
0 & 6 & 0 \\
0 & 0 & 2
\end{pmatrix}$
$A^s =
\begin{pmatrix}
0 & 0 & \sqrt{3} \\
0 & 6 & 0 \\
\sqrt{3} & 0 & 2
\end{pmatrix}$$
A^a =
\begin{pmatrix}
0 & 0 & \sqrt{3} \\
0 & 0 & 0 \\
-\sqrt{3} & 0 & 0
\end{pmatrix}$
I have no idea what "simplifying" in this case means. Maybe there is some kind of relationship between $A^s$ and $A^a$ which I am not aware of and that's why it seems so unclear. Can you help me with the answer of provide tips on how to crack this problem?
| I went to ask about this and got a solution. As all of the tensors here are $T^2$, using tensor properties, the whole equation can be simplified to:
$$(A^s \cdot A^s) \cdot A^a \cdot (I \cdot A^s) = \text{||using associative property||} = (A^s \cdot A^s)(I \cdot A^s) \cdot A^a$$
Because, as I said above, the tensors are $T^2$, then the "tensor multiplication" will reduce the row of the answer to $T^{|2-2|} = T^0 = R$ - it will be a scalar (where the result is $A\cdot B = A_{ij}B_{ij}$ - using summing convention). At the end we get multiplication of 2 scalars and multiplication of the product by the $A^a$ tensor:
$$(3+36+3+4)(6+2)\begin{pmatrix}
0 & 0 & \sqrt{3} \\
0 & 0 & 0 \\
-\sqrt{3} & 0 & 0
\end{pmatrix}=368\begin{pmatrix}
0 & 0 & \sqrt{3} \\
0 & 0 & 0 \\
-\sqrt{3} & 0 & 0
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using integration to solve a formula for the area of a ellipse Problem:
Set up a definite integral to find the area of an ellipse with axis lengths $a$ and $b$. Use a trigonometric substitution to find a formula for the area. What happens if $a=b$? Does this agree with a Geometry formula for a circle? Explain.
$$\frac {x^2}{a^2} + \frac{y^2}{b^{2}} = 1$$
where $a$ & $b$ are positive constants.
$B$ = Area of the first quadrant of the ellipse.
Total area = $4B$
\begin{align}
& y^{2} = \left(\frac{b^{2}}{a^{2}}\right)({a^{2}-x^{2}}) \\
\implies & y = \frac{b}{a}\sqrt{a^{2}-x^{2}}
\end{align}
Hence,
\begin{align}
B & = \int_0^a{\frac{b}{a}\sqrt{a^{2}-x^{2}}}\;dx = \frac{b}{a}\int_0^a{\sqrt{a^{2}-x^{2}}}\;dx
\end{align}
To remove $\sqrt{\quad}$, make a trig sub.
$$1 - \sin^{2} \Theta = \cos^{2}\Theta$$
$$x = a\,\sin\Theta$$
\begin{align}
B & = \frac{b}{a}\int_0^a{\sqrt{a^{2}-a^{2}\sin^{2}\Theta}\;dx} = \frac{b}{a}\int_0^a{\sqrt{a^{2}(\cos^{2}\Theta)}}\;dx\\
& = \frac{b}{a}\int_0^a{\sqrt{a^{2}}\,\sqrt{\cos^{2}\Theta}}\;dx = \frac{b}{a}\int_0^a{a\,\cos\Theta}\;dx
\end{align}
\begin{align}
\frac{dx}{dΘ} = (a \, \sin\Theta)' \implies dx = a \, \cos \Theta \, d\Theta
\end{align}
Therefore,
\begin{align}
B & = \frac{b}{a}\int_0^a{a \, \cos\Theta(a \, \cos \Theta \, d \Theta)} = \frac{a^{2} \, b}{a}\int_0^a{\cos^{2}\Theta}\;d\Theta = (a \, b)\int_0^a{\cos^{2}\Theta}\;d\Theta
\end{align}
I am a little lost up until this point and the formula doesn't seem to be going in the direction it needs to so that it will become the area of an ellipse. I feel like I made a mistake somewhere along the way. Any help is much appreciated!
| Hint...You just need to change the limits to $0$ and $\frac {\pi}{2}$ and use the identity $\cos^2\theta=\frac 12(1+\cos2\theta)$ and you will be finished
| {
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Problem on row Echelon form Consider a $3\times 3$ matrix $$A =\begin{bmatrix}1 & 2 & -1\\2&1&0\\ 3&0&1\\\end{bmatrix}.$$ I have to find nonsingular matrix $P$ such that $PA$ is in row reduced Echelon form.
I am not able to get any idea to solve this problem. I understand Echelon form of a matrix. But what exactly should I do to solve this problem?
Thanks
| We can obtain P in this way by left multiplication
$$P(I|A)=(P|PA)$$
thus consider $(I|A)$
$$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &2 & -1\\ 0 & 1 & 0 & 2 &1 &0 \\ 0 & 0 & 1 & 3 &0 &1\end{array}\right]$$
and by row operations
$$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &2 & -1\\ -2 & 1 & 0 & 0 &-3 &2 \\ -3 & 0 & 1 & 0 &-6 &4\end{array}\right]$$
$$\left[\begin{array}{rrr|rrr} 1 & 0 & 0 & 1 &2 & -1\\ -2 & 1 & 0 & 0 &-3 &2 \\ 1 & -2 & 1 & 0 &0 &0\end{array}\right]$$
thus
$$P =\begin{bmatrix}1 & 0 & 0\\-2&1&0\\ 1&-2&1\\\end{bmatrix}$$
| {
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Inequality Proof $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$ Let $a,b,c\in \mathbb{R}^+$, and $a^2+b^2+c^2=1$, show that:
$$ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq \frac{3\sqrt{3}}{2}$$
| Here is a hint: it is enough to show for $x\in (0, 1)$,
$$f(x) = \left(\frac{x}{1-x^2}-\frac{\sqrt3}2\right)-\frac{3\sqrt3}2\left(x^2-\frac13 \right) \geqslant 0$$
$$\iff \frac{x(\sqrt3x+2)(3x-\sqrt3)^2}{6(1-x^2)} \geqslant 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding value of Quadratic If the quadratic equations $3x^2+ax+1=0$ and $2x^2+bx+1=0$ have a common root, then the value of $5ab-2a^2-3b^2$ has to be find.
I tried by eliminating the terms but ended with $(2a-b)x=1$. Can you please suggest how to proceed further?
| The common root is where two curves of functions $y_1=3x^2+ax+1$ and $y_2=2x^2+bx+1$ intersect; so we may write:
$3x^2+ax+1=2x^2+bx+1$ ⇒ $x^2 +(a-b)x=0$ ⇒ $ x[x+(a-b)]=0$
$x=0$ is not acceptable
$x= b-a$ can be a solution; plugging this in one of equations we get:
$3(b-a)^2 +a(b-a) +1=0$
Reducing this relation we get:
$5ab -2a^2-3b^2=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the inequality analytically?
Solve the inequality analytically.
$$ \log_x{(\sqrt{x^2 + 2x - 3} + 2)} \cdot \log_5{(x^2 + 2x - 2)} \ge \log_x{4} $$
My solution
$$
\left.
\left\{
\begin{array}{l}
x > 0 \\
x \ne 1 \\
x^2 + 2x - 3 \ge 0 \\
x^2 + 2x - 2 > 0
\end{array}
\right.
\right\vert \Rightarrow x \in (1; + \infty).
$$
Since $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$,
$$
\frac{\log_5(\sqrt{x^2 + 2x - 3} + 2) \cdot \log_5{(x^2 + 2x -2)}}{\log_5{x}}
\ge
\frac{\log_5{4}}{\log_5{x}}
$$
$$
\log_5{(\sqrt{x^2 + 2x - 3} + 2)} \cdot \log_5{(x^2 + 2x - 2)} \ge \log_5{4}.
$$
Let $b = x^2 + 2x$,
$$
\log_5{(\sqrt{b - 3} + 2)} \cdot \log_5{(b - 2)} \ge \log_5{4}.
$$
Unfortunately, that's all that came to mind.
P.S. Excuse me for my bad English! It isn't my native language.
| You have correctly eliminated $x \leqslant 1$. For $x> 1$, we may take powers of $x$ on both sides and have the equivalent:
$$(x^2+2x-3)^{\log_5(x^2+2x-2)} \geqslant 4$$
Now if $b=x^2+2x-2 \in (1, 5)$, then in the LHS, the exponent is $\in (0, 1)$, while the base is $\in (0, 4)$. Clearly the LHS will be less than RHS here.
We are left with $b \geqslant 5$ which gives exponent $ \geqslant 1$ and base $\geqslant 4$, which clearly, will always work. Thus what is left to solve is $x^2+2x-2\geqslant 5$, which is left for you...
| {
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Bayes's rule and unfair coin | Solution Explanation
There are three coins in a bag. Two of them are fair. One has heads on
both sides. A coin selected at random shows heads in two successive
tosses.
What is the conditional probability of obtaining another head in the
third trial given the fact that the first two trials showed heads.
I think this problem should be solved in the following way
$$P(one\ more\ head) = \frac{1}{3}\cdot 1 +\frac{1}{3}\cdot \frac{1}{2}+ \frac{1}{3}\cdot \frac{1}{2} = \frac{2}{3} $$
but my book says the right solution is
$$P(HHH|HH) = \frac{5}{6}$$
But the first two trials do not affect the third trial, so I should only have to calculate the probability of getting one more head, since I already have two.
Can anyone explain me what is going on?
| Bayes' Rule says
$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$
so in your question, it is
$$P(HHH|HH) = \frac{P(HH|HHH)\cdot P(HHH)}{P(HH)}$$
Now, notice that
$$P(HH|HHH) = 1$$
because it simply says "if we know that three successive tosses are resulted in $HHH$, what is the probability that first two of them are resulted in $2H$".
For $P(HHH)$, we have
$$P(HHH) = \frac{1}{3}\cdot1+\frac{2}{3}\cdot \frac{1}{8} = \frac{5}{12}$$
because if we choose unfair coin with $\frac{1}{3}$ probability, we have $HHH$ with probability $1$ and if we don't choose it with $\frac{2}{3}$, we have $\frac{1}{2} \cdot\frac{1}{2} \cdot\frac{1}{2} = \frac{1}{8}$ probability to have $HHH$.
