an idea i had: factorizing taylor polynomials Can any taylor polynomial be factorized into an infinite product representation? I think so. I was able to do this(kinda) with sin(x), i did it this way. because sin(0)=0, there must be an x in the factorization. because every x of sin(x)=0 is n*pi where n is any number, there is (x+pi)(x-pi)(x+2pi)(x-2pi)... in the factorization the (x+n*pi)(x-n*pi) can be simplified to (x^2-n^2*pi^2) i know therefore that the factorized polynomial of sin(x) is in the form of ax(x^2-pi^2)(x^2-4pi^2)(x^2-9pi^2)(x^2-16pi^2)(x^2-25pi^2)... heres where i got messed up: a= the coefficient of the highest exponent in the series, which is n*2+1 where n is the number of factors. a=1/(2n+1)! which comes from the infinite sum representation. but as it turns out this is incorrect.(which after reasoning through it i realize that this the factorization is totally different from the infinite sum representation. and therefore i was wrong) but with help from my graphing calculator i was able use guess and check to decide the value of a. also, it turns out that if n even, the factorization is sinx, while if n is odd, the factorization is -sinx. this is strange. heres one factorization i came up with. but its only accurate between -pi and pi. graph it and see what i mean. sinx=(1/500000000000)x(x^2-pi^2)(x^2-4pi^2)(x^2-9pi^2)(x^2-16pi^2)(x^2-25pi^2)(x^2-36pi^2) it seems as though regardless of the number of factors, the polynomial is only accurate between -pi and pi. with the moderate success of factorizing sin(x), i decided to try it with e^x e^x=x^0/0!+x^1/1!+x^2/2!+x^3/3!+... however, the only solution of e^x=0 is negative infinity; there are not even any complex roots that i can work with. so does this mean that e^x non-factorizable? i cant imagine why.