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Related rates of a cone I am in an intro calculus class and I have a problem which I am unsure about. It reads:
Water is being poured into a conical reservoir at a rate of pi cubic feet per second. The reservoir has a radius of 6 fees across the top and a height of 12 feet. At what rate is the depth of the water increasing when the depth is 6 feet?
Here's what I came up with for the problem:
h = height of cone
r = radius of cone
V = volume of cone
$\frac{r}{h}$ = $\frac{1}{2}$
V = $ \frac{\pi r^2h}{3}$
$\frac{dV}{dt}$ = $\frac {2\pi r \frac{dr}{dt}h}{3} + $$\frac{1}{3} \pi r^2
\frac {dh}{dt}$
(Using product rule)
Because r = $\frac{h}{2},
$$\frac{dv}{dt}$ = $\frac{2\pi h}{2}\frac{dh}{dt}\frac{h}{6} + \frac{\pi (\frac{h}{2})^2}{3}
\frac{dh}{dt}$
I solved for $\frac{dh}{dt}$
$\frac{dh}{dt}$ = $\frac{12\frac{dv}{dt}}{3\pi h^2}$
I plugged in $\pi$ for $\frac{dv}{dt}$ and 6 for h and got
1/9 or 0.111
Can anyone tell me if this is correct? Thanks if advance!
|
Since the dimension of the cone is such that $r = 6$ft and $h=12$ft, we have $\dfrac{r}{h} = \dfrac{6}{12}$, which means $r = \dfrac{h}{2}$. We must substitute this into the equation of the volume of a cone before differentiating. It follows that
$V=\pi r^2 \dfrac{h}{3}$
$V=\pi \bigg(\dfrac{h}{2}\bigg)^2 \dfrac{h}{3}$
$V=\pi \dfrac{h^2}{4} \dfrac{h}{3}$
$V= \pi \dfrac{h^3}{12}$
$\dfrac{dV}{dt} = \dfrac{\pi}{12} 3h^2 \dfrac{dh}{dt}$
$\pi = \dfrac{\pi}{12} 3(6)^2 \dfrac{dh}{dt}$
$\pi = 9\pi \dfrac{dh}{dt}$
$\dfrac{dh}{dt} = \dfrac{1}{9}$
The height of the cone is increasing at a rate of $\dfrac{1}{9}$ feet per second .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Extremely ugly integral $\int_{-\pi}^{\pi} \frac{\operatorname{sign}(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$
Evaluate: $\DeclareMathOperator{\sign}{sign}$ $$\int_{-\pi}^{\pi} \dfrac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$$
My idea:
\begin{align*}I=\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)| }\,dx &=\int_{-\pi}^{\pi} \dfrac{\sign(-x)\arctan((-x)^2)-|-x|}{\sin^2(x)+|\cos(x)|}\,dx\\
&=\int_{-\pi}^{\pi} \frac{-\sign(x)\arctan(x^2)-|x|}{ \sin^2(x)+|\cos(x)| }\,dx.
\end{align*}
So, it means that the integral $$\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)}{\sin^2(x)+|\cos(x)| }\,dx=0.$$
Therefore, my integral now looks like
$$I=\int_{-\pi}^{\pi} \frac{-|x|}{\sin^2(x)+|\cos(x)|}\,dx,$$
and since my integrand is even function, I have:
$$I=-2\int_{0}^{\pi} \frac{|x|}{ \sin^2(x)+|\cos(x)| }\,dx=-2\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx.$$
And so,
\begin{align*}
\frac{I}{2}&=-\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\int_{0}^{\pi} \frac{\pi-x}{ \sin^2(\pi-x)+|\cos(\pi-x)|}\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)|}+\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)| }-\frac{I}{2}.
\end{align*}
Hence,
$$I=-\pi\int_{0}^{\pi} \dfrac{dx}{ \sin^2(x)+|\cos(x)| }.$$
That is the end of the road. I tried to eliminate absolute value over the cosine but always get some divergent integral. Is this computation correct and if it is, what next? Bonus question: Where can I find more problems like this?
|
Noting
$$ \int_0^{\pi/2}\frac{1}{\sin^2x+\cos x}dx=\int_0^{\pi/2}\frac{1}{\cos^2x+\sin x}dx=\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx $$
and using $u=\tan(\frac{x}{2})$, one has
\begin{eqnarray}
\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx&=&\int_0^{\pi/2}\frac{1}{1-(\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})})^2+\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}dx\\
&=&\int_0^1\frac{1}{1-(\frac{2u}{1+u^2})^2+\frac{2u}{1+u^2}}\frac{2du}{1+u^2}\\
&=&\int_0^1\frac{u^2+1}{u^4+2u^3-2u^2+2u+1}du\\
&=&\int_0^1\frac{u^2+1}{[(u-\phi)^2+\phi][(u+\frac1\phi)^2-\frac1\phi]}du
\end{eqnarray}
Here $\phi=\frac{-1+\sqrt5}{2}$. Noting
$$ \frac{u^2+1}{[(u-\phi)^2+\phi][(u+1/\phi)^2-1/\phi]}=-\frac{\phi}{\sqrt5}\frac{1}{(u-\phi)^2+\phi}+\frac{1}{\sqrt5\phi}\frac{1}{(u+\frac1\phi)^2-\frac1\phi} $$
one has
\begin{eqnarray}
\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx
&=&\int_0^1\frac{u^2+1}{[(u-\phi)^2+\phi][(u+\frac1\phi)^2-\frac1\phi]}du\\
&=&-\frac{\phi}{\sqrt5}\int_0^1\frac{1}{(u-\phi)^2+\phi}du+\frac{1}{\sqrt5\phi}\int_0^1\frac{1}{(u+\frac1\phi)^2-\frac1\phi}du\\
&=&-\frac{\sqrt\phi}{\sqrt5}\bigg[\arctan(\frac{u-\phi}{\sqrt\phi})-\text{arctanh}(\frac{1+u\phi}{\sqrt\phi})\bigg]\bigg|_0^1\\
&=&-\frac{\sqrt\phi}{\sqrt5}\bigg[\arctan(\frac{1-\phi}{\sqrt\phi})+\arctan\sqrt\phi-\text{arctanh}(\frac{1}{\phi\sqrt\phi})+\text{arctanh}\sqrt\phi\bigg].
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{7}{2}$$
$$\int \frac{1}{\sqrt{u^2 - a^2} +3}\, \,du$$
Attempt I - By Trigonometric substitution
$$\sqrt{u^2 - a^2} = \sqrt{a\sec^2\varTheta - a^2} = \sqrt{a^2(\sec^2\varTheta - 1)} = \sqrt{a^2\tan^2\varTheta} = a\tan\varTheta $$
$$u = a\sec\varTheta \implies du = a\sec\varTheta \tan\varTheta d\varTheta$$
$$\int \frac{1}{a\tan\varTheta + 3}\, a\sec\varTheta \tan\varTheta d\varTheta$$
$$a\int \frac{\sec\varTheta \tan\varTheta}{a\tan\varTheta + 3}\, d\varTheta$$
How can I continue here ?
Attempt II - By First Euler Substitution
$$\sqrt{u^2 - a^2} = u - t$$
$\sqrt{u^2 - a^2} = u - t$
$ u^2 - a^2 = (u - t)^2$
$u^2 - a^2 = u^2 -2ut + t^2$
$- a^2 = -2ut + t^2$
$2ut = t^2 + a^2$
$$u = \frac{t^2 + a^2}{2t} \implies du = \frac{1}{2} - \frac{a^2}{2t^2}\, dt$$
Thus the integral takes the form
$$\int \frac{1}{u - t +3}\, \left( \frac{1}{2} - \frac{a^2}{2t^2}\right) \,du$$
Since $u = \frac{t^2 + a^2}{2t}$ and $\left( \frac{1}{2} - \frac{a^2}{2t^2}\right) = -\frac{a^2 - t^2}{2t^2}$ then
$$I = \int \frac{1}{ \frac{t^2 + a^2}{2t} - t +3} \left( -\frac{a^2 - t^2}{2t^2} \right) \, dt$$
$$-\int \frac{a^2 - t^2}{ ta^2 - t^3 +6t^2}\, dt$$
$$-\int \frac{a^2}{ ta^2 - t^3 +6t^2}\, dt + \int \frac{t^2}{ ta^2 - t^3 +6t^2}\, dt$$
How can I continue?
Attempt III -By First Euler Substitution
$$\sqrt{x^2 + 5x} = x + t$$
$\sqrt{x^2 + 5x} = x + t$
$ x^2 + 5x = (x + t)^2$
$x^2 + 5x = x^2 + 2ut + t^2$
$x^2 + 5x = x^2 + 2xt + t^2$
$5x = 2xt + t^2$
$5x -2xt = t^2$
$x(5 -2t) = t^2$
$$x = \frac{t^2}{(5 -2t)} \implies \frac{dt}{dx} = -\frac{2(t - 5)t}{(5 - 2t)^2 } \iff dx = -\frac{(5 - 2t)^2 }{2(t - 5)t}\,dt $$
$$\int \frac{1}{x + t + 3}\, \left( -\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int \frac{1}{\frac{t^2}{(5 -2t)} + t + 3}\, \left(\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int\frac{(5 - 2t)^2}{\frac{2(t - 5)t^3}{5 -2t} + 2(t - 5)t^2 + 6(t - 5)t}\,dt$$
How can I continue here?
|
Setting $$\sqrt{x^2+7x}=t+x$$ then we have $$x=\frac{t^2}{7-2t}$$ and $$dx=-\frac{2 (t-7) t}{(2 t-7)^2}$$ and $$t+x=\frac{(t-7) t}{2 t-7}$$
and our integral will be $$\int -\frac{2 (t-7) t(2t-7)}{9\left(21-t+t^2\right)}dt$$
|
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|
Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
|
Let $v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$
$v^2=a^2+b^2+2\sqrt{\left(a^2\cos^2(x)+b^2\sin^2(x)\right)\left(b^2\cos^2(x)+a^2\sin^2(x)\right)}$
$v^2=a^2+b^2+2\sqrt{a^2b^2\left(\cos^4(x)+\sin^4(x)\right)+\left(a^4+b^4\right)\cos^2(x)\sin^2(x)}$
$v^2=a^2+b^2+\sqrt{4a^2b^2\left(1-2\sin^2(x)\cos^2(x)\right)+\left(a^4+b^4\right)\sin^2(2x)}$
$v^2=a^2+b^2+\sqrt{4a^2b^2-2a^2b^2\sin^2(2x)+\left(a^4+b^4\right)\sin^2(2x)}$
$v^2=|a|^2+|b|^2+\sqrt{4a^2b^2+\left(a^2-b^2\right)^2\sin^2(2x)}$
To find minimumvalue of $v^2$
put $\sin^2(2x)=0$
to obtain $v^2=\left(|a|+|b|\right)^2$. So
$v_{min}=|a|+|b|$
And to obtain maximum value of $v^2$ put $\sin^2(2x)=1$
$v_{max}=\sqrt{2(a^2+b^2)}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
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|
How to get the limit of this series I need to solve this series:
$$\sum _{ k=2 }^{ \infty } (k-1)k \left( \frac{ 1 }{ 3 } \right) ^{ k+1 }$$
I converted it into $$\sum _{ k=0 }^{ \infty } \frac { { k }^{ 2 }-k }{ 3 } \left(\frac { 1 }{ 3 } \right)^{ k } -\frac { 4 }{ 3 } $$ with the idea, that $$\sum _{ k=0 }^{ \infty }{ { q }^{ k } } =\frac { 1 }{ 1-q } $$but I don't know how to get rid of $\frac { { k }^{ 2 }-k }{ 3 }$.
Can someone please tell me how to go on?
|
By stars and bars the coefficient of $x^n$ in $\frac{1}{(1-x)^3}$, that is the number of ways of writing $n$ as the sum of three natural numbers (or the number of ways for writing $n+3$ as the sum of three positive natural numbers), equals $\binom{n+2}{2}$. It follows that
$$ \frac{1}{(1-x)^3} = \sum_{n\geq 0}\binom{n+2}{2}x^n,\qquad \frac{x^2}{(1-x)^3}=\sum_{m\geq 2}\binom{m}{2}x^m $$
and by evaluating $\frac{2x^2}{3(1-x)^3}$ at $x=\frac{1}{3}$ it follows that
$$ \sum_{k\geq 2}k(k-1)\frac{1}{3^{k+1}}=\color{red}{\frac{1}{4}}.$$
|
{
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"url": "https://math.stackexchange.com/questions/2264435",
"timestamp": "2023-03-29T00:00:00",
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|
Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$
Let $h$ be $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
{]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.
I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(\pi)-\cos(\beta)\sin(\pi) = -\sin(\beta)$$
$$\\$$
$$-\sin(\beta) = \frac{1}{3} \Leftrightarrow \\
\sin(\beta) = - \frac{1}{3}$$
$$\\$$
$$\cos \beta = \sqrt{1-\sin^2(\beta)} = \sqrt{1-(-\frac{1}{3})^2} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$
And so $h(\beta)$ is :
$$\frac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} = \\
\frac{\sqrt{2}}{2\sqrt{2}-3}$$
But my book says the solution is $6\sqrt{2}-8$. What went wrong?
|
3 things
$\dfrac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} =
\dfrac{2\sqrt{2}}{2\sqrt{2}-3}$
you dropped a 2.
$\dfrac{2\sqrt{2}}{2\sqrt{2}-3} = -6\sqrt 2 - 8$
With a little bit of simplification you would see that you were "only" off by a minus sign.
and here is where you lost it.
$\cos \beta = -\frac {\sqrt {2}}{3}$ as $\beta$ is in QIII
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the signal and zeros of $\cos x - \sin x$
Let $f$ be a function of domain $\mathbb{R}$ defined by:
$$f(x) = \cos x - \sin x$$
Find the signal and zeros of the function.
First I tried to find the zeros of $f(x)$:
$$0 = \cos x - \sin x \Leftrightarrow \\
0 = \sqrt{1-\sin^2x}-\sin x \Leftrightarrow \\
\sin x = \sqrt{1-\sin^2x} \Leftrightarrow \\
\sin^2x=1-\sin^2x\Leftrightarrow \\
2\sin^2x=1 \Leftrightarrow \\
\sin x = \frac{1}{\sqrt{2}} \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \pi - \frac{\pi}{4}+2k\pi \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \frac{3\pi}{4}+2k\pi $$
But then when I check the value of $f(x)$ for the answers above, $\frac{\pi}{4}+2k\pi$ gives me a zero while $\frac{3\pi}{4}+2k\pi$ gives me a minimum. Why is that?
As for the signal, $\frac{\pi}{4}+2k\pi$ gives me the zero on the left side when the function becomes negative.
My book states the solution is
Zeros:
$x = \frac{\pi}{4}+k\pi, k\in \mathbb{R}$
$f$ is negative in the intervals
$$]\frac{\pi}{4}+k2\pi,\frac{5\pi}{4}+k2\pi[,k\in\mathbb{Z};$$
$f$ is positive in the intervals
$$]-\frac{3\pi}{4}+k2\pi,\frac{\pi}{4}+k2\pi[,k\in\mathbb{Z}$$
How do I solve this?
|
There is a trigonometric identity $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$. Put $y = 3\pi/4$ to show $f(x) = \sqrt2\sin(x+3\pi/4)$. This quickly gives the answer you want.
You asked about some spurious extra solutions. I think these come from using $\cos(x) = \sqrt{1-\sin^2(x)}$. This is true only if the square root has the right sign.
|
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|
$\cos\gamma+\sin\gamma$ in a 3-5-7 triangle ($\gamma$ is the greatest angle) Clearly $\gamma$ has to be the angle oposite to the side of lenght 7. The law of cosines gives me $$7^2 = 5^2+3^2 - 2\cdot 5\cdot3\cos{\gamma}\Longleftrightarrow\cos{\gamma}=-\frac{7^2-5^2-3^2}{2\cdot5\cdot 3}=-\frac{1}{2}.$$
This value of $\cos{\gamma}$ means there can be two values of $\gamma$ satisfying it in $0\leq\gamma \leq 2\pi$. So $\sin{\gamma}$ can be either negative or positive. Using the trigonometric identity $\sin^2{\theta}+\cos^2{\theta}=1$, I obtain $$\sin{\gamma}=\pm\sqrt{1-\cos{\gamma}}=\pm\sqrt{1-\left(-\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}.$$
Using the positive value I get the correct answer of $$\cos{\gamma} + \sin{\gamma}=-\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-1}{2}.$$
Why is it incorrect to use the negative value of $\sin{\gamma}$ ?
|
As $0<\gamma<\pi$ (actually $\frac{\pi}{2}<\gamma<\pi$), we have $\sin\gamma>0$
|
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|
What are the values of $a$ that satisfy the equation $\frac{ax^2+ax-1}{2(x+\frac{5}{4})^2+\frac{47}{8}} < 0$? I got this question in a test today, I was stumped on how to do it. I am right to say that since $2(x+\frac{5}{4})^2+\frac{47}{8}$ is always positive $ax^2+ax-1$ is always negative. So, $$ax^2+ax-1<0$$$$Discriminant < 0$$$$(a)^2-4(a)(-1)<0$$$$a(a+4)<0$$$$-4<a<0$$ or is it $$ax^2+ax-1<2x^2+5x+9$$$$(a-2)x^2+(a-5)x-10<0$$$$Discriminant < 0$$$$(a-5)^2-4(a-2)(-10)<0$$
|
Hint: You want $a < 0$, and $a^2 - 4a(-1) < 0$
|
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|
Show monotonicity Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \ \ \text{for} \ \ x, a,b,c > 0 \ .
$$
To do so, I figured it is sufficient to show that
$$
f(x) = \sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)
$$
is monotonically increasing for $x > 0$.
Of course, I took the derivative $f'(x)$ and proceeded with the demand
$$
\frac{\frac{b}{a}x}{\sqrt{a \cdot b + \frac{b}{a}x^2}} \arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) - \frac{c \frac{b}{a} x}{a \cdot b + \frac{b}{a} x^2 + c^2} > 0 \ .
$$
In the end, I got stuck with
$$
\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) > \frac{c \sqrt{a \cdot b + \frac{b}{a}x^2}}{a \cdot b + \frac{b}{a} x^2 + c^2} \ .
$$
Inserting values for a,b, and c seems to work perfectly, but I can't manage to analytically solve the inequation.
Does anyone have an idea of how to approach this problem?
Edit:
I came across the Shafer-Fink inequality stating
$$
\frac{3y}{1+2\sqrt{1 + y^2}} < \arctan y < \frac{\pi y}{1 + 2\sqrt{1 + y^2}} \ .
$$
Can I substitute
$$
y = \frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}
$$
and therefore receive
$$
\arctan (y) > \frac{3y}{1+2\sqrt{1 + y^2}} > \frac{y}{1 + y^2} ?
$$
Is that a proper way?
|
There is an integral for arctan(x):
\[
\arctan(x) = \int_{0}^{x} \frac{dt}{t^2 + 1}, \quad x > 0.
\]
Perhaps this can help.
|
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|
Two six-sided dice are rolled. Find the probability distribution of $\max[X_1,X_2]$ I am not sure if I am understanding this problem correctly. It says;
Two six-sided dice are rolled. Let X1 and X2
denote the outcomes of the first and second die roll respectively.
Define a new random variable Y such that:
Y = max{X1 , X2}
Does this mean I am supposed to roll the two dice and find the largest roll?
From my own understanding, if I am supposed to find the largest roll from the two dice roll, is this example of mine correct?
ex: the outcomes are {2,3} then Y=3
Moving on to what is asked of me:
Find the probability distribution of Y
For this part, do I need to find all the possible outcomes of the two dice? For example;
{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}
{2,1} {2,2} {2,3} {2,4} {2,5} {2,6}
{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}
etc....
And do I need to find the probability of the sum of two dice? For example;
Sum = 2 , Prob = 1/36
Sum = 3 , Prob = 2/36
Sum = 4 , Prob = 3/36
etc....
After this, what do I do from here to find the probability distribution of Y?
If I am totally off track, explanation from you guys will be great. Sorry if it looks like I'm asking too much. Please feel free to correct me on anything!
|
It may be useful to construct a table, with $X_1$ down the left, and $X_2$ along the top, and each cell gets the maximum of the two values. The table below has been started for you; fill in the rest of the values.
$$
\begin{array}{|c|c|c|c|c|c|c|} \hline
& \textbf{1} & \textbf{2} & \textbf{3}
& \textbf{4} & \textbf{5} & \textbf{6} \\ \hline
\textbf{1} & 1 & 2 & 3 & & & \\ \hline
\textbf{2} & 2 & 2 & 3 & & & \\ \hline
\textbf{3} & 3 & 3 & 3 & & & \\ \hline
\textbf{4} & & & & & & \\ \hline
\textbf{5} & & & & & & \\ \hline
\textbf{6} & & & & & & \\ \hline
\end{array}
$$
The probability distribution for $Y = \max(X_1, X_2)$ can then be obtained by counting up the proportion of entries that are $1, 2, 3, \ldots$.
|
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|
Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$.
......
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
It takes $\infty - \infty $ form when $x=\infty $.
