proofs

Below are selected mathematical proofs rendered with LaTeX.

Problem 13 – Taylor Series and Arcsin Function

Problem. Let

\[f(x) = x + \frac{2}{3} x^3 + \frac{2 \cdot 4}{3 \cdot 5} x^5 + \dots + \frac{2 \cdot 4 \dotsm 2n}{3 \cdot 5 \dotsm (2n + 1)} x^{2n + 1} + \dotsb\]

on the interval \((-1,1)\).

(a) Prove that \((1 - x^2) f'(x) = 1 + xf(x).\)

(b) Prove that \(f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.\)

Solution.

Part (a)

We let \(f(x)\) be a Taylor Series on the interval \((-1, 1)\) by denoting

\[f(x)=\sum_{n=0}^{\infty}c_nx^{2n+1},\]

such that \(c_n=\frac{2\cdot4\cdot6\cdots2n}{3\cdot5\cdot7\cdots(2n+1)}\). We take the derivative of our sum with respect to \(x\), giving us

\[f'(x)=\sum_{n=0}^{\infty}c_n(2n+1)x^{2n}.\]

We then multiply both sides by \((1-x^2)\) to get

\[\begin{align*}(1-x^2)f'(x) &= (1-x^2)\sum_{n=0}^{\infty}c_n(2n+1)x^{2n}\\ &= \sum_{n=0}^{\infty}c_n(2n+1)x^{2n} - \sum_{n=0}^{\infty}c_n(2n+1)x^{2n+2}\\ &= \sum_{n=0}^{\infty}c_n(2n+1)x^{2n} - \sum_{n=1}^{\infty}c_{n-1}(2n-1)x^{2n}\\ &= c_0 + \sum_{n=1}^{\infty}[c_n(2n+1) - c_{n-1}(2n-1)]x^{2n}. \end{align*}\]

Now we simplify the coefficient of \(x^{2n}\) for \(n \geq 1\). After algebraic manipulation, we find that \(c_n(2n+1) - c_{n-1}(2n-1) = c_{n-1}\).

Therefore,

\[\begin{align*}(1-x^2)f'(x) &= c_0 + \sum_{n=1}^{\infty}c_{n-1}x^{2n}\\ &= 1 + \sum_{k=0}^{\infty}c_kx^{2k+2}\\ &= 1 + x\sum_{k=0}^{\infty}c_kx^{2k+1}\\ &= 1 + xf(x). \end{align*}\]

Part (b)

Let \(g(x) = \sqrt{1 - x^2} f(x).\) Then by the product rule:

\[\begin{align*}g'(x) &= \frac{d}{dx}[\sqrt{1 - x^2}] \cdot f(x) + \sqrt{1 - x^2} \cdot f'(x)\\ &= \frac{-x}{\sqrt{1 - x^2}} f(x) + \sqrt{1 - x^2} f'(x)\\ &= \frac{-xf(x) + (1-x^2)f'(x)}{\sqrt{1 - x^2}}\\ &= \frac{-xf(x) + 1 + xf(x)}{\sqrt{1 - x^2}}\\ &= \frac{1}{\sqrt{1 - x^2}}. \end{align*}\]

Since \(\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}\), we have that \(g(x) = \arcsin x + C\) for some constant \(C.\)

To find \(C\), we evaluate at \(x = 0\):

\[g(0) = \sqrt{1 - 0^2} f(0) = 1 \cdot f(0) = f(0) = 0.\]

Also, \(\arcsin(0) = 0\), so \(g(0) = 0 + C\), which gives us \(C = 0.\)

Therefore, \(g(x) = \arcsin x,\) which means

\[\sqrt{1 - x^2} f(x) = \arcsin x,\]

\[f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.\quad \blacksquare\]

Series Convergence with Powers

Theorem. Let \(\{ a_n \}_{n=0}^{\infty}\) be a sequence of positive real numbers. If \(\sum_{n=0}^{\infty} a_n\) converges, then so does \(\sum_{n=0}^{\infty} {a_n}^r\), where \(r\) is a positive integer.

Proof. By the problem statement, we conclude that \(r \geq 1\) since \(1\) is the smallest positive integer.

We begin by considering

\[0< a_n\leq a_n^r \quad \forall a_n \geq 1\]

by the problem statement. However, we know that a sequence \(a_n\) converges to zero if and only if

\[\exists \epsilon > 0 \text{ such that } |a_n-L|<\epsilon,\]

where we set \(L=0\). When we set \(\epsilon=1\), we see that \(0\leq a_k\leq 1\) beginning at some point \(k\) and continuing onwards for all \(n\geq k\).

Since \(a_n\leq1\) for all \(n\geq k\), \(a_n^r\) must also be less than or equal to \(1\) for all \(n\geq k\) since a positive real less than or equal to \(1\) raised to a positive integer power will always be less than or equal to the original number before raising it to a power \(r\).

