34 Elementary Functions
34.1 Logarithms
Use substitution to show that \(\int_{[1,x]}\frac{1}{t}\) is a logarithm. Show this is the natural log.
Get a power series for this!
34.2 Inverse Trigonometric Functions
34.2.1 \(\bigstar\) The ArcSine
Thus, we find ourselves interested in calculating these functions. Inspired by our previous treatment of logarithms (where we were able to find the derivative of \(L(x)\) using that it was the inverse of an exponential, without actually knowing a formula for \(L\)) we seek to begin our study of inverse trigonometry via differentiation:
Proposition 34.1 The derivative of the inverse sine function is \[(\arcsin x)^\prime = \frac{1}{\sqrt{1-x^2}}\]
Proof. Let \(f(x)=\arcsin(x)\). Then where defined, \(f(\sin(\theta))=\theta\) by definition, and we may differentiate via the chain rule: on the left side \[\frac{d}{d\theta}f(\sin(\theta))=f^\prime(\sin(\theta))\cos(\theta)\] and on the right \(\frac{d}{d\theta} \theta=1\). Equating these and solving for \(f^\prime\) yields \[f^\prime(\sin(\theta))=\frac{1}{\cos(\theta)}\] The only remaining problem is that we want to know \(f^\prime\) as a function of \(x\) and we only know its value implicitly, as a function of \(\sin(\theta)\). But setting \(x=\sin\theta\) we can express \(\cos\theta = \sqrt{1-x^2}\) via the pythagorean identity \(\sin^2\theta+\cos^2\theta =1\). Thus
\[f^\prime(x)=\frac{1}{\sqrt{1-x^2}}\]
Before integration this would have been a mere curiosity. But, armed wtih the fundamental theorem this is an extremely powerful fact: indeed, it directly gives us a representation as an integral:
Corollary 34.1 The inverse sine function is defined on the interval \([0,1]\) by the integral \[\arcsin(x)=\int_{[0,x]}\frac{1}{\sqrt{1-x^2}}\]
Proof. Since \((\arcsin x)^\prime =\frac{1}{\sqrt{1-x^2}}\), the inverse sine is an antiderivative of \(\frac{1}{\sqrt{1-x^2}}\), and also \(\sin(0)=0\) implies \(\arcsin(0)=0\), so it is zero at \(x=0\). Thus, it is exactly the area function \[\arcsin(x)=\int_{[0,x]}\frac{1}{\sqrt{1-t^2}}\]
34.2.2 The ArcTangent
Proposition 34.2 \[(\arctan x)^\prime=\frac{1}{1+x^2}\]
Proof. We again proceed by differentiating the identity \(\arctan(\tan\theta)=\theta\). This yields \(\arctan^\prime(\tan\theta)\frac{1}{\cos^2\theta}=1\) and multiplying through by \(\cos^2\) we can solve for the derivative of arctangent: \[\arctan^\prime(\tan\theta)=\cos^2\theta\]
The only problem is again we have the derivative as a function implicitly of of \(\tan\theta\), and we need it in terms of just an abstract variable \(x\). Setting \(x=\tan\theta\) we see that \(x^2=\tan^2\theta\) and (using the pythagorean identity) \(x^2+1=\tan^2\theta+1 =\frac{1}{\cos^2\theta}\). Thus \[\cos^2\theta = \frac{1}{1+x^2}\] and putting these two together, we reach what we are after
\[\arctan^\prime(x)=\frac{1}{1+x^2}\]
Proposition 34.3 The inverse function \(\arctan(x)\) to the tangent \(\tan(x)=\sin(x)/\cos(x)\) admits an integral representation \[\arctan(x)=\int_{[0,x]}\frac{1}{1+t^2}\]
Proof. This follows as \(\arctan^\prime(x)=1/(1+x^2)\), so both \(\arctan\) and this integral have the same derivative. As antiderivatives of the same function this means that they differ by a constant. Finally, this constant is equal to zero as \(\arctan(0)=0\) and \(\int_{[0,0]}\frac{1}{1+x^2} =0\) as it is an integral over a degenerate interval.
This integral expression is quite nice - the arctangent like the logarithm is shown to be the integral of a rather simple rational function. But like arcsine, an integral expression is rather difficult to use for computing actual values: we’d need to actually compute (or estimate) some Riemann sums. So it’s helpful to look for other expressions as well, and here arctan has a particularly nice power series.
Recall the geometric series \[\frac{1}{1-x}=\sum_{n\geq 0}x^n\]
We can substitute \(-x^2\) for the variable here to get a series for \(1/(1+x^2)\):
\[\frac{1}{1+x^2}=\sum_{n\geq 0}(-x^2)^n=\sum_{n\geq 0}(-1)^nx^{2n}\] \[=1-x^2+x^4-x^6+x^8-\cdots\]
This power series has radius of convergence \(1\) (inherited from the original geometric series) and converges at neither endpoint. We know from the above that this function is the derivative of the arctangent, so we should integrate it!
\[\arctan(x)=\int_{[0,x]}\frac{1}{1+t^2}\,dt = \int_{[0,x]} \sum_{n\geq 0}(-1)^nt^{2n}\,dt\]
Inside its radius of convergence we can exchange the order of the sum and the integral:
\[\begin{align*} \int_{[0,x]} \left(\sum_{n\geq 0}(-1)^nt^{2n}\right)\,dt&=\sum_{n\geq 0}\int_{[0,x]}(-1)^nt^{2n}\,dt\\ &=\sum_{n\geq 0}(-1)^n\int_{[0,x]}t^{2n}dt\\ &=\sum_{n\geq 0}(-1)^n \frac{x^{2n+1}}{2n+1} \end{align*}\]
Theorem 34.1 For \(x\in(-1,1)\), \[\arctan(x)=\sum_{n\geq 0}(-1)^n\frac{x^{2n+1}}{2n+1}\] \[=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\cdots\]
34.3 Elementary Functions via Calculus
Discovering the elementary functions as solutions to simple ODEs. (SHOULD WE SAVE THIS FOR FUNCTION-SPACE CHAPTER IN THE FUTURE? YES PROBABLY!)
