Trigonometric Functions and Applications

Abstract

We investigate the sine and cosine functions, show the Weierstrass function is continuous and nowhere differentiable, construct the number π, establish that π is an irrational number, and give a brief treatment of polar coordinates.

Appendices

Problems

1.1 Problems for Sect. 11.1

• 1. Let $f(x):=\log \left (1+x\right )$ for $x>-1.$ Let

  $$ T\left(x\right):=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k} $$

  be the Taylor series of $f.$ Use the series to define a function

  $$ g\left(z\right):=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}z^{k} $$

  of a complex variable $z.$

  • a. Find the radius of convergence $R$ of the power series.

  • b. Show $f(x)=T\left (x\right )$ for $-R<x<R.$

  • c. Conclude $f(x)=g(x)$ for $-R<x<R.$

1.2 Problems for Sect. 11.2

• 1. Prove that $\int _{0}^{\infty }\frac {sin(xt)}{1+t^{2}}\, dt$ converges uniformly. That is prove the improper integral exists and

  $$ \forall\varepsilon,\exists M,\forall R,R>M\implies\left[\forall x,\int_{0}^{\infty}\frac{\sin(xt)}{1+t^{2}}\, dt-\int_{0}^{R}\frac{\sin(xt)}{1+t^{2}}\, dt<\varepsilon\right]. $$

• 2. Prove $e^{\pi }-\pi \neq 20.$

• 3. Suppose $f$ is continuous. Let $g(x):=\int _{0}^{x}\sin (x-t)f(t)dt.$ Prove $g''+g=f.$ [Hint: Begin by showing that $g'$ exists and finding a formula for $g'.$]

• 4. Let $f(x):=\frac {\cos (x)}{1+x^{2}}.$ Prove $\int _{0}^{1}f<1.$

1.3 Problems for Sect. 11.3

• 1. Fix $0\leq a<b\leq 2\pi .$ Find the length of the curve $\phi (t):=e^{it},$ $t\in \left [a,b\right ].$

1.4 Problems for Sect. 11.4

• 1. Does $\left (\sum _{n=0}^{N}b^{n}\cos (a^{n}x\pi )\right )'$ converge as $N\to \infty$ for all $x$?

• 2. Use Corollary 9.2.9. to find a sufficient condition for the Weierstrass function to be differentiable.

• 3. If $x_{0}=0,$ the proof in the text simplifies. Write out this simplification. [For example, $\beta _{m}=1$ for all $m$ and $\cos \left (a^{n}x_{0}\pi \right )=1.$]

• 4. Continuation of Problem 3. Find integers $\beta _{m}$ such that

  $$ (-1)^{\beta_{m}}\frac{f\left(\frac{\beta_{m}}{a^{m}}\right)-f\left(0\right)}{\frac{\beta_{m}}{a_{m}}-0}\to-\infty. $$

  [Hint: Equation (11.12) must be replaced.]

• 5. Show it is possible to choose $\beta _{m}$ such that $\beta _{m}/a^{m}\to x_{0}$ and

  $$ (-1)^{\beta_{m}}\frac{f\left(\frac{\beta_{m}}{a^{m}}\right)-f\left(x_{0}\right)}{\frac{\beta_{m}}{a_{m}}-x_{0}}\to-\infty. $$

  Thus, $f'(x_{0})$ cannot be a finite or even an infinite value.

1.5 Problems for Sect. 11.5

An alternative proof that $\pi$ is irrational is outlined as sequence of problems below. This proof is due to Ivan Morton Niven (25 October 1915, Vancouver to 9 May 1999, Eugene).

