Question

Consider the function F(x)=integral from 0 to x of ln(1+t^4)dt.
a. find the 2nd derivative of F(x)
b.find a power series representation of a.?

Answer 1

the second derivative is just extra verbiage ...
\[F(x)=\int_{a}^{b}f(t)~dt=F(b)-F(a)\]
\[\frac d{dx}F(x)=\frac d{dx}F(b)-F(a)=f(b)b'-f(a)a'\]

Answer 2

Fundamental theorem of calc for the first part (at least the first part of the first part...):\[\frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\cdot g'(x)\]

Answer 3

the 2nd derivative is just the derivative of the first ...

Answer 4

as Sith points out, a constant goes to zero so this is just teh b stuff

Answer 5

product rule is next :)

Answer 6

since x'=1 this is really just the derivative of: ln(1+x^4)

Answer 7

ok so its just 1/u du

Answer 8

yep, and then if you work out the division you form a power series

Answer 9

ok i might need some help with the power series i am really confused with them

Answer 10

if you know a power series for ln(u), then taking its derivative and subbing in specifics might be "simpler"

Answer 11

id rather run the division tho :)


4 ( x^3 - x^7+x^10 ...)
--------------
1 + x^4 | x^3
-(x^3 + x^7)
-------------
-x^7
-(-x^7 -x^10)
--------------
x^10

Answer 12

Actually, I don't think you need to know the power series for \(\ln u\). The second derivative that you get in the first part gives you something fairly easy to work with, in the sense that you use the fact that
\[\frac{1}{1-x}=\sum_{n=0}^\infty x^n~~~~\text{for }|x|<1\]

Answer 13

Since I don't want to give out an answer right away, why don't you tell use what you get for part (a) @jcaruso?

Answer 14

\[4t ^{3}\div(1+t ^{4})\]

Answer 15

Good. Think you can write that as a geometric-looking series?

Answer 16

its not t, but yeah

Answer 17

I have no idea

Answer 18

if forgot how to add 7 and 4 on mine lol

Answer 19

\[\frac{4x^3}{1+x^4}=4x^3\left(\frac{1}{1-(-x^4)}\right)\]
Using the fact that I posted earlier, you can write
\[4x^3\sum_{n=0}^\infty\left(-x^4\right)^n=4x^3\sum_{n=0}^\infty(-1)^nx^{4n}\]
which converges for \(|x|<1\).

Answer 20

ok thank you very much. they also want me to now get the power series for F(x)and determine any integration constants

Answer 21

integrate your power series up 2 times to F(x)

Answer 22

might want to merge your xs for that

Answer 23

so do i just integrate the 4x^3 outside

Answer 24

i wouldnt, id merge the x^3 inside to create x^(3+4n) which represents a polynomial .... and barring any cosmic lapses in reasoning, we should be able to work that by a power rule

Answer 25

well just do like Foureir and Euler and ignore any consequences of our actions in the process :)

Answer 26

Wait, do you need the power series for \(F(x)\) or \(F''(x)\)? Part (b) appears to want the one for \(F''(x)\).

Answer 27

F''(x) = 4 ( x^3 - x^7+x^11 ...)

F'(x) = B + x^4 - x^8/2 + x^12/3 ...

F(x) = A + Bx + x^5/5 - x^9/18 + x^13/39 ...

Answer 28

i need both

Answer 29

i dont get what youre trying to figure out

Answer 30

@SYNDICATE0612
We're given a function of \(x\) in terms of an integral with respect to t:
\[F(x)=\int_0^x \ln(1+t^4)~dt\]
The first part wants the second derivative. To find the first, you'd use the fundamental theorem of calculus (the first formula I posted above). Finding the second one involves using the log and chain rules, etc.

The next part asks for the power series representation of what we found in (a), \(F''(x)\).

And apparently we also need to find the power series representation for \(F(x)\).

Using the method I suggested, \(F''(x)\) can be modeled by a geometric-esque series:
\[\large F''(x)=4\sum_{n=0}^\infty (-1)^nx^{4n+3}\]
To find the power series representation of \(F(x)\), you would simply integrate the series rep of \(F''(x)\) twice. (Term-by-term integration, as amistre points out in his most recent post, makes it easier.)