For $P(HH)$, by similar argument to $P(HHH)$, we have
$$P(HH) = \frac{1}{3}\cdot 1+\frac{2}{3} \cdot\frac{1}{4} = \frac{1}{2}$$
So the answer is
$$P(HHH|HH) = \frac{1\cdot \frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$$
Now, when it comes to your argument "the first two trials do not affect the third trial", it is wrong in some manner. Because although it seems third trial is not affected by the first two trials, notice that whether we have fair or unfair coin affects probability of having $HH$ in the first two tosses. In other words, since $P(HH)$ differs from fair coin to unfair coins, probability of having a third $H$ is also affected by $P(HH)$, hence the first two tosses.
| {
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Probability Problem based on the a set of numbers. Determine if the number is divisible by three Three distinct numbers are selected at random from the set $\{1,2,3,4,5,6\}$. What is the probability that their product is divisible by $3$?
I think that since because $3$ and $6$ are the only numbers that divide into three with an integer remainder, the number would have to have $3$ or $6$ in the unit's digit. Since this is the case the probability would be $2/6$ or $1/3$. Have I done anything wrong?
| We need to find the probability that at least one of the numbers is divisible by three. This is the same as
$$\begin{align*}
1-P(\text{none of the numbers are divisible by three})
&=1-\left(\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)\\\\
&=1-0.2\\\\
&=0.8
\end{align*}$$
In order to get three numbers that are not $3$ or $6$, we must first select $1,2,4,$ or $5$ out of the six numbers giving a probability of $\frac{4}{6}$.
Take that number away. We must now select one of the remaining three numbers of the five with probability $\frac{3}{5}$.
Finally, we must now select one of the two remaining numbers of the four with probability $\frac{2}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove determinant is negative Prove that the determinant $\Delta$ is negative
$$
\Delta=\begin{vmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{vmatrix}<0
$$
where $a,b,c$ are positive and $a\neq b\neq c$.
My Attempt:
Applying Sarrus' rule,
$$
\begin{matrix}
a&b&c&a&b\\
b&c&a&b&c\\
c&a&b&c&a
\end{matrix}
$$
$$
\Delta=acb+bac+cba-c^3-a^3-b^3=3abc-(a^3+b^3+c^3)\\
=-\Big[(a^3+b^3+c^3)-3abc\Big]
$$
How do I prove that $(a^3+b^3+c^3)-3abc>0$ thus prove $\Delta<0$ ?
| By the AM-GM inequality:
$$
\frac{a^3+b^3+c^3}{3} \ge \sqrt[3]{a^3b^3c^3} = abc
$$
The strict inequality holds unless $\,a=b=c\,$.
| {
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Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$ Related (but different) question: Is it true that $76^n=76\pmod{100}$ for all $n>0$?
If you raise $76$ to an integer power ($n\ge2$) and ignore the last two digits, it appears that you always have an integer that is divisible by $57$, because the prime factorisation always seems to involve $3$ and $19$:
$76^2$ = 5776, and 57 = 3 x 19
$76^3$ = 438976, and 4389 = 3 x 7 x 11 x 19
$76^4$ = 33362176, and 333621 = 3 x 3 x 19 x 1951
$76^5$ = 2535525376, and 25355253 = 3 x 7 x 11 x 19 x 53 x 109
$76^6$ = 192699928576, and 1926999285 = 3 x 5 x 19 x 71 x 95231
$76^7$ = 14645194571776, and 146451945717= 3 x 3 x 7 x 11 x 19 x 1951 x 5701
| Another way to look at it:
$\frac {76^n - 76}{100} = \frac {76(76^{n-1} - 1)}{100} = \frac {19(76^{n-1}-1)}{25} = 19* \frac {(76 -1)(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} =$
$19*\frac {75(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} = 3*19*(76^{n-1} + 76^{n-2} + ...... + 76 + 1)$
So that's that.
But there so some interesting observations. If $n = m + 1$ is odd and $m$ is even then: $(76^{n-1=m} - 1) = (76^{2} - 1)(76^{m-2}+76^{m-4} + ... + 1) = (76-1)(76 + 1)(76^{m-2}+76^{m-4} + ... + 1)$ so $7=7*11|7^{n-1} - 1$ so for ever odd $n$, $57*77$ will divide it.
Likewise if $n = 3m + 1$ for some $m$ then $(76^{3m} - 1) = (76-1)(76^2 + 76 + 1)(76^{3m -3} + 76^{3m-6}.... + 1) = 7*(3*1951)*(76^{3m -3} + 76^{3m-6}.... + 1)$.
So for $n = 4,7,10, etc.$ you will get $3*19*3*1951$ will divide the values.
And we can find these patterns all day long.
$n = km + 1$ will always mean $76^{m-1}+ 76^{m-2} + ... + 76+ 1$ will divide the result.
For $n = 4m + 1$ we well always have $\frac{(76^2 + 1)(76^2 - 1)}{25}=3*53*109*7*11$ will always divide the results.
And so on.
| {
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How many positive integer solutions satisfy the condition $y_1+y_2+y_3+y_4 < 100$ In preparation for an upcoming test I have come across the following problem and I am looking for some help with it just in case a question of its kind comes up on a evaluation. Thanks!
How many positive integer solutions satisfy the condition:$$y_1+y_2+y_3+y_4 < 100$$
| Lets say the question was:
$$y_1+y_2+y_3+y_4 = 4$$
There is only $1$ solution:
$$1+1+1+1 = 4$$
Now lets assume it was:
$$y_1+y_2+y_3+y_4 = 5$$
There are $4$ solutions:
$$1+1+1+2=5$$
$$1+1+2+1 = 5$$
$$1+2+1+1 = 5$$
$$2+1+1+1 = 5$$
Next assume it was:
$$y_1+y_2+y_3+y_4 = 5 $$
There are $10$ solutions:
$$1+1+1+3 = 6$$
$$1+1+2+2 = 6$$
$$1+1+3+1 = 6$$
$$1+2+2+1 = 6$$
$$1+2+1+2 = 6$$
$$1+3+1+1 = 6$$
$$2+1+1+2 = 6$$
$$2+1+2+1 = 6$$
$$2+2+1+1 = 6$$
$$3+1+1+1 = 6$$
There's a pattern:
$$n=4 \implies \binom{3}{3} = 1$$
$$n=5 \implies\binom{4}{3} = 4$$
$$n=6 \implies\binom{5}{3} = 10$$
$$n=7 \implies\binom{6}{3} = 20$$
$$n=8 \implies\binom{7}{3} = 35$$
Then by adding up all the terms up to $98$ you get the answer:
$$\sum_{i=3}^{98}\binom{i}{3} =\binom{99}{4}= 3764376$$
Indeed by Hockey-stick identity we have
$$\sum^n_{i=r}{i\choose r}={n+1\choose r+1} \qquad \text{ for } n,r\in\mathbb{N}, \quad n>r$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility of a Polynomial Prove that for all $n$, $121\mid n^2+3n+5$.
I thought proving $n^2 +3n+5\pmod{121}\equiv 0$ had no solutions would be a good start, so I completed the square of $n^2+3n+5($, and arrived at $(n+62)^2 \pmod{121} \equiv 3718$. Replacing $n+62$ by $x$, I lastly arrived at $x^2 \equiv 88\pmod{121}$. Now, Im stuck. Any hints would be appreciated.
| I think you want to prove that $121\nmid(n^2+3n+5)$, that is, $121$ does not divide $n^2+3n+5$.
First of all ask yourself whether $n^2+3n+5$ is a multiple of $11$. The equation $n^2+3n+5\equiv0\pmod{11}$ can be rewritten
$$
n^2-8n+16=(n-4)^2\equiv0\pmod{11}
$$
which only holds for $n\equiv4\pmod{11}$.
Now we know that a possible solution of $n^2+3n+5\equiv0\pmod{121}$ should have $n=11k+4$, so
$$
n^2+3n+5=121k^2+88k+16+33k+12+5=121(k^2+k)+33
$$
which is never $\equiv0\pmod{121}$
| {
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How to find the sum of the following series from $n=0$ to $n=99$ Find the summation of the series from $n=0$ to $n=99$. The question was given in the format of
$$(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100).$$
I was able to generalise it but could not solve it. Help!! the general formula is summation (2n+2)!/(2n)!
| We have that
$$n(n-1)=\frac{n^3-(n-1)^3}{3}-\frac{1}{3}.$$
Therefore
$$\begin{align}
(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100)&=\sum_{n=1}^{50}(2n-1)(2n)=4\sum_{n=1}^{50}n(n-1)+2\sum_{n=1}^{50}n\\
&=4\sum_{n=1}^{50}\frac{n^3-(n-1)^3}{3}-\frac{4\cdot 50}{3}+50\cdot 51\\
&=\frac{4}{3}\left(50^3-0^3\right)-\frac{200}{3}+50\cdot 51=169150
\end{align}$$
where we noted that the last sum on the right is telescopic.
| {
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Formulating LPP from given optimal table in Simplex method I have a problem set in my assignment sheet that asks to formulate a Linear Programming Problem from a given optimal solution and we know that method used is Simplex method.
What is the correct procedure for going in reverse order? I know our basic thinking to go for such a problem is to make s1 and s2 equal to 0 in z row.
Here is the snapshot of the problem: Question in Assignment
| To explain why the basis matrix can be read from the given simplex tableau, I introduce these additional symbols:
*
*$A \in M_{m\times n}(\Bbb R)$ ($m\le n$) has rank $n$ and current basis matrix $B$.
*$x_B$ denotes the current basic solution.
*$c_B$ denotes the reduced objective function so that $c^T x=c_B^T x_B$. (So the order/arrangement of basic variables is very important.)
As the comment from the asker of the question Need help finding unknowns in simplex tableau. suggests, $B^{-1}$ can be directly read under the columns for slack variables because the initial tableau has the form
\begin{array}{cc|c}
-c^T & 0 & 0 \\ \hline
A & I & b
\end{array}
Multiplying $B^{-1}$ on both sides gives
\begin{array}{cc|c}
c_B^T B^{-1}A-c^T & c_B^T B^{-1} & c_B^T B^{-1}b \\ \hline
B^{-1}A & B^{-1} & B^{-1}b \tag1 \label1
\end{array}
It's given that $B^{-1} = \begin{bmatrix} 2/7 & -1/7 \\ -1/7 & 4/7 \end{bmatrix}$. It's easy to (mentally) calculate $B = \begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix}$.