Assume $\pi =a/b$ for some positive integers $a$ and $b.$ For any natural number $n\geq 1,$ let

$$ f_{n}(x):=\frac{x^{n}(a-bx)^{n}}{n!}\text{ and }F_{n}(x):=\sum_{j=0}^{n}(-1)^{j}f_{n}^{(2j)}(x), $$

where $f^{(k)}$ is the $k$th derivative of $f.$ Note that $a-b\pi =0.$ Expanding the product in the definition of $f_{n}$ we see that $f_{n}(x)=\frac {1}{n!}\sum _{k=n}^{2n}c_{k}x^{k}$ for some integers $c_{k}.$ Differentiating this polynomial we get

$$ (**)\quad f_{n}^{(j)}(x)=\frac{1}{n!}\sum_{k=\max\{j,n\}}^{2n}c_{k}\frac{k!}{(k-j)!}x^{k-j}, $$

for $0\leq j\leq 2n.$

The proof is now completed in 11 easy steps:

• 1. $0\leq f_{n}(x)\leq \pi ^{n}a^{n}/n!$ for $0\leq x\leq \pi$ and all $N$ [Directly from the definition of $F_{n}.$]

• 2. $0<\int _{0}^{\pi }f_{n}\sin$ for all $n.$ [Since both $f_{n}$ and $\sin$ are $>0$ on $]0,\pi [.$]

• 3. There is an $N$ such that $\int _{0}^{\pi }f_{n}\sin <1.$ [Is a consequence of 1]

• 4. $\frac {k!}{n!(k-j)!}$ is an integer for all $n\leq j\leq k.$ [$\frac {k!}{n!(k-n)!}$ is a binomial coefficient, hence an integer. $\frac {k!}{n!(k-(n+1))!}=(k-n)\frac {k!}{n!(k-n)!},$ etc.]

• 5. $f_{n}(x)=f_{n}(\pi -x)$ for all $x$ and all $n.$ [Directly from the definition of $F_{n},$ since $a-b(\pi -x)=x.$]

• 6. $f_{n}^{j}(0)=f_{n}^{(j)}(\pi )=0,$ for all $n,j$ such that $0\leq j<n$. [By 5 and $(**).$]

• 7. $f_{n}^{(j)}(0)$ and $f_{n}^{(j)}(\pi )$ are integers, for all $n,j$ such that $n\leq j\leq 2n.$ [By 4, 5, and $(**).$]

• 8. $F_{n}(0)$ and $F_{n}(\pi )$ are integers for all $n.$ [By the definition of $F_{n},$ 6, and 7]

• 9. $F_{n}+F_{n}''=f_{n}$ for all $n.$ [The point of the $(-1)^{j}$ in the definition of $F_{n}.$]

• 10. $(F_{n}'\sin -F_{n}\cos )'=f_{n}\sin$ for all $n.$ [Derived using 9]

• 11. $\int _{0}^{\pi }f_{n}\sin =F_{n}(0)-F_{n}(\pi )$ is an integer for all $n.$ [By 8 and 10]

By 2, 3, and 11 $\int _{0}^{\pi }f_{N}\sin$ is an integer in $]0,1[.$ This contradiction completes the proof that $\pi$ is irrational.

Solutions and Hints for the Exercises

Exercise 11.1.1. The ratio test show that the series is absolutely convergent for all $z.$

Exercise 11.2.1. $e^{-iy}=\overline {e^{iy}.}$

Exercise 11.2.2. $\sum _{k=0}^{\infty }\frac {(iy)^{k}}{k!}=\cdots +i\cdots .$

Exercise 11.2.4. Similar to our proof of (11.3).

Exercise 11.2.6. $\sqrt {6-2\sqrt {3}}<\sqrt {3}$ $\,\Leftarrow \,$ $6-2\sqrt {3}<3$ $\,\Leftarrow \,$ $3<2\sqrt {3}$ $\,\Leftarrow \,$ $9<12.$

Exercise 11.2.8. $1=\cos ^{2}(\pi /2)+\sin ^{2}(\pi /2)=0+\sin ^{2}(\pi /2)$ and $\sin (\pi /2)>0.$

Exercise 11.2.9. $\sin (2\pi )=2\sin (\pi )\cos (\pi )=2\cdot 0\cdot (-1)=0$ and $\cos ^{2}(2\pi )=\cos ^{2}(\pi )-\sin ^{2}(\pi )=\left (-1\right )^{2}-0^{2}=1.$

Exercise 11.2.10. Similar to $\cos (x+2\pi )=\cos (x).$

Exercise 11.2.11. Similar to $\cos (x+2\pi )=\cos (x)$ and $\sin (x+2\pi )=\sin (x).$

Exercise 11.2.12. Use Exercise 11.2.11.