Limits, Sequences, Series: Learning Objectives 5, 6, 7

Differentiate the power series term-by-term several times. For each derivative, substitute $x=a$ and observe the pattern that emerges for the coefficients $c_n$.

By successively differentiating $f(x)$ and evaluating at $x=a$, we find $f(a) = c_0$, $f'(a) = c_1$, $f''(a) = 2c_2$, and $f'''(a) = 6c_3$. This generalizes to $f^{(n)}(a) = n!c_n$, which can be solved for $c_n$.

Start with the power series: $$ f(x) = c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3 + \dots $$ If we substitute $x=a$, every term except the first becomes zero, so: $$ f(a) = c_0 $$ Differentiate the series term-by-term: $$ f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + 4c_4(x-a)^3 + \dots $$ Substitute $x=a$: $$ f'(a) = c_1 $$ Differentiate again: $$ f''(x) = 2c_2 + 3 \cdot 2 c_3(x-a) + 4 \cdot 3 c_4(x-a)^2 + \dots $$ Substitute $x=a$: $$ f''(a) = 2c_2 $$ Differentiate a third time: $$ f'''(x) = 3 \cdot 2 \cdot 1 c_3 + 4 \cdot 3 \cdot 2 c_4(x-a) + \dots $$ Substitute $x=a$: $$ f'''(a) = 3!c_3 $$ By continuing this process, we see a pattern emerge: $$ f^{(n)}(a) = n(n-1)\dots(2)(1)c_n = n!c_n $$ Solving for $c_n$, we get the desired formula: $$ c_n = \frac{f^{(n)}(a)}{n!} $$

Take the general form of a power series, $f(x) = \sum_{n=0}^{\infty}c_n(x-a)^n$, and substitute the expression for $c_n$ that was derived in the previous question.

By substituting $c_n = \frac{f^{(n)}(a)}{n!}$ into the general power series definition, we arrive at the formula for the Taylor series of the function $f$ at (or about) $a$.

From Question 1, we know that if a function $f$ has a power series representation at $a$, its coefficients must be $c_n = \frac{f^{(n)}(a)}{n!}$.

The general form of a power series centered at $a$ is: $$ f(x) = \sum_{n=0}^{\infty}c_n(x-a)^n $$ Substituting our formula for the coefficients $c_n$ into this general form gives: $$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n $$ Expanding this series gives: $$ f(x) = \frac{f^{(0)}(a)}{0!}(x-a)^0 + \frac{f^{(1)}(a)}{1!}(x-a)^1 + \frac{f^{(2)}(a)}{2!}(x-a)^2 + \dots $$ $$ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots $$ This is the definition of the Taylor series for the function $f$ about $a$.

The derivative of $e^x$ is always $e^x$. For part (c) and (d), remember that Taylor series generally provide better approximations closer to the center of expansion ($a$), and higher-degree polynomials are generally more accurate.

1. $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ (This is also known as the Maclaurin series for $e^x$).
2. $\sum_{n=0}^{\infty} \frac{e^3}{n!}(x-3)^n$.
3. The Taylor series approximation using $a=0$ is more accurate for estimating $f(1)$.
4. The 4th-degree approximation provides a better estimate of $f(1)$.

For $f(x)=e^x$, we have $f^{(n)}(x) = e^x$ for all $n$.

(a) About $a=0$: $f^{(n)}(0) = e^0 = 1$. The Taylor series is $\sum_{n=0}^{\infty} \frac{1}{n!} (x-0)^n = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

(b) About $a=3$: $f^{(n)}(3) = e^3$. The Taylor series is $\sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n$.

(c) Estimating $f(1) = e^1 \approx 2.718$:
The 4th-degree polynomial at $a=0$ is $T_4(x) = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}$. $T_4(1) = 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24} = \frac{24+24+12+4+1}{24} = \frac{65}{24} \approx 2.7083$.
The 4th-degree polynomial at $a=3$ is $T_4(x) = e^3 + e^3(x-3) + \frac{e^3}{2}(x-3)^2 + \frac{e^3}{6}(x-3)^3 + \frac{e^3}{24}(x-3)^4$. $T_4(1) = e^3 + e^3(1-3) + \frac{e^3}{2}(1-3)^2 + \frac{e^3}{6}(1-3)^3 + \frac{e^3}{24}(1-3)^4$. $ = e^3 - 2e^3 + 2e^3 - \frac{8e^3}{6} + \frac{16e^3}{24} = e^3 - \frac{4}{3}e^3 + \frac{2}{3}e^3 = \frac{1}{3}e^3 \approx \frac{20.0855}{3} \approx 6.695$. The approximation centered at $a=0$ is closer to the true value of $e$ because the point of estimation ($x=1$) is closer to the center of expansion ($a=0$) than it is to $a=3$.

