Exercise Set 11 Answers

1. Write the series $$\begin{equation*} \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6} \end{equation*}$$ using Sigma notation in the forms $\displaystyle\sum_{k=1}^{n}(\dots)$ and $\displaystyle\sum_{l=0}^{m}(\dots)$.

   Answers:

   $$\sum_{k=1}^{5}\frac{1}{k(k+1)}$$ $$\sum_{l=0}^{4}\frac{1}{(l+1)(l+2)}$$

2. Evaluate $\displaystyle\sum_{n=1}^{10}\frac{2}{3^n}$ and $\displaystyle\sum_{n=0}^{9}\frac{2}{3^n}$ using the formula for the $n^{\text{th}}$ partial sum of a geometric series.

   Answers:

   1. The geometric series $\displaystyle\sum_{n=1}^{10}\dfrac{2}{3^n}$ has first term $a=\dfrac{2}{3^1}=\dfrac{2}{3}$ and second term $ar=\dfrac{2}{3^2}$ and therefore the common ratio is $r=\dfrac{ar}{a}=\dfrac{1}{3}$. Since the sum is for $n=1$ to $n=10$, it is the sum of the first $10$ terms, $$S_{10}=\frac{a(1-r^{10})}{1-r}=\frac{2\left (1-\dfrac{1}{3^{10}}\right)}{3\left (1-\dfrac{1}{3}\right )}=1-\dfrac{1}{3^{10}}.$$

   2. The geometric series $\displaystyle\sum_{n=0}^{9}\frac{2}{3^n}$ has first term $a=\dfrac{2}{3^0}=2$ and second term $ar=\dfrac{2}{3^1}$, and therefore the common ratio is $r=\dfrac{ar}{a}=\dfrac{1}{3}$. Since the sum is for $n=0$ to $n=9$, it is the sum of the first $10$ terms, $$S_{10}=\dfrac{a(1-r^{10})}{1-r}=\dfrac{2\left(1-\dfrac{1}{3^{10}}\right)}{\left(1-\dfrac{1}{3}\right)}=3\left(1-\dfrac{1}{3^{10}}\right).$$

3. Do the following geometric series converge or diverge? If they converge, find their sum.

   1. $\displaystyle\sum_{n=1}^{\infty}\frac{2}{3^n}$
   2. $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2^{n-1}}$
   3. $\displaystyle\sum_{n=1}^{\infty}\left( \dfrac{\pi}{2}\right)^n$

   Answers:

   1. We already know that the first term is $a=\dfrac{2}{3}$ and the common ratio is $r=\dfrac{1}{3}$. Since $|r|<1$, the geometric series converges, with sum $$\sum_{n=1}^{\infty}\frac{2}{3^n}=\frac{a}{1-r}=\frac{\frac{2}{3}}{1-\frac{1}{3}}=1.$$

   2. The first term of the series is $a=1$ and the common ratio is $r=-\dfrac{1}{2}$. Since $|r|<1$ the series converges, and the sum is $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2^{n-1}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}.$$

   3. The first term of the series is $a=\dfrac{\pi}{2}$, and the common ratio is also $r=\dfrac{\pi}{2}$. Since $|r|>1$, the series diverges.

4. Find the $3^{\text{rd}}$ degree Taylor polynomials of the following functions.

   1. $f(x)=2x^3-x^2+4x+5$ at $a=0$
   2. $f(x)=\dfrac{1}{1+x}$ at $a=0$.
   3. $f(x)=\sqrt{x}$ at $a=1$

   Answers:

   1. The $3^{\text{rd}}$ degree polynomial of this polynomial at $a=0$ will simply be equal to the polynomial itself. For completeness, we still derive the Taylor series to demonstrate this $$\begin{align} f(x)&=2x^3 - x^2 + 4x + 5 & f(0)&=5\\ f'(x)&= 6x^2 - 2x + 4 & f'(0)&=4\\ f''(x)&= 12x - 2 & f''(0) &= -2\\ f'''(x) &= 12 & f'''(0) &= 12 \end{align}$$ Substituting these into the Taylor series about $x=0$ $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x\\ f(x) = f(0) + \frac{f'(0)}{1}x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3\\ f(x) = 5 + 4x - x^2 + 2 x^3$$ since the Taylor series terminates after the $x^3$ term. The original polynomial has been recovered.

   2. We start by finding the first 3 derivatives. $$\begin{align} f(x) &= \frac{1}{x+1} & f(0) &= 1\\ f'(x) &= \frac{-1}{(x+1)^2} & f'(0) &= -1\\ f''(x) & = \frac{2}{(x+1)^3} & f''(0) &= 2\\ f'''(x) &= \frac{-6}{(x+1)^4} & f'''(0) &= -6 \end{align}$$ Substituting these terms into the Taylor series expansion about $x=0$, up to the $3^\text{rd}$ degree term.: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x\\ f(x) = 1 - x + x^2 - x^3 + \cdots$$ Having just included the first few terms.

