Taylor Theorem

In calculus, Taylor's Theorem gives a way to approximate a function near a point using its derivative at that point. This makes it easier to work with complicated functions like e^x, sin(x) or ln(x), especially when we need a quick approximation. The more terms we use, the better the estimate becomes.

Formula used in Taylor's Theorem:

f(x) = f(a) + f'(a)(x - a) + \frac{f"(a)}{2!}(x - a)^2 + .........+ \frac{f^{(n)}(a)}{n!}(x - a)^n + R_n(x)

Where:

a - The point around which the expansion is done.
R_n(x) - The remainder term (error term).

Statement

Suppose the function f(x) satisfies the following condition:

f(x) and its derivatives up to order (N+1 )exist and are continuous in some open interval around a, i.e., x ∈ I
For any x ≠ in I, there exists a number z between a and x such that:

f(x) = f(a) + f'(a)(x - a) + \frac{f"(a)}{2!}(x - a)^2 + .........+ \frac{f^{(n)}(a)}{n!}(x - a)^n + \frac{f^{(n + 1)}(a)}{n + 1!}(x - a)^{n+1}

Proof:

We know that a power series for the function f(x) can be written as:

f(x) = \sum_{n=0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3+ ....

Evaluate the function and its derivatives at x = 0

Value x = 0; f(x) = a₀
First Derivative:

f'(x) = a₁ + 2a₂x + 3a₃x² + 4a₄x³ + .......
At x = 0: f′(0) = a₁

Second Derivative:

f′′(x)= 2a₂+ 6a₃x + 12a₄x²+ ⋯
At x = 0: f''(0) = 2a₂

So we get: a_{2 =} \frac{f''(0)}{2!} = a₂

By generalising the equation, we get

a_n = \frac{f^{(n)}(0)}{n!}

Now substitute the equation in the power series we get,

f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots

This is the Maclaurin expansion (Taylor series about a = 0).

Generalise f in a more general form, and it becomes

f(x) = b + b₁(x - a) + b₂(x - a)²+ b₃(x - a)³ +...

Differentiating the above series n times and evaluating at x = a, we obtain

f^{(n)}(a)=n!\,b_n

Hence,

b_n=\frac{f^{(n)}(a)}{n!}

Now, substitute b_n in a general form.

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

Hence, the Taylor theorem is proved.

Applications

Approximating complex functions like sin(x), e^x, and log(x) with simpler polynomials for easier calculation and analysis.
Forming the basis for algorithms like Simpson's Rule by approximating the integrand as a polynomial.
Approximating non-linear systems (e.g., pendulum motion, circuit behavior) as linear systems near an equilibrium point for simplified analysis.
Deriving equations of motion by expanding the position function.
How computers and calculators actually compute functions (e.g., sin, cos, exp) is by using optimized, truncated Taylor or Maclaurin series.
The foundation for many algorithms, including Newton's Method for finding roots of functions, which uses a first-order Taylor expansion to make a better guess.

Taylor Series
Maclaurin Series
Quiz

Solved Questions on Taylor's Theorem

Question 1: Find the first two terms of the Taylor series of f(x) = e^x around a = 1.

Solution:

The Taylor series of f(x) around x = a is:
f(x) = f(a) + f'(a)(x - a)
at a = 1
f(x) = e^x
f'(x) = e^x
Evalutate at x = 1:
f(1) = e¹ = e
f'(1) = e¹ = e
Substitute into formula at a = 1
f(x) = f(1) + f'(1)(x-1)
f(x) = e + e(x - 1)

Question 2: Find the first two terms of the Taylor series of f(x) = e^x around a = 1.

Solution:

The Taylor series of f(x) around x = a is:
f(x) = f(a) + f'(a)(x - a) + \frac{f"(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f''''(a)}{4!}(x - a)^4 + .........
Compute Derivative at a = 1
f(x) = ln x, f′(x) = x¹, f′′(x) = −\frac{1}{-x^2}, f′′′(x) = \frac{2}{x^3} , f⁽⁴⁾(x)= \frac{6}{-x^4} , f⁽⁵⁾(x) = \frac{24}{x^5}
At x = 1
f(1) = 0 , f′(1) = 1, f′′(1) = −1, f′′′(1) = 2, f⁽⁴⁾(1)=−6, f⁽⁵⁾(1)=24
Substitute into formula
f(x) = f(0) + 1(x - 1) + \frac{1}{2!}(x - 1)^2 + \frac{2}{3!}(x - 1)^3 + \frac{16}{4!}(x - 1)^4 + \frac{24}{5!}(x - 1)^5 .........
Simplify Fcatorial
f(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 + \frac{1}{5}(x - 1)^5
ln x = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3 }{3} - \frac{(x - 1)^4}{4} + \frac{(x - 1)^5 }{5}

Question 3: Find the first two non-zero terms of the Maclaurin series of f(x) = cos⁡x.

Solution:

The Maclaurin series of a function f(x) is:
f(x) = f(0) + f'(0)x + \frac{f"(0)}{3!}x ^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^4(0)}{4!}x^4
Derivatives of f(x) = cosx
f(x) = cos⁡x
f′(x) = −sin⁡x
f′′(x)= −cos⁡x
f′′′(x) = sin⁡x
f⁽⁴⁾(x) = cos⁡x
Evaluate at x=0
f(0) = cos⁡0 = 1
f′(0) = −sin⁡0 = 0
f′′(0)= −cos⁡0 = −1
f′′′(0) = sin⁡0 = 0
f⁽⁴⁾(0) = cos⁡0 = 1
Using the Maclaurin formula:
f(x) = f(0) + f \frac{f''(0)}{2!}x^2
Substitute values:
f(x) = 1 - \frac{x^2}{2}

Unsolved Questions on Taylor's Theorem

Question 1: Find the first three terms of the Taylor series of f(x) =\frac{1}{1+x} around a=0.

Question 1: Find the first 5 terms of the Maclaurin series of f(x) = sin⁡x.

Question 2: Find the first 5 terms of the Maclaurin series of f(x) = e^x.

Question 4: Find the first five terms of the Taylor series of f(x) = x around a = 4.

Statement

Proof:

Applications

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Solved Questions on Taylor's Theorem

Unsolved Questions on Taylor's Theorem

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