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Understanding kinematics formula in two dimensions Determine the angle of projection of a projectile if its speed at maximum height is $\sqrt{\frac{2}{5}}$ of its speed at half the maximum height.
My solution:
$$H_{max}=\frac{{v_0}^2\sin^2(\theta)}{2g}\implies \frac{1}{2}H_{max}=\frac{{v_0}^2\sin^2(\theta)}{4g}\\v_{x}=v_0\cos(\theta)\quad {v_{\frac{H}{2}y}}^2={v_0}^2\sin^2(\theta)-2g\left(\frac{{v_0}^2\sin^2(\theta)}{4g}\right)=\frac{1}{2}{v_0}^2\sin^2(\theta)\\v_0\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{{v_0}^2\cos^2(\theta)+\frac{1}{2}{v_0}^2\sin^2(\theta)}\\\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{\cos^2(\theta)+\frac{1}{2}-\frac{1}{2}\cos^2(\theta)}\\\cos^2(\theta)=\frac{1}{5}\cos^2(\theta)+\frac{1}{5}\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$
Solution found on another website:
$$gH_{max}=\frac{{v_0}^2\sin^2(\theta)}{2}\quad {v_x}^2={v_0}^2\cos^2(\theta)\\{v_{\frac{H}{2}}}^2={v_0}^2-2g\left(\frac{1}{2}H_{max}\right)={v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\\v_{0}\cos(\theta)=\sqrt{\frac{2}{5}}v_{\frac{H}{2}}\implies {v_{0}}^2\cos^2(\theta)=\frac{2}{5}{v_{\frac{H}{2}}}^2=\frac{2}{5}\left({v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\right)\\5\cos^2(\theta)=2-\sin^2(\theta)=1+\cos^2(\theta)\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$
What I don't quite understand in the second solution is the application of the kinematics formula $v^2={v_0}^2+2a\Delta d$ (second line). I thought the formula held only for one dimensional kinematics, but its usage here would imply two dimensional vector addition since the initial velocity and gravity aren't parallel vectors. Can someone help clarify this for me?
| You know that
$$v_y^2 = v_{0y}^2 + 2a_y\Delta y$$ and that
$$v_x^2 = v_{0x}^2 + 2a_x\Delta x.$$
First,
$$v^2 = \textbf{v}\cdot\textbf{v} = \left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right)\cdot\left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right) = v_x^2 + v_y^2.$$
Second,
$$v_0^2 = \textbf{v}_0\cdot\textbf{v}_0 = v_{0x}^2 + v_{0y}^2,$$
and
$$2a_x\Delta x = 0.$$
Last, $$v^2 = v_0^2 - 2g\Delta y.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $8x-3+\sqrt{x+2}-\sqrt{x-1}=7 \sqrt{x^2+x-2}$
Solve the equation
$$8x-3+\sqrt{x+2}-\sqrt{x-1}=7 \sqrt{x^2+x-2}$$
I have this idea: set $$\sqrt{x+2}=a , x+2=a^2 , \sqrt{x-1}=b.$$
So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve.
Any hint is appreciated.
| The domain gives $x\geq1$ and we need to solve that
$$\sqrt{x+2}-\sqrt{x-1}=7\sqrt{x^2+x-2}-8x+3.$$
Now, since $\sqrt{x+2}-\sqrt{x-1}\geq0,$ we obtain
$$7\sqrt{x^2+x-2}-8x+3\geq0$$ or
$$\frac{97-\sqrt{2989}}{30}\leq x\leq\frac{97+\sqrt{2989}}{30}.$$
Thus, we need to solve
$$\left(\sqrt{x+2}-\sqrt{x-1}\right)^2=\left(7\sqrt{x^2+x-2}-(8x-3)\right)^2$$ or
$$(112x-44)\sqrt{x^2+x-2}=113x^2-x-90$$ or
$$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
$$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
$$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
$$\left\{2,\frac{23+4\sqrt{10}}{9}\right\}.$$
Also, your idea helps.
Indeed, we got the following system:
$$a^2-b^2=3$$ and
$$8(a^2-2)-3+a-b=7ab.$$
From the second equation we obtain:
$$b=\frac{8a^2+a-19}{7a+1},$$ which gives
$$a^2-\left(\frac{8a^2+a-19}{7a+1}\right)^2=3$$ or
$$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.
| {
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"url": "https://math.stackexchange.com/questions/3072817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$
Find any local max or min of
\begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
\end{align}
such that
\begin{align}
x+y+z=1 && (2)\\
3x+y+z=5 && (3)
\end{align}
My attempt. Let
$L(x,y,z,\lambda_1, \lambda_2)= f(x,y,z)+\lambda_2 (x+y+z-1) + \lambda_1 (3x+y+z-5)$
$L_x=2x+ 3 \lambda_1 + \lambda_2 =0$
$L_y=2y+ \lambda_1 + \lambda_2=0$
$L_z=2z+\lambda_1 + \lambda_2=0$
Solve for $x,y,z$ we get:
$x=\frac{-3 \lambda_1 - \lambda_2}{2}$
$z=y=\frac{-\lambda_1 - \lambda_2}{2}$
with the use of $(2)$ and $(3)$ $\implies$
$x=2$
$y=z= \frac{-1}{2}$
so the stationary point is $(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$
The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus
$(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.
Is this correct?
| An option:
1) $x+y+z=1$; and
2) $3x+y+z=5$;
$2$ planes , their intersection is a straight line.
Subtract: 2)-1):
$2x=4$; $x=2$ ;and
$y+z=-1$;
$d^2=x^2+y^2+z^2$ .
Minimal distance of line from origin:
$d^2= 4 +y^2+z^2.$
2D problem:
Minimize $y^2+z^2$ with constraint $y+z=-1$.
$d_2^2= $
$[-(1+z)]^2+z^2=2z^2+2z+1=$
$2(z^2+z)+1= $
$2[(z+1/2)^2]-1/2+1\ge 1/2$.
Equality at $z=-1/2$;
Finally:
Minimum at :
$x=2$ ; $y=-1/2$; $z=-1/2$;
$d^2_{\min}= 4+1/2=9/2;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
minimum value of $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2\geq 18\sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
| Hint: Apply $AM \ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM \le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
\frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $\le$ quadratic mean”:
$$
\frac{4}{\frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c}}
\le \sqrt{\frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of geometric series where $x=\sin^{-1}{\frac{7}{8}}$ This problem is from a local contest.
The series $\sin{x}$, $\sin{2x}$, $4\sin{x}-4\sin^3{x}$, $...$ is a geometric series if $x=\sin^{-1}{\frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning.
For the first term, $\sin{\sin^{-1}{\frac{7}{8}}}=\frac{7}{8}$
I was not able to find $\sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4\sin{\sin^{-1}{\frac{7}{8}}}=\frac{28}{8}$ and $4\sin^3{\sin^{-1}{\frac{7}{8}}}=4(\frac{7}{8})^3=\frac{1372}{512}$. The third term is the difference of $\frac{28}{8}$ and $\frac{1372}{512}$.
My question is, how can I compute the second term, $\sin{2x}$?
| A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=\sin x,\qquad a_1=\sin2x=2\sin x\cos x
$$
Then
$$
c=\frac{a_1}{a_0}=2\cos x
$$
and indeed
$$
a_2=4\sin x-4\sin^3x=4\sin x\cos^2x=a_1\cdot 2\cos x
$$
Thus you have
$$
a_n=2^n\cos^nx\sin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2\cos x=2\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{4}<1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integrate $\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx$ I had to integrate the following integral:
\begin{equation}
\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx
\end{equation}
but I can't find a suitable substitution to find a solution. Nothing I try works out and only seems to make it more complicated. Does anyone have an idea as to how to solve this?
I also tried to get help from WolframAlpha but it just says that there is no step-by-step solution available.
The sollution by wolfram alpha is:
\begin{equation}
\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx = \frac{x\cos(x)}{x+\cos(x)} + c
\end{equation}
| Note that the derivative of $x + \cos(x)$ in the denominator is $1-\sin(x)$ . We can try to make this term appear in the numerator and then integrate by parts. We have
$$ \frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2} = \frac{\cos^2(x) - x^2 + x^2(1-\sin(x))}{(x+\cos(x))^2} = \frac{\cos(x) - x}{x+\cos(x)} + x^2 \frac{1-\sin(x)}{(x+\cos(x))^2} \, ,$$
so
\begin{align}
\int \frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2} \, \mathrm{d}x &= \int \frac{\cos(x) - x}{x+\cos(x)} \, \mathrm{d} x - \frac{x^2}{x+\cos(x)} + \int \, \frac{2 x}{x+\cos(x)}\mathrm{d} x \\
&= x - \frac{x^2}{x+\cos(x)} = \frac{x \cos(x)}{x+\cos(x)}
\end{align}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Solving $(\ln(x)-1)y'' - \frac{1}{x}y' + \frac{1}{x^2}y = \frac{(\ln(x) - 1)^2}{x}$ On my exam I had to solve the following differential equation.
\begin{equation}
(\ln(x)-1)y'' - \frac{1}{x}y' + \frac{1}{x^2}y = \frac{(\ln(x) - 1)^2}{x^2}
\end{equation}
Which is a differential equation of the form:
\begin{equation}
y'' + a(x)y' + b(x)y = R(x)
\end{equation}
The only method we've seen to solve this kind of differential equations is:
If the differential equation is of the form:
\begin{equation}
y'' + a(x)y' + b(x)y = 0
\end{equation}
First find a solution of the characteristic equation, being $\varphi_1$. Then:
\begin{equation}
\varphi_2(x) = \varphi_1(x)\int\frac{dx}{A(x)(\varphi_1(x))^2}
\end{equation}
With $A(x) = e^{\int a(x) dx}$
Then the homogenous solution is given by:
\begin{equation}
y(x) = c_1\varphi_1(x) + c_2\varphi_2(x)
\end{equation}
The first problem is that this doesn't satisfy the requirements for this method since the differential equation is not homogenous, but since this is the only fitting method, I'd still try to use it. My guess would be to start with the characteristic equation which gives:
\begin{equation}
(\ln(x)-1)x^2 - 1 + \frac{1}{x^2}y = 0
\end{equation}
or
\begin{equation}
x^2 - \frac{1}{(\ln(x)-1)} + \frac{1}{x^2(\ln(x)-1)} = 0
\end{equation}
but i wouldn't even know how to start solving this equation to find the roots of the equation. Does anyone have an idea as to how to tackle this problem.
Note the only other ways of solving linear differential equations that we have seen are ways to solve first order differential equation or ways to solve second order differential equations in the form:
\begin{equation}
y'' + py' + qy = R(x)\;\;\;\text{with}\;\; p,q\in\mathbb{R}
\end{equation}
| $$(\ln(x)-1)\frac{d^2y}{dx^2} - \frac{1}{x}\frac{dy}{dx} + \frac{1}{x^2}y = \frac{(\ln(x) - 1)^2}{x^2}$$
Change of variable : $t=\ln(x)-1\quad;\quad x=e^{t+1}\quad;\quad dx=x\:dt$
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}$
$\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x}\frac{d^2y}{dt^2}\frac{dt}{dx}= -\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x^2}\frac{d^2y}{dt^2}$
$$t(-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x^2}\frac{d^2y}{dt^2}) -\frac{1}{x}(\frac{1}{x}\frac{dy}{dt})+ \frac{1}{x^2}y = \frac{t^2}{x^2}$$
$$t\frac{d^2y}{dt^2}-(t+1)\frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{\ln(x)-1}+c_2(\ln(x)-1+1)-(\ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2\ln(x)-(\ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2\ln(x)-(\ln(x))^2-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Error in calculation of intersection of three cylinders Set $A:=\{(x,y,z)\in \mathbb R^{3}: x^2+y^2\leq 1, x^2+z^2\leq 1, y^2+z^2 \leq 1\}$
I want to find the volume. I have to use symmetry and without polar coordinates.
My idea: Using symmetry we can look at the first octant and can restrict $x,y,z \geq 0$
$\lambda^{d}(A)=\int_{A}dxdydz$
And we then set $0\leq z\leq 1$ and $0 \leq x \leq \sqrt{1-z^2}$ and $0 \leq y \leq \sqrt{1-z^2}$
So $\int_{A}dxdydz=\int_{0}^{1}\int_{0}^{\sqrt{1-z^2}}\int_{0}^{\sqrt{1-z^2}}dxdydz=\int_{0}^{1}2\sqrt{1-z^2}dz$
And then I would use substitution, namely, $z = \sin{x}\Rightarrow dz =\cos{x}dx$, so $\int_{0}^{1}2\sqrt{1-z^2}dz=\int_{0}^{\frac{\pi}{2}}2\sqrt{1-\sin{x}^2}\cos{x}dx=2\int_{0}^{\frac{\pi}{2}}\cos^{2}{x}dx=2\int_{0}^{\frac{\pi}{2}}\frac{1}{2}+\frac{1}{2}\cos{2x}dx=\int_{0}^{\frac{\pi}{2}}1+\cos{2x}dx=x+\frac{\sin{2x}}{2}\vert_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}$
So getting back to eight octants, we'd get $\lambda^{d}(A)=4\pi$ but I have been told this is incorrect. I do not understand where I went wrong
| $4\pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $\frac{4}{3}\pi$. Assume that the value of $z\in[-1,1]$ has been fixed. The $z$-section of our body is shaped as
$$ \left\{\begin{array}{rcl}x^2+y^2&\leq& 1\\x^2,y^2&\leq &1-z^2\end{array}\right.$$
hence it is a square with side length $2\sqrt{1-z^2}$ for any $|z|\geq \frac{1}{\sqrt{2}}$ and a circle with four circle segments being removed for any $|z|\leq\frac{1}{\sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $\pi + 4|z|\sqrt{1-z^2}-4\arcsin(|z|)$. The wanted volume is so
$$ 2\left[\int_{0}^{1/\sqrt{2}}4(1-z^2)\,dz + \int_{1/\sqrt{2}}^{1}\pi + 4z\sqrt{1-z^2}-4\arcsin(z)\,dz\right]$$
i.e. $\color{blue}{8\sqrt{2}-2\pi}\approx 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with Maclaurin series of $\cos(\ln(x+1))$? Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:
Find the Maclaurin series of $g(x) = \cos(\ln(x+1))$ up to order 3.
I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:
$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$
$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$
I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks
| Since you also don't need very high orders, the straightforward calculation of derivatives is tractable, though not preferable computationally. If $f(x) = \cos \log(1+x)$, then $$f'(x) = -\frac{\sin \log (1+x)}{1+x}, \quad f'(0) = 0.$$ Then $$f''(x) = -\frac{\cos \log (1+x)}{(1+x)^2} + \frac{\sin \log(1+x)}{(1+x)^2}, \quad f''(0) = -1.$$ Finally, $$f'''(x) = \frac{\sin \log(1+x)}{(1+x)^3} + \frac{2\cos \log(1+x)}{(1+x)^3} + \frac{\cos \log(1+x)}{(1+x)^3} - \frac{2 \sin \log (1+x)}{(1+x)^3}, \quad f'''(0) = 3.$$ Then $$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + O(x^4) = 1 - \frac{x^2}{2} + \frac{x^3}{2} + O(x^4).$$
It is worth noting that $$f^{(n)}(x) = \frac{A_n \cos \log (1+x) + B_n \sin \log (1+x)}{(1+x)^n},$$ for suitable constants $A_n$, $B_n$. A proof by induction is straightforward and yields insight into the recursion relations defining $\{(A_n, B_n)\}_{n \ge 1}$: $$\begin{align*} f^{(n+1)}(x) &= -A_n \frac{\sin \log (1+x)}{(1+x)^{n+1}} - nA_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} - n B_n \frac{\sin \log(1+x)}{(1+x)^{n+1}} + B_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} \\
&= \frac{(B_n - nA_n) \cos \log(1+x) + (-A_n - nB_n)\sin \log(1+x)}{(1+x)^{n+1}}.
\end{align*}.$$ Therefore, $$\begin{align*} A_{n+1} & = -nA_n + B_n, \\ B_{n+1} &= -A_n - nB_n, \\ A_0 &= 1, \\ B_0 &= 0. \end{align*}$$ In matrix form, this recurrence is equivalent to $$\begin{bmatrix}A_{n+1} \\ B_{n+1}\end{bmatrix} = \begin{bmatrix} -n & 1 \\ -1 & -n \end{bmatrix} \begin{bmatrix} A_n \\ B_n \end{bmatrix},$$ consequently $$\begin{bmatrix}A_n \\ B_n\end{bmatrix} = M_n \begin{bmatrix} 1 \\ 0 \end{bmatrix},$$ where $$M_n = \prod_{k=0}^{n-1} \begin{bmatrix} -k & 1 \\ -1 & -k \end{bmatrix}$$ and $f^{(n)}(0) = A_n$. This lets us continue our calculation of higher orders with relative ease if we keep track of the matrix product $M_n$, which is always of the form $$M_n = \begin{bmatrix} a & b \\ -b & a \end{bmatrix};$$ for example, $$M_3 = \begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix}, \quad M_4 = \begin{bmatrix} -3 & 1 \\ -1 & -3\end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix} = \begin{bmatrix} -10 & 0 \\ 0 & -10 \end{bmatrix},$$ so that $A_4 = -10$ and the next coefficient is $-10/4! = -5/12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiation under the integral sign - what transformations to use? Need some help with this integral
$$I (\alpha) = \int_1^\infty {\arctan(\alpha x) \over x^2\sqrt{x^2-1}} dx$$
Taking the first derivative with respect to $\alpha$
$$I' (\alpha) = \int_1^\infty { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }$$
What transformations to use in order to solve $I'(\alpha)$?
| Substitute
$$u=\sqrt{x^2-1}\implies du=\frac{x}{\sqrt{x^2-1}}dx\implies dx=\frac{\sqrt{x^2-1}}{x}du$$
Then
$$\int { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }=\int { du\over x^2(1+\alpha^2 x^2)}=\int { du\over (u^2+1)(a^2u^2+a^2+1) }$$
Perform partial fraction decomposition
$$\int { du\over (u^2+1)(a^2u^2+a^2+1) }=\int\frac{du}{u^2+1}-a^2\int\frac{du}{a^2u^2+a^2+1}$$
Sure you know that
$$\int\frac{du}{u^2+1}=\arctan(u)+C$$
To solve for
$$\int\frac{du}{a^2u^2+a^2+1}$$
Use substitution
$$v=\frac{au}{\sqrt{a^2+1}}\implies du=\frac{a^2+1}{a}$$
$$\int\frac{du}{a^2u^2+a^2+1}=\int\frac{\sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=\frac{1}{a\sqrt{a^2+1}}\int\frac{dv}{v^2+1}=\frac{\arctan(v)}{a\sqrt{a^2+1}}+C$$
Now plug in back $x$, you would get
$$\int_{1}^{\infty} { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }=\left[\arctan\left(\sqrt{x^2-1}\right)-\frac{a\arctan\left(\frac{a\sqrt{x^2-1}}{a^2+1}\right)}{\sqrt{a^2+1}}\right]_{1}^{\infty}$$
I think you can handle the rest of the calculation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.
When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately appear, as a function of $x$?
It seems to be around ~ $c \sqrt x$.
Example: $x = 2500$:
2500, 1250, 833, 625, 500, 416, 357, 312, 277, 250, 227, 208, 192, 178, 166, 156, 147, 138, 131, 125, 119, 113, 108, 104, 100, 96, 92, 89, 86, 83, 80, 78, 75, 73, 71, 69, 67, 65, 64, 62, 60, 59, 58, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 43, 42, 41, 40, 40, 39, 39, 38, 37, 37, 36, 36, 35, 35, ...
| It's essentially the same as Jyrki Lahtonen's answer, but they invited me, so here's mine. Well, it's the same until the part where I go into detail about estimating where in that interval of potential values we actually get the first pair of equal values.
Let the sequence $a_n$, for $n=1,2,\dots$, be defined as $\left\lfloor \frac xn\right\rfloor$ for some positive real $x$. We seek the least $n$ for which $a_n=a_{n+1}$, or equivalently the greatest $a$ which appears twice in the sequence. Also, for convenience, define $b_n=\frac xn$. Note that the differences $b_n-b_{n+1}=\frac{x}{n(n+1)}$ form a decreasing sequence.
First, a fact about the floor and ceiling function: $\lfloor u\rfloor + \lfloor v\rfloor\le \lfloor u+v\rfloor \le \lfloor u\rfloor + \lceil v\rceil$, with equality when $v$ is an integer. What does this mean for our sequence $a_n$? When $\frac xn - \frac x{n+1} \ge 1$, $a_n-a_{n+1}\ge 1$ as well; we can't have two consecutive entries equal until $b_n$ has two consecutive entries that differ by less than $1$. From that, we will get our lower bound: if $a_n=a_{n+1}$, $b_n-b_{n+1}=\frac{x}{n(n+1)}<1$ and $x < n^2+n=(n+\frac12)^2-\frac14$. Solving for $n$ in terms of $x$, $n > \sqrt{x+\frac14}-\frac12$. Let $$N(x)=\left\lfloor\sqrt{x+\frac14}+\frac12\right\rfloor$$
be the first integer value that get us the inequality.
Now, for the upper bound. No matter how far we go, until $a_n$ drops all the way to zero, we still have the chance of $a_n$ and $a_{n+1}$ being different. Looking at one difference just won't be enough. Instead, we stack differences together; if $a_n-a_{n+k} < k$, then since $a_n$ is a decreasing sequence of integers, some two consecutive values in that range must be zero. By the other half of our key inequality, this is guaranteed to happen when $b_n-b_{n+k} \le k-1$. Start at the first possible place for two values to be equal; we're looking for the least $k$ such that $b_{N(x)}-b_{N(x)+k} \le k-1$. This inequality becomes
$$k-1 \ge \frac{x}{N(x)}-\frac{x}{N(x)+k} = \frac{kx}{N(x)(N(x)+k)}$$
$$(k-1)N^2(x)+k(k-1)N(x) \ge kx$$
This is - well, it's a mess, because of the floor in the definition of $N$. So, we approximate - $N(x) \le \sqrt{x+\frac14}+\frac12$, so $x \ge (N(x)-\frac12)^2-\frac14=N^2(x)-N(x)$. Oops - we actually need a lower bound for $x$ here (see comments). If
$$(k-1)N^2(x)+k(k-1)N(x) \ge k(N^2(x)+N(x))$$
$$(k^2+2k)N(x) \ge N^2(x)$$
$$N(X)\le (k+1)^2-1$$
then, since $k(N^2(x)+N(x)) > kx$, $(k-1)N^2(x)+k(k-1)N(x) > kx$ and we have a $k$ that works. This is true precisely when $k>\sqrt{N(x)+1}-1$; we will, of course, take the first successful value. The least $n$ with $a_n=a_{n+1}$ must satisfy
$$\sqrt{x}\approx N(x) \le n \le N(x)+\lfloor\sqrt{N(x)+1}\rfloor-1 $$
$$\le \left\lfloor\sqrt{x+\frac14}+\frac12\right\rfloor + \left\lfloor\sqrt{\sqrt{x+\frac14}+\frac32}\right\rfloor-1\approx \sqrt{x}+\sqrt[4]{x}$$
And now, for something new.
Where in that interval will it happen? For a randomly chosen $x$, it's essentially random - but biased. The deviation $1-(b_n-b_{n+1})$ increases approximately linearly with $n$ starting at zero for $n=N(x)$, so the sum of $j$ of them grows like $j^2$. The probability of our first duplication coming in the first $j$ chances is thus approximately proportional to $j^2$, and the location follows a wedge distribution; the probability of it being at $N(x)+j$ is approximately $\frac{2j+1}{N(x)}$ for $0\le j<\sqrt{N(x)}$.
