22 Birthday Problems

\[I = \int^3_0 \sqrt{9-x^2}\,\, \mathrm{d}x\]

Thus the mathematics simplifies from using a trigonometric substitution to simply \[\begin{align*} A_{\text{circle}} &= \pi r^2\\ &=9\pi\\ \implies I &= \frac{9\pi}{4}\end{align*}\]

We continue with the theme of visualising integrals in the multivariate case.

\[2\iiint\limits_{V} \,\mathrm{d}V, V : \{(r, \theta, \phi) \,|\, 0 \leq r \leq 1,\, 0 \leq \theta \leq 2\pi,\, 0\leq \phi \leq \pi\}\]

Here we realise that we are being asked to find the volume of 2 hemispheres using polar coordinates.

The radius \(r\) varies from 0 to 1, whilst the \(\theta\) value traces out a flat disk in the \(x-y\) plane. Then the \(\phi\) varies up to \(\pi\), tracing out the volume of a hemisphere. Then this value is multiplied by 2, forming a sphere:

Our mathematics then again simplifies significantly to just \[\begin{align*} V_{\text{sphere}} &= \frac{4}{3}\pi r^3\\ &=\frac{4}{3}\pi\end{align*}\]

Regrettably a mistake in this question steered the crowd in the wrong direction¹. The question asked was regarding an indefinite integral which can be correctly solved by using the Weierstrass \(t\) substitution of \[\cos{t} = \frac{1-t^2}{1+t^2}\quad .\]

Churning out the algebra from the substitution, performing a partial fraction decomposition, back-substituting in terms of the original variable and rationalising yields:

\[\begin{align*} I &= \int \frac{\cos{x}}{3+2\cos{x}} \, \mathrm{d}x\\ &= \int \frac{2-2t^2}{(5+t^2)(1+t^2)} \mathrm{d}t\\ &= \int -\frac{3}{5+t^2} + \frac{1}{1+t^2} \mathrm{d}t\\ &= \frac{x}{2} - \frac{3\sqrt{5}\arctan{\frac{\sqrt{5}\tan{\frac{x}{2}}}{5}}}{5} + C \end{align*}\]

However, this was not my intention. My intention was to ask for a value of the definite integral: \[J = \int_0^{2\pi} \frac{\cos{x}}{3+2\cos{x}} \, \mathrm{d}x\] Which I wanted solved using the theory of residues and complex analysis — oops!

Letting \[\begin{align*} \cos{x} &= \frac{z+z^{-1}}{2},\qquad \text{ where }z=e^{ix};\\ J &= \oint_\mathcal{C} \frac{z+z^{-1}}{2(3+z+z^{-1})} \\ &= -\frac{i}{2} \oint_\mathcal{C} \frac{z^2+1}{z^2(3+z+z^{-1}}\mathrm{d}z\\ &= -\frac{i}{2} \oint_\mathcal{C} \frac{z^2+1}{z(3z+z^2+1}\mathrm{d}z\\ &= -\frac{i}{2} \,2\,\pi\, i\, (B_0 + B_2),\qquad\text{note the exclusion of }B_1 \end{align*}\]

At this point we now take off our brain-dead algebra hats, and put on some more careful complex analysis hats: Our function \(\large f(z) = \frac{z^2+1}{z(3z+z^2+1}\) has poles at \(z=0, \frac{1}{2}(-3\pm\sqrt{5})\) with corresponding residues of \(B_0 = 1, B_1 = \frac{3}{\sqrt{5}}, B_2 = -\frac{3}{\sqrt{5}}\).

However, because the pole at \(\frac{1}{2}(-3-\sqrt{5})\) falls outside of the unit disk (the region of integration), we find that the total contribution to the integral is found by

\[\begin{align*} J &= \frac{\rlap{-i}\textcolor{red}{\cancel{\phantom{-i}}}}{\rlap{\,2}\textcolor{red}{\cancel{\phantom{2}}}}\rlap{\,2}\textcolor{red}{\cancel{\phantom{2}}} \pi \rlap{\,i}\textcolor{red}{\cancel{\phantom{i}}} (B_0 + B_2)\\ &= \pi(1 - \frac{3}{\sqrt{5}})\\ &= \pi - \frac{3\sqrt{5}\pi}{5}\quad\square \end{align*}\]

We must think outside of the proverbial box for this one:

\[\begin{align*} \left ( \frac{3}{2} \right ){\large !} &= \Gamma\left(\frac{3}{2}+1\right)\\ &= \Gamma\left(\frac{5}{2}\right) \end{align*}\]

Recall the fact that \[\Gamma(x) = \int^\infty_0 t^{x-1}e^{-t}\mathrm{d}t\] and the property that \[\Gamma(x+1) = x\,\Gamma(x)\] Thus \[\begin{align*} \Gamma\left(\frac{5}{2}\right) &= \frac{3}{2}\Gamma\left(\frac{3}{2}\right)\\ &= \frac{3}{2}\frac{1}{2}\Gamma\left(\frac{1}{2}\right) \end{align*}\]

and \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\), either by doing it manually yourself via IBP (integration by parts), or by having done it previously and remembering it.

\[\therefore \Gamma\left (\frac{5}{2}\right ) =\left( \frac{3\pi}{4}\right )\]

Recall that \[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\]

Our task is to find the derivative of the sine function, so \(f(x) = \sin(x)\).

