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Exercises: Maclaurin Series

Theory

What is a Maclaurin Series?

A Maclaurin series is simply a Taylor series centered at a=0a=0.

If a function f(x)f(x) is infinitely differentiable at 00, then

f(x)=∑n=0∞f(n)(0)n!xnf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n

In practice, we usually:

  1. Start from a known basic Maclaurin series.
  2. Replace xx if needed.
  3. Multiply or combine series when necessary.
  4. Determine the radius of convergence from the known basic series.

Exercises

Exercise 1
Question
f(x)=e3xf(x)=e^{3x}
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Solution

We start from the well-known Maclaurin series:

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

To find the series for e3xe^{3x}, replace xx by 3x3x:

e3x=∑n=0∞(3x)nn!e^{3x} = \sum_{n=0}^{\infty}\frac{(3x)^n}{n!}

Now simplify the power:

(3x)n=3nxn(3x)^n = 3^n x^n

Therefore,

e3x=∑n=0∞3nn!xne^{3x} = \sum_{n=0}^{\infty}\frac{3^n}{n!}x^n

Since the exponential function converges for all real numbers, the radius of convergence is infinite.

Conclusion:

e3x=∑n=0∞3nn!xn,R=∞\boxed{ e^{3x} = \sum_{n=0}^{\infty}\frac{3^n}{n!}x^n, \quad R=\infty }
Exercise 2
Question
f(x)=x2cos⁡xf(x)=x^2\cos x
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Solution

We begin with the known series:

cos⁡x=∑n=0∞(−1)nx2n(2n)!\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

Now multiply the entire series by x2x^2:

x2cos⁡x=x2∑n=0∞(−1)nx2n(2n)!x^2\cos x = x^2 \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

Combine the powers of xx:

x2cos⁡x=∑n=0∞(−1)nx2n+2(2n)!x^2\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n)!}

Since the cosine series converges for all real numbers, multiplying by x2x^2 does not change the radius of convergence.

Conclusion:

x2cos⁡x=∑n=0∞(−1)n(2n)!x2n+2,R=∞\boxed{ x^2\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n+2}, \quad R=\infty }
Exercise 3
Question
f(x)=xsin⁡x2f(x)=x\sin\frac{x}{2}
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Solution

We start with:

sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Now replace xx by x2\frac{x}{2}:

sin⁡x2=∑n=0∞(−1)n(2n+1)!(x2)2n+1\sin\frac{x}{2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{x}{2}\right)^{2n+1}

Simplify the power:

(x2)2n+1=x2n+122n+1\left(\frac{x}{2}\right)^{2n+1} = \frac{x^{2n+1}}{2^{2n+1}}

So,

sin⁡x2=∑n=0∞(−1)n22n+1(2n+1)!x2n+1\sin\frac{x}{2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{2n+1}(2n+1)!} x^{2n+1}

Now multiply by xx:

xsin⁡x2=∑n=0∞(−1)n22n+1(2n+1)!x2n+2x\sin\frac{x}{2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{2n+1}(2n+1)!} x^{2n+2}

Again, since sine converges for all real numbers, the radius remains infinite.

Conclusion:

xsin⁡x2=∑n=0∞(−1)n22n+1(2n+1)!x2n+2,R=∞\boxed{ x\sin\frac{x}{2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{2n+1}(2n+1)!} x^{2n+2}, \quad R=\infty }
Exercise 4
Question
f(x)=sin⁡2x(Hint: sin⁡2x=12(1−cos⁡2x))f(x)=\sin^2 x \quad (\text{Hint: } \sin^2 x=\tfrac12(1-\cos 2x))
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Solution

Using the identity:

sin⁡2x=12(1−cos⁡2x)\sin^2 x = \frac12(1-\cos 2x)

Now expand cos⁡2x\cos 2x:

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

Simplify the power:

(2x)2n=22nx2n(2x)^{2n} = 2^{2n}x^{2n}

So,

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

Now substitute into the identity:

sin⁡2x=12−12∑n=0∞(−1)n22n(2n)!x2n\sin^2 x = \frac12 - \frac12 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n}}{(2n)!} x^{2n}

Notice the constant term cancels when n=0n=0. Therefore the series starts from n=1n=1:

sin⁡2x=∑n=1∞(−1)n−122n−1(2n)!x2n\sin^2 x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}2^{2n-1}}{(2n)!} x^{2n}

Since cosine converges for all real numbers, the radius is infinite.