But we can do better than that. Write $N^2(x)=x+c$; rearranging our inequalities $N^2-N\le x< N^2+N$,
$$x-N(x) < N^2(x) \le x+N(x)$$
Then $\frac{x}{N(x)}=\frac{N^2(x)+c}{N(x)}=N(x)+\frac{c}{N(x)}$. This fractional part $\frac{c}{N(x)}$, between $-1$ and $1$, is what actually determines where in the interval we finally reach a spot with two consecutive $a_n$ equal. As we repeatedly subtract quantities slightly less than $1$ from $b_n$, its fractional part increases until it ticks over an integer - and when that happens, we get our first repeat in the $a_n$.
Let $\frac{x}{N(x)+k}=N(x)-k+e_k$. As already noted, $e_0=\frac{c}{N(x)}\in [-1,1)$. For $e_0\in [-1,0)$, we seek the first $k$ such that $e_k \ge 0$. For $e_0\in [0,1)$, we seek the first $k$ such that $e_k \ge 1$. We will then have $a_{N(x)+k-1}=a_{N(x)+k}$. Clear the denominator to get
$$x = N^2(x) - k^2 + N(x)e_k + ke_k$$
$$0 = N(x)(e_k-e_0) + ke_k - k^2$$
$$k = \frac{e_k +\sqrt{k^2+4(e_k-e_0)N(x)}}{2}$$
For negative $e_0$, the key point comes when $e_k\approx 0$, and $2k\approx \sqrt{k^2-4e_0 N(x)}$, or $3k^2\approx -4e_0 N(x)$ and $k\approx \frac{2}{\sqrt{3}}\sqrt{-e_0 N(x)}$. For positive $e_0$, the key point comes when $e_k\approx 1$, and $2k-1\approx \sqrt{k^2+4(1-e_0) N(x)}$. Solve that to $k\approx \frac{2}{\sqrt{3}}\sqrt{(1-e_0)N(x)}+\frac23$.
So then, the amount $k$ we need to add to $N(x)$ is about $\frac{2}{\sqrt{3}}\sqrt{N(x)}$ times the square root of either $-e_0$ or $1-e_0$. It takes longest when $e_0$ is equal to $-1$ or $0$, at $x=N^2-N$ or $x=N^2$, and shortest when $x$ is slightly less than one of those values. And that's all I have to say on this one.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the textbook solution wrong by a sign? Laurent Series
Find the Laurent series of $\frac{e^z}{z^2 -1}$ about $z = 1$.
Here is my solution:
Factor denominator
$\frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$\frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$\frac{e^{w+1}}{w(w+2)}$ = $\frac{A}{w} + \frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $\frac{1}{e} = -2B$, $B = \frac{-1}
{2e}$
Thus our new equation is
$\frac{e}{2w} + \frac{-1}{2e(w+2)}$
Because the first term $\frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$\frac{-1}{4e} \cdot \frac{1}{1 - (\frac{-w}{2})}$
and so the Laurent series we get is
$$\frac{e}{2w} - \frac{1}{4e} \cdot \{1 - \frac{w}{2} + \frac{w^2}{4} - \frac{w^3}{8} ... \} $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?
| You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=e\cdot e^w=e \left(1+w+\frac{w^2}{2}+\frac{w^3}{6}+\dots\right).$$
Hence, after decomposing the rational function $\frac{1}{w(w+2)}$,
$$\frac{e^{w+1}}{w(w+2)}=e^{w+1}\left(\frac{1}{2w} - \frac{1/4}{1+w/2}\right)\\=
e\left(1+w+\frac{w^2}{2}+\frac{w^3}{6}+\dots\right)\left(\frac{1}{2w} - \frac{1}{4}\left(1-\frac{w}{2}+\frac{w^2}{4}+\dots\right)\right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$\frac{e}{2w} + \frac{e}{4} \left(1 + \frac{w}{2} + \frac{w^2}{12} +\dots\right).$$
| {
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Inequality. ${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$ Question. If ${{a} \over {1+a}}+{{b} \over {1+b}}+{{c} \over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that
$${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$$
My approach. If we let
$$x:=\frac{1}{1+a}, y:=\frac{1}{1+b}, z:=\frac{1}{1+c}$$
Then we can know that
$${{a} \over {1+a}}+{{b} \over {1+b}}+{{c} \over {1+c}}=2 \Leftrightarrow abc=a+b+c+2 \Leftrightarrow x+y+z=1$$ (By definition of x, y, z)
And,
$$a=\frac{1}{x}-1=\frac{1-x}{x}=\frac{y+z}{x} (\because x+y+z=1)$$
$$\therefore (a, b, c)=(\frac{y+z}{x}, \frac{z+x}{y}, \frac{x+y}{z})$$
T.S. $$\sum_{cyc}\frac{\sqrt{a}}{2}>\sum_{cyc}\frac{1}{\sqrt{a}}$$
$$ \Leftrightarrow \sum_{cyc}(\sqrt{a}-\frac{2}{\sqrt{a}}) \ge 0$$
$$ \Leftrightarrow \sum_{cyc}(\sqrt{\frac{y+z}{x}}-2\sqrt{\frac{x}{y+z}}) \ge 0$$
$$ \Leftrightarrow \sum_{cyc}\frac{(y-x)+(z-x)}{\sqrt{x(y+z)}}\ge 0$$
$$ \Leftrightarrow \sum_{cyc}(x-y)(\frac{1}{\sqrt{y(z+x)}}-\frac{1}{\sqrt{x(y+z)}})\ge 0$$
$$ \Leftrightarrow \sum_{cyc}(x-y)\frac{\sqrt{x(y+z)}-\sqrt{y(z+x)}}{\sqrt{xy(x+z)(y+z)}} \ge 0$$
But I don't know the next stage.
What should I do?
| Now, use $$\sqrt{x(y+z)}-\sqrt{y(x+z)}=\frac{x(y+z)-y(x+z)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}=\frac{z(x-y)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the recurrence relation $a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$, $a_0=0$, $a_1=1$ My task:
$a_n=6a_{n-1}-9a_{n-2}-8$ for $n\geq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-6x+9$
$\Delta=0$
$x_0=3 $
So I am gonna use following formula:
$a_n=ar^{n}+bnr^{n}$
$a_n=a*(3)^{n}+bn*3^{n}$
$-8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=6b_{n-1}-9b_{n-2}$
$$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+6c$$ I am setting $c=\frac{4}{3}$ so
$$b_n=6b_{n-1}-9b_{n-2}\implies\exists a,\,b:\,b_n=a*3^{n}+bn*3^{n}.$$From $b_0=\frac{4}{3},\,b_1=\frac{7}{3}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.
$$a=\frac{4}{3}$$
$$b=-\frac{5}{9}$$
$b_2=22$
$a_2=22-\frac{4}{3}=\frac{62}{3}$
Actual $a_2=-2$
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
Can I use this $b_n$ method if delta equals 0? or should $c=-\frac{4}{3}$?
| I think that your choice of $c$ is wrong
From $$b_n=6(b_{n-1}-c)-9(b_{n-2}-c)-8+c=6b_{n-1}-9b_{n-2}-8+4c$$
this gives you $c=2$ and $b_n=a_n+2$
$$b_0=2, b_1=3$$ after solving $$b_n=a*3^{n}+bn*3^{n}$$ you get $a=2 ,b=-1$
| {
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The set of all possible values of $AC$ is an open interval $(m,n)$
Side $\overline{AB}$ of $\triangle ABC$ has length $10$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20
Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $\frac{30}{x}$. Using triangular inequality, we get that $$7 < x+\frac{30}{x}$$ $$x<7+\frac{30}{x}$$ and $$\frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.
| I believe you have the correct range for $x$ of $2 \lt x \lt 10$, but that is for $\overline{BD}$. However, the question asks for $\overline{AC}$. Since $\overline{AC} = \frac{30}{x}$, this means it's range is $3 \lt \overline{AC} \lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
| {
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General solution to linear system with general form I am trying to find the general solution of the system
$$X' = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}X$$
where $a+d \not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong?
| We are given
$$X' = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}X\\\text{where}~ ~~a+d \ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - \lambda I| = 0$, yielding
$$\lambda_{1,2} = \frac{1}{2} \left(-\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right),\frac{1}{2} \left(\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$$
We can then find the associated eigenvectors by solving $[A-\lambda_i]v_i = 0$, yielding
$$v_1 = \begin{pmatrix}
-\dfrac{\sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\\1
\end{pmatrix}, v_2 = \begin{pmatrix} -\dfrac{-\sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\\1\end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d \not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$\lambda_1 = 0, v_1 = \begin{pmatrix} -\dfrac{d}{c} \\ 1 \end{pmatrix} \\ \lambda_2 = a + d , v_2 = \begin{pmatrix} \dfrac{a}{c} \\ 1 \end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{\lambda_1 t} ~v_1 + c_2 ~e^{\lambda_2 t}~ v_2 = c_1 \begin{pmatrix} -\dfrac{d}{c} \\ 1 \end{pmatrix} + c_2~ e^{(a+d)t} \begin{pmatrix} \dfrac{a}{c} \\ 1 \end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum involving $\ln{(2)}$ I got this sum. How do can this sum be equal to $8\ln{(2)}?$
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{35n-37}{(2n-1)(n-1)^2}+\frac{35n+37}{(2n+1)(n+1)^2}\right]=8\ln{(2)}$$
I have try to expand out the sum but it is too messy. Dealing the sum in this form, I haven't got any idea.
Any help.
| First note that
\begin{eqnarray*}
&&\frac{1}{n}\left[\frac{35n-37}{(2n-1)(n-1)^2}+\frac{35n+37}{(2n+1)(n+1)^2}\right]\\
&=&\frac{74}{n}+\frac2{(n+1)^2}-\frac2{(n-1)^2}+\frac{41}{n+1}+\frac{41}{n-1}-\frac{156}{2n+1}-\frac{156}{2n-1}
\end{eqnarray*}
and
\begin{eqnarray*}
&&\sum_{n=2}^{\infty}\frac{(-1)^n}{n-1}=\ln2,\sum_{n=2}^{\infty}\frac{(-1)^n}{n}=1-\ln2,\sum_{n=2}^{\infty}\frac{(-1)^n}{n+1}=-\frac12+\ln2,\\
&&\sum_{n=2}^{\infty}\frac{(-1)^n}{2n-1}=\frac{4-\pi}{4},\sum_{n=2}^{\infty}\frac{(-1)^n}{2n+1}=\frac{-8+3\pi}{12},\\
&&\sum_{n=2}^{\infty}(-1)^n\bigg[\frac2{(n+1)^2}-\frac2{(n-1)^2}\bigg]\\
&=&2\sum_{n=1}^{\infty}\bigg[\frac1{(2n+1)^2}-\frac1{(2n-1)^2}\bigg]-2\sum_{n=2}^{\infty}\bigg[\frac1{(2n)^2}-\frac1{(2n-2)^2}\bigg]\\
&=&-2+\frac12=-\frac32.
\end{eqnarray*}
Then you can put them to give the answer.
| {
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Simplifying $\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$ In simplifying
$$\frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
| $$\dfrac{2x^2-5x-3}{6x^3-2x^4}=\dfrac{(2x+1)(x-3)}{2x^3(3-x)}=\dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=\dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=\dfrac{2x+1}{-2x^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| $$2x\cos (2x)+(x^2-1)\sin 2x=(x^2+1)\bigg[\frac{2x}{x^2+1}\cos 2x+\frac{x^2-1}{x^2+1}\sin 2x\bigg]$$
$$=(x^2+1)\cos\bigg(2x-2\alpha\bigg)$$
$\displaystyle \sin(2\alpha)=\frac{x^2-1}{x^2+1}$ and $\displaystyle \cos(2\alpha)=\frac{2x}{x^2+1}$
integration $$=\int\sec\bigg(2x-\tan^{-1}\frac{x^2-1}{2x}\bigg)\frac{2x^2}{x^2+1}dx$$
put $\displaystyle 2x-\tan^{-1}\bigg(\frac{x^2-1}{2x}\bigg)=t$
And $\displaystyle \frac{2x^2}{x^2+1}dx=dt$
Integral $$=\int \sec(t)dt=\ln\bigg|\sec (t)+\tan (t)\bigg|+C$$
| {
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$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$ Show $\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
from LS
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(3/2) + \log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2\big(\frac{6(n+1)}{6}\big)}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
is this right?
| I would write $$\log_{2}\frac{3}{2}+\log_{2}\frac{2}{3}+\log_{2}(n+1)=\log_{2}(n+1)$$ since $$\log_{2}{\frac{2}{3}}=-\log_{2}{\frac{3}{2}}$$
| {
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"source": "stackexchange",
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Find Basis and Dimension of specific subspace of $M_{2 \times 2}$. Let $V=M_{2 \times 2}(\mathbb{R})$ be the set of all $2 \times 2$ real-valued matrices, and let the field $K = \mathbb{R}$. Then $V$ is a vector space under matrix addition and Scaler multiplication.
Let $$A = \begin{bmatrix}1&0\\1&2\end{bmatrix}$$
If $C(A) = \{B \mid B \in V$, such that $AB=BA$} find the dimension of the subspace $C(A)$ and determine a basis.
First of all I believe I would have to prove this is a subspace. So I would have to show that the subspace is closed under vector addition and scaler multiplication. Then I would have to construct a basis. The number of vectors in the basis is the dimension of the subspace.
It is the condition that is tripping me up. How do show all this with the condition that these $2 \times 2$ matrices are commutative?
| To show that $C(A)$ is a subspace, we need to show it closed under both addition and scalar multiplication; so let
$B_1, B_2 \in C(A); \tag 1$
then
$B_1A = AB_1, \; B_2A = AB_2; \tag 2$
thus,
$(B_1 + B_2)A = B_1A + B_2A = AB_1 + AB_2 = A(B_1 + B_2), \tag 3$
which of course implies
$B_1 + B_2 \in C(A); \tag 4$
likewise if $\alpha$ is any scalar, we have
$(\alpha B)A = \alpha (BA) = \alpha (AB) = A(\alpha B); \tag 5$
thus
$\alpha B \in C(A) \tag 6$
as well. Also, it is pretty easy to see that (6) implies
$0 \in C(A), \tag 7$
and
$B \in C(A) \Longleftrightarrow -B \in C(A); \tag 8$
since the rest of the vector space axioms are inherited by $C(A)$ from $M_{2 \times 2}(\Bbb R)$, it follows that $C(A)$ is indeed a subspace.
So, what do the elements of $C(A)$ look like? If
$B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}, \tag 9$
then the condition
$AB = BA \tag{10}$
reads, with
$A = \begin{bmatrix}1&0\\1&2\end{bmatrix}, \tag{10}$
$\begin{bmatrix}1&0\\1&2\end{bmatrix}\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}\begin{bmatrix}1&0\\1&2\end{bmatrix}; \tag{11}$
at this point, before proceeding further, we observe that the computations specified in (11) may be considerably simplified if we a priori write $A$ in the form
$A = \begin{bmatrix}1&0\\1&2\end{bmatrix} = I + \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \tag{12}$
since
$IB = BI; \tag{13}$
we are left with finding those $B$ such that
$ \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}; \tag{14}$
that is,
$\begin{bmatrix} 0 & 0 \\ b_{11} + b_{21} & b_{12} + b_{22} \end{bmatrix} = \begin{bmatrix} b_{12} & b_{12} \\ b_{22} & b_{22} \end{bmatrix}, \tag{15}$
we thus find that
$b_{12} = 0, \; b_{11} + b_{21} = b_{22} = b_{12} + b_{22}; \tag{16}$
by virtue of these equations, we see we may write $B$ in the form
$B = \begin{bmatrix} b_{11} & 0 \\ b_{22} - b_{11} & b_{22} \end{bmatrix} = b_{11} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} + b_{22} \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}; \tag{17}$
it is now clear that we may take $b_{11}$ and $b_{22}$ as free parameters, and that $C(A)$ is two dimensional, being spanned by matrices
$\begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}, \; \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \tag{18}$
which form a basis for $C(A)$.
| {
"language": "en",
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Proving $\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$
I would like to prove using Mean Value theorem for $n \ge 1$
$$\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$$
RHS can be proved by rationalizing the square root term, not sure about the LHS.
| Why MVT? For the LHS you are asking about you have
\begin{eqnarray*} \sqrt{n^2+n} -n
& = & \frac{n^2 + n - n^2 }{\sqrt{n^2+n} + n} \\
& \color{blue}{>} & \frac{n}{\sqrt{n^2+2n+1}+n} \\
& = & \frac{n}{\sqrt{(n+1)^2}+n} \\
& = & \frac{n}{2n+1} \\
\end{eqnarray*}
| {
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} |
find the sum of$\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!}$ $\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!} = \sum_{n=1}^{\infty } \frac{4^{n}}{ \left ( 2n+1 \right )!B\left ( n+1,n+1 \right ) \left ( n+1 \right ) \left ( n+2 \right )}$
Given, $F(n) = \frac{4^{n}}{ \left ( 2n+1 \right )!B\left ( n+1,n+1 \right ) }$
then, $\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!} = \sum_{n=1}^{\infty } \frac{F\left ( n \right )}{ \left ( n+1 \right )\left ( n+2 \right ) } = \frac{F(1)}{2} + \frac{F(2) - F(1)}{3} + \frac{F(3) - F(2)}{4} + \frac{F(4) - F(3)}{5} + \cdots \cdots $
finally, I couldn't solve it. Please guide me how to deal with this problem. Thank you in advance.
| The sum is $$-\frac{1}{2}+\frac{{I}_{2}(4)}{4}$$
where $I_2$ is a modified Bessel function of order $2$. More generally,
$$ \sum_{n=0}^\infty \frac{x^n}{n!(n+k)!} = \frac{I_k(2 \sqrt{x})}{x^{k/2}}$$
for nonnegative integers $k$.
| {
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How can I simplify $\sqrt{\frac{5+\sqrt{5}}{2}}$? I've tried to see the root as
$\sqrt{\frac{5+\sqrt{5}}{2}} = \sqrt{a}+\sqrt{b},$
but this method doesn't give me something good.
| We can write it so:
$$\sqrt{\frac{5+\sqrt{5}}{2}}=\sqrt{\frac{5+\sqrt{5}}{2}}+\sqrt0.$$
Let there be rationals $a$ and $b$ for which $\sqrt{\frac{5+\sqrt{5}}{2}}=\sqrt{a}+\sqrt{b}.$
Thus, $$\frac{5+\sqrt5}{2}=a+b+2\sqrt{ab},$$ which gives
$$a+b=\frac{5}{2}$$ and
$$ab=\frac{5}{16},$$ which says that $a$ and $b$ are roots of the equation
$$x^2-\frac{5}{2}x+\frac{5}{16}=0,$$ which is a contradiction because it is easy to see that this equation has no rational roots.
| {
"language": "en",
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What is the area of $\triangle ABC$ where $\triangle ADC$ is cyclic, point $P$ is on the circumference and $AD = AP$?
$\triangle ABC$ is a right angled triangle. The perpendicular drawn form $A$ on $BC$ intersects $BC$ at point $D$. A point $P$ is chosen on the circle drawn through the vertices of $\triangle ADC$ such that $CP$ $\perp$ $BC$ and $AP$ = $AD$.A square is drawn on the side $BP$ and its area is 350 unit$^\text{2}$. What is the area of $\triangle ABC$?
My Attempt:
Here, $ADCP$ is a square because $AD = AP$ and $\angle ADC$ = $\angle DCP$ = 90$^\circ$. So, by expressing $AD = x$ and $BD$ = $y$,
From the right angled $\triangle ADB$,
$AB^\text{2}$ = $x^\text{2} + y^\text{2}$
And now, from the $\triangle ABC$,
$x^\text{2} + y^\text{2}$ + ($\sqrt 2 x$)$^\text{2}$ = $(x + y)$$^\text{2}$.....($AC$ is the diagonal of the square $ADCP$)
$x^\text{2}$ + $y^\text{2}$ + 2x$^\text{2}$ = $x^\text{2}$ + $y^\text{2}$ + 2$xy$ $\implies$ $x$ =$y$
So, $BD$ = $x$ and $AB$ = $\sqrt(2x^\text{2})$ $\implies$ $AB$ = $\sqrt2 x$. And now from $\triangle BCP$,
($2x$)$^\text{2}$ + $x$$^\text{2}$ = ($\sqrt350$)$^\text{2}$
$5x$$^\text{2}$ = $350$ $\implies$ $x$$^\text{2}$ = $70$ $\implies$ $x$ = $\sqrt70$
After that, the area of $\triangle ABC$ = $\frac{1}{2}$×$\sqrt2x$×$\sqrt2x$
= $\frac {1}{2}$×$2x$ $^\text{2}$ = $x^\text{2}$ = ($\sqrt70$)$^\text{2}$ = $70$.
Is my answer correct? If not so, can anyone please provide me with another solution or method for better learning process? Or simply any kind of clue or hint will be so much helpful. Thanks in advance.
| Since $AC$ is the diagonal of square $APCD$, then $\angle{C}=45^{\circ}$. Then, we know $\triangle{ABC}$ is a 45-45-90 right triangle. Notice that $ABC$ is isosceles, so $BD=DC=CB=x$. Then, you can use Pythagorean Theorem to find $x$. Then, it is easily seen that $[ABC]=\frac{(\sqrt{2\cdot 70})^2}{2}=70$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Constrained (and non) extrema of $f(x,y)=2x^2-2xy^3+3y^2$ I need to find the critical points on the boundary,inside $D$,outside $D$, and find the image of this function (constrained on $D$).
$f(x,y)=2x^2-2xy^3+3y^2$
$D=\{2x^2+3y^2\le 9\}$
Critical points non-constrained:
$f_x=4x-2y^3=0$
$f_y=-6xy^2+6y=0$ --> $6y(-xy+1)=0$
$f_y$ is equal to $0$ if $y=0$ or $x=\frac{1}{y}$, plugging it in $f_x$ brings to these critical points : $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$
using the second derivate test: $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$ seem to be saddle points (?).
Constrained Critical points (extremas):
$$
\left\{
\begin{array}{c}
4x-2y^3=4x\lambda\\
-6xy^2+6y=6y\lambda\\
2x^2+3y^2-9=0
\end{array}
\right.
$$
$-6xy^2+6y=6y\lambda$ --> $6y(-x+1-\lambda)=0$ from this one I can see that it nullifies when $y=0$ or $x=1-\lambda$.
If $y=0$ the first equation gets to $4x(1-\lambda)=0$ which means that it nullifies when $x=0$ or $\lambda = 1$. Looking at the third equation tells us that $(x,y)=(0,0)$ can't be used;
Plugging $y=0$ in the third equation : possible critical points $(\pm \frac{3}{\sqrt(2)},0)$
What I can say about $(\pm \frac{3}{\sqrt(2)},0)$ ?