Now,
$$=\lim_{x\to \infty } (\sqrt {x-a} - \sqrt {bx})\times \dfrac {\sqrt {x-a}+\sqrt {bx}}{\sqrt {x-a}+\sqrt {bx}}$$
$$=\lim_{x\to \infty } \dfrac {x-a-bx}{\sqrt {x-a}+\sqrt {bx}}$$
|
I'll add to Eugen's comment:
If $b=1$ then we have:
$$ \lim_{x\to \infty} \frac{-a}{\sqrt{x-a}+\sqrt{x}} = \frac{-a}{\infty} = 0$$
otherwise,
$$ \lim_{x\to \infty} \frac{(1-b)x-a}{\sqrt{x-a}+\sqrt{x}} =\lim_{x\to \infty} \frac{(1-b)\sqrt{x} - \frac{a}{\sqrt{x}}}{\sqrt{1-\frac{a}{x}}+1} = \lim_{x\to \infty} \frac{(1-b)\sqrt{x}-0}{\sqrt{1-0}+1}=\lim_{x\to\infty} \frac{1-b}{2}\sqrt{x}$$
Meaning that the answer will be $\pm \infty$ dependeing on $\operatorname{sign}(1-b)$
|
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|
A proof of the approximate expression about totient summatory function I'm a high school student in Korea.
I am preparing for a presentation. so I prove an approximate expression about totient summatory function , but I'm not sure that the proof is correct.
If the proof is incorrect , please tell me what is incorrect.
The following is the proof :
(Prove $ \Phi \left ( x \right ) \sim \frac{x^{2}}{2\zeta \left ( 2 \right )}$)
$\Phi \left ( x \right )= \sum_{k<x}\varphi \left ( k \right ) = \sum_{dm<x}m\mu (d)$
$=\sum_{d<x}\mu\left ( d \right )\sum_{m<x/d}m$
$=\sum_{d<x}\mu\left ( d \right ) \cdot \frac{1}{2} \cdot (\left [ \frac{x}{d} \right ]^{2} + \left [ \frac{x}{d} \right ] )$
$\sum_{d<x}\mu\left ( d \right )\left [\frac{x}{d} \right ] = 1 $ so $\Phi \left ( x \right ) - \frac{1}{2} = \sum_{d<x}\mu\left ( d \right ) \cdot \frac{1}{2} \cdot \left [ \frac{x}{d} \right ]^{2} $
$\mu\left ( d \right )\frac{1}{2}(\frac{x^{2} - 2xd + d^{2} }{d^{2}}) < \mu\left ( d \right )\frac{1}{2}\left [ \frac{x}{d} \right ]^{2} < \mu\left ( d \right )\frac{1}{2}(\frac{x^{2} }{d^{2}})
$
(because $\frac{x}{d} - 1 < \left [ \frac{x}{d} \right ] \leq \frac{x}{d}$)
so
$\sum_{d<x}\mu\left ( d \right )\cdot\frac{1}{2}\cdot(\frac{x^{2} - 2xd + d^{2} }{d^{2}}) <\Phi \left ( x \right ) - \frac{1}{2} < \sum_{d<x}\mu\left ( d \right )\cdot\frac{1}{2}\cdot\frac{x^{2} }{d^{2}} $
$\Rightarrow \frac{x^{2}}{2}\sum_{d<x}\frac{\mu\left ( d \right )}{d^{2}} - x \sum_{d<x}\frac{\mu\left ( d \right ) }{d} + \frac{1}{2} \sum_{d<x} \mu \left ( d \right )
<\Phi \left ( x \right ) <\frac{x^{2}}{2}\sum_{d<x}\frac{\mu\left ( d \right )}{d^{2}} $
we know $ \left | \sum_{d<x}\frac{\mu\left ( d \right ) }{d} \right |\leq 1$ and $\sum_{k=1}^{\infty}\frac{\mu \left ( s \right )}{n^{-s}} = \frac{1}{\zeta \left ( s \right )}$
Finally,
$\lim_{x->\infty }\frac{\Phi \left ( x \right )}{x^{2}} = \frac{1}{2\zeta \left ( 2 \right )} $
$\Rightarrow \Phi \left ( x \right ) \sim \frac{x^{2}}{2\zeta \left ( 2 \right )}$
|
Since $\varphi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$ and $\sum_{d < x} \mu(d) \lfloor x/d \rfloor = 1$ $$\sum_{n < x} \varphi(n) = \sum_{n < x}\sum_{d | n} \mu(d) \frac{n}{d}=\sum_{md < x} \mu(d) m$$
$$=\frac{1}{2} \sum_{d<x}\mu(d) (\left\lfloor \frac{x}{d} \right\rfloor^2 + \left\lfloor \frac{x}{d} \right\rfloor )=\frac{1}{2} +\frac{1}{2} \sum_{d<x}\mu(d) \left\lfloor \frac{x}{d} \right\rfloor^2$$
$\frac12\left|\left\lfloor \frac{x}{d} \right\rfloor^{2}-\frac{x^2}{d^2}\right| < 1+\frac{x}{d}$ so that $$\frac12\left|\sum_{d<x}\mu( d) \left(\left\lfloor \frac{x}{d} \right\rfloor^{2}-\frac{x^2}{d^2}\right)\right| <\sum_{d<x} (1+\frac{x}{d})< x+x \ln (x+1)$$
Also
$$\left|\sum_{d < x} \frac{\mu(d)}{d^2} -\frac{1}{\zeta(2)}\right| < \sum_{d \ge x} \frac{1}{d^2} < \frac{1}{x-1}$$
Therefore
$$\sum_{n < x} \varphi(n) \sim \frac12\sum_{d<x}\mu( d) \left\lfloor \frac{x}{d} \right\rfloor^{2} \sim \frac12\sum_{d<x}\mu( d) \frac{x^2}{d^2} \sim \frac{x^2}{2\zeta(2)}$$
|
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|
Evaluate $\int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$
Evaluate
$$I(\eta)=2 a \int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$$
where $a,p,c,t>0,N\in{\mathbb{N}}, \alpha>0$.
Any idea how to evaluate this integral? This comes from the expected value of $\frac{p/X^2+c}{(1/X^2-t)^2}$, where the pdf of $X$ is $f_X(x)=2 a x e^{-a x^2}$ (Rayleigh distribution).
|
\begin{align}
I &= \frac{2ap}{p^2} \int_\alpha^2 xe^{-ax^2} \frac{\frac{1}{x^2+c}}{\frac{1}{(x^2-t)}}dx = \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{(x^2-t)^2}{x^2+c}dx\\ &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{x^4-2tx^2+t^2}{x^2+c}dx \\
&= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \left(x^2-(2t+c) + \frac{t^2+c(2t+c)}{x^2+c} \right)dx \\
\end{align}
We use long division. Now, let $M = t^2 + c(2t+c)$ and $N = (2t+c)$ then
\begin{align}
I &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \left(x^2+N + \frac{M}{x^2+c} \right)dx \\
&= \frac{2a}{p} \left( \int_\alpha^2 x^2 (x e^{-ax^2})dx + N\int_\alpha^2 (x e^{-ax^2})dx + M \int_\alpha^2 \frac{1}{x^2+c} (x e^{-ax^2})dx\right)
\end{align}
After this you can use integration by parts and integration by substitution to get the value of the integral.
|
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|
Equivalent $\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}dx=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx?$ Proposed:
Is $(1)$ equivalent to $(2)$ in term of transformation?
$$\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx={5\pi^2\over 2\cdot 6^3}\tag1$$
and
$$\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx={5\pi^2\over 2\cdot 6^3}\tag2$$
|
Let
$$I=\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx$$
and
$$J=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx.$$
Note
\begin{eqnarray}
J&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{1}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{0}^{1}{\frac{1}{x^2}-\frac{1}{x}\over \frac{1}{x^2}+1}\cdot{\ln(1+\frac1x)\over \frac{1}{x^3}+1}\frac{1}{x^2}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{0}^{1}{x-x^2\over x^2+1}\cdot{\ln(\frac{1+x}x)\over x^3+1}\mathrm dx\\
&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx\\
&=&I
\end{eqnarray}
and
\begin{eqnarray}
I&=&\int_0^1\bigg[\frac{x^2}{1+x^3}-\frac{x}{1+x^2}\bigg]\ln x\mathrm dx\\
&=&\int_0^1\frac{x^2}{1+x^3}\ln x\mathrm dx-\int_0^1\frac{x}{1+x^2}\ln x\mathrm dx\\
&=&\frac19\int_0^1\frac{1}{1+x}\ln x\mathrm dx-\frac14\int_0^1\frac{1}{1+x}\ln x\mathrm dx\\
&=&-\frac{5}{36}\int_0^1\frac{1}{1+x}\ln x\mathrm dx\\
&=&-\frac{5}{36}(-\frac{\pi^2}{12})\\
&=&\frac{5\pi^2}{432}.
\end{eqnarray}
|
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|
How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$ let $x>1$ show that:
$$(2+5x)\ln{x}-6(x-1)>0.\forall x>1$$
Let
$$f(x)=(2+5x)\ln{x}-6(x-1),~~~f'(x)=\dfrac{2}{x}+5\ln{x}-1$$
since
$f(1)=1$.so it must prove
$$f'(x)=\dfrac{2}{x}+5\ln{x}-1>0?$$
|
Let $f(x)=\ln{x}-\frac{6(x-1)}{5x+2}$.
Hence,
$$f'(x)=\frac{1}{x}-6\cdot\frac{5x+2-5(x-1)}{(5x+2)^2}=\frac{1}{x}-\frac{42}{(5x+2)^2}=$$
$$=\frac{25x^2-22x+4}{x(5x+2)^2}=\frac{25x^2-25x+3x+4}{x(5x+2)^2}>0$$
for all $x>1$.
Thus, $f(x)>f(1)=0$ and we are done!
|
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|
What is the value of this infinite fraction, where each successive row counts until a power of two? Here is the fraction
$$\frac{1}{\cfrac{2}{\cfrac{4}{8\cdots+9\cdots}+\cfrac{5}{10\cdots+11\cdots}} + \cfrac{3}{\cfrac{6}{12\cdots+13\cdots}+\cfrac{7}{14\cdots+15\cdots}}}$$
I have tried iterating row by row, seeing if the fraction converges, however it seems to jump back and forth between a small and slightly larger number between each iteration.
For $(n,x)$, where $n$ is the number of rows calculated, and $x$ is the value of the entire fraction, I have found the points $(1,1)$, $(2,0.2)$, $(3,2.207547169)$, $(4,0.0956302360)$, $(5,4.5138490341)$, and
$(6,0.04730978991)$. Is there any way to find a value for the entire fraction, continuing to infinity?
|
Let $a_n$ denote your sequence:
$$ a_n = \dfrac{1}{\dfrac{2}{\dfrac{4}{\vdots} + \dfrac{5}{\vdots}} + \dfrac{3}{\dfrac{6}{\vdots} + \dfrac{7}{\vdots}}}. $$
Let us make an observation. The general rule is that each bottom-most integer $m$ is replaced by $\frac{m}{2m + (2m+1)}$ in the next step. By applying this rule twice, this $m$ is replaced by
$$ \phi(m) := \frac{m}{\frac{2m}{4m + (4m+1)} + \frac{2m+1}{(4m+2)+(4m+3)}} = \frac{m(8m+1)(8m+5)}{32m^2 + 20m + 1} \geq 2m. \tag{$\diamond$} $$
Now it is easy to notice that replacing bottom-most terms by larger value increases the value for $a_{2k+1}$ and decreases the value for $a_{2k}$. From this relation, it is not hard to check that
$$a_{2k+1} \geq 2^k \quad \text{and} \quad a_{2k} \leq \frac{1}{5 \cdot 2^{k-1}}. $$
Therefore $a_{2k+1} \to \infty$ and $a_{2k} \to 0$ as $k \to \infty$.
More formally, define $f_n : (0, \infty)^{2^{n-1}} \to \Bbb{R}$ inductively by
$$f_1(x_1) = x_1, \qquad f_{n+1}(x_1, \cdots, x_{2^{n}}) = f_n \left( \frac{2^{n-1}}{x_1 + x_2}, \frac{2^{n-1} + 1}{x_3 + x_4}, \cdots, \frac{2^n - 1}{x_{2^n - 1} + x_{2^n}} \right). \tag{*}$$
Then we can write $a_n = f_n (2^{n-1}, \cdots, 2^n - 1)$ and utilizing the observation $(\diamond)$ shows that
$$ a_{n+2} = f_n(\phi(2^{n-1}), \cdots, \phi(2^n - 1)). $$
Now by the construction $\text{(*)}$, we can inductively prove that the followings are true:
*
*$f_{2k+1}$ is increasing in each variable and $f_{2k+1}(\lambda \mathrm{x}) = \lambda f_{2k+1}(\mathrm{x})$ for $\lambda > 0$,
*$f_{2k}$ is decreasing in each variables and $f_{2k}(\lambda \mathrm{x}) = \lambda^{-1} f_{2k}(\mathrm{x})$ for $\lambda > 0$.
So we have
\begin{align*}
a_{2k+1}
&= f_{2k-1}(\phi(2^{2k-2}), \cdots, \phi(2^{2k-1} - 1)) \\
&\geq f_{2k-1}(2\cdot 2^{2k-2}, \cdots, 2 \cdot (2^{2k-1} - 1) ) \\
&= 2 f_{2k-1}(2^{2k-2}, \cdots, 2^{2k-1} - 1 ) \\
&= 2 a_{2k-1}
\end{align*}
and hence $a_{2k+1} \geq 2^k a_1 = 2^k$. Similarly,
\begin{align*}
a_{2k+2}
&= f_{2k}(\phi(2^{2k-1}), \cdots, \phi(2^{2k} - 1)) \\
&\leq f_{2k}(2\cdot 2^{2k-1}, \cdots, 2 \cdot (2^{2k} - 1) ) \\
&= \frac{1}{2} f_{2k}(2^{2k-1}, \cdots, 2^{2k} - 1 ) \\
&= \frac{1}{2} a_{2k}
\end{align*}
and therefore $a_{2k} \leq \frac{1}{2^{k-1}} a_2 = \frac{1}{5\cdot 2^{k-1}}$.
|
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|
Solving intersection of translated spiric curve The spiric section is the curve obtained by slicing a torus along a plane parallel to its axis. Assume the torus midplane lies in the $z=0$ plane, $a$ is the radius of the torus tube and $c$ is the distance from the origin of the center of the tube.
Now cut the torus with the plane $y = r$ and we get the curve:
$$z^2=a^2-(c-\sqrt{x^2+r^2})^2$$
I'm interested only in the bottom part:
$$f(x)=z=-\sqrt{a^2-(c-\sqrt{x^2+r^2})^2}$$
And more specifically I'm trying to find the horizontal position of the intersection between $f(x)$ and a translated copy $q+f(x+p)$ (with $p$ and $q$ integers), in other words:
I want to solve the following equation for $x$, symbolically if possible:
$$q - \sqrt{a^2 - (c - \sqrt{(p - x)^2 + r^2})^2} + \sqrt{a^2 - (c - \sqrt{r^2 + x^2})^2} = 0$$
(in the appropriate domain where square roots are real)
Note: I don't have a lot of experience with math software, but I have tried both SageMath and WolframAlpha, and wasn't able to get a closed form solution, although WolframAlpha can solve the case where $c=0$:.
|
The implicit equation of the torus is
$$(x^2+y^2+z^2+R^2-r^2)^2=4r^2(x^2+y^2)$$
and the parametric equation of the shifted one is
$$\begin{cases}x=(R+r\cos\theta)\cos\phi+p\\y=(R+r\cos\theta)\sin\phi\\z=r\sin\theta+q.\end{cases}$$
The sections are given by
$$y=Y$$ and by eliminating $\theta$,
$$R+r\cos\theta=Y\csc\phi,\\
\\
x=Y\cot\phi+p,\\z=q\pm\sqrt{r^2-\left(Y\csc\phi-R\right)^2}.$$
Then plugging one into the other,
$$\left((Y\cot\phi+p)^2+Y^2+\left(q\pm\sqrt{r^2-\left(Y\csc\phi-R\right)^2}\right)^2+R^2-r^2\right)^2=4r^2((Y\cot\phi+p)^2+Y^2).$$
We can rationalize with $\csc\theta=\dfrac{1+t^2}{2t}$ and $\cot\theta=\dfrac{1-t^2}{2t}$, giving
$$\left(\left(Y\dfrac{1-t^2}{2t}+p\right)^2+Y^2+\left(q\pm\sqrt{r^2-\left(Y\dfrac{1+t^2}{2t}-R\right)^2}\right)^2+R^2-r^2\right)^2=4r^2\left(\left(Y\dfrac{1-t^2}{2t}+p\right)^2+Y^2\right),$$
or, multiplying by $16t^4$,
$$\left(\left(Y(1-t^2)+2pt^2\right)^2+4(Y^2+R^2-r^2)t^2+\left(2qt\pm\sqrt{4r^2t^2-\left(Y(1+t^2)-2Rt\right)^2}\right)^2\right)^2=16r^2\left(\left(Y(1-t^2)+2pt\right)^2+4Y^2t^2\right)t^2.$$
It is possible to turn this equation in a polynomial one after leeennnngthy computation to eliminate the square root.
|
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|
Put fraction in "arctan-friendly" form I would like to put $\int\frac{1}{(2x^2+x+1)}dx$ into something like $\int\frac{1}{(u^2+1)}dx$. What is the quickest way to proceed? I know that previous fraction can be rewritten as $2t^2+t+1 = \frac{7}{8}\left( \left( \frac{4t+1}{\sqrt{7}} \right)^2 +1 \right)$, but I don't have any explaination from where this comes from.
Finally, the integral yields
$$\int_b^a \frac{7}{8} \left( \left(\frac{4t+1}{\sqrt{7}} \right)^2+1 \right)dt = \frac{2}{\sqrt{7}}\left[\arctan \left(\frac{4t+1}{\sqrt{7}}\right)\right]^a_b $$
|
Multiply numerator and denominator by $4\cdot 2=8$ (the $2$ is the coefficient of $x^2$) and “complete the square”:
$$
\frac{1}{2x^2+x+1}=
\frac{8}{16x^2+8x+8}=
\frac{8}{16x^2+8x+1+7}=
\frac{8}{(4x+1)^2+7}
$$
Now you know that you should set $4x+1=u\sqrt{7}$, so you get
$$
\frac{8}{7}\frac{1}{u^2+1}
$$
Moreover, $4\,dx=\sqrt{7}\,du$ and the integral becomes
$$
\int\frac{8}{7}\frac{1}{u^2+1}\frac{\sqrt{7}}{4}\,du=
\frac{2}{\sqrt{7}}\int\frac{1}{u^2+1}\,du=
\frac{2}{\sqrt{7}}\arctan u+c=
\frac{2}{\sqrt{7}}\arctan\left(\frac{4x+1}{\sqrt{7}}\right)+c$$
|
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|
If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc+c} \\B^2=abc+c=c(ab+1) $$
Now what?
|
The given inequality is equivalent to
$$ \sqrt{a-1} + \sqrt{b-1} \leq \sqrt{abc + c} - \sqrt{c-1}.$$
Now set $f(c) := \gamma \sqrt{c} - \sqrt{c-1}$ where $ \gamma := \sqrt{ab+1} $. Then we see that $f'(c) = \gamma \frac{1}{2\sqrt{c}} - \frac{1}{2\sqrt{c-1}}$, from which it is easy to check $f$ takes its minimum when $c=\frac{1}{\gamma^2-1} + 1 = \frac{1}{ab} + 1$. So what we have to prove is
$$ \sqrt{a-1} + \sqrt{b-1} \leq f(\frac{1}{ab} + 1) = \sqrt{ab}$$
which is equivalent to
$$ a-1 + b-1 + 2 \sqrt{(a-1)(b-1)} \leq ab$$
or, $2\sqrt{(a-1)(b-1)} \leq (a-1)(b-1) + 1$. The last inequality can be easily derived from AM-GM applied to $(a-1)(b-1)$ and $1$. In addition we see that the equality holds iff $(a-1)(b-1) = 1$ and $c = \frac{1}{ab} + 1$.
|
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|
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^n$ begins with a $1$, we have $10^m \leq 5^n < 2 \cdot 10^m$ for some positive integer $m$. Then multiplying by $\dfrac{2^{n+1}}{5^n}$ we get $10^m \cdot \dfrac{2^{n+1}}{5^n} \leq 2^{n+1} < 10^m \cdot \dfrac{2^{n+2}}{5^n}$, but I didn't see how this helped.
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First, note that the inequalities must be strict, as we cannot have $5^n = 10^m$ for natural numbers $n,m$. Since $10^m = 2^m5^m$, your last equation becomes
$$2^{n+m+1}5^{m-n} < 2^{n+1} < 2^{n+m+2}5^{m-n}$$
$$\implies 2^{1+2n}10^{m-n} < 2^{n+1} < 2\cdot 2^{1+2n}10^{m-n}$$
Dividing each term by $2^{1+2n}$ gives us
$$10^{m-n} < 2^{-n} < 2\cdot 10^{m-n} $$
Taking the reciprocal of each term,
$$10^{n-m} > 2^n > \frac 1 2 10^{n-m} $$
Finally, multiplying by $2$ gives us
$$2\cdot 10^{n-m} > 2^{n+1} > 10^{n-m}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $|x^2-2x|+|x-4|>|x^2-3x+4|$
How do I solve $|x^2-2x|+|x-4|>|x^2-3x+4|?$
I can see the difference of the two terms on the left gives the term on the right. Now,what should I do? Is there any general method for solving $$|a|+|b|>|a-b|?$$
Thanks for any help!!
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Solving this is just a matter of using the definition of |x|.