We begin with \[\begin{align*} f'(x) &= \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}\\ &= \lim_{h\to 0} \frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\ &= \lim_{h\to 0} \sin(x)\frac{\cos(h)}{h} + \cos(x)\frac{\sin(h)}{h} - \frac{\sin(x)}{h}\\ \end{align*}\]

At this point we get stuck because it is not immediately obvious what the limit of \(\frac{\sin(h)}{h}\) is.

Resuming our initial proof: \[\lim_{h\to 0}\,\, \sin(x)\frac{\cos(h)}{h} + \cos(x)\frac{\sin(h)}{h} - \frac{\sin(x)}{h} = 0 + \cos(x)\times 1 - 0\\ = \cos(x)\,\,\square .\]

Again I make a mistake and ask the wrong question¹. \[\rlap{\int^\infty_\infty e^{-x^2}\mathrm{d}x}\textcolor{red}{\xcancel{\phantom{\int^\infty_\infty e^{-x^2}\mathrm{d}x}}} = 0\] but obviously what I intended to ask was \[I =\int^\infty_{-\infty} e^{-x^2}\mathrm{d}x.\]

We solve this by squaring the integral, changing variables and square rooting the multivariate answer: \[\begin{align*} I^2 &= \left( \int_{-\infty}^\infty e^{-x^2}~ \mathrm{d}x \right )^2\\ &= \int^\infty_{-\infty} e^{-x^2}\mathrm{d}x \times \int^\infty_{-\infty} e^{-y^2}~\mathrm{d}y \\ &= \int^\infty_{-\infty}\int^\infty_{-\infty} e^{-x^2} e^{-y^2}~\mathrm{d}x~\mathrm{d}y\\ &= \int^\infty_{-\infty}\int^\infty_{-\infty} e^{-x^2-y^2}~\mathrm{d}x~\mathrm{d}y\\ &= \int^{2\pi}_{0}\int^\infty_{0} e^{-r^2}r~\mathrm{d}r~\mathrm{d}\theta\qquad\text{by converting to polar coordinates}\\ &= -\frac{1}{2} \int^{2\pi}_0 e^{-r^2} ~\mathrm{d}\theta ~ \bigg\rvert^\infty_0 \\ &= -\frac{1}{2} \bigg[ -\theta ~\bigg\rvert^{2\pi}_0 \bigg] \\ &= \pi\\ \therefore I^2&=\pi \\ \implies I &= \sqrt{\pi} \end{align*}\]

For reference the curve we are integrating looks like:

Recall that \[\frac{1}{1+x^2} = 1 - x^2 + x^4 - \dots + \dots \]

Then upon integration we obtain: \[\begin{align*} \int^1_0 \frac{1}{1+x^2} &= 1 - \frac{x^3}{3} + \frac{x^5}{5} - \dots \bigg\rvert^1_0 \\ &= 1 - \frac{1}{3} + \frac{1}{5} -\frac{1}{7}\dots \end{align*}\]

Also, we can directly integrate: \[\begin{align*} \int^1_0 \frac{1}{1+x^2} &= \arctan(x) \bigg\rvert^1_0\\ &= \frac{\pi}{4} \end{align*}\]

Thus the infinite series \(1 - \frac{1}{3} + \frac{1}{5} -\frac{1}{7}+\dots \) indeed converges to \(\frac{\pi}{4}\).

Calculus is for children whilst analysis is for adults.

Given that \(f(x) = x^4 -5x^3\) and \(g(x) = 8x-40\) we can find their points of intersection by equating the functions and factoring the polynomial:

\[x^4-5x^3-8x+40 = 0 \\ \implies (x-5)(x^3-8) = 0\]

Taking derivatives and substituting the x-intercept of 5 to find the gradients of \(f\) and \(g\) at those points, we find that the angle difference between the two lines is \[6.67^\circ.\]

\[S\rightarrow V_2 \rightarrow V_4 \rightarrow t \implies \text{path cost of }31\]

To solve

\begin{equation} \label{eq:1} \sin(z) = 2 \end{equation}

we use the substitution \(z=e^{i\theta}\) such that \( \sin(z) = {\large\frac{z-z^{-1}}{2i}}\) and \ref{eq:1} becomes:

\[\begin{align*} \frac{z-z^{-1}}{2i} &= 2\\ z^2 - 1 &= 4i\\ z^2-4iz-1&=0\\ z &= i(2\pm\sqrt{3}) \qquad\text{(by the quadratic formula)} \end{align*}\]

We begin by considering \(X^2-X-1=0\label{eq:2}\) and letting its solution be \(\phi\).