Therefore, we can conclude that

\[\forall n \geq k, \quad 0 < a_n^r \leq a_n \leq 1.\]

Finally, by the Series Comparison Test, we know that \(\sum_{n=0}^{\infty} {a_n}^r\) must converge since \(\sum_{n=0}^{\infty} a_n\) also converges from the given condition and \(a_n^r\leq a_n\) as \(n\) approaches \(\infty\). \(\quad \blacksquare\)

Integral of Arctan

Problem. Compute \(\int \tan^{-1}x \,dx\).

Solution. We must use integration by parts to solve this integral, but we notice that there are no obvious choices for \(dv\) and \(u\) since our integrand appears to consist of only one term. We set \(u=\tan^{-1}x\) because we know \(\frac{d}{dx}\tan^{-1}x=\frac{1}{x^2+1}\), forcing us to set \(dv=dx\) since it is the only remaining term. We set \(du=\frac{1}{x^2+1}dx\) and \(v=x\) by the integral power rule.

Next, by integration by parts, we write

\[\int u\,dv=uv-\int v\,du \quad \rightarrow \quad \int\tan^{-1}x\,dx = x\tan^{-1}x-\int\frac{x}{x^2+1}dx.\]

We recognize that making a substitution will make our integral significantly easier to solve; we set \(x=\tan\theta\) by substitution.

After our clever substitution, we have

\[\int\tan^{-1}x\, dx = \theta\tan\theta-\int\frac{\tan\theta}{\tan^2\theta+1} \sec^2\theta \,d\theta.\]

We may initially seem stuck, but we notice that when we divide our Fundamental Pythagorean Identity by \(\cos^2\theta\), we get

\[\sin^2\theta+\cos^2\theta=1 \quad \rightarrow \quad \tan^2\theta + 1 = \sec^2\theta.\]

When we make this substitution using our trigonometric identity, we notice that our integrand simplifies nicely; our integral term can be written as

\[\int\frac{\tan\theta\sec^2\theta}{\sec^2\theta}d\theta=\int\tan\theta\,d\theta.\]

We will omit the proof for the antiderivative of tangent because we have proven it previously, but it’s important to note that \(\int\tan\theta\,d\theta=\log\lvert\sec\theta\rvert+C\) by substitution.

Combining this with our original equation, we have

\[\int\tan^{-1}x\, dx = \theta\tan\theta-\log\lvert\sec\theta\rvert +C.\]

We may think we’re done, but we must remember that our original integral was written in terms of \(x\), so our answer must also be in terms of \(x\), not \(\theta\). Fortunately, we can resolve this issue quickly.

We previously made the substitution \(x=\tan\theta\). If we take the \(\arctan\) of each side of the expression, we get \(\theta=\tan^{-1}x\). Our final step is to rewrite our answer in terms of \(x\):

\[\int\tan^{-1}x\, dx = x\tan^{-1}x-\log\lvert\sec(\tan^{-1}x)\rvert + C. \quad \blacksquare\]

Orthogonal Curves

Problem. Find a set of curves that is orthogonal to the set \(\mathcal{F} = \{y = kx^2 : k \text{ is a constant}\}\).

Solution. We let \(\mathcal{F}\) and \(\mathcal{G}\) denote a set of orthogonal curves, meaning that one or more of the curves in the sets intersect at a right angle.

We let \(\mathcal{F}=\{y=kx^2 :k \text{ is a constant}\}\). Since \(k\) is a constant that depends on the curves in \(\mathcal{F}\), we want to express \(k\) as a function of \(y\) and \(x\).

We rearrange \(y=kx^2\) in terms of \(k\), giving us

\[k=\frac{y}{x^2}.\]

Next, we take the derivative of \(y=kx^2\), giving us

\[\frac{dy}{dx}=2kx\]

by implicit differentiation.

Next, we make our substitution for \(k\), giving us

\[\frac{dy}{dx}=\frac{2y}{x}.\]

To find the orthogonal set, we take the negative reciprocal of our slope \(\frac{dy}{dx}\):

\[\frac{dy}{dx}=-\frac{x}{2y}.\]

We solve this first-order differential equation by separation of variables:

\[\begin{align*} 2y\,dy&=-x\,dx \\ 2\int y\,dy&=-\int x\,dx \\ y^2&=-\frac{1}{2}x^2+C \\ C&=y^2+\frac{1}{2}x^2 \\ C&=2y^2+x^2 \\ y&=\sqrt{\frac{C-x^2}{2}} \end{align*}\]

for a variable constant \(C\).

Therefore, because the set \(\mathcal{G} = \left\{y=\sqrt{\frac{C-x^2}{2}}\right\}\) is orthogonal to \(\mathcal{F}\) at a minimum of one point, we conclude that

\[\mathcal{G} = \left\{y=\sqrt{\frac{C-x^2}{2}}\right\}. \quad \blacksquare\]