This is |x||x- 2|+ |x+ 4|> |x^2- 3x+ 4|.
x^2- 3x+ 4 is positive for all x so we need to consider three cases:
x< -4. Then all of x, x- 2, and x+ 4 are negative. This is (-x)(2- x)- (x+ 4)= x^2- 3x- 4> x^2- 3x+ 4. That reduces to -4> 4 which is never true. There is no x< -4 that satisfies this.
-4< x< 0. x+ 4 is positive but x and x+ 2 are still negative so this is (-x)(2- x)+ (x+ 4)= x^2- x+ 4> x^2- 3x+ 4. That reduces to 2x> 0 but this case is for x< 0 so there is no x between -4 and 0 that satisfies this.
0< x< 2. Now both x+ 4 and x are positive while x- 2 is negative so this is (x)(2- x)+ (x+ 4)= -x^2+ 3x+ 4> x^2- 3x+ 4. That reduces to -2x^2+ 6x= x(6- 2x)> 0. A product is positive if both numbers are positive or if both numbers are negative. Either x> 0 and 6> 2x so x> 3 or x< 0 and 6< 2x. The first is satisfied for x> 3 and the second for x< 0. Neither of those is true for 0< x< 2 so there is no x that satisfies this.
2< x. Now all three terms are positive so this is (x)(x- 2)+ x+ 4= x^2- x+ 4> x^2- 3x+ 4. That reduces to 2x> 0 or x> 0. Since, here, x> 2 x is also greater than 0. This inequality is satisfied for x> 2..
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|
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$$
Note that:
$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$$
Now we have $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$, i.e.$\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ and $\tan h \sim_{0}h.$
I'm stuck.
Update: here is another way :
\begin{aligned}
\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\
&=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
\end{aligned}
Thus
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left(\dfrac{\pi x}{2a}\right)} =e^{\dfrac{2}{\pi}} $$
*
*Am i right beside i'm intersted in way which use equivalents
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$$A=\left(2-\frac{x}{a} \right)^{\tan( \frac{\pi x}{2a})}\implies \log(A)=\tan( \frac{\pi x}{2a})\log\left(2-\frac{x}{a} \right)$$ Now, using Taylor expansion around $x=a$
$$\log\left(2-\frac{x}{a} \right)=-\frac{x-a}{a}-\frac{(x-a)^2}{2 a^2}+O\left((x-a)^3\right)$$ Using Laurent expansion around $x=a$
$$\tan( \frac{\pi x}{2a})=-\frac{2 a}{\pi (x-a)}+\frac{\pi (x-a)}{6 a}+O\left((x-a)^3\right)$$ Combining the product
$$\log(A)=\frac{2}{\pi }+\frac{x-a}{\pi a}+O\left((x-a)^2\right)$$ Taylor again $$A=e^{\log(A)}=e^{\frac{2}{\pi }}\left(1+\frac{x-a}{\pi a}+O\left((x-a)^2\right)\right)$$ which shows the limit and also how it is approached.
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|
Simultaneous triangularization of matrices Let $\mathcal{F}=\{A_1,A_2,\ldots,A_r\}$ be a triangulable commuting family of $n\times n$ matrices (that is, each $A_i$ is triangulable and $A_iA_j=A_jA_i$ for every $i,j$). I know that $\mathcal{F}$ can be simultaneous triangularization, but what is the algorithm of finding the invertible matrix $P$ such that $P^{-1}A_iP$ is triangular? As a working example consider the matrices
$$
A=
\begin{pmatrix}
-3 & 2 & -4 \\
-1 & 0 & -1\\
2 & -2 & 3
\end{pmatrix}\qquad
B=
\begin{pmatrix}
3 & -2 & 2 \\
-1 & 2 & -1\\
-2 & 2 & -1
\end{pmatrix}\qquad
C=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}
$$
Here $AB=BA$, $AC=CA$ and $BC=CB$. In addition, the characteristic polynomials of these matrices are
$$
f_A(x)=x(x-1)(x+1)\\
f_B(x)=(x-2)(x-1)^2\\
f_C(x)=x^2(x-1)
$$
so each one of them is triangulable. Thanks!
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Here, it is easy. Since $A$ has $3$ distinct eigenvalues and $B,C$ commute with $A$, we can deduce that $B,C$ are polynomials in $A$ and it suffices to triangularize $A$.
In the general case.
Step 1. Find a common eigenvector of $A,B,C$.
Step 2. Proceed by recurrence.
Moreover, you can choose $P$ as an orthogonal matrix.
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|
Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$ Find $\tan x$ if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan x}{9+\tan^2x}\right)=\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$ Now using above result $(1)$ can be written as
$$2x=2 \arctan(2 \tan^2 x)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right) \tag{2}$$
But $$2\arctan (\theta)=\arctan \left(\frac{2 \theta}{1-\theta^2}\right)$$
So $(2)$ becomes
$$2x=\arctan \left(\frac{4 \tan^2x}{1-4 \tan^4x}\right)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$
Now using $$\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)$$ i am getting a sixth degree polynomial in $\tan x$.
is there any better approach?
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Let $\alpha=\arctan(2\tan^2x)$ and $\beta=\arcsin(\frac{3\sin2x}{5+4\cos2x})$.
\begin{align}
\sin\beta&=\frac{3\sin2x}{5+4\cos2x}\\
\frac{2\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3(\frac{2\tan x}{1+\tan^2x})}{5+4(\frac{1-\tan^2x}{1+\tan^2x})}\\
\frac{\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3\tan x}{9+\tan^2x}\\
3\tan x\tan^2\frac{\beta}{2}-(9+\tan^2x)\tan\frac{\beta}{2}+3\tan x&=0\\
\left(3\tan \frac{\beta}{2}-\tan x\right)\left(\tan \frac{\beta}{2}\tan x-3\right)&=0\\
\tan\frac{\beta}{2}&=\frac{1}{3}\tan x \quad\textrm{or}\quad \frac{3}{\tan x}
\end{align}
Note that $x=\alpha-\frac{1}{2}\beta$.
\begin{align}
\tan x&=\tan\left(\alpha-\frac{1}{2}\beta\right)\\
&=\frac{\tan \alpha-\tan\frac{\beta}{2}}{1+\tan\alpha\tan\frac{\beta}{2}}\\
&=\frac{2\tan^2 x-\frac{1}{3}\tan x}{1+2\tan^2 x(\frac{1}{3}\tan x)} \quad\textrm{or}\quad \frac{2\tan^2 x-\frac{3}{\tan x}}{1+2\tan^2 x(\frac{3}{\tan x})} \\
&=\frac{\tan x(6\tan x-1)}{3+2\tan^3 x} \quad\textrm{or}\quad \frac{2\tan^3 x-3}{\tan x(1+6\tan x)} \\
\end{align}
So we have $\tan x=0$, $\tan^3x-3\tan x+2=0$ or $4\tan^3x+\tan^2x+3=0$.
Solving, we have $\tan x=0$, $1$, $-1$ or $-2$.
Note that $-1$ should be rejected.
$\tan x=-1$ is corresponding to $\tan\frac{\beta}{2}=\frac{3}{\tan x}$. So $\tan\frac{\beta}{2}=-3$, which is impossible as $\beta\in[\frac{-\pi}{2},\frac{\pi}{2}]$.
The answers are $0$, $1$ and $-2$.
|
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|
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
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HINT:
From the Right Hand Side, if $f(n)=\dfrac{n(n+1)(n+2)(n+3)(n+4)}5$
$$f(m+1)-f(m)=?$$
|
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|
Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$
Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??
I'm sorry for my typo, I have edited it.
I have now edited in my steps, per request of commenters.
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\begin{align}
(b-2)\sqrt{ac} & = \sqrt{(b-2)^2} \sqrt{ac} & & \text{ if } b-2\ge 0 \\[10pt]
& = \sqrt{(b-2)^2 ac}.
\end{align}
Is $(b-2)^2ac$ the same as $b^2-4ac\,$?
If one were to think that $(b-2)^2$ is the same as $b^2-4$ (and it is not) then one would have $(b-2)^2ac = b^2ac-4ac,$ so that is still not the same as $b^2-4ac.$
Notice that $\sqrt{5^2 - 3^2} = 4$ and $\sqrt{5^2}-\sqrt{3^2} = 5 - 3 = 2,$ so
$\sqrt{5^2-3^2}$ is different from $\sqrt{5^2}-\sqrt{3^2}.$
One should not generally ask why one cannot do things like this; but rather whether one can. Don't start from the presumption that it can be done. That puts the burden of proof in the wrong place.
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|
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmatrix} a^2+bc &b(a+d) \\ c(a+d)&d^2+bc \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
So
$$a^2+bc=1 \tag{1}$$
$$b(a+d)=2 \tag{2}$$
$$c(a+d)=3 \tag{3}$$
$$d^2+bc=4 \tag{4}$$ From $(2)$ and $(3)$ we get
$$\frac{b}{c}=\frac{2}{3}$$
Let $b=2k $ and $c=3k$ Then
$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$ So
$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\ 3k&\sqrt{4-6k^2} \end{bmatrix}$$
So
$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\ 3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4 \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
Now we have to solve for $k$ using equation
$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also
$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$
$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$
Subtracting $(6)$ from $(5)$ we get
$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$ Squaring Both sides we get
$$33k^4-10k^2+1=0$$ we get
$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?
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Square roots suck, so I'll do this with polynomials instead. Any matrix $B$ with $B^2=A$ must commute with $A$. Since there exist vectors $e$ such that $(e,Ae)$ form a basis of $\Bbb R^2$ (in fact the first standard basis vector will do, as will almost any other vector), any matrix commuting with $A$ must be a polynomial in$~A$, as shown for instance elementarily in this answer.
Since $A^2=2I_2+5A$ (check it) any polynomial in$~A$ can be rewritten to a polynomial of degree${}<2$ in$~A$. So let $B=sI_2+tA$ then $B^2=s^2I_2+2stA+t^2A^2=(s^2+2t^2)I_2+(2st+5t^2)A$. You need that to be equal to $A$ and since $I_2$ and $A$ are linearly independent, it will hold if and only if $s^2+2t^2=0$ and $2st+5t^2=1$. Clearly there are no solutions over $\Bbb R$, so if the question was to find a real square root of$~A$, the answer is there are none.
Working over the complex numbers, we can however continue. We must have $t\neq0$, and then the first equation says $(s/t)^2=-2$. First take $s/t=\sqrt2\,\mathbf i$, then the second equation becomes $\sqrt8\,\mathbf it^2+5t^2=1$ or $t^2=\frac{5-\sqrt8\,\mathbf i}{33}$. Taking complex square roots gives you two solutions. Taking instead $s/t=-\sqrt2\,\mathbf i$ one gets $t^2=\frac{5+\sqrt8\,\mathbf i}{33}$ with again two solutions. These appear to be the ones you found in the question. I don't think it will clarify anything to detail these solutions any further (it just becomes a mess, though it will work if you persevere).
|
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|
Find an upperbound for the rational function I know
$$\lim_{(x, y)\to (0,0)} \frac{x^3 + y^4}{x^2 + y^2} = 0$$ so
$$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert \le f(x, y)$$ for some simpler $f(x, y)$ whose limit is also $0$.
How do I find the function $f(x, y)$? In other words, how do I get the upper bound?
|
Using the inequalities $x^2+y^2\ge x^2$ and $x^2+y^2\ge y^2$, we can write
$$\begin{align}
\left|\frac{x^3+y^4}{x^2+y^2}\right|&\le \frac{|x|^3}{x^2+y^2}+\frac{y^4}{x^2+y^2}\\\\
&\le |x|+y^2
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration $\int \frac{\sec(x)}{\tan^2(x) + \tan(x)}$ How to evaluate the following integral $$\int \frac{\sec(x)}{\tan^2(x) + \tan(x)} \ dx$$
Thank you.
|
Note that:
\begin{align}
I = \frac{\sec x}{\tan^2x + \tan x} &= \frac{\frac{1}{\cos x}}{\frac{\sin^2x}{\cos^2x}+\frac{\sin x}{\cos x}}\\
&= \frac{\cos x}{\sin^2x+\sin x\cos x}\\
&=\frac{(\cos x + \sin x) - \sin x}{\sin x (\sin x + \cos x)}\\
&=\frac{1}{\sin x} - \frac{1}{\sin x + \cos x} \\
&= \csc x - \frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}\\
&= \csc x - \frac{1}{\sqrt{2}}\csc(x+\frac{\pi}{4})
\end{align}
So that
\begin{align}
\int \frac{\sec x}{\tan^2x + \tan x} dx&= \int \csc x - \frac{1}{\sqrt{2}}\csc(x+\frac{\pi}{4}) ~dx\\ &= -\ln|\csc x+\cot x| + \frac{1}{\sqrt{2}} \ln|\csc (x + \frac{\pi}{4})+\cot (x+\frac{\pi}{4})|+C
\end{align}
|
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|
Is there any way to simplify the product of cosines? I recently saw a problem: Estimate the following: $cos(\frac{\pi}{15})cos(\frac{2\pi}{15})\ldots cos(\frac{7\pi}{15})$, and the options were between different consecutive power of ten.
How would I do this, no calculator of course, and is there's any way to shorten any general products of sines or cosines or other trig functions?
|
We have $\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$. Let's write this as $p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$
We have to multiply both sides of the equation with
$$q= \sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})$$
Now, $$p.q = \sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$$
Multiply both sides with $2^7$
$$2^7 p.q =[2 \sin(\frac{\pi}{15})\cos(\frac{\pi}{15})][2\sin(\frac{2\pi}{15})\cos(\frac{2\pi}{15})]\ldots [2\sin(\frac{7\pi}{15})\cos(\frac{7\pi}{15})]$$
$$2^7 p.q=\sin(\frac{2\pi}{15})\sin(\frac{4\pi}{15})\ldots \sin(\frac{14\pi}{15})$$
Now we have to apply the identity $\sin\theta=\sin(\pi-\theta)$
$$2^7 p.q=\sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})$$
Which means $$2^7 p.q=q$$
Therefore,
$$p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})=\frac1{2^7}$$
|
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|
What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$? The methods that I can think of are:
1) Chain rule
2) Binomial series of $f(x)$
3) Through the formula $f\prime(g(x))=\frac{1}{g\prime (x)}$ at a particular point
What are the other methods to determine the derivative of $f(x)$?
|
From first principles.
\begin{align}
f'(x)&=\lim_{h\to0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h}\\
&=\lim_{h\to0}\frac{(\sqrt{1+x+h}-\sqrt{1+x})(\sqrt{1+x+h}+\sqrt{1+x})}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1+x+h-1-x}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1}{\sqrt{1+x+h}+\sqrt{1+x}}\\
&=\frac{1}{2\sqrt{1+x}}
\end{align}
$[f(x)]^2=1+x$. So
\begin{align}
2f(x)f'(x)&=1\\
f'(x)&=\frac{1}{2f(x)}\\
&=\frac{1}{2\sqrt{1+x}}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2309074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Basis Transformation Matrix I have a problem related to transformation matrix.
The problem is like that;
Given basis $B$: $b_1 = \begin{pmatrix}1&2\end{pmatrix}^T$ and $b_2 =\begin{pmatrix}2&1\end{pmatrix}^T$ and $A$: $a_1=\begin{pmatrix}1&2\end{pmatrix}^T$, $a_2=\begin{pmatrix}2&7\end{pmatrix}^T$.
Find the transformation matrix $T$ from the basis $B$ into basis $A$?
I would appreciate if anyone help me with this.
|
The solution by provided by @NDewolf. Let's fill out the details.
Vectors will be colored according to the basis membership, and named based upon color:
$$
\color{blue}{\mathbf{B}\ (standard)}, \qquad
\color{red}{\mathbf{R}\ (a)}, \qquad
\color{green}{\mathbf{G}\ (b)}.
$$
$$
\mathbf{T}_{\color{red}{R}\to \color{blue}{B}}=
\color{black}{\left[
\begin{array}{r|r}
1 & 2 \\
2 & 7 \\
\end{array}
\right]}
$$
The matrix $\mathbf{T}_{\color{red}{B}\to \color{blue}{A}}$ is an operator which maps a $\color{blue}{blue}$ vector to a $\color{red}{red}$ vector.
The inverse matrix is a map which connects vectors in the $\color{blue}{standard}$ basis to vectors in the $\color{red}{\mathbf{R}}$ basis:
$$
\mathbf{T}^{-1}_{\color{blue}{B}\to \color{red}{R}} =
%
\color{blue}{
\frac{1}{3}
\left(
\begin{array}{rr}
7 & -2 \\
-2 & 1 \\
\end{array}
\right)}
$$
For the green basis $\color{green}{\mathbf{G}}$, the maps are
$$
\mathbf{G}_{\color{green}{G}\to \color{blue}{A}}=
\color{black}{\left[
\begin{array}{r|r}
1 & 2 \\
2 & 1 \\
\end{array}
\right]}, \qquad
%
\mathbf{G}^{-1}_{\color{blue}{A}\to \color{green}{G}}=
\color{black}{\left[
\begin{array}{rr}
-1 & 2 \\
2 & -1 \\
\end{array}
\right]}
$$
Transition from red $\color{red}{\mathbf{R}\ (a)}$ to green $\color{green}{\mathbf{G}\ (b)}$
As so succinctly expressed by @Bernard, connect all bases through the hub of the blue basis, $\color{blue}{standard}$. Start with a vector in the $\color{red}{\mathbf{R}}$ basis, map that to a vector in the $\color{blue}{standard}$ basis, then map that to a vector in the $\color{green}{\mathbf{G}}$ basis:
$$
\color{red}{
\left[
\begin{array}{c}
x_{1} \\
y_{1} \\
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{blue}{
\left[
\begin{array}{c}
x_{2} \\
y_{2} \\
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{green}{
\left[
\begin{array}{c}
x_{3} \\
y_{3} \\
\end{array}
\right]}
%
$$
The solution is
$$
\begin{align}
T_{\color{red}{R}\to\color{green}{G}} &= \color{green}{G^{-1}} \color{red}{R} \\
& =
\left[
\begin{array}{rr}
-1 & 2 \\
2 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
1 & 2 \\
2 & 7 \\
\end{array}
\right] \\
%
&=
\left[
\begin{array}{cr}
1 & 4 \\
0 & -1 \\
\end{array}
\right]
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2309910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The value of k is given to be 4, but I am unable to derive that result.
Kindly provide me with some insight, or with a step-by-step solution.
Thanks in advance,
Abhigyan
|
We need to prove that
$$3-\tan^2\frac{\pi}{7}=4(1-\tan^2\frac{\pi}{7})\cos\frac{\pi}{7}$$ or
$$\frac{3}{2}+\frac{3}{2}\cos\frac{2\pi}{7}-\frac{1}{2}+\frac{1}{2}\cos\frac{2\pi}{7}=4\cos\frac{2\pi}{7}\cos\frac{\pi}{7}$$ or
$$1+\frac{3}{2}\cos\frac{2\pi}{7}+\frac{1}{2}\cos\frac{2\pi}{7}=2\cos\frac{3\pi}{7}+2\cos\frac{\pi}{7}$$ or
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$ or
$$2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}=-2\sin\frac{\pi}{7}$$ or
$$\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}=-\sin\frac{\pi}{7}.$$
Id est, indeed, $k=4$.
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
|
We can do better than hypergeometric function and elliptic integral:
$$\color{blue}{\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{{{\Gamma ^2}(\frac{1}{4})}}{{12\sqrt \pi }}}$$
Firstly, integration by part gives
$$\int_0^1 {\sqrt {1 + {x^4}} dx} = \sqrt 2 - 2\int_0^1 {\frac{{{x^4}}}{{\sqrt {1 + {x^4}} }}dx} = \sqrt 2 - 2\int_0^1 {\left( {\sqrt {1 + {x^4}} - \frac{1}{{\sqrt {1 + {x^4}} }}} \right)dx} $$
Hence
$$\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{2}{3}\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} $$
Making $x=1/u$ in the last integral gives
$$\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} = \frac{1}{2}\int_0^\infty {\frac{1}{{\sqrt {1 + {x^4}} }}} dx = \frac{1}{{8\sqrt \pi }}{\Gamma ^2}(\frac{1}{4})$$
which can be evaluated by using some formula for beta function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2311583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
}
|
Solving the limit: $\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$ Find the value of the limit $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$$ where $[\cdot]$ denotes the greatest integer function or the box function.
My attempt: I am aware of the standard limits $$\lim_{x\to0}\left(\frac{\sin^{-1}(x)}{x} \right) = 1 $$ and $$\lim_{x\to0}\left(\frac{\tan^{-1}(x)}{x} \right) = 1 $$ but am not sure how will I apply the box function on this limit.
