### Taylor Series

If a function can be expressed in the form of,

f(x) = f(a) + f'(a)(x-a) + f"(a)(x-a)^{2}/2! + ...,

it is considered as **Taylor's Series.**

**Taylor Series:** f(x) = f(a) + f'(a)(x-a) + f"(a)(x-a)^{2}/2! + ...

If a = 0, then it leads to another series, known as **Maclaurin Series**

So, Maclaurin Series is as follows:

**Maclaurin Series:** f(x) = f(0) + f'(0)x/1! + f"(0)x^{2}/2!....

Taylor series leads to the following power series:

**E.g.1**

If f(x)=e^{x}, then,

dy/dx=e^{x}, d^{2}y/dx^{2} = e^{x}; so,

f(0)=1;

f'(0)=1;

f"(0)=1;

So,

e^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3!....

In the same way,

e^{-x} = 1 - x/1! + x^{2}/2! - x^{3}/3!....

**E.g.2**

If f(x) = sin x, then,

dy/dx=cos x, d^{2}y/dx^{2} = -sin x; so,

f(0) = 0;

f'(0)=1;

f"(0)=0;

So,

sin x = 1 - x^{3}/3! + x^{5}/5!....

**E.g.3**

If f(x) = cos x, then,

dy/dx= -sin x, d^{2}y/dx^{2} = -cos x; so,

f(0) = 1;

f'(0)=0;

f"(0)=-1;

cos x = 1 - x^{2}/2! + x^{4}/4! - x^{6}/6!....

**E.g.4**

Show that e^{sin(x)} = 1 + x + x^{2}/2 -x^{4}/8 +...

From Maclauren's Series,

sin(x) = x - x^{3}/3!...

e^{sin(x)} = e^{x - x3}

e^{x} x e^{-x3/6}

(e^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3!...)(e^{-x3/6} = 1 + x^{-6}/3!...)

So, e^{sin(x)} = 1 + x + x^{2}/2 -x^{4}/8 +...

Now work out the following:

- Show Taylor Series to prove that 1/(1 + x) = 1 -x + x2 - x3 + x4 + ...
- Find an expression for tan x, using Taylor Series.
- Show that ln(1 + x) = x - x
^{2}/2 + x^{3}/3 - x^{4}/4... - Use Taylor Series to find e
^{sin x.} - Use Taylor Series to find e
^{cos x.}