I think I missed some points.
| Critical points non-constrained:
$$(0,0),\quad \left(\pm2^{-1/4}, \pm2^{1/4}\right)$$
Global minimum:
$$f_{min}=f(0,0)=0$$
Saddle points:
$$f\left(\pm2^{-1/4}, \pm2^{1/4}\right)=2\sqrt2$$
Constrained Critical points: from system
$$\left\{
\begin{array}{c}
4x-2y^3=4x\lambda\\
-6xy^2+6y=6y\lambda\\
2x^2+3y^2-9=0
\end{array}
\right.$$
eliminate $\lambda$. We get
$$
\left\{
\begin{array}{c}
(2x^2-y^2)y^2=0\\
2x^2+3y^2-9=0
\end{array}
\right.
$$
Solutions is
$$\left(\pm\frac{3}{\sqrt2},0\right),\;
\left(\frac{3}{2\sqrt2},-\frac32\right),\;
\left(-\frac{3}{2\sqrt2},\frac32\right),\;
\left(\frac{3}{2\sqrt2},\frac32\right),\;
\left(-\frac{3}{2\sqrt2},-\frac32\right)
$$
Global maxima:
$$f_{max}=f\left(\frac{3}{2\sqrt2},-\frac32\right)=f\left(-\frac{3}{2\sqrt2},\frac32\right)=9+\frac{81}{{{2}^{\frac{7}{2}}}}\approx16.159456$$
Local minimum:
$$f_{min}=f\left(\frac{3}{2\sqrt2},\frac32\right)=f\left(-\frac{3}{2\sqrt2},-\frac32\right)=9-\frac{81}{{{2}^{\frac{7}{2}}}}\approx1.84054384$$
Saddle points:
$$f\left(\pm\frac{3}{\sqrt2},0\right)=9$$
WolframAlpha return in this case "local maxima".
Other method: parametric equations of ellipse is $x=\frac{3 \cos{(\phi)}}{\sqrt{2}},y=\sqrt{3} \sin{(\phi)}$. We get
$$f=9-9 \sqrt{6} \cos{(\phi)} {{\sin{(\phi)}}^{3}}$$
with critical points on $[0, 2\pi]$:
$$0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi }{3}$$
| {
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"url": "https://math.stackexchange.com/questions/3108567",
"timestamp": "2023-03-29T00:00:00",
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How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6$ in $(x+(1+i))^{10}(x+(1-i))^{10}$, then dividing by $2^{10}$.
Then, we find that the coefficient of $x^6$ would be:
$$\sum_{i=0}^{6} \binom{10}{6-i} \binom{10}{i} (1-i)^{10-(6-i)} (1+i)^{10-i}$$
with the knowledge that $(1-i)(1+i)=2$, I simplified to
$$\binom{10}{6}\binom{10}{0}2^4((1-i)^6+(1+i)^6)+\binom{10}{5}\binom{10}{1}2^5((1-i)^4+(1+i)^4)+\binom{10}{4}\binom{10}{2}2^6((1-i)^2+(1+i)^2)+\binom{10}{3}\binom{10}{3}2^7$$
Note: the formula $(1+i)^x+(1-i)^x$ gives:
$ 2(2^{\frac{x}{2}}) \cos(\frac{x\pi}{4})$
After simplifying and reapplying the division by $2^{10}$, I get $(\frac{0}{1024}) + (-8)(2520)(\frac{32}{1024}) + (\frac{0}{1024}) + (120)(120)(2^7)$, which gives $0-630+0+1800,$ which is 1170, and I checked this over with an expression expansion calculator.
If the original equation was $(1+x+x^2)^{10}$, I would have used binomials to find the answer, however, the $x^2$ was replaced by $\frac{x^2}{2}$.
My question is whether anyone has a combinatorics solution to this question, rather than just algebra. It would be nice if the solution did not require complex numbers.
| Write the expression as $\frac{1}{2^{10}}\left((x+1)^2+1\right)^{10}$.
Hint if you would like to do it using combinatorial arguments : Coefficient of $x^6$ in $(1+x^2)^{n}$ would give you the number of tuples such that $a_1 + a_2 + \cdots a_{n} = 6$ such that $a_i\in \{0,2\}$. This should be easy- just note that any three of the $a_i$s should be $2$ which gives $n\choose 2$ as the answer to the example here.
To add to the hint, note that if you write $x+1=t$, then you need to calculate the coefficients of $x^6$ which you get directly from the coeff of $t^6$ and another from the coeff of $t^4$ because $t^4 = (1+x)^4.
| {
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If $\cos\theta=\frac{\cos\alpha+\cos \beta}{1+\cos\alpha\cos\beta}$, then prove that one value of $\tan(\theta/2)$ is $\tan(\alpha/2)\tan(\beta/2)$
If $$\cos\theta = \frac{\cos\alpha + \cos \beta}{1 + \cos\alpha\cos\beta}$$ then prove that one of the values of $\tan{\frac{\theta}{2}}$ is $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$.
I don't even know how to start this question. Pls help
| I think i have found out how to approach it. I hope this proof is satifactory.
Using the half angle formula,
$$\tan{\frac{\theta}{2}} = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \longrightarrow \text{eq.1}$$
Evaluating $\frac{1 - \cos\theta}{1 + \cos\theta}$ first,
$$\frac{1 - \cos\theta}{1 + \cos\theta} = \frac{1 - (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}{1 + (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}\text{ }[\text{since } \cos\theta= \frac{\cos\alpha +\cos\beta}{1 + \cos\alpha\cos\beta}]$$
$$=\frac{1+\cos\alpha\cos\beta - \cos\alpha -\cos\beta}{1+\cos\alpha\cos\beta + \cos\alpha+ \cos\beta}$$
$$=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}$$
Substituting this value of $\frac{1 - \cos\theta}{1 + \cos\theta}$ into equation 1,
$$\tan{\frac{\theta}{2}} = \sqrt{\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}}$$
$$=\sqrt{\frac{(1-\cos\alpha)^2(1-\cos\beta)^2}{(1-\cos^2\alpha)(1-\cos^2\beta)}}$$
$$=\pm\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta}$$
Taking the positive value of $\tan{\frac{\theta}{2}}$,
$$\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta} = \frac{4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}}{4\sin\frac{\alpha}{2}\cos{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}} \text{(using half angle and double angle formula)}$$
$$=\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$$
Therefore, $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$ is one of the values of $\tan{\frac{\theta}{2}}$.
| {
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How to find $\lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x}$ when $x\to 0^+$ and when $x\to 0^-$? I'm trying to find:
$$ \lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x} $$
Since there is a discontinuity at $x=0$ I know that I have to take the limits from both sides, $x \to 0^+$ and $x \to 0^-$, and check if they're equal.
If I factor it I get:
$$ \lim \limits_{x \to 0} \left(\frac{\sqrt{x+4}} {x-1}\right) = - 2$$
Is this the same as $x \to 0^+$?
If so, how do I approach the problem for $x \to 0^-$?
If not, how do I do I do it from both sides?
| Limit from right side is
$
\lim \limits_{x \to 0^+} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
= \lim \limits_{x \to 0^+} \left(\frac{ |x| \sqrt{x+4}} { x(x-1) }\right) \\
= \lim \limits_{\delta \to 0} \left(\frac{ |0+\delta| \sqrt{ (0+\delta) +4}}{ (0+\delta)( (0+\delta) -1 ) }\right) \ [ \ \text{substituting} \ x = 0 + \delta \ , \delta > 0 \ ] \\
= \lim \limits_{\delta \to 0} \left(\frac{ \delta \sqrt{ \delta+4}}{ \delta (\delta-1) }\right) \\
= -2
$
Limit from left side is
$
\lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
= \lim \limits_{x \to 0^-} \left(\frac{ |x| \sqrt{x+4}} { x(x-1) }\right) \\
= \lim \limits_{\delta \to 0} \left(\frac{ |0-\delta| \sqrt{ (0-\delta) +4}}{ (0-\delta)( (0-\delta) -1 ) }\right) \ [ \ \text{substituting} \ x = 0 - \delta \ , \delta > 0 \ ] \\
= \lim \limits_{\delta \to 0} \left(\frac{ -\delta \sqrt{ 4 - \delta }}{ \delta (-1 - \delta) }\right) \\
= 2
$
$
\therefore \
\lim \limits_{x \to 0^+} \frac{\sqrt{x^3+4x^2}} {x^2-x}
\neq \lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
\Rightarrow \lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x} \ \text{does not exist}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of the shaded region of two circle with the radius of $r_1$ and $r_2$ In the given figure , $O$ is the center of the circle and $r_1 =7cm$,$r_2=14cm,$ $\angle AOC =40^{\circ}$. Find the area of the shaded region
My attempt: Area of shaded region $=\pi r^2_2 - \pi r^2_1= \pi( 196-49)= 147\pi$
Is it true ?
| Use the formula for the area of a sector (the angle is measured in radians) and the formula for the area of a circle:
$$
A=\frac{1}{2}r^2\theta
$$
$$
A=\pi r^2
$$
$40^\circ$ in radians would be:
$$
40^\circ=\frac{40\pi}{180}
$$
The area of the top peice:
$$
A_1=\frac{1}{2}r_2^2\cdot \frac{40\pi}{180} - \frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{\pi}{9}\left(r_2^2-r_1^2\right)
$$
The area of the bottom piece:
$$
A_2=\pi r_1^2-\frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{8\pi}{9}r_1^2
$$
Thus, the area of the shaded region would be:
$$
A_1+A_2=\frac{\pi}{9}\left(r_2^2-r_1^2\right)+\frac{8\pi}{9}r_1^2=\frac{\pi}{9}\left(r_2^2+7r_1^2\right)=\frac{\pi}{9}\left(14^2+7\cdot7^2\right)=\frac{539\pi}{9}
$$
Answer: $A=\frac{539\pi}{9}$ square units.
| {
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Determinant of a particular matrix. What is the best way to find determinant of the following matrix?
$$A=\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right)$$
I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.
| Note that
$$
\det\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right) =\det \left(\begin{matrix}
1&ax&x^2\\1&ay&y^2\\ 1&az&z^2
\end{matrix}\right)=a\cdot\det \left(\begin{matrix}
1&x&x^2\\1&y&y^2\\ 1&z&z^2
\end{matrix}\right)
$$ which boils down to Vandermonde determinant
$$
a(x-y)(y-z)(z-x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that the product of any two numbers between two consecutive squares is never a perfect square In essence, I want to prove that the product of any two distinct elements in the set $\{n^2, n^2+1, ... , (n+1)^2-1\}$ is never a perfect square for a positive integers $n$. I have no idea on how to prove it, but I've also yet to find a counterexample to this statement. Can anyone help?
| For any two numbers $n^2+a,\ n^2+b;\ 0<a<b<(2n+1)$, their product will satisfy $n^4<n^4+(a+b)n^2+ab<(n^2+2n+1)^2$.
All of the squares between $n^4$ and $(n^2+2n+1)^2$ will have the form $(n^2+m)^2=n^4+2mn^2+m^2;\ 1\le m\le 2n$
If $n^4+(a+b)n^2+ab$ is a perfect square, it will be one of the squares between $n^4$ and $(n^2+2n+1)^2$ and hence equal to $n^4+2mn^2+m^2$ for some $m$.
Thus $(a+b)=2m;\ ab=m^2$ for some $m$
Rearranging, we get $m^2=\frac{a^2+2ab+b^2}{4}=ab=\frac{4ab}{4}$, or $a^2+b^2=2ab$.
This implies both $a\mid b$ and $b\mid a$, meaning $b=a$ and the numbers being multiplied to obtain a perfect square are not distinct.
Added by edit: John Omielan comments (for the specific case $k=1$) that my original answer fails to consider possible solutions of the form $a+b=2m-k;\ ab=m^2+kn^2$. He separately provides a more complete answer that addresses those cases. Marty Cohen comments that I can only properly conclude $n^4+(a+b)n^2+ab=n^4+2mn^2+m^2 \Rightarrow n^2(a+b-2m)=(m^2-ab)$. Let me address those shortcomings.
If $(n^2+a)(n^2+b)=(n^2+m)(n^2+m)$ then either $a=b=m$ (addressed in my original answer) or $a<m<b$ (which this edit will address). If $n^2(a+b-2m)=(m^2-ab)$, then $(m^2-ab)$ is divisible by $n^2$, so $(m^2-ab)=kn^2=n^2(a+b-2m)$, or $(a+b-2m)=k\Rightarrow a+b=2m+k$.
$m,a,b$ have limits on their sizes. $m<b\le 2n\Rightarrow m^2< 4n^2$. Also $a<b\le2n \Rightarrow ab<4n^2$ Hence, $|(m^2-ab)|<4n^2 \Rightarrow |k|=(a+b-2m)<4$. For $(a+b)=2m+k,\ |k|=0,1,2,3$, where $k=0$ corresponds to the case where $a=b=m$.
The midpoint $t$ between $a$ and $b$ is the average $t=\frac{a+b}{2}$. Let $b-t=r,\ t-a=r$. Note that if $a$ and $b$ have different parity, $t$ and $r$ may have half integral values. Finally, $b-a=2r$, but since $b\le2n,\ a\ge 1$, then $2r=b-a<2n\Rightarrow r<n$.
$(n^2+a)(n^2+b)=((n^2+t)-r)((n^2+t)+r)=(n^2+t)^2-r^2$. Therefore, unless $m>t$, $(n^2+a)(n^2+b)<(n^2+m)^2$. But $t=\frac{a+b}{2}=m-\frac{k}{2}$. $m-t=\frac{1}{2},1,\frac{3}{2}$. Within the constraints of the original question, $m$ must be larger than the average of $a$ and $b$, but it must be very close to that average.
Substituting $t-\frac{k}{2}$ for $m$ in $(n^2+m)^2$, we get $((n^2+t)-\frac{k}{2})^2$. Can that equal $(n^2+a)(n^2+b)=((n^2+t)-r)((n^2+t)+r)=(n^2+t)^2-r^2$? Letting $s=(n^2+t)$ for simplicity in keeping track during expansion, we ask whether $(s-\frac{k}{2})^2=(s-r)(s+r)=s^2-r^2\Rightarrow ks-\frac{k^2}{4}=r^2\Rightarrow k(n^2+t)-\frac{k^2}{4}=r^2$?
$r<n\Rightarrow r^2<n^2$. It is obvious for $3\ge k>1$ that $k(n^2+t)-\frac{k^2}{4}>n^2>r^2$. For $k=1$, $n^2+t-\frac{1}{4}>n^2 \iff t>\frac{1}{4}$. For $a<m<b$, $\min(a+b)=4 \Rightarrow \min(t)=2; 2>\frac{1}{4}$.
There are no solutions to the question for $a<m<b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Limiting value of a sequence when n tends to infinity Q) Let, $a_{n} \;=\; \left ( 1-\frac{1}{\sqrt{2}} \right ) ... \left ( 1- \frac{1}{\sqrt{n+1}} \right )$ , $n \geq 1$. Then $\lim_{n\rightarrow \infty } a_{n}$
(A) equals $1$
(B) does not exist
(C) equals $\frac{1}{\sqrt{\pi }}$
(D) equals $0$
My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.
So, I tried like this simple way to substitute values and trying to find the limiting value :-
$\left ( 1-\frac{1}{\sqrt{1+1}} \right ) * \left ( 1-\frac{1}{\sqrt{2+1}} \right )*\left ( 1-\frac{1}{\sqrt{3+1}} \right )*\left ( 1-\frac{1}{\sqrt{4+1}} \right )*\left ( 1-\frac{1}{\sqrt{5+1}} \right )*\left ( 1-\frac{1}{\sqrt{6+1}} \right )*\left ( 1-\frac{1}{\sqrt{7+1}} \right )*\left ( 1-\frac{1}{\sqrt{8+1}} \right )*.........*\left ( 1-\frac{1}{\sqrt{n+1}} \right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$\left ( \frac{\sqrt{2}-1}{\sqrt{2}} \right )*\left ( \frac{\sqrt{3}-1}{\sqrt{3}} \right )*\left ( \frac{\sqrt{4}-1}{\sqrt{4}} \right )*.......\left ( \frac{\sqrt{(n+1)}-1}{\sqrt{n+1}} \right )$
= $\left ( \frac{(\sqrt{2}-1)*(\sqrt{3}-1)*(\sqrt{4}-1)*.......*(\sqrt{n+1}-1)}{{\sqrt{(n+1)!}}} \right )$
Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.
| As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.
If we multiply $a_n$ by $b_n:=\left(1+\frac1{\sqrt 2}\right)\cdots \left(1+\frac1{\sqrt {n+1}}\right)$, note that the product of corresponding factors is $\left(1-\frac1{\sqrt {k}}\right)\left(1+\frac1{\sqrt {k}}\right)=1-\frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms
$$b_n\ge 1+\frac1{\sqrt 2}+\frac1{\sqrt 3}+\ldots+\frac1{\sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_n\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Wave Equation with Initial Conditions on Characteristic Curves I am trying to solve the initial value problem:
$$\begin{cases}
u_{tt}-u_{xx} =0\\
u|_{t = x^2/2} = x^3, \quad |x| \leq 1 \\
u_{t}|_{t = x^2 / 2} = 2x, \quad |x| \leq 1 \\
\end{cases}
$$
I'm unsure if we can use D'Alambert's formula in this problem, so what I have attempted is just using the general solution and going from there: $u(x, t) = f(x - t) + g(x + t)$, means (after plugging in initial conditions): $$f(x - \frac{x^2}{2}) + g(x + \frac{x^2}{2}) = x^3, \ \mathrm{and}$$ $$-f'(x - \frac{x^2}{2}) + g'(x + \frac{x^2}{2}) = 2x.$$
I am unsure where to go from here, or if my approach is correct. Could someone help me out?
| The general solution is $u(x,t)=f(x-t)+g(x+t)$
$u|_{t=\frac{x^2}{2}}=x^3$ :
$f\left(x-\dfrac{x^2}{2}\right)+g\left(x+\dfrac{x^2}{2}\right)=x^3......(1)$
$u_t(x,t)=f_t(x-t)+g_t(x+t)=g_x(x+t)-f_x(x-t)$
$u_t|_{t=\frac{x^2}{2}}=2x$ :
$g_x\left(x+\dfrac{x^2}{2}\right)-f_x\left(x-\dfrac{x^2}{2}\right)=2x$
$g\left(x+\dfrac{x^2}{2}\right)-f\left(x-\dfrac{x^2}{2}\right)=x^2+c......(2)$
$\therefore f\left(x-\dfrac{x^2}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(x+\dfrac{x^2}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{x^2-2x}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(\dfrac{x^2+2x}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{(x-1)^2-1}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(\dfrac{(x+1)^2-1}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{(x+1)^3-(x+1)^2-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{(x-1)^3+(x-1)^2+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{x^3+2x^2+x-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{x^3-2x^2+x+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{x(x+1)^2-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{x(x-1)^2+c}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $\frac{4^{-2}x^3y^{-3}}{2x^0}$ to $\frac{x^3}{32y^3}$ I am to simplify $\frac{4^{-2}x^3y^{-3}}{2x^0}$ and I know that the solution is $\frac{x^3}{32y^3}$
I understand how to apply rules of exponents to individual components of this expression but not as a whole.
For example, I know that $4^{-2}$ = $\frac{1}{4^2}$ = $1/16$
But how can I integrate this 1/16 to the expression?
Do I remove the original $4^{-2}$ and replace with 1 to the numerator and a 16 to the denominator like this?
$\frac{1x^3y^{-3}}{16*2x^0}$
How can I simplify the above expression to $\frac{x^3}{32y^3}$? Would be grateful for a granular set of in between steps, even if they are most basic to others.
| Do it step by step :
First, as you said $4^{-2} = \frac{1}{16}$, replace it in the given expression :
$$ \dfrac{4^{-2}x^3y^{-3}}{2x^0} = \dfrac{x^3y^{-3}}{16\cdot 2 x^0}$$
Then, simplify the denominator, $16\cdot 2 = 32$ and $x^0 = 1$ so that :
$$\dfrac{4^{-2}x^3y^{-3}}{2x^0} = \dfrac{x^3\cdot y^{-3}}{32} $$
Then, because $y^{-3} = \dfrac{1}{y^3}$, you can find the final expression :
$$ \dfrac{4^{-2}x^3y^{-3}}{2x^0} = \dfrac{x^3}{32y^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \, + \, \sqrt{x+6} = 9 \, - \, \sqrt{x-1} $$
On Squaring both sides, we get:
$$(x-6) \, + \, (x+6) + 2 \,\,. \sqrt{(x^2-36)} = 81 + (x-1)\, - 18.\, \sqrt{x-1}$$
$$\implies 2x + 2 \,\,. \sqrt{(x^2-36)}= 80 + x - 18.\, \sqrt{x-1}$$
$$\implies x + 2 \,\,. \sqrt{(x^2-36)}= 80 - 18.\, \sqrt{x-1} \tag{ii}$$
Again we are getting equation in radical form.
But, in Wolfram app, I am getting its answer as $x=10$, see it in WolframAlpha.
So, how to solve this equation? Please help...
| I'm not entirely sure if this approach will find all possible $x$ in any such question, nevertheless here it is giving the correct answer and is relatively short.
Since the RHS $=9$ is an integer, all three terms in the RHS under the radicals must be perfect squares. $$\therefore, x-6=p^2$$
$$x-1=q^2$$ $$x+6=r^2$$
where $p,q,r$ are integers. $$\Rightarrow x=p^2+6=q^2+1=r^2-6$$
Here we need value of only one of $p,q,r$ to find $x$
$\Rightarrow (p+q)(p-q)=-5$
$\Rightarrow (p+q),(p-q)=(-1)(5), (1,-5), (5,-1), (-5,1)$
$\Rightarrow p=±2$
Since $x=p^2+6$,
$$\therefore \fbox{x=10}$$
| {
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"url": "https://math.stackexchange.com/questions/3130395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for these polynomials? I have a recurrence relation for these polynomials $p_i(x) $:
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
I've been unable to find a closed form, but the recurrence is simple so it seems there might be one. However the polynomials arise in a deep context, so maybe not. Is it possible to find one?
The context:
Let $G=\mathbb Z_2*\mathbb Z_2$, with generators $s, t$. Let $M$ be the free module over the ring $\mathbb Z[c, d] $ of rank $2$ with basis $x_1,x_2$. Define
\begin{align}
s(x_1)&=-x_1 \\
t(x_2)&=-x_2 \\
s(x_2)&=cx_1+x_2 \\
t(x_1)&=x_1+dx_2 \\
\end{align}
Define elements
\begin{align}
y_0&=x_1 \\
y_{2i+1}&=t(y_{2i}) \\
y_{2i}&=s(y_{2i-1}) \\
\end{align}
Then
$$y_i=p_{2\lfloor i/2\rfloor+1}(cd)x_1+dp_{2\lceil i/2\rceil}(cd)x_2$$
| Writing $v_n=\begin{pmatrix} p_{2n+1} \\ p_{2n}\end{pmatrix}$, the recurrence can be written as $v_{n+1}=Av_n$ where $$A=\begin{pmatrix} x-1 & -x \\ 1 & -1\end{pmatrix}.$$ So, we have $v_n=A^nv_0$ for all $n$, and the question is just about computing powers of the matrix $A$. This can be done by diagonalizing: the characteristic polynomial of $A$ is $q(t)=t^2-(x-2)t+1$. Picking a root $r$ of $q$ over an algebraic closure of $\mathbb{Q}(x)$, then $r$ and $s=x-2-r$ are the eigenvalues of $A$ with eigenvectors $\begin{pmatrix}r+1\\ 1\end{pmatrix}$ and $\begin{pmatrix}s+1\\ 1\end{pmatrix}$. So we can write $$A=B\begin{pmatrix} r & 0 \\ 0 & s\end{pmatrix}B^{-1}$$ where $B=\begin{pmatrix} r+1 & s+1 \\ 1 & 1\end{pmatrix}$. If I have not screwed up the calculation, this gives the following "explicit" formula for $A^n$: $$A^n=\frac{1}{r-s}\begin{pmatrix} r^n(r+1)-s^n(s+1) & (r+1)(s+1)(s^n-r^n) \\ r^n-s^n & s^n(r+1) - r^n(s+1)\end{pmatrix}.$$ Applying this to $v_0=\begin{pmatrix} 1 \\ 0\end{pmatrix}$ we get $$p_{2n+1}=\frac{r^n(r+1)-s^n(s+1)}{r-s}$$ and $$p_{2n}=\frac{r^n-s^n}{r-s}.$$ Note that by the quadratic formula, we can write $$r=\frac{x-2+\sqrt{x^2-4x}}{2}$$ and $$s=\frac{x-2-\sqrt{x^2-4x}}{2}$$ and so substituting these into the formulas above gives a closed form for $p_{2n+1}$ and $p_{2n}$ in terms of $x$. Or, for an explicit formula without any radicals, we can expand out $r^n$ and $s^n$ to get $$p_{2n}=\frac{1}{2^{n-1}}\sum_{0\leq 2k<n}\binom{n}{2k+1}(x^2-4x)^k(x-2)^{n-2k-1}$$ and then $p_{2n+1}$ is just $p_{2n}+p_{2n+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the number of ways to express 1050 as sum of consecutive integers I have to solve this task:
Find the number of ways to present $1050$ as sum of consecutive
positive integers.