Any detailed explanation to help me understand this concept will be appreciated.
|
When $x\to 0 $
$$\sin x\sim x-\frac{x^3}{6} \space , \space \space\space\sin^{-1} x\sim x+\frac{x^3}{6}\\\tan x\sim x+\frac{x^3}{3}\space ,\space\space \space\tan^{-1} x\sim x-\frac{x^3}{3} $$so
$$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\
\lim_{x\to0}\left[100\frac{x+\frac{x^3}{6}}{x}\right]+\left[100\frac{x-\frac{x^3}{3}}{x}\right]=\\
\lim_{x\to0}\left[100(1+\frac{x^2}{6})\right]+\left[100(1-\frac{x^2}{3})\right]=\\
\lim_{x\to0}\left[100+\frac{100x^2}{6}\right]+\left[100-\frac{100x^2}{3})\right]=\\
\left[100^+\right]+\left[100^-\right]=100+99$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2311795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Find a point $P$ on a curve given the gradient of the normal at $P$. The point $P$ lies on curve $y=(x-5)^2$. It is given that the gradient of the normal at $P$ is $-\frac 14$
Find the coordinates of $P$.
|
First find the gradient function of the curve:
$$\frac{dy}{dx}=2(x-5)$$
We want to find the point where the gradient is $-\frac{1}{-\frac14}=4$
So we solve \begin{align}2(x-5)&=4\\
x-5&=2\\
x&=7\end{align}
And therefore the point $P$ has an $x$ coordinate of $7$, and we can find its $y$ coordinate:
\begin{align}y&=(x-5)^2\\
&=(7-5)^2\\
&=2^2\\
&=4\end{align}
So $P$ has coordinates $(7,4)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2314841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
find the limit of $ (1/\sin\ (x)-1/x)^x$ I am trying to find the limit of $\lim_{x \downarrow 0} (\frac{1}{sinx}- \frac{1}{x})^x$.
My current progress:
$\lim_{x \downarrow 0} (\frac{1}{\sin x}- \frac{1}{x})^x = \lim_{x \downarrow 0} (\frac{x-\sin x}{x\sin x})^x = \lim_{x \downarrow 0} e^{xln(\frac{x-\sin x}{x \sin x})} = e^{\lim_{x \downarrow 0}xln(\frac{x-\sin x}{x \sin x})} = e^{\lim_{x \downarrow 0}\frac{ln(\frac{x-\sin x}{x \sin x})}{1/x}}$
This is where I used l'Hospital, but after 2 iterations of l'Hospital it doesn't look like I'll get anywhere with it.
Expression after first l'Hospital:
(I'll not write the e as the base, so that the limit is easier to read)
$\lim_{x \downarrow 0}\frac{-x^3\sin x\cos x - x \sin^3x}{x\sin^2x - \sin^3x} = \lim_{x \downarrow 0}\frac{-x^3\cos x - x \sin^2x}{x\sin x - \sin^2x}$
Am I missing something or is it just tedious to find the limit of this expression?
After the second iteration of l'Hospital, I've got this term:
$\lim_{x \downarrow 0}\frac{-3x^2\cos x + x^3\sin x - \sin^2x -2x\sin x\cos x}{x\cos x + \sin x -2\sin x\cos x} = "\frac{0}{0}"$
is that term correct? If so, I'll try to apply l'Hospital a third time.
|
If $ \lim\ (\frac{1}{\sin\ x} - \frac{1}{x} )^x=L$, then $$ \lim\ x\cdot
\ln\ (\frac{1}{\sin\ x} - \frac{1}{x} )=\ln\ L$$
Here by Taylor series, $$ x
\ln\ ( \frac{1}{\sin\ x} - \frac{1}{x} )= x \ln\ \frac{x-\sin\
x}{x\sin\ x} =x\ln\ \frac{x/6 + O(x^2)}{1 + O(x^2)} =\frac{\ln\
(\frac{x}{6} + O(x^2) ) }{1/x}$$
L'Hospital implies that $$ \ln\ L = \lim\ \frac{ \frac{ 1/6 + O(x) }{
x/6 + O(x^2) } }{ -1/x^2} =\lim\ - \frac{ 1/6 + O(x) }{
1/6 + O(x) } x =0 $$
Hence $L=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2314973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
|
Using partial fractions we have
\begin{eqnarray*}
\frac{3n+7}{n(n+1)(n+2)}= \frac{\frac{7}{2}}{n}+\frac{-4}{n+1}+\frac{\frac{1}{2}}{n+2}
\end{eqnarray*}
It is a telescoping sum ...
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{3n+7}{n(n+1)(n+2)} = \frac{\frac{7}{2}}{1}&+&\color{green}{\frac{-4}{2}} &+&\color{red}{\frac{\frac{1}{2}}{3}} \\&+&\color{green}{\frac{\frac{7}{2}}{2}}&+&\color{red}{\frac{-4}{3}+\frac{\frac{1}{2}}{4}} \\
&& &+&\color{red}{\frac{\frac{7}{2}}{3}}+\color{red}{\frac{-4}{4}+\frac{\frac{1}{2}}{5}} \\
&&&& \ddots
\end{eqnarray*}
After the dust settles ... we have $\color{red}{\frac{13}{4}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ I'm looking for assistance with the following problem:
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$.
My attempt:
We need $\delta$ such that $\vert f(x)-2 \vert < .2$. That is, we need $\vert x^2 - 1 - 3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < .2$. We can control $\vert x - 2 \vert$ by $\delta$. Let us then restrict $x$ to the interval $(1,3)$. Thus $\vert x+2 \vert < 5$. Then we will let $\delta = \frac{.2}{5}$.
Now then, if $0 < \vert x - 2 \vert < \delta = \frac{.2}{5}$ then $\vert f(x)-3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < 5 \cdot \frac{.2}{5}=.2.$ $\Box$
Does this work for a valid delta? I'm basically taking the more general case where $\delta$ would be $\min(\frac{\epsilon}{5},1)$ and trying to fit it to this specific case.
|
Since
\begin{eqnarray}
|f(x)-3|&=&|x^2-4|\\
&=&|x-2||(x-2)+4|\\
&\le& |x-2|(|x-2|+4)\\
&=&|x-2|^2+4|x-2|
\end{eqnarray}
if we solve the inequality
$$
y^2+4y\le 0.2
$$
we get
$$
a\le y\le b,
$$
with
\begin{eqnarray}
a&=&-2-\sqrt{4.2}\approx-4.0494\\
b&=&-2+\sqrt{4.2}\approx0.0494\\
\end{eqnarray}
Hence, choosing $\delta \in (0,b)$, we have
$$
0<|x-2|<\delta \implies a<|x-2|<b \implies |f(x)-1|\le |x-2|^2+4|x-2|<0.2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2319006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]{b}=\frac{a\pm b}{\sqrt[3]{a^2}\mp \sqrt[3]{ab}+\sqrt[3]{b^2}}$ :
$\sqrt[3]{x+2}-ax+\sqrt[3]{2x-1}-bx=4-x$
Need find $ax$, $bx$ but it's very bad, too
|
to solve this equation use $$(a+b)^3=a^3+b^3+3ab(a+b)$$
$$(\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x)^3\\x+2+2x-1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3$$ put $\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x \\$ so
$$3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3\\
3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3\\3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3-3x-1\\
27(x+2)(2x-1)(4-x)^3=((4-x)^3-3x-1)^3$$ now you must go to the power of 3
and solve $ax^9+....=0$ degree=9
but it will be polynomial .
if you solve the eqaution with numerical or graphical method ,it has a root $x=1.325$ (the only root )
are you sure that type the equation correct ?
***why there is one real root ?
if you take $f(x)=\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x-4 $ so $$f'(x)=\dfrac{1}{3\sqrt[3]{(x+2)^2}}+\dfrac{2}{3\sqrt[3]{(2x-1)^2}}+1 >0$$ so f(x) is strictly increasing $\to $
so $$f(x)=0$$ has only one root , It can be checked by
$$f(1)<0 ,f(1.5)>0 \to x_0 \in (1,1.5)$$ so you can go on
split $(1,1.5) \to (1,1.25) ,(1.25,1.5)$
$$f(1)f(1.25)>0 \\f(1.25)f(1.5)<0 \to x_0 \in (1.25,1.5)$$ and take over
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2319277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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|
Evaluate the following Determinant Find $$ \Delta=\begin{vmatrix}
\frac{1}{a+x} &\frac{1}{b+x} &\frac{1}{c+x} \\
\frac{1}{a+y} &\frac{1}{b+y} &\frac{1}{c+y} \\
\frac{1}{a+z} &\frac{1}{b+z} &\frac{1}{c+z}
\end{vmatrix}$$
I applied $C_1 \to C_1-C_2$ and $C_2 \to C_2-C_3$ we get
$$ \Delta=\begin{vmatrix}
\frac{b-a}{(a+x)(b+x)} &\frac{c-b}{(b+x)(c+x)} &\frac{1}{c+x} \\
\frac{b-a}{(a+y)(b+y)} &\frac{c-b}{(b+y)(c+y)}&\frac{1}{c+y} \\
\frac{b-a}{(a+z)(b+z)} &\frac{c-b}{(b+z)(c+z)} &\frac{1}{c+z}
\end{vmatrix}$$
Now taking $b-a$,$\:$$c-b$ common and taking $$\frac{1}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)}$$ outside Determinant we get
$$\Delta=\frac{(b-a)(c-b)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix}
c+x &a+x &(a+x)(b+x) \\
c+y &a+y &(a+y)(b+y)\\
c+z &a+z &(a+z)(b+z)
\end{vmatrix}$$
Now apply $C_1 \to C_1-C_2$ we get
$$\Delta=\frac{(b-a)(c-b)(c-a)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix}
1 &a+x &(a+x)(b+x) \\
1 &a+y &(a+y)(b+y)\\
1 &a+z &(a+z)(b+z)
\end{vmatrix}$$
any clue here?
|
\begin{align}
\Delta&=\begin{vmatrix} \frac{1}{a+x} & \frac{1}{b+x} & \frac{1}{c+x} \\ \frac{1}{a+y} & \frac{1}{b+y} & \frac{1}{c+y} \\ \frac{1}{a+z} & \frac{1}{b+z} & \frac{1}{c+z}\end{vmatrix}\\
\left[\prod_{q\in\{x,y,z\}}(a+q)\right]\Delta&=\begin{vmatrix} 1 & \frac{a+x}{b+x} & \frac{a+x}{c+x} \\ 1 & \frac{a+y}{b+y} & \frac{a+y}{c+y} \\ 1 & \frac{a+z}{b+z} & \frac{a+z}{c+z}\end{vmatrix}\\
\left[\prod_{p\in\{a,b\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} b+x & a+x & \frac{(a+x)(b+x)}{c+x} \\ b+y & a+y & \frac{(a+y)(b+y)}{c+y} \\ b+z & a+z & \frac{(a+z)(b+z)}{c+z}\end{vmatrix}\\
\left[\prod_{p\in\{a,b,c\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\ (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix}
\end{align}
Let $\displaystyle \Delta_0=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\ (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix}$.
Note that $\Delta_0=0$ when $a=b$, $b=c$, $c=a$, $x=y$, $y=z$ and $z=x$.
Since $\Delta_0$ is a polynomial of degree $6$ in $a,b,c,x,y,z$,
$$\Delta_0=k(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$$
for some constant $k$.
Put $a=x=1$, $b=y=0$ and $c=z=-1$.
\begin{align}
\begin{vmatrix} 0 & 0 & 2 \\ 0 & -1 & 0 \\ 2 & 0 & 0\end{vmatrix}&=k(1)(1)(-2)(1)(1)(-2)\\
k&=1
\end{align}
Therefore,
$$\Delta=\frac{(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)}{(a+x)(a+y)(a+z)(b+x)(b+y)(b+z)(c+x)(c+y)(c+z)}$$
|
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|
Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
|
We prove that $a = 1009, b = 1$ is the only solution.
First we restrict attention to the case in which $a, b \geq 3$.
We can rewrite the equation as
$$\frac{\log a}{a} - \frac{\log b}{b} = \frac{\log\left(1 + \frac{1008}{b^a}\right)}{ab}.$$
Let $f(x) = \frac{\log x}{x}$. We have $f'(x) = \frac{1 - \log x}{x^2} < 0$, so $f(x)$ is decreasing on $[3,+\infty)$. Therefore we have $b > a$.
By the mean value theorem, there is some $c \in (a,b)$ such that
$$\frac{\log a}{a} - \frac{\log b}{b} = (b - a)\frac{\log c - 1}{c^2} \geq \frac{\log 3 - 1}{b^2}.$$
On the other hand, we have
$$\frac{\log\left(1 + \frac{1008}{b^a}\right)}{ab} \leq \frac{1008/b^a}{ab} = \frac{1008}{ab^{a+1}}.$$
Therefore
$$ab^{a-1} \leq \frac{1008}{\log 3 - 1} < 10222.$$
Consequently, we must have $a \leq 5$, and:
*
*If $a = 3$, then $4 \leq b \leq 58$.
*If $a = 4$, then $5 \leq b \leq 13$
*If $a = 5$, then $b = 6$.
But all of these cases are impossible.
*
*$5^6 - 6^5 \ne 1008$, so $a \ne 5$.
*When $a = 4$, we have $4^b - b^4 = 1008$. $1008$ is equal to $16$ modulo $32$, so we must have $b = 2$ modulo 4. But $b = 6$ and $b = 10$ are not solutions.
*When $a = 3$, we have $3^b - b^3 = 1008$. Since $b > 3$, we have $b^3 = 18$ modulo 27, but this equation has no solutions.
Other cases
When $b = 1$, we have only the solution $a = 1009$.
When $b = 2$, we have $a^2 > 2^a + 1$, which never occurs.
When $a = 1$, there are obviously no solutions.
When $a = 2$, we have $2^b - b^2 = 1008$. If $x_n = 2^n - n^2$, then $x_{n + 1} - x_n = 2^n - (2n + 1) > 0$ for $n \geq 3$, so the sequence $x_n$ is increasing from $n \geq 3$. We have $x_{10} < 1008 < x_{11}$, so there are no solutions for $a = 2$.
|
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|
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
|
Assume a polynomial function of the form $f(x)=ax^3+bx^2+cx+d$ then $f^{'}(x)=3ax^2+2bx+c$
Thus the equations becomes $ax^3+(3a+b)x^2+(c+2b)x+d+c=x^3+5x^2+x+2$.
Now you can compare the coefficients to get the solution.
|
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|
Trigonometric Equation : $\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$ Please help solving this equation:
$\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$
I used numerical method to solve it and got $x=6^\circ$ but I am not able to solve it by trigonmetry.
Thank!
Best Regards, Michael.
|
Observation
$$
\sin 96^{\circ} \sin 12^{\circ} = \sin 18^{\circ} \sin 42^{\circ}
$$
Implication
$$
\sin x = \sin \left(12^{\circ} - x \right) \qquad \Rightarrow \qquad x = 12^{\circ} - x
$$
Conclusion
$$x=6^{\circ}$$
|
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|
Find area of curve $(x^2+y^2)^2=2a^2xy$ How can I find the area of this curve?
$(x^2+y^2)^2=2a^2xy$
Should I first assume that $x=a\cos t$, or $y=\sin t$?
And then use the formula with $x$ and $y$ derivative?
|
Let's convert to polar coordinates.
We get $\displaystyle (r^2)^2 = 2(a^2)(r\cos(\theta))(r\sin(\theta))$.
That boils to $r^4 = a^2r^2\sin(2\theta)$.
We then get $r^2 = a^2\sin(2\theta)$.
So that means $r = \pm a\sqrt{\sin(2\theta)}$.
Graph analysis.
Notice that it actually doesn't matter which root we take, because $-r$ corresponds to a reflection about the origin and $\sin(2(\theta+π))=\sin(x)$, since $π$ is the period.
What does that mean? That means that taking the negative root traces out the same region in reverse.
Now to the integration.
We have two intervals of $\sin(2\theta)$, where it is positive and the root is defined. Because these two portions of sine are identical, so is the graph. We will therefore compute the area enclosed by one petal and mutliply by $2$.
We have that $\sin(2\theta)$ has roots at $\theta=0$ and $\displaystyle \theta = \frac{π}{2}$.
Then, $A_{\text{petal}} = \displaystyle \frac{1}{2}\int_0^{\frac{π}{2}} r^2\,d\theta= \frac{1}{2}\int_0^{\frac{π}{2}} (a\sqrt{\sin(2\theta}))^2\,d\theta= \frac{a^2}{2}\int_0^{\frac{π}{2}} \sin(2\theta)\,d\theta=\frac{a^2}{2}$.
Therefore, two petals enclose an area of $\boxed{a^2}$.
|
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|
Prove that each divisor of $5^{(4n+1)}+5^{(3n+1)}+1 =$ $1 \pmod {10}$ Prove that if $d$ divides $5^{(4n+1)}+5^{(3n+1)}+1$ for any $n$, then $d =$ $1 \pmod {10}$.
A similar statement can be proven, where $d$ divides $(3^{(2n+1)}+1)/4$, for any $n$, then $d =$ $1 \pmod {6}$ by first showing that if $-3$ is a quadratic residue $\pmod p$, then $p = 1 \pmod 3$. For the base $5$ expression I would start by showing that if $-5$ is a quartic residue $\pmod p$, then $p = 1 \pmod {10}$. Any help on proving the original statement with base $5$ expression? Thanks!!!
|
We can restrict to prime divisors $p$. These must be odd, so
all we need to prove is $p\equiv1\pmod5$. We can't have $p=5$ either.
Now
$$5a^4+5a^3+1\equiv0\pmod p$$
where $a=5^n$ and so
$$b^4+5b+5\equiv0\pmod p$$
where $b$ is the inverse of $a$ modulo $p$.
Consider $f(x)=x^4+5x+5$ and let $\zeta$ be a primitive fifth root
of unity. Let $\eta=\zeta^2-\zeta$. Then
\begin{align}
\eta^4&=\zeta^3-4\zeta^2+6\zeta-4+\zeta^4\\
&=-5\zeta^2+5\zeta-5=-5-5\eta.
\end{align}
Thus $f(\eta)=0$ and any zero of $f$ generates the cyclotomic
field $K=\Bbb Q(\zeta)$.
If $f(a)\equiv0\pmod p$ then $p$ splits in $K$. As the primes that
split in $K$ are those congruent to $1$ modulo $5$ then $p\equiv1\pmod5$.
|
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|
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$ if$\ δ=ε$ we have to prove that:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|<ε$$
But I'm having a hard time trying to prove the last part, I tried:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$
|
Since
$$
0\leq \frac{y^2}{x^4+y^2} \leq 1 \qquad \forall (x,y)\neq (0,0),
$$
you get
$$
|f(x,y)| = |x|^3 \left[\frac{y^2}{x^4+y^2}\right]^2 \leq |x|^3,
\qquad \forall (x,y)\neq (0,0),
$$
hence $f(x,y) \to 0$ as $(x,y)\to (0,0)$.
|
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|
Complex Integration - $\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx$
Exercise :
Show that :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$
Attempt :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$
$$x^4 + x^3 + x^2 + x + 1=0 \Leftrightarrow x = \{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$
So, the function $f(z) = \frac{1}{x^4 + x^3 + x^2 + x + 1}$ has poles at the points :
$$\{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$
Now, I know you have to integrate through a closed curve $C$ and on a line $γ_R$ and then continue on with residues for the poles that reside in this curve, but I am stuck on how to apply it here and I also miss it a bit on how to split the integral for the curve and the line. Most examples I've saw get simpler due to the even function trick, but that cannot be applied here.
I would really appreciate a thorough solution and explanation, since I've just started working on generalized integrals and I have to clear my mind on them.
Thanks for your time !
|
Using the upper part of the complex plane we get:
$\displaystyle \int_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=i2\pi Res(\frac{x-1}{x^5-1},e^{i2\pi/5})+ i2\pi Res(\frac{x-1}{x^5-1},e^{i4\pi/5})$
$\displaystyle = i2\pi\frac{x-1}{x^5-1}(x-e^{i2\pi/5})|_{x\to e^{i2\pi/5}}+ i2\pi\frac{x-1}{x^5-1}(x-e^{i4\pi/5})|_{x\to e^{i4\pi/5}} $
$\displaystyle =\frac{\pi}{4}\frac{1}{\sin(\frac{\pi}{5})\sin(\frac{3\pi}{5})\sin(\frac{4\pi}{5})}=\frac{4\pi}{5}\sin\frac{2\pi}{5}$
|
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|
A Conditional Probability problem with Biased Coins I need explanation for solving the following problem - exactly how one goes about solving it ? Any pointers or hints will certainly be appreciated:
A box contains 5 fair coins and 5 biased coins. Each biased coin has probability of a head as $\frac{4}{5}$. A coin is drawn at random from the box and tossed. Then a second coin is drawn at random from the box (without replacing the first one). Given that the first coin has shown head, the conditional probability that the second coin is fair, is:
A) $\frac{20}{39}$ B) $\frac{20}{37}$ C) $\frac{1}{2}$ D) $\frac{7}{13}$
|
$$P(\text{second is fair}\mid \text{first is head}) = \frac{P(\text{second is fair and first is head})}{P(\text{1st is head})}.$$
$$P(\text{first is head}) = P(\text{first is biased}) \times P(\text{head}\mid\text{biased}) + P(\text{first is unbiased}) \times P(\text{head}\mid\text{unbiased}) $$ $$= \left(\frac{4}{5} \times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\right)=\frac{4}{10}+\frac{1}{4}=\frac{13}{20}.$$
$$P(\text{second is fair and first is head})$$ $$= P(\text{second is fair and first is fair and is head}) + P(\text{second is fair and first is unfair and is head}) $$ $$=
\left(\frac{1}{2}\times\frac{1}{2}\times\frac{4}{9}\right)+\left(\frac{4}{5}\times\frac{1}{2}\times\frac{5}{9}\right) = \frac{4}{36}+\frac{20}{90}=\frac{60}{180}=\frac{1}{3}.$$
Hence, the given probability equals $$\frac{1}{3} \div \frac{13}{20}= \boxed{\frac{20}{39}}.$$
|
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|
Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$? First of all, note that $\frac{n^{n+1}}{(n+1)^n} \sim \frac{n}{e}$.
