Taylor Series Expansion: Formula, Remainder, Solved Problems

In this page, we will learn Taylor series expansion formula along with its remainder form and solved examples.

Taylor Series Formula

The Taylor series expansion formula of f(x) about x=a is given by

f(x) = f(a) + (x-a)f'(a) + $\dfrac{(x-a)^2}{2!}f”(a)$ + $\dfrac{(x-a)^3}{3!}f”'(a) + \cdots$)f'(a) + $\dfrac{(x-a)^2}{2!}f”(a)$ + $\dfrac{(x-a)^3}{3!}f”'(a) + \cdots$

Maclaurin Series Formula

Taylor series expansion of f(x) about x=0 is known as the Maclaurin series of f(x). So the Maclaurin series of f(x) is given by

f(x) = f(0) + xf'(0) + $\dfrac{x^2}{2!}f”(0)$ + $\dfrac{x^3}{3!}f”'(0) + \cdots$

Solved Problems:

Taylor Series of sinx at x=0

Question 1: Find the Taylor series expansion of f(x)=sinx at x=0.

Solution:

Given f(x)=sinx ⇒ f(0) = sin0 = 0.

So f$’$(x) = cosx ⇒ f$’$(0) = cos0 = 1.

f$”$(x) = -sinx ⇒ f$”$(0) = -sin0 = 0.

f$^{\prime\prime\prime}$(x) = -cosx ⇒ f$^{\prime\prime\prime}$(0) = -cos0 = -1.

$\vdots$

Putting these values in the Taylor series formula of f(x)=sinx at x=0

f(x) = f(0) + (x-0)f'(0) + $\dfrac{(x-0)^2}{2!}f”(0)$ + $\dfrac{(x-0)^3}{3!}f”'(0) + \cdots$

we have that

sinx = 0 + x⋅1 + $\dfrac{x^2}{2!}$ ⋅ 0 + $\dfrac{x^3}{3!}$ ⋅ (-1) + $\cdots$

⇒ sinx = x – $\dfrac{x^3}{3!}$ + $\cdots$

As this is the Taylor series expansion of sinx at x=0, it is the Maclaurin series expansion of sinx.

Maclaurin series of e^x

Question 2: Find the Maclaurin series expansion of f(x)=ex.

Solution:

The Maclaurin series expansion of f(x)=e^x is the Taylor series expansion at x=0.

We have f(x) = e^x.

So f(0) = e⁰ = 1.

f$^{(n)}$(x) = e^x ⇒ f$^{(n)}$(0) = e⁰ = 1.

Therefore, the Maclaurin series of f(x)=e^x is given by

f(x) = f(0) + xf'(0) + $\dfrac{x^2}{2!}f”(0)$ + $\dfrac{x^3}{3!}f”'(0) + \cdots$

⇒ e^x = 1 + x⋅1 + $\dfrac{x^2}{2!} \cdot 1$ + $\dfrac{x^3}{3!}\cdot 1 + \cdots$

⇒ e^x = 1 + x + $\dfrac{x^2}{2!}$ + $\dfrac{x^3}{3!} + \cdots$

⇒ e^x = $\sum_{n=0}^\infty \dfrac{x^n}{n!}$.

It is the Taylor series expansion of e^x at x=0.

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Lagrange’s Form of Remainder

The Taylor’s series expansion of f(x) about x=a with Lagrange’s form of remainder is given as follows:

$f(x) = f(a) + \dfrac{(x-a)}{1!}f'(a)$ $+\dfrac{(x-a)^2}{2!}f”(a)$ $+\dfrac{(x-a)^3}{3!}f”'(a) + \cdots$ $+\dfrac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)}(a)$ $+\dfrac{(x-a)^{n}}{n!}f^{(n)} \Big(a+\theta(x-a) \Big)$

where 0<θ<1.

In this formula, putting a=0 we get the Maclaurin’s series expansion of f(x) with Lagrange’s form of remainder which is given below.

$f(x) = f(0) + \dfrac{x}{1!}f'(0)$ $+\dfrac{x^2}{2!}f”(0)$ $+\dfrac{x^3}{3!}f”'(0) + \cdots$ $+\dfrac{x^{n-1}}{(n-1)!}f^{(n-1)}(0)$ $+\dfrac{x^{n}}{n!}f^{(n)} \big(\theta x \big)$ …(∗)

where 0<θ<1.

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Question 1: Find the Maclaurin’s series of e^x with Lagrange’s form of remainder.

Solution:

Let f(x) = e^x.

Its nth derivative is f⁽ⁿ⁾(x) = e^x ⇒ f⁽ⁿ⁾(0) = e⁰ = 1.

∴ Using (∗) the Maclaurin’s series of e^x with Lagrange’s form of remainder is

f(x) = $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}$ $+\cdots \dfrac{x^{(n-1)}}{(n-1)!}+\dfrac{x^n}{n!}f^{(n)}(\theta x)$

⇒ e^x = $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}$ $+\cdots \dfrac{x^{(n-1)}}{(n-1)!}+\dfrac{x^n}{n!}e^{\theta x}$

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Question 2: Find the Maclaurin’s series of cosx with Lagrange’s form of remainder.

Solution:

Let f(x) = cos x.

Its nth derivative is f⁽ⁿ⁾(x) = cos($\frac{n\pi}{2}$+x).

Therefore, f(0) = cos0 = 1. f'(0) = cos($\frac{\pi}{2}$) = 0.

f”(0) = cos($2\cdot\frac{\pi}{2}$) = cos($\pi$) = -1.

f'”(0) = cos($3\cdot\frac{\pi}{2}$) = 0.

f⁽⁴⁾(0) = cos($4\cdot\frac{\pi}{2}$) = cos($2\pi$) = 1.

$\vdots$

∴ Using (∗) the Maclaurin’s series of cosx with Lagrange’s form of remainder is

f(x) = $1+0-\dfrac{x^2}{2!}+0+\dfrac{x^4}{4!}\cdots$ $+\dfrac{x^n}{n!}f^{(n)}(\theta x)$

⇒ cosx = $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!} \cdots$ $+\dfrac{x^n}{n!}\cos(\theta x+\frac{n\pi}{2})$

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