Question

expand (x-k)^6 with taylors theorem

Answer 1

Is Newton also allowed?

Answer 2

I haven't learned that yet...

Answer 3

Let \(f(x)=(x-k)^6\). Then \(f(0)=k^6\)
\(f'(x)=6(x-k)^5\quad f'(0)/1!=-6k^5\)
\(f''(x)=30(x-k)^4\quad f''(0)/2!=15k^4\)
\(f^{(3)}(x)=120(x-k)^3\quad f^{(3)}(0)/3!=-20k^3\)
\(f^{(4)}(x)=360(x-k)^2\quad f^{(4)}(0)=15k^2\)
\(f^{(5)}(x)=720(x-k)\quad f^{(5)}(0)/5!=-6k\)
\(f^{(6)}(x)=720\quad f^{(6)}(0)/6!=1\)
\(f^{(n)}(x)=0\text{ for all } n\geq 7\)
So \((x-k)^6=k^6-6k^5x+15k^4x^2-20k^3x^3+15k^2x^4-6kx^5+x^6\)

Answer 4

I was trying to use f(k), f'(k)) ect. so the center is at 0 though?

Answer 5

Yes, of course :). If the center at \(k\) then the Taylor expansion of \((x-k)^6\) is just itself.

Answer 6

Ok cool that's right, I get confused over the simple stuff sometimes. Thank You!