To calculate $A$ in the initial tableau, we left multiply $B^{-1}A$ by $B$. ($A = B(B^{-1}A)$). The third column of $A$ is
$$\begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix} 1/7 \\ 17/7 \end{bmatrix}
= \begin{bmatrix} 3 \\ 5 \end{bmatrix}.$$
$$\therefore A = \begin{bmatrix} 1 & 4 & 3 \\ 2 & 1 & 5 \end{bmatrix}$$
$$b = B(B^{-1}b) = \begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
To compute the objective function $c$, note that
\begin{align}
c^T &= c_B^T B^{-1}A-(c_B^T B^{-1}A-c^T) \\
&= \begin{bmatrix} 4\\2 \end{bmatrix}^T
\begin{bmatrix} 0 & 1 & 1/7 \\ 1 & 0 & 17/7 \end{bmatrix}
- \begin{bmatrix} 0 & 0 & 17/7 \end{bmatrix} \\
&= \begin{bmatrix} 2 & 4 & 3 \end{bmatrix}
\end{align}
Therefore, the LPP is to maximize $2x_1+4x_2+3x_3$ subject to
$$x_1+4x_2+3x_3 \le 1 \\
2x_1+x_2+5x_3 \le 2 \\
x_1,x_2,x_3 \ge 0.$$
| {
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>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)^2-4(2y-1)(8y-5)\geq 0$$
$$4(2y-1)^2-(2y-1)(8y-5)\geq 0$$
$y\geq 0.5$
Could some help me where I have wrong, thanks
| You can simplify the expression as
$$\frac12+\frac1{2(\sin x+2)^2}$$ and the extreme values are
$$\frac12+\frac1{2\cdot3^2},\\\frac12+\frac1{2\cdot1^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$
However obviously if you replace it directly to calculate $S$, it would be extremely time-consuming.
Is there another way? I think the first equation can be changed (without needing to solve it) to calculate $S$, as when I use brute force (with calculator), I found out that $S\approx 1.414213562\approx\sqrt 2$.
| Hint: Use
$$4a^2+\sqrt 2 a-\sqrt 2=0 \implies a^2=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a$$
and
$$a^4 = \dfrac{2}{16}-2\dfrac{\sqrt{2}}{4}\dfrac{\sqrt{2}}{4}a+\dfrac{2}{16}a^2$$
$$= 1/8-a+1/8\left[\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a\right].$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $ \int_{0}^{\pi/2} \frac{\sin(nx)}{\sin(x)}\,dx $ For every $odd$ $n \geq 1$ the answer should be $\pi/2$
For every $even$ $n \geq 2$ the possible answers are :
$A )$ $ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} $
$B )$ $ 3(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$C )$ $ 2(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n/2+1}\frac{1}{n-1} )$
$D )$ $ \frac{1}{2}(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots + (-1)^{n+1}\frac{1}{n-1} )$
$E )$ $ 1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\cdots - \frac{1}{n-1} $
Does the recurrence $ (n-1)(I_{n}-I_{n-2})=2sin(n-1)x $ help?
| For $n$ even, say $2m$, we can write
$$\begin{align}
\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx&=\text{Im}\int_0^{\pi/2}\frac{e^{i2mx}}{\sin(x)}\,dx\\\\
&=2\text{Re}\int_0^{\pi/2}\frac{e^{i2mx}}{e^{ix}-e^{-ix}}\,dx\tag1
\end{align}$$
Then, letting $z=e^{ix}$ and using long division reveals that
$$\begin{align}
\text{Re}\left(\frac{e^{i2mx}}{e^{ix}-e^{-ix}}\right)&=\text{Re}\left(\frac{z^{2m+1}}{z^2-1}\right)\\\\
&=\text{Re}\left(\sum_{k=1}^{m}z^{2k-1}+\frac{1}{z-1/z}\right)\\\\
&=\sum_{k=1}^{m} \cos((2k-1)x)\tag 2
\end{align}$$
Then, substituting $(2)$ into $(1)$ we find that
$$\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx=2\sum_{k=1}^{m} \frac{(-1)^{k-1}}{2k-1}$$
If $n=2m+1$, then we see that
$$\begin{align}
\int_0^{\pi/2}\frac{\sin((2m+1)x)}{\sin(x)}\,dx&=\text{Im}\int_0^{\pi/2}\frac{e^{i(2m+1)x}}{\sin(x)}\,dx\\\\
&=2\text{Re}\int_0^{\pi/2}\frac{e^{i(2m+1)x}}{e^{ix}-e^{-ix}}\,dx\tag1
\end{align}$$
Then, letting $z=e^{ix}$ and using long division reveals that
$$\begin{align}
\text{Re}\left(\frac{e^{i(2m+1)x}}{e^{ix}-e^{-ix}}\right)&=\text{Re}\left(\frac{z^{2m+2}}{z^2-1}\right)\\\\
&=\text{Re}\left(\sum_{k=0}^{m}z^{2k}+\frac{1/z}{z-1/z}\right)\\\\
&=\frac12+\sum_{k=1}^{m} \cos(2kx)\tag 2
\end{align}$$
Then, substituting $(2)$ into $(1)$ we find that
$$\int_0^{\pi/2}\frac{\sin(2mx)}{\sin(x)}\,dx=\frac{\pi}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/2675397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Odds of winning a superlottery The problem:
There's a lottery (or a "superlottery", I'm not really sure how that's different). In order to play, I select 8 numbers from the first 90 positive integers (so, 1-90 inclusive). Also, a computer selects 12 numbers from the first 90 positive integers. If all of my 8 numbers are in the set of 12 selected by the computer, I win. Assuming all numbers are randomly selected, what are the odds of winning?
My solution:
Once the computer has chosen its 12 numbers, there are $\binom{12}{8}$ ways for me to pick a winning set of numbers, and $\binom{90}{8}$ ways to pick a set of 8 numbers overall. So, the odds of winning are $\frac{\binom{12}{8}}{\binom{90}{8}}=\frac{1}{156597013}\approx6.38*10^{-9}$.
My friend's solution:
There are $\binom{12}{8}$ ways to choose winning numbers. The odds of the first number matching are $\frac{12}{90}$, the odds of the second number matching are $\frac{11}{89}$, and so on. So the odds of all the numbers matching are $\binom{12}{8}*\frac{12}{90}*...*\frac{5}{83}=\frac{495}{156597013}\approx2.107*10^{-6}$.
Which of us is right (if either), and what mistake is the other person making?
| Your solution is correct and your friend made a common mistake.
Suppose the winning numbers are $\{1,2,3,4,5,6,7,8,9,10,11,12\}$.
The first number selected must be in this set with probability $\frac{12}{90}$.
The second number selected must be one of the remaining numbers with probability $\frac{11}{89}$.
and so forth giving
$$\begin{align*}
p=\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}
&\approx6.386\cdot10^{-9}
\end{align*}$$
which agrees with your solution.
Using this method, there is no need to account for how the numbers are arranged since we arbitrarily let the first number selected to be any of the $12$, the second number selected to be any of the remaining $11$, etc.
Note:
It's easy to show that
$$\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}=\frac{12 \choose 8}{90 \choose 8}$$
We have that
$$\begin{align*}
\frac{12 \choose 8}{90 \choose 8}
&=\frac{\frac{12!}{8!4!}}{\frac{90!}{82!8!}}\\\\
&=\frac{12\cdot11\cdot10\cdot9}{\color{red}{{4\cdot3\cdot2\cdot1}}}\cdot\frac{8\cdot7\cdot6\cdot5\cdot\color{red}{4\cdot3\cdot2\cdot1}}{90\cdot89\cdot88\cdot87\cdot86\cdot85\cdot84\cdot83}\\\\
&=\frac{12}{90}\cdot\frac{11}{89}\cdot\frac{10}{88}\cdot\frac{9}{87}\cdot\frac{8}{86}\cdot\frac{7}{85}\cdot\frac{6}{84}\cdot\frac{5}{83}
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to the equation $x^4+3y^4=z^2$ It was proven that the equation $x^4+2y^4=z^2$ has no non-trivial solutions in integers. What about the equation
$x^4+3y^4=z^2$? It has a solution $x=1,y=1, z=1$. Can we find all solutions?
| Yes, the equation $x^4+3y^4 = z^2$ is birationally equivalent to an elliptic curve, hence we can find all its solutions.
However, if an easy proof there are infinitely many integer solutions with $\gcd(x,y)=1$ will suffice, then given an initial solution to,
$$x^4+dy^4=z^2$$
then, in general, further ones can be generated as,
$$S_n=\frac{x_n}{y_n}=\frac{-x^4+dy^4}{2xyz}$$
Example: Let $d=3$, and initial $x,y,z = 1,1,2$, then,
$$S_1 = \frac11\\
S_2 = \frac12\\
S_3 = \frac{47}{28}\\
S_4 = \frac{3035713}{6824776}
$$
such that,
$$47^4+3\times28^4=2593^2$$
and so on.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?
So this track seems to difficult, another approach. Can I square the area first, find the derivative of that to solve for x?
So the Area = $4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$Area^2 = 16x^2 \cdot (4 - \frac{4x^2}{9}$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} Area^2 = 128x - \frac{256x^3}{9}$$
$$128x(1-\frac{2x^2}{9}$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals 0 when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of x into y we get that $y = \sqrt{2}$ so the Area is 3.
Is this valid? If so why? Does squaring not cause any problems?
| You are over-complicating it. Affine maps preserve the ratios of areas and, in a circle, the inscribed squares are pretty obviously the largest inscribed quadrilaterals. Their area is $\frac{2}{\pi}$ times the area of the circle. By applying $\varphi:(x,y)\mapsto(ax,by)$ the ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is mapped into a circle. By applying $\varphi^{-1}$ we get that
A. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the largest inscribed rectangle (which is also one of the largest inscribed quadrilaterals) is symmetric with respect to the axis of the ellipse and its area equals $2ab$.
By the very same principle,
B. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the area of the largest inscribed triangle equals $\frac{3\sqrt{3}}{4}ab$.