Thus \(\phi^2 - \phi - 1 = 0 \implies \phi^2 = \phi + 1\). \[\begin{align*} \text{Multiplying by this by }\quad\phi: \phi^3 = \phi^2 +\phi &=2\phi +1\\ \text{similarly, }\quad \phi^4 = \phi^3 +\phi^2 &= 3\phi+2\\ \text{continuing in this fashion }\quad \phi^5 &= 5\phi + 3\\ \phi^6 &= 8\phi + 5 \end{align*}\]

Clearly a pattern is emerging; the coefficient of \(\phi^1\) is \(F_n\) (where \(F_n\) is the \(n^\text{th}\) value in the Fibonacci series) and that of \(\phi^0\) is \(F_{n-1}\).

Thus we can write \[\phi^n = F_n \phi + F_{n-1}\]

Then we note that our original equation \(X^2-X-1=0\) was quadratic and so would admit 2 such solutions:

\begin{align} \phi_1^n &= F_n \phi_1 + F_{n-1}\\ \phi_2^n &= F_n \phi_2 + F_{n-1} \end{align}

Then subtracting (3) from (2):

\begin{align} \label{eq:4} \phi_1^n - \phi_2^n &= F_n(\phi_1 - \phi_2)\nonumber \\ \implies F_n &= \frac{\phi_1^n-\phi_2^n}{\phi_1-\phi_2} \end{align}

Zooming out and solving the quadratic \(X^2-X-1=0\) we get the following results. \[\phi_1 = \frac{1+\sqrt{5}}{2}, \qquad\phi_2 = \frac{1-\sqrt{5}}{2}, \qquad \phi_1 - \phi_2 = \sqrt{5}\] Substituting these \(\phi_1\) and \(\phi_2\) into \ref{eq:4} yields \[F_n = \frac{1}{\sqrt{5}}\left[ \bigg(\frac{1+\sqrt{5}}{2}\bigg)^n - \bigg(\frac{1-\sqrt{5}}{2}\bigg)^n\right].\]

We solve the ODE by combining solutions to the homogenous case \[y_h = Ae^{-x} + Bxe^{-x}\] with the solutions to the particular case \[y_p = e^{-x}-e^{-x}\cos(x) \] And solving the initial conditions to find our constants and ultimately produce: \[y = e^{-x}-e^{-x}\cos(x)\]

\[\begin{align*} \langle\sin(x),\cos(x)\rangle &= \int^{2\pi}_0\sin(x)\cos(x)\mathrm{d}x\\ &= \frac{1}{2}\int^{2\pi}_0 \sin(2x)~\mathrm{d}x\\ &= 0\text{, by the oddness of the sine function} \end{align*}\]

There are 8 corners with 3 permutations each: \(3^8\). They can all be arranged in \(8!\) ways. Then similarly, we have 12 edges with 2 permutations each: \(2^{12}\); these can be arranged in \(12!\) ways.

However, because we are not simply calculating permutations of an abstract 3 by 3 object with corner and edges pieces, but are rather trying to find how many possible permutations of the Rubik's cube there are, we have another 3 restrictions:

1. You cannot turn a corner piece in place
2. You cannot turn an edge piece in place
3. You cannot swap 2 pieces at will

As a consequence of these physical restrictions on the puzzle, we have the corresponding combinatoric limitations:

1. The permutation of the last corner will be predetermined by all other corners
2. The permutation of the last edge will be predetermined by all other edges
3. The corner positions of the last 2 corners will be predetermined by all the previous edges and corners

Thus \[3^8\times 8! \times 3^{12}\times 12!\] Reduces to \[3^7\times 8! \times 3^{11}\times 11! \times \frac{1}{2} = 43,252,003,274,489,856,000\]

Rome was not built in a day

We take an infinitesimal summation (an integral) of all the length components of the curve with: \[\int\sqrt{1+[f'(x)]^2} \mathrm{d}x\] In this case \(f(x) = x^2\), so \(f'(x) = 2x\). We solve the following integral by way of a trigonometric substitution \(x = \frac{\tan(u)}{2} \implies \frac{\mathrm{d}x}{\mathrm{d}u} = \frac{\sec^2(u)}{2} \).

After the algebraic dust settles we are left with: \[\frac{x\sqrt{1+4x^2}}{2} + \frac{\ln(\sqrt{1+4x^2}+2x)}{4} \bigg\rvert^4_0 \approx 16.82\]

Someone who is majoring in mathematics, has a friend who needs help with his or her homework.

\[D(x) = \begin{cases} 1 & \text{if } x \text{ is rational}, \\ 0 & \text{if } x \text{ is irrational} \end{cases}\\I = \int_0^1 D(x)\, \mathrm{d}x\]

To solve this we usually need measure theory, but we can intepret the integral probabilistically as:

What is the probability that you pick a rational number number? Which is equivalent to 1 - "what is the probability you pick an irrational number". Clearly the probability of picking an irrational number is low, considering \(0.12345\ldots\) ad infinitum. At first you would need to pick 1, then next 2, then 3, etc. each with a probability of 1/10th:

\[P(irrational) = \frac{1}{10}^\infty = 0 \]

This can be visualised as

Thus \(I = \int_0^1 D(x)\, \mathrm{d} = 1 - 0 = 1\).