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12

$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1+ \sqrt x$$ Squaring $$ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = 1+ 2\sqrt x$$ Squaring $$ \sqrt {16x + \sqrt {64x + 5}} = 1+ 4\sqrt x$$ Squaring $$ \sqrt {64x + 5} = 1+ 8\sqrt x$$ Squaring $$ 5 = 1+ 16\sqrt x$$


7

This is a very intuitive approach based on the fact that I assume to be in the exam room with no computer and even no calculator. If there is a simple root $\sqrt{64x+5}$ should reduce to a whole number and $x=\frac1{16}$ is obvious (since $9$ is the closest square to $5$). From here, we can go backward (verify that every time we get another square) and ...


6

You get rid of the square roots by successive squarings and changes of side. $$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1,$$ $$\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = -x+1,$$ $$\sqrt {16x + \sqrt {64x + 5}} = x^2-6x+1,$$ $$\sqrt {64x + 5} = x^4-12x^3+38x^2-28x+1,$$ $$0=x^8-24x^7+220x^6-968x^5+2118x^4-2152x^3+860x^2-120x-4.$$ Using a ...


4

The best solution I can imagine is to square repeatedly and then use numerical methods to find roots of the resulting polynomial and then checking for extranneous answers from our squaring.


4

\begin{align*} \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x & = 1\\ \sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\ x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\ 4x + \sqrt {16x + \sqrt {64x + 5}} & = ...


3

With the extra $\sqrt x$ there, you just square both side and keep doing it, miraculously some terms cancel out :)


3

There may be some cases where complex variables, the residue theorem and the residue at infinity are helpful. Suppose your OGF is $f(z)$ and the desired EGF is $g(w).$ Then we have $$g(w) = \sum_{n\ge 0} \frac{w^n}{n!} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$ This will simplify together with some conditions on convergence ...


2

(a) Fucntion $y$ describes strainght line iff coefficient of $x^2$ is $0$, so $$1-2\cos \theta=0\implies\cos\theta=\frac12$$ From this we get $$\sin \theta =\pm\sqrt{1-\cos^2\theta}=\pm\frac{\sqrt3}2$$ So, equations of these lines are $$y=\pm\frac{\sqrt3}2x+\frac12$$ (b) Let $y=0$. Discriminant of quadratic equation for $x$ is ...


2

Ok let's see: There is a famous integral representation for the Hurwitz function due to Hermite which reads as follows: $$ \zeta(s,a)= \frac{a^{-s}}{2}+\frac{a^{1-s}}{1-s}+2\int_0^{\infty}dt \frac{\sin(s\arctan(t/a))}{(e^{2 \pi t}-1)(t^2+a^2)^{s/2}} \quad (1) $$ You may check for yourself that it is allowed to switch differentiation w.r.t to $s$ with ...


2

I'd say it's quite plausible to find the roots! :) Rewrite as $\cos x=4^{x-x^{\cos x}}$. For $x\ge1$, $x-x^{\cos x}\ge0$ is true because $x^{\cos x}\le x^1$, thus $4^{x-x^{\cos x}}\ge4^0=1$, with equality if and only if $\cos x=1\iff x=2\pi k$ for $k\in\mathbb N$. For $x<0$, the problem is nonsense. For $x=0$, equality holds. For $0<x<1$, we have ...


1

Numerically, I get $$ S_1+S_2 = 0.14836252987273216621 $$ which agrees with $$ \frac{\pi^6}{6480} $$ Also numerically, $$ S_1 = 0.074181264936366083104 \\ S_2 = 0.074181264936366083104 $$ are seemingly equal.


1

You can try the following $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$ Substitution now: $$u:=\frac t{1-t}=\frac1{1-t}-1\implies du=\frac{dt}{(1-t)^2}$$ so we get $$B(x,y)=\int_0^\infty \frac{u^{x-1}}{(u+1)^{x-1}}\frac1{(u+1)^{y-1}} (u+1)^2du=\int_0^\infty\frac{u^{x-1}}{(u+1)^{x+y}}du$$ Now, we have $$\Gamma(x):=\int_0^\infty ...


1

Assuming the variable in your equation is $x$ and $\theta$ is a fixed real number (which the statement of your problem fails to specify), then you're looking to get only first or zeroth powers of $x$, as in : $y = ax + b, (a,b) \in \mathbb{R}^2$ where $ax$ is the first-power term and $b = bx^0$ the zeroth-power term. This means that terms of higher power ...


1

In each interval $\left(\pi n - \frac{\pi}{2}, \pi n + \frac{\pi}{2}\right)$ there is exactly one solution $x_n$ (i.e. $\tan x_n = \ln x_n$), and, when $n$ is large, it appears that $x_n$ is approximately $\pi n + \frac{\pi}{2}$. Let's show this. Since $\tan$ is $\pi$-periodic we have $$\tan\left(\pi n + \frac{\pi}{2} - x_n\right) = ...


1

There exists an addition formula for Bessel functions due to Graf. In the case you are interested in this addition formula takes the form $$H^{(1)}_{\nu}\left(R_1+R_2\right)=\sum_{k\in\mathbb{Z}}H^{(1)}_{\nu-k}\left(R_1\right)J_k\left(R_2\right),$$ where it is assumed that $|R_2|<|R_1|$. I'm not sure whether this can be called a simplification, but there ...


1

The integral $$ \int_{0}^{+\infty} x\, J_n(x)\,dx $$ is not converging for any $n\in\mathbb{N}$, since $J_n(x)$ decays like $\frac{1}{\sqrt{x}}$ as $x\to +\infty$. On the other hand, for any $n\in\mathbb{N}$ we have: $$ \int_{0}^{+\infty} J_n(x)\,dx = 1 $$ since: $$ \mathcal{L}(J_n(x)) = \frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^n}.$$



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