Recent content by deagledoubleg

Does the above look better?

I think that I have f^n(x)=\frac{(-1)^(n)*\frac{(n-1)!}{6}}{(1+x)^(n+4)}

True, I can see that now. I think the issue is the (n-1)! in the numerator. I can't find something that works because I want it to be n! but to stop at 4. like 7*6*5*4, but not include *3*2*1. To do this would I just divide that by 6?

Fixed

f0(x)= \frac{(1)}{(1+x)^4} f1(x)= \frac{(-4)}{(1+x)^5} f2(x)= \frac{(20)}{(1+x)^6} f3(x)= \frac{(-120)}{(1+x)^7} f4(x)= \frac{(840)}{(1+x)^8} I think that I got the Taylor Series of f^n(x)=\frac{(-1)^(n+1)*(n-1)!}{2^(n+4)}

Let f(x) = (1+x)-4 Find the Taylor Series of f centered at x=1 and its interval of convergence. \sum_{n=0}^\infty f^n(c)\frac{(x-c)^n}{n!} is general Taylor series form My attempt I found the first 4 derivatives of f(x) and their values at fn(1). Yet from here I do not know how to find the...