(d) 2nd vs 4th degree at $a=0$:
The 2nd-degree polynomial at $a=0$ is $T_2(x) = 1+x+\frac{x^2}{2}$. $T_2(1) = 1+1+\frac{1}{2} = 2.5$. As shown above, $T_4(1) \approx 2.7083$. The 4th-degree polynomial gives a better approximation because, in general, higher-degree Taylor polynomials provide more accuracy.

A Maclaurin series is a Taylor series centered at $a=0$. Find the derivatives of $f(x)=\sin(x)$, evaluate them at $x=0$, and find the pattern.

$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

Let $f(x) = \sin(x)$. We find the derivatives and evaluate at $a=0$:

• $f(x) = \sin(x) \implies f(0) = 0$
• $f'(x) = \cos(x) \implies f'(0) = 1$
• $f''(x) = -\sin(x) \implies f''(0) = 0$
• $f'''(x) = -\cos(x) \implies f'''(0) = -1$
• $f^{(4)}(x) = \sin(x) \implies f^{(4)}(0) = 0$

The pattern of the derivatives evaluated at 0 is $0, 1, 0, -1, 0, 1, \dots$. The non-zero terms occur for odd $n$. The Taylor series is: $$ \frac{f(0)}{0!}x^0 + \frac{f'(0)}{1!}x^1 + \frac{f''(0)}{2!}x^2 + \dots $$ $$ = 0 + \frac{1}{1!}x + 0 + \frac{-1}{3!}x^3 + 0 + \frac{1}{5!}x^5 + \dots = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots $$ In sigma notation, this is $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$.

You can differentiate the Maclaurin series for $\sin(x)$ term-by-term, since $\frac{d}{dx}(\sin x) = \cos x$.

$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$

We know that $\frac{d}{dx}\sin(x) = \cos(x)$. We can differentiate the Maclaurin series for $\sin(x)$ to find the series for $\cos(x)$. $$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots $$ Differentiating term-by-term: $$ \cos(x) = \frac{d}{dx}\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots\right) $$ $$ = 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \frac{7x^6}{7!} + \dots $$ $$ = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots $$ In sigma notation, this is $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$.

Multiply the known Maclaurin series for $\cos(x)$ by $x$.

$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n)!}$

We start with the Maclaurin series for $\cos(x)$: $$ \cos(x) = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots $$ Now, multiply the entire series by $x$: $$ x\cos(x) = x \left( \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} \right) = \sum_{n=0}^{\infty} x \cdot (-1)^n\frac{x^{2n}}{(2n)!} $$ $$ = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n)!} = x - \frac{x^3}{2!} + \frac{x^5}{4!} - \dots $$

Find the first few derivatives of $f(x)=(1+x)^k$ and evaluate them at $a=0$. Look for the pattern in the coefficients.

$\sum_{n=0}^{\infty} \binom{k}{n}x^n$, where $\binom{k}{n} = \frac{k(k-1)\dots(k-n+1)}{n!}$.

Let $f(x) = (1+x)^k$. We find the derivatives at $a=0$:

• $f(x) = (1+x)^k \implies f(0) = 1$
• $f'(x) = k(1+x)^{k-1} \implies f'(0) = k$
• $f''(x) = k(k-1)(1+x)^{k-2} \implies f''(0) = k(k-1)$
• $f'''(x) = k(k-1)(k-2)(1+x)^{k-3} \implies f'''(0) = k(k-1)(k-2)$
• $f^{(n)}(x) = k(k-1)\dots(k-n+1)(1+x)^{k-n} \implies f^{(n)}(0) = k(k-1)\dots(k-n+1)$

The general coefficient $c_n$ is $\frac{f^{(n)}(0)}{n!} = \frac{k(k-1)\dots(k-n+1)}{n!}$. This is the definition of the binomial coefficient, denoted $\binom{k}{n}$.

The Maclaurin series is therefore: $$ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty} \binom{k}{n}x^n $$ $$ = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \dots $$