I was thinking if factorization can help there:
$$1050 = 2 \cdot 3 \cdot 5^2 \cdot 7 $$
but I am not sure how to use that information (if there is a sense)
example
I can solve something similar but on smaller scale:
\begin{align} 15 &= 15 \\ &= 7+8 \\ &=4+5+6 \\ &= 1+2+3+4+5 \end{align}
($4$ ways)
| For the sum of next integer we may use formula for the sum of arithmetic sequence
$$\sum_{k=1}^na_k=\frac n2(a_1+a_n)$$
So
\begin{aligned}
1050 &= \frac n2(a_1+a_n) \\
2100 &= n(a_1+a_n) \\
2100&= n(a_1+a_n) \\
2100&= n(a_1+a1+(n-1)) \\
2^2\cdot3\cdot5^2\cdot7&= n(2a_1+n-1)
\end{aligned}
Now:
*
*If $n$ is even, then $(2a_1+n-1)$ is odd, so
$$n=2^2\cdot3^x\cdot5^y\cdot7^z$$
where $x \in \{0,1\},\; y \in \{0,1,2\}, \;z \in \{0,1\}$,
so there are $2 \times 3 \times 2 = 12$ possibilities for $n$.
*If $n$ is odd, then similarly
$$n=3^x\cdot5^y\cdot7^z$$
and we obtain other $12$ possibilities for $n$.
So there are $24$ solutions altogether, a half o them, i. e. $\color{red}{12}$, for only positive integers, because for positive integers must be $a_1 \ge 1$, and consequently $(2a_1+n-1) >n$, so in the product $n(2a_1+n-1)$ the first multiplier have be smaller than the second one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Find all positive integers $a, b, c$ such that $21^a+ 28^b= 35^c$ .
Find all positive integers $a, b, c$ such that $21^a+ 28^b= 35^c$.
It is clear that the equation can be rewritten as follows:
$$
(3 \times 7)^a+(4 \times 7)^b=(5 \times 7)^c
$$
If $a=b=c=2$ then this is the first possible answer to this issue. It is also obvious that the sum of $(3*7)^a+(4*7)^b$ must end and be divisible by $5$. Since $21^a$ always ends at $1$, then $28^b$ should end at $4$ . Defined $b$ as $b=2+4k$ -- even positive integer.
| Note that $21=3\times7$, $28=4\times7$ and $35=5\times7$, and so by unique factorization the numbers $21^a$, $28^b$ and $35^c$ are all distinct for all positive integers $a$, $b$ and $c$.
By unique factorization we see that the left hand side of
$$21^a+28^b=35^c,$$
is divisible by $7^{\min\{a,b\}}$ and hence $c\geq\min\{a,b\}$. Moreover the right hand sides of
$$21^a=35^c-28^b
\qquad\text{ and }\qquad
28^b=35^c-21^a,$$
are divisible by $7^{\min\{b,c\}}$ and $7^{\min\{a,c\}}$, respectively, because $28^b\neq35^c\neq21^a$. This implies
$$a\geq\min\{b,c\}
\qquad\text{ and }\qquad
b\geq\min\{a,c\},$$
from which it follows that $a=b=c$. Dividing out the factor $7^a$ leaves us with
$$3^a+4^a=5^a,$$
which clearly has the unique solution $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ I am trying to integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ via trig substitution. I decided to substitute $x = \sec\theta$ into the square root and $dx = \sec\theta \tan\theta\,d\theta$.
$$\int \frac{\sqrt{\sec^2 \theta-1^2}}{\sec^4\theta} \,dx
= \int \frac{\sqrt{\tan^2\theta + 1 - 1}}{\sec^4\theta}\,dx
= \int \frac{\tan\theta}{\sec^4\theta} \sec\theta \tan\theta\,d\theta = \int \dfrac{\tan^2\theta}{\sec^3\theta}\,d\theta$$
Here is where I am currently stuck. I attempted substitution with $u = \sec\theta, du = \sec x \tan x dx$ but that didn't seem to work out. I wasn't able to get an integration by parts strategy working either.
I think the answer lies in some sort of trigonometry regarding $\int \frac{\tan^2\theta}{\sec^3\theta}\,d\theta$ that I am overlooking to further simplify the problem, but no idea what it is
| $$
\begin{aligned}
& \int \frac{\sqrt{x^{2}-1}}{x^{4}} d x \\\stackrel{y=\frac{1}{x}}{=} &\int\frac{\sqrt{\frac{1}{y^{2}}-1}}{\frac{1}{y^{4}}}\left(-\frac{1}{y^{2}} d y\right)\\
=& -\int y \sqrt{1-y^{2}}d y \\
=&\frac{\left(1-y^{2}\right)^{\frac{3}{2}}}{3}+C
\\=&\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate :$\int_{-1}^{1} 2\sqrt{1-x^2} dx $
Evaluate: $$\int_{-1}^{1} 2\sqrt{1-x^2} dx $$
The answer is $\pi$
My attempt
$x = \sin(u), dx = \cos(u)du$
$$\int_{-1}^{1} 2 \sqrt{1-\sin^2(u)}\cos(u)du = \int_{-1}^{1} 2 \cos^2(u)du =\int_{-1}^{1} \frac{1}{2}(1+\cos(2u))du = \bigg(\frac{u}{2} + \frac{1}{2}\sin(2u) \bigg)\Bigg|_{-1}^{1}$$
confused how to proceed ?
| Geometrically , the unit circle can be represented as $$x^2+y^2=1$$
so $$y=\pm \sqrt{1-x^2}$$ and your case $y=+ \sqrt{1-x^2}$
So $\int_{-1}^1\sqrt{1-x^2} dx $ is the area of a (upper )semi circle, which is $\frac{\pi}{2}$. So $$2 \int_{-1}^1 \sqrt{1-x^2}dx =2 \frac{\pi}{2}=\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Find the coefficient of $x^{13}$ in the convolution of two generating functions
Four thiefs have stolen a collection of 13 identical diamonds. After the
theft, they decided how to distribute them.
3 of them have special requests:
*
*One of them doesn't want more than 2 diamonds ($\leq2$).
*The other one only wants a number of diamonds that's a multiple of 3.
*And the other one wants an odd number of diamonds greater or equal than 3.
Find in how many ways they can distribute the diamonds.
My first thought was to use generating functions to find the coefficient of $x^{13}$, for this problem it would be:
$f(x)=(1+x+x^2+x^3+...)(1+x+x^2)(1+x^3+x^6+...)(x^3+x^5+x^7+...)=\frac{1}{1-x} \frac{1-x^3}{1-x} \frac{1}{1-x^3}\frac{x^3}{1-x^3}=\frac{1}{(1-x)^{2}}\frac{x^3}{(1-x^2)}$
and that would be equivalent to finding the coefficient of $x^{10}$ in $\frac{1}{(1-x)^{2}}\frac{1}{(1-x^2)}$.
I know that I could use the binomial theorem, but the solution I have says I should be using convolution of these two generating functions, but I have no idea how to use it to find the coefficient of $x^{10}$.
| It is a laborious work.
$$\frac{d^n}{dx^n}\frac{1}{(1-x)^2(1-x^2)}=-\frac{1}{2}(-2-n)_n(-1+x)^{-3-n}+\frac{1}{4}(-1-n)_n(-1+x)^{-2-n}+\frac{1}{8}(-n)_n(1+x)^{-1-n}-\frac{1}{8}(-n)_n(-1+x)^{-1-n}$$
The bracket symbol is the Pochhammer symbol.
$$f^{(10)}(x)=\frac{d^{10}}{dx^{10}}\frac{1}{(1-x)^2(1-x^2)}=$$
$$\frac{-7257600(143x^{10}+715x^9+2574x^8+5148x^7+7722x^6+7722x^5+5720x^4+2860x^3+975x^2+195x+18)}{(x+1)^{11}(x-1)^{13}}$$
$$c_{10}=\frac{f^{(10)}(0)}{10!}=36$$
Or collect together all coefficients of $x^{13}$ from your series representation in your question, e.g. by a computer program.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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If $\sqrt{a}-\sqrt{b}$ is a root of a polynomial with integer coefficients, then so is $\sqrt{a}+\sqrt{b}$.
If $\sqrt{a} - \sqrt{b}$, where $a$ and $b$ are positive integers and non-perfect squares, is a root of a polynomial with integer coefficients, then $\sqrt{a} + \sqrt{b}$ also is.
It seems to hold some relationship with the quadratic formula. However, I have no idea on how to prove it.
| Returning to this question to complete my work in my previous answer - the crucial insight to the following argument was provided by the answer of @robjohn.
Supposing $a$, $b$ are positive integers, $a \neq b$ such that none of $a$, $b$, $ab$ are perfect squares, then for any polynomial with integer coefficients $p(x)$, it holds that $p(\sqrt a - \sqrt b) = 0 \implies p(\sqrt a + \sqrt b) = 0$
The necessary lemma, already argued by @robjohn, is that $1$, $\sqrt a$, $\sqrt b$, and $\sqrt {ab}$ are all linearly independent over $\mathbb Q$. That is, if $A,B,C,D$ are rationals such that $A + B\sqrt a + C\sqrt b + D\sqrt{ab} = 0$, then $A = B = C = D = 0$.
To complete the work I presented in my previous answer, I shall prove
There is no nonzero polynomial $r(x)$ of degree at most $3$ with rational coefficients such that $r(\sqrt a - \sqrt b) = 0$.
In which case $H(x) = (x^2 - a - b)^2 - 4ab$ is the unique monic polynomial of minimal degree for which $H(\sqrt a - \sqrt b) = 0$. Then, the observation that $H(\sqrt a + \sqrt b) = 0$ and an argument via the polynomial division algorithm finishes the problem.
I have shown before that there is no nonzero degree $1$ polynomial $r(x)$ such that $r(\sqrt a - \sqrt b) = 0$. For the quadratic case, suppose $r(x) = Ax^2 + Bx + C$ is a polynomial with integer coefficients such that $r(\sqrt a - \sqrt b) = 0$. By expanding the equation $A(\sqrt a - \sqrt b)^2 + B(\sqrt a - \sqrt b) + C = 0$, we can conclude
$$(Aa + Ab + C) + B \sqrt a - B \sqrt b - 2A \sqrt{ab} = 0$$
From the lemma above, we conclude in particular that $B = 0$. However this also implies
$$2A\sqrt{ab} = Aa + Ab + C$$
which implies $\sqrt{ab}$ is rational (and must therefore be an integer), contradicting the assumption that $ab$ is not a square.
Finally, turning to the cubic case. Suppose $r(x)$ is a polynomial of degree $3$ such that $r(\sqrt a - \sqrt b) = 0$. Without loss of generality $r(x)$ has rational coefficients and is monic, in which case $r(x) = x^3 + Ax^2 + Bx + C$ for rational $A$, $B$, $C$. Expanding
$$(\sqrt a - \sqrt b)^3+A(\sqrt a - \sqrt b)^2 + B(\sqrt a - \sqrt b) + C = 0$$
and applying the linear independence of $1$, $\sqrt a$, $\sqrt b$, and $\sqrt {ab}$ over $\mathbb Q$ leads to the following linear system of equations for $A$, $B$, and $C$.
\begin{array}{rlc}
Aa + Ab + C & =0 \quad & (1)\\
a + 3b + B &=0 & (\sqrt a) \\
-3a - b - B &=0 & (\sqrt b) \\
-2A &=0 & (\sqrt{ab})
\end{array}
From the fourth equation $A = 0$, which implies by the first equation $C = 0$. But if $C = 0$ then $\frac{r(x)}{x}$ is a rational polynomial of degree $2$ with a root at $x = \sqrt a - \sqrt b \neq 0$, contradicting the previous argument. This finishes the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove that $(a_1+a_2+\dotsb+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n})\geq n^2$ I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+\dotsb+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n}\right)\geq n^2$.
Where each $a_i$'s are positive and $n$ is a natural number.
First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.
Suppose the inequality holds true when $n=k$, i.e.,
$(a_1+a_2+\dotsb+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_k}\right)\geq n^2$.
This is true if and only if
$(a_1+a_2+\dotsb+a_k+a_{k+1})\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_k}+\frac{1}{a_{k+1}}\right) -a_{k+1}\left(\frac{1}{a_1}+\dotsb+\frac{1}{a_k}\right)-\frac{1}{a_{k+1}}(a_1+\dotsb+a_k)-\frac{a_{k+1}}{a_{k+1}} \geq n^2$.
And I got stuck here.
The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.
| Here is the proof by induction
that you wanted.
I added a more exact
version of
the identity used
in the proof
at the end.
Let
$s_n
=u_nv_n
$
where
$u_n=\sum_{k=1}^n a_k,
v_n= \sum_{k=1}^n \dfrac1{a_k}
$.
Then,
assuming
$s_n \ge n^2$,
$\begin{array}\\
s_{n+1}
&=u_{n+1}v_{n+1}\\
&=(u_n+a_{n+1}) (v_n+\dfrac1{a_{n+1}})\\
&=u_nv_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\
&=s_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\
&\ge n^2+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\
\end{array}
$
So it is sufficient
to show that
$u_n\dfrac1{a_{n+1}}+v_na_{n+1}
\ge 2n
$.
By simple algebra,
if $a, b \ge 0$ then
$a+b
\ge 2\sqrt{ab}
$.
(Rewrite as
$(\sqrt{a}-\sqrt{b})^2\ge 0$
or,
as an identity,
$a+b
=2\sqrt{ab}+(\sqrt{a}-\sqrt{b})^2$.)
Therefore
$\begin{array}\\
u_n\dfrac1{a_{n+1}}+v_na_{n+1}
&\ge \sqrt{(u_n\dfrac1{a_{n+1}})(v_na_{n+1})}\\
&= \sqrt{u_nv_n}\\
&=2\sqrt{s_n}\\
&\ge 2\sqrt{n^2}
\qquad\text{by the induction hypothesis}\\
&=2n\\
\end{array}
$
and we are done.
I find it interesting that
$s_n \ge n^2$
is used twice in the
induction step.
Note that,
if we use the identity above,
$a+b
=2\sqrt{ab}+(\sqrt{a}-\sqrt{b})^2$,
we get this:
$\begin{array}\\
s_{n+1}
&=s_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\
&=s_n+2\sqrt{u_n\dfrac1{a_{n+1}}a_{n+1}v_n}+1+(\sqrt{u_n\dfrac1{a_{n+1}}}-\sqrt{a_{n+1}v_n})^2\\
&=s_n+2\sqrt{s_n}+1+\dfrac1{a_{n+1}}(\sqrt{u_n}-a_{n+1}\sqrt{v_n})^2\\
&=(\sqrt{s_n}+1)^2+\dfrac1{a_{n+1}}(\sqrt{u_n}-a_{n+1}\sqrt{v_n})^2\\
&\ge(\sqrt{s_n}+1)^2\\
\end{array}
$
with equality
if and only if
$a_{n+1}
=\sqrt{\dfrac{u_n}{v_n}}
=\sqrt{\dfrac{\sum_{k=1}^n a_k}{\sum_{k=1}^n \dfrac1{a_k}}}
$.
| {
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"url": "https://math.stackexchange.com/questions/3145187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place:
$$ \frac{x}{y^3(1+y^2x)} + \frac{y}{z^3(1+z^2y) } + \frac{z}{x^3(1+x^2z)} > \frac{1}{3}(xy+zx+yz) $$
I've tried using the fact that $(xy+yz+zx)^2 \ge xyz(x+y+z) $ or $xy+yz+zx \le \frac{(x+y+z)^2}{3} $
I've also arrived to the fact that the inequality is equivalent to
$$ \sum_{cyc}{\frac{(xz)^{7/3}}{y^{5/3}(z+y)} > \frac{1}{3}(xy+yz+zx)} $$ which is homogenous.
I can't seem to find a nice way of using the given conditions for the sum and their order, thank you.
| Note: I have found a solution.
Shall we observe that each term of the LHS sum is of the form $\frac{x^4z^4}{y+z}$, the inequality is equivalent to
$$\sum_{cyc}{\frac{x^4z^4}{y+z}} > \frac{1}{3}{(xy+yz+zx)} $$
But from Titu's Lemma, we have
$$ \sum_{cyc}{\frac{x^4z^4}{y+z}} = \sum_{cyc}{\frac{(x^2z^2)^2}{y+z}} \ge \frac{({x^2z^2+y^2x^2+z^2y^2})^2}{2(x+y+z)} \ge^{(Quadratic Mean\ge AM)} \frac{(xy+yz+xz)^4}{18(x+y+z)} $$
Hence it suffices to prove $$\frac{(xy+yz+yz)^4}{18(x+y+z)} > \frac{1}{3}{(xy+yz+zx)} $$which is equivalent to $$(xy+yz+xz)^3 > 6(x+y+z)=27 $$
or, equivalently, $$ xy+yz+xz > 3$$ which is true by AM-GM inequality:
$$xy+yz+xz \ge 3(x^2y^2z^2)^{\frac{1}{3}}=3$$
With the equality case being impossible, since it would imply $x=y=z$, implying both $x=y=z=1$ ( from $xyz=1$) and $x=y=z=\frac{3}{2}$ (from $x+y+z=\frac{9}{2}$), we have only the strict version taking place: $ xy+yz+xz > 3$
Q. E. D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?
I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!
| There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
| {
"language": "en",
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"source": "stackexchange",
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$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
I'm getting a wrong answer even though my solution looks valid to me:
Inverse of $(1+\frac{1}{x^2})$ is $(\frac{1}{{\sqrt{x-1}}})$, so I plug this expression in wherever I see $x$'s
$$f(x)=(\frac{1}{{\sqrt{x-1}}})^2+(\sqrt{x-1})^2-2$$
Finally, I get $f(x)=\frac{(x-2)^2}{x-1}$ which is wrong according to the answer key. The answer should have been $x^2-4$
What am I doing wrong?
| I think your answer is correct. I solved the problem like this:
$$f(\frac{x^2+1}{x^2})=f(1+\frac{1}{x^2})=x^2+\frac{1}{x^2}-2$$
Now $f(1+\frac{1}{x^2})$ can be written as a function of just $\frac{1}{x^2}$ so,
$$f(1+\frac{1}{x^2})=g(\frac{1}{x^2})=x^2+\frac{1}{x^2}-2$$
Put $t=\frac{1}{x^2}$, to get:
$$f(1+t)=g(t)$$
$$g(t)=t+\frac{1}{t}-2$$
Or, $$g(x)=x+\frac{1}{x}-2$$
Now, $g(x)=f(1+x)$
Put, $1+x=\alpha$ to get,
$$f(\alpha)=g(\alpha-1)=\alpha -1 + \frac{1}{\alpha-1} -2$$
$$f(x)=x - 1 + \frac{1}{x-1} -2$$
$$f(x)=\frac{(x-2)^2}{x-1}$$ where $x\neq 0$,$x\neq1$
Hope this helps...
| {
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"url": "https://math.stackexchange.com/questions/3150399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3-a^3\end{vmatrix}\\&=\begin{vmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3-a^3\end{vmatrix}\\&=(b-a)(c-a)\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}\\&=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}\\\end{align}$$
I don't know what should I do next, maybe I've made a mistake but I didn't notice it
| Consider the matrix
\begin{bmatrix} 1& 1 &1 &1 \\ X & a&b&c \\X^2 & a^2 & b^2 & c^2 \\ X^3 & a^3 & b^3 & c^3 \end{bmatrix}
Its determinant is a Vandermonde determinant, equal to
$$(a-X)(b-X)(c-X)(b-a)(c-a)(c-b)$$
But developping with respect to the first column, you see that the determinant you are looking for is just equal to the coefficient (with a sign minus) of the $X$ term in this polynomial, which is
$$(-bc-ac-ab)(b-a)(c-a)(c-b)$$
Finally you get that the $K$ you are looking for is equal to
$$K = ab + ac + bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve tan(x)+cos(x)=1/2 Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
\frac{\sqrt{1-c^2}}{c}+c=\frac{1}{2}
$$
$$
\sqrt{1-c^2}+c^2=\frac{1}{2}c
$$
$$
1-c^2=c^2(\frac{1}{2}-c)^2=c^2(\frac{1}{4}-c+c^2)
$$
$$
c^4-c^3+\frac{5}{4}c^2-1=0
$$
| If we set $X=\cos x$ and $Y=\sin x$, the equation becomes
$$
Y=\frac{1}{2}X-X^2
$$
so the problem becomes intersecting the parabola with the circle $X^2+Y^2=1$.
This is generally a degree four problem. The image suggests there is no really elementary way to find the intersections.
The equation becomes
$$
X^4-X^3+\frac{5}{4}X^2-1=0
$$
as you found out. The two real roots are approximately
$$
-0.654665139167 \qquad 0.921490878816
$$
These correspond to $x=\pm2.284535877184578$ and $x=\pm0.39889463967156$, that correspond to what WolframAlpha finds.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.
| Note that you have
$$
7^n - 1 = 6a\\
5^n - 1 = 4b
$$
Now we're interested in whether $a$ and $b$ are even or odd. Which is to say we want to know when $7^n - 1$ is divisible by $4$ (so that when you divide it by $6$ you get an even number), and when $5^n-1$ is divisible by $8$ (so that when you divide it by $4$, you get an even number).
The binomial theorem gives
$$
7^n - 1 = (8-1)^n - 1\\
= 8^n - \binom n18^{n-1} + \cdots + (-1)^{n-1}\binom{n}{n-1}8 + (-1)^n - 1
$$
We see that this is divisible by $4$ exactly when $(-1)^n - 1$ is, which is to say when $n$ is even.
Then we have
$$
5^n - 1 = (4 + 1)^n - 1\\
= 4^n + \binom n14^{n-1} + \cdots + \binom{n}{n-1}4 + 1 - 1
$$
and we see that this is divisible by $8$ precisely when $\binom{n}{n-1} = n$ is even.
So $a$ and $b$ are both even for even $n$, and both odd for odd $n$, proving that $a+b$ is always even, meaning $12(a+b)$ is divisible by $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155303",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$ There's a math competition I participated yesterday (19/3/2019).
In these kinds of competitions, there will always be at least one problem about inequalities.
Now this year's problem about inequality is very easy. I am more interested in last year's problem, which goes by the following:
If $a$ and $b$ are positives then prove that $$ \frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$$
Now I know what you are thinking. It's a simple problem.
By the Cauchy - Schwarz inequality, we have that $\dfrac{a^2}{b} + \dfrac{b^2}{a} \ge \dfrac{(a + b)^2}{a + b} = a + b$. $$\implies \dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) \ge (a + b) + 7(a + b) = 8(a + b) \le 8\sqrt{2(a^2 + b^2)}$$
And you have fallen into the traps of the people who created the test. Almost everyone in last year's competition did too.
But someone came up with an elegant solution to the problem. He was also the one winning the contest.
You should have 15 minutes to solve the problem. That's what I also did yesterday.
| Let $a^2+b^2=2k^2ab,$ where $k>0$.
