A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
expand (xk)^6 with taylors theorem
anonymous
 5 years ago
expand (xk)^6 with taylors theorem

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is Newton also allowed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't learned that yet...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Let \(f(x)=(xk)^6\). Then \(f(0)=k^6\) \(f'(x)=6(xk)^5\quad f'(0)/1!=6k^5\) \(f''(x)=30(xk)^4\quad f''(0)/2!=15k^4\) \(f^{(3)}(x)=120(xk)^3\quad f^{(3)}(0)/3!=20k^3\) \(f^{(4)}(x)=360(xk)^2\quad f^{(4)}(0)=15k^2\) \(f^{(5)}(x)=720(xk)\quad f^{(5)}(0)/5!=6k\) \(f^{(6)}(x)=720\quad f^{(6)}(0)/6!=1\) \(f^{(n)}(x)=0\text{ for all } n\geq 7\) So \((xk)^6=k^66k^5x+15k^4x^220k^3x^3+15k^2x^46kx^5+x^6\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was trying to use f(k), f'(k)) ect. so the center is at 0 though?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, of course :). If the center at \(k\) then the Taylor expansion of \((xk)^6\) is just itself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok cool that's right, I get confused over the simple stuff sometimes. Thank You!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.