Conclusion:

R=∞\boxed{ R=\infty }
Exercise 5
Question
f(x)=1+xf(x)=\sqrt{1+x}
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Solution

We use the binomial series formula:

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

where

(αn)=α(α−1)(α−2)⋯(α−n+1)n!\binom{\alpha}{n} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}

Here,

α=12\alpha=\frac12

So,

1+x=(1+x)1/2=∑n=0∞(12n)xn\sqrt{1+x} = (1+x)^{1/2} = \sum_{n=0}^{\infty} \binom{\frac12}{n}x^n

Compute the first few coefficients:

(120)=1\binom{\frac12}{0}=1(121)=12\binom{\frac12}{1}=\frac12(122)=12(−12)2!=−18\binom{\frac12}{2}=\frac{\frac12(-\frac12)}{2!} =-\frac18(123)=12(−12)(−32)3!=116\binom{\frac12}{3} = \frac{\frac12(-\frac12)(-\frac32)}{3!} = \frac1{16}

Therefore,

1+x=1+12x−18x2+116x3−⋯\sqrt{1+x} = 1+\frac12x-\frac18x^2+\frac1{16}x^3-\cdots

The binomial series converges when

|x|<1

Conclusion:

1+x=∑n=0∞(12n)xn,R=1\boxed{ \sqrt{1+x} = \sum_{n=0}^{\infty} \binom{\frac12}{n}x^n, \quad R=1 }
Exercise 6
Question
f(x)=11−x3=(1−x)−1/3f(x)=\frac{1}{\sqrt[3]{1-x}}=(1-x)^{-1/3}
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Solution

Again we use the binomial series:

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

Here,

α=−13,u=−x\alpha=-\frac13, \quad u=-x

So,

(1−x)−1/3=∑n=0∞(−13n)(−x)n(1-x)^{-1/3} = \sum_{n=0}^{\infty} \binom{-\frac13}{n}(-x)^n

Using the formula for binomial coefficients and simplifying, we obtain

(1−x)−1/3=1+∑n=1∞1⋅4⋅7⋯(3n−2)3nn!xn(1-x)^{-1/3} = 1 + \sum_{n=1}^{\infty} \frac{1\cdot4\cdot7\cdots(3n-2)}{3^n n!}x^n

The binomial series converges when

|x|<1

Conclusion:

(1−x)−1/3=∑n=0∞(−13n)(−x)n,R=1\boxed{ (1-x)^{-1/3} = \sum_{n=0}^{\infty} \binom{-\frac13}{n}(-x)^n, \quad R=1 }
Exercise 7
Question
f(x)=(1+x)−3f(x)=(1+x)^{-3}
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Solution

We use the binomial formula:

(1+x)^{-3} = \sum_{n=0}^{\infty} \binom{-3}{n}x^n, \quad |x|<1

Now simplify the binomial coefficient.

Recall:

(−3n)=(−1)n(n+22)\binom{-3}{n} = (-1)^n\binom{n+2}{2}

And

(n+22)=(n+2)(n+1)2\binom{n+2}{2} = \frac{(n+2)(n+1)}{2}

Therefore,

(1+x)−3=∑n=0∞(−1)n(n+1)(n+2)2xn(1+x)^{-3} = \sum_{n=0}^{\infty} (-1)^n \frac{(n+1)(n+2)}{2} x^n

The binomial series converges when

|x|<1

Conclusion:

(1+x)−3=∑n=0∞(−1)n(n+1)(n+2)2xn,R=1\boxed{ (1+x)^{-3} = \sum_{n=0}^{\infty} (-1)^n \frac{(n+1)(n+2)}{2} x^n, \quad R=1 }
Exercise 8
Question
f(x)=ln⁡∣5+x∣f(x)=\ln|5+x|
Show solution ↓
Solution

First rewrite the function in a convenient form.

ln⁡∣5+x∣=ln⁡(5(1+x5))\ln|5+x| = \ln\left(5\left(1+\frac{x}{5}\right)\right)

Using logarithm properties:

ln⁡∣5+x∣=ln⁡5+ln⁡(1+x5)\ln|5+x| = \ln5 + \ln\left(1+\frac{x}{5}\right)

Now use the known Maclaurin series:

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

Let

u=x5u=\frac{x}{5}

Then

ln⁡(1+x5)=∑n=1∞(−1)n−1n(x5)n\ln\left(1+\frac{x}{5}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \left(\frac{x}{5}\right)^n

So,

ln⁡∣5+x∣=ln⁡5+∑n=1∞(−1)n−1n5nxn\ln|5+x| = \ln5 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n} x^n

For convergence:

\left|\frac{x}{5}\right|<1

which gives

|x|<5

Conclusion:

ln⁡∣5+x∣=ln⁡5+∑n=1∞(−1)n−1n5nxn,R=5\boxed{ \ln|5+x| = \ln5 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n}x^n, \quad R=5 }