Help me in math?

Question

How is infinite series of sin, cos and also exponent series derived, could you tell me detailed about it, i know it come from the taylor polynomial but how the taylor polynomial is derived..

Answer 1

Try this: http://en.wikipedia.org/wiki/Taylor_series

The taylor series looks at a function and its derivatives at a certain point. We can calculate the taylor series for sin, cos etc quite easily. For other function, they are often listed as derivatives get messy.

sinx = x - x^3/3! + x^5/5! - x^7/7! + ......

cosx = 1 - x^2/2! + x^4/4! - x^6/6! + ......

Notice that powers of x are odd for sin (odd function), and even for cos (even function).

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

Answer 2

ill just do one for you...
you need to know some calculus...

the easiest way to explain...

assume cos x is a polynomial cox x = ax^5 + b x^4... + ex + f

let x = 0 cos 0 = 1 = f

the differentiate...

-sin x = 5a x^4 + 4b x^3.... + e

x = 0
-sin 0 = 0 = e

differentiate again...

-cos x = 20 a x^3 .... + 2d

x = 0

d = -1/2

repeat to get

sin x = 6c
c = 0

cos x = 24b
b= 1/24

-sin x = 120a
a = 0

and you get

cos x = 1/24 x^4 - 1/2 x^2 + 1

and if you continued this you would notice the pattern

cos x = 1 - 1/2! x^2 + 1/4! x^4 - 1/6! x^6 + 1/8! x^8 ...

more generally

f(x) = f(0) + f'(0)/1! + f''(0)/2! + f'''(0)/3! .... etc

Answer 3

1) If you have a polynomial f(x)= anx^n +(an-1)x^n-1+++++ao as
bo+b1(x-a) +b2(x-a)^2+++++bn(x-a)^nf(x)
If you put in a in both expressions you get bo=f(a)
If you derivate both expression
f´(x) =n*an*x^n-1++++++ a1= b1+2b2(x-a)+ 3b3(x-a)^2+++
so f´(a) =b1 and so on
so you can write a polynomial exactly as
P(x) = P(a) +P´(a) (x-a) ++++P^n(a)/n! (x-a)^n
so you use the same type of expression for other than polynomials
adding a complimentary term Tn to make the equation true
and you have to study under what conditions Tn==>0 as n
==> to infinity.
If is it so you get the series for the give function