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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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

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

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

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

AustralopithecusBest ResponseYou've already chosen the best response.0
I 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
 one year ago

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

AustralopithecusBest ResponseYou've already chosen the best response.0
yes but I have to use the Mclaurin series method
 one year ago

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

eliassaabBest ResponseYou've already chosen the best response.1
I think @91 is leading you to the easiest method.
 one year ago

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

AustralopithecusBest ResponseYou've already chosen the best response.0
this question is more about learning how to use the Mclaurin series method than anything else
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
We should add that the convergence is only valid for x<1
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
I know the method he is using
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
Since the McLaurin Series is unique. It does not matter which method to use to find it.
 one year ago

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