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Australopithecus
 4 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?
Australopithecus
 4 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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think there is easier way but let me study it for a bit

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will teach you a better way

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

Australopithecus
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you want to find series for \[\frac{1}{(1x)^2}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0notice \[\frac{D}{\text{Dx}}\left[\frac{1}{(1x)}\right]=\frac{1}{(1x)^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think @91 is leading you to the easiest method.

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

Australopithecus
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We should add that the convergence is only valid for x<1

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since the McLaurin Series is unique. It does not matter which method to use to find it.

Australopithecus
 4 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
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