C. In a triangle $ABC$ the largest inscribed ellipse is tangent to the sides at their midpoints and the center of such ellipse lies at the centroid of $ABC$. This is the Steiner inellipse of $ABC$.
| {
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Finding equation and centre of circle through 3 points using matrices Based on answer given here: Get the equation of a circle when given 3 points. We can find equation of circle through points $(1,1), (2,4), (5,3)$ by taking:
$\left|\begin{array}{cccc}
x^2+y^2&x&y&1\\
1^2+1^2&1&1&1\\
2^2+4^2&2&4&1\\
5^2+3^2&5&3&1\\
\end{array}\right|=0$
Can anyone explain why this works? I know generally what determinant equal zero means but can't see why doing this works.
| This comes from the general equation of a circle
$$x^2+y^2+Ax+By+E=0$$ which gives since $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are points in the circle
$$x^2+y^2+Ax+By+E=0\\x_1^2+y_1^2+Ax_1+By_1+E=0\\x_2^2+y_2^2+Ax_2+By_2+E=0\\x_3^2+y_3^2+Ax_3+By_3+E=0$$ This is equivalent to
$$\left (\begin{matrix}x^2+y^2&x&y&1\\x_1^2+y_1^2&x_1&y_1&1\\x_2^2+y_2^2&x_2&y_2&1\\x_3^2+y_3^2&x_3&y_3&1\end{matrix}\right)*\left(\begin{matrix}1\\A\\B\\C\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\\0\end{matrix}\right)$$
Because, clearly, there is not unique solution ( in particular the zero matrix in the RHS is solution) the determinant must be equal to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
\end{align*}
and I'm stuck, should I divide $63$ by $9$ to get $7$? but it does not comply to the given equation as when $x=7$, it would become $16 \pmod{47}$.
| you can write $$x\equiv \frac{3}{9}=\frac{1}{3}\equiv \frac{48}{3}\equiv 16\mod 47$$
| {
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.
| You forgot a minus sign.
$\displaystyle P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(-1)^k(k+1)^2\\=(-1)^{k-1}\dfrac{k.(k+1)}{2}+(-1)^k(k+1)^2\\=\dfrac{(k+1)}{2} [(-1)^{k-1}.k +2(-1)^k (k+1)]\\=\dfrac{(k+1)}{2} [-(-1)^{k}.k +2(-1)^k (k+1)]\\=(-1)^k\dfrac{(k+1)}{2}[-k+2k+2]\\=(-1)^k\dfrac{(k+1)}{2}[k+2]\\=(-1)^k\dfrac{(k+1)(k+2)}{2}$
(Proved)
| {
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
| Using your method, which probably isn't the best:
$2079000 =2^3*3^3*5^3*7*11$ so all factors are of form $2^a3^b*5^c*7^d*11^e$ where $a,b,c$ are between $0$ and $3$ and $d,e$ are between $0$ and $1$ exclusively.
So there $4*4*4*2*2$ factors.
But to be even the $a$ must be at least $1$ and divisible by $15$ they must be divisible by $3$ (why did you overlook $3$????) and $5$ so $b$ and $c$ must be at least $1$. So There are only $3$ options for $a,b,c$; $1$ to $3$ inclusively.
So there are $3*3*3*2*2$ such factors.
But perhaps a better way to do it is $30*k$ is a factor of $207900$ if and only if $k$ is a factor of $\frac {207900}{30}=69300 = 2^2*3^2*5^2*7*11$ so there are $3*3*3*2*2$ such factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fraction and simplification
solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$
What are the possible answers ?
(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0
The answer from where i've referred this is (B), but when i simplify it I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$$
$$ \frac{x +x(x-1)}{x(x-1)\cdot x} = \frac{1}{x-1} \text{ (took l.c.m on l.h.s)}$$
$$ \frac{x + (x^2 -x)}{(x^2 - x)\cdot x}= \frac{1}{x-1}$$
$$\frac{x^2}{x^3 - x^2} = \frac{1}{x-1}$$
$$ x^2(x-1) = x^3 - x^2$$
$$ x^3 - x^2 = x^3 - x^2$$
Have I simplified it correctly?
| You have got an identity, L.H.S. = R.H.S. which will hold for all values of X in domain of equation.
This implies Infinite solution as domain of the equation is infinite.
In this case $X ={0,1}$ are not in domain of equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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general equation to find cubic polynomial from two minimums? I tried researching and found that I can use a system of linear equations and solve by an inverse matrix to find the cubic equation given 4 points which satisfy the function f(x) of the general form $f(x)=ax^3+bx^2+cx+d$
I can also find a cubic of the form $ax^3 + d$ with no $x^2$ or $x$ term from 2 points, however I was wondering how one would go about finding a full general form cubic given only the minimum and maximum.
Example minimums could be (-1,4) and (2,3)
| First find a quadratic that has zeros at the $x$-values of your points. Then integrate. Then adjust the constant to get the correct $y$-values. Using your example, we need a quadratic that passes through the points $(-1,0)$ and $(2,0)$.
So it must have the form $g(x+1)(x-2) = gx^2-gx -2g$.
Integrate this to get $f(x) = \frac{g}{3}x^3-\frac{g}{2}x^2 - 2gx +C.$
The $y$-values of your points insist that $f(-1) = 4$ and $f(2) = 3$. This
gives two linear equations:
$$f(-1) = -\frac{g}{3} - \frac{g}{2}+2g +C = 4$$
$$f(2) = \frac{8g}{3} -2g -4g+C = 3$$
which we can solve to get $g=\frac{1}{9}$ and $C=\frac{91}{27}$. This gives
$$f(x) = \frac{1}{27}x^3 -\frac{1}{18}x^2-\frac{2}{9}x +\frac{91}{27}.$$
| {
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Find all positives integers $n$ such that $n^3+1$ is a perfect square A solution as follows:
$n^3+1=x^2$
$n^3=x^2-1$
$n^3=(x-1)(x+1)$
$x-1=(x+1)^2~~or~~x+1=(x-1)^2$
$x^2+x+2=0~~or~~x^2-3x=0$
$x(x-3)=0$
$x=0~~or~~x=3~~\Longrightarrow~~n=2$
Does it cover all possible solutions? How to prove that 2 is the only which solves the problem.
| Hint: use that $n^3+1=(n+1)(n^2-n+1)$ so $n+1=n^2-n+1$.
As @Arthur pointed out, it's possible that both factors are not equal. Let's say $n+1=km^2$ and $n^2-n+1=k$. Then we have $n=km^2-1$, $n^2-n+1=k^2m^4-2km^2+1-km^2+1+1=k^2m^4-3km^2+3=k$ or $km^2(km^2-3)=k-3$. The last equation does not have a solution for positive $k>1$, $m>1$ as $km^2-3>k-3$
| {
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Evaluating $\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ I have this series:
$$\sum_{n=1}^\infty s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}+...$$
The sum of the first N terms is:
$$S_N=\sum_{n=1}^N s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}$$
It converges for Leibniz's test to a certain values
$$\sigma=\sum_{n=1}^\infty s_n=\lim_{N\rightarrow \infty} S_N$$
I want to proof that $S_N \ge\frac{1}{2}$ and then $\sigma \ge\frac{1}{2}$ (in my book the author says that it is possible , but he doesn't show that)
$P(1)$ is true (because $s_1=1\ge \frac{1}{2}$).
We suppose that $P(N)$ is true (because $s_N\ge \frac{1}{2}$) and we have to proof that for $N+1$.
$$S_{N+1}=\sum_{n=1}^N s_n+\frac{(-1)^{N+2}}{N+1}=S_N+\frac{(-1)^{N+2}}{N+1} \ge \frac{1}{2}+\frac{(-1)^{N+2}}{N+1}= \frac{N+1+2(-1)^{N+2}}{2*(N+1)}
$$
If n is even $\frac{N+3}{2*(N+1)} \ge \frac {1}{2}$
If n is odd $\frac{N-1}{2*(N+1)} \le \frac {1}{2}$
where am I wrong?
| HINT:
Note that
$$S_{2N}=\sum_{n=1}^{2N}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$
and
$$S_{2N+1}=\sum_{n=1}^{2N+1}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\frac1{2N+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $23x \equiv 1 \mod 120$ using Euclidean Algorithm I want to do inverse modulo to solve the equation
$$23x \equiv 1 \mod 120$$
And to do that I used the extended Euclidean Algorithm.
$$120=23\times5+5$$
$$23=5 \times4+3$$
$$5=3\times1+2$$
$$3=2\times1+1$$
Which makes $1 \equiv3-2\times1=3-2$. I then substituted the other values to get the following
$$2=5-3\times1=5-3$$
$$1=3-(5-3)$$
$$3=23-5\times4$$
$$1=(23-5\times4)-(5-(23-5\times4))$$
$$5 \equiv120-23 \times5$$
$$1 \equiv (23-((120-23\times5)\times4)-(120-23\times5(23-((120-23\times5)\times4))$$
But now I'm not sure how much I need to simplify this down in order to find the answer.
| We wish to solve the congruence $23x \equiv 1 \pmod{120}$.
You correctly obtained
\begin{align*}
120 & = 5 \cdot 23 + 5\\
23 & = 4 \cdot 5 + 3\\
5 & = 1 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Now, if we work backwards, we can obtain $1$ as a linear combination of $23$ and $120$.
\begin{align*}
1 & = 3 - 2\\
& = 3 - (5 - 3)\\
& = 2 \cdot 3 - 5\\
& = 2(23 - 4 \cdot 5) - 5\\
& = 2 \cdot 23 - 9 \cdot 5\\
& = 2 \cdot 23 - 9 \cdot (120 - 5 \cdot 23)\\
& = 47 \cdot 23 - 9 \cdot 120
\end{align*}
Thus,
$$23 \cdot 47 \equiv 1 \pmod{120}$$
Hence, $x \equiv 47 \pmod{120}$, so $47$ is the multiplicative inverse of $23$ modulo $120$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$? Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$?
I started by assuming there exists some $k\in\Bbb N$ s.t. $4^n-1=3k\iff \dfrac{4^n}3-\dfrac 13=k$, so for $k$ to be a natural number, $4^n\equiv 1\mod 3$ must be true, but this tells us no new information, and I don't know how to follow from there. Any help would be appreciated.
| The simples way is using modular arithmetic, under which
$$4^n-1\equiv 1^n-1\equiv 1-1\equiv 0\mod 3$$
Another way is to write
$$4^n = (3+1)^n$$
and use the binomial formula to get
$$(3+1)^n = 3^n +{n\choose 1} 3^{n-1} + \cdots + {n\choose n-1} 3^1 + 1\\=3\left(3^{n-1} +{n\choose 1} 3^{n-2} + \cdots + {n\choose n-1} 3^0 \right) + 1$$
A third way would be to use induction. The step $n=1$ is simple, as $4^n-1=4-1=3$ is clearly divisible by $3$.
