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Exercises: Taylor's Series

Theory

Basic Idea of Taylor Series (Beginner Explanation)

If a function f(x)f(x) is infinitely differentiable near x=ax=a, then its Taylor series about x=ax=a is

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

This series represents the function near x=ax=a.

To find a Taylor series in practice:

  1. If possible, start from a known Maclaurin series (centered at 00).
  2. If the center is a≠0a \neq 0, rewrite the function in terms of (x−a)(x-a).
  3. The radius of convergence RR is determined by the nearest point where the function is undefined (or from a known standard result).

Exercises

Exercise 1
Question
f(x)=e2x,a=0f(x)=e^{2x}, \quad a=0
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Solution

We know the Maclaurin series of exe^x:

ex=∑n=0∞xnn!e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}

To find the series for e2xe^{2x}, we replace xx by 2x2x:

e2x=∑n=0∞(2x)nn!e^{2x}=\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}

Now simplify:

(2x)n=2nxn(2x)^n = 2^n x^n

So,

e2x=∑n=0∞2nxnn!e^{2x}=\sum_{n=0}^{\infty}\frac{2^n x^n}{n!}

The exponential function is defined for all real numbers, so the series converges everywhere.

Conclusion:

e2x=∑n=0∞2nxnn!,R=∞\boxed{ e^{2x}=\sum_{n=0}^{\infty}\frac{2^n x^n}{n!}, \quad R=\infty }
Exercise 2
Question
f(x)=sin⁡x,a=π4f(x)=\sin x, \quad a=\frac{\pi}{4}
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Solution

Since the center is not 00, we must use the Taylor formula:

sin⁡x=∑n=0∞f(n)(a)n!(x−a)n\sin x=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n

First compute derivatives:

sin⁡x\sin xcos⁡x\cos x−sin⁡x-\sin x−cos⁡x-\cos x

These four derivatives repeat in a cycle.

Now evaluate them at a=π4a=\frac{\pi}{4}:

sin⁡(π4)=12,cos⁡(π4)=12\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}

So the derivatives alternate between

12,12,−12,−12,…\frac{1}{\sqrt{2}}, \quad \frac{1}{\sqrt{2}}, \quad -\frac{1}{\sqrt{2}}, \quad -\frac{1}{\sqrt{2}}, \dots

Therefore the Taylor series becomes

sin⁡x=12∑n=0∞(−1)n(n−1)2n!(x−π4)n\sin x = \frac{1}{\sqrt{2}} \sum_{n=0}^{\infty} \frac{(-1)^{\frac{n(n-1)}{2}}}{n!} \left(x-\frac{\pi}{4}\right)^n

Since sine is defined for all real numbers, the series converges everywhere.

Conclusion:

R=∞\boxed{R=\infty}
Exercise 3
Question
f(x)=1x+1,a=0f(x)=\frac{1}{x+1}, \quad a=0
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Solution

Rewrite the function:

1x+1=11+x\frac{1}{x+1}=\frac{1}{1+x}

We know the geometric series formula:

\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n, \quad |x|<1

To match this form, replace xx by −x-x:

\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^n x^n, \quad |x|<1

The function is undefined at x=−1x=-1. The distance from the center a=0a=0 to −1-1 is 11.

Conclusion:

1x+1=∑n=0∞(−1)nxn,R=1\boxed{ \frac{1}{x+1} = \sum_{n=0}^{\infty}(-1)^n x^n, \quad R=1 }
Exercise 4
Question
f(x)=1(x+1)2,a=−2f(x)=\frac{1}{(x+1)^2}, \quad a=-2
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Solution

Since the center is a=−2a=-2, we rewrite everything in terms of (x+2)(x+2).

Let

u=x+2u = x+2

Then

x+1=(x+2)−1=u−1x+1 = (x+2)-1 = u-1

So

1(x+1)2=1(u−1)2=1(1−u)2\frac{1}{(x+1)^2} = \frac{1}{(u-1)^2} = \frac{1}{(1-u)^2}

Now recall the known formula:

\frac{1}{(1-u)^2} = \sum_{n=0}^{\infty} (n+1)u^n, \quad |u|<1

Substitute back u=x+2u=x+2:

1(x+1)2=∑n=0∞(n+1)(x+2)n\frac{1}{(x+1)^2} = \sum_{n=0}^{\infty} (n+1)(x+2)^n

The function is undefined at x=−1x=-1. Distance from center −2-2 to −1-1 is

∣−1−(−2)∣=1|-1 - (-2)| = 1

Conclusion:

1(x+1)2=∑n=0∞(n+1)(x+2)n,R=1\boxed{ \frac{1}{(x+1)^2} = \sum_{n=0}^{\infty} (n+1)(x+2)^n, \quad R=1 }
Exercise 5
Question
f(x)=1x,a=1f(x)=\frac{1}{x}, \quad a=1
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Solution

Since the center is a=1a=1, we rewrite the function in terms of (x−1)(x-1).