- \(y^\prime=c\): affine
- \(y^\prime=cy\): exponential
- \(y^{\prime\prime}=cy\): trigonometric for \(c<0\) (new functions - hyperbolic trig for \(c>0\))
Theorem 34.2 (Nullspace of the \(n^{th}\) derivative) Polynomials
Theorem 34.3 (Eigenspaces of the \(n^{th}\) derivative) Exponentials and trigonometric functions
34.3.1 Complexity of Differentiation and Integration
Differentiating an elementary function always gives another elementary function. Integration doesn’t work this way!
34.4 Euler’s Infinite Product for \(\sin(x)\)
Here we will use our ability to integrate trigonometric functions to provide one proof of Euler’s famous infinite product for sine:
Theorem 34.4 (Infinite Product for \(\sin\)) For all \(x\neq 0\), \[\frac{\sin \pi x}{\pi x}=\prod_{k\geq 1}\left(1-\frac{x^2}{k^2}\right)\]
To begin we make the following auxiliary definition: for \(n\geq 0\) and \(c\in\RR\) let
\[I_n(c):=\int_{[0,\pi/2]}\cos^n(t)\cos(c t)\]
Exercise 34.1 Show that \(I_0(0)=\pi/2\) and \(I_0(2x)=\sin(\pi x)/(2x)\). Thus \[\frac{I_0(2x)}{I_0(0)}=\frac{\sin \pi x}{\pi x}\]
Our approach will be to show \(I_0(2x)/I_0(0)\) is also equal to the desired infinite product. To do so, we need to run some computation with
Proposition 34.4 For \(n\geq 2\), the following recursion formula holds: \[(n^2-c^2)I_n(c)=(n^2-n)I_{n-2}(c)\]
Proof. https://math.stackexchange.com/questions/786046/infinite-product-prod-limits-k-1-infty-left1-fracx2k2-pi2-right
Taking a copy of this recursive expression at \(c\) and at \(0\) and dividing yields
\[\frac{(n^2-c^2)I_n(c)}{(n^2-0^2)I_n(0)}=\frac{(n^2-n)I_{n-2}(c)}{(n^2-n)I_{n-2}(0)}\]
which simplifies to \[\frac{I_{n-2}(c)}{I_{n-2}(0)}=\frac{n^2-c^2}{n^2}\frac{I_n(c)}{I_n(0)}\]
This gives us a means of computing the ratio \(I_0(c)/I_0(0)\) inductively:
\[\begin{align*} \frac{I_{0}(c)}{I_{0}(0)}&=\frac{2^2-c^2}{2^2}\frac{I_2(c)}{I_2(0)}\\ &=\frac{2^2-c^2}{2^2}\frac{4^2-c^2}{4^2}\frac{I_4(c)}{I_4(0)}\\ &=\frac{2^2-c^2}{2^2}\frac{4^2-c^2}{4^2}\frac{6^2-c^2}{6^2}\frac{I_6(c)}{I_6(0)}\\ &=\frac{2^2-c^2}{2^2}\frac{4^2-c^2}{4^2}\frac{6^2-c^2}{6^2}\frac{8^2-c^2}{8^2}\frac{I_8(c)}{I_8(0)}\\ \end{align*}\]
Exercise 34.2 For any \(m\geq 1\), \[\frac{I_0(c)}{I_0(0)}=\prod_{1\leq k\leq m}\frac{(2k)^2-c^2}{(2k)^2}\frac{I_{2m}(c)}{I_{2m}(0)}\]
Plugging in \(c=2x\) gives an expression for \(\sin(\pi x)/(\pi x)\) by Exercise 34.1:
Corollary 34.2 For any \(m\geq 1\), \[ \frac{\sin\pi x}{\pi x}=\frac{I_0(2x)}{I_0(0)} =\prod_{1\leq k\leq m}\left(1-\frac{x^2}{k^2}\right)\frac{I_{2m}(2x)}{I_{2m}(0)} \]
Thus it only remains to show the ‘error term’ \(I_{2m}(2x)/I_{2m}(0)\) goes to \(1\) as \(n\to\infty\). Note that \(\sin(\pi x)/(\pi x)\) is an even periodic function of period \(1\), so it suffices to prove the claim for \(x\in[0,1/2]\).
Exercise 34.3 Prove for \(n\geq 0\) and \(x\in[0,1/2]\) that \[I_{n+2}(0)<I_n(2x)<I_n(0)\]
Hint: use that \(\cos(2xt)<\cos^2(x)\) for \(x\in[0,1/2]\) for the first inequality, and that \(I_{n+2}(0)/I_n(0)=\tfrac{n+1}{n+2}\)
Together this does it; we have a proof of Euler’s product for sine.