The induction step uses the fac that since $4^n-1$ is divisible by $3$, there exists some $k$ such that $4^n-1=3k$, therefore $4^n=3k+1$ and
$$4^{n+1} - 1=4\cdot 4^n - 1 = 4\cdot (3\cdot k + 1) - 1 = 12 k + 3 = 3(4k+1)$$
which is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expressing $\tan 20°$ in terms of $\tan 35°$ If $\tan 35^\circ = a$, we are required to express $\left(\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ}\right)$ in terms of $a$.
Here's one way to solve this: $$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan (145^\circ - 125^\circ) = \tan 20^\circ = \tan (90^\circ - 70^\circ) = \cot 70^\circ = \frac{1}{\tan 70^\circ} = \frac{1}{\tan (2 \times 35^\circ)} = \frac{1 - \tan^2 35^\circ}{2\tan 35^\circ} = \frac{1 - a^2}{2a}$$
However, I tried to solve it using another method as described below, and faced a problem:
$$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan 20^\circ = \tan (35^\circ - 15^\circ) = \frac{\tan 35^\circ - \tan15^\circ}{1 + \tan 35^\circ\tan 15^\circ} = \frac{a - (2 - \sqrt3)}{1 + a(2 - \sqrt3)} = \frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$$
I tried to simplify it to get $\frac{1 - a^2}{2a}$, but I couldn't. So my question is, is there any way to show that $\frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$ is equal to $\frac{1 - a^2}{2a}$? If not, why are we getting two different answers?
| If $\dfrac{1-a^2}{2a}=\dfrac{a-b}{1+ab}$
$$\iff\dfrac{a^3-3a}{1-3a^2}=\dfrac1b$$
If $a=\tan x,b=2-\sqrt3$
$$-\tan3x=\tan(-3x)=2+\sqrt3=\csc30^\circ+\cot30^\circ=\cdots=\cot15^\circ=\tan75^\circ$$
$\implies-3x=180^\circ n+75^\circ$ where $n$ is any integer
$\implies-x\equiv25^\circ,(60+25)^\circ,(120+25)^\circ\pmod{180^\circ}$
$\implies x\equiv-25^\circ\equiv155^\circ,-85^\circ\equiv95^\circ,-145^\circ\equiv35^\circ\pmod{180^\circ}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| A completely informal approach:
consider $x^2+bx+c$, now differentiate with respect to $x$ twice. This gives a positive value,which indicates that the parabola is upturned. Now consider the fact that $b^2<4c$, clearly gives imaginary solutions. Now if the parabola doesn't touch the $X$ axis and is upturned it must lie above the $X$ axis thus proving the fact that $f(x)>0$,because an upturned parabola whose minima lies below the '$X$' axis will always intersect the $X$ axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Compute the intersection of perpendicular lines Lines Ax + By = C and Bx – Ay = 0 are perpendicular.
I am trying to understand how their intersection is computed below:
$$ x = \frac{AC}{A^2 + B^2}$$
$$ y= \frac{BC}{A^2 + B^2}$$
I tried solving for y first but I am not sure how to continue:
$$ Ax+By-C= Bx-Ay $$
$$By+Ay =-Ax +C +Bx $$
$$y(B+A) = -Ax +C +Bx$$
$$y= \frac{-Ax + Bx +C}{ B +A} $$
$$y= \frac{x(-A +B) +C}{ B +A} $$
| Assuming that $(A,B)\neq(0,0)$:\begin{align}\left\{\begin{array}{l}Ax+By=C\\Bx-Ay=0\end{array}\right.&\implies A(Ax+BY)+B(Bx-Ay)=AC\\&\iff(A^2+B^2)x=AC\\&\iff x=\frac{AC}{A^2+B^2}.\end{align}From this and from $Bx=Ay$, you get that$$y=\frac{BC}{A^2+B^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\sqrt{1-\frac{1}{x^3}}+\frac{1}{x^2}}$$
which evaluates to$\frac{1}{1+0}$.
For the second term $\frac{\sqrt[3]{x+1}}{{\sqrt{x^{3}-1}+x}}$ I can't get the manipulation right. Help is much apreciated!
| In the numerator, the terms are of order $x^{3/2}$ and $x^{1/3}$, so that the first dominates (the terms are added, there is no cancellation). In the denominator, $x^{3/2}+x^1$.
So the expression is virtually $$\frac{x^{3/2}}{x^{3/2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ Prove that $x_{0}= \cos \frac{2\pi }{21}+ \cos \frac{8\pi }{21}+ \cos\frac{10\pi }{21}$ is a solution of the equation
$$4x^{3}+ 2x^{2}- 7x- 5= 0$$
My try:
If $x_{0}$, $x_{1}$, $x_{2}$ be the solutions of the equation then
$$\left\{\begin{matrix}
x_{0}+ x_{1}+ x_{2} = -\frac{1}{2} \\
x_{0}x_{1}+ x_{1}x_{2}+ x_{2}x_{0}= -\frac{7}{4} \\
x_{0}x_{1}x_{2} = -\frac{5}{4}
\end{matrix}\right.$$
I can' t continue! Help me!
| As above $ 4x^{3}+ 2x^{2}- 7x- 5$ has a rational root of -1 so the problem reduces after polynomial division to proving $\cos(\frac{2 \pi}{21}) + \cos(\frac{8 \pi}{21}) + \cos(\frac{10 \pi}{21})$ is a root of $4x^{2} - 2x - 5$
First some observations since these are all 21st roots of unity
Let $\omega = \frac{2 \pi}{21}$ the first root.
Then since the sum of all the roots is 0, the sum of all the cosines of the roots must also be 0. Also in the odd roots of unity half the of roots are symmetrical to the other ones across the x axis. And from this 3 things fall out:
From the thirds: $cos(7 \omega) = -\frac{1}{2}$
From the sevenths: $cos(3 \omega) + cos(6 \omega) + cos(9 \omega) = -\frac{1}{2}$
And left over: $cos(\omega) + cos(2 \omega) + cos(4 \omega) + cos(5\omega) + cos(8 \omega) + cos(10 \omega) = \frac{1}{2}$
(Note: using the vieta formula $x_0 + x_1 = \frac{1}{2}$ the 2nd root must be $cos(2 \omega) +cos(8 \omega) + cos(10 \omega)$ from this plus above.)
Now you just have to plug $cos( \omega) + cos(4 \omega) + cos(5 \omega)$ into the polynomial $4x^2 - 2x$ and see what simplifies.
To start with $4x^2$ term:
$4[cos( \omega) + cos(4 \omega) + cos(5 \omega)]^2 = 4[cos( \omega)^2 + cos(4 \omega)^2 + cos(5 \omega)^2 + 2cos(\omega)cos(4\omega) + 2cos(\omega)cos(5\omega) + 2cos(4 \omega)cos(5 \omega) ]$
Eliminate the squares via the identity $cos^2x = \frac{1 + \cos(2x)}{2}$:
$6 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[2cos(\omega)cos(4\omega) + 2cos(\omega)cos(5\omega) + 2cos(4 \omega)cos(5 \omega)]$
Next use the trig product identify on the the last 3 terms to get:
$6 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(3\omega) + cos(4 \omega) + cos(5 \omega) + cos(6 \omega) + cos(9 \omega)]$
From the initial observations $cos(3 \omega) + cos(6 \omega) + cos(9 \omega) = -\frac{1}{2}$
We get:
$4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(4 \omega) + cos(5 \omega)]$
Now back to the original equation add in the -2x term:
$4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega)] + 4[cos(\omega) + cos(4 \omega) + cos(5 \omega)] - 2[cos(\omega) + cos(4 \omega) + cos(5 \omega)]=
4 + 2[cos( 2\omega) + cos(8 \omega) + cos(10 \omega) + cos(\omega) + cos(4 \omega) + cos(5 \omega)]$
But from our original observations the second sum = $\frac{1}{2}$
So we get $4 + 2[\frac{1}{2}] = 5$ which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving $\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$ $$a, b, c \in \left [ 1, 4 \right ] \text{ }
\frac{a}{a+ b}+ \frac{b}{b+ c}+ \frac{c}{2c+ 5a}\geq \frac{38}{39}$$
This is an [old problem of mine in AoPS]
(https://artofproblemsolving.com/community/u372289h1606772p10020940).
First solution
$$ab\geq 1\Leftrightarrow \frac{1}{1+ a}+ \frac{1}{1+ b}- \frac{2}{1+ \sqrt{ab}}= \frac{\left ( \sqrt{ab}- 1 \right )\left ( \sqrt{a}- \sqrt{b} \right )^{2}}{\left ( 1+ a \right )\left ( 1+ b \right )\left ( 1+ \sqrt{ab} \right )}\geq 0$$
then
$$\frac{1}{1+ a}+ \frac{1}{1+ b}\geq \frac{2}{1+ \sqrt{ab}}$$
Thus, we have
$$P= \frac{1}{1+ \frac{b}{a}}+ \frac{1}{1+ \frac{c}{b}}+ \frac{c}{2c+ 5a}\geq \frac{2}{1+ \sqrt{\frac{c}{a}}}+ \frac{\frac{c}{a}}{2\frac{c}{a}+ 5}\geq \frac{38}{39}$$
which is true by $\frac{c}{a}\leq 4$
How about another solution? I hope to see more. Thanks!
| We can use also the following way.