Observe that

x=1+(x−1)x = 1 + (x-1)

So,

1x=11+(x−1)\frac{1}{x} = \frac{1}{1+(x-1)}

Now we use the geometric series formula:

\frac{1}{1+u} = \sum_{n=0}^{\infty}(-1)^n u^n, \quad |u|<1

Let u=x−1u = x-1. Then,

1x=∑n=0∞(−1)n(x−1)n\frac{1}{x} = \sum_{n=0}^{\infty}(-1)^n (x-1)^n

The function 1x\frac{1}{x} is undefined at x=0x=0.

The distance from the center a=1a=1 to 00 is

∣0−1∣=1|0-1|=1

So the radius of convergence is 11.

Conclusion:

1x=∑n=0∞(−1)n(x−1)n,R=1\boxed{ \frac{1}{x} = \sum_{n=0}^{\infty}(-1)^n (x-1)^n, \quad R=1 }
Exercise 6
Question
f(x)=ex,a=3f(x)=e^x, \quad a=3
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Solution

We know the Maclaurin series of exe^x:

ex=∑n=0∞xnn!e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}

But now the center is a=3a=3, so we rewrite:

ex=e3+(x−3)e^x = e^{3+(x-3)}

Using properties of exponents:

ex=e3⋅ex−3e^x = e^3 \cdot e^{x-3}

Now expand ex−3e^{x-3} using the known series:

ex−3=∑n=0∞(x−3)nn!e^{x-3} = \sum_{n=0}^{\infty} \frac{(x-3)^n}{n!}

Multiply by e3e^3:

ex=∑n=0∞e3n!(x−3)ne^x = \sum_{n=0}^{\infty} \frac{e^3}{n!}(x-3)^n

Since the exponential function is defined for all real numbers, the series converges everywhere.

Conclusion:

ex=∑n=0∞e3n!(x−3)n,R=∞\boxed{ e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!}(x-3)^n, \quad R=\infty }
Exercise 7
Question
f(x)=x,a=4f(x)=\sqrt{x}, \quad a=4
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Solution

Since the center is a=4a=4, we rewrite x\sqrt{x} in terms of (x−4)(x-4).

Write:

x=4+(x−4)x = 4 + (x-4)

So,

x=4+(x−4)\sqrt{x} = \sqrt{4+(x-4)}

Factor out 44:

x=4(1+x−44)\sqrt{x} = \sqrt{4\left(1+\frac{x-4}{4}\right)}=21+x−44= 2\sqrt{1+\frac{x-4}{4}}

Now use the binomial series formula:

(1+u)^{1/2} = \sum_{n=0}^{\infty} \binom{\frac12}{n} u^n, \quad |u|<1

where

(12n)=12(12−1)(12−2)⋯(12−(n−1))n!\binom{\frac12}{n} = \frac{\frac12\left(\frac12-1\right)\left(\frac12-2\right)\cdots\left(\frac12-(n-1)\right)}{n!}

Let

u=x−44u=\frac{x-4}{4}

Then,

x=2∑n=0∞(12n)(x−44)n\sqrt{x} = 2 \sum_{n=0}^{\infty} \binom{\frac12}{n} \left(\frac{x-4}{4}\right)^n

The first few terms are:

x=2+14(x−4)−164(x−4)2+⋯\sqrt{x} = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 + \cdots

Now determine the radius of convergence.

The binomial series converges when

\left|\frac{x-4}{4}\right|<1

Multiply both sides by 44:

|x-4|<4

The function x\sqrt{x} is undefined for x<0. The nearest problematic point to 44 is x=0x=0.

Distance:

∣0−4∣=4|0-4|=4

Conclusion:

R=4\boxed{ R=4 }
Exercise 8
Question
f(x)=tan⁡−1(2x),a=0f(x)=\tan^{-1}(2x), \quad a=0
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Solution

We use the known Maclaurin series:

tan⁡−1x=∑n=0∞(−1)nx2n+12n+1,∣x∣≤1\tan^{-1}x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x|\le1

Now replace xx by 2x2x:

tan⁡−1(2x)=∑n=0∞(−1)n(2x)2n+12n+1\tan^{-1}(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{2n+1}

Simplify the power:

(2x)2n+1=22n+1x2n+1(2x)^{2n+1} = 2^{2n+1} x^{2n+1}

So,

tan⁡−1(2x)=∑n=0∞(−1)n22n+12n+1x2n+1\tan^{-1}(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{2n+1} x^{2n+1}

For convergence we need:

∣2x∣≤1|2x| \le 1∣x∣≤12|x| \le \frac{1}{2}

So the radius of convergence is 12\frac{1}{2}.

Conclusion:

tan⁡−1(2x)=∑n=0∞(−1)n22n+12n+1x2n+1,R=12\boxed{ \tan^{-1}(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{2n+1} x^{2n+1}, \quad R=\frac{1}{2} }