Thus, by AM-GM $k\geq1$ and we need to prove that:
$$\frac{\sqrt{(a+b)^2}(a^2-ab+b^2)}{ab}+7\sqrt{(a+b)^2}\geq8\sqrt{2(a^2+b^2)}$$ or
$$\sqrt{2k^2+2}(2k^2-1)+7\sqrt{2k^2+2}\geq16k$$ or
$$\sqrt{2k^2+2}(k^2+3)\geq8k,$$ which is true by AM-GM:
$$\sqrt{2k^2+2}(k^2+3)\geq\sqrt{4k}\cdot4\sqrt[4]{k^2\cdot1^3}=8k.$$
Done!
| {
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How to integrate $\frac{1}{\sqrt{x^2+x+1}}$
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$
I tried to solve this integral as follows
$\displaystyle
\int \frac{1}{\sqrt{x^2+x+1}} \ dx=
\int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{\frac{3}{4}((\frac{2x+1}{\sqrt{3}})^2+1)}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx$
Substitution $t=\frac{2x+1}{\sqrt{3}} ;dt=\frac{2}{\sqrt{3}} \ dx$
$\displaystyle\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx=
\int \frac{1}{\sqrt{t^2+1}} \ dt$
Substitution $\sqrt{u-1}= t;\frac{1}{2\sqrt{u-1}} \ du= dt$
$\displaystyle
\int \frac{1}{\sqrt{t^2+1}} \ dt=
\int \frac{1}{2 \sqrt{u(u-1)}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}} \ du=
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du$
Substitution $g= 2u-1;dg= 2 \ du$
$\displaystyle
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{2}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{g^2-1}} \ dg=
\frac{1}{2}\arcsin g +C=\frac{1}{2}\arcsin (2u-1) +C= \frac{1}{2}\arcsin (2(t^2+1)-1) +C=\frac{1}{2}\arcsin (2((\frac{2x+1}{\sqrt{3}})^2+1)-1) +C$
However when I tried to graph it using desmos there was no result, and when i used https://www.integral-calculator.com/ on thi problem it got the result $$\ln\left(\left|2\left(\sqrt{x^2+x+1}+x\right)+1\right|\right)$$
where have I made a mistake?
| As you did, by completing the square and with $y:=\dfrac2{\sqrt3}\left(x+\dfrac12\right)$,
$$I:=\int\frac{dx}{\sqrt{x^2+x+1}}=\int\frac{dy}{\sqrt{y^2+1}}.$$
Then by some magic $y:=\dfrac12\left(t-\dfrac1t\right)$ yields $dy=\dfrac12\left(1+\dfrac1{t^2}\right)$ and
$$\int\frac{dy}{\sqrt{y^2+1}}=\int\frac{\dfrac12\left(1+\dfrac1{t^2}\right)}{\dfrac12\left(t+\dfrac1{t}\right)}dt=\log t.$$
Now we need to invert,
$$y=\dfrac12\left(t-\dfrac1t\right)\iff t^2-2yt-1=0\iff t=y\pm{\sqrt{y^2+1}}.$$
Finally, choosing the plus sign and ignoring the additive constant,
$$I=\log\left(x+\dfrac12+\sqrt{x^2+x+1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int\frac{dx}{\sqrt{4-9x^2}}$ with different trig substitutions ($\sin$ vs $\cos$) gives different results I am trying to solve the following integral with trig substitutions. However, I get a different answer for two substitutions that should yield the same result.
$$\int\frac{dx}{\sqrt{4-9x^2}}$$
*
*For the first trig sub, I set $9x^2 = 4\cos^2\theta$. This simplifies to: $x = \frac{2}{3}\cos\theta$, and $dx = -\frac{2}{3}\sin\theta$. Substituting in, I get:
$$\int\frac{-2\sin\theta}{6\sin\theta} = -\frac{\theta}{3} = -\frac{1}{3}\,\cos^{-1}\left(\frac{3x}{2}\right)+C \tag{1}$$
*For the second trig sub, I set $9x^2 = 4\sin^2\theta$. This simplifies to: $x = \frac{2}{3}\sin\theta$, and $dx = \frac{2}{3}\cos\theta$. Substituting in, I get:
$$\int\frac{2\cos\theta}{6\cos\theta} = \frac{\theta}{3} = \frac{1}{3}\,\sin^{-1}\left(\frac{3x}{2}\right)+C \tag{2}$$
My question is:
Why do these two trig substitutions yield different results graphically? Shouldn't they result in the same graph?
| Since $(-\arcsin)'(x)=\arccos'(x)=-\dfrac1{\sqrt{1-x^2}}$, you got twice the same thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is
What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\displaystyle \sin (5x)-\sin (3x)=2\sin (4x)\cos (x)=0$
$\displaystyle 4x= m\pi$ and $\displaystyle x= 2m\pi\pm \frac{\pi}{2}$
$\displaystyle \frac{4\pi}{n}=m\pi\Rightarrow n=\frac{4}{m}\in \mathbb{Z}$ for $m=\pm 1,\pm 2\pm 3,\pm 4$
put into $\displaystyle x=2m\pi\pm \frac{\pi}{2}$
How do i solve my sum in some easy way Help me please
| Since $-n$ is a solution if $n$ is a solution, it suffices to look for solutions with $n\gt1$. Since $\sin x$ is strictly increasing for $0\le x\le\pi/2$, we cannot have $\sin(3\pi/n)=\sin(5\pi/n)$ if $n\ge10$, so it suffices to consider $2\le n\le9$.
From $\sin x=\sin(\pi-x)$, we have
$$\sin\left(5\pi\over n\right)=\sin\left(\pi-{5\pi\over n}\right)=\sin\left((n-5)\pi\over n\right)$$
For $2\le n\le5$, the signs of $\sin(3\pi/n)$ and $\sin((n-5)\pi/n)$ do not agree (e.g., $\sin(3\pi/2)=-1$ while $\sin(-3\pi/2)=1$). For $6\le n\le9$, we have $0\le{3\pi\over n},{(n-5)\pi\over n}\le{\pi\over2}$, in which case
$$\sin\left(3\pi\over n\right)=\sin\left((n-5)\pi\over n\right)\iff{3\pi\over n}={(n-5)\pi\over n}\iff3=n-5\iff n=8$$
Thus we have two solutions, $n=8$ and $n=-8$.
Remark: The cases $n=3$ and $n=5$ could have been rejected outright, since $\sin(3\pi/n)\sin(5\pi/n)=0$ in those cases, guaranteeing a forbidden $0$ in one of the denominators in the original expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Can $5^n+1$ be sum of two squares? I want to determine whether or not $5^n+1$ , $n\in\mathbb{N}$ can be written as sum of two squares.
Obviously, the real problem is when $n$ is odd. I am aware of the known results about numbers written as sum if squares, but I couldn't apply them here. We can see that when $n$ is odd, $5^n+1$ ia divisible by 6 and gives reminder 2 when divided by 4 which means that the two squares we would have to add must end in 1 and 9 or 5 and 5.
| Say $x^2 + y^2 = 5^n + 1$.
Looking mod 2, we get that $x^2 + y^2 = 0 \mod 2$, so $x$ and $y$ have the same parity, and looking mod 4, we get that $x^2 + y^2 = 2 \mod 4$, so $x$ and $y$ must both be odd.
Say $x = 2r + 1$ and $y = 2s+ 1$, so $$x^2 + y^2 = (2r + 1)^2 + (2s+1)^2 = 4(r^2 + r + s^2 + s) + 2 = 5^n + 1$$
or
$$4(r(r+1) + s(s+1)) = 5^n - 1$$
Now, the left side must be divisible by $8$, so $n$ must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
An integration that wolfram cannot help me. $$\int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx$$
I noted the fact that $\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$ but I cannot apply the substitution on it.
| Here is the best solution I can do
Compute the following:
\begin{align}
& \int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx \\
& =\int e^{x\sin x+\cos x} \cdot x^2 \cos x dx+ \int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx \\
\end{align}
Remark:
$$\frac{d(x\sin x+\cos x)}{dx}=x \cos x$$
$$\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$$
Therefore, using integration by parts, the first term is
\begin{align}
\int e^{x\sin x+\cos x} \cdot x^2 \cos x dx &= \int x d(e^{x\sin x+\cos x}) \\
&= x \cdot e^{x\sin x+\cos x} - \int e^{x\sin x+\cos x} dx \qquad (1)
\end{align}
Also, the second term is
\begin{align}
\int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= \int \frac{e^{x\sin x+\cos x}}{x^2\cos^2x}d(x \cos x) \\
&= \frac{e^{x\sin x+\cos x}}{x \cos x}-\int x \cos x d(\frac{e^{x\sin x+\cos x}}{x^2\cos^2x}) \qquad (2)
\end{align}
And derive:
\begin{align}
\int x \cos x d(\frac{e^{x\sin x+\cos x}}{x^2\cos^2x}) &= \int \frac{e^{x\sin x+\cos x} \cdot (-2 \cos x+x^2 \cos^2 x+2x \sin x)}{x^2 \cos^2 x}dx \\
&= \int e^{x\sin x+\cos x} dx - 2 \cdot \int \frac{(-x\sin x+\cos x)e^{x\sin x+\cos x}}{x^2 \cos^2 x} dx
\end{align}
Thus, the equation (2) becomes:
\begin{align}
\int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx = \frac{e^{x\sin x+\cos x}}{x \cos x} - \int e^{x\sin x+\cos x} dx \\ + 2 \cdot \int \frac{(-x\sin x+\cos x)e^{x\sin x+\cos x}}{x^2 \cos^2 x} dx
\end{align}
By rearranging the terms in the equation above, we have:
\begin{align}
-\int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= \frac{e^{x\sin x+\cos x}}{x \cos x} - \int e^{x\sin x+\cos x} dx \\
\int e^{x\sin x+\cos x} \cdot \frac{-x\sin x+\cos x}{x^2\cos^2x}dx &= -\frac{e^{x\sin x+\cos x}}{x \cos x} + \int e^{x\sin x+\cos x} dx \qquad (3)
\end{align}
Combine the equation (1) and (3):
\begin{align}
& \int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx \\
& = x \cdot e^{x\sin x+\cos x} - \int e^{x\sin x+\cos x} dx -\frac{e^{x\sin x+\cos x}}{x \cos x} + \int e^{x\sin x+\cos x} dx \\
& = x \cdot e^{x\sin x+\cos x} -\frac{e^{x\sin x+\cos x}}{x \cos x}
\end{align}
as above
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Partial fractions, disagreement with Wolfram Alpha On my math homework, I have this problem, and WolframAlpha says that $A=-\frac{1}{7}$ and $B=\frac{1}{7}$. However, while solving the problem on my own, I found a restriction that $A \neq -B$. How is it that this answer works? The problem in question is:
$$\dfrac{1}{x^2 - 3 x - 10} = \dfrac{A}{x+2} + \dfrac{B}{x - 5}$$
| Here's the full derivation: $$\frac{1}{x^2-3x-10}=\frac{A}{x+2}+\frac{B}{x-5}\implies$$ $$(x+2)(x-5)\left(\frac{1}{x^2-3x-10}\right)=(x+2)(x-5)\left(\frac{A}{x+2}+\frac{B}{x-5}\right)\implies$$ $$1=A(x-5)+B(x+2)=A(x)-5A+B(x)+2B=x(A+B)+1(2B-5A)$$
Hence we need $A+B=0\space\text{and}\space2B-5A=1$. So substitute $A=-B$ into the second equation: $$2B-5(-B)=1\implies7B=1\implies B=\frac{1}{7}$$
But we have the condition that $A+B=0$ and we know $B=\frac{1}{7},$ hence $$A+\frac{1}{7}=0\implies A=-\frac{1}{7}$$ Thus $$\frac{1}{x^2-3x-10}=\frac{A}{x+2}+\frac{B}{x-5}=\frac{1}{7(x-5)}-\frac{1}{7(x+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integral roots of cubic equation $x^3-27x+k=0$ The number of integers $k$ for which the equation $x^3-27x+k=0$ has atleast two distinct integer roots is
(A)$1$
(B)$2$
(C)$3 $
(D)$4$
My Attempt: The condition for cubic $x^3+ax+b=0$to have $3$ real roots happens to be $4a^3+27b^2\leq0$. But how to go about finding condition for integer roots.
| Suppose $x^3 - 27x + k = 0$ has distinct integer roots $a$ and $b$; then
$$
a^3 - 27a = b^3 - 27b,
$$
or
$$
a^3 - b^3 = 27(a - b).
$$
Since, by hypothesis, $a\ne b$, a factor of $a-b$ can be removed, resulting in
$$
a^2 + ab + b^2 = 27.
$$
After multiplying by $4$, this can be rearranged into
$$
(2a + b)^2 + 3b^2 = 108.
$$
It follows that the integer $2a+b$ is a multiple of $3$, and has a square $\le 108$; thus $2a+b = 0,\pm3,\pm6$ or $\pm9$.
*
*If $2a+b = 0$, then $b^2 = 36$, so $b = \pm6$.
*If $2a+b = \pm3$, then $b^2 = 33$, so this has no integral solution.
*If $2a+b = \pm6$, then $b^2 = 24$, so this has no integral solution.
*If $2a+b = \pm9$, then $b^2 = 9$, so $b = \pm3$.
In the first case, we find $(a,b) = (-3,6)$ or $(3,-6)$. In the fourth case,
the four possible combinations of signs result in $(a,b) = (3,3), (6,-3), (-3,-3)$ or $(-6,3)$. Rejecting the cases with $a=b$, $(a,b) = (-3,6)$ or $(6,-3)$ results in $k = 27a - a^3 = -54$ and $(a,b) = (3,-6)$ or $(-6,3)$ results in $k = 54$. Thus there are two possible values of $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the integrals $\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx$ Playing with integrals of type,
$$
I(n)=\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx,
$$
$$
n \in \mathbb{N}
$$
I got two interesting results for the limiting cases $n=1$ and $n \to \infty$:
$$
\lim_{n \to \infty} I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \approx \frac {\pi}{\sqrt {27}}
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.
| For $n \in \mathbb{N}$ we have
\begin{align}
I (n) &= \int \limits_{-1}^0 \left[x - x^{\frac{1}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} x \stackrel{x = -y}{=} \int \limits_0^1 \left[y^{\frac{1}{2n+1}} - y\right]^{\frac{1}{2n+1}} \mathrm{d} y = \int \limits_0^1 y^{\frac{1}{(2n+1)^2}}\left[1 - y^{\frac{2n}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} y \\
&\hspace{-10pt}\stackrel{y = t^{\frac{2n+1}{2n}}}{=} \frac{2n+1}{2n} \int \limits_0^1 t^{\frac{n+1}{n (2n+1)}} (1-t)^{\frac{1}{2n+1}} \mathrm{d}t = \frac{2n+1}{2n} \operatorname{B}\left(\frac{n+1}{n(2n+1)}+1,\frac{1}{2n+1} + 1\right) \, .
\end{align}
Using $\Gamma(x+1) = x \Gamma(x)$ we can rewrite this result to find
$$ I (n) = \frac{\operatorname{B} \left(\frac{1}{n} - \frac{1}{2n+1}, \frac{1}{2n+1}\right)}{2 (2n+1)} = \frac{1}{2(2n+1)} \frac{\operatorname{\Gamma}\left(\frac{1}{n} - \frac{1}{2n+1}\right) \operatorname{\Gamma}\left(\frac{1}{2n+1}\right)}{\operatorname{\Gamma}\left(\frac{1}{n}\right)}$$
for $n \in \mathbb{N}$. In particular,
$$ I(1) = \frac{\operatorname{\Gamma}\left(\frac{2}{3}\right) \operatorname{\Gamma}\left(\frac{1}{3}\right)}{6} = \frac{\pi}{6 \sin \left(\frac{\pi}{3}\right)} = \frac{\pi}{3 \sqrt{3}} \,.$$
Moreover, we obtain
$$ \lim_{n \to \infty} I (n) \stackrel{\Gamma(x) \, \stackrel{x \to 0}{\sim} \, \frac{1}{x}}{=} \lim_{n \to \infty} \frac{1}{2(2n+1)} \frac{\frac{n(2n+1)}{n+1} (2n+1)}{n} = \lim_{n \to \infty} \frac{2n+1}{2(n+1)} = 1 \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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An interesting inequality with condition If $a,b,c$ positive reals and $\frac{a}{b+c} \ge 2$ I have to prove that
$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)\geq \frac{49}{18}$
We may assume that $a\geq b \geq c.$ Firstly, let's show that
$\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2}\geq \frac{1}{4ab}+\frac{2}{(a+c)(b+c)}.$ This can be rewritten as
$\left(\frac{1}{a+c}-\frac{1}{b+c}\right)^2\geq \frac{(a-b)^2}{4ab(a+b)^2},$ or equivalently $4ab(a+b)^2\geq (a+c)^2(b+c)^2.$ This is obvious, since $4ab\geq (b+c)^2$ and $(a+b)^2\geq (a+c)^2.$
Thus, it remains to prove that
$(ab+bc+ca)\left(\frac{1}{4ab}+\frac{2}{(a+c)(b+c)} \right)\geq \frac{49}{18}.$ Using the identities
$\frac{ab+bc+ca}{4ab}=\frac{1}{4}+\frac{c(a+b)}{4ab}, \quad \frac{2(ab+bc+ca)}{(a+c)(b+c)}=2 -\frac{2c^2}{(a+c)(b+c)},$ this becomes
$\frac{c(a+b)}{4ab}\geq \frac{2c^2}{(a+c)(b+c)}+\frac{17}{36}.$
Then I stuck. Any idea please?
| This can be solved in a brute force way:
$$\frac{a}{b+c}\ge2\implies a=2b+2c+x$$
..where $x$ is some non-negative value. The inequality:
$$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)-\frac{49}{18}\ge0$$
...becomes:
$$((2b+2c+x)b+bc+c(2b+2c+x))\left(\frac{1}{(b+c)^2}+\frac{1}{(2b+3c+x)^2}+\frac{1}{(3b+2c+x)^2} \right)-\frac{49}{18}\ge0$$
This can be writen as:
$$\frac AB\ge 0\tag{1}$$
where
$$A=654 b^5 c+462 b^5 x+2783 b^4 c^2+3620 b^4 c x+851 b^4 x^2+4276 b^3 c^3+8748 b^3 c^2 x+4260 b^3 c x^2+572 b^3 x^3+2783 b^2 c^4+8748 b^2 c^3 x+6854 b^2 c^2 x^2+1932 b^2 c x^3+167 b^2 x^4+654 b c^5+3620 b c^4 x+4260 b c^3 x^2+1932 b c^2 x^3+352 b c x^4+18 b x^5+462 c^5 x+851 c^4 x^2+572 c^3 x^3+167 c^2 x^4+18 c x^5$$
$$B=18 (b+c)^2 (3 b+2 c+x)^2 (2 b+3 c+x)^2$$
$A,B$ are positive so (1) is obviously true.
There is a similar problem which I find to be much more interesting:
For all positive $a,b,c$:$$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)\ge\frac94$$
| {
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"url": "https://math.stackexchange.com/questions/3173875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to simplify the the binomial coefficient in the binomial series? Use binomial series to expand the function $\frac{5}{(6+x)^3}$ as a power series.
I understand the process to get the following summation:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {-3 \choose n} (\frac{x}{6})^n $
However, I am stuck on seeing what's going on with ${-3 \choose n}$.
From the Stewart Calculus textbook, it says that ${k \choose n} = \frac{k(k-1)(k-2)...(k-n+1)}{n!}$.
By applying that, I would get:
${-3 \choose n}=\frac {(-3)(-4)(-5)...[-(n+2)]}{n!}$.
I think the next step would be to extract out the negative, so I will get $(-1)^n$.
The summation would then be:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {\frac {(-1)^n(3)(4)(5)...[(n+2)]}{n!}} (\frac{x}{6})^n $
The solution to this problem is $\frac{5}{2}\sum_{n=0}^{\infty} {\frac {(-1)^n(n+1)(n+2)x^n}{6^{n+3}}}$
I am not sure what has happened to the factorial. Was it cancelled out due to the (3)(4)(5)... in the numerator? Where did (n+1) come from?
| $$\frac {(3)(4)(5)\cdots(n+2)}{n!} = \frac {(3)(4)(5)\cdots(n)(n+1)(n+2)}{(1)(2)(3)(4)(5)\cdots(n)} = \frac {(n+1)(n+2)}{(1)(2)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating with different methods leads to different results? Acceleration of a particle is given as $a = 0.1(t-5)^2$. The particle is initially at rest.
Here is the way I initially attempted it. (I didn't use substitution as the question in the textbook appears before substitution is officially taught.)
$$a\:=\:0.1t^2-t+2.5$$
so
$$v = \frac{1}{30}t^3-0.5t^2+2.5t+c$$
$c$ evaluates to zero.
But my textbook uses a substitution and gets:
$$\frac{1}{30}\left(t-5\right)^3 +c$$
In this case, c has a value when t= 0 and so the end result is different. Wouldn't this lead to different values in general?
| Note that you ended up with
$$
v(t) = \frac{1}{30} t^3 - \frac12 t^2 + \frac52 t
$$
But the book's answer must have $c = 5^3/30$, so the complete answer is
$$
\begin{split}
v(t) &= \frac{1}{30} (t-5)^3 + \frac{5^3}{30}\\
&= \frac{(t-5)^3-5^3}{30} \\
&= \frac{t^3 - 3\cdot 5 t^2 + 3 \cdot 5^2t - 5^3 + 5^3}{30} \\
&= \frac{t^3}{30} - \frac{t^2}{2} + \frac{5t}{2}\\
\end{split}
$$
which is the same answer as the one you ended up with.
| {
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Evaluate $\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}$
Evaluate:$$
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}
$$
I'm trying to spot an error in my calculations. It is known that $x^n - 1$ may be factored out as $(x-1)(1+x+x^2+\cdots+x^{n-1})$. Using that fact consecutively for all the brackets one may obtain:
$$
k\ \text{times}\begin{cases}
x^n-1 = (x-1)(1+x+\cdots+x^{n-1})\\
x^{n-1}-1 = (x-1)(1+x+\cdots+x^{n-2})\\
\cdots\\
x^{n-k+1}-1 = (x-1)(1+x+\cdots+x^{n-k})
\end{cases}
$$
For the denominator:
$$
k\ \text{times}\begin{cases}
(x-1) = (x-1)\\
(x^2-1) = (x-1)(1+x)\\
\cdots \\
(x^k-1) = (x-1)(1+x+\cdots+x^{k-1})
\end{cases}
$$
So if we denote the expression under the limit as $f(x)$ we get:
$$
f(x) = \frac{(x-1)^k\prod\sum\cdots}{(x-1)^k\prod \sum\cdots}
$$
Now if we let $x\to1$ we get:
$$
\lim_{x\to1}f(x) = \frac{(n-1)(n-2)\cdots (n-k)}{1\cdot 2\cdot 3\cdots (k-1)} = \frac{(n-1)(n-2)\cdots (n-k)}{(k-1)!}
$$
But this doesn't match the keys section which has $n\choose k$ as an answer. I've checked several times but couldn't spot a mistake. Looks like I'm missing a $+1$ somewhere.
| A quick way to evaluate the limit:
First consider
\begin{align}
\lim_{x \to 1} \frac{x^{n-j+1} -1}{x^j -1} &= \lim_{x \to 1} \frac{(n-j+1) x^{n-j}}{j \, x^{j-1}} = \frac{n-j+1}{j}
\end{align}
now,
\begin{align}
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)} &=
\lim_{x \to 1} \prod_{j=1}^{k} \frac{x^{n-j+1} -1}{x^j -1} \\
&= \prod_{j=1}^{k} \, \lim_{x \to 1} \frac{x^{n-j+1} -1}{x^j -1} \\
&= \prod_{j=1}^{k} \frac{n-j+1}{j} \\
&= \frac{(n)(n-1) \cdots (n-k+1)}{k!} = \frac{n!}{k! \, (n-k)!} \\
&= \binom{n}{k}.
