IMPORTANT

[Punicher1990]

, i need to transform a basic taylor series function with integer coefficients

,to an equation called infinite product which has the following formula

= 1/(1-x)^a1 * (1-x^2)^a2......

and how to find the value of the a's

[Thomas Huxhorn]

Is this a programming task?

[Punicher1990]

yes

[lastchance]

Write a clearer question @Muhammed1990.

That isn't a Taylor series ... and we can't see the one you are referring to. From the little information you have given it would be a matter of changing the + operations in the power series to * operations in the product.

You have to give the a's: they won't magically appear.

[Punicher1990]

@lastchance

it is a mathematical task , we got for example the taylor series for e^x which is 1 + x + x^2/2! + x^3/3! + ........ with int coefficients where x^2 is the second derivative of X etc.
it is needed to be changed into an "infinite product" with the given formula
1/(1-x)^a1 * (1-x^2)^a2...... where X^2 is the second derivative of X et cetera

the question is how to find the a's , and a program is needed to compute that

[lastchance]

Your Taylor series doesn't have integer coefficients.
I can assure you that 1/2!, 1/3! etc. are not whole numbers and lie between 0 and 1. You have some integer exponents, but that is different. I can equally assure you that x^2 is not the second derivative of x. What you have written makes no sense.

Are you sure that you know what your assignment is about? But function are you trying to write a product for and WHAT OTHER INFORMATION ARE YOU GIVEN?

[Punicher1990]

is it possible to transform the taylor expansion of e^x to the following formula , which is named as infinite product formula :

1/(1-x)^a1 * (1-x^2)^a2......

i am asking how to find the value of the a's

the aim is that it is just not e^x, it can be e^x plus an another term, the infinite product function aim is to save the steps made in between for an expression like that

[lastchance]

https://math.stackexchange.com/questions/502160/exponential-function-as-an-infinite-product

http://mathworld.wolfram.com/MoebiusFunction.html (equation 16)


Apparently:
exp(x) = prod_(n=1)^infinity (1-x^n)^(-mu(n)/n)
for abs(x) < 1, where mu(n) is the Mobius function:
mu(n) = 0 (if n has one or more repeated prime factors)
mu(1) = 1
mu(n) = (-1)^(number of prime factors) otherwise

Punicher1990 wrote:

it can be e^x plus an another term, the infinite product function aim is to save the steps made in between for an expression like that

That makes zero grammatical sense; please reword.