Full expanding gives:
$$(5a+41c)b^2+(200a^2-71ac-37c^2)b+5a^2c+41ac^2\geq0,$$ which is obvious for
$200a^2-71ac-37c^2\geq0.$
But for $200a^2-71ac-37c^2<0$ we obtain $a<c$ and it's enough to prove that
$$(200a^2-71ac-37c^2)^2-4(5a+41c)(5a^2c+41ac^2)\leq0$$ or
$$(c-a)(4a-c)(10000a^2+5375ac+1369c^2)\geq0,$$ which is true because
$$4a-c\geq4\cdot1-4=0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3)^2}}$ , but still didn't get an answer. I got -$\infty$, and everytime I am getting $\infty -\infty$ .
| Use the identity
$$
a^3+b^3=(a+b)(a^2-ab+b^2)
$$
with $a=x$ and $b=\sqrt[3]{1-x^3}$, to get that
$$
1=(x+\sqrt[3]{1-x^3})(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3}).
$$
Thus
$$
\lim_{x\to\infty} (x+\sqrt[3]{1-x^3})=\lim_{x\to\infty}\frac{1}{(x^2-x\sqrt[3]{1-x^3}+(1-x^3)^{2/3})}
=\lim_{x\to\infty}\frac{1}{x^2(1-(x^{-3}-1)^{1/3}+(x^{-3}-1)^{2/3})}=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Explain how to solve this trigonometric limit without L'Hôpital's rule? In my previous class our professor let us the following limit:
$$
\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)}
$$
He solved it by applying L'Hôpital's rule as follow:
$
\lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^2(x) + \cos(x) + 1}{\cos^2(x)} = \frac{3}{1} = 3
$
He only wrote that in the blackboard and then told us to solve it but without using L'Hôpital's rule. So I proceeded in this way:
$
\lim_{x\to0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{x - \sin(x)} = \lim_{x\to0} \frac{\frac{\sin(x)(1 - \cos(x))}{\cos(x)}}{x - \sin(x)} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))}
$
From there I multiply by its conjugate $(1 - \cos(x))$ but then I get more confused:
$
\lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} \frac{(1 + \cos(x))}{(1 + \cos(x))} = \lim_{x\to0} \frac{\sin^3(x)}{(x - \sin(x))(\cos(x))(1 + \cos(x))}
$ ...
Can someone give me a better advice on how to get the right result.
| HINT
By Taylor's expansion
$$ \frac{\tan(x)-\sin(x)}{x-\sin(x)}= \frac{x+\frac{x^3}3-x+\frac{x^3}6+o(x^3)}{x-x+\frac{x^3}6+o(x^3)}=\frac{\frac{x^3}2+o(x^3)}{\frac{x^3}6+o(x^3)}=\frac{\frac{1}2+o(1)}{\frac{1}6+o(1)}\to 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\left ( b+ c \right )^{2}\geq 4bc$$
$$\left ( c+ a \right )^{2}\geq 4ca$$
So I want to prove $abc\geq 1$. I need to the help. Thanks!
| Consider the polynomial
$$p(x) = (x-a)(x-b)(x-c) = x^3 - Ax^2 + Bx - C
\quad\text{ where }\quad
\begin{cases}
A &= a + b + c\\
B &= ab + bc + ca\\
C &= abc
\end{cases}
$$
We are given $A = 3$. By AM $\ge$ GM, we have
$$1 \ge C \implies \sqrt{C} \ge C$$
Since all the roots of $p(x)$ is real. By Newton's inequalities, we have
$$\left(\frac{B}{3}\right)^2 \ge \left(\frac{A}{3}\right)C \iff B^2 \ge 3AC
\implies B \ge \sqrt{3AC} = 3\sqrt{C}$$
Combine these two inequalities, we find
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C
\ge 3(3\sqrt{C}) - \sqrt{C} = 8\sqrt{C}\\
\implies
(a+b)^2(b+c)^2(c+a)^2 \ge 64C = 64abc
$$
| {
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"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\int_{0}^{1} \sqrt[]{1+t^4}dt$ I tried by using $t^2$ = $tan\theta$ and then by inserting $t^4$ by $tan^2\theta$ in $\int_{0}^{1} \sqrt[]{1+t^4}dt$, I get $dt$ = $\frac{sec^2\theta\times d\theta}{2\times \sqrt[]{tan\theta}}$ & $\sqrt[]{1+t^4}$ = $sec\theta$. Thus the integration becomes $\int_{0}^{\frac{\pi}{4}} \frac{sec^3 \theta \times d\theta}{2\times \sqrt{tan\theta}}$
What to do after this step?
I have tried to do this which I have mistakenly written in the answer column.
\begin{align}
\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\int_{0}^{1}\frac{1+t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}\frac{t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}t\cdot\frac{t^3}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\left[\frac12t\sqrt{1+t^4}\right]_{0}^{1}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac121\sqrt{1+1^4}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\
\implies \frac32\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\frac{1}{2}\sqrt{1+1^4}+\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\
\implies \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\frac{1}{3}\sqrt{1+1^4}+\frac23\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}.\\
\end{align}
| \begin{align}
\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\int_{0}^{1}\frac{1+t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}\frac{t^4}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\int_{0}^{1}t\cdot\frac{t^3}{\sqrt{1+t^4}}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}+\left[\frac12t\sqrt{1+t^4}\right]_{0}^{1}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac121\sqrt{1+1^4}-\frac12\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t\\
\implies \frac32\int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\frac{1}{2}\sqrt{1+1^4}+\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\
\implies \int_{0}^{1}\sqrt{1+t^4}\,\mathrm{d}t
&=\frac{1}{3}\sqrt{1+1^4}+\frac23\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+t^4}}.\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues of matrix with ijth entry as $i+j$
Let $A$ be a matrix with entries as $(a_{ij})=i+j$. I would like to find out eigenvalues of $A$.
I noticed that after reducing the matrix to Echelon form only two non zero rows will be left and thus rank(A)=2, that means $n-2$ eigenvalues are zero. Let $\lambda_1, \lambda_2$ be two non zero eigenvalues, then as trace(A)=$n(n+1)$, we get $\lambda_1+ \lambda_2= n(n+1)$. How do I find another relation in $\lambda_1, \lambda_2$
| The conjecture of Carl Schildkraut concercing $\lambda_1 \lambda_2$ is correct up to a minus sign. I assume $1$-indexed rows and columns, however. Note that $\lambda_1^2$ and $\lambda_2^2$ are the two non-zero eigenvalues of $A^2$, and since $A$ is symmetric, $AA = A^T A$. Using the formula $\mathrm{Tr}(X^TY) = \sum_{i, j = 1}^n X_{ij}Y_{ij}$ for $(n \times n)$-matrices (see for example, this Wikipedia page), we find that
\begin{align*}
\lambda_1^2 + \lambda_2^2 = \mathrm{Tr}(A^T A) &= \sum_{i, j = 1}^n (i + j)^2 = \sum_{i, j = 1}^n (i^2 + 2ij + j^2) \\
&= 2n \cdot \frac{n (n + 1)(2n + 1)}{6} + 2 \cdot \left(\frac{n (n + 1)}{2}\right)^2 \\
&= n^2 (n + 1) \frac{7n + 5}{6}.
\end{align*}
Using this expression and the fact that $\lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2 \lambda_1 \lambda_2$, we find that
\begin{align*}
\lambda_1 \lambda_2 &= \frac{n^2 (n + 1)^2}{2} - \frac{n^2 (n + 1) (7n + 5)}{12} \\
&= \frac{n^2 (n + 1)}{12} (6n + 6 - 7n - 5)\\
&= - \frac{n^2 (n + 1)(n - 1)}{12} = - \frac{n^2(n^2 - 1)}{12}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| You could also have used Talor series around $x=0$ and even got more than the limit itself
$$\sin(a+k x)=\sin (a)+k \cos (a)\,x-\frac{1}{2} k^2 \sin (a)\,x^2+O\left(x^3\right)$$ making
$$\sin(a+ x)\sin(a+2 x)=\sin ^2(a)+3 \sin (a) \cos (a)\,x+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin
^2(a)\right)\,x^2+O\left(x^3\right)$$
$$\frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x=3 \sin (a) \cos (a)\,+ \left(2 \cos ^2(a)-\frac{5 }{2}\sin
^2(a)\right)\,x+O\left(x^2\right)$$ which shows the limit and how it is approached.
If you simplify, you should notice that the end result is just
$$\frac 32 \sin(2a)+\frac{1}{4} (9 \cos (2 a)-1)\,x+O\left(x^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity.
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix}
1-\lambda && 7 && -2 \\
0 && 3-\lambda && -1 \\
0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix}
-2 && 7 && -2 \\
0 && 0 && -1 \\
0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && -\frac{7}{2} && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
\frac{7}{2}\\
1 \\
0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix}
-1 && 7 && -2 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && 0 && -5 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
5\\
1 \\
1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally,
$ A - I = \begin{bmatrix}
0 && 7 && -2 \\
0 && 2 && -1 \\
0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting $\begin{bmatrix}
x \\
y \\
z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix}$ ? or is that wrong?
| $$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\iff \begin{bmatrix}y\\z\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}$$
meaning that $y=z=0$, while leaving $x$ to be whatever you want, results in a kernel vector.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve pair of equations : $ x(y-\sin{y}) = -\pi$ and $ x(1-\cos{y}) = 2$ I was trying to solve the following pair of equations :
$$ x(y-\sin{y}) = -\pi$$
$$ x(1-\cos{y}) = 2$$
My attempt :
We get, $x \sin{y} = xy + \pi$ and $x \cos y = x - 2$ , then by squaring and eliminating $x^2$ from both sides we get, $x^2 y^2 + 2 xy \pi + {\pi}^2 - 4x + 4 = 0$ and so on. But couldn't solve the equation.
It would also be enough to solve $\sqrt{x} y$ for the problem I am trying . Thanks in advance for help.
| I find it easier to use $\sin(x)$ instead of $\sin(y)$ so lets change $y\leftrightarrow x$ to get the system:
$$y=\frac{-\pi}{x-\sin(x)} \tag{1}$$
$$y = \frac{2}{1-\cos(x)} \tag{2}$$
cross multiply noting that $x\neq 2n\pi$, let
$$f(x)=2(x-\sin x)+\pi(1-\cos x ) $$
Where we need to solve $f(x) = 0$. We know for $x>0$, $x \ge \sin(x)$ and $1 \ge \cos(x)$. So for $x > 0$ there are no solutions.
For $x< 0$ analyse the derivative
$$\begin{align}f'(x) &= 2-2\cos x +\pi \sin x \\
&= 4 \sin^2(\tfrac{x}{2})+2\pi \sin(\tfrac{x}{2})\cos(\tfrac{x}{2})\\
&= 2\sin(\tfrac{x}{2})(2\sin(\tfrac{x}{2})+\pi \cos(\tfrac{x}{2})) \\
& = 2\sqrt{\pi^2+4}\sin(\tfrac{x}{2})\sin(\tfrac{x}{2}+\arcsin(\tfrac{\pi}{\sqrt{\pi^2+4}}))
\end{align}$$
From here for $f'(x) = 0$ we have $x=2n\pi$ or $2n\pi -\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}})$. This provides us with some intervals of monotonicity.