\end{align}
| {
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How to solve $x^{x^x}=(x^x)^x$? How can we solve the equation :
$x^{x^x}=(x^x)^x$
with $x \in {\mathbb{R}+}^*$
Thanks for heping me :)
| We can check when the exponents are equal.
It is $x^{x^x}=(x^x)^x\Leftrightarrow x^{(x^x)}=x^{x^2}\Leftrightarrow x^x=x^2$
Now x^x-x^2=0\Leftrightarrow $x^2(x^{x-2}-1)=0$.
So $x^2=0$ or $x^{x-2}=1$.
Since $x\neq 0$ we have $x^{x-2}=1$ left, which holds if $x-2=0$. So $x=2$
And $2^{2^2}=2^{2\cdot 2}$
Edit: And x=1 is an obvious solution...
| {
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Complex impedence between 2 terminals I have this problem to solve.
I have this so far. I am struggling with this so looking for some help please.
$$\frac{1}{5j}+\frac{1}{5+8.66j}+\frac{1}{15}+\frac{1}{-10j}=-\frac{j}{5}+\frac{5-8.66j}{5^2+8.66^2}+\frac{1}{15}+\frac{j}{10}=a-bj$$
$$a:=\frac{5}{5+8.66}+\frac{1}{15},\,b=\frac{1}{5}+\frac{8.66}{5^2+8.66^2}-\frac{1}{10}=\frac{1}{10}+\frac{8.66}{5^2+8.66^2},$$
$$\frac{1}{a-bj}=\frac{a+bj}{a^2+b^2}.$$
I know i need a imaginary number for the x and real number for the y axis.
I an struggling because thier is more than 2 branches. I know:
$$ A + B = (4 + j1) + (2 + j3)
A + B = ( 4+2 ) + j(1+3)$$
So i added like this:
$$ j + 5 - 8.66j + 1 + j = -2.66j $$
But i guess this isn't correct?
This is for cartesian form.
| Impedances sum in series; admittances (reciprocals of impedances) sum in parallel. Thus the total admittance in $\Omega^{-1}$ is $$\frac{1}{5j}+\frac{1}{5+8.66j}+\frac{1}{15}+\frac{1}{-10j}=-\frac{j}{5}+\frac{5-8.66j}{5^2+8.66^2}+\frac{1}{15}+\frac{j}{10}=a-bj$$with $$a:=\frac{5}{5+8.66}+\frac{1}{15},\,b=\frac{1}{5}+\frac{8.66}{5^2+8.66^2}-\frac{1}{10}=\frac{1}{10}+\frac{8.66}{5^2+8.66^2},$$while the total impedance in $\Omega$ is $$\frac{1}{a-bj}=\frac{a+bj}{a^2+b^2}.$$I'll leave arithmetic to you.
| {
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If $f(x,y)=9-x^2-y^2$ if $x^2+y^2\leq9$ and $f(x,y)=0$ if $x^2+y^2>9$ study what happens at $(3,0)$
If$$f(x,y)=\begin{cases}9-x^2-y^2&\text{if }x^2+y^2\leq9\\0&\text{if }x^2+y^2>9\end{cases}$$study the continuity and existence of partial derivative with respect to $y$ at point $(3,0)$.
The graph of the domain of $f$ is:
Continuity study at $(3,0)$:
$f(3,0)=9-3^2-0=0$, but I do not know how to find $$\lim_{(x,y)\to(3,0)}f(x,y).$$ I tried the following: $$\lim_{(x,y)\to(3,0)}f(x,y)=\left\{\begin{array}{l}\displaystyle\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}(9-(x^2+y^2))=9-9=0\\\displaystyle\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}\text{??}=\text{??????}\end{array}\right.$$ but then I realized that the "curve" $C_2$ is actually NOT a curve but a set of infinite points, as shown in the previous image.
Existence of partial derivative with respect to $y$ at $(3,0)$:
I know that I need to study whether $$\frac\partial{\partial y}f(3,0)=f'((3,0);(0,1))=f_y(3,0)=\lim_{h\to0}\frac{f((3,0)+h(0,1))-f(3,0)}{h}=\lim_{h\to0}\frac{f(3,h)}{h}$$ exists or not, but I am not able to even find that limit.
Any help? Thanks!!
| Continuty at (3,0) : When $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$
Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at $(3,0)$ wrt $y$ and the partial derivative is $0.$
| {
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
| Note: $90\mod37 =16$, so the answer you found was correct, just not fully reduced.
| {
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Polynomial Roots with no complex roots Let $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_n,b_n)$ be the ordered pairs $(a,b)$ of real numbers such that the polynomial
$$p(x) = (x^2 + ax + b)^2 +a(x^2 + ax + b) - b$$has exactly one real root and no nonreal complex roots. Find $a_1 + b_1 + a_2 + b_2 + \dots + a_n + b_n.$
I have no idea how to do this. Can someone help please?
| We expand the OP's polynomial and write it as
$\tag 1 p(x) = (a b + b^2 - b) + (a^2 + 2 a b) x + (a^2 + a + 2 b) x^2 + 2 a x^3 + x^4 $
We also have (see Robert Israel's answer)
$\tag 2 (x-r)^4 = r^4 - 4 r^3 x + 6 r^2 x^2 - 4 r x^3 + x^4$
Fortunately, we can find an easy 'coefficient hook', and so $2a = -4r$, or
$$\tag 3 r = -\frac{a}{2}$$
Plugging this back into $\text{(2)}$, we get
$$\tag 4 (x+a/2)^4 = \frac{a^4}{16} + \frac{a^3}{2} x + \frac{3 a^2}{2} x^2 + 2 a x^3 + x^4$$
Once again, there is an easy way to proceed, and we find
$$\tag 5 b = \frac{a^2 - 2a}{4}$$
Once again, there is a (relatively) easy way to proceed, and we find that
$$\tag 6 \frac{a^4}{16} - \frac{a^2}{2} + \frac{a}{2} = \frac{a^4}{16}$$
must be true.
So there are at most two possible solutions:
$(a_1, b_1) = (0, 0)$
and
$(a_2, b_2) = (1, -\frac{1}{4})$
You will find that by plugging into $p(x)$ that they both work - the polynomial will only have one real root with multiplicity $4$.
I have no idea why we are interested in the sum of these coordinates. Perhaps I need to review algebra-precalculus.
| {
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How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For the left hand side expression, I can find 2 functions. Consider;
$$f_1 (x) = \frac{1}{(1-3x)} \\ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + \cdots + 3^nx^n + \cdots \\ f_2(x) = \frac{1}{(1-x)^{n+1}} \\ = {n \choose n} + {n+1 \choose n}x + {n+2 \choose n}x^2 + \cdots + {2n-1 \choose n}x^{n-1} + {2n \choose n}x^n + \cdots + $$
Consider the coefficient of $x^n$ in the expansion of $f_1 (x) . f_2 (x)$. Then the coefficient will be the expression on the left hand side.
Now we further consider 2 functions for the right-hand side expression.
Consider;
$$f_3 (x) = \frac {1}{(1-2x)} \\ = 1 + 2^1x + 2^2x^2 + \cdots + 2^{n-1}x^{n-1} + 2^nx^n + \cdots \\ f_4 (x) = (1+x)^{2n+1} \\= 1 + {2n+1 \choose 1}x + \cdots + {2n+1 \choose n-1}x^{n-1} + {2n+1 \choose n}x^n +\cdots + {2n+1 \choose 0}x^{2n +1} \\ = {2n+1 \choose 2n+1} + {2n+1 \choose 2n}x + {2n+1 \choose 2n-1}x^2 + \cdots + {2n+1 \choose n+2}x^{n-1} + {2n+1 \choose n+1}x^{n} + \\
+ {2n+1 \choose n}x^{n+1} +\cdots + {2n+1 \choose 0}x^{2n +1}$$
Hence the coefficient of $x^n$ is the coefficient of $x^n$ in the expansion of $f_3(x) . f_4(x)$
This is what I managed to do so far. I'm not sure if $f_1(x) .f_2(x) = f_3(x).f_4(x)$. If the two functions are indeed equal, then I can conclude that their coefficient of $x^n$ must be equal, which will immediately answer the question. If they are equal, how do I show that they are?
If the two functions are not equal? How do I proceed to show this question?
Edit: It might not be true that the product of the two functions are equal. I tried substituting $x=0.1, n=1$. Seems like the two values are not equal. How do I proceed with this question?
| We seek to show that
$$\sum_{q=0}^n {2n-q\choose n} 3^q
= \sum_{q=0}^n {2n+1\choose n+1+q} 2^q.$$
We have for the LHS
$$\sum_{q=0}^n {2n-q\choose n-q} 3^q
= \sum_{q=0}^n 3^q [z^{n-q}] (1+z)^{2n-q}
\\ = [z^n] (1+z)^{2n} \sum_{q=0}^n 3^q z^q (1+z)^{-q}.$$
The coefficient extractor controls the range and we obtain
$$[z^n] (1+z)^{2n} \sum_{q\ge 0} 3^q z^q (1+z)^{-q}
= [z^n] (1+z)^{2n} \frac{1}{1-3z/(1+z)}
\\ = [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
We could conclude at this point by inspection. Continuing anyway we
get for the RHS
$$\sum_{q=0}^n {2n+1\choose n-q} 2^q
= \sum_{q=0}^n 2^q [z^{n-q}] (1+z)^{2n+1}
\\ = [z^n] (1+z)^{2n+1} \sum_{q=0}^n 2^q z^q.$$
The coefficient extractor once more controls the range and we obtain
$$[z^n] (1+z)^{2n+1} \sum_{q\ge 0} 2^q z^q
= [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
The two generating functions are the same and we have equality for LHS
and RHS.
| {
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Finding minimum value of x such that GCD(A+x,B+x) = C where A , B ,C are given I need to add Minimum non-negative Integer such that I can get the desired GCD(a+x,b+x)
Let say A=12 & B=26
For GCD(12+x,26+x) = 1 , x should be 1
For GCD(12+x,26+x) = 2 , x should be 0
For GCD(12+x,26+x) = 7 , x should be 9
For GCD(12+x,26+x) = 14 , x should be 2
| Your question title says you want to find the minimum value of a non-negative integer $x$ such that $\gcd(A + x, B + x) = C$, where $A, B, C$ are given. First, note there's no such integer in certain cases. In particular, since $C$ divides $A + x$ and $B + x$, it must also divide their difference, i.e., $C \mid (A + x) - (B + x) = A - B$. Thus, a minimum requirement is that
$$A \equiv B \pmod C \tag{1}\label{eq1}$$
Next, consider the special case of $A = B$. If $A \le -C$, then $x = -C - A$ works. Otherwise, if $A \le C$, then $x = C - A$ works. Finally, if $A \gt C$, then the gcd would always be $A + x \gt C$, so there are no solutions.
If \eqref{eq1} is satisfied and $A \neq B$, next note $A$ and $B$ can be rewritten as
$$A = mC + r, B = nC + r, \; \text{ where } \; m,n,r \in \mathbb{Z}, \; m \neq n, \; 0 \le r \lt C \tag{2}\label{eq2}$$
Thus, $A + x = mC + (r + x)$, so $C \mid A + x$ requires $C \mid r + x$. In general, this requires
$$x = kC - r, \; k \ge 0 \tag{3}\label{eq3}$$
The smallest non-negative value of $x$ which might work is $0$ if $r = 0$, else it's $C - r$. For $B + x = nC + (r + x)$, it's also true that $C \mid B + x$ in any general case.
The one thing you need to be careful of is that there are no other common factors so the GCD would then be a multiple of $C$. In particular, \eqref{eq3}, gives $A + x = C(m + k)$ and $B + x = C(n + k)$. Thus, you want to find the smallest non-negative integer $k$ such that $x$ from \eqref{eq3} is non-negative and $\gcd(m + k, n + k) = 1$. Note there'll always be such a $k$. To see this, the gcd of $m + k$ and $n + k$ must divide the difference, i.e., $m - n$. Since $m \neq n$, then $\left|m - n\right|$ is either $1$, in which case the gcd is always $1$, or it's a product of $1$ or more primes, i.e., $\left|m - n\right| = \prod_{i = 1}^j p_i^{e_i}$ for some $j \ge 1$, primes $p_i$ and integers $e_i \ge 1$. In the latter case, for each $i$, note that $m \equiv n \equiv r_i \pmod p_i$ for some $0 \le r_i \le p_i - 1$. Since all primes are $\ge 2$, there are always at least $2$ possible values for each $r_i$. For each one, choose any value such its sum with $r_i$ is not congruent to $0$. No $p_i$ would divides $n + k$ or $m + k$. Thus, any prime factors of $\gcd(n + k, m + k)$ must be different than any $p_i$. However, since any prime factors of the gcd must divide the difference, this means there are no prime factors of the gcd, i.e., $\gcd(n + k, m + k) = 1$. For each possible combination of these congruences, the method described to handle more than $2$ numbers section in Extended Euclidean algorithm shows you can always find non-negative integers $k$ satisfying those congruences. Thus, choose the smallest non-negative integer $k$ among all those values so that $x$ given in \eqref{eq3} is non-negative.
To demonstrate how this works, here's how to apply this to your examples with $A = 12$ and $B = 26$. First, as $26 - 12 = 14 = 2 \times 7$, the only possible values of $C$ are $1, 2, 7, 14$, as you've shown. For $C = 1$, since $\gcd(12, 26) = 2$, as $12 = 12 \times 1 + 0$ and $26 = 26 \times 1 + 0$, but $\gcd(12 + 1, 26 + 1) = 1$, you need to use $x = 1$. For $C = 2$, just $x = 0$ works. For $C = 7$, since $12 = 1 \times 7 + 5$, $26 = 3 \times 7 + 5$, and $\gcd(1 + 1, 3 + 1) = 2$, you can't use $x = 7 - 5 = 2$. Instead, since $\gcd(1 + 2, 3 + 2) = 1$, you need to use $x = 14 - 5 = 9$ instead. Finally, for $C = 14$, since $12 = 0 \times 14 + 12$, $26 = 1 \times 14 + 12$ and $\gcd(0 + 1, 1 + 1) = 1$, you can just use $x = 14 - 12 = 2$.
| {
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Stuck on candy bowl fraction I am really stuck on this problem because I'm not even sure where to start.
Larissa has a bowl of candies. On the first day, she eats 1/2 of the candies plus one more. On the second day, she eats 1/3 of the remaining candies plus one more. On the third day, she eats 1/5 of the remaining candies plus one more. On the fourth day she eats the three remaining candies. How many candies did she have at the start?
| Hints
Let $N$ be the number of candies.
Day 1: she eats $\frac{1}{2}N + 1$: remainder $N_{1}$
Day 2: she eats $\frac{1}{3}N_{1} + 1$: remainder $N_{2}$
Day 3: she eats $\frac{1}{5}N_{2} + 1$: remainder $N_{3}$
Day 4: she eats $3$: $N_{3} = 3$
Can you go from here? (answer $=20$)
Step 1: English $\to$ algebra
Create a table which defined the key variables of $e_{k}$, the amount of candies eaten on day $k$, and $a_{k}$, the amount available at the end of day $k$.
Step 2: Solution criteria
We need to find $a_{0}$, the initial amount of candies, such that $a_{3}=3$.
Step 3: Solution strategy
We all agree that the solution involves expressing the intermediate variables $a_{k}$, and $e_{k}$, $k=1,2,3$ in terms of $a_{0}$.
The are a few ways to do this. We chose a top down approach.
Day 1
$$a_{1} = a_{0} - e_{1} = \tfrac{1}{2}a_{0}-1$$
Day 2
$$e_{2} = \tfrac{1}{3}a_{1} + 1 = \tfrac{1}{6}a_{0}+\tfrac{2}{3}$$
$$a_{2} = a_{1} - e_{2} = \tfrac{1}{3}a_{0}-\tfrac{5}{3}$$
Day 3
$$e_{3} = \tfrac{1}{5}a_{2} + 1 = \tfrac{1}{15}a_{0}+\tfrac{2}{3}$$
$$a_{3} = a_{2} - e_{3} = \tfrac{4}{15}a_{0}-\tfrac{7}{3}$$
Step 4: Solve
$$
a_{3} = 3
$$
implies
$$
\tfrac{4}{15}a_{0}-\tfrac{7}{3} = 3 \quad \implies \quad a_{0} = 20
$$
Hopefully, this is reflected in your class notes.
Check the solution
| {
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Another combinatorial identity involving product of binomial coefficients While answering a question I came in a side calculation across the following identity valid by numerical evidence for integer $m$ and $n$:
$$
\sum_{k=m}^{\left\lfloor\frac n2\right\rfloor} \binom km\binom n{2k}=\frac{n(n-m-1)!}{m!(n-2m)!}2^{n-2m-1}.
$$
I would appreciate any hints and suggestions on combinatorial as well as algebraic proofs of the identity.
| Here is an algebraic proof. With $n\ge 2m$ we claim that
$$\sum_{k=m}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose m} =
\frac{n}{m} {n-m-1\choose m-1} 2^{n-2m-1}.$$
The LHS is
$$\sum_{k=m}^{\lfloor n/2 \rfloor}
{k\choose m} {n\choose n-2k}
= [z^n] (1+z)^n \sum_{k=m}^{\lfloor n/2 \rfloor}
{k\choose m} z^{2k}.$$
The coefficient extractor enforces the upper limit and we get
$$[z^n] (1+z)^n \sum_{k\ge m}
{k\choose m} z^{2k}
= [z^n] (1+z)^n z^{2m} \sum_{k\ge 0}
{k+m\choose m} z^{2k}
\\ = [z^{n-2m}] (1+z)^n \frac{1}{(1-z^2)^{m+1}}
= [z^{n-2m}] (1+z)^{n-m-1} \frac{1}{(1-z)^{m+1}}
\\ = \sum_{q=0}^{n-2m} {n-m-1\choose q}
{n-m-q\choose m}.$$
Now we have
$${n-m-1\choose q} {n-m-q\choose m}
= \frac{n-m-q}{m} {n-m-1\choose q} {n-m-1-q\choose m-1}
\\ = \frac{n-m-q}{m}
\frac{(n-m-1)!}{q! \times (m-1)! \times (n-2m-q)!}
\\ = \frac{n-m-q}{m}
{n-m-1\choose m-1} {n-2m\choose q}.$$
We get for our sum
$$\frac{1}{m} {n-m-1\choose m-1}
\sum_{q=0}^{n-2m}
(n-m-q) {n-2m\choose q}
\\ = \frac{1}{m} {n-m-1\choose m-1}
\sum_{q=0}^{n-2m}
(m+q) {n-2m\choose q}.$$
For the sum without the scalar we obtain
$$m 2^{n-2m} + \sum_{q=1}^{n-2m}
q {n-2m\choose q}
= m 2^{n-2m} + (n-2m) \sum_{q=1}^{n-2m}
{n-2m-1\choose q-1}
\\ = m 2^{n-2m} + (n-2m) 2^{n-2m-1}
= n 2^{n-2m-1}.$$
Collecting everything we find
$$\bbox[5px,border:2px solid #00A000]{
\frac{n}{m} {n-m-1\choose m-1} 2^{n-2m-1}.}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?
| Another method:
$$\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2=\\
\sin ^4x+\cos ^4x+4+\frac1{\sin^4 x}+\frac1{\cos^4 x}=\\
(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x+4+\frac{(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x}{\sin^4x\cos^4x}=\\
1-\frac12\sin^2 2x+4+\frac{1-\frac12\sin 2x}{\frac1{16}\sin^4 2x}=\\
\left(1-\frac12\sin^2 2x\right)\left(1+\frac{16}{\sin^4 2x}\right)+4\ge \\
\left(1-\frac12\cdot 1\right)\left(1+\frac{16}{1^2}\right)+4=12.5.
$$
Note: $1-\frac12\sin ^2 2x>0$, so $\sin^2 2x$ should be maximized to $1$.
| {
"language": "en",
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"answer_count": 5,
"answer_id": 2
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What formula could generate this sequence related to the Collatz conjecture The collatz conjecture states that every number eventually reaches $1$ under the repeated iteration of
$$ f_0(n) =
\begin{cases}
n/2, & \text{if $n$ even} \\
3n+1, & \text{else}
\end{cases}$$
As a number is guaranteed to be even after the $3n+1$ step, one can replace $f_0(n)$ with
$$ f_1(n) =
\begin{cases}
n/2, & \text{if $n$ even} \\
\frac{3n+1}{2}, & \text{else}
\end{cases}$$
and obtain an equivalent conjecture.
One can tabulate the possible expressions that can arise from applying $f_1$ to $n$ $x$ times, as is shown in the following table.
\begin{array}{|c|c|c|c|}
\hline
\frac{n}{2^1}& \frac{n}{2^2} & \frac{n}{2^3} & \frac{n}{2^4} \\ \hline
\frac{3^1n+1}{2^1}& \frac{3^1n+1\cdot2^1}{2^2} & \frac{3^1n+1\cdot2^2}{2^3} & \frac{3^1n+1\cdot2^3}{2^4}\\ \hline
& \frac{3^1n+1\cdot2^0}{2^2} & \frac{3^1n+1\cdot2^1}{2^3} &\frac{3^1n+1\cdot2^2}{2^4}\\ \hline
& \frac{3^2n+5\cdot2^0}{2^2} & \frac{3^2n+5\cdot2^1}{2^3}& \frac{3^2n+5\cdot2^2}{2^4}\\ \hline
& &\frac{3^1n+1\cdot2^0}{2^3} & \frac{3^1n+1\cdot2^1}{2^4}
\end{array}
Let $x$ be the column index and the number of iterations, starting at $1$.
Let $y$ be the row index, starting at $0$.
The content of cell $x,y$ is $f_1$ applied to the content of cell $x-1,\lfloor\frac{y}{2}\rfloor$. The parity of $y$ decides which 'path' of $f_1$ (even or odd) is taken. Or formulated another way: the table is built up recursively. Column 1 row 0 was the input for column 2 rows 0 and 1. Column 1 row 1 was the input for column 2 rows 2 and 3 and so on. The resulting large fractions are then factored into this form.
I've pasted the html of a table for the first 8 iterations to this page.
When written this way, the expressions exhibit some nice pattern, namely:
Every expression (apart from row 0) is of the form
$$ \frac{3^bn+q\cdot2^d}{2^x} $$
where $b$ is the hamming weight of $y$, i.e. the number of 1-bits in it's binary representation and $d = x - \log_2 (c) + 1$ with $c$ being the greatest power of 2 $\leq y$.
$q$ is the only thing which seems not be as easily parameterisable. However, it seems to be somehow similar to A035109, which is defined as
$$ \frac{1}{n}\sum_{d \mid n}{\mu\left(\frac{n}{d}\right)\sum_{e\mid d} e\sum_{\substack{e\mid d \\ e \text{ odd}}}e} $$
The first values of q are:
$0,1,1,5,1,7,5,19,1,11,7,29,5,23,19,65,1,19,\dots$ More can be read out from the linked table.
My question is: what formula could generate this sequence?
| This is not an exact answer to your question, but an simple observation.