Question: Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$?
There is an OEIS sequence for $n^{n+1}\mod (n+1)^n$: https://oeis.org/A176823.
$0, 1, 8, 17, 399, 73, 44638, 1570497, 5077565, 486784401, 22187726197,
166394893969, 13800864889148, 762517292682713, 9603465430859099,
803800832678655745, 3180753925351614970, 947615093635545799201$
|
We first prove that it is impossible when $n\not\equiv 1\pmod 4$.
Notice how:
$$n^{n+1}=1+(n-1)\sum_{k=0}^{n}n^k$$
So, we would have $n^{n+1}\equiv 1\pmod{(n+1)^n}$ if and only if:
$$(n+1)^n\mid (n-1)\sum_{k=0}^{n}n^k$$
And since the RHS won't be equal to $0$, we'd have:
$$(n-1)\sum_{k=0}^{n}n^k\ge(n+1)^n$$
but, since $4\nmid n-1$, we have $\gcd(n-1,(n+1)^n)\le 2$, so this becomes:
$$2\sum_{k=0}^{n}n^k\ge(n+1)^n$$
Multiplying both sides with $(n-1)$ yields:
$$2n^{n+1}>2(n^{n+1}-1)\ge (n-1)(n+1)^n=\frac{n-1}{n+1}\cdot(n+1)^{n+1}$$
or, assuming $n>1$:
$$\frac{2n+2}{n-1}>\left(\frac{n+1}{n}\right)^{n+1}=\left(1+\frac1n\right)^{n+1}>\left(1+\frac1n\right)^n$$
However, as $n$ tends to infinity, the LHS tends to $2$, while the RHS tends to $e$. Using induction, we first prove that for $n\ge 11$, we have $2.4\ge\frac{2n+2}{n-1}$. This is quite easy and I'll leave it out.
For $n=11$, we also have that the RHS is greater than $2.6$ and since $(1+\frac1n)^n$ keeps increasing as $n$ keeps increasing, this shows that there are no $n\not\equiv 1\pmod 4$ with $n\ge 11$ with $n^{n+1}\equiv1\pmod{(n+1)^n}$. Some quick testing reveals that there are no solutions at all for $n\not\equiv 1\pmod 4$.
As per @san's request, I'll also provide his solution for the case $n\equiv 1\pmod 4$, so that there is one complete answer.
Assume by contradiction that $n=4j+1$ for some positive integer $j$, and $n^{n+1} \equiv 1 \mod (n+1)^n$. Then
$$
n+1=2(2j+1)\quad\text{and}\quad n-1=2^ra
$$
for some $r\ge 2$ and some odd $a$.
There exists some $k$ such that $n^{n+1} - 1 =k\cdot (n+1)^n=k\cdot 2^n(2j+1)^n$.
But
\begin{eqnarray*}
n^{n+1} - 1&=&\sum_{s=0}^{n+1}\binom{n+1}{s}(n-1)^s-1\\
&=& \sum_{s=1}^{n+1}\binom{n+1}{s}(2^r a)^s\\
&=& (n+1)2^r a+2^{2r}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\\
&=& (2j+1)2^{r+1}a+2^{2r}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\\
&=& 2^{r+1}\left((2j+1)a+2^{r-1}a^2 \sum_{s=2}^{n+1}\binom{n+1}{s}(2^r a)^{s-2}\right)\\
\end{eqnarray*}
and $2^n$ divides $n^{n+1}-1$, hence $2^n$ divides $2^{r+1}$, and so $r\ge n-1$, which contradicts the fact that $n-1=2^ra$, since in general $n-1<2^{n-1}$.
|
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|
$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$ or $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!
|
\begin{eqnarray}
\sin x&=&\sin\left(x-\frac{\pi}{3}\right)\\
&=&\sin x\cos\frac{\pi}{3}-\cos x\sin\frac{\pi}{3}\\
&=&\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\\
\frac{1}{2}\sin x&=&-\frac{\sqrt{3}}{2}\cos x\\
\tan x&=&-\sqrt{3}\tag{1}\\
x&=&-\frac{\pi}{3},-\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}
\end{eqnarray}
This answer the title of your question.
To answer the second question contained within the text of your question, once one has equation $(1)$ above, one knows that the general solution will be
$$ x=\tan^{-1}\left(-\sqrt{3}\right)+\pi k=-\frac{\pi}{3}+\pi k$$
which is equivalent to
$$ x=\frac{2\pi}{3}+\pi k$$
As for the results which you give, see the following:
Using the identities
$$ \sin x=\sin(\pi-x) $$
and
$$\sin x=\sin(x+2\pi k)$$
if $\sin(x)=\sin\left(x-\frac{\pi}{3}\right)$ then either
$$ x=x-\frac{\pi}{3}+2\pi k\quad\text{ for some integer }k\tag{2} $$
or
$$ \pi-x=x-\frac{\pi}{3}+2\pi k\quad\text{ for some integer }k\tag{3} $$
Equation $(2)$ gives the nonsensical result $k=\frac{1}{6}$ and equation $(3)$ gives
$$x=\frac{2\pi}{3}-\pi k$$
which is equivalent to
$$x=\frac{2\pi}{3}+\pi k$$
COMMENT: Your answer sheet did the same thing I did in equation $(2)$ and $(3)$ except they replaced the $x-\frac{\pi}{3}$ on the right side of the equation with $\pi-(x-\pi/3)$ whereas I replaced the $x$ on the left side with $\pi-x$ which I think is a little less confusing.
|
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|
Unexpected outcome of integral A given method for calculating $\int^1_{-1} \frac 1 {1+x^2} \, dx$ is
\begin{align}
& \int^1_{-1} \frac 1 {1+x^2} \, dx=\int^1_{-1} \frac 1 {x^2(1+\frac 1 {x^2})} \, dx = -\int^1_{-1} \frac 1 {1+(\frac 1 x)^2} \,d(1/x) = \left.-\arctan\left(\frac 1 x\right)\right|_{x=-1}^1 \\[10pt]
= {} & -\arctan(1) + \arctan(-1)=-\frac{\pi}{2}.
\end{align}
This differs from the expected outcome $$\int^1_{-1} \frac 1 {1+x^{2}}dx= \arctan x \Big|_{x=-1}^1=\arctan(1)-\arctan(-1)=\frac \pi 2.$$
Is there something wrong with the first calculation? I noticed that $\frac{1}{1+x^{2}}-1=\frac{1}{1+(\frac{1}{x})^{2}}$, but I don't yet see how this can be of any help.
|
Upon enforcing the substitution $x=1/t$, the domain transforms from $[-1,1]$ to $(-\infty,-1)\cup [1,\infty)$. Hence we have
$$\begin{align}
\int_{-1}^1\frac{1}{1+x^2}\,dx&= \int_{-\infty}^{-1}\frac{1}{1+t^2}\,dt+\int_1^\infty \frac{1}{1+t^2}\,dt\\\\
&=\left(\arctan(-1)+\frac\pi2\right)+\left(\frac\pi2 -\arctan(1)\right)\\\\
&=\frac{\pi}{2}
\end{align}$$
|
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|
Chinese Remainder theorem polynomials Is there a program, small script, or easy method I can use to find the polynomial P(x) such that:
$P(x) = x+1$ $\pmod {x^2+1}$
$P(x) = x+2$ $\pmod {x^2+2}$
$P(x) = x+3$ $\pmod {x^2+3}$
$P(x) = x+4$ $\pmod {x^2+4}$
.........
This is easy to do with integers, but I can't get across this with polynomials. Thanks for help!
|
Hint Since $k=-X^2 \pmod{X^2+k}$, you have
$$P(X) \equiv X-X^2 \pmod{X^2+1} \\
P(X) \equiv X-X^2 \pmod{X^2+2} \\
P(X) \equiv X-X^2 \pmod{X^2+3} \\
P(X) \equiv X-X^2 \pmod{X^2+4} \\$$
|
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|
show this inequality $f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$ let $$f(x)=\dfrac{1}{3^x}-\dfrac{1}{4^x},x\ge 1$$
For any $x_{1},x_{2}\ge 1$,show that
$$f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$$
or
$$\left(\dfrac{1}{3^{\sqrt{x_{1}x_{2}}}}-\dfrac{1}{4^{\sqrt{x_{1}x_{2}}}}\right)^2 \ge \left(\dfrac{1}{3^{x_{1}}}-\dfrac{1}{4^{x_{1}}}\right)\left(\dfrac{1}{3^{x_{2}}}-\dfrac{1}{4^{x_{2}}}\right)$$Any ideas?
Thanks.
|
$f$ is nonincreasing since
\begin{align}
f^\prime(x)&=-3^{-x}\log 3+4^{-x}\log 4<0,\\
f^{\prime\prime}(x)&=3^{-x}(\log 3)^2-4^{-x}(\log 4)^2.
\end{align}
Similarly, denoting with $g(x)=\log(f(x))$, you have
\begin{align}
g^\prime(x)&=f^\prime(x)/f(x)<0,\\
g^{\prime\prime}(x)&=\frac{f^{\prime\prime}(x)f(x)-(f^\prime(x))^2}{f^2(x)}=-\frac{12^x\log^2(4/3)}{(4^x-3^x)^2}<0,
\end{align}
hence $g$ is concave and nonincreasing.
It follows by AM-GM that
$$
\log f(\sqrt{xy})\ge \log f\left(\frac{x+y}{2}\right)\ge \frac{\log f(x)+\log f(y)}{2}.
$$
The claim follows taking the exponentiation to both sides.
|
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|
Solve this Differential Equation in y and x.
Solve the following DE:
$$y'=\frac{y+y^2}{x+x^2}$$
in particular
$$y(2)=1$$$$\frac{y'}{y+y^2}=\frac{1}{x+x^2}$$$$\int \frac{dy}{y+y^2}=\int \frac{dx}{x+x^2}$$$$y'dx=dy$$$$ln(y)-ln(y+1)=ln(x)-ln(x+1)+C$$
but where to from here?
|
Well, when we solve:
$$\text{y}\space'\left(x\right)=\frac{\text{y}\left(x\right)+\text{y}\left(x\right)^2}{x+x^2}\space\Longleftrightarrow\space\int\frac{\text{y}\space'\left(x\right)}{\text{y}\left(x\right)+\text{y}\left(x\right)^2}\space\text{d}x=\int\frac{1}{x+x^2}\space\text{d}x\tag1$$
So, for the intergals we get:
$$\ln\left|\frac{\text{y}\left(x\right)}{1+\text{y}\left(x\right)}\right|=\ln\left|1+\frac{1}{x}\right|+\text{C}\tag2$$
Using $\text{y}\left(2\right)=1$ we get:
$$\ln\left|\frac{1}{1+1}\right|=\ln\left|1+\frac{1}{2}\right|+\text{C}\space\Longleftrightarrow\space\text{C}=-\ln\left(3\right)\tag3$$
So, we get:
$$\ln\left|\frac{\text{y}\left(x\right)}{1+\text{y}\left(x\right)}\right|=\ln\left|1+\frac{1}{x}\right|-\ln\left(3\right)\tag4$$
Take the $\exp$ of both sides:
$$\left|\frac{1}{1+\frac{1}{\text{y}\left(x\right)}}\right|=\frac{1}{3}\cdot\left|1+\frac{1}{x}\right|\space\Longleftrightarrow\space\left|1+\frac{1}{\text{y}\left(x\right)}\right|=\frac{3}{\left|1+\frac{1}{x}\right|}\tag5$$
|
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|
How do I find the term of a recursive sequence? I have $\{a_n\}$ the following sequence:
$a_1 =-1$
$a_k a_{k+1} = - a_k - \frac{1}{4}$
How can I find $a_1 a_2 \cdots a_n$?
|
$$a_{k+1} = \frac{-4a_k - 1}{4a_k + 0}$$
Let $p_k / q_k = a_k$ :
$$\frac{p_{k+1}}{q_{k+1}} = \frac{-4 \frac{p_k}{q_k} - 1}{4\frac{p_k}{q_k} + 0} = \frac{-4 p_k - 1q_k}{4p_k + 0q_k}$$
$$\begin{align}
%
\begin{bmatrix} p_{k+1} \\ q_{k+1} \end{bmatrix}
%
& = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} p_{k} \\ q_{k} \end{bmatrix}
\\ %
& = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix}^k \begin{bmatrix} p_{1} \\ q_{1} \end{bmatrix}
\\ %
& \text{ Jordan Decomposition...}
\\ %
& = \left(
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\right)^k
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ %
& =
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}^k
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ %
& =
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} (-2)^k & k~(-2)^{k - 1} \\ 0 & (-2)^k \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ %
& =
(-2)^k
\begin{bmatrix} -k/2 - 1 \\ k + 1\end{bmatrix}
\\ %
\end{align}$$
So
$$a_{k+1} = \frac{p_{k+1}}{q_{k+1}} = \frac{-k/2 - 1}{k + 1}$$
$$a_k = \frac{-k - 1}{2k}$$
|
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|
What is the limit of the following function? $\lim_{n\rightarrow \infty }\left ( n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right )$
|
$$
\begin{aligned}
\lim_{n\rightarrow \infty }\left (n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right)
& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\frac{n}{2}\right)!}{n}}{\frac{2\left(\frac{n-1}{2}\right)!}{n-1}}\right)^2\right)
\\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\sqrt{2\pi \:\left(\frac{n}{2}\right)}\left(\frac{\left(\frac{n}{2}\right)}{e}\right)^{\frac{n}{2}}\right)}{n}}{\frac{2\left(\sqrt{2\pi \:\:\left(\frac{n-1}{2}\right)}\left(\frac{\left(\frac{n-1}{2}\right)}{e}\right)^{\frac{n-1}{2}}\right)}{n-1}}\right)^2\right)
\\& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-\frac{n^{n-1}\left(n-1\right)^{-n+2}}{e}\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{1}{t}-1-\frac{\left(\frac{1}{t}\right)^{\left(\frac{1}{t}\right)-1}\left(\frac{1}{t}-1\right)^{-\frac{1}{t}+2}}{e}\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(-t+1\right)^{\frac{2t-1}{t}}}{et}-1\right)
\\& = \lim _{t\rightarrow 0\:}\:\left(\frac{e-\left(e-\frac{3et}{2}+o\left(t\right)\right)}{et}-1\right)
\\& = \color{red}{\frac{1}{2}}
\end{aligned}
$$
Solved with Stirling's approximation and Taylor expansion
|
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|
Prove that there do not exist positive integers $a,b$ such that $p \mid a+b+1,a^2+b^2+1,a^3+b^3+1$ where $p > 3$ is a prime
Prove that if $p > 3$ is a prime then there do not exist positive integers $a,b$ such that \begin{align*}a+b+1 &\equiv 0 \pmod{p}\\a^2+b^2+1 &\equiv 0 \pmod{p}\\a^3+b^3+1 &\equiv 0 \pmod{p}.\end{align*}
We have $a+b \equiv a^2+b^2 \equiv a^3+b^3 \pmod{p}$. Thus $a(a-1)+b(b-1) \equiv 0 \pmod{p}$ and $a^2(a-1)+b^2(b-1) \equiv 0 \pmod{p}$.Then $$a(a-1)+b(b-1) \equiv a^2(a-1)+b^2(b-1) \pmod{p}.$$ How can we continue from here to show that the equation doesn't have solutions in positive integers?
|
Let $p\gt 3$ be a prime. In the field $\mathbb F_p$ we have $$\begin{align*}a+b=-1\\a^2+b^2=-1\\a^3+b^3=-1\end{align*}$$ It follows from the first equation $$a^2+b^2+2ab=1 \Rightarrow ab=1$$
$$a^3+b^3+3ab(a+b)=-1 \Rightarrow 3ab(a+b)=0 \Rightarrow ab=0$$ Thus $$ 0=-1\text{ contradiction}.$$
|
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|
how to enumerate and index partial permutations with repeats I know how to count permutations of, say, p red balls and q white balls and r blue balls. I also know how to count permutations of part of a set of distinct objects. But I don't know an efficient way to count partial permutations with repetition.
To illustrate what I mean: suppose I have a supply of eight As, seven Bs and six Cs, and want to know how many sequences of length 12 can be made from these. I can do it by adding up the permutations of 8 As and 4 Bs, 8 As and 3 Bs and one C, ... but that's tedious (this example has 45 cases). Is there a better way?
Finally, is there a procedure to generate the nth such sequence in lexicographic order, without cranking out all of them?
|
I assume your $45$ cases result from solving the equation
$$a + b + c = 12$$
in the nonnegative integers subject to the restrictions that $a \leq 8$, $b \leq 7$, and $c \leq 6$.
We can reduce the number of cases by considering the number of A's in the sequence.
*
*$8$ A's: There are $\binom{12}{8}$ ways to choose the positions of the A's. Since we have $7$ B's and $6$ C's, we are free to fill each of the remaining four positions in $2$ ways. Hence, there are
$$\binom{12}{8}2^4$$
such sequences.
*$7$ A's: By similar reasoning, there are
$$\binom{12}{7}2^5$$
such sequences.
*$6$ A's: By similar reasoning, there are
$$\binom{12}{6}2^6$$
such sequences.
*$5$ A's: There are $\binom{12}{5}$ ways to choose the positions of the A's. If we had at least $7$ B's and $7$ C's, we could fill the remaining seven positions in $2^7$ ways. However, we cannot have a sequence with $7$ C's. Hence, there are
$$\binom{12}{5}(2^7 - 1)$$
such sequences.
*$4$ A's: There are $\binom{12}{4}$ ways to choose the positions of the A's. We cannot have a sequence with eight B's, eight C's, or seven C's. Hence, there are
$$\binom{12}{4}\left[2^8 - 2\binom{8}{8} - \binom{8}{7}\right]$$
such sequences.
*$3$ A's: There are $\binom{12}{3}$ ways to choose the positions of the A's. We cannot have a sequence with nine B's, nine C's, eight B's, eight C's, or 7 C's. Hence, there are
$$\binom{12}{3}\left[2^9 - 2\binom{9}{9} - 2\binom{9}{8} - \binom{9}{7}\right]$$
such sequences.
*$2$ A's: There are $\binom{12}{2}$ ways to choose the positions of the A's. Since we cannot have more than seven B's or six C's, there are
$$\binom{12}{2}\left[2^{10} - 2\binom{10}{10} - 2\binom{10}{9} - 2\binom{10}{8} - \binom{10}{7}\right]$$
such sequences. Alternatively, we note that since we can fill at most six of the remaining ten positions with C's, we must use at least $10 - 6 = 4$ B's. Since we can use at most seven B's, there are
$$\binom{12}{2}\left[\binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \binom{10}{7}\right]$$
such sequences.
*$1$ A: There are $12$ ways to choose the position of the A. Since we can fill at most six of the remaining eleven positions with C's, we must use at least $11 - 6 = 5$ B's. Since we can use at most seven B's, there are
$$\binom{12}{1}\left[\binom{11}{5} + \binom{11}{6} + \binom{11}{7}\right]$$
such sequences.
*$0$ A's: Since we can fill at most six of the twelve positions with C's, we must use at least $12 - 6 = 6$ B's. Since we can use at most seven B's, there are
$$\binom{12}{6} + \binom{12}{7}$$
such sequences.
Since the above cases are disjoint, the total number of sequences is found by adding the above results.
Since the problem lacks symmetry, determining the position of a sequence in the lexicographic ordering is tedious. That said, it makes sense to pivot on the number of A's at the beginning of the sequence. Consider a sequence such as AABACBBAABCA. Any sequence with more than two A's at the beginning of the sequence precedes it. Add those cases. Since the next two letters are BA, it is not possible to introduce any more sequences before this sequence until we reach the fifth letter. At this point, we have to count how many sequences begin AABAA given that we have $8 - 4 = 4$ A's, $7 - 1 = 6$ B's, and $6$ C's left and how may sequences begin AABAB given that we have $8 - 3 = 5$ A's, $7 - 2 = 5$ B's, and $6$ C's left. Once, we have done that, we have to add the number of sequences that begin $AABACA$ given that we have $8 - 4 = 4$ A's, $7 - 1 = 6$ B's, and $6 - 1 = 5$ C's left. Continue from there.
|
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|
How do I solve $\int_0^\infty \frac{x^2}{x^4+1} \, dx$?
$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$
All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$
|
note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$
|
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|
Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$. Question:
Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.
We can write $ \left|x^2-4x+3\right|<3 $ as $ |x-3||x-1| < 3$. If I start with $ \left|x-1\right|<1$, how can I show that $ |x-3| < 3$ ?
|
This is true for
complex numbers as well as
real numbers.
The reason is that,
if $a$, $b$, and $c$
are positive reals,
then
$b \lt c$
implies
$ab \lt ac$.
(Do you see why this is true?)
Since
$|x^2-4x+3|
= |x-1||x-3|
$,
if
$|x-1| < 1$,
then
$|x^2-4x+3|
\lt |x-3|
$.
So,
we are done
if we can show that
$|x-3| < 3$.
But,
applying the
triangle inequality
($|u+v| \le |u|+|v|$),
$|x-3|
=|x-1-2|
\le |x-1|+|-2|
\lt 1+2
=3
$.
|
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|
If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!
