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# Taylor Series Calculator

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A statement of a function into an infinite polynomial at a point is known as Taylor series. Taylor series Calculator determines the taylor series of a given function using its formula. If f(x) is the function, the Taylor series expansion is given by,

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

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

Here, n is the degree of polynomial for a given function.

If a = 0 in the Taylor Series, the series becomes Maclaurin series.

## Steps

Step 1 : Observe the problem and note the degree of the polynomial and differentiate the function with respect to its degree.

Step 2 : Substitute the differentiated value into the Taylor series formula given above and you can see the respected Taylor series.

## Problems

Given below are some of the examples based on Taylor Series .

### Solved Examples

Question 1: Find the Taylor series for the function $f(x) = x^3 - 10x^2 + 6$ at x = 2 for the third degree of polynomial.
Solution:

Step 1 : Given : $f(x) = x^3 - 10x^2 + 6$ at x = 2

$f(2) = (2)^3 - 10(2)^2 + 6 = -26$

degree n = 3,

For n = 1, $f'(x) = 3x^2 - 20x$
at x = 2, $f'(2) = 3(2)^2 - 20(2) = -28$

For n = 2, $f''(x) = 6x - 20$
at x = 2, $f''(2) = 6(2) - 20 = -8$

For n = 3, $f'''(x) = 6$
at x = 2, $f'''(2) = 6$

Step 2 : Taylor series for the given function is,
f(x)=$f(a)+f'(a)(x-a)$+$\frac{f''(a)(x-a)^2}{2!}$+$\frac{f'''(a)(x-a)^3}{3!}$

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

=$-26+(-28)(x-2)$+$\frac{(-8)(x-2)^2}{2!}$+$\frac{6(x-2)^3}{3!}$

=$-26-28(x-2)-4(x-2)^2+(x-2)^3$

Question 2: Find the Taylor series for the function $f(x) = cos x$ at x = 0 for the fourth degree of polynomial.
Solution:

Step 1 : Given : $f(x) = cos x$ at x = 0

$f(0) = cos (0) = 1$

degree n = 4,
For n = 1, $f'(x) = -sin x$
at x = 0, $f'(0) = 0$

For n = 2, $f''(x) = -cos x$
at x = 0, $f''(0) = -1$

For n = 3, $f'''(x) = sin x$
at x = 0, $f'''(0) = 0$

For n = 4, $f^4(x) = cos x$
at x = 0, $f^4(0) = 1$

Step 2 : Taylor series for the given function is,
f(x)=$f(a)+f'(a)(x-a)$+$\frac{f''(a)(x-a)^2}{2!}$+$\frac{f'''(a)(x-a)^3}{3!}$+$\frac{f^4(a)(x-a)^4}{4!}$

=$f(0)+f'(0)(x-0)$+$\frac{f''(0)(x-0)^2}{2!}$+$\frac{f'''(0)(x-0)^3}{3!}$+$\frac{f^4(0)(x-0)^4}{4!}$

=1+0(x-0)+$\frac{(-1)(x-0)^2}{2!}$+$\frac{0(x-0)^3}{3!}$+$\frac{1(x-0)^4}{4!}$

=1-$\frac{x^2}{2}$+$\frac{x^4}{24}$