A community for students.
Here's the question you clicked on:
 0 viewing
 2 years ago
find the Mclaurin series for f(x) = (1x)^(2)
So I took the derivative of
f(x) = (1x)^(2)
f(x) = (1x)^(2)
f'(x) = 2(1x)^(3)
f''(x) = 6(1x)^(4)
f'''(x) = 24(1x)^(5)
f''''(x) = 120(1x)^(6)
then I took f(0) for all the derivatives
f(0) = 1
f''(0) = 2
f'''(0) = 6
f''''(0) = 24
f'''''(0) = 120
Can anyone show me how to get a series from this?
 2 years ago
find the Mclaurin series for f(x) = (1x)^(2) So I took the derivative of f(x) = (1x)^(2) f(x) = (1x)^(2) f'(x) = 2(1x)^(3) f''(x) = 6(1x)^(4) f'''(x) = 24(1x)^(5) f''''(x) = 120(1x)^(6) then I took f(0) for all the derivatives f(0) = 1 f''(0) = 2 f'''(0) = 6 f''''(0) = 24 f'''''(0) = 120 Can anyone show me how to get a series from this?

This Question is Closed

91
 2 years ago
Best ResponseYou've already chosen the best response.1I think there is easier way but let me study it for a bit

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0I assume I use the formula \[\frac{f^{n}(0)(x)^{n}}{n!}\]

91
 2 years ago
Best ResponseYou've already chosen the best response.1I will teach you a better way

91
 2 years ago
Best ResponseYou've already chosen the best response.1now , it is common knowlege(back of the book) \[\frac{1}{1x}=\sum _{n=0}^{\infty } x^n\]

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0I need to use Mclaurin series, I know that method at least I think I do I still need to seek help to ensure I'm doing it correclty

91
 2 years ago
Best ResponseYou've already chosen the best response.1you want to find series for \[\frac{1}{(1x)^2}\]

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0yes but I have to use the Mclaurin series method

91
 2 years ago
Best ResponseYou've already chosen the best response.1notice \[\frac{D}{\text{Dx}}\left[\frac{1}{(1x)}\right]=\frac{1}{(1x)^2}\]

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1I think @91 is leading you to the easiest method.

91
 2 years ago
Best ResponseYou've already chosen the best response.1remember \[\frac{1}{(1x)}=\sum _{n=0}^{\infty } x^n\] so take derivative of other side to get \[\frac{1}{(1x)^2}=\sum _{n=0}^{\infty } n x^{n1}\]

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0this question is more about learning how to use the Mclaurin series method than anything else

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1We should add that the convergence is only valid for x<1

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0I know the method he is using

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1Since the McLaurin Series is unique. It does not matter which method to use to find it.

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0ugh, I should have asked a question with a trig function or e^(x) something
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.