Looks like a fractal or reccurence sequence, similar to fibonacci, or one think it might have recurrent relationships. Most likely this is a sequence which can not be shortcut, i.e. one might have to compute every step along the way to get the numbers in the sequence (this is a guess though).
$$0,0,1,1,5,1,7,5,19,1,11,7,29,5,23,19,65,1,19,…$$
I've highlighted it's fractal behavior by the numbers in red color:
$$0,\color{red}0,1,\color{red}1,5,\color{red}1,7,\color{red}5,19,\color{red}1,11,\color{red}7,29,\color{red}5,23,\color{red}{19},65,\color{red}1,19,…$$
$$\color{red}0,\color{red}0,\color{red}1,\color{red}1,\color{red}5,\color{red}1,\color{red}7,\color{red}5,\color{red}{19},\color{red}1,11,7,29,5,23,19,65,1,19,…$$
Now if we take the other sequence (it's quite different!), iv'e highlighted that in blue):
$$\color{blue}0,0,\color{blue}1,1,\color{blue}5,1,\color{blue}7,5,\color{blue}{19},1,\color{blue}{11},7,\color{blue}{29},5,\color{blue}{23},19,\color{blue}{65},1,\color{blue}{19},…$$
$$\color{blue}0,\color{blue}1,\color{blue}5,\color{blue}7,\color{blue}{19},\color{blue}{11},\color{blue}{29},\color{blue}{23},\color{blue}{65},\color{blue}{19},…$$
The sequence in blue which is part of your sequence is not in the OEIS. The question for the red sequence is wether one can take every fourth, eight, and so on to state that it is fractal. In your sequence one could try other simple techniques to find out if there are self recurring similarities.
As a sidenote: the ruler sequence has the same fractal behaviour, which is also related to the largest power of $p$ that divides $2^n$ in the Reduced Collatz Function that is applied only to the odd integers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Computing $\int_{|z-i|=\frac{3}{2}}\frac{e^{\frac{1}{z^2}}}{z^2+1}$
Compute the integral using residues: $\int_{|z-i|=\frac{3}{2}}\frac{e^{\frac{1}{z^2}}}{z^2+1}$
Inside the circumference there are the following singular points $-i$ which is a pole of order 1 and $0$ which is essential.
So:
$\int_{|z-i|=\frac{3}{2}}
\frac{e^{\frac{1}{z^2}}}{z^2+1}=res_{z_0=i}\frac{e^{\frac{1}{z^2}}}{z^2+1}+res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}=\frac{\pi}{e}+res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}$
$res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}=c_{-1}$ which is the first negative coefficient of the Laurent series. So developing the Laurent series:
$\frac{e^{\frac{1}{z^2}}}{z^2+1}=\sum_\limits{n=0}^{\infty}\frac{1}{n!z^{2n}}\sum_\limits{m=0}^{\infty}(-1)^m z^{2m}$
However I am not seeing the coefficient $c_{-1}$
Question:
How should I compute the coefficient $c_{-1}$?
Thanks in advance!
| Here is a way to calculate the integral without Laurent series to find the residue at $z = 0$. It uses that the sum of all residues is $0$ including the residue at infinity:
*
*$f(z)= \frac{e^{\frac{1}{z^2}}}{z^2+1} \Rightarrow$
$$ \operatorname{Res}_{z=0}f(z) + \operatorname{Res}_{z=i}f(z) = - \left( \operatorname{Res}_{z=-i}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) $$
Calculating the residues:
$$\operatorname{Res}_{z=-i}f(z) =\lim_{z\to -i}\frac{(z+i) e^{\frac{1}{z^2}}}{(z+i)(z-i)} = -\frac{e^{-1}}{2i}$$
$$\operatorname{Res}_{z=\infty}f(z) = -\operatorname{Res}_{w=0}\frac{1}{w^2}f\left(\frac{1}{w} \right)=-\operatorname{Res}_{w=0}\frac{1}{w^2}\frac{ e^{w^2}}{\frac{1}{w^2}+1} = -\operatorname{Res}_{w=0}\frac{ e^{w^2}}{1+w^2} = 0$$
The last residue is $0$ since $\frac{ e^{w^2}}{1+w^2}$ is holomorphic in a neighbourhood of $0$.
So, all together
$\int_{|z-i|=\frac{3}{2}}
\frac{e^{\frac{1}{z^2}}}{z^2+1}dz = -2\pi i\left(\operatorname{Res}_{z=-i}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) = -2\pi i\left( -\frac{e^{-1}}{2i}+0 \right) = \frac{\pi}{e}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find maximum of function $A=\sum _{cyc}\frac{1}{a^2+2}$ Let $a,b,c\in R^+$ such that $ab+bc+ca=1$. Find the maximum value of $$A=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}$$
I will prove $A\le \dfrac{9}{7}$ and the equality occurs when $a=b=c=\dfrac{1}{\sqrt3 }$
$$\frac{\sum _{cyc}\left(b^2+2\right)\left(c^2+2\right)}{\Pi _{cyc}\left(a^2+2\right)}\le \frac{9}{7}$$
$$\Leftrightarrow 7\sum _{cyc}a^2b^2+28\sum _{cyc}a^2+84\le 9a^2b^2c^2+18\sum _{cyc}a^2b^2+36\sum _{cyc}a^2+72 (1)$$
Let $a+b+c=3u;ab+bc+ca=3v^2=1;abc=w^3$ then we need to prove
$$9w^6+11\left(9v^4-6uw^3\right)+8\left(9u^2-6v^2\right)-12\ge 0$$
$$\Leftrightarrow 9w^6+11\left(27v^6-18uv^2w^3\right)+8\left(81u^2v^4-54v^6\right)-324v^6\ge 0$$
$$\Leftrightarrow 9w^6-198uv^2w^3-459v^6+648u^2v^4\ge 0$$
We have: $$f'\left(w^3\right)=18\left(w^3-11uv^2\right)\le 0$$
So $f$ is a decreasing function of $w^3$ it's enough to prove it for an equality case of two variables. Assume $a=b\rightarrow c=\dfrac{1-a^2}{2a}$
$$(1)\Leftrightarrow \frac{(a^2+2)(a^2+4)(3a^2-1)^2}{4a^2}\ge0$$
Please check my solution. It's the first time i use $u,v,w$, if i have some mistakes, pls fix for me. Thanks!
| $$a=\tan \left(\frac {\alpha}{2}\right), b=\tan \left(\frac {\beta}{2}\right), c=\tan \left(\frac {\gamma}{2}\right)\ \ \ \ (\alpha+\beta+\gamma= \pi)$$
$$A=\dfrac { \cos^2 \left(\frac {\alpha}{2}\right) }{1+ \cos^2 \left(\frac {\alpha}{2}\right) }+ \dfrac { \cos^2 \left(\frac {\beta}{2}\right) }{1+ \cos^2 \left(\frac {\beta}{2}\right) }+ \dfrac { \cos^2 \left(\frac {\gamma}{2}\right) }{1+ \cos^2 \left(\frac {\gamma}{2}\right) } \le \dfrac {3t}{3+t}\le \dfrac {9}{7} $$
$$t= \cos^2 \left(\frac {\alpha}{2}\right) + \cos^2 \left(\frac {\beta}{2}\right) + \cos^2 \left(\frac {\gamma}{2}\right)=\dfrac {3}{2} + \dfrac {1}{2} (\cos (\alpha)+\cos (\beta)+\cos (\gamma)) =\dfrac {3}{2}+\dfrac {R+r}{2R} \le \dfrac {9}{4}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to tell when a limit diverges? I have $\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$ which doesn't exist because it diverges. I spent awhile trying to remove the division by zero before plugging it into an online math calculator, and I want to know how I could have known that it diverges. I'm not skilled enough to just look at it and figure out there's no way for me to reform it so it doesn't have zero division when $x=-2$. Is there a way for me to find out whether it diverges other than just working on it for awhile and then guessing that the limit diverges if I can't get rid of the zero division?
My Steps:
$$\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$$
$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{2x+4(2x-4)}$$
$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{4x^2-16}$$
$$=\underset{x \to -2} \lim \frac{(x^2-1)(2x-4)}{4(x^2-4)}$$
$$=\underset{x \to -2} \lim \frac{(x-1)(x+1)2(x-2)}{4(x-2)(x+2)}$$
$$=\underset{x \to -2} \lim \frac{2(x-1)(x+1)}{4(x+2)}$$
$$=\underset{x \to -2} \lim \frac{(x-1)(x+1)}{2(x+2)}$$
$$=\underset{x \to -2} \lim \frac{x^2-1}{2x+4}$$
| $$
\frac{x^2}{2x+4}=
\frac{x^2-4+4}{2(x+2)}=
\frac{(x-2)(x+2)}{2(x+2)}+\frac{4}{2(x+2)}=\\
\frac{x-2}{2}+2\frac{1}{x+2}.
$$
It's not difficult to see that as $x$ approaches $-2$ from the left, $\frac{1}{x+2}$ goes to negative infinity: $\lim_{x\to-2^-}\frac{1}{x+2}=-\infty$. And as $x$ approaches $-2$ from the right, $\frac{1}{x+2}$ goes to positive infinity: $\lim_{x\to-2^+}\frac{1}{x+2}=+\infty$. So, that limit clearly does not exist and by extension it must be the case that the entire limit does not exist either. In more precise mathematical language, it all looks like this:
$$
\lim_{x\to-2^-}\frac{x^2}{2x+4}=\lim_{x\to-2^-}\left(\frac{x-2}{2}+2\frac{1}{x+2}\right)=-2+2\cdot(-\infty)=-2-\infty=-\infty,\\
\lim_{x\to-2^+}\frac{x^2}{2x+4}=\lim_{x\to-2^-}\left(\frac{x-2}{2}+2\frac{1}{x+2}\right)=-2+2\cdot(+\infty)=-2+\infty=+\infty.
$$
For a limit to exist, the two one-sided limits should be equal the same value. This is clearly not the case.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sqrt{3}\sin(x)+\cos(x)-2=0$ I need to solve the equation $$\sqrt{3}\sin(x)+\cos(x)-2=0$$
My try: I separated the radical then I squared and I noted $\cos(x)=t$ and I got a quadratic equation with $t=\frac{1}{2}$
To solve $\cos(x)=\frac{1}{2}$ I used formula $x\in\left \{ +-\arccos(\frac{1}{2})+2k\pi \right \}k\in\mathbb{Z}$ and I got $x\in\left \{ -+ \frac{\pi}{3}+2k\pi \right \}$ but the right answer is $x=(-1)^{k}\cdot\frac{\pi}{2}+k\pi-\frac{\pi}{6}$
Where's my mistake?How to get the right answer?
| Your mistake was forgetting that when you square a non-$0$ equality, the result satisfies two possible equations. In particular, both $x = y$ and $x = -y$ gives that $x^2 = y^2$. From what you describe, you did the following:
$$\sqrt{3}\sin(x) = 2 - \cos(x) \tag{1}\label{eq1}$$
then square both sides to get
$$3\sin^2(x) = 4 - 4\cos(x) + \cos^2(x) \tag{2}\label{eq2}$$
Next, you made certain manipulations to get
$3 - 3\cos^2(x) = 4 - 4\cos(x) + \cos^2(x) \; \Rightarrow \; 4\cos^2(x) - 4\cos(x) + 1 = 0$
Using $\cos(x) = t$ then gives
$$4t^2 - 4t + 1 = 0 \; \Rightarrow \; (2t - 1)^2 = 0 \; \Rightarrow \; t = \frac{1}{2} \tag{3}\label{eq3}$$
The full set of solutions for $\cos(x) = \frac{1}{2}$ is $x = \pm\frac{\pi}{3} + 2k\pi, k \in \mathbb{Z}$. However, note the RHS of \eqref{eq1} is always $\frac{3}{2}$, but for $x = \frac{\pi}{3} + 2k\pi$ the LHS is also $\frac{3}{2}$, but for $x = -\frac{\pi}{3} + 2k\pi$ the LHS is $-\frac{3}{2}$ instead. Squaring with either case still gives \eqref{eq2}.
Whenever you use a non-reversible operation, like squaring, it's important you check to remove any extraneous results you may have got from your manipulations. However, checking by substituting your results into the original equation is generally always a good idea in case you made some mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2)$ inequality I need to prove that $$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2) \leq 1/20$$ given $u + v + w = 9$ and $u,v,w$ positive reals.
Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(51 +w^2))$, $$or,$$
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2) \leq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$
In order to take the next step, I need to estimate the RHS with an upper bound. Here is where I face the problem. With $(u+v+w)^2 \leq 3(u^2 + v^2 + w^2)$ from C-S inequality and $u+v+w =9$ we get $27 \leq (u^2 + v^2 + w^2)$ which does not provide an upper bound. I tried several other inequalities, but each lead to a bound on the "wrong" side. Am I at all looking in the right direction?
| You're doing well. Note that you have $u^2+v^2+w^2$ in the denominator, and $1/x$ is a decreasing function. That means that
$$ \left( u^2+v^2+w^2 \ge \frac{(u+v+w)^2}{3} \right) \Rightarrow \left( \frac{9}{153+u^2+v^2+w^2} \le \frac{9}{153+\frac{(u+v+w)^2}{3}} \right)$$
EDIT: The answer is not correct, because the inequality $$\frac{1}{51+u^2} +\frac{1}{51+v^2}+ \frac{1}{51+w^2} \le \frac{9}{153+u^2+v^2+w^2} $$ is not correct. The correct one has $\ge$ sign, which doesn't help.
| {
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| The reason why the 1st method is wrong :
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan A $ and $ \tan B $
Settings discriminant greater than equal to zero.
It seems that you meant "the sum and product of $\color{red}{\tan C}$ and $\tan B$".
Considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt \frac 34\pi$.
This means that we have to consider the condition that $f(x)=0$ has at least one solution such that $x\lt -1$ or $x\gt 0$.
Therefore, the 1st method is wrong.
The reason why the 2nd method is wrong :
In the solution in Quora,
tan B + tan C = tanB tanC - 1
tanB+tan(3pi/4 - B) = p
This step is wrong. This should be
$$\tan B+\tan\left(\frac 34\pi-B\right)=\color{red}{p-1}$$
Then, we get
$$\tan^2B-(p-1)\tan B+p=0$$
But, again, similarly as the 1st method, considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt\frac 34\pi$.
Therefore, the 2nd method is wrong.
To make the two methods correct, we have to consider the condition that
$$x^2-(p-1)x+p=0$$
has at least one solution such that $x\lt -1$ or $x\gt 0$.
Solving this gives $$p\in (-\infty,0)\cup [3+2\sqrt 2,\infty)$$
which is the same as the answer in the third method.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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In a triangle, prove that $ \sin A + \sin B + \sin C \leq 3 \sin \left(\frac{A+B+C}{3}\right) $
Prove that for any $\Delta ABC$ we have the following inequality:
$$ \sin A + \sin B + \sin C \le 3 \sin \left(\frac{A+B+C}{3}\right) $$
Could you use AM-GM to prove that?
| I'm sure there's a different way to approach this question, but here's one way using the graph of $\sin x $:
Consider $3$ points on the graph for $x\in(0,\pi)$. They are $(A,\sin A)$, $(B,\sin B)$, and $(C,\sin C)$ as shown. $A,B,C$ are such that $A+B+C=\pi$.
(Ignore the fact that $A,B$ and $C$ are angles of a triangle, they are just some values of $x$ that satisfy the condition $A+B+C=\pi$)
$A, B$ and $C$ are plotted on the graph of $y=\sin x$ and are joined to form a triangle, as shown.
Consider the centroid of the triangle, $G$, given by:
$$G=\left(\dfrac{A+B+C}{3}, \dfrac{\sin A+\sin B+\sin C}{3}\right)$$
Draw a line $PG$ as shown at $x=\dfrac{A+B+C}{3}$ (or $\dfrac{\pi}{3}$). This line intersects the curve at the point $P$ given by:
$$P=\left(\dfrac{A+B+C}{3}, \sin \left(\dfrac {A+B+C}{3}\right)\right)$$
From the figure, it is evident that the $y$-value of $P$ $>$ the $y$-value of $G$.
So we obtain the inequality $$\sin \left(\dfrac {A+B+C}{3}\right) >\dfrac{\sin A+\sin B+\sin C}{3}$$
Note that, for the case where $A=B=C$, we can deduce that $A=B=C= \dfrac{\pi}{3}$ and therefore $A,B$ and $C$ will coincide with point $P$ on the curve. For this particular case, we can deduce that $$\sin \left(\dfrac {A+B+C}{3}\right) =\dfrac{\sin A+\sin B+\sin C}{3}$$
Combining both the inequalities obtained, we get the desired result:
$$\sin \left(\dfrac {A+B+C}{3}\right) \geq \dfrac{\sin A+\sin B+\sin C}{3}$$
or
$$3 \sin \left(\dfrac {A+B+C}{3}\right) \geq \sin A+\sin B+\sin C$$
Credit for the answer: "Play with Graphs" by Amit Agarwal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples] The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that,
$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$
and,
$$Q(x)=x^4+x^3+x^2+x+1$$
what is the remainder of $P(x)$ divided by $Q(x)$?
The given answer was:
Let $Q(x)=0$. Multiplying both sides by $x-1$:
$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$
Substituting $x^5=1$ in $P(x)$ gives $x^4+x^3+x^2+x+1$. Thus,
$$P(x)\equiv\mathbf0\pmod{Q(x)}$$
Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
| I would have thought that bright students, who knew $1+x+x^2+\cdots +x^{n-1}= \frac{x^n-1}{x-1}$ as a geometric series formula, could say
$$\dfrac{P(x)}{Q(x)} =\dfrac{x^{104}+x^{93}+x^{82}+x^{71}+1}{x^4+x^3+x^2+x+1}$$
$$=\dfrac{(x^{104}+x^{93}+x^{82}+x^{71}+1)(x-1)}{(x^4+x^3+x^2+x+1)(x-1)}$$
$$=\dfrac{x^{105}-x^{104}+x^{94}-x^{93}+x^{83}-x^{82}+x^{72}-x^{71}+x-1}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-\dfrac{x^{104}-x^{94}}{x^5-1}-\dfrac{x^{93}-x^{83}}{x^5-1}-\dfrac{x^{82}-x^{72}}{x^5-1}-\dfrac{x^{71}-x}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-x^{94}\dfrac{x^{10}-1}{x^5-1}-x^{83}\dfrac{x^{10}-1}{x^5-1}-x^{72}\dfrac{x^{10}-1}{x^5-1}-x\dfrac{x^{70}-1}{x^5-1}$$
and that each division at the end would leave zero remainder for the same reason, replacing the original $x$ by $x^5$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where were shown how with mathematical induction prove equations like $(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$.
Now I'm trying to complete exercise where I have to find sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$.
I tried to do it with mathematical induction. Like this:
n=1: $\frac{1}{1+x^2}$
n=2: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}=\frac{1+1+x^2}{(1+x^2)^2}$
n=3: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}=\frac{(1+x^2)^2+2+x^2}{(1+x^2)^3}$
...
And so on (I've calculated til n=5). But I don't see any consistent pattern to evaluate sum of progression.
After that I found formulas of geometrical progression:
$q=\frac{b_{n+1}}{b_n}$ and $S_n=\frac{b_1(1-q^n)}{1-q}$, so I've evaluated:
$q=\frac{\frac{1}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{1}{1+x^2}$
and
$S_n=\left(\frac{1}{1+x^2}\left[1-\left(\frac{1}{1+X^2}\right)^n\right]\right):\left(1-\left[\frac{1}{1+x^2}\right]\right)=\left[\frac{1}{1+x^2}-\left(\frac{1}{1+x^2}\right)^{n+1}\right]\frac{1+x^2}{x^2}=\frac{\left[1-\left(\frac{\sqrt[n+1]{1+x^2}}{1+x^2}\right)^{n+1}\right]}{x^2}$
First of all, I'd like to know how to find sum of geometrical progression with mathematical induction.
Secondly, I'd like to know what is wrong with my evaluations.
| Hint:
For a geometric series with ratio $q$:
$$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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System of first order linear differential equations
The solutions of a homogeneous system of linear differential equations of the first order $$\frac{d \vec x}{d t} = A \vec x$$ are $$\vec x = e^{ \lambda t} \vec v $$ where $ \lambda $ are the eigenvalues of A and $\vec v$ are the associated eigenvectors.
So the first and second options are true. As for the others I'm not sure.
| Consider the matrix
\begin{align}
A =&\
\begin{pmatrix}
1 & -1 & 1 & 0\\
-1 & 2 & 0 & 1\\
1 & 2 & 0 & 0 \\
-2 & -1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
-2 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
1 & -1 & 1 & 0\\
-1 & 2 & 0 & 1\\
1 & 2 & 0 & 0 \\
-2 & -1 & 0 & 0
\end{pmatrix}^{-1}\\
=&\ \frac{1}{3}\begin{pmatrix}
0 & 0 & 4 & 5\\
0 & 0 & -6 & -6\\
0 & 0 & -2 & 2 \\
0 & 0 &-2 & -7
\end{pmatrix}
\end{align}
then it's clear that $v=(-1, 2 ,2, -1)^T$ is an eigenvector with eigenvalue $-1$. Likewise, $w=(1, -1, 1, -2)^T$ is an eigenvector with eigenvalue $-2$. However, 3 is not an eigenvalue of $A$ so $(A-3I)\mathbf{x} = \mathbf{0}$ has only a trivial solution. It's also easy to check that the last statement is false.
| {
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"timestamp": "2023-03-29T00:00:00",
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How much ways there are to produce sum = 21 with 4 different natural numbers? 0 isn't natural number , and the sum way is important (e.g. 3+5+6+7 is different from 5+6+7+3).
I got that I have for a+b+c+d = 21 , 2024 options.
I think I need to sub the invalid numbers , so I need to relate for the cases :
a = 0 , b = 0, c = 0, d = 0 ,
a = b , a = c , a = d, b = c, b = d, c = d?
| $$
\eqalign{
& N = {\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a \ne b \ne c \ne d} \right.} \right) \cr
& \quad \Downarrow \cr
& N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c + d = 21\quad \left| {\;0 < a < b < c < d} \right.} \right) \cr
& \quad \Downarrow \cr
& 0 < a = x_{\,1} \quad b = x_{\,2} + 1\quad c = x_{\,3} + 2\quad d = x_{\,4} + 3 \cr
& \quad \Downarrow \cr
& N = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1} + x_{\,2} + 1 + x_{\,3} + 2 + x_{\,4} + 3 = 21\quad
\left| {\;0 < x_{\,1} \le x_{\,2} \le x_{\,3} \le x_{\,4} } \right.} \right) = \cr
& = 4!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 15\quad
\left| {\;0 < x_{\,1} \le x_{\,2} \le x_{\,3} \le x_{\,4} } \right.} \right) = \cr
& = 4!\;{\rm N}{\rm .}\,{\rm of}\,{\rm partitions}\,{\rm of}\,15\;{\rm into}\,4{\rm parts} = \cr
& = 4!\;27 = 648 \cr}
$$
which checks with direct computation.
To exemplify that, let's take a case with smaller values
$$
\eqalign{
& N = {\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c = 9\quad \left| {\;0 < a \ne b \ne c} \right.} \right) \cr
& \quad \Downarrow \cr
& N = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {a + b + c = 9\quad \left| {\;0 < a < b < c} \right.} \right) \cr
& \quad \Downarrow \cr
& N = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1} + x_{\,2} + 1 + x_{\,3} + 2 = 9\quad \left| {\;0 < x_{\,1} \le x_{\,2} \le x_{\,3} } \right.} \right) = \cr
& = 3!\;{\rm No}{\rm .}\,{\rm of}\,{\rm sol}{\rm .}\left( {x_{\,1} + x_{\,2} + x_{\,3} = 6\quad \left| {\;0 < x_{\,1} \le x_{\,2} \le x_{\,3} } \right.} \right) = \cr
& = 3!\;{\rm N}{\rm .}\,{\rm of}\,{\rm partitions}\,{\rm of}\,6\;{\rm into}\,3\;{\rm parts} = \cr
& = 3!\;3 = 18 \cr}
$$
and in fact
$$
\underbrace {\left[ {1,1,4} \right],\left[ {1,2,3} \right],\left[ {2,2,2} \right]}_{partit.\;x_{\,1} \le x_{\,2} \le x_{\,3} }\quad \Rightarrow \quad
\underbrace {\left[ {1,2,6} \right],\left[ {1,3,5} \right],\left[ {2,3,4} \right]}_{ordered\;0 < a < b < c}
$$
are the only ordered triplets of different positive integers that sums to $9$.