Please suggest a simpler process.
|
$$3a^2+2a+1=0 \to 3a^2+3a+1=a\\3b^2+2b+1=0\to 3b^2+3b+1=b\\a-1=3a(a+1)\\b-1=3b(b+1)$$
so $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\\
\left(\dfrac{-3a(a+1)}{1+a}\right)^3+\left(\dfrac{-3b(b+1)}{1+b}\right)^3\\=-27(a^3+b^3)=-27(s^3-3ps)\\=-27\left(\left(\frac{-2}{3}\right)^3-3\left(\frac{1}{3}\times\left(\frac{-2}{3}\right)\right)\right)\\=+8-18\\=-10$$where $$s=a+b\\p=ab$$
|
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|
Flux through sphere I wish to find the flux of $\mathbf{F}=(x^2,y^2,z^2)$ through $S: (x-1)^2+(y-3)^2+(z+1)^2$
Here is what I tried:
I "moved" the sphere to $(0,0)$ by changing the variables to:
$u=x-1$ , $v=y-3$ , $w=z+1$
so now we have $F=((u+1)^2,(v+3)^2,(w-1)^2)$ and $S$ is the unit sphere.
So my calculation is (after switching to polar):
$$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}{\rm div}\,\mathbf{F}\,r\,{\rm d}z\,{\rm d}r\,{\rm d}\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)r\,{\rm d}z\,{\rm d}r\,{\rm d}\theta$$
but I got $0$ instead $8\pi$
What did I do wrong?
|
thanks to zack I found my miscalculation:
from where i stopped:
$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}divFrdzdrd\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)rdzdrd\theta=
\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}[r(2r\sin\theta-2r\cos\theta+(2z+6)\theta]_0^{2\pi}=2\pi\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}2zr+6r=2\pi\int_0^1[rz^2+6rz]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}=2\pi\int_0^112r\sqrt{1-r^2}=[-4(1-r^2)]_0^1=8\pi$
|
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|
Are there infinitely many positive integer solutions to the equation $\overline{x^2y^2} = z^2\,$?
Do there exist infinitely many solutions to the equation $$\overline{x^2y^2} = z^2$$ where $x,y,z$ are positive integers? Note: the notation $\overline{xy}$ means a concatenation of the numbers $x$ and $y$ to form one number.
One solution to this equation is $(x,y,z) = (4,9,41)$. Then we have $\overline{4^29^2} = 1681 = 41^2$. If $y = \overline{a_1 \ldots a_n}$, then $\overline{x^2y^2} = 10^{2n} x^2+y^2 = z^2$, but I didn't see how to find infinitely many integer solutions to this.
|
$$\overline{2^2(3\cdot10^k)^2}=(7\cdot10^k)^2$$
and in general, if $(a,b,c)$ is a solution, then so is $(a,b\cdot10^k,c\cdot10^k)$
|
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|
Issue with Baby Rudin Example 7.3 Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$. This confuses me. As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{n=0}^{\infty}x^n,
\end{align}
and such a series converges to $\frac{1}{1 - x}$ if $x \in [0,1)$. Now, it is certainly true that if we put $y = \frac{x^2}{1 + x^2}$, then $\frac{1}{1 - y} = 1 + x^2$. But in this case, shouldn't the series we're discussing be
\begin{align}
\sum_{n=0}^{\infty}\left[\frac{x^2}{1 + x^2}\right]^n?
\end{align}
What am I missing?
|
$$\begin{align}
\sum_{n=0}^\infty \frac{x^2}{(1+x^2)^n}&=x^2\sum_{n=0}^\infty \left((1+x^2)^{-1}\right)^n\\\\
&=\frac{x^2}{1-(1+x^2)^{-1}}\\\\
&=\frac{x^2(1+x^2)}{1+x^2-1}\\\\
&=1+x^2
\end{align}$$
for $x\ne0$.
|
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|
Maximum integer $n$ satisfying Taylor approximation
What is the maximum $n \in \mathbb N$ satisfying the formula
$$\sinh(x^4) = x^4 + \frac 1 6 x^{12} + \mathcal o(x^n) \quad \mathrm{for}\ x\to0 $$
The Taylor's theorem assures me that
$$\sinh x = x + \frac{x^3}{3!} + o(x^3)\ \mathrm{or}\ o(x^4) \quad \mathrm{for}\ x \to 0$$
which should mean
$$\sinh x^4 = x^4 + \frac{x^{12}}{3!} + o(x^{12})\ \mathrm{or}\ o(x^{16}) \quad \mathrm{for}\ x \to 0$$
However, the answer seems to be $n= 19$ rather than $16$. What am I missing? Is there a faster and more rigorous way to do these?
|
As you know the taylor series of the $\sinh(x)$ is:
$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!}+\cdots$$
Therfore,
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + \frac{x^{20}}{5!}+\cdots$$
Hence for integer powers of $x$:
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + O(x^{20}) = x^4 + \frac{x^{12}}{3!} + o(x^{19})$$
If you know the difference of big-$O$ and small-$o$ notation the above equality makes sense for you when $x\rightarrow 0$.
|
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|
How to apply Parseval's equation and checking to see if we have convergence in $L^2$
On a previous assignment, you found the sine series on the interval $[0,l]$:
$$x(l-x) = \frac{8l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n \pi x}{l}\right)$$
Note that the sum includes odd terms only.
a.) Show that this series converges in $L^2$. Then, apply Parseval's equation to find the value of $\sum_{n=0}^{\infty}\frac{1}{(2n +1)^6}$.
b.) Show that this series converges pointwise. Then, plug in $x = \frac{l}{2}$ to find the value of $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}$.
c.) Does the series converge uniformly? Why or why not?
Attempted solution a.) We have
$$\Bigg| \frac{1}{n^3}\sin\left(\frac{n \pi x}{l}\right)\Bigg| \leq \frac{1}{n^3} $$
Now take $M_n = \frac{1}{n^3}$ since $\sum_{n=1}^{\infty}\frac{1}{n^3}$ is convergent then by the Weirerstrass $M$-test the original series is uniformly convergent and thus convergent in $L^2$.
Attempted solution b.) Since we have shown that the series is uniformly convergent then we automatically have pointwise convergence since uniform convergence implies pointwise convergence. \
\noindent
Now, letting $x = \frac{l}{2}$, we have
\begin{align*}
&\frac{l}{2}\left(l - \frac{l}{2} \right) = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n\pi (l/2)}{l} \right)\\
&\Rightarrow \frac{l^2}{4} = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n\pi}{2}\right)
\end{align*}
Note that
$$\sin\left(\frac{n\pi}{2}\right) = \begin{cases}
0 \ \ &\text{if} \ \ n \ \ \text{even}, n = 2m, m = 0,1,2,\ldots\\
(-1)^n \ \ &\text{if} \ \ n \ \ \text{odd}, n = 2m + 1, m = 0,1,2,\ldots
\end{cases}$$
Therefore we have
\begin{align*}
&\frac{l^2}{4} = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}(-1)^n\\
&\Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3} = \frac{\pi^3}{32}
\end{align*}
Attempted solution c.) - Yes, the series converges uniformly by the Weirerstrass $M$-test, see the solution to part a.).
First off I want to know if these solutions are sufficient. Second, I want to know how to apply Pareseval's equation to find the value of $$\sum_{n=1}^{infty}\frac{1}{(2n+1)^6}$$
|
Parseval's equation,$$\sum_{n=1}^{\infty}|\hat{f}_n|^2 = \|f\|^2$$ We have
$$\sum_{n=1}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{n^3}\right)^2 = \sum_{k=0}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{(2k+1)^3}\right)^2 = \frac{2}{l}\int_{0}^{l}(x(l-x))^2dx = \frac{l^4}{15}$$
Therefore, $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^6} = \frac{\pi^6}{64}\frac{1}{15} = \frac{\pi^6}{960}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?
|
If
*
*$f(x) = x^3 + ax^2 + bx + c$
*$f(1) = 1$
*$f(2) = 4$
*$f(3) = 9$
then
\begin{align}
f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\
\hline
4-1 &= f(2)-f(1) \\
3 &= 7 + 3a + b \\
\hline
9-4 &= f(3) - f(2) \\
5 &= 19 + 5a + b \\
\hline
3a + b &= -4 \\
5a + b &= -14 \\
\hline
a &= -5 \\
b &= 11 \\
c &= -6
\end{align}
S0 $f(x) = x^3 - 5x^2 + 11x - 6$.
You could make a difference table
\begin{array}{c}
1 && 4 && 9 \\
& 3 && 5 \\
&& 2
\end{array}
Using $f(x) = x^3 + ax^2 + bx + c$, this corresponds to
\begin{array}{c}
1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\
& 7+3a+b && 19 + 5a + b \\
&& 12+2a
\end{array}
Then, comparing entries...
\begin{align}
12+2a=2 &\implies a = -5 \\
7+3a+b = 3 &\implies b=11 \\
1+a+b+c = 1 &\implies c = -6
\end{align}
|
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|
Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:
$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$
The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:
$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$
Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:
$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$
However, using an online calculator, the answer turned out to be:
$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$
It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that
$$\sqrt{\frac{x^2}{9} - 9}$$
becomes $$\sqrt{u^2 - 1}$$
And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?
|
Your mistake:
$$\sec\theta=\frac x3$$ yields
$$\cos\theta=\frac 3x,\\\sin\theta=\sqrt{1-\frac9{x^2}},\\\tan\theta=\sqrt{\frac{x^2}9-1}.$$
Not $\sqrt{x^2-9}$.
This said, you can rescale the variable with $x=3t$ to get
$$\int\frac{dt}{\sqrt{t^2-1}}.$$
You may recognize another familiar integral,
$$\int\frac{dt}{\sqrt{1-t^2}}=-\arccos t+C.$$
What you have here is just the hyperbolic equivalent,
$$\int\frac{dt}{\sqrt{t^2-1}}=\text{arcosh }t+C,$$
which can also be written
$$\log\left(t+\sqrt{t^2-1}\right)+C.$$
|
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|
Heaviside Step Function In studying properties of 1/x, its derivatives and its integral, I came across the following "apparent" identity. Plot it, and found that it appears to be the same as the Heaviside function. Any post with reference to the RHS, or verifying/disproving would be appreciated.
$$\frac{\arctan(1/x)+\arctan(x+1/2)+\arctan(2(x^2+1)+x)}{\pi} = H(x)$$
Thanks in advance.
|
$$\frac{d}{dx}\arctan(X)=\frac{1}{X^2+1}\quad\to\quad \begin{cases}
\frac{d}{dx}\arctan(\frac{1}{x} )=\frac{1}{\frac{1}{x^2}+1}\left(-\frac{1}{x^2} \right) \\
\frac{d}{dx}\arctan(x+\frac{1}{2} )=\frac{1}{\left(x+\frac{1}{2} \right)^2+1} \\
\frac{d}{dx}\arctan(2x^2+x+2)=\frac{1}{(2x^2+x+2)^2+1}(4x+1)
\end{cases}$$
$$f(x)=\arctan(\frac{1}{x} )+\arctan(x+\frac{1}{2} )+\arctan(2x^2+x+2)$$
$f(x)$ is continuous except at $x=0$
$$\begin{cases}
x\to 0^+ \quad f(x)\to \frac{\pi}{2}+\arctan(\frac{1}{2} )+\arctan(2)=\pi \\
x\to 0^- \quad f(x)\to -\frac{\pi}{2}+\arctan(\frac{1}{2} )+\arctan(2)=0
\end{cases} \tag 1$$
$$\frac{d}{dx}f(x)=\frac{1}{\frac{1}{x^2}+1}\left(-\frac{1}{x^2} \right)+\frac{1}{\left(x+\frac{1}{2} \right)^2+1}+\frac{1}{(2x^2+x+2)^2+1}(4x+1)\qquad x\neq 0$$
Expanding and simplifying leads to
$$\frac{d}{dx}f(x)=0\qquad x\neq 0$$
Hence
$$f(x)=\text{constant}\qquad x\neq 0$$
Determination of the constant in each continuous range:
$$x>0\qquad x\to+\infty \qquad \begin{cases}
\arctan(\frac{1}{x} )\to 0 \\
\arctan(x+\frac{1}{2} )\to \frac{\pi}{2}\\
\arctan(2x^2+x+2)\to \frac{\pi}{2}
\end{cases}
\qquad f(x)\to \frac{\pi}{2}+\frac{\pi}{2}=\pi$$
$$x<0\qquad x\to-\infty \qquad \begin{cases}
\arctan(\frac{1}{x} )\to 0 \\
\arctan(x+\frac{1}{2} )\to -\frac{\pi}{2}\\
\arctan(2x^2+x+2)\to \frac{\pi}{2}
\end{cases}
\qquad f(x)\to -\frac{\pi}{2}+\frac{\pi}{2}=0$$
$$f(x)=\begin{cases}0\qquad x<0 \\ 1\qquad x>0 \end{cases} \quad\text{and with }(1)\quad\implies\quad f(x)=H(x)$$
$$\frac{f(x)}{\pi}=\frac{\arctan(1/x)+\arctan(x+1/2)+\arctan(2(x^2+1)+x)}{\pi} = H(x)$$
|
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|
How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me in finding the value.
Any good hint is welcomed.
|
Some assumptions:
$c, \alpha, \beta \neq 0$ (the answer will be a bit modified without these assumptions)
Since $\alpha$ is a root therefore
\begin{align*}
a\alpha^2+b\alpha+c &=0\\
a\alpha+b & = \frac{-c}{\alpha}\\
\frac{1}{(a\alpha+b)^2} & = \frac{\alpha^2}{c^2}
\end{align*}
Thus
\begin{align*}
\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}& = \frac{\alpha^2+\beta^2}{c^2}\\
&= \frac{(\alpha+\beta)^2-2\alpha\beta}{c^2}
\end{align*}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but clearly this will not work because for $n=3$ we have $3\cdot2^3+1=3\cdot8+1=24+1=25$ which is clearly not divisible by $3$.
Any ideas how to evaluate a formula that generates all the $n$'s for which $n2^n+1$ is divisible by $3$?
|
This is an approach which does not use (directly) modular arithmetic.
Set $a_n = n 2^n + 1$.
Let's define $S:=\{ n\in \mathbb{Z} \mid n \geq 0 \text{ and $3$ divides $a_n$} \}$. Observe that for integers $n,k \geq 0$:
$$ a_{n+k} - a_n = (n+k) 2^{n+k} - n 2^n = 2^n((n+k)2^k - n) = 2^n((2^k-1)n +k2^k) $$
For $k=6$ we get:
$$ a_{n+6} - a_n =2^n( 63n +6\cdot 64) =3 \cdot 2^n( 21n +128) \quad \Rightarrow \quad 3 \text{ divides } (a_{n+6} - a_n)$$
If two integers $a,b$ are multiple of $3$ then also $a+b$ and $a-b$ are multiple of $3$. Since clearly:
$$a_{n+6}=(a_{n+6} - a_n)+a_n \quad\text{ and }\quad a_n=a_{n+6} -(a_{n+6} - a_n)$$
then $3$ divides $a_n$ if and only if $3$ divides $a_{n+6}$. In other words, $n\in S$ if and only if $n+6 \in S$.
This can be used to prove by induction that the general element of $S$ can be written as $n=x+6k$, where $k$ is any integer satisfying $k\geq 0$ and $x$ is an element of $S$ satisfying $0\leq x <6$.
Now let's see which $x$ satisfying $0\leq x <6$ are in $ S$ :
$$
\begin{align}
a_{0}&= 1 & &\Rightarrow 0 \notin S\\
a_{1}&= 2 + 1 = 3 & &\Rightarrow 1 \in S\\
a_{2}&= 2\cdot 2^2 + 1= 9 & &\Rightarrow 2\in S\\
a_{3}&= 3\cdot 2^3 + 1= 25 & &\Rightarrow 3\notin S\\
a_{4}&= 4\cdot 2^4 + 1= 65 & &\Rightarrow 4\notin S\\
a_{5}&= 5\cdot 2^5 + 1= 161 & &\Rightarrow 5\notin S
\end{align}$$
Therefore:
$$ S=\{1+6k,2+6k\; \text{ s.t. }\; k \geq 0\} $$
|
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|
Canonical form of a quadratic form
Find the canonical form of $2x^2+y^2+3y=0$
What are the way to approach this?
I know that the eigenvalues are positive and that
$$2x^2+y^2+3y=0\iff2x^2+(y+\frac{3}{2})^2-\frac{9}{4}=0$$
For: $x'=2x$, $y'=y+\frac{3}{2}$
We get: $$x'^2+y'^2-\frac{9}{4}=0$$
How do I categorize this quadratic form?
|
As you noticed the equation
$$2x^2+y^2+3y=0\tag{1}$$
is equivalent to
\begin{equation*}
2x^{2}+\left[ y-\left( -\frac{3}{2}\right) \right] ^{2}-\left( \frac{3}{2}
\right) ^{2}=0.\tag{2}
\end{equation*}
Dividing by $(3/2)^2 $ and writting $2$ as $1/(\sqrt{2}/2)^2$, we obtain
\begin{equation*}
\frac{x^{2}}{\left( \frac{3}{4}\sqrt{2}\right) ^{2}}+\frac{\left[ y-\left( -\frac{3}{2}\right) \right] ^{2}}{\left( \frac{3}{2}\right) ^{2}}=1,\tag{3}
\end{equation*}
which is the equation of the shifted ellipse centered at $\left( h,k\right)
=\left( 0,-\frac{3}{2}\right) $ and semiaxes $a=\frac{3}{4}\sqrt{2}$ and $ b=\frac{3}{2}>a$:
\begin{equation*}
\frac{\left( x-h\right) ^{2}}{a^{2}}+\frac{\left( y-k\right) ^{2}}{b^{2}}=1.\tag{4}
\end{equation*}
$\qquad\quad\qquad\quad\qquad\quad$
Comments:
*
*The general equation of a conic is
\begin{equation*}
Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0
\end{equation*}
If $B^{2}-4AC<0$, the conic is an ellipse. Equation $(1)$ is the particular case
\begin{equation*}
A=2,\quad C=1,\quad E=3,\quad B=D=F=0;
\end{equation*}
with $B^{2}-4AC=-4\times 2\times 1=-8<0$.
*You only get
$$x'^2+y'^2-\frac{9}{4}=0$$
for $x'=\sqrt{2}x$, $y'=y+\frac{3}{2}$.
|
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|
Express the differential equation as power series
My attempt: $y=\sum^{\infty}_{n=0}a_n x^n \rightarrow y'=\sum^{\infty}_{n=0}na_n x^{n-1}\rightarrow y''=\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}\\
(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n\\
2\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+\sum^{\infty}_{n=0}n(n-1)a_n x^{n}+\sum^{\infty}_{n=0}na_n x^{n+1}+3\sum^{\infty}_{n=0}a_n x^{n}$
from here can any help me
|
You are asked to put the series on the form :
$$(2+x^2)y"+x^2y'+3y=\sum^{\infty}_{n=0}c_n x^n$$
ant to find the coefficients $c_n$.
$$(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n$$
So, you have to gather the terms of common power $n$.
$$---------------------$$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 2\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+\sum^{\infty}_{n=0}n(n-1)a_{n} x^{n} $$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 2\sum^{\infty}_{n=0}(n+2)(n+1)a_{n+2} x^{n}+\sum^{\infty}_{n=0}n(n-1)a_n x^{n} $$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 4a_{2}+ 12a_{3}x+\sum^{\infty}_{n=2}\left( 2(n+2)(n+1)a_{n+2}+n(n-1)a_n \right)x^n $$
$$---------------------$$
$$x^2\sum^{\infty}_{n=0}na_n x^{n-1}=\sum^{\infty}_{n=0}na_n x^{n+1}=\sum^{\infty}_{n=2}(n-1)a_{n-1} x^{n} $$
$$---------------------$$
$$(2+x^2)y"+x^2y'+3y=4a_{2}+ 12a_{3}x+\sum^{\infty}_{n=2}\left( 2(n+2)(n+1)a_{n+2}+n(n-1)a_n \right)x^n +\sum^{\infty}_{n=2}(n-1)a_{n-1} x^{n} +3a_0+3a_1x+\sum^{\infty}_{n=2}3a_n x^n$$
$$---------------------$$
$$(2+x^2)y"+x^2y'+3y=\sum^{\infty}_{n=0}c_n x^n$$
$$c_0=4a_{2}+3a_0$$
$$c_1=12a_{3}+3a_1$$
$$c_n=2(n+2)(n+1)a_{n+2}+n(n-1)a_n +(n-1)a_{n-1}+3a_n\qquad n\geq 2$$
|
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|
$\int (x^2+1)/(x^4+1)\ dx$ I have Divided the numerator and Denominator by $x^2$
to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:
$$
\int\frac{x^2+1}{x^4+1} \, dx = \int (u^2+2)^{-1} \, du
$$
After this I need help!
|
Express MichaelRozenberg's denominator, $x^2 + \sqrt{2} x + 1$ in the form $u^2 + 1$ so you can use the $\tan^{-1}$ form.
Specifically:
$x^2 + \sqrt{2} x + 1$
Let $\sqrt{2}x + 1 = u$,
so
$u^2 = 2 x^2 + 2 \sqrt{2} x + 1$
so
$u^2 + 1 = 2 x^2 + 2 \sqrt{2} x + 2$
or
${1 \over u^2 + 1} = {1 \over x^2 + \sqrt{2} x + 1}$.
And likewise for the other term with the minus sign.
Now you have the arc tan integral formula, giving:
$$\frac{\tan ^{-1}\left(\sqrt{2} x+1\right)-\tan ^{-1}\left(1-\sqrt{2} x\right)}{\sqrt{2}}$$
|
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|
If $A∆B=A∆C$, then $B=C$ If $A\Delta B=A\Delta C$, then $B=C$.
