# Taylor Polynomial Formula

Before going to learn the Taylor polynomial formula, let us recall what is Taylor's polynomial. Taylor polynomial of degree "n" is the function formed by the partial sum of first n terms of a Taylor series. Taylor Polynomial Formula helps in the calculation of nth degree Taylor polynomials using the Taylor series. Let us study the Taylor polynomial formula using a few solved examples at the end of the page.

## What Is Taylor Polynomial Formula?

Taylor polynomials are approximations of a function, which generally becomes more accurate as n increases. The Taylor polynomial formula can be represented as,

### P\(_n\)(x) = f(a) + f'(a) (x − a) + [ \(\frac{f''(a)}{2!}\) (x − a)^{2}] + [\( \frac{f'''(a)}{3!}\) (x − a)^{3}] + ….. + [ \(\frac{f^{(n)}(a)}{n!}\) (x − a)^{n}]

OR

### \( P_n(x) = \sum_{n = 0}^\infty \frac{ f^{(n)} a }{ n! } \times (x - a)^n \)

- \(P_n\)(x) = Taylor polynomial, which is real or complex-valued function, that is infinitely differentiable at a real or complex number “a” is the power series
- n = Total number of terms in the series or the degree of the Taylor polynomial

Let us see the applications of the Taylor polynomial formula in the following section.

## Solved Examples Using Taylor Polynomial Formula

**Example 1:** Find the Taylor polynomial for the function, f(x) = 3x - 2x^{3} centered at a = -3.

**Solution: **

To find: Taylor polynomial for the given function

Given:

Function, f(x) = 3x - 2x^{3}

Center at a = -3

Using the aylor polynomial formula,

\(P_n\)(x) = f(a) + f′(a)(x − a) + f′′(a)/2! × (x − a)^{2 }+ f′′′(a)/3! × (x − a)^{3 }+ f^{(4)}(a)/4! × (x − a)^{4 }+ ... + f^{(n) }(a)/n! × (x − a)^{n}

The function and its derivatives are:

f(x) = 3x − 2x^{3}

f′(x) = 3 − 6x^{2}

f′′(x) = −12x

f′′′(x) = −12

Since a = −3 and n = 3, the required polynomial is:

\(P_3\)(x) = f(−3) + f′(−3)(x − (−3)) + f′′(−3)/2! × (x − (−3))^{2 }+ f′′′(−3)/3! × (x − (−3))^{3 }

\(P_3\)(x) = f(−3) + f′(−3)(x + 3) + f′′(−3)/2! × (x + 3)^{2 }+ f′′′(−3)/3! × (x + 3)^{3}

We evaluate the function and its derivatives at x = a = −3:

f(−3) = 3(−3) − 2(−3)^{3}^{ }= 45

f′(−3) = 3 - 6(-3)^{2 }= -51

f′′(−3) = −12(−3) = 36

f′′′(−3) = −12

Taylor Polynomial, \(P_3\)(x) = 45 - 51(x + 3) + 36/2 (x + 3)^{2 }- 12(x + 3)^{3}

**Answer: **The Taylor polynomial of degree n = 3 around a = −3 for the function f(x) = 3x − 2x^{3} is \(P_3\)(x) = 45 - 51(x + 3) + 18 (x + 3)^{2 }- 12(x + 3)^{3}.

**Example 2:**

### Find the Taylor polynomial for function, f(x) = cos x, centred at x = 0.

**Solution:**

To find: Taylor polynomial

Given:

Function, f(x) = Cos x

Using Taylor polynomial formula,

\(P_n\)(x) = f(a) + f′(a)(x − a) + f′′(a)/2! × (x − a)^{2 }+ f′′′(a)/3! × (x − a)^{3 }+ f^{(4)}(a)/4! × (x − a)^{4 }+ ... + f^{(n) }(a)/n! × (x − a)^{n}

We evaluate the function and its derivatives:

f(x) = cos(x)

f'(x) = −sin(x)

f''(x) = −cos(x)

f'''(x) = sin(x)

Thus,

cos(x) = cos(a) − sin(a)/1! (x - a) − cos(a)/2! (x - a)^{2} + sin(a)/3! (x - a)^{3} + ...

Now, put a = 0.

P(x) = 1 − 0/1! (x - 0) − 1/2! (x - 0)^{2} + 0/3! (x - 0)^{3} + 1/4! (x - 0)^{4} + ...

P(x) = 1 − x^{2}/2! + x^{4}/4! − ...

**Answer: **Taylor polynomial for given function,cos(x) is P(x) = 1 − x^{2}/2! + x^{4}/4! − ...

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