Since they contain different integers, you can permute each of them
to obtain that the number of unordered triplets is $18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reasoning behind integrating f(x)/g(x)? I understand the method to integrate this function would be:
$\int{\frac{x^2+1}{x^4-x^2+1} \thinspace dx}$
Divide all terms by $x^2$:
$= \int{\frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x^4}{x^2}-\frac{x^2}{x^2}+\frac{1}{x^2}} \thinspace dx}$
=$ \int{\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}} \thinspace dx}$
Factor the denominator:
=$ \int{\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2 + 1} \thinspace dx}$
Use $u$-substitution:
$ u = x -\frac{1}{x}, du = 1 +\frac{1}{x^2} \thinspace dx$
$ \int{\frac{du}{u^2+1}} $
$= \tan^{-1}{(x-\frac{1}{x})}+C$
I can verify that this is the correct method using the chain rule:
$u = x-\frac{1}{x}, \frac{du}{dx} = 1 +\frac{1}{x^2} $
$y = \tan^{-1}{u}, \frac{dy}{du} = \frac{1}{u^2+1} $
$ \frac{dy}{dx} = \frac{1}{u^2+1} \times (1 +\frac{1}{x^2})$
$= \frac{1 +\frac{1}{x^2}}{u^2+1} $
$ = \frac{1 +\frac{1}{x^2}}{u^2+1} $
$ = \frac{\frac{x^2 +1}{x^2}}{(x-\frac{1}{x})^2+1} $
$ = \frac{\frac{x^2 +1}{x^2}}{x^2-2+\frac{1}{x^2}+1} $
$ = \frac{\frac{x^2 +1}{x^2}}{\frac{x^4-x^2+1}{x^2}} $
$ = \frac{x^2 +1}{x^4-x^2+1} $
Only after differentiating the resulting integral, does it make sense where the $x^2$ comes from.
As some terms are fractions, distributing the denominator and dividing all terms by the same common denominator simplifies the function.
$$ \frac{\frac{x^2 +1}{x^2}}{\frac{x^4-x^2+1}{x^2}} = \frac{x^2 +1}{x^4-x^2+1} $$
So it makes sense then that this is the term we will multiply with the original integral in order to make it more useful to work with.
Firstly, is there a name for this technique, that is, multiply all terms by some function of x in order to make it easier to factor into something integratable?
Is there a telltale way to recognise when to use this method, and what the divisor should be (obviously without knowing the answer and working backwards)?
Are there any other ways to solve this integral?
Sometimes I don't feel this method is intuitive enough to remember in exam conditions.
Thanks
| Here's another way to do it. I would go for this way if I don't immediately see a trick.
It's a big theorem that all rational functions have elementary antiderivatives. The general way to integrate a rational function is to factor it into quadratics and linears (this is always possible by FTA), and use partial fractions decomposition.
For our specific example, we have to factor $x^4-x^2+1$. The following is a somewhat common trick:
$$x^4-x^2+1 = x^4+2x^2+1 - 3x^2=(x^2+1)^2-(\sqrt3x)^2 = (x^2-\sqrt3x+1)(x^2+\sqrt3x+1)$$
For the partial fraction decomposition, it seems reasonable that the numerators should be constants; fortunately, we can even guess them
$$\dfrac {x^2+1}{(x^2-\sqrt3x+1)(x^2+\sqrt3x+1)}= \dfrac {1/2}{x^2-\sqrt3x+1}+ \dfrac {1/2}{x^2+\sqrt3x+1}$$
And now the way you integrate $\frac {1}{ \text{quadratic}}$ is you complete the square on the bottom.
Completing the Square: From $x^2+bx+c$ we can complete the square to $\left(x+\dfrac {b}{2}\right)^2+ \left( c-\dfrac {b^2}{4} \right)$. This means that
$$\dfrac {1}{x^2+\sqrt3x+1} = \dfrac {1}{(x+\frac {\sqrt 3}{2})^2+\frac 14}$$
Let me know if you need more help.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
| First, this is stated as an equation to solve (for $x$) rather than an identity to be shown.
So with $a=\sqrt {1+x}$ and $b=\sqrt {1-x}$ we have $$a^2+b^2=2$$ and $$(a+b)^2=a^2+2ab+b^2=2(1+ab)$$ and $$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(2-ab)$$
Then $$\sqrt {1+ab}\cdot (a^3+b^3)=\frac {\sqrt 2}2(a+b)(a+b)(2-ab)=\sqrt 2(1+ab)(2-ab)$$
If we then put $c=ab$ the equation to solve is then $$\sqrt 2(1+c)(2-c)=2+c$$ which is a straightforward quadratic in $c$. Then solve for $x$ by noting $c^2=1-x^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Modular exponentitation (Finding the remainder) I want to find the remainder of $8^{119}$ divided by $20$ and as for as i do is follows:
$8^2=64\equiv 4 \pmod {20} \\
8^4\equiv 16 \pmod {20} \\
8^8\equiv 16 \pmod {20}\\
8^{16}\equiv 16 \pmod {20}$
from this i see the pattern as follows $8^{4\cdot 2^{n-1}} \text{is always} \equiv 16 \pmod {20} \,\forall n \ge 1$
So,
$\begin{aligned} 8^{64}.8^{32}.8^{16}.8^7 &\equiv 16.8^7 \pmod{20}\\
&\equiv 16.8^4.8^3 \pmod{20} \\
&\equiv 16.8^3 \pmod {20}\end{aligned}$
And i'm stuck. Actually i've checked in to calculator and i got the answer that the remainder is $12$. But i'm not satisfied cz i have to calculate $16.8^3$
Is there any other way to solve this without calculator. I mean consider my condisition if i'm not allowed to use calculator.
Thanks and i will appreciate the answer.
| Although $8^{4\cdot 2^{n-1}}\equiv 16\pmod {20} $ for all positive integer $n $, it is actually much simpler:
$8^{4k}\equiv 16\pmod {20} $ for all $k>0$. That is for any multiple of $4$, not just $4$times powers of $2$.
And furthermore $8^{4k+1}\equiv 16*8\equiv -4*8\equiv-32\equiv 8\pmod {20} $ for $k>0$
And $8^{4k+2}\equiv 8*8\equiv 4\pmod {20} $
And $8^{4k+3}\equiv 4*8\equiv 32\equiv 12\pmod {20} $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inverse of an upper bidiagonal Toeplitz matrix I have a matrix with the following structure
$$\left[\begin{array}{cccccc|c}
-1 & 1-b & 0 & \dots & 0 & 0 & b \\
0 & -1 & 1-b & \dots & 0 & 0 & b \\
\cdots \\
0 & 0 & 0 & \dots &-1 & 1-b & b \\
0 & 0 & 0 & \dots & 0 & -1 & 1 \\ \hline
0 & 0 & 0 & \dots & 0 & 0 & -1
\end{array}\right]$$
I have to find the inverse of this matrix. I begin by using the block matrix inversion formula
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&-\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\end{bmatrix}$$
where
$$\mathbf{A} = \begin{bmatrix}
-1 & 1-b & 0 & \dots & 0 & 0 \\
0 & -1 & 1-b & \dots & 0 & 0 \\
\cdots \\
0 & 0 & 0 & \dots &-1 & 1-b \\
0 & 0 & 0 & \dots & 0 & -1
\end{bmatrix}$$
$$\mathbf{B} = \begin{bmatrix} b \\ b \\ \vdots \\ b \\ 1 \end{bmatrix}$$
$$\mathbf{C} = \begin{bmatrix} 0 & 0 & 0 & \dots & 0 & 0 \end{bmatrix}$$
$$\mathbf{D} = \begin{bmatrix} -1 \end{bmatrix}$$
Given that $\mathbf{C}$ and $\mathbf{D}$ I can simplify the formula as
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\mathbf{A}^{-1} & \mathbf {A}^{-1} \mathbf {B}\\ \mathbf{0} & -1\end{bmatrix}$$
The only part left is to solve for $\mathbf{A}$ which I believe can be called an upper bidiagonal Toeplitz matrix. Unfortunately, I have not been able to find a formula to compute the inverse for the same.
| Elements $m_{ij}$ of the inverse matrix $M=A^{-1}$ are:
\begin{align}
m_{ij}
&=
\begin{cases}
-(1-b)^{j-i}, &j\ge i,
\\
0, &j<i.
\end{cases}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility by 3 Maybe this is a duplicate question (if so, can the moderators be kind enough to merge this appropriately?), but what is the condition on $a$ and $b$ for an expression $am+b$ to be divisible by $3$ ($a$ and $b$ are integers)? For example, I can say $16m+3$ is divisible by $3$ since $b=3$ is divisible by $3$ and for $m$ a multiple of $3$, $3|16m$. In the case of $176m+23$, I don't think it is divisible by $3$. Is this the case?
| A basic principle is that if $3|v$, then $3|w \iff 3|v+w$. Furthermore, if $3|v$ then
$3|nv$.
So $3|ma+b\iff3|ra+s,$ where $r$ is the remainder when $m$ is divided by $3,$
and $s$ is the remainder when $b$ is divided by $3$.
Now we have only $9$ possibilities to consider:
$r=0, 1, $ or $2, $ and $s=0, 1, $ or $2$.
When $r=0$ and $s=0$, $ra+s=0$ is divisible by $3$.
When $r=0$ and $s=1$ or $2$, $ra+s=1$ or $2$ is not divisible by $3$.
When $s=0$ and $r=1$ or $2$, then $ra+s=ra$ is divisible by $3$ when $a$ is and not when $a$ is not.
When $r=1$ and $s=1,$ then $ra+s=a+1,$ so $ra+s$ is divisible by $3$ when $a$ leaves remainder $2$ when divided by $3$.
When $r=1$ and $s=2,$ then $ra+s=a+2,$ so $ra+s$ is divisible by $3$ when $a$ leaves remainder $1$ when divided by $3$.
When $r=2$ and $s=1,$ then $ra+s=2a+1,$ so $ra+s$ is divisible by $3$ when $a$ leaves remainder $1$ when divided by $3$.
When $r=2$ and $s=2,$ then $ra+s=2a+2,$ so $ra+s$ is divisible by $3$ when $a$ leaves remainder $2$ when divided by $3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given three postive numbers $a,b,c$ so that $a\geqq b\geqq c$. Prove that $\sum\limits_{cyc}\frac{a+bW}{aW+b}\geqq 3$ .
Given three postive numbers $a, b, c$ so that $a\geqq b\geqq c$. Prove that
$$\sum\limits_{cyc}\frac{a+ b\sqrt{\frac{b}{c}}}{a\sqrt{\frac{b}{c}}+ b}\geqq 3$$
I make it
Firstly, we need to have one general inequality
$$\sum\limits_{cyc}\frac{a+ bW}{aW+ b}\geqq 3$$
By buffalo-way, let $c= 1, b= 1+ u, a= 1+ u+ v$ so $W= \sqrt{1+ u}$, be homogeneous. I guess that
$${\rm W}= \sqrt{\frac{b}{c}}$$
An inspiration-one Given three positive numbers $a,b,c$ so that $a\leqq b\leqq c$. Prove that $\sum\limits_{cyc}\frac{a+1.4b}{1.4a+b}\geqq 3$ .
| Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, we need to prove that
$$\sum_{cyc}\frac{a+\sqrt{\frac{b}{c}}b}{\sqrt{\frac{b}{c}}a+b}\geq3$$ or
$$\sum_{cyc}\frac{x^2z+y^3}{x^2y+y^2z}\geq3.$$
Now, by AM-GM
$$\sum_{cyc}\frac{x^2z+y^3}{x^2y+y^2z}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(y^3+x^2z)}{\prod\limits_{cyc}(x^2y+y^2z)}}$$ and it's enough to prove that
$$\prod\limits_{cyc}(x^3+z^2y)\geq\prod\limits_{cyc}(x^2y+y^2z)$$ or
$$\sum_{cyc}(x^5z^4+x^6y^2z)\geq xyz\sum_{cyc}(x^3y^3+x^4yz),$$ which is true by Rearrangement twice:
$$\sum_{cyc}x^5z^4=x^4y^4z^4\sum_{cyc}\frac{x}{y^4}\geq x^4y^4z^4\sum_{cyc}\frac{x}{x^4}=xyz\sum_{cyc}x^3y^3$$ and
$$\sum_{cyc}x^6y^2z=x^2y^2z^2\sum_{cyc}\frac{x^4}{z}\geq x^2y^2z^2\sum_{cyc}\frac{x^4}{x}=xyz\sum_{cyc}x^4yz.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{4^p - 1}{3}$ is a Fermat pseudoprime with respect to 2 I have to prove that $n = \frac{4^p - 1}{3}$ is a Fermat pseudoprime with respect to $2$ when $p \geq 5$ is a prime number. I have proved that $n$ is not prime because $4^p - 1 = (2^p-1)(2^p+1)$ and $(2^p + 1)$ is divisible by $ 3$. But now I can't show that $2^{n-1} \equiv 1\bmod n$.
I calculated that $2^{n-1} = 2^{(2^p + 2)(2^p-2)/3}$ but I don't know if I can deduce anything from this.
| $n=\dfrac{4^p-1}3=\dfrac{2^{2p}-1}3,$ so $n\mid 2^{2p}-1,\,$ so $\,\color{#c00}{2^{2p}\equiv 1}\pmod{\!n}$
Further, $2p$ divides $2\times\dfrac{(2^{p-1}-1)}3\times{(2^{p}+2)}=\dfrac{2^{2p}-4}3=n-1,\,$ so $n-1 = 2pk$
Therefore, $\!\bmod n,\,$ we have $\,2^{n-1}\equiv (\color{#c00}{2^{2p}})^{k}\equiv \color{#c00}1^k\equiv 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| $4(\cos^6\theta-\sin^6\theta)$
$=4((\cos^2\theta)^3-(\sin^2\theta)^3)$
$=4(\cos^2\theta-\sin^2\theta)(\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta)$
$=4\cos 2\theta[\{(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$
$=4\cos 2\theta[\{1-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$
$=4\cos 2\theta(1-\cos^2\theta\sin^2\theta)$
$=4\cos2\theta-\cos2\theta\sin^22\theta$
$=4\cos2\theta-\cos2\theta(1-\cos^22\theta)$
$=\cos^32\theta+3\cos2\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of equation based on some geometry hints Plots of the equations $y = 8 - x^2$ and $|y|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$. We have to solve the equation $$8-x^2=\sqrt{8+x}$$
| If $y=8-x^2$ and $|y|=\sqrt{8+x}$, then $y^2-y=(8+x)-(8-x^2)=x+x^2,$
so $y(y-1)+x(-1-x)=0,$ so $y(y-1-x)+x(y-1-x),$ so $(y+x)(y-1-x)=0,$
i.e., $y=-x$ or $y=1+x$. Therefore $x$ must be a solution of $8-x^2=-x$ or $8-x^2=1+x$.
Can you solve these quadratic equations? Note that you want solutions where $y=8-x^2\ge0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
| Hint:
For real $x,$ $x^2\ge0$, so $x^2+2>0$, so $|x^2+2|=x^2+2$. $|x^2-11|=x^2-11$ or $-(x^2-11).$ Equate them and see what you get.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate limit as $n$ tends to infinity of $\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$? This question stems from and old revision of this question, in which an upper bound for $n!$ was asked for.
The original bound was incorrect. In fact, I want to show that the given expression divided by $n!$ goes to $0$ as $n$ tends to $\infty$.
I thus want to show:
$$\lim_{n\to\infty}\frac{(n+1)^{n^2+n+1}}{n!(n+2)^{n^2+1}}=0.$$
Using Stirling's approximation, I found that this is equivalent to showing that
$$\lim_{n\to\infty} \frac{\exp(n)}{\sqrt n}\cdot\left(\frac{n+1}{n+2}\right)^{n^2+1}\cdot\left(\frac{n+1}{n}\right)^n=0.$$
However, I don't see how to prove the latter equation.
EDIT: It would already be enough to determine the limit of $$\exp(n)\left(\frac{n+1}{n+2}\right)^{(n^2)}\left(\frac{n+1}{n}\right)^n$$ as $n$ goes to $\infty$.
| $$\frac{(n+1)^{n^2+n+1}}{n! (n+2)^{n^2+1}}$$
= $$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^{n^2+n+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} \frac{n^nn^{n^2+1}}{n^{n^2+1}}$$
=$$\frac{(1+\frac{1}{n})^{n^2+n+1}}{n! (1+\frac{2}{n})^{n^2+1}} n^n$$
=$$\frac{((1+\frac{1}{n})^{n})^n(1+\frac{1}{n})^{n}(1+\frac{1}{n})}{n! ((1+\frac{2}{n})^{n})^n (1+\frac{2}{n}) } n^n$$
at the limit n tends to infinity, using the standard definition of log(z) = $\lim_{x\to\infty}(1 + 1/z)^z$
=$$\frac{e^n.e.1}{n!e^{2n}e^2} n^n$$
=$$\frac{ n^n}{n!e^{n+2}} $$
is how far I got really
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $\{a_n\}$ be a sequence of positive real numbers such that $\sum_{n=1}^\infty a_n$ is divergent. Which of the following series are convergent? Let $\{a_n\}$ be a sequence of positive real numbers such that
$\sum_{n=1}^\infty a_n$ is
divergent. Which of the following series are convergent?
a.$\sum_{n=1}^\infty \frac{a_n}{1+a_n}$
b.$\sum_{n=1}^\infty \frac{a_n}{1+n a_n}$
c.
$\sum_{n=1}^\infty \frac{a_n}{1+ n^2a_n}$
My Solution:-
(a)Taking $a_n=n$, then $\sum_{n=1}^\infty \frac{n}{1+n}$ diverges.
(b) Taking $a_n=n$, $\sum_{n=1}^\infty \frac{n}{1+n^2}$ diverges using limit comparison test with $\sum_{n=1}^\infty \frac{1}{n}$
(c) $\frac{a_n}{1+n^2a_n}\leq \frac{a_n}{n^2a_n}=\frac{1}{n^2} $. Using comparison test. Series converges. I am not able to conclude for general case for (a) and (b)?
| If $\sum_{n=1}^{\infty} a_n$ be a divergent series of positive real numbers prove that the
series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$ is divergent.
Proof:
Let $S_n = a_1 +a_2 + ... +a_n$ .
Since the series $\sum_{n=1}^{\infty} a_n$ is a divergent series of positive real numbers, the
sequence $\{S_n\}$ is a monotone increasing sequence and $\lim S_n = \infty.$
Therefore for every natural number $n$ we can choose a natural number
$p$ such that $S_{n+p} > 1 + 2S_{n}$ . Now, $\frac{a_{n+1}}{1+a_{n+1}}+\frac{a_{n+2}}{1+a_{n+2}}+....+\frac{a_{n+p}}{1+a_{n+p}} >\frac{a_{n+1}}{1+S_{n+1}}+\frac{a_{n+2}}{1+S_{n+2}}+....+\frac{a_{n+p}}{1+S_{n+p}}$ (since $S_{n+p}\ge S_{n+1}\ge a_{n+1}$....$S_{n+p}\ge a_{n+p}$)
Now $\frac{a_{n+1}}{1+S_{n+1}}+\frac{a_{n+2}}{1+S_{n+2}}+....+\frac{a_{n+p}}{1+S_{n+p}}= \frac{S_{n+p}-S_n}{1+S_{n+p}}>\frac{\frac{1}{2}(1+S_{n+p})}{1+S_{n+p}}=\frac{1}{2}$
This shows that Cauchy's principle of convergence is not satisfied by
the series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$ Hence the series is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
| $$x=\pm i$$
so
$$a(\pm i)^3-a(\pm i)^2+b(\pm i)+4=0$$
so
$$ai+a\pm bi+4=0$$
$$ai+a+bi+4=0\tag 1$$
or
$$ai+a-bi+4=0\tag 2$$
now we will solve the first equation
$$a+4=0\rightarrow a=-4$$
$$a-b=0\rightarrow b=-4$$
hence $$f(x)=-4x^3+4x^2-4x+4$$
at $x=4$
$$f(4)=-204=-4(51)$$
the second equation gives
$$a=-4$$
$$b=4$$
hence $$f(x)=-4x^3+4x^2+4x+4$$
at $x=4$
$$f(4)=-176=-4(43)$$
as @marty said the problem is wrong
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to use L Hospital's rule for $\lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) $ I want to find the following limit using L Hospital's rule:
$$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) $$
I know that this can be solved using squeezed theorem from Cal 1:
$$ 0 < \sqrt{x}\sin( \frac{1}{x} ) < \frac{1}{x} $$
since $0 < \sin( \frac{1}{x}) < \frac{1}{x} $.
What I have done so far is trying to convert it to fraction form
$$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) = \lim_{x \to \infty} \frac{\sin( \frac{1}{x})}{\frac{1}{\sqrt{x}} }$$
But what next?
| Consider the Taylor series expansion for $\sin(x)$.
$$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
For $x^{-1}$, this series is
$$\frac{1}{x} - \frac{1}{3! *x^3} + \frac{1}{5! * x^5} - \ldots$$
This is equivalent to $O\Big(\frac{1}{x}\Big)$ (meaning $\sin(\frac{1}{x}) \to \frac{1}{x}$ as $x \to \infty$). So rewrite your equation as
$$\lim_{x \to \infty} \sqrt{x} *O\Big(\frac{1}{x}\Big) = O\Big(\frac{1}{\sqrt{x}}\Big)$$
Which goes to $0$ as x goes to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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challenging sum $\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}$ How to prove that
\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)
\end{align}
where $H_n^{(m)}=1+\frac1{2^m}+\frac1{3^m}+...+\frac1{n^m}$ is the $n$th harmonic number of order $m$.
This problem was proposed by Cornel Valean.
Here is integral expression of the sum $\displaystyle -\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx\quad $.
| Different approach:
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^2}\\
&=\sum_{n=1}^\infty H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\
&=-\int_0^1\ln x\sum_{n=1}^\infty(x^2)^nH_n^{(2)}\\
&=-\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx,\quad \operatorname{Li}_2(x^2)=2\operatorname{Li}_2(x)+2\operatorname{Li}_2(-x)\\
&=-2\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x^2}\ dx-2\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x^2}\ dx\\
&=-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\
&=-I_1-I_2-I_3-I_4
\end{align}
\begin{align}
I_1&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}\\
&=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}+\zeta(4)
\end{align}
\begin{align}
I_2&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}\ dx\\
&=-\sum_{n=1}^\infty (-1)^n\int_0^1\ x^{n-1}\ln x\operatorname{Li}_2(x)\ dx\\
&=-\sum_{n=1}^\infty (-1)^n\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n}{n^3}-\frac{2\zeta(2)}{n^2}\right)
\end{align}
\begin{align}
I_3&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^n \ln x}{1-x}\ dx\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(H_n^{(2)}-\zeta(2)\right)
\end{align}
\begin{align}
I_4&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\
&=-\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{(n+1)^2}\\
&=\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n^2}+\frac78\zeta(4)
\end{align}
Combine the four integrals we get
$$S=\frac98\zeta(4)+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-\sum_{n=1}^\infty(-1)^n\frac{H_n^{(2)}}{n^2}$$
Plugging the two sums we get
$$S=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)
$$
| {
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"url": "https://math.stackexchange.com/questions/3259011",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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