   3. Note that it is not possible to take the Taylor series about $x=0$ because the derivative of $\sqrt{x}$ does not exist at $0$. Taking the expansion about $x=1$. $$\begin{align} f(x) &= \sqrt{x} = x^\frac{1}{2} & f(1) &= 1\\ f'(x) &= \frac{1}{2}x^{-\frac{1}{2}} & f'(1) &= \frac{1}{2}\\ f''(x) & = -\frac{1}{4}x^{-\frac{3}{2}} & f''(1) &= -\frac{1}{4}\\ f'''(x) &= \frac{3}{8}x^{-\frac{5}{2}} & f'''(1) &= \frac{3}{8} \end{align}$$ Substituting these into the first few terms of the Taylor series about $x=a$ $$f(x-1) = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n\\ f(x-1) = 1 + \frac{1}{2}(x-1) -\frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} + \cdots\\ f(x-1) = 1 + \frac{1}{2}(x-1) - \frac{(x-1)^2}{8} + \frac{(x-1)^3}{16} + \cdots$$

5. Derive the Taylor series for $e^x$ at $a=0$ (this converges for all $x$).

   Answers:

   Let $f(x)=e^x$. $$\begin{align*} f(x)&=e^x & f(0)&=1\\ f'(x)&=e^x & f'(0)&=1\\ f^{(n)}(x)&=e^x & f^{(n)}(0)&=1 \end{align*}$$ The $n^{\text{th}}$ degree Taylor polynomial is $$P_{n,0}(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb+\frac{x^n}{n!}$$ and the Taylor series is $$\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

6. Derive the Taylor series for $\sin(x+a)$ at $a=\dfrac{\pi}{2}$ (this converges for all $x$).

   Answers:

   The derivatives at $\dfrac{\pi}{2}$ form a periodic sequence of period $4$ $$\begin{align*} f(x)&=\sin(x) & f\left(\frac{\pi}{2}\right)&=1\\ f^{(1)}(x)&=\cos(x) & f^{(1)}\left(\frac{\pi}{2}\right)&=0\\ f^{(2)}(x)&=-\sin(x) & f^{(2)}\left(\frac{\pi}{2}\right)&=-1\\ f^{(3)}(x)&=-\cos(x) & f^{(3)}\left(\frac{\pi}{2}\right)&=0\\ f^{(4)}(x)&=\sin(x) & f^{(4)}\left(\frac{\pi}{2}\right)&=1 \end{align*}$$

   and the Taylor series is $$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}$$ which we recognise as the Taylor series for $\cos$ about $0$, which should be unsurprising since $\sin\left(x+\frac{\pi}{2}\right)=\cos(x)$.

7. From the Taylor series of $f(x)=\ln(x+1)$ at zero (this converges for $-1 < x \le 1$), show that: $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\ln(2).$$

   Answers:

   Since we are told that the Taylor series equals $f(x)$ for $-1 < x\le 1$ and $f(1)=\ln(2)$, we have $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}=\ln(2).$$

8. The Taylor series for $\tan^{-1}$ at zero is $$\tan^{-1}(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$$ which converges for $|x|\leq 1$. Use this to approximate $\pi$ (you could do this in MATLAB/Python using a for loop to easily experiment with the number of terms in the sum).

   Answer:

   We have that $\tan^{-1}(1)=\frac{\pi}{4}$, so can find $\pi$ as $4\tan^{-1}(1)=4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$. We can get an approximation by summing over the first $N$ terms of this infinite series.

   For N=1000 we get $… $

9. Prove Euler’s formula for complex numbers using the Taylor series for $e^x$, $\cos(x)$ and $\sin(x)$ at $0$, .

   Answer: First, recall the Taylor series for $e^x$ $$e_x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ And now let $x = i\theta$. The odd terms will become imaginary, while the even powers will have changing signs with $i^2$:

   $$\begin{align} e^{i\theta} &= 1 + i\theta + \frac{i^2 \theta^2}{2!} + \frac{i^3 \theta^3}{3!} + \frac{i^4 \theta^4}{4!} + \cdots \\ &= 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots \end{align}$$

   Now, group every second term together to identify the Tarlor series of $\cos(x)$ and $i\sin(x)$.

   $$e^x = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + \cdots \right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{x^5}{5!} + \cdots\right)\\ e^x = \cos(\theta) + i \sin(\theta)$$

10. Find the Taylor series for $\cos(x)$ at zero by differentiating the Taylor series for $\sin(x)$ at zero term by term (we can differentiate any power series term by term for values of $x$ where the series converges).

    Answers: Taking the derivative of the power series, and noting that $\frac{n}{n!} = \frac{1}{(n-1)!}$ $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\\ \frac{d}{dx}\sin (x)= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \cos(x)$$

11. Find the Taylor series of $\dfrac{1}{1-x}$ at $a=0$ (this converges for $|x|<1$). Hence find the Taylor series of $\dfrac{1}{(1-x)^2}$ at $a=0$.

    Answers:

    $$\begin{align*} f(x)&=(1-x)^{-1} & f(0)&=1\\ f^{(1)}(x)&=(1-x)^{-2} & f^{(1)}(0)&=1\\ f^{(2)}(x)&=2(1-x)^{-3} & f^{(2)}(0)&=2\\ f^{(3)}(x)&=6(1-x)^{-4} & f^{(3)}(0)&=6=3!\\ f^{(4)}(x)&=24(1-x)^{-5} & f^{(4)}(0)&=24=4! \end{align*}$$ so the Taylor Series is $$\begin{equation*} \sum_{n=0}^{\infty}x^n. \end{equation*}$$

    Now to find the Taylor series of $\frac{1}{(1-x)^2}$ at $0$, we note that $\frac{d}{dx}\frac{1}{(1-x)}=\frac{1}{(1-x)^2}$, hence we can obtain the Taylor series by differentiating the series for $\frac{1}{(1-x)}$ term by term, obtaining: $$\begin{equation*} \sum_{n=1}^{\infty}nx^{n-1}. \end{equation*}$$