I use the formula $X∆Y=(X\setminus Y)\cup(Y\setminus X)$ in $A∆B=A∆C$. And further use the formula $X\setminus Y=X\cap Y^c$. I am not able to reach the conclusion. Please help me.
|
You can do element chasing.
Or you can manipulate symbols and properties and identities.
But you can also divide and conquer.
Divide the space into $8$ disjoint sets determined by the elements that are are not in $A,B,C$.
They are:
1) $A^c \cap B^c \cap C^c$ (the elements that are in none)
2) $A^c \cap B^c \cap C$
3) $A^c \cap B \cap C^c$
4) $A^c \cap B \cap C$
5) $A \cap B^c \cap C^c$
6) $A \cap B^c \cap C$
7) $A \cap B \cap C^c$
8) $A \cap B \cap C$
$A\Delta B$ consists of $3$ and $4$ and $5$ and $6$ but none of the others $1,2,7,8$.
$A \Delta C$ consists of $2$ and $4$ and $5$ and $7$ but none of the others $1,3,6,8$.
So we get the seemingly paradoxical result that sets $2,3,6,7$ are all both subsets of $A \Delta B = A \Delta C$, and completely disjoint from $A \Delta B = A \Delta C$. The only way a set can be both a subset and disjoint from another set, is if the set is empty. So $2,3,6,7$ are all empty.
$B = 3 \cup 4\cup 7 \cup 8 = 4 \cup 8$.
$C = 2 \cup 4 \cup 6 \cup 8 = 4 \cup 8 = B$.
|
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|
$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$
I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^{\circ} \cos 60^{\circ}-\cos 76^{\circ} \, \cos16^{\circ}.$$
I couldn't simplify after that.
|
$$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}=$$
$$=\frac{1}{2}\left(1+\cos152^{\circ}+1+\cos32^{\circ}-\cos60^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(\cos152^{\circ}+\cos32^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(2\cos92^{\circ}\cos60^{\circ}-\cos92^{\circ}\right)=\frac{3}{4}$$
|
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|
Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$
|
$$2\sum \sin \frac{A}{2}\sin \frac{B}{2}\leq \frac{2}{3}\left(\sum \sin \frac{A}{2}\right)^2\leq\sum \sin \frac{A}{2}$$
$$=\sum \cos \frac{A}{2}\cos \frac{B}{2}-\sum \sin \frac{A}{2}\sin \frac{B}{2},$$
$$ \frac{\sqrt{3}}{2}\sum \cos \frac{A-B}{2}=\frac{\sqrt{3}}{2} \left(\sum \cos \frac{A}{2}\cos \frac{B}{2}+\sum \sin \frac{A}{2}\sin \frac{B}{2}\right)$$
$$\leq \frac{2\sqrt{3}}{3} \sum \cos \frac{A}{2}\cos \frac{B}{2}\leq \frac{2\sqrt{3}}{9}\left (\sum \cos \frac{A}{2}\right)^2\leq \sum \cos \frac{A}{2}.$$
|
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|
Cross ratio of 4 points on circle I'm in a bit of trouble I want to calculate the cross ratio of 4 points $ A B C D $ that are on a circle
Sadly "officially" it has to be calculated with A B C D as complex numbers and geometers sketchpad ( the geomerty program I am used to) don't know about complex numbers
Now I am wondering
The cross ratio of 4 points on a circle is a real number
And is there so much difference between
$$ \frac { (A-C)(B-D) }{ (A-D)(B-C) } \text{ (complex method) }$$
And
$$ \frac { |AC||BD| }{ |AD||BC| } \text{ (distance method). }$$
Where X-Y is the complex difference between X and Y
and |XY| is the distance between X andY
If allready is given that the four points are on a circle?
I think that the absolute values of the calculations are the same but am I right (and can we prove it)
ADDED LATER :
on Cave 's suggestion I used a circle inversion to move the four points on the circle to four points on a line and then the formulas give the same value ( if I did it correctly)
What I did:
I took a diameter of the circle
This diameter intersects the circle in O and U ,
u is the line through U perpendicular to OU
Project the 4 points onto line u with centre O (draw a ray through O and the point, the new point is where this ray intersects line u)
And calculate the cross ratio of the four new points.
This "projection" method and the earlier "distance" method give the same value but does this prove anything?
|
For $A,B,C,D$ on the unit circle,
$$
\begin{align}
\frac{(A-C)(B-D)}{(A-D)(B-C)}
&=\frac{\left(e^{ia}-e^{ic}\right)\left(e^{ib}-e^{id}\right)}{\left(e^{ia}-e^{id}\right)\left(e^{ib}-e^{ic}\right)}\\
&=\frac{\left(e^{i(a-c)}-1\right)\left(e^{i(b-d)}-1\right)}{\left(e^{i(a-d)}-1\right)\left(e^{i(b-c)}-1\right)}\\
&=\frac{e^{i\frac{a-c}2}2i\sin\left(\frac{a-c}2\right)e^{i\frac{b-d}2}2i\sin\left(\frac{b-d}2\right)}{e^{i\frac{a-d}2}2i\sin\left(\frac{a-d}2\right)e^{i\frac{b-c}2}2i\sin\left(\frac{b-c}2\right)}\\
&=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{1}
\end{align}
$$
which is indeed real.
For $A,C$ on the unit circle,
$$
\begin{align}
|A-C|
&=\sqrt{\left(e^{ia}-e^{ic}\right)\left(e^{-ia}-e^{-ic}\right)}\\[3pt]
&=\sqrt{2-2\cos(a-c)}\\
&=2\sin\left(\frac{a-c}2\right)\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{|A-C||B-D|}{|A-D||B-C|}
&=\frac{2\sin\left(\frac{a-c}2\right)2\sin\left(\frac{b-d}2\right)}{2\sin\left(\frac{a-d}2\right)2\sin\left(\frac{b-c}2\right)}\\
&=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{3}
\end{align}
$$
which is obviously real, but not so obviously, the same as $(1)$.
When is the Complex Cross-Ratio Real?
$$
\begin{align}
\frac{A-C}{A-D}
&=\frac{C-A}{D-A}\\
&=\frac{|C-A|}{|D-A|}\,e^{-i\alpha}
\end{align}
$$
$$
\begin{align}
\frac{B-D}{B-C}
&=\frac{D-B}{C-B}\\
&=\frac{|D-B|}{|C-B|}\,e^{-i\beta}
\end{align}
$$
Therefore,
$$
\frac{(A-C)(B-D)}{(A-D)(B-C)}
=\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha+\beta)}
$$
$\alpha+\beta$ is an integer multiple of $\pi$ ($\pi$) if and only if $A$ and $B$ are on opposing arcs of a circle with $C$ and $D$.
$$
\begin{align}
\frac{A-C}{A-D}
&=\frac{C-A}{D-A}\\
&=\frac{|C-A|}{|D-A|}\,e^{-i\alpha}
\end{align}
$$
$$
\begin{align}
\frac{B-D}{B-C}
&=\frac{D-B}{C-B}\\
&=\frac{|D-B|}{|C-B|}\,e^{i\beta}
\end{align}
$$
Therefore,
$$
\frac{(A-C)(B-D)}{(A-D)(B-C)}
=\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha-\beta)}
$$
$\alpha-\beta$ is an integer multiple of $\pi$ ($0$) if and only if $A$ and $B$ are on the same arc of a circle with $C$ and $D$.
|
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|
What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that Cathy
wins?
This was I question I got when competing at a Mathematics competition earlier this week. I thought the answer was $\frac{1}{18}$ but it turns out the actual answer is $\frac{1}{13}$...
This is my working out of why I thought it was $\frac{1}{18}$:
For Cathy to win:
*
*Alice needs to get a 4, 5 or 6 and there is a $\frac{3}{6}$ chance of that happening when she rolls the die
*Bob needs to get a 1, 2, 3 or 6 and there is a $\frac{4}{6}$ probability of that happening when he rolls the die
*Cathy needs to get 6 and there is a $\frac{1}{6}$ probability of that happening when she rolls the die
Those three need to happen for Cathy to win so:
$$\frac{3}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{1}{18}$$
However, the answer was $\frac{1}{13}$ so something must be wrong! Could someone please demonstrate why it was $\frac{1}{13}$?
|
You are wrong because that is not the only way in which Cathy can win! Note that if Cathy rolls anything other than a 6, the game repeats itself until someone wins. Using very similar reasoning to yours, show that the probability of her winning is the infinite sum
$$
\sum_{n=0}^\infty\Big(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}\Big)^n\Big(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{1}{6}\Big)
$$
Can you compute this sum now?
Edit: Here is a slicker way to compute the probability without an infinite sum. Either Cathy wins on the first round, which as you correctly computed, has probability $\frac{1}{18}$. Otherwise, with probability $\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}$, nobody wins. In this case, the game continues in exactly the same way as if the first round had not happened.
Thus
$$
P=\frac{1}{18}+\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}P,
$$
i.e. $P=\frac{1}{13}$.
|
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|
Difficult Homogeneous Differential Equation Solve the differential equation:
$$\frac{dy}{dx}=\frac{\sqrt{x^2+3xy+4y^2}}{x+2y}$$
I tried to solve it by putting $t=x+2y$ but that lead to a very complicated integral. The hint given is that equation is reducible to homogeneous form.
|
Partial answer :
Reformulate it as $$\frac{dy}{dx} = \frac{\sqrt{1+3\frac{y}{x}+4\left(\frac{y}{x}\right)^2}}{1+2\frac{y}{x}}$$
The you have $dy/dx = f(y/x)$.
Let $u(x) = \frac{y(x)}{x}$. Then $xu = y$. Then $dy = xdu + udx$. Then $dy/dx = xdu/dx + u$. Then you have
$$x\frac{du}{dx}+u = f(u)\quad \text{where} \quad f(u) = \frac{\sqrt{1+3u+4u^2}}{1+2u}$$
Then $x\frac{du}{dx} = f(u)-u$. Then
$$\int \frac{du}{f(u)-u} = \int\frac{dx}{x}$$
i.e.
$$\int \frac{1+2u}{\sqrt{4u^2+3u+1}-u-2u^2} = \int\frac{dx}{x}$$
So this must be solved (to get further from here). It'd give you $u(x)$ as a function of $x$ and from there you get $y(x)$ using $y(x)=x u(x)$. Here's a beginning :
$$\frac{1+2u}{\sqrt{4u^2+3u+1}-(u+2u^2)} = \frac{(1+2u)(\sqrt{4u^2+3u+1}+u+2u^2)}{4u^2+3u+1-(u+2u^2)^2}
\\ = -\frac{(1+2u)(\sqrt{4u^2+3u+1}+u+2u^2)}{4u^4 + 4u^3 - 3u^2 - 3u -1}
\\ = -\frac{(\sqrt{4u^2+3u+1}+u+2u^2)}{2u^3 + 2u^2 - 2u - \frac{u+1}{2u+1}}
\\ = \text{ugly...}$$
Then : either goto Claude Leibovici's highly plausible answer or please post yours if you know how to solve this.
|
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|
Find A Polynomial Solution For The Legendre equation.
The Legendre equation
$$(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0$$
has a polynomial solution $P_n$ when $\alpha$ is a non-negative number $n$. Find $P_1$ satisfying $P_1(1)=1$ and then find the general solution of the D.E. with $\alpha=1$
$$y=Ax+B$$$$y'=A,y''=0$$$$B=0, y=Ax$$$$y=x$$$$W=e^{\int \frac{-2x}{1-x^2}}=x^2-1$$$$y_2=x\int \frac{x^2-1}{x^2}dx=x^2+1$$$$y=A_1x+A_2x(x+\frac{1}{x})$$
But where to from here?
|
$$\alpha=1\qquad\to\qquad (1-x^2)y''-2xy'+2y=0 \tag 1$$
You found a particular solution $y=x$ which is correct.
Or, more general, a family of solutions : $\quad y=C\:x\quad$ where $C$ is a constant.
In order to find the general solution of the ODE, one can use the method of variation of parameter.
In the present case, remplace the parameter $C$ by an unknown function $u(x)$:
$y=u(x)\:x \quad\to\quad y'=xu'+u \quad\to\quad y''=xu''+2u'$
Putting them into $(1)$ leads to :
$$x(1-x^2)u''+2(1-2x^2)u'=0$$
$$\frac{u''}{u'}=2\frac{2x^2-1}{x(1-x^2)}$$
$$\ln|u'|=2\int \frac{2x^2-1}{x(1-x^2)}dx =-\ln|1-x^2|-2\ln|x|+\text{constant}$$
$$u'=\frac{c_1}{x^2(1-x^2)}$$
$$u=c_1\int \frac{dx}{x^2(1-x^2)} = c_1\left(-\frac{1}{x}+\frac{1}{2}\ln|1+x|-\frac{1}{2}\ln|1-x| \right)+c_2$$
The general solution of $(1)$ is :
$$y(x)=c_1\int \frac{dx}{x^2(1-x^2)} = c_1\left(-1+\frac{1}{2}x\ln|1+x|-\frac{1}{2}x\ln|1-x| \right)+c_2x$$
|
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|
In showing integer sum $(1+2+3+...+n)$ by l'Hopital rule why they take lim as $r$ approaches to $1$ in one of steps? Why $1$? So why lim as $r \to 1$ (why $1$?) Here's the method:
|
(Arithmetic - geometric progression)
If we differentiate both sides with respect to $r$, we have
\begin{equation}
1+2r+3r^2+\dots+nr^{n-1}=\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}
\end{equation}
Letting $r \to 1$ and applying L'Hospital's rule on the fraction, we end up with
\begin{equation}
\frac{n(n+1)}2=1+2+3+\dots+n
\end{equation}
$$\lim_{r\to 1}\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=\lim_{r\to 1}1+2r+3r^2+\dots+nr^{n-1}\\
\lim_{r\to 1}\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=\frac{0}{0} =\\
\lim_{r\to 1}\frac{n(n+2)r^{n+1}-(n+1)^2r^{n}+1}{(2(1-r)(-1))}=\frac{0}{0}\\\
\lim_{r\to 1}\frac{n(n+2)(n+1)r^{n}-(n+1)^2nr^{n-1}+0}{(2(-1)(-1))}=\\
\frac{n(n+2)(n+1)-n(n+1)^2}{+2}=\\
\frac{n((n^2+3n+2-(n+1)^2)}{+2}=\\
\frac{n(n+1)}{2}$$R..H.S. is $$\lim_{r\to 1}1+2r+3r^2+\dots+nr^{n-1}=1+2+3+...+n$$
|
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|
Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$ Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$
I am not allowed to use induction. I do not know how to approach this. Any help would be appreciated.
|
Approach 1: When we calculate the first few derivatives, we have:
$$y'=-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}=-\frac{1}{x^2}(\ln{x}-1).$$
$$y''=\frac{2}{x^3}\left(\ln{x}-1-\frac12 \right).$$
$$y^{(3)} = -\frac{3!}{x^4}\left(\ln{x}-1-\frac12 -\frac13\right).$$
so we can recognize the desired pattern.
Approach 2: We use the generalised Leibniz rule:
$$(fg)^{n} = \sum_{k= 0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x)$$
with $f(x) = 1/x, g(x) = \log x$
to obtain:
$$\left(\frac{\log x}{x}\right)^{(n)}= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \left(\frac{1}{x}\right)^{(n-k)}(x)$$
$$= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \binom{n}{k} \left(\frac{1}{x}\right)^{(k-1)} \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \binom{n}{k} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k(k-1)!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{k-1}}{x^{k}}\frac{(-1)^{n-k}}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{n-1}}{x^{n+1}}$$
$$ = \log(x) \frac{(-1)^{n}n!}{x^{n+1}} - \frac{n!(-1)^n}{x^{n+1}} \sum_{k= 1}^n \frac{1}{k}$$
$$= \frac{n!(-1)^n}{x^{n+1}} \left(\log(x) - 1 - \frac{1}{2} - \dots - \frac{1}{n}\right)$$
Approach 3
Induction: For $n=0$, the formula holds.
Assume it holds for $n-1$, and then show it holds for $n$, by taking the derivative and rearranging terms.
Other/similar approaches: n-th derivative of $\frac{\ln x}{x}$.
|
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|
Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM and CS inequality but failed to get a result. Please give me some hint in how to solve this question.
|
Write $c=-a-b$.
Then
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{2a^2-ab-b^2}=\frac{a^2}{(2a+b)(a-b)}$$
and
$$\frac{c^2}{2c^2+ab}=\frac{(a+b)^2}{2a^2+5ab+2b^2}=\frac{(a+b)^2}{(2a+b)(a+2b)}.$$
The sum therefore equals
$$\frac{a^2(a+2b)-b^2(2a+b)+(a+b)^2(a-b)}{(2a+b)(a+2b)(a-b)}=\cdots$$
etc. (as long as the denominator is nonzero).
|
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|
Floor functions of powers increase What is the smallest real number $x$ such that $\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor$ for all positive integers $n$?
In particular, we have $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<\dots$. For the first inequality to hold it must be that $x\geq\sqrt{2}$. But if $x=\sqrt{2}$ then $\lfloor x^2\rfloor=\lfloor x^3\rfloor=2$, so we need $x\geq \sqrt[3]{3}\approx 1.44$. From the first few powers it looks like this is the answer, but how can we prove it?
|
You have correctly shown that $x \ge \sqrt[3]3.\ $ Now if $x^n \ge 3$ and $x \gt \sqrt [3]3, \lfloor x^{n+1}\rfloor= \lfloor x^n\cdot x\rfloor\ge\lfloor x^n\rfloor+\lfloor x^n(x-1)\rfloor \ge \lfloor x^n \rfloor +1$ so any higher $x$ will work as well, so $\sqrt[3]3$ is the smallest $x$
|
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show that $7\mid p^3-p$ if $p$ is a prime divisor of $n^3+n^2-2n-1$
Let $p$ be a prime number, and $n$ a positive integer such
$$p\mid n^3+n^2-2n-1, \quad n\ge 2.$$
Show that $$7\mid p^3-p.$$
It maybe can use Fermat's little theorem?
|
A blind approach, although it will give the same solution.
The polynomial $f(x)=x^3+ x^2 - 2 x -1$ has discriminant $\Delta=49$, a square. Therefore, given a root $x_1$ of $f(x)$, the other roots can be expressed in terms of $x_1$ from the equality
$$(x_2-x_3) = \frac{(x_1-x_2)(x_1-x_3)(x_2-x_3)}{(x_1-x_3)(x_2-x_3)}= \frac{\pm\sqrt{\Delta}}{f'(x_1)}$$
and combined with the equality $x_2+x_3= s-x_1= -1-x_1$ give the expression for the other roots.
Note that we have a GCD formula for $f$ and $f'$ involving the discriminant. In our case we have
$$(-6 x -1)(x^3+x^2- 2 x -1)+ (2x^2 + x- 3)(3 x^2 + 2 x - 2)=7$$
and so if $\alpha $ is a root of $f$ then
$$\frac{\sqrt{\Delta}}{f'(\alpha)}= \frac{7}{f'(\alpha)} = 2\alpha^2 + \alpha- 3$$
Therefore, if $\alpha$ is a root of $f$ in a field $F$ then
$$\frac{1}{2}(-1-\alpha \pm (2\alpha^2 + \alpha- 3))= \alpha^2 - 2, - \alpha^2 - \alpha + 1$$
are also roots of $f$. Also,the product of their differences is $7$, so they are distinct if $7 \ne \operatorname{char} (F)$. In that case, starting from $\alpha$ a root we get all the roots in this way.
This works for any cubic with discriminant a square. This one is special, because we notice:
$$\alpha \text{ root of }\ f \implies \alpha^2-2 \text{ root of }\ f$$
and we might recall the polynomial $x^2-2$ from $t^2 +\frac{1}{t^2}=(t+\frac{1}{t})^2-2$. Now we check that
$$- \alpha^2 - \alpha -1 = (\alpha^2-2)^2 -2$$ if $\alpha$ is a root.
So we may rephrase the above statement as (assume $7 \ne \operatorname{char}(F)$
$$t+\frac{1}{t} \text{root of }\ f \implies t^2 +\frac{1}{t^2} , t^4 +\frac{1}{t^4}\text{the other roots of }\ f$$
Now consider $p\ne 7$ such that $f$ has a root $\alpha$ in $F_{p}$. Consider then $t$ in $F_{p^2}$ such that $t+\frac{1}{t}=\alpha$. Then
$$t+\frac{1}{t}, t^2+\frac{1}{t^2}, t^4+\frac{1}{t^4}$$ are the distinct roots of $f$ , all in $F_{p}$. But then $t^8 + \frac{1}{t^8}$ must get us back to $t + \frac{1}{t}$. So... $t^8=t$ or $t^8=\frac{1}{t}$. First case gives us an element of order $7$ in $F^{\times}_{p^2}$ , so $7\mid p^2-1$. Second case we hope to exclude. Short of other ideas, we substitute:
$$f(t+\frac{1}{t})=\frac{t^6 + t^5 + t^4 + t^3 + t^2 + 1}{t^3}$$
Therefore
$$t^6 + t^5 + t^4 + t^3 + t^2 + 1 = 0$$ and
we get $t^7=1$, $t\ne 1$, so $t$ element of order $7$ in $\mathbb{F}_{p^2}^{\times}$.
Added: another approach: we have
$$1+ (t+ \frac{1}{t}) + (t^2+ \frac{1}{t^2}) + (t^4 + \frac{1}{t^4}) = 0$$
so
$$\frac{(t+ \frac{1}{t}-1)(t^6 + t^5 + t^4 + t^3 + t^2 + 1)}{t^3}=0$$
|
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|
Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $[x,y,z]$ and applying the property of orthogonal matrix.