*
*Now for $x \in (-2\pi, -\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}))$, function is monotonous, and $f(-2\pi) < 0$ and $f(\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}})) > 0$ so we will give only one root, the trivially found $x=-\pi$.
*Rewrite $f(x) = 2(x-\sin(x))+2\pi \sin^2(\tfrac{x}{2})$. Here first term is negative for $x < 0$. For $x \in (-\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}),0)$, $f(x)$ is monotone, but $f(0) = 0$ and $\arcsin(\tfrac{\pi}{\sqrt{\pi^2 + 4}}) > 0$ so no roots.
*For $x < -2\pi$, $2x < -4\pi$ but $2\sin(x)-2\pi \sin^2(x)$ can be said to have a lower bound of $-2-2\pi$, that too not achievable. Thus here $2x < 2\sin x -2\pi\sin^2(\tfrac{x}{2})$. So there are no roots here also.
Thus the only root of our system is $(x,y) = (-\pi,1)$ or original equations solutions are $(x,y) = \color{red}{(1,-\pi)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $X^2$ of a Maxwell-distributed RV I'm trying to calculate $X^2$, where $X$ is a Maxwell-distributed RV; which means it has the density function $\phi$ with
$$\phi(x) = \frac{2}{d^3 \sqrt{2\pi}} x^2 \exp\left(- \frac{x^2}{2d^2}\right) \text{ for } x \ge 0.$$
In the lecture we proved that the density function $\psi$ of $X^2$ can be calculated via the following formula:
$$\psi(x) = \frac{\phi(\sqrt{x}) + \phi(-\sqrt{x})}{2\sqrt{x}} \text{ for } x \ge 0$$
But when I apply this formula I get:
\begin{align}
\psi(x) & = \frac{\phi(\sqrt{x}) + \phi(-\sqrt{x})}{2\sqrt{x}} \\[10pt]
& = \frac{\frac{2}{d^3\sqrt{2\pi}} (\sqrt{x})^2 \exp\left(- \frac{(\sqrt x)^2}{2d^2} \right) + \frac{2}{d^3\sqrt{2\pi}} (-\sqrt{x})^2 \exp\left(- \frac{(-\sqrt{x})^2}{2d^2}\right)}{2\sqrt{x}} \\[10pt]
& = \frac{\frac{4}{d^3 \sqrt{2\pi}} x \exp\left(- \frac{x}{2d^2}\right)}{2\sqrt{x}} \\[10pt]
& = \frac{2}{d^3 \sqrt{2\pi}} \sqrt{x} \exp\left(- \frac{x}{2d^2}\right)
\end{align}
But according to the solution I should get:
$$ \frac 1 {d^3 \sqrt{2\pi}} \sqrt{x} \exp\left(- \frac{x}{2d^2}\right)$$
Could you tell me what I'm doing wrong?
| When the distribution of $X$ has support $X \ge 0$, the transformation $Y = X^2$ is already one-to-one, thus we can simply write $$f_Y(y) = \frac{f_X(\sqrt{y})}{2\sqrt{y}}, \quad y > 0.$$ That is why you have an extra factor of $2$: you have presumed that $X$ has negative support when it does not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the value of $x$ that satisfies $3x=4$ in $\mathbb Z/5\mathbb Z$? Let $\mathbb Z_5 = \mathbb Z/5\mathbb Z$.
The value of $x$ which satisfies the equation $3x = 4\bmod 5$ is...?
The answer is $3$. I understand why the answer is $3$, but not how it was derived. Is there an equation or process I can use that will give me the correct answer no matter how large the numbers in the equation?
| You need to find the inverse of $3$ modulo $5$. By Bézout’s identity, there are integers $x$ and $y$ such that $3x+5y=1$, which you can find with the (reverse) Euclidean algorithm:
\begin{align}
\color{red}{5}&=\color{red}{3}\cdot1+\color{red}{2}\\
\color{red}{3}&=\color{red}{2}\cdot1+\color{red}{1}\\
\color{red}{2}&=\color{red}{1}\cdot2+0
\end{align}
The reverse algorithm gives
\begin{align}
\color{red}{1}&=\color{red}{3}+\color{red}{2}\cdot(-1)\\
&=\color{red}{3}+(\color{red}{5}+\color{red}{3}\cdot(-1))\cdot(-1)\\
&=\color{red}{3}\cdot(1+1)+\color{red}{5}\cdot(-1)\\
&=\color{red}{3}\cdot2+\color{red}{5}\cdot(-1)
\end{align}
which shows we can choose $x=2$ and $y=-1$. Since $3\cdot2+5\cdot(-1)=1$, we have
$$
2\cdot3\equiv 1\pmod{5}
$$
and so
$$
2\cdot3x\equiv2\cdot4\pmod{5}
$$
which gives
$$
x\equiv3\pmod{5}
$$
Of course, for congruences modulo $5$ it's much easier to do trial and error, that is, trying $3\cdot m$ for $m=1,2,3,4$ and stop when the hit is found.
On the other hand, the above method will work for every problem of the form
$$
ax\equiv b\pmod{n}
$$
where $\gcd(a,n)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the surface area of the unit sphere $x^2+y^2+z^2=1$ Compute the surface area of the unit sphere $x^2+y^2+z^2=1$
The following is a solution the book suggests:
The upper hemisphere is the graph of the function $\varphi(x,y)=\sqrt{1-x^2-y^2}.$
A little calculation yields $$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac 1{\sqrt{1-x^2-y^2}} \tag {1}$$
Note that $d A=\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}d x d y \tag {*} $
So, by $(*)$, the area of the hemisphere is obtained by integrating the function in $(1)$ over the unit disc. Switching to polar cordinates yields $$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi \tag{2}$$
Firstly, in $(1)$ how does this "$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}$ " become equal to $\sqrt{1-x^2-y^2}$? Also how do we get $(2)$ , aren't we supposed to compute $d A=\sin\varphi d\theta d\varphi $?
| Note that
$$(\partial_x\varphi)^2=\frac{x^2}{\varphi^2}$$
$$(\partial_y\varphi)^2=\frac{y^2}{\varphi^2}$$
and then
$$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac1{\sqrt{1-x^2-y^2}}$$
and thus
$$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Value of $k$ in ratio of two definite integration
If $\displaystyle I_{1}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{12}dx$ and $\displaystyle I_{2}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ and $I_{1}=k(108\sqrt{3})I_{2}$. Then $k$ is
Try: put $x=\sin^2\theta$. Then $dx=2\sin\theta \cos\theta d\theta$.
so $$I_{1}=\frac{1}{6}\int^{\frac{\pi}{2}}_{0}\sin^6\theta \cos ^{8}\theta d\theta=\frac{5\cdot 3\cdot 7\cdot 5 \cdot 3}{14\cdot 12\cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2}\times \frac{\pi}{2}$$
Could some helo me to solve second one , Thanks
| Just try to convert $I_1$ into $I_2$ form
Substituting $x = \cfrac{4y}{3+y}$
$$I_1 =\int_0^1 \cfrac{4^{\cfrac{5}2}3^{\cfrac72} x^{\cfrac52}(1-x)^{\cfrac72}}{(3+x)^6} \cfrac{4(3+x)-4x}{(3+x)^2}dx\\ \hspace{-1cm}= 32*27*\sqrt 3 *12*I_1 \\ \hspace{-1cm} =10368\sqrt 3*I_1 \\ \hspace{0cm}= 96*108*\sqrt 3 I_1$$
$\implies k = 96$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Using the $5$-th order Maclaurin polynomial of $f(x) = e^x$ to approximate $f(-1)$
Find the 5th-order Maclaurin polynomial $P_5(x)$ for $f(x) = e^x$.
I got
$$P_5(x) = 1 + x +\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + O(x^6) $$
From this answer, I'm supposed to approximate $f(-1)$, correct to the fifth decimal place. Is it right to just put in $x= -1$ like below?
$$\begin{align} & 1 + (-1) + \frac{(-1)^2}{2} + \frac{(-1)^3}{6} + \frac{(-1)^4}{24} + \frac{(-1)^5}{120} + O((-1)^6) \\
=\; & 1 -1 + \frac12 - \frac16 + \frac1{24} - \frac1{120} + O((-1)^6) \\
=\; & 0.5 - 0.166666 + 0.041666 - 0.008333 + O((-1)^6) \\
=\; & 0.366667 \\
=\; & 0.36667 + O((-1)^6)
\end{align}$$
Is this done right?
| Yes, it looks like your method for evaluating $f(-1)$ there is correct. I'd avoid the notation $O((-1)^6)$, and instead just write $$f(-1)\approx1 + (-1) + {(-1)^2\over2} + {(-1)^3\over6} + {(-1)^4\over24} + {(-1)^5\over120}$$
and so on.
However it's worth pointing out that this isn't an approximation within 5 decimals, and you ought to use more terms in the approximating polynomial to get close enough.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
| Let $b_n:=a_{2n-1}$ and $c_n:=a_{2n}$ for $n\in\mathbb{N}$. Since $a_n$ is bounded because $a_n\leqslant 1/2$ and $a_n>0$ for all $n$, so are the sequences $b_n$ and $c_n$. Now we show that $b_n$ is monotonic decreasing while $c_n$ is monotonic increasing.
First note that $b_n>\sqrt{2}-1$ for all $n$. For $n=1$ it is clear since $b_1=a_1=1/2>\sqrt{2}-1$. Then
$$b_{n+1}=a_{2n+1}=\frac{1}{2+a_{2n}}=\frac{1}{2+\frac{1}{2+a_{2n-1}}}=\frac{2+a_{2n-1}}{5+2a_{2n-1}}=\frac{2+b_{n-1}}{5+2b_{n-1}}\\=\frac{2+b_{n-1}}{1+2(2+b_{n-1})}>\frac{2+(\sqrt{2}-1)}{1+2(2+\sqrt{2}-1)}\\=\frac{1+\sqrt{2}}{1+2(1+\sqrt{2})}=\frac{1}{(\sqrt{2}-1)+2}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$$
where we have used the fact that the function $$f(x):=\frac{x}{1+2x}$$ is increasing on $\mathbb{R}^+$ together with the inductive hypothesis $b_{n-1}>\sqrt{2}-1$. Next we show that $b_n$ is monotone decreasing
Note
$$b_{n+1}-b_n=a_{2n+1}-a_{2n-1}=\frac{1}{2+a_{2n}}-a_{2n-1}=\frac{1}{2+\frac{1}{2+a_{2n-1}}}-a_{2n-1}\\=\frac{2+a_{2n-1}}{5+2a_{2n-1}}-a_{2n-1}=\frac{2(1-2a_{2n-1}-a_{2n-1}^2)}{5+2a_{2n-1}}\\=\frac{2(1-2b_{n-1}-b_{n-1}^2)}{5+2b_{n-1}}=\frac{2(\sqrt{2}-1-b_{n-1})(\sqrt{2}+1+b_{n-1})}{5+2b_{n-1}}<0$$
since by previous observation $b_n>\sqrt{2}-1$ for all $n$. Therefore $b_n$ is bounded and monotone so by Bolzano-Weierstrass we obtain that $b_n$ has a limit point $b$. Hence
$$\lim_nb_{n+1}=\lim_n\frac{2+b_{n-1}}{5+2b_{n-1}}\Rightarrow b=\frac{2+b}{5+2b}\Rightarrow b^2+2b-1=0\Rightarrow b\in\{\sqrt{2}-1,-\sqrt{2}-1\}$$
Since $b_n>0$ for all $n$ then $b=\sqrt{2}-1$.
