Question

Find the value of the 150th derivative of e^(x^50) at x = 0.
I think I should use a Maclaurin series but I don't know how.
Thanks

Answer 1

\[e^u=\sum_0\frac 1{n!}u^n\]
\[e^{x^a}=\sum_0\frac 1{n!}x^{an}\]

Answer 2

\[\sum_0\frac 1{n!}x^{an}\]
\[\sum_1\frac 1{n!}~an~x^{an-1}\]
\[\sum_2\frac 1{n!}~an(an-1)~x^{an-2}\]
...
\[\sum_{150}\frac 1{n!}~an(an-1)...(an-149)~x^{an-150}\]

Answer 3

we can adjust the index now to 0 by -150 +150

\[\sum_{150-150}\frac 1{(n+150)!}~a(n+150)(a(n+150)-1)...(a(n+150)-149)~x^{a(n+150)-150}\]

Answer 4

almost forgot that a = 150 too :)

Answer 5

er ..a = 50 that is

Answer 6

It should somehow be 150! / 6, I can't see how.

Answer 7

well, at x=0, everything but the first term goes zero, so lets test out my idea

Answer 8

\[\frac 1{(150)!}~50(150)(50(150)-1)...(50(150)-149)\]
the wolf might work for a quick determination :)

Answer 9

Well wolfram agrees but this was an old exam question so there must be a simpler way.

Answer 10

\[\frac 1{(150)!}~k(k-1)(k-2)...(k-149)\]
k=50(150) = 7500

Answer 11

yeah, i think im off in my coding .... but that is the process i would attempt

Answer 12

http://www.wolframalpha.com/input/?i=150!%2F6%2C+1%2F150!+*+Product [7500-x%2C+{x%2C0%2C149}]

Seems to be a bit off, but I like the approach style!

Answer 13

:)

Answer 14

0: e^u
1: u' e^(u)
2: u'' e^(u) + (u')^2 e^(u)
3: u''' e^(u) + 2(u')u'' e^(u) + u'' u' e^(u) + (u')^3 e^(u)

hmmm this idea doesnt seem to be evening out too well ... mac series is the way to go fer sure

Answer 15

Seems to be that way! I'll keep the question open to see if someone else has an idea!
Thank you for all your help!

Answer 16

youre welcome, and good luck :)