In each approach I got a different solution, which is not correct.
Even I suspect there is an error in given problem.
Please help me in finding the third row.
Adding cross product calculation below.
\begin{vmatrix}
\bar{\imath}&\bar{\jmath}&\bar{k}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}
\end{vmatrix}
\begin{align}
&=\bar{\imath}(0)-\bar{\jmath}\left(-\frac{1}{3\sqrt{2}}\right)+\bar{k}\left(\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\frac{1}{\sqrt{\frac{1}{18}+\frac{1}{18}}}\left(0,\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)
\end{align}
|
Let $[x~y~z]$ be the third row. Due to the orthogonality, we have $$\frac{x}{3}+\frac{2y}{3}+\frac{2z}{3}=0,$$ and $$\frac{y}{\sqrt{2}}-\frac{z}{\sqrt{2}}=0.$$ Consequently, $y=z$, and $x=-4y$. Using $x^2+y^2+z^2=1$, the values of $x$, $y$, and $z$ can easily be found. The last condition ensures that the vector has unit norm.
If there is no constraint on the norm of the third vector, there will be infinite solutions. One such is $[-4~1~1]$. But, if the norm of the third vector is 1 (as is the case with the other two vectors), there are only two such solutions, which are given below.
$$\begin{bmatrix}\frac{-4}{\sqrt{18}} & \frac{1}{\sqrt{18}} & \frac{1}{\sqrt{18}}\end{bmatrix},$$ and $$\begin{bmatrix}\frac{4}{\sqrt{18}} & \frac{-1}{\sqrt{18}} & \frac{-1}{\sqrt{18}}\end{bmatrix}.$$
|
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|
Geometry : $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$ Let $G$ be centriod of $\triangle ABC$ and GA, GB, GC cut the circumcircle of $\triangle ABC$ again at A', B', C' respectively. Prove that $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$.
My attempt :
Let $D, E, F$ be the midpoints of $BC, CA, AB$ respectively.
$\angle GFE = \angle FCD = \angle C'B'G$ so $C'FEB'$ concyclic.
Similarly, $DEB'A'$ concyclic.
By Power of point, $C'G\cdot FG=GE\cdot GB'=GD\cdot GA'$
|
Say $AD =2x$, $BG = 2y$, $CG=2z$ and $A'D = x'$, $B'E=y'$, $C'F=z'$.
Finally $AB = 2c$, $BC= 2a$ and $AC=2b$.
By the PoP we have: $x'\cdot 3x= a^2$, $y'\cdot 3y=b^2$, $z'\cdot 3z=c^2$ and $$x(x+x') = y(y+y')= z(z+z') =:k$$
Then $$3x^2+a^2 = 3y^2+b^2= 3y^2+c^2=3k$$
So $3(x^2+y^2+z^2)+a^2+b^2+c^2 =9k$
We are interested in $$I= {2x\over x+x'}+ {2y\over y+y'}+ {2z\over z+z'}= {2x^2+2y^2+2z^2\over k}$$
By parallelogram identity we have
\begin{eqnarray*}
(6x)^2+(2a)^2 &=& 2(2b)^2+2(2c)^2\\
(6y)^2+(2b)^2 &=& 2(2a)^2+2(2c)^2\\
(6z)^2+(2c)^2 &=& 2(2b)^2+2(2a)^2\\
\end{eqnarray*}
and so $$3(x^2+y^2+z^2)=a^2+b^2+c^2$$
Thus $2(a^2+b^2+c^2) =9k$ and thus $I= 3$
|
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|
Find the distance of the centroid of the triangle ABC from the origin? A sphere of radius $r$ passes through the origin and the other points where it meets the coordinate axes are A, B and C. Find the distance of the centroid of the triangle ABC from the origin (in terms of $r$).
i was taking the coordinate of the centroid ($\frac{x+y+z}{3}$, $\frac{X+Y+Z}{3}$). but i don't know the distance of the centroid of triangle ABC from the origin. From my point of view i was using the distance formula from the origin,, but i didn't get it
if anybody help me me i would be very thankful to him
thank you
|
If the sphere intersects the three axes at the points A, B and C, it means that the sphere passes through these three points:
$(A,0,0)$; $(0,B,0)$; $(0,0,C)$
i.e., the sphere intersects x-axis at $(A,0,0)$ and y-axis at $(0,B,0)$ and z-axis at $(0,0,C)$
Now, coordinates of the centroid of triangle formed by these three co-ordinates will be:
$(\frac{A+0+0}{3},\frac{0+B+0}{3},\frac{0+0+C}{3})$ = $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$
To find distance of this centroid from origin $(0,0,0)$ we simply use the two point distance formula in 3d:
Distance between $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$ and $(0,0,0)$ =
$\sqrt{(\frac{A}{3}-0)^2 + (\frac{B}{3}-0)^2 + (\frac{C}{3}-0)^2}$ = $\frac{\sqrt{A^2 + B^2 + C^2}}{3}$
In our case, the radius of sphere is given as $r$.
Let's assume the centre of sphere to be at $(x_0,y_0,z_0)$. Then the equation of the sphere can be written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = r^2$
Since it's also given that the sphere passes through the origin $(0,0,0)$, this point must satisfy the above equation of sphere.
Or, $(0-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$
Or, $x_0^2 + y_0^2 + z_0^2 = r^2$ --- $(1)$
Now, the sphere also passes through the point $(A,0,0)$. Substituting this point in the equation of sphere, we have:
$(A-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$
Or, $(A-x_0)^2+y_0^2+z_0^2 = r^2$
From $(1)$, let's also substitute $r^2$ with $x_0^2 + y_0^2 + z_0^2$
$(A-x_0)^2+y_0^2+z_0^2 = x_0^2 + y_0^2 + z_0^2$
Or, $(A-x_0)^2 = x_0^2$
Or, $A-x_0 = x_0$
Thus we have $A = 2x_0$.
Similarly by substituting $(0,B,0)$ and $(0,0,C)$ in the equation of sphere we get $B = 2y_0$ and $C= 2z_0$.
Hence, the three points A, B and C are $(2x_0,0,0)$, $(0,2y_0,0)$ and $(0,0,2z_0)$ respectively.
Thus, by using the formula we arrived at in section 1, we can say the distance of centroid of triangle ABC from origin =
$\frac{\sqrt{A^2 + B^2 + C^2}}{3}$ = $\frac{\sqrt{(2x_0)^2 + (2y_0)^2 + (2z_0)^2}}{3}$ = $\frac{2\sqrt{x_0^2 + y_0^2 + z_0^2}}{3}$
But from equation $(1)$ we know that $x_0^2 + y_0^2 + z_0^2 = r^2$.
Substituting this we have the required distance as $\frac{2\sqrt{r^2}}{3} = \frac{2r}{3}$
|
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|
Integrating $\frac{1}{(a+b\cos x)}$
Question: How do you prove this integral$$\int\frac {dx}{a+b\cos x}=\frac 2{\sqrt{a^2-b^2}}\arctan\left\{\tan\frac x2\sqrt{\frac {a-b}{a+b}}\right\}$$or$$\int\frac {dx}{a+b\cos x}=\frac 1{\sqrt{b^2-a^2}}\log\frac {\sqrt{b+a}+\sqrt{b-a}\tan\tfrac x2}{\sqrt{b+a}-\sqrt{b-a}\tan\tfrac x2}$$According as $a> b$ and $a<b$.
I'm not entirely sure how to prove the two integrals. I started off with the identity$$\cos x=\frac {1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2}$$And substituted to get$$\begin{align*}\int\frac {dx}{a+b\cos x} & =\int\frac {dx}{a+b\left(\tfrac {1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2}\right)}\\\\ & =\int\frac {1+\tan^2\tfrac x2}{(a+b)+(a-b)\tan^2\tfrac x2}\, dx\\ & =\int\frac {dz}{(a+b)+(a-b)z^2}\end{align*}$$where $z=\tan\tfrac x2$. Using the rule$$\int\frac 1{x^2+a^2}\, dx=\frac 1a\arctan\frac xa$$I get the integral as$$\frac 1{\sqrt{a+b}}\arctan\left\{\frac z{\sqrt{a+b}}\sqrt{a-b}\right\}=\frac 1{\sqrt{a+b}}\arctan\left(\tan\frac x2\sqrt{\frac {a-b}{a+b}}\right)$$Which doesn't match up with the solutions given. I'm also not sure how the second solution comes up.
|
Lets compute the integral using the half tangent angle sub where
$$
\cos x = \frac{1-t^2}{1+t^2}
$$
where $t = \tan\left(\frac{x}{2}\right)$
we obtain and integral of the form
$$
\int \frac{1}{a+b\left[\frac{1-t^2}{1+t^2}\right]}\frac{2}{1+t^2}dt
$$
or
$$
\int \frac{2}{a(1+t^2)+b(1-t^2) }dt
$$
or
$$
\int \frac{2}{a+b + (a-b)t^2}dt
$$
you can re-arrange to obtain
$$
\frac{2}{a+b}\int\frac{1}{1 + \lambda^2t^2}dt
$$
where $\lambda^2 = \frac{a-b}{a+b}$
so we have
or using $\lambda t = u\to dt = \frac{1}{\lambda }du$ we find
which is
$$
\frac{2}{\lambda (a+b)}\int\frac{1}{1 + u^2}du
$$
which is easily integrated.
For the case where $b > a$ we have to look at
$$
-\lambda^2 = \frac{a-b}{a+b} < 0 \to
$$
so we have
you can re-arrange to obtain
$$
\frac{2}{a+b}\int\frac{1}{1 - \lambda^2t^2}dt = \frac{2}{a+b} \int \frac{1}{(1-\lambda t)(1+\lambda t)}dt
$$
we can split the final integrand as follows
$$
\frac{1}{(1-\lambda t)(1+\lambda t)} = \frac{1}{2}\left[\frac{1}{1-\lambda t} + \frac{1}{1+\lambda t}\right]
$$
so you end up with an integral of the form
$$
\frac{2}{a+b}\int\frac{1}{1 - \lambda^2t^2}dt = \frac{1}{a+b} \int \left[\frac{1}{1-\lambda t} + \frac{1}{1+\lambda t}\right] dt
$$
which is once again sdtraightforward to integrate.
|
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|
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.
The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1-
\dfrac{x^2}{a^2}\right) \tag{1}.$$
Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$
We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.
Substituting back in $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.
The given answer is $3A$.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?
|
The given answer is correct. Your mistake is in the formula for the area,
which should be $A/2 = 1/2\ {\rm base} \times {\rm height}$ for each half half triangle or
$A = (a-x) b \sqrt{1 - x^2/a^2}$
for the full triangle.
The derivative is then:
$${d A \over d x} = -\frac{b (a+2 x) \sqrt{1-\frac{x^2}{a^2}}}{a+x} .$$
Set it to $0$ to find $x = -a/2$.
Then compute the area:
$$A = {3 \sqrt{3} \over 4} a b$$
|
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|
Find the Laurent Series of $\frac{z^2}{(z-1)(z^2+4)}$ I have attempted to split this into partial fractions of the form $\frac{A}{(z-1)} + \frac{B}{(z-2i)} + \frac{C}{(z+2i)}$ as well as by $\frac{A}{(z-1)} + \frac{Bz+C}{(z^2+4)}$, but things get very complex (ha!) with i's and z's showing up in the numerator. How should I proceed?
I am told to find the series about 0 and 1, taking into account the different values of the radii.
|
Since you are told to find the series about $0$ and $1$, the first one does not make much trouble using long division.
For the second one, define first $z=x+1$ which makes
$$f=\frac{z^2}{(z-1)(z^2+4)}=\frac{(x+1)^2}{x \left(x^2+2 x+5\right)}=\frac{x^2+2x+1}{x \left(x^2+2 x+5\right)}$$ Perform the long division for $$x \times f=\frac{x^2+2x+1}{x^2+2 x+5}$$ When done, divide the result by $x$ and replace $x$ by $(z-1)$
|
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|
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$.
$(x-1)+(y+3)+(z-1)+(w-2)=x+y+z+w-1=28$
Now $x-1=y+3=z-1=w-2=7$ since product is maximum when numbers are equal
My answer came out to be as $6*10*6*5=1800$ but the answer is $2401$. What am I doing wrong? And also, how we will get the answer $2401$
|
Let $$f(x,y,z,w) = (x-1)(y+3)(z-1)(w-2)$$ and $$g(x,y,z,w) = x + y + w - 29$$
We want to $$\max\{f(x,y,x,w)\}$$
subject to:
$$g(x,y,z,w) = 0, \ \ \ x,y,z > 2$$
Let
\begin{align*}
\mathcal{L}(x,y,z,\lambda) &= f(x,y,z,w) + \lambda g(x,y,z,w)\\ &= (x-1)(y+3)(z-1)(w-2) + \lambda(x + y + w - 29)
\end{align*}
Then
$$\nabla \mathcal{L}(x,y,z,\lambda) = 0$$ yields $4$ equations:
\begin{equation}{\tag{1}}
(y+3)(z-1)(w-2) + \lambda = 0
\end{equation}
\begin{equation}{\tag{2}}
(x-1)(z-1)(w-2) + \lambda = 0
\end{equation}
\begin{equation}{\tag{3}}
(x-1)(y+3)(w-2) + \lambda = 0
\end{equation}
\begin{equation}{\tag{4}}
(x-1)(y+3)(z-1) + \lambda = 0
\end{equation}
Now when you set each one equal to each other and find another four equations where $y$ will be a linear combination of $x,y,w$ we find another $4$ equations:
\begin{equation}{\tag{5}}
y = x - 4
\end{equation}
\begin{equation}{\tag{6}}
y = z - 4
\end{equation}
\begin{equation}{\tag{7}}
y = y
\end{equation}
\begin{equation}{\tag{8}}
y = w - 5
\end{equation}
Now, we see that
$$(y+4) + y + (y + 4) + (y+5) = 29 \Rightarrow y = 4$$
Then it is trivial to find $x,z,w$ by plugging in $y=4$ to the equations above. Hope that helps!
|
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|
Find $ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$
Find the limit of
$$ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$$
Let $$u= 1 +\frac{x}{n} \implies du =\frac{1}{n} dx \implies n \cdot du = dx$$
at $x=0$ $u=1$ and at $x=1$ $u=1+\frac{1}{n}$ so now limit will change from $1$ to $1+\frac{1}{n}$
Back to the integral
$$ \lim\limits_{n \rightarrow \infty} \left( n \cdot \int_{1}^{1+\frac{1}{n}} u^n du \right)= \lim\limits_{n \rightarrow \infty} \left( n \cdot \left[ \frac{nu^{n+1}}{n+1} \right]_1^{1+\frac{1}{n}} \right) = \lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1} \left[ u^{n+1} \right]_1^{1+\frac{1}{n}} \right) $$
$$\implies\lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1} \left[ \left(1+\frac{1}{n} \right)^{n+1}-1 \right] \right)=\infty$$
Is my finding correct? Is the procedure of taking the limit before completing the integration correct?
Much appreciated
|
You forgot to change the limits of the integral. The integral limits should be $1$ and $1+\frac{1}{n}$.
In particular, $$n\int_{1}^{1+1/n} u^n du = \left(1+\frac{1}{n}\right)^n-\frac{n}{n+1}.$$
|
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|
Maclaurin Series $ \int \frac{\sin(x)}{5x} $ I am supposed to evaluate the indefinite integral as an infinite series centered at $ x=0 $ and give the first five non-zero terms of the series.
$ \int \frac{\sin(x)}{5x} $
Here is what I have done so far:
$ g(x) = \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} $
$ \int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{5x} $
$ \frac{1}{5}\int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} x^{2n} $
$ \frac{1}{5} \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{2n+1} $
For $ n=0 $, $ \frac{1}{5}(x)$
$ n=1 $, $ \frac{1}{5}(\frac{-1}{3!} \frac{x^3}{3})$
$ n=2 $, $ \frac{1}{5}(\frac{1}{5!} \frac{x^5}{5})$
$ n=3 $, $ \frac{1}{5}(\frac{-1}{7!} \frac{x^7}{7})$
$ n=4 $, $ \frac{1}{5}(\frac{1}{9!} \frac{x^9}{9})$
And sum them.
I've done this a few different ways and am getting the same result. Can anyone shed some light on where I am going wrong?
|
So, what happened was when I took the indefinite integral, to cover all potential functions which satisfy the indefinite integral, + C needs to be included.
|
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|
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:
A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$
I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.
|
$$\cot x =\frac{-7}{24}=\frac{\cos x}{\sin x}$$
$$\frac{49}{576}=\frac{\cos^2 x}{1-\cos^2 x}$$
It gives $$\cos x=\frac{-7}{25}$$
And
$$\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\frac{3}{5}=0.6 $$
Hence (A) is correct
EDIT:
After concerning with experts , I concluded that $\cos \frac{x}{2} $ is negative and hence (D) is correct
|
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Determine IFS of the von Koch Curve I am (still) reading Falconer's Fractal Geometry book and I have reached the section on Iterated Function Systems (IFSs). He defines the IFS for the middle third Cantor set:
as $S_{1}(x)=\frac{1}{3}x$, $S_{2}(x)=\frac{1}{3}x+\frac{1}{3}$ since, at each iteration of the construction, you shorten the line to a third of its length and then do the same to get a second interval of a third the length and then move it along to the right by $\frac{2}{3}$.
I am trying to determine the IFS for the von Koch curve:
So far I have worked out that $S_{1}(x,y)=(\frac{1}{3}x,y)$ and $S_{2}(x,y)=(\frac{1}{3}x+\frac{2}{3},y)$ since we are now working in $\mathbb{R}^{2}$. I have also partially worked out $S_{3}$ and $S_{4}$ to be $S_{3}(x,y)=(\frac{1}{5}x+\frac{1}{3},Y_{3})$ and $S_{4}(x,y)=(\frac{1}{5}x+\frac{2}{3},Y_{4})$.
Does anyone know how to determine the $Y_{3}$ and $Y_{4}$?
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TL;DR version: you are not taking rotation into account. This is basically irrelevant in $\mathbb{R}$, but makes a huge difference in $\mathbb{R}^d$, where $d\ge 2$.
In Detail:
A contracting similitude can be characterized as a map
$$ S : \mathbb{R}^{d} \to \mathbb{R}^d $$
which takes the form
$$ S(\vec{x}) = c U \vec{x} + \vec{b}, $$
where $c \in (0,1)$ is the scaling ratio of $S$, $U$ is a unitary matrix (e.g. a rotation in $\mathbb{R}^2$, and $\vec{b}\in\mathbb{R}^n$ is a translation. In this language, the maps $S_1$ and $S_4$ which you describe can be written as
$$ S_1(\vec{x}) = \frac{1}{3} \vec{x}
\qquad\text{and}\qquad
S_4(\vec{x}) = \frac{1}{3} \vec{x} + \begin{bmatrix} \frac{2}{3} \\ 0 \end{bmatrix}. $$
In each of these cases, the we can take $U = I$ to be the identity. For the other two maps, you are going to need to rotate. In general, we can describe a rotation by $\theta$ in $\mathbb{R}^2$ by the matrix
$$
R_{\theta} := \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \end{bmatrix}.
$$
To get the piece of the von Koch curve that slopes up from left to right, we require a rotation of $\frac{\pi}{3}$, then a translation by $\frac{1}{3}$, which gives the map
$$ S_2(\vec{x})
= \frac{1}{3} R_{\pi/3} \vec{x} + \begin{bmatrix} \frac{1}{3} \\ 0 \end{bmatrix}
= \frac{1}{3} \begin{bmatrix}
\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \vec{x} + \begin{bmatrix} \frac{1}{3} \\ 0 \end{bmatrix}.
$$
Finally, for the last piece of the curve, we will need to rotate by $-\frac{\pi}{3}$, translate right by $\frac{1}{2}$, and translate up by $\frac{\sqrt{3}}{6}$ (the origin gets mapped to the "top" of the snowflake, which is the vertex of an equilateral triangle sitting on the $x$-axis, centered at $x=\frac{1}{2}$, with sides of length $\frac{1}{3}$). This gives
$$ S_3(\vec{x})
= \frac{1}{3} R_{-\pi/3} \vec{x} + \begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{6} \end{bmatrix}
= \frac{1}{3} \begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
-\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \vec{x} + \begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{6} \end{bmatrix}.
$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/2393507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\lambda(x^2+y^2+z^2-2x+2y+6z+9)=0$$
Differentiated partially wrt $x$, $y$ and $z$, and equated to $0$.
I get the following equations
$$(2\lambda+2)x-y-z=2\lambda$$
$$-x+(2\lambda+2)y-z=-2\lambda$$
$$-x-y+(2\lambda+2)z=6\lambda$$
In matrix form:
$$\begin{bmatrix} 2\lambda+2 & -1 & -1 \\-1 & 2\lambda+2 & -1 \\ -1 & -1 & 2\lambda+2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}2 \\ -2 \\ -6 \end{bmatrix} $$
When I try to find $x$, $y$ and $z$ from these equations, it gets complex. Is there an easier way to ascertain their values?
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You already got the system of equations. So the solution simply follows from there.
You should get $(x,y,z)=(2,-1,-4)$ and $(x,y,z)=(0,-1,-2)$ which gives you the maximum (27) and the minimum (3), respectively.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2395581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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|
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