Exaclty analogue arguments for the sequence $c_n$ one finds that it is bounded and monotone increasing satisfying $c_n<\sqrt{2}-1$ for all $n$. The limit point of $c_n$ we denote by $c$ can be shown to be $\sqrt{2}-1$.
Finally note by construction of $b_n$ and $c_n$ it must be the case that
$$\lim_n b_n=\lim\sup_n a_n\hspace{0.2cm}\text{and}\hspace{0.2cm} \lim_n c_n=\lim\inf_n a_n$$
Since $\lim_n b_n=\sqrt{2}-1=\lim_n c_n$ we get
$$\lim_n a_n=\lim\sup_n a_n=\lim\inf_n a_n=\sqrt{2}-1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding probability -picking at least one red, one blue and one green ball from an urn when six balls are selected Six balls are to be randomly chosen from an urn containing $8$ red, $10$ green, and 12 blue balls.
What is the probability at least one red ball, one blue and one green ball is chosen?
Sample space = $\binom{30}{6}$
P = 1 - P(All red + All Green + All Blue + Only red and Green + Only Red and Blue + Only Green and Blue )
$$P = \Large 1 - \frac{\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}}{\binom{30}{6}}$$
According to this, I got $\large 1 - \frac{133099}{593775}$,
which is $0.7758$?
Is my approach correct?
| Your approach is not correct.
In 'Only red and green" you don't exclude situations "Only red" and "only green".
Also each of the situations "Only x" is counted twice - in situation "Only x and y" and "only x and z"
Thus the number of situations where there is a color missing should be:
$$\left(\binom{8}{6} + \binom{10}{6} + \binom{12}{6} + \binom{18}{6} + \binom{22}{6} + \binom{20}{6}\right)-2\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)=$$
$$= \binom{18}{6} + \binom{22}{6} + \binom{20}{6}-\left( \binom{8}{6} + \binom{10}{6} + \binom{12}{6} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Stuck on a recursively defined converging sequence problem. I am given a simple quadratic equation
$$x^2-x-c=0, x>0, c>0$$
and then we define a sequence $\{x_n\}$ with $x_1>0$ fixed and then, if $n$ is an index for which $x_n$ has been defined, we define
$$x_{n+1}=\sqrt{c+x_n}$$.
With that I am asked to prove that $\{x_n\}$ converges monotonically to the solution of the polynomial.
I've done quite a bit of scratch work. Obviously we can solve the quadratic and the positive solution is $\frac{1+\sqrt{5}}{2}$. I have an inkling that the equation is decreasing and so I tried working with $x_{n+1}-x_{n+2}$ to show that the difference is positive but I didn't come up with anything useful.
I did realize that if I simply write out the limit we see that
$$\lim_{n\rightarrow\infty}[(\sqrt{c+x_{n+1}})^2)-\sqrt{c+x_{n+1}}-c]=\lim_{n\rightarrow\infty}[x_{n+1}-\sqrt{c+x_{n+1}}]$$
So if we want this final limit to go to zero then all I really need is $x_n$ to be monotonically decreasing since it is clearly bounded below by zero since $c$ and $x_1$ were taken to be positive.
| $\begin{array}\\
x_{n+1}-x_n
&=\sqrt{x_n+c}-x_n\\
&=(\sqrt{x_n+c}-x_n)\dfrac{\sqrt{x_n+c}+x_n}{\sqrt{x_n+c}+x_n}\\
&=\dfrac{x_n+c-x_n^2}{\sqrt{x_n+c}+x_n}\\
\end{array}
$
If
$f(x) = x^2-x-c$,
$f'(x) = 2x-1$,
so $f'(x) > 0$
for $x > \frac12$.
The roots of $f(x)$
are
$x
=\dfrac{1\pm\sqrt{1+4c}}{2}
$,
so the positive root
$x_c$ satisfies
$1 < x_c
\lt 1+c$.
If
$\frac12 < x_n < x_c$,
then
$x_n^2-x_n-c < 0$
so
$x_{n+1} > x_n$.
If
$ x_n > x_c$,
then
$x_n^2-x_n-c > 0$
so
$x_{n+1} < x_n$.
Similarly,
$\begin{array}\\
x_{n+1}-x_c
&=\sqrt{x_n+c}-x_c\\
&=(\sqrt{x_n+c}-x_c)\dfrac{\sqrt{x_n+c}+x_c}{\sqrt{x_n+c}+x_c}\\
&=\dfrac{x_n+c-x_c^2}{\sqrt{x_n+c}+x_c}\\
&=\dfrac{x_n+c-(x_c+c)}{\sqrt{x_n+c}+x_c}
\qquad\text{since } x^c_2 = x_c+c\\
&=\dfrac{x_n-x_c}{\sqrt{x_n+c}+x_c}\\
\end{array}
$
Therefore
$x_{n+1}-x_c$
has the same sign
and is smaller in absolute value
than
$x_n-x_c$.
Therefore
$x_n \to x_c$
since
$\sqrt{x_n+c}+x_c
\gt 1+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far
| Method I
Using integration by parts twice,
\begin{align}
I &=\int \sin ax \sinh bx \, dx \\
&= \frac{\sin ax \cosh bx}{b}-\int \frac{\cosh bx}{b} d(\sin ax) \\
&= \frac{\sin ax \cosh bx}{b}-\frac{a}{b} \int \cos ax \cosh bx \, dx \\
&= \frac{\sin ax \cosh bx}{b}-\frac{a}{b}
\left[
\frac{\cos ax \sinh bx}{b}-\int \frac{\sinh bx}{b} d(\cos ax)
\right] \\
&= \frac{\sin ax \cosh bx}{b}-
\frac{a\cos ax \sinh bx}{b^2}-
\frac{a^2}{b^2} \int \sin ax \sinh bx \, dx \\
I &=
\frac{\sin ax \cosh bx}{b}-
\frac{a\cos ax \sinh bx}{b^2}-\frac{a^2}{b^2}I+C \\
I &= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C
\end{align}
Method II
\begin{align}
\cos (u+iv) &= \cos u \cosh v-i\sin u \sinh v \\
\sin (u+iv) &= \sin u \cosh v+i\cos u \sinh v \\
\int \cos [(a+ib)x] \, dx &=
\int \cos ax \cosh bx \, dx-
i\int \sin ax \sinh bx \, dx \\
\frac{\sin [(a+ib)x]}{a+ib} &=
\int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx \\
\frac{(a-ib)(\sin ax \cosh bx+i\cos ax \sinh bx)}{a^2+b^2} &=
\int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx
\end{align}
Equating the imaginary parts,
$$\fbox{$
\int \sin ax \sinh bx \, dx=
\frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C
$}$$
Method III
By brute force,
\begin{align}
& \quad \int \sin ax \sinh bx \, dx \\[10pt]
&= \int
\left[
\frac{(e^{iax}-e^{-iax})(e^{bx}-e^{-bx})}{4i}
\right] \, dx \\[10pt]
&= \int
\left[
\frac{e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}-e^{(-b+ia)x}}{4i}
\right] \, dx \\[10pt]
&= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4i(b+ia)}
+\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4i(b-ia)} \\[10pt]
&= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4(-a+ib)}
+\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4(a+ib)} \\[10pt]
&= \frac{(-a-ib)[e^{(b+ia)x}-e^{-(b+ia)x}]}{4(a^2+b^2)}
+\frac{(a-ib)[-e^{(b-ia)x}+e^{-(b-ia)x}]}{4(a^2+b^2)} \\[10pt]
&= \frac{bi}{4(a^2+b^2)}
[-e^{(b+ia)x}+e^{-(b+ia)x}+e^{(b-ia)x}-e^{-(b-ia)x}] \\
& \quad +
\frac{a}{4(a^2+b^2)}
[-e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}+e^{-(b-ia)x}] \\[10pt]
&= \frac{bi}{4(a^2+b^2)}
[e^{bx}(-e^{iax}+e^{-iax})+e^{-bx}(e^{-iax}-e^{iax})] \\
& \quad +
\frac{a}{4(a^2+b^2)}
[e^{bx}(-e^{iax}-e^{-iax})+e^{-bx}(e^{-iax}+e^{iax})] \\[10pt]
&= \frac{b}{a^2+b^2} \times
\frac{e^{bx}+e^{-bx}}{2} \times
\frac{e^{iax}-e^{-iax}}{2i}-
\frac{a}{a^2+b^2} \times
\frac{e^{bx}-e^{-bx}}{2} \times
\frac{e^{iax}+e^{-iax}}{2} \\[10pt]
&= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing Minima for degree 8 polynomial Is there any better method than derivative or trial and error to calculate minima of $$f(x)=x^8-8x^6+19x^4-12x^3+14x^2-8x+9$$
The minima occurs at $x=2$ and $f(2)=1$
We were given option too $-1,9,6,1$ if that helps
| By splitting conveniently the terms (by completing squares from left to right) we obtain
$$\begin{align}f(x)&=x^8-8x^6+(16+3)x^4-12x^3+(12+2)x^2-8x+(8+1)
\\&=(x^8-8x^6+16x^4)+(3x^4-12x^3+12x^2)+(2x^2-8x+8)+1\\
&=x^4(x^2-4)^2+3x^2(x-2)^2+2(x-2)^2+1\geq 1=f(2).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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