## anonymous 5 years ago expand (x-k)^6 with taylors theorem

1. anonymous

Is Newton also allowed?

2. anonymous

I haven't learned that yet...

3. watchmath

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 $$(x-k)^6=k^6-6k^5x+15k^4x^2-20k^3x^3+15k^2x^4-6kx^5+x^6$$

4. anonymous

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

